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HYPOTHYROIDISM | Harrison's Manual of Medicine
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HYPOTHYROIDISM
ETIOLOGY --------ETIOLOGY --------
ETIOLOGY
Deficient thyroid hormone production can be due to thyroid failure (primary hypothyroidism) or, less commonly, pituitary or hypothalamic disease (secondary hypothyroidism) (Table 173-1). Congenital hypothyroidism is present in 1 of 4000 newborns; the importance of its recognition and prompt treatment for child development has led to the adoption of neonatal screening programs. Transient hypothyroidism may occur in silent or subacute thyroiditis. Subclinical (or mild) hypothyroidism is a state of normal free thyroid hormone levels and mild elevation of thyroid-stimulating hormone (TSH). With higher TSH levels and low free T 4 levels, symptoms become more readily apparent in clinical (or overt) hypothyroidism. In areas of iodine sufficiency, autoimmune disease and iatrogenic causes are the most common etiologies of hypothyroidism. The peak age of occurrence is around 60 years, and prevalence increases with age. Novel anticancer and immunomodulatory treatments, such as tyrosine kinase inhibitors and alemtuzumab, can also induce thyroid autoimmunity via their effects on T cell regulation.
TABLE 173-1: Causes of HypothyroidismPrimary
Autoimmune hypothyroidism: Hashimoto’s thyroiditis, atrophic thyroiditis
Iatrogenic: 131 I treatment, subtotal or total thyroidectomy, external irradiation of neck for lymphoma or cancer
Drugs: iodine excess (including iodine-containing contrast media and amiodarone), lithium, antithyroid drugs, p-aminosalicylic acid, interferon α and other cytokines, aminoglutethimide, sunitinib
Congenital hypothyroidism: absent or ectopic thyroid gland, dyshormonogenesis, TSH-R mutation
Iodine deficiency
Infiltrative disorders: amyloidosis, sarcoidosis, hemochromatosis, scleroderma, cystinosis, Riedel’s thyroiditis
Overexpression of type 3 deiodinase in infantile hemangioma
Transient
Silent thyroiditis, including postpartum thyroiditis
Subacute thyroiditis
Withdrawal of thyroxine treatment in individuals with an intact thyroid
After 131 I treatment or subtotal thyroidectomy for Graves’ disease
Secondary
Hypopituitarism: tumors, pituitary surgery or irradiation, infiltrative disorders, Sheehan’s syndrome, trauma, genetic forms of combined pituitary hormone deficiencies
Isolated TSH deficiency or inactivity
Bexarotene treatment
Hypothalamic disease: tumors, trauma, infiltrative disorders, idiopathic
Abbreviations: TSH, thyroid-stimulating hormone; TSH-R, TSH receptor.
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Citation
Kasper, Dennis L., et al., editors. "HYPOTHYROIDISM." Harrison's Manual of Medicine, 20th ed., McGraw Hill Inc., 2020. harrisons.unboundmedicine.com/harrisons/view/Harrisons-Manual-of-Medicine/623416/all/HYPOTHYROIDISM.
HYPOTHYROIDISM. In: Kasper DLD, Fauci ASA, Hauser SLS, et al, eds. Harrison's Manual of Medicine. McGraw Hill Inc.; 2020. Accessed September 28, 2025.
HYPOTHYROIDISM. (2020). In Kasper, D. L., Fauci, A. S., Hauser, S. L., Longo, D. L., Jameson, J. L., & Loscalzo, J. (Eds.), Harrison's Manual of Medicine (20th ed.). McGraw Hill Inc..
HYPOTHYROIDISM [Internet]. In: Kasper DLD, Fauci ASA, Hauser SLS, Longo DLD, Jameson JLJ, Loscalzo JJ, editors. Harrison's Manual of Medicine. McGraw Hill Inc.; 2020. [cited 2025 September 28]. Available from:
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TY - ELEC T1 - HYPOTHYROIDISM ID - 623416 ED - Jameson,J Larry, ED - Loscalzo,Joseph, ED - Kasper,Dennis L, ED - Fauci,Anthony S, ED - Hauser,Stephen L, ED - Longo,Dan L, BT - Harrison's Manual of Medicine UR - PB - McGraw Hill Inc. ET - 20 DB - Harrison's Manual of Medicine DP - Unbound Medicine ER -
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Related Topics
THYROTOXICOSIS
AMIODARONE
NONTOXIC GOITER
THYROID NEOPLASMS
SICK EUTHYROID SYNDROME
Chapter 173: Thyroid Gland Disorders
TOXIC MULTINODULAR GOITER AND TOXIC ADENOMA
PITUITARY HORMONE HYPERSECRETION SYNDROMES
SINOATRIAL NODE DYSFUNCTION
more...
HYPOTHYROIDISM
ETIOLOGY
CLINICAL FEATURES
DIAGNOSIS
Treatment: Hypothyroidism
Outline
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25.09.24-25.09.24_05
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201
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https://www.youtube.com/watch?v=Tuzn9WOoKQg
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Comparing Fractions for Kids | Easy Math Lessons
Doodles and Digits | Educational Math Videos
22300 subscribers
1883 likes
Description
289750 views
Posted: 17 Jan 2023
Grab the worksheet here:
Hey kids! Are you learning about fractions in math class? Comparing fractions can be tricky, but we're here to help! Find the worksheets that go along with this video in the link below! This video will show you some easy and fun ways to compare fractions, including using a number line. We'll use models and visuals to help you understand the concept and show you a fraction chart that can make it even easier. So if you want to ace your math exams and impress your teacher, be sure to watch this video and learn how to compare fractions like a pro!
Want more information? Check out our website at
Transcript:
Hi, I'm Caroline from Doodles and Digits. Did you know that there are multiple ways
to compare fractions? That's what we're going to learn
today. Let's get into it. Doodly doodle, doodly doodle doodly Doodles and Digits! Area symmetry, fractions too! It’s all here for you! The first strategy we're going to learn is we're going to actually draw models
to help us compare fractions. Let's look at these two fractions here. What do you notice? Which one is greater? That's right. We can actually look at visuals or models
to help us compare fractions. Let's try a few more. Let's try these two. 5/8 or 3/4? That's right. All right, let's look at one more. 3/8 or 1/2? Yeah. These actually
have different denominators, but you can easily see with the visual
which one is greater. You can also use a fraction chart
to help you compare fractions. Let's look at this fraction chart here. All right. What do you notice about the fraction
chart? Let's try comparing two numbers. Which one's larger, 1/2 or 2/5? By looking at the chart, you can see that one
half is greater than 2/5. All right, let's try another one. Which one's larger? One eighth or 2/12. Now that we've tried a few together, let's see if you can compare
a few on the number chart right now. 2/3 or 4/5? 2/3 is less than 4/5. 4/6 or 10/12? 4/6 is less than 10/12 5/8 or 3/5 5/8 is greater than 3/5. A number line is another great tool
to compare fractions. Let's look at a number line. Oh number line! All right. What do you notice? What's going on with this number line? Yeah. You might notice that there's actually a different fraction
being shown on the top of the number line than the bottom of the number line.
What two fractions are being shown? So when we
compare fractions, we can actually place both fractions on this
same number line to compare them. Let's try these two fractions. Which one is greater? All right, let's try another one. Let's compare these two fractions
with the number line. 2/3 and 3/4. Two thirds is less than three fourths. All right. Now that we've tried a few together, let's
see if you can compare a few fractions using a number line on your own 2/5 and 3/7. 2/5 is less than 3/7. 4/5 and 5/7. 4/5 is greater than 5/7. The last strategy is you can also use
reasoning to compare fractions. So for example,
you can actually use benchmark fractions and then just looking and thinking
through it to know which one is greater. So for example, if we're looking at these two fractions
here, they both have the same denominator. So we can think,
What do I know about this? Well, we know if you had the same
denominator, it's the same size piece. So whichever one has more
pieces is going to be greater. Let's look at this here. You've got 2/5 and 3/5. They both have fifths in the denominator. So we can look at the numerator
and know that the 3/5 is greater. Let's try another one. What about 4/8 and 6/8? Since they have the same denominator, we know that the larger numerator
is going to be greater. You can also use benchmark fractions
like one half. So let's say we have these two fractions
here. You can think, okay, is this greater
than one half or less than one half? This one here is greater than one half. This one over here is less than one half. So we know the one that'll be greater
than one half is going to be greater. Review time. What is the greater or less than symbol? This is the greater than symbol. And this is the less than symbol. How can you remember
which one is greater than or less than? Some people like to think of the
opening as a mouth and eat the bigger number. Now that we've learned a bunch
of different strategies, and also we've done a few together, let's
see if you can solve these on your own. You can choose whatever strategy works
best for you. Remember, you can pause this video
in case you need more time 4/8 or 2/8? 4/8 is greater than 2/8. 2/6 or 6/10? 2/6 is less than 6/10. 2/3 or 3/6? 2/3 is greater than 3/6. How did you do? Did you get them all correct? What strategy is your favorite? Do you need other videos
talking about fractions? Check out our other videos on our channel. We also put new videos out every month,
so make sure you're subscribed! Bye!
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Computer Science
Standard Libraries
Number Sets
C++ (C plus plus)
General Programming
Data Structure Basics
Multisets (programming)
Programming and Design
STL (C++)
5
What is the difference between set and multiset in STL?
Brian Bi
uses math
·
Author has 4.8K answers and 65.1M answer views
·
8y
A multiset can contain multiple elements that are equivalent, in the sense that neither one is less than the other. A set cannot; an attempt to insert the second equivalent element will fail since the first is already present.
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Related questions
How is std::multiset implemented?
What is the advantage of using set instead of multiset in C++?
How does multiset work in C++?
What is the best way to learn C++ STL for programming contests?
Should I learn STL or something else?
Addy Radwan
I do mathematics since childhood
·
Author has 505 answers and 2.4M answer views
·
8y
The set in STL in general, keeps the unique elements only and keep them sorted., and this is the default set.
set - C++ Reference
there are many versions of the set, like the multiset, and it loses the feature that “keep one unique instance” so it can hold many instances of the same value, and also keep them sorted.
multiset - C++ Reference
another type of sets is the unordered_set, and it loses the features of “keep them sorted”, but keep using the feature of “keep them unique”.
unordered_set - C++ Reference
Assistant
Bot
·
1y
In the Standard Template Library (STL) of C++, both set and multiset are associative containers that store elements in a sorted order. However, they have key differences in terms of how they handle duplicate elements:
set
Definition: A set is a collection of unique elements.
Duplicates: It does not allow duplicate values. If you try to insert a duplicate element, the insertion will fail (or have no effect).
Ordering: Elements are automatically sorted based on their values.
Complexity: Typical operations like insertion, deletion, and lookup have logarithmic complexity, O(log n).
Example:
```
include <se
```
In the Standard Template Library (STL) of C++, both set and multiset are associative containers that store elements in a sorted order. However, they have key differences in terms of how they handle duplicate elements:
set
Definition: A set is a collection of unique elements.
Duplicates: It does not allow duplicate values. If you try to insert a duplicate element, the insertion will fail (or have no effect).
Ordering: Elements are automatically sorted based on their values.
Complexity: Typical operations like insertion, deletion, and lookup have logarithmic complexity, O(log n).
Example:
```
include #include int main() { std::set mySet; mySet.insert(1); mySet.insert(2); mySet.insert(2); // Duplicate, will not be added for (int num : mySet) { std::cout << num << " "; // Output: 1 2 } return 0;}
```
multiset
Definition: A multiset is a collection that allows multiple occurrences of the same element.
Duplicates: It allows duplicate values. You can insert the same element multiple times, and each insertion will be counted.
Ordering: Like set, elements are also automatically sorted.
Complexity: Similar to set, typical operations have logarithmic complexity, O(log n).
Example:
```
include #include int main() { std::multiset myMultiSet; myMultiSet.insert(1); myMultiSet.insert(2); myMultiSet.insert(2); // Duplicate, will be added for (int num : myMultiSet) { std::cout << num << " "; // Output: 1 2 2 } return 0;}
```
Summary
Set: Unique elements, no duplicates.
Multiset: Allows duplicates, multiple instances of the same element are stored.
These differences make set suitable for scenarios where uniqueness is required, while multiset is useful when you need to count occurrences of elements.
Tanvi Jain
Incoming CS Grad in Computer Science, Stony Brook University (Graduated 2025)
·
2y
The only difference in set and multiset is that multiset can store duplicate elements while set only stores unique elements.
similarity: both stores elements in linear ascending fashion.
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Related questions
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Blaze
Lives in Solapur, Maharashtra, India (2001–present)
·
11mo
Originally Answered: What is the difference between set and multiset in C++ STL?
·
Here’s the scoop:
Set
: Stores unique elements in a sorted order. No duplicates allowed.
Multiset
: Also stores elements in a sorted order but allows duplicates.
So, use
set
for unique items and
multiset
when you need duplicates. Easy peasy!
Jerry Coffin
Guy who writes some code.
·
Author has 5.3K answers and 19.8M answer views
·
Updated 3y
Related
How is std::multiset implemented?
It's most often implemented as a threaded red-black tree, but almost any other form of balanced tree (e.g., AVL tree, B-tree) could be used instead1 (and implementations using both of these have been written). Regardless of the form the balancing takes, the tree does have to be threaded (i.e., incrementing or decrementing an iterator is required to have constant complexity, so you need some way to get from one node to its predecessor/successor without traversing an arbitrary number of nodes in between.
A B+ tree supports that quite directly as part of its basic design. Most other trees need a
It's most often implemented as a threaded red-black tree, but almost any other form of balanced tree (e.g., AVL tree, B-tree) could be used instead1 (and implementations using both of these have been written). Regardless of the form the balancing takes, the tree does have to be threaded (i.e., incrementing or decrementing an iterator is required to have constant complexity, so you need some way to get from one node to its predecessor/successor without traversing an arbitrary number of nodes in between.
A B+ tree supports that quite directly as part of its basic design. Most other trees need a little extra help (so to speak) in the form of pointers directly to the predecessor/successor nodes to do that.
I suspect as the importance of cache usage becomes more apparent to more programmers, we'll see increased use of B trees and similar structures that combine the (approximately) logarithmic complexity of a tree with nodes large enough to improve locality of reference so they'll make better use of caching.
Classically, B trees (and variants) have been thought of primarily as being used for secondary storage (e.g., for indexing files on disks), but as caches get faster compared to main memory, the costs of transactions with main memory follow roughly similar patterns to what we saw with external storage: there's a relatively high initial cost for initiating a read/write at a particular address. Having started that transaction, we can read or write quite a bit more consecutive data at relatively low cost. The similarity in pattern makes many of the structures originally designed for external storage quite useful for transactions with main memory.
Technically, some balanced trees would probably still have a minor complexity problem in a few cases, but not very many (and they would be minor).
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Kurt Guntheroth
Software Engineer and Writer
·
Upvoted by
John Kramer
, MS Computer Science, University of Rochester (2004) and
John Grout
, Master of Science Computer Science & Computer Architecture, University of Illinois at Urbana-Champaign (1995) · Author has 26.7K answers and 286.2M answer views
·
4y
Related
What is the benefit of STL in C++?
In C, if you wanted a linked-list, you had to write it yourself. Over and over for each next program. You had to decide if you wanted a head node to make the insertion and removal code easier. You had to write the insertion and deletion code, which isn’t hard by itself, but you had to debug it too. Then if you wanted a tree, you had to write that too. Yeah, you could reuse last year’s code if you still worked at the same place, or find someone else’s code. but you might have to fiddle it. Did you want the element type to be a value or a pointer? Who deletes removed nodes…
The STL was the very f
In C, if you wanted a linked-list, you had to write it yourself. Over and over for each next program. You had to decide if you wanted a head node to make the insertion and removal code easier. You had to write the insertion and deletion code, which isn’t hard by itself, but you had to debug it too. Then if you wanted a tree, you had to write that too. Yeah, you could reuse last year’s code if you still worked at the same place, or find someone else’s code. but you might have to fiddle it. Did you want the element type to be a value or a pointer? Who deletes removed nodes…
The STL was the very first library of pre-written data structures that was really reusable. It used C++ templates to let you parameterize the code with your favorite element type. The STL came along before the internet too, so that it was harder to find pre-written code.
Kartik Sharma
Studied at Chitkara University (Graduated 2023)
·
5y
Related
What is the difference between set, multiset, unordered_set, and unordered multi-set in C++?
Difference between set, multiset, unordered_set, unordered_multiset - GeeksforGeeks
Read the above article let me know if you still don't get that.
If someone don't wants to open the link, here is a screenshot :
Moreover, insertion time complexity of set and multiset is O(logn) while insertion time complexity of unordered_set and unordered_multiset is O(1).
Difference between set, multiset, unordered_set, unordered_multiset - GeeksforGeeks
Read the above article let me know if you still don't get that.
If someone don't wants to open the link, here is a screenshot :
Moreover, insertion time complexity of set and multiset is O(logn) while insertion time complexity of unordered_set and unordered_multiset is O(1).
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Christopher Susie
Software Engineer
·
Author has 3.8K answers and 4.8M answer views
·
6y
Related
What's the difference between “STL” and “C++ Standard Library”?
The Standard Template Library was a library that came out (I believe in the early 90’s) that contained templates (containers, algorithms, etc) similar in name and functionality to the ones that are currently in the C++ standard library. When C++ became standardized, much of the functionality from the STL was included into the ISO standard for C++. There were a few changes and modifications — and a lot of improvements. The C++ standard library now also contains a lot of stuff that was never in the STL.
So the C++ standard library is an integral part of the C++ language and conforming compilers m
The Standard Template Library was a library that came out (I believe in the early 90’s) that contained templates (containers, algorithms, etc) similar in name and functionality to the ones that are currently in the C++ standard library. When C++ became standardized, much of the functionality from the STL was included into the ISO standard for C++. There were a few changes and modifications — and a lot of improvements. The C++ standard library now also contains a lot of stuff that was never in the STL.
So the C++ standard library is an integral part of the C++ language and conforming compilers must implement all of it as specified by the ISO standard. It contains most but not all of the stuff that was in the STL (generally vastly improved) and frequently it has the same or a similar name to what its STL ancestor had.
The STL itself is no longer actively developed and is not widely used. The C++ Standard Library is required to be included with C++ implementations that wish to conform to the ISO C++ language standard. A lot of people still (somewhat mistakenly) the C++ standard library the “STL” — particularly those parts of the C++ standard library that were inspired by the STL.
If you want to call the C++ Standard Library the STL, people will know what you mean unless perhaps they still work with the actual (nearly defunct) STL libary. Pedantic people may correct you even if they know what you mean. The C++ standard library is part of the C++ language standard. Working knowledge of the standard library (at least parts of it) is mandatory for any programmer claiming general proficiency at C++. The STL is an ancient library that inspired parts of the C++ standard library.
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https://www.tiger-algebra.com/drill/r~2-4r-40=0/
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring r2-4r-40
The first term is, r2 its coefficient is 1 .
The middle term is, -4r its coefficient is -4 .
The last term, "the constant", is -40
Step-1 : Multiply the coefficient of the first term by the constant 1 • -40 = -40
Step-2 : Find two factors of -40 whose sum equals the coefficient of the middle term, which is -4 .
| | | | | | | |
--- --- ---
| | -40 | + | 1 | = | -39 | |
| | -20 | + | 2 | = | -18 | |
| | -10 | + | 4 | = | -6 | |
| | -8 | + | 5 | = | -3 | |
| | -5 | + | 8 | = | 3 | |
| | -4 | + | 10 | = | 6 | |
| | -2 | + | 20 | = | 18 | |
| | -1 | + | 40 | = | 39 | |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 1 :
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = r2-4r-40
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ar2+Br+C,the r -coordinate of the vertex is given by -B/(2A) . In our case the r coordinate is 2.0000
Plugging into the parabola formula 2.0000 for r we can calculate the y -coordinate :
y = 1.0 2.00 2.00 - 4.0 2.00 - 40.0
or y = -44.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = r2-4r-40
Axis of Symmetry (dashed) {r}={ 2.00}
Vertex at {r,y} = { 2.00,-44.00}
r -Intercepts (Roots) :
Root 1 at {r,y} = {-4.63, 0.00}
Root 2 at {r,y} = { 8.63, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving r2-4r-40 = 0 by Completing The Square .
Add 40 to both side of the equation :
r2-4r = 40
Now the clever bit: Take the coefficient of r , which is 4 , divide by two, giving 2 , and finally square it giving 4
Add 4 to both sides of the equation :
On the right hand side we have :
40 + 4 or, (40/1)+(4/1)
The common denominator of the two fractions is 1 Adding (40/1)+(4/1) gives 44/1
So adding to both sides we finally get :
r2-4r+4 = 44
Adding 4 has completed the left hand side into a perfect square :
r2-4r+4 =
(r-2) • (r-2) =
(r-2)2
Things which are equal to the same thing are also equal to one another. Since
r2-4r+4 = 44 and
r2-4r+4 = (r-2)2
then, according to the law of transitivity,
(r-2)2 = 44
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(r-2)2 is
(r-2)2/2 =
(r-2)1 =
r-2
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
r-2 = √ 44
Add 2 to both sides to obtain:
r = 2 + √ 44
Since a square root has two values, one positive and the other negative
r2 - 4r - 40 = 0
has two solutions:
r = 2 + √ 44
or
r = 2 - √ 44
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving r2-4r-40 = 0 by the Quadratic Formula .
According to the Quadratic Formula, r , the solution for Ar2+Br+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
r = ————————
2A
In our case, A = 1
B = -4
C = -40
Accordingly, B2 - 4AC =
16 - (-160) =
176
Applying the quadratic formula :
4 ± √ 176
r = —————
2
Can √ 176 be simplified ?
Yes! The prime factorization of 176 is
2•2•2•2•11
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 176 = √ 2•2•2•2•11 =2•2•√ 11 =
± 4 • √ 11
√ 11 , rounded to 4 decimal digits, is 3.3166
So now we are looking at:
r = ( 4 ± 4 • 3.317 ) / 2
Two real solutions:
r =(4+√176)/2=2+2√ 11 = 8.633
or:
r =(4-√176)/2=2-2√ 11 = -4.633
Two solutions were found :
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WOAH Terrestrial Manual 2023 1 S E C T I O N 3 . 1 . MULTIPLE SPECIES C H A P T E R 3 . 1 . 1 . ANTHRAX SUMMARY Description and importance of the disease: Anthrax is primarily a disease of herbivorous animals, although all mammals, including humans, and some avian species can contract it. Mortality can be very high, especially in herbivores. The aetiological agent is the spore-forming, Gram-positive rod-shaped bacterium, Bacillus anthracis. The disease has world-wide distribution and is a zoonosis. The disease is mediated mainly by exotoxins. Peracute, acute, subacute and, rarely, chronic forms of the disease are reported. Ante-mortem clinical signs may be virtually absent in peracute and acute forms of the disease. Subacute disease may be accompanied by progressive fever, depression, inappetence, weakness, prostration and death. Acute, subacute, and chronic disease may show localised swelling and fever. In chronic disease, the only sign may be enlarged lymph glands. Detection and identification of the agent: Bacillus anthracis is readily isolated in relatively high numbers from blood or tissues of a recently dead animal that died of anthrax, and colony morphology of B. anthracis is quite characteristic after overnight incubation on blood agar. The colony is relatively large, measuring approximately 0.3–0.5 cm in diameter. It is grey-white to white, non-haemolytic with a rough, ground-glass appearance and has a very tacky, butyrous consistency. The vegetative cells of B. anthracis are large, measuring 3–5 µm in length and approximately 1 µm in width. Ellipsoidal central spores, which do not swell the sporangium, are formed at the end of the exponential cell growth phase. The cells stain strongly Gram positive, and long chains are often seen in vitro, while paired or short chains are seen in vivo. Visualisation of the encapsulated bacilli, usually in large numbers, in a blood smear stained with azure B or polychrome methylene blue (M’Fadyean’s reaction) is fully diagnostic. Serological tests: Antibody detection in serum from infected animals is rarely used for diagnostic purposes and is essentially a research tool. The predominant procedure used is the enzyme-linked immunosorbent assay. Requirements for vaccines: The most widely used livestock anthrax vaccine developed by Max Sterne in 1937 is a live, non-encapsulated, spore former held in suspension. In Russia and Eastern Europe, an equivalent type of vaccine is used (strain 55). A list of producers is given in the World Health Organization anthrax guidelines. A. INTRODUCTION Anthrax, an acute bacterial disease primarily of herbivores, is transmissible to humans. The aetiological agent, Bacillus anthracis, is a Gram-positive spore-forming rod-shaped bacterium. Anthrax is known by many names around the world including charbon, woolsorters’ disease, ragpickers’ disease, malignant carbuncle, malignant pustule and Siberian ulcer. Animals become infected by ingesting spores, or possibly by being bitten by blood feeding insects such as stable flies (Stomoxys calcitrans) and mosquitoes (Aedes aegypti and Aedes taeniorhynchus) that have fed on an infected Chapter 3.1.1. – Anthrax 2 WOAH Terrestrial Manual 2023 animal or carcass. Infected animals are usually found dead as death can occur within 24 hours. To avoid environmental contamination with spores, post-mortem examinations of carcasses of animals suspected to have died of anthrax are discouraged and natural orifices such as mouth, nostril and anus should be closed (for example plugged with cotton wool or other suitable material soaked in an approved disinfectant) to prevent the formation of spores. There are regulations in most countries that prohibit post-mortem examination when anthrax is suspected. Recently dead animals may show any number of lesions, none of which is pathognomonic or entirely consistent. Lesions most commonly seen are those of a generalised septicaemia often accompanied by an enlarged spleen with a dark semi-fluid pulp (‘blackberry jam’ consistency) and poorly clotted blood. Haemorrhage from the nose, mouth, vagina and/or anus at death may be found. Gram-positive rod-shaped B. anthracis is an obligate pathogen. Most of the other species of Bacillus are common ubiquitous environmental saprophytes, although a number, notably B. cereus, B. licheniformis and B. subtilis, are occasionally associated with food poisoning in humans and with other clinical manifestations in both humans and animals. 1. Zoonotic risk and biosafety requirements More than 95% of human anthrax cases take the cutaneous form and result from handling infected carcasses or hides, hair, meat or bones from such carcasses. Bacillus anthracis is not invasive and requires a lesion to infect. Protection for veterinarians and other animal handlers involves wearing gloves and other protective clothing (including full personal protective equipment [PPE] depending on the situation) when handling specimens from suspected anthrax carcasses and never rubbing the face or eyes. The risk of gastrointestinal anthrax may arise if individuals eat meat from animals infected with anthrax. The risk of inhaling infectious doses becomes significant in occupations involving the processing of animal by-products for manufacturing goods (industrial anthrax). These include the tanning, woollen, animal hair, carpet, bone processing, and other such industries, where the potential for aerosolisation of substantial numbers of spores increases the risk of exposure to infectious doses. It is important that industrial workers use appropriate personal protective clothing and equipment and follow standard operating procedures that minimise the risk of transmission. Efficient air extraction equipment should be positioned over picking, combing, carding and spinning machines. Air blowing machinery should never be used for cleaning equipment due to the risk of spore dispersal. Clinical specimens and cultures of B. anthracis should be handled with appropriate biosafety and containment procedures as determined by biorisk analysis (see Chapter 1.1.4 Biosafety and biosecurity: Standard for managing biological risk in the veterinary laboratory and animal facilities). Vaccination of laboratory personnel is recommended. B. DIAGNOSTIC TECHNIQUES Table 1. Test methods available for the diagnosis of anthrax and their purpose Method Purpose Population freedom from infection Individual animal freedom from infection prior to movement Contribute to eradication policies Confirmation of clinical cases Prevalence of infection – surveillance Immune status in individual animals or populations post-vaccination Identification of the agent(a) Demonstration of capsule – – – ++ – – Demonstration of lack of motility – – – ++ – – Gamma phage lysis – – – ++ – – Chapter 3.1.1. – Anthrax WOAH Terrestrial Manual 2023 3 Method Purpose Population freedom from infection Individual animal freedom from infection prior to movement Contribute to eradication policies Confirmation of clinical cases Prevalence of infection – surveillance Immune status in individual animals or populations post-vaccination Penicillin susceptibility – – – ++ – – PCR – – – ++ +++/++ – Key: +++ = recommended for this purpose; ++ recommended but has limitations; + = suitable in very limited circumstances; – = not appropriate for this purpose. PCR = polymerase chain reaction. (a)A combination of agent identification methods applied on the same clinical sample is recommended. 1. Identification of the agent Demonstration of encapsulated B. anthracis in smears of blood or tissues from fresh anthrax-infected carcasses and growth of the organism on blood agar plates is relatively uncomplicated and within the capability of most bacteriology laboratories. Difficulty may be encountered in the case of pigs and carnivores in which the terminal bacteraemia is frequently not marked, or in animals that received antibiotics before death. Recovery of B. anthracis from old decomposed carcasses, processed specimens (bone meal, hides), or environmental samples (contaminated soil) is often difficult, requiring demanding and labour-intensive procedures. However live spores may be recovered from the turbinate bones of dead livestock and wildlife for an extended period after death (M. Hugh-Jones, personal communication). 1.1. Culture and identification of Bacillus anthracis 1.1.1. Fresh specimens Bacillus anthracis grows readily on most types of nutrient agar, however, 5–7% horse or sheep blood agar is the diagnostic medium of choice. Blood is the primary clinical material to examine. Swabs of blood, other body fluids or swabs taken from incisions in tissues or organs can be spread over blood agar plates. After overnight incubation at 37°C, B. anthracis colonies are grey-white to white, 0.3–0.5 cm in diameter, non-haemolytic, with a ground-glass surface, and very tacky when teased with an inoculating loop. Tailing and prominent wisps of growth trailing back toward the parent colony, all in the same direction, are sometimes seen. This characteristic has been described as a ‘medusa head’ or ‘curled hair’ appearance. Confirmation of B. anthracis should be accomplished by the demonstration of a capsulated, spore-forming, Gram-positive rod in blood culture. Absence of motility is an additional test that can be done. Anthrax-specific phages were first isolated in the 1950s, and the specifically named gamma phage was first reported in 1955 (Brown & Cherry, 1955) and quickly became the standard diagnostic phage for anthrax. Gamma phage belongs to a family of closely related anthrax phages (World Health Organization [WHO], 2008). Two tests for confirming the identity of B. anthracis are gamma phage lysis and penicillin susceptibility. The typical procedure for these tests is to plate a lawn of suspect B. anthracis on a blood or nutrient agar plate and place a 10–15 µl drop of the phage suspension on one side of the lawn and a 10-unit penicillin disk to the other side. Allow the drop of phage suspension to soak into the agar before incubating the plate at 37°C. A control culture, e.g. the Sterne vaccine or the NCTC strain 10340, should be tested at the same time as the suspect culture to demonstrate the expected reaction for gamma phage lysis and penicillin susceptibility. If the suspect culture is B. anthracis, the area under the phage will be devoid of bacterial growth, because of lysis, and a clear zone will be seen around the penicillin disk indicating antibiotic susceptibility. Note that some field isolates of B. anthracis may be phage resistant or penicillin resistant. As the performance of the gamma phage lysis assay may be affected by the density of bacterial inoculum, Abshire et al. (2005) recommend streaking the suspect culture on the agar plate over several quadrants instead of using a lawn format and inoculating a drop of gamma phage on the Chapter 3.1.1. – Anthrax 4 WOAH Terrestrial Manual 2023 first and second quadrants on the plate. If antibiotic or phage resistant B. anthracis is suspected then polymerase chain reaction (PCR) diagnostic methods may be applied. Phage suspensions may be obtained from central veterinary laboratories or central public health laboratories. The phage can be propagated and concentrated by the following protocol. Store phage at 2–4°C and do not freeze phage as it will quickly become non-viable. 1.1.1.1. Stage one i) Spread a blood agar (BA) plate of the Sterne vaccine strain of B. anthracis. Incubate overnight at 37°C. ii) Inoculate approximately 10 ml of nutrient broth (NB) with growth from the BA plate and incubate at 37°C for approximately 4 hours or until just cloudy, then refrigerate. iii) Spread 100 µl of the culture from step ii on three pre-dried BA plates and incubate at 37°C for 30–60 minutes. iv) Spread 100 µl of the phage suspension to be amplified over the same plates. Incubate at 37°C overnight. v) Harvest the phage-lysed growth on the BA plate in 5 ml of NB followed by a second ‘wash’ of 5 ml NB. Incubate at 37°C overnight. vi) Filter (0.45 µm) and count by dropping 20 µl drops (three drops per dilution) of tenfold dilutions of the filtrate in saline onto lawns of the B. anthracis culture prepared as in step iii. 1.1.1.2. Stage two This is essentially the same procedure as Stage one, only uses the filtrate from step vi to harvest the phage from the plates. vii) Prepare three Sterne strain lawns on BA, as in step iii. Incubate at 37°C for 30–60 minutes. viii) Spread 100 µl phage from step vi. Incubate at 37°C overnight. ix) To 9 ml of filtrate from step vi, add 1 ml of 10× concentrated NB. x) Harvest the phage from step viii with 5 ml of the solution from step ix, followed by a second 5 ml wash with the rest of the solution from step vi. xi) Add 10 ml of 1× NB. xii) Incubate at 37°C overnight, filter and count. 1.1.1.3. Stage three xiii) Inoculate 100 ml of brain–heart infusion broth with approximately 2.5 ml of the culture from step ii. Incubate on a rotary shaker at 37°C until just turbid. xiv) Add the 20 ml of filtrate from step xii and continue incubation overnight. xv) The resultant filtrate is checked for sterility and titrated in tenfold dilutions on lawns of the vaccine strain as in step vi to determine the concentration of the phage. This should be of the order of 108–109 plaque forming units per ml. 1.1.2. Capsule visualisation Virulent encapsulated B. anthracis is present in tissues and blood and other body fluids from animals that have died from anthrax. Thin smears may be prepared from blood from ear veins or other peripheral veins, exudate from orifices and, for horses and pigs, from oedematous fluid or superficial lymph nodes in the neck region. However if the animal has been dead more than 24 hours, the capsule may be difficult to detect. The bacteria should be looked for in smears of these specimens that have been dried, fixed and then stained with azure B (M’Fadyean’s reaction). The capsule stains pink, whereas the bacillus cells stain dark blue. The cells are found in pairs or short chains and are often square-ended (the chains are sometimes likened to a set of railway Chapter 3.1.1. – Anthrax WOAH Terrestrial Manual 2023 5 carriages – so-called ‘box-car’ or ‘jointed bamboo-rod’ appearance). Gram stain does not reveal the capsule. The capsule is not present on B. anthracis grown aerobically on nutrient agar or in nutrient broths but can be seen when the virulent bacterium is cultured for a few hours in a few millilitres of blood (defibrinated horse or sheep blood seems to work best). Alternatively, the capsule is produced when the virulent B. anthracis is cultured on nutrient agar containing 0.7% sodium bicarbonate and incubated in the presence of CO2 (20% is optimal, but a candle jar works well). The agar is prepared by weighing nutrient agar base powder required for a final volume of 100 ml but reconstituting the measured agar in only 90 ml of water. Autoclave and cool to 50°C in a water bath. Add 10 ml of a filter-sterilised (0.22–0.45 µm filter) 7% solution of sodium bicarbonate. Mix and pour into Petri dishes. The encapsulated B. anthracis will form mucoid colonies and the capsule can be visualised by making thin smears on microscope slides, fixing, and staining with azure B or polychrome methylene blue (M’Fadyean’s stain). Polychrome methylene blue can be prepared as follows: 0.3 g of methylene blue is dissolved in 30 ml of 95% ethanol; 100 ml of 0.01% potassium hydroxide (KOH) is mixed with the methylene blue solution. Ideally, this should be allowed to stand exposed to the air, with occasional shaking, for at least 1 year to oxidise and mature. Addition of K2CO3 (to a final concentration of 1%) hastens the ‘ripening’ of the stain, but before it is regarded as diagnostically reliable, its efficacy should be established by testing it in parallel with an earlier, functional batch of stain on bona fide samples. It has been found that stains that give positive reactions with cultures of B. anthracis cultured artificially in horse blood sometimes do not give positive results in the field. The method for capsule visualisation has been revised according to the description by Owen et al. (2013). The established simple polychrome methylene blue (PMB) staining method for blood or tissue smears from dead animals (M’Fadyean’s reaction) is therefore replaced by azure B staining. It must be noted that the rarity of anthrax disease worldwide due to improvements in the control of the disease has led to quality controlled commercially produced PMB being difficult to obtain. In addition, reports of inaccurate results using alternative methylene blue-based stains has also become a concern. Hence, for laboratories requiring a reliable M’Fadyean stain for rapid detection, the recommended approach is to use commercially pure azure B (Owen et al., 2013). This microscopy method was validated by Aminu et al. (2020), for the detection of B. anthracis in field-collected blood smears. Four capsule-staining techniques were evaluated in an investigation of livestock mortalities suspected to be caused by anthrax. Field-prepared blood samples were tested by microscopy and the results indicated that the median sensitivity and specificity of microscopy using azure B were found to be comparable with those of the recommended standard, polychrome methylene blue (Aminu et al. 2020). Azure B stain is prepared by constituting into a solution of 0.03 g azure B in 3 ml of 95% ethanol or methanol to which is then added 10 ml of 0.01% KOH (0.23% final azure B concentration). This can then be used immediately and throughout the tests. If stored in the dark at room temperature, the shelf life is at least 12 months. Smears to be stained should be fixed with ethanol or methanol (95–100%), not by heat, and the stain left for 5 minutes before washing off for optimum effect (Owen et al., 2013). In making smears for staining, only small drops of blood or tissue fluid are needed, and a thin, small smear is best. After fixing by dipping the smear in 95–100% alcohol for about 5 minutes and drying, a small (approximately 20 µl) drop of stain is placed on the smear and spread over it with an inoculating loop. After 5 minutes, the stain is washed with water, blotted, air-dried, and observed initially using the ×10 objective lens under which the short chains appear like short hairs; once found, these can be observed under oil immersion (×1000) for the presence of the pink capsule surrounding the blue/black-staining bacilli. To avoid laboratory contamination, the slide and blotting paper should be autoclaved or left for some hours in a 10% sodium hypochlorite solution. 1.1.3. Other specimens Identification of B. anthracis from old, decomposed specimens, processed materials, and environmental samples, including soil, is possible but these samples often have saprophytic contaminants that outgrow and obscure B. anthracis on non-selective agars. The following procedure is suggested: Chapter 3.1.1. – Anthrax 6 WOAH Terrestrial Manual 2023 i) The sample is blended in two volumes of sterile distilled or deionised water and placed in a water bath at 62.5 ± 0.5°C for 15–30 minutes. Turnbull et al. (2007) have demonstrated that heat activation of spores can be conducted at a temperature range of 60–70°C with holding times not exceeding 15–30 minutes for best recovery. ii) Tenfold dilutions to 10–2 or 10–3 are then prepared. From each dilution, 10–100 µl are plated on to blood agar and optionally 250–300 µl on to PLET agar (polymyxin, lysozyme, EDTA [ethylene diamine tetra-acetic acid], thallous acetate) (Knisely, 1966; WHO, 2008). All plates are incubated at 37°C. For combined selection and differentiation, blood agar-based media can provide advantages. In addition, PLET agar contains the toxic thallium acetate. An alternative for this is the so-called TSPB agar, a blood agar that has high selectivity against Gram-negative bacteria due to the addition of trimethoprim, sulfamethoxazole and polymyxin B (Hudson et al., 2007; Rosenblatt & Stewart, 1974). TSPB agar is prepared by dissolving 40 g/litre nutrient agar base (e.g. SIFIN). The mixture is autoclaved and uniformly cooled to 45°C before adding 50 ml/litre sterile sheep blood (5%), trimethoprim (13.1 mg/litre), sulfamethoxazole (20 mg/litre) and polymyxin B (30,000 IU/litre). After mixing thoroughly, the agar is dispensed into Petri dishes. iii) Blood agar plates are examined for typical colonies as previously described after overnight incubation, and the PLET plates are examined after 40–48 hours. Confirmation of the identity of suspect colonies as B. anthracis is done as described above. PLET medium (Knisely, 1966; WHO, 2008) is prepared by using heart-infusion agar base (DIFCO) made up to the manufacturer’s instructions with the addition of 0.25–0.3 g/litre EDTA and 0.04 g/litre thallous acetate. The mixture is autoclaved and uniformly cooled to 50°C before adding the polymyxin at o30,000 units/litre and lysozyme at 300,000 units/litre. After mixing thoroughly, the agar is dispensed into Petri dishes. Reports of procedures for direct detection of B. anthracis in soils and other environmental specimens using the PCR are emerging. None of these has become routinely applicable at the present time. Animal inoculation may be considered for recovery of B. anthracis if all other methods fail. Examples of when this might occur are specimens from animals that received antibiotic therapy before death or environmental samples containing sporostatic chemicals. Due to the increasing concern to eliminate the use of animals for biological testing, this approach should be used as a last resort and only if justified. Adult mice or guinea-pigs are the animals of choice. If the samples involved are soils, the animals should be pretreated, the day before testing, with both tetanus and gas gangrene antiserum. The samples are prepared as described for culturing, including heat-shocking at 62.5°C for 15 minutes. Mice are injected subcutaneously with 0.05–0.1 ml; guinea-pigs are inoculated intramuscularly with up to 0.4 ml (0.2 ml in each thigh muscle). Any B. anthracis present will result in death in 48–72 hours and the organism can be cultured from the blood as described above. 1.2. Immunological detection and diagnosis It needs to be borne in mind that B. anthracis is antigenically very closely related to B. cereus, which is considered a ubiquitous component of the environmental microflora. The only unshared antigens that lend themselves to differentiating these two species by immunological approaches are the anthrax toxin antigens, produced during the exponential phase of growth, and the capsule of B. anthracis. This places considerable constraints on the extent to which immunological methods can be used in routine detection methodology. 1.2.1. Ascoli test Ascoli (1911) published a procedure for the detection of thermostable anthrax antigen in animal tissue being used for by-products. This uses antiserum raised in rabbits to produce a precipitin reaction. The test lacks high specificity, in that the thermostable antigens of B. anthracis are shared by other Bacillus spp., and is dependent on the probability that only B. anthracis would proliferate throughout the animal and deposit sufficient antigen to give a positive reaction. This test appears to be used only in Eastern Europe. Chapter 3.1.1. – Anthrax WOAH Terrestrial Manual 2023 7 To perform the Ascoli test, put approximately 2 g of sample in 5 ml of saline containing 1/100 final concentration of acetic acid and boil for 5 minutes. The resultant solution is cooled and filtered through filter paper. A few drops of rabbit antiserum (see preparation below) are placed in a small test tube. The filtrate from the previous step is gently layered over the top of the antiserum. A positive test is the formation of a visible precipitin band in under 15 minutes. Positive and negative control specimen suspensions should be included. Antiserum is prepared in rabbits by the subcutaneous inoculation of Sterne anthrax vaccine on days 1 and 14. On days 28 and 35, the rabbits receive 0.5 ml of a mixture of several strains of virulent B. anthracis not exceeding 105 colony-forming units (CFU)/ml suspended in saline. Alternatively, the live virulent bacteria can be inactivated by prolonged suspension in 0.2% formalised saline, but the antigen mass needs to be increased to 108–109 CFU/ml. The suspension should be checked for inactivation of the B. anthracis before animal inoculation by culture of 0.1 ml into 100 ml of nutrient broth containing 0.1% histidine and, after incubation at 37°C for 7 days, subculture on to blood or nutrient agar. The dose regimen for the formalised suspension after initial vaccination on days 1 and 14 is increasing doses of 0.1, 0.5, 1, and 2 ml given intravenously at intervals of 4–5 days. Following either procedure, a test bleed at 10 days after the last injection should determine whether additional 2 ml doses should be administered to boost the precipitin titre. 1.2.2. Immunofluorescence While some success has been achieved with immunofluorescence for capsule observation in the research situation (Ezzell & Abshire, 1996), it does not lend itself to routine diagnosis. 1.3. Confirmation of virulence with the polymerase chain reaction Confirmation of virulence can be carried out using the PCR. The following instructions are taken from the WHO (2008). Template DNA for PCR can be prepared from a fresh colony of B. anthracis on nutrient agar by suspension of a loop of growth in 25 µl sterile deionised (or distilled) water and heating to 95°C for 20 minutes. Following cooling to approximately 4°C, and brief centrifugation, the supernatant can be used for the PCR reaction. Examples of suitable primers (Beyer et al., 1996; Hutson et al., 1993) for confirming the presence of the pXO1 and pXO2 plasmids are given in the table below. Target Primer ID Sequence (5’ → 3’) Product size Concentration Protective antigen (PA) PA 5 3048–3029 TCC-TAA-CAC-TAA-CGA-AGT-CG 596 bp 1 mM PA 8 2452–2471 GAG-GTA-GAA-GGA-TAT-ACG-GT Capsule 1234 1411–1430 CTG-AGC-CAT-TAA-TCG-ATA-TG 846 bp 0.2 mM 1301 2257–2238 TCC-CAC-TTA-CGT-AAT-CTG-AG PCR can be carried out in 50 µl volumes using the above primers, 200 µM each of dATP, dCTP, dTTP and dGTP, 1.5 mM MgCl2 and 2.5 units of DNA polymerase, all in NH4 buffer, followed by the addition of 5 µl of template DNA. A 2% agarose gel has been found to work best with these small fragments. Alternatively, premixed, predispensed, dried beads available commercially can be used. These are stable at room temperature, containing all the necessary reagents, except primer and template, for performing 25 µl PCR reactions. The template can be added in a 2.5 µl volume. The following PCR cycle can be used: 1 × 95°C for 5 minutes; 30 × 95°C for 0.5 minute followed by 55°C for 0.5 minute followed by 72°C for 0.5 minute; 1 × 72°C for 5 minutes; cool to 4°C. Chapter 3.1.1. – Anthrax 8 WOAH Terrestrial Manual 2023 It should be noted that the primers given in the table above have proved successful for confirming the presence or absence of pXO1 and/or pXO2 in pure cultures of isolates from animal (including human) specimens or environmental samples. They may be unsuitable, however, for direct detection of B. anthracis in such specimens or samples. A choice of alternatives can be found in Jackson et al. (1998) and Ramisse et al. (1996). For the rare possibility that an isolate may lack both pXO1 and pXO2, a chromosomal marker should also be run; primers for these are also described in Jackson et al. (1998) and Ramisse et al. (1996). Ågren et al. (2013) published a very comprehensive study on the in-silico and in-vitro evaluation of 35 PCR-based methods for 20 chromosomal markers of B. anthracis. The PL3 (target: part of pro-phage type 3) assay (Wielinga et al., 2011) was identified as one of the best performing assays in this study and could be used for routine diagnostics. Real-time PCR assays have been developed for enhanced speed, sensitivity and specificity of detection of pXO1, pXO2 and chromosomal genes of Bacillus anthracis and other closely related Bacillus spp. (e.g. Hadjinicolaou et al., 2009; Hoffmaster et al., 2002; Irenge et al., 2010; Qi et al., 2001; Rao et al., 2010). Selection of a particular assay will be dependent on the fitness for purpose and source of starting material (e.g. isolates, clinical specimen, environmental sample), requirement to differentiate from other Bacillus spp. or vaccine strains, demonstration of genetic diversity or confirmation of isolate identity. It is important that the laboratory conducting real-time PCR evaluate the performance of the test for their purpose and complete a validation analysis to ensure that it has been optimised and standardised for its intended use (see Chapter 1.1.6 Validation of diagnostic assays for infectious diseases of terrestrial animals). Molecular typing techniques for B. anthracis like canonical single-nucleotide polymorphisms typing (e.g. Van Ert et al., 2007) and variable number tandem repeat analysis (e.g. Keim et al., 2000) are appropriate for use in specialised laboratories. Molecular typing based on whole genome sequencing e.g. core genome multilocus sequence typing may be useful to elucidate the diversity of B. anthracis genotypes circulating, to determine connections between outbreak events and supports infection chain tracing (Abdel-Glil et al., 2021). C. REQUIREMENTS FOR VACCINES 1. Background 1.1. Rationale and intended use of the product The most widely used vaccine for prevention of anthrax in animals was developed by Sterne (1937). He derived a rough variant of virulent B. anthracis from culture on serum agar in an elevated CO2 atmosphere. This variant, named 34F2, was incapable of forming a capsule and was subsequently found to have lost the pXO2 plasmid, which codes for capsule formation. It has become the most widely used strain world-wide for animal anthrax vaccine production. In Central and Eastern Europe, an equivalent pXO2– derivative, Strain 55, is the active ingredient of the current livestock vaccine. A list of manufacturers of anthrax vaccine for use in animals is given in Annex 5 of WHO (2008). The following information concerning preparation of the anthrax vaccine for use in animals is based on Misra (1991) and the WHO (1967). Generalised procedures are given; national regulatory authorities should be consulted in relation to Standard Operating Procedures that may pertain locally. 2. Outline of production and minimum requirements for conventional vaccines 2.1. Characteristics of the seed 2.1.1. Biological characteristics Anthrax vaccine production is based on the seed-lot system. A seed lot is a quantity of spores of uniform composition processed at one time and maintained for the purpose of vaccine preparation. Each seed lot is no more than three passages from the parent culture and must produce a vaccine that is efficacious and safe for use in animals. It is recommended that a large Chapter 3.1.1. – Anthrax WOAH Terrestrial Manual 2023 9 seed lot be prepared from the parent strain and preserved by lyophilisation for future production lots. The parent culture can be purchased1. 2.1.2. Quality criteria The seed lot is acceptable for anthrax vaccine if a vaccine prepared from the seed lot or a suspension harvested from a culture derived from a seed lot meets the requirements for control of final bulk with respect to freedom from bacterial contamination, safety and efficacy (immunogenicity). 2.2. Method of manufacture 2.2.1. Procedure i) Preparation of the master seed Seed lots are cultured on solid media formulated to promote sporulation of the organism. The solid medium formula for casein digest agar (sporulation agar) given in Misra (1991) is: 50 g tryptic digest of casein; 10 g yeast extract; 0.1 g CaCl2.6H2O; 0.01 g FeSO4.7H2O; 0.05 g MgSO4.7H2O; 0.03 g MnSO4.4H2O; 5.0 g K2HPO4; 1.0 g KH2PO4; 22 g agar; 1000 ml deionised or distilled water. The ingredients are dissolved in the water with the appropriate amount of heating; the solution is adjusted to pH 7.4, distributed into Roux bottles (120 ml per bottle) or other appropriate container, sterilised by autoclaving and cooled in the horizontal position. After the agar has solidified, excess liquid should be removed aseptically and the bottles left in an incubator (37°C) for at least 2 days to dry and to check the sterility. Volumes of 2 ml of vaccine seed should be spread across the agar in Roux bottles, which should be incubated at 37°C until at least 80% sporulation is apparent by microscopic examination of aseptically extracted loopfuls (at least 72 hours). The growth is harvested with 10 ml per bottle of sterile deionised or distilled water and checked for purity. After washing three times in sterile deionised or distilled water with final suspension, also in sterile deionised or distilled water, sterilised lyophilisation stabiliser is added and the suspension is dispensed into lyophilisation vials and freeze-dried. Attenuated vaccine strains can gradually lose their antigenicity over repeated subculturing conditions. Therefore, it is recommended that master seed lots be made in bulk and kept within three passages from the original seed culture. A large number of master seed stocks should be prepared. ii) Preparation and testing of the working seed Reconstitute a vial of seed stock and inoculate several slants (approximately 10 ml) of sporulation (casein digest) agar. Incubate at 37°C for 72 hours and store in a refrigerator. Test the slants for purity by culture on to nutrient agar plates and in nutrient broth (0.1 ml in 100 ml of nutrient broth). The latter should be subcultured on to nutrient agar after incubation at 37°C for 7 days and should be a pure culture of B. anthracis. A sample of the broth culture should also be checked for lack of motility. Volumes of seed needed for a production run should be calculated on the basis of harvesting the spores from each slant with 10 ml of sterile deionised or distilled water and using this to inoculate five Roux bottles. iii) Preparation of vaccine concentrate Roux bottles with casein digest agar are prepared as for the master seed in Section C.2.2.1.i above. One Roux bottle can be expected to yield about 2000 doses of vaccine. Each Roux bottle is inoculated with 2 ml of working seed suspension and incubated at 37°C with porous plugs for several days until small loopfuls of culture from randomly selected bottles show at least 90% of the organisms to be in sporulated forms when examined in wet mounts by phase contrast (phase bright spores) or following staining for spores. The growth from each 1 UK Health Security Agency, Microbiology Services, Porton Down, Salisbury SP4 0JG, UK (www.culturecollections.org.uk/) Chapter 3.1.1. – Anthrax 10 WOAH Terrestrial Manual 2023 bottle is then harvested with 20 ml of physiological saline. Tests for contaminants should be carried out by subculture to nutrient agar plates and inoculation of 100 ml nutrient broth with 0.1 ml of harvested spores followed by subculture to nutrient agar after 7 days at 37°C and by tests for motility. Acceptable harvests (i.e. those showing no evidence of contaminants) are pooled. iv) Glycerination Twice the volume of sterile, pure, neutral glycerol should be added to the bulk pool of vaccine concentrate. Saponin (0.1% final concentration) may also be added at this point if it is to be included as an adjuvant. Mix thoroughly (the inclusion of sterilised glass beads may be helpful). Carry out a purity test and hold for 3 weeks at ambient temperature to allow lysis of any vegetative bacteria, determine the viable spore count and store under refrigeration thereafter. v) Determining titre and dilution for use The number of culturable spores in the product is then calculated by spreading tenfold dilutions on nutrient agar plates. The suspension is diluted so that the final bulk contains the number of culturable spores desired. The diluent should contain the same proportions of saline, glycerol and (if being included) saponin as present in the vaccine concentrate. The vaccine should contain a minimum of 2–10 × 106 culturable spores per dose for cattle, buffaloes and horses, and not less than 1–5 × 106 culturable spores per dose for sheep, goats and pigs. vi) Filling the containers Distribution of aliquots of vaccine into single and multidose containers is performed as outlined in WHO (1965). Basically, the final bulk is distributed to containers in an aseptic manner in an area not used for production, and any contamination or alteration of the product must be avoided. The vaccine may be lyophilised after distribution into appropriate dosage containers. Containers are sealed as soon as possible with a material that is not detrimental to the product and that is capable of maintaining a hermetic seal for the life of the vaccine. 2.3. Requirements for substrates and media Please refer to Misra (1991) for detailed information on substrates and media used for anthrax vaccine production. 2.3.1. In-process controls i) Purity of the seed lot Purity tests consist of microscopic examination of stained smears with culture and motility tests as described in Section C.2.2. ii) Safety of the seed lot Not less than 5 × 109 culturable spores should be injected subcutaneously into each of three healthy, 1–2-year-old, unvaccinated sheep, which must survive an observation period of at least 10 days. iii) Immunogenicity of the seed lot At least 10 healthy guinea-pigs, 300–500 g in weight should be inoculated with 5 × 106 viable spores and observed for 21 days. At least 80% of the animals should survive. The immunised animals, together with three unimmunised controls, should then be challenged with 10 median lethal doses (LD50) of the strain 17 JB of B. anthracis. During a 10-day observation period, none of the immunised animals should succumb to the challenge while all the controls should die from anthrax. The test should be repeated if one of the immunised animals dies. Chapter 3.1.1. – Anthrax WOAH Terrestrial Manual 2023 11 2.3.2. Final product batch tests i) Sterility and purity The vaccine is a live culture of B. anthracis spores; sterility does not apply, but the batches must be tested for freedom from contamination (see Chapter 1.1.9 Tests for sterility and freedom from contamination of biological materials intended for veterinary use). ii) Safety Safety testing is performed on two healthy sheep or goats and consists of inoculating subcutaneously twice the recommended vaccination dose. The animals are observed for 10 days. The final bulk passes the test if no systemic reactions develop and if not more than a transient oedema is observed at the injection site. If the test is carried out in sheep only, a progressive oedema indicates that the vaccine may be unsuitable for goats. iii) Batch potency Efficacy or immunogenicity is tested on the final bulk as follows: at least ten healthy 300– 500 g guinea-pigs are inoculated with a sheep dose of the vaccine. The guinea-pigs are observed for 21 days, and at least 80% of the animals must survive the observation period. Surviving immunised guinea-pigs and three non-vaccinated controls are challenged with an appropriate dose of virulent B. anthracis. A recommended challenge is 200 LD50 of the Pasteur II strain (17JB). If, by 10 days after challenge, all vaccinated guinea-pigs survive and control animals die, the final bulk is deemed to be satisfactory. If any vaccinated animals die during the post-challenge observation period from a cause other than anthrax, and death is not associated with the vaccine, the test may be repeated. 2.4. Requirements for authorisation 2.4.1. Safety requirements i) Target and non-target animal safety The vaccine has been shown to cause disease in some goats and llamas; this may be related to the saponin adjuvant. The vaccine is not recommended for use in pregnant animals, nor in animals destined for slaughter within 2–3 weeks of vaccination. Local regulations may specify other time periods in some countries or regions, but there is no scientific reason for regarding meat from clinically healthy animals as unfit for human handling or consumption after a holding period of 2 weeks following vaccination. Concurrent administration of antibiotics to vaccinated animals is contraindicated as the antibiotic will interfere with the vaccine. Antibiotics should not be given for several days before and after vaccination. Accidental human inoculation is treated by expressing as much of the inoculum as possible from the injection site and washing the wound thoroughly with soap and water. Medical attention should be sought if infection develops. ii) Reversion–to-virulence for attenuated/live vaccines The 34F2 strain of B. anthracis is known to be stable and cannot produce capsule in vitro. iii) Environmental consideration Leftover vaccine, empty vials, and equipment used for vaccinating are contaminated with the live spores and should be autoclaved, disinfected, or incinerated. 2.4.2. Efficacy requirements i) For animal production Not applicable. ii) For control and eradication The recommended dose for cattle and horses is a minimum of 2–10 × 106 culturable spores; for sheep, goats and pigs, it is 1–5 × 106 culturable spores. The vaccine should contain these Chapter 3.1.1. – Anthrax 12 WOAH Terrestrial Manual 2023 spores in an appropriate volume, e.g. 2 × 106/ml. Immunity should be good for at least 1 year and it is recommended that an annual booster be given. Horses may be slow to develop immunity following initial vaccination; some manufacturers therefore recommend a two-dose initial vaccination, administered 1 month apart, followed by a single annual booster. Bacillus anthracis spores are stable in unlyophilised or lyophilised vaccine and preservatives are not required. Storage under refrigeration is recommended (4°C). As there is no generally acceptable test for stability of anthrax vaccines, it is recommended that, in each filling lot, the number of culturable spores be determined before and after holding at an appropriate temperature for an appropriate period. There should be no evidence of a fall in the number of culturable spores. 3. Vaccines based on biotechnology 3.1. Vaccines available and their advantage There are no vaccines based on biotechnology available for anthrax. 3.2. Special requirements for biotechnological vaccines, if any Not applicable. REFERENCES ABDEL-GLIL M.Y., CHIAVERINI A., GAROFOLO G., FASANELLA A., PARISI A., HARMSEN D., JOLLEY K.A., ELSCHNER M.C., TOMASO H., LINDE J. & GALANTE D. (2021). A Whole-Genome-Based Gene-by-Gene Typing System for Standardized High-Resolution Strain Typing of Bacillus anthracis. J. Clin. Microbiol., 59(7):e0288920. doi: 10.1128/JCM.02889-20 ABSHIRE T.G., BROWN J.E. & EZZELL J.W. (2005). Production and validation of the use of gamma phage for identification of Bacillus anthracis. J. Clin. Microbiol., 43, 4780–4788. ÅGREN J., HAMIDJAJA R.A., HANSEN T., RUULS R., THIERRY S., VIGRE H. & DERZELLE S. (2013). In silico and in vitro evaluation of PCR-based assays for the detection of Bacillus anthracis chromosomal signature sequences. Virulence, 4, 671– 685. AMINU O.R, LEMBO T., ZADOKS R.N., BIEK R., LEWIS S., KIWELU I., MMBAGA B.T., MSHANGA D., SHIRIMA G., DENWOOD M. & FORDE T.L. (2020). Practical and effective diagnosis of animal anthrax in endemic low-resource settings. PLoS Negl. Trop. Dis., 14(9):e0008655. doi: 10.1371/journal.pntd.0008655. eCollection 2020 Sep. ASCOLI A. (1911). Die Präzipitindiagnose bei Milzbrand. Centralbl. Bakt. Parasit. Infeckt., 58, 63–70. BEYER W., GLOCKNER P., OTTO J. & BOHM R. (1996). A nested PCR and DNA-amplification-fingerprinting method for detection and identification of Bacillus anthracis in soil samples from former tanneries. Salisbury Med. Bull., No. 87, Special Suppl., 47–49. BROWN E.R. & CHERRY W.B. (1955). Specific identification of Bacillus anthracis by means of a variant bacteriophage. J. Infect. Dis., 96, 34–39. EZZELL J.W. & ABSHIRE T.G. (1996). Encapsulation of Bacillus anthracis spores and spore identification. Salisbury Med. Bull., No 87, Special Suppl., 42. HADJINICOLAOU A.V., DEMETRIOU V.L., HEZKA J., BEYER W., HADFIELD T.L. & KOSTRIKIS L.G. (2009). Use of molecular beacons and multi-allelic real-time PCR for detection of and discrimination between virulent Bacillus anthracis and other Bacillus isolates. J. Microbiol. Methods, 78, 45–53. HOFFMASTER A.R., MEYER R.F., BOWEN M.P., MARSTON C.K., WEYANT R.S., THURMAN K., MESSENGER S.L., MINOR E.E., WINCHELL J.M., RASSMUSSEN M.V., NEWTON B.R., PARKER J.T., MORRILL W.E., MCKINNEY N., BARNETT G.A., SEJVAR J.J., Chapter 3.1.1. – Anthrax WOAH Terrestrial Manual 2023 13 JERNIGAN J.A., PERKINS B.A. & POPOVIC T. (2002). Evaluation and validation of a real-time polymerase chain reaction assay for rapid identification of Bacillus anthracis. Emerg. Infect. Dis., 8, 1178–1182. HUDSON M.J, BEYER W., BOHM R., FASANELLA A., GAROFOLO G., GOLINSKI R., GOOSSENS P.L., HAHN U., HALLIS B., KING A., MOCK M., MONTECUCCO C., OZIN A., TONELLO F. & KAUFMANN S.H.E. (2007). Bacillus anthracis: balancing innocent research with dual-use potential. Int. J. Med. Microbiol., 298, 345–364. HUTSON R.A., DUGGLEBY C.J., LOWE J.R., MANCHEE R.J. & TURNBULL P.C.B. (1993). The development and assessment of DNA and oligonucleotide probes for the specific detection of Bacillus anthracis. J. Appl. Bacteriol., 75, 463–472. IRENGE L.M., DURANT J.-F., TOMASO H., PILO P., OLSEN J.S., RAMISSE V., MAHILLON J. & GALA J.-L. (2010). Development and validation of a real-time quantitative PCR assay for rapid identification of Bacillus anthracis in environmental samples. Appl. Microbiol. Biotechnol., 88, 1179–1192. JACKSON P.J., HUGH-JONES M.E., ADAIR D.M., GREEN G., HILL K.K., KUSKE C.R., GRINBERG L.M., ABRAMOVA F.A. & KEIM P. (1998). PCR analysis of tissue samples from the 1979 Sverdlovsk anthrax victims: The presence of multiple Bacillus anthracis strains in different victims. Proc. Natl Acad. Sci. USA, 95, 1224–1229. KEIM P., PRICE L.B., KLEVYTSKA A.M., SMITH K.L., SCHUPP J.M., OKINAKA R., ET AL. (2000). Multiple-locus variable-number tandem repeat analysis reveals genetic relationships within Bacillus anthracis. J. Bacteriol., 182, 2928–1936. KNISELY R.F. (1966). Selective medium for Bacillus anthracis. J. Bacteriol., 92, 784–786. MISRA R.P. (1991). Manual for the Production of Anthrax and Blackleg Vaccines. Food and Agriculture Organisation of the United Nations (FAO) Animal Production and Health Paper 87, FAO, Rome, Italy. OWEN M.P., SCHAUWERS W., HUGH-JONES M.E., KIERNAN J.A., TURNBULL P.C.B. & BEYER W. (2013). A simple, reliable M’Fadyean stain for visualizing the Bacillus anthracis capsule. J. Microbiol. Methods, 92, 264–269. QI Y., PATRA G., LIANG X., WILLIAMS L.E., ROSE S., REDKAR R.J. & DELVECCHIO V.G. (2001). Utilization of the rpoB gene as a specific chromosomal marker for real-time PCR detection of Bacillus anthracis. Appl. Environ. Microbiol., 67, 3720–3727. RAO S.S., MOHAN K.V.K. & ATREYA C.D. (2010). Detection technologies for Bacillus anthracis: prospects and challenges. J. Microbiol. Methods, 82, 1–10. RAMISSE V., PATRA G., GARRIGUE H., GUESDON J.L. & MOCK M. (1996). Identification and characterization of Bacillus anthracis by multiplex PCR analysis of sequences on plasmids pXO1 and pXO2 and chromosomal DNA. FEMS Microbiol. Lett., 145, 9–16. ROSENBLATT J.E. & STEWART P.R. (1974). Combined activity of sulfamethoxazole, trimethoprim, and polymyxin b against gram-negative bacilli. Antimicrob. Agents Chemother., 6, 84–92. STERNE M. (1937). The effect of different carbon dioxide concentrations on the growth of virulent anthrax strains. Onderstepoort J. Vet. Sci. Anim. Ind., 9, 49–67. TURNBULL P.C.B., FRAWLEY D.A. & BULL R.L. (2007). Heat activation/shock temperatures for Bacillus anthracis spores and the issue of spore plate counts versus true numbers of spores. J. Microbiol. Methods, 68, 353–357. VAN ERT M.N., EASTERDAY W.R., HUYNH L.Y., OKINAKA R.T., HUGH-JONES M.E., RAVEL J., ZANECKI S.R., PEARSON T., SIMONSON T.S., U’REN J.M., KACHUR S.M., LEADEM-DOUGHERTY R.R., RHOTON S.D., ZINSER G., FARLOW J., COKER P.R, SMITH K.L., WANG B., KENEFIC LG., FRASER-LIGGETT C.M., WAGNER D.M. & KEIM P. (2007). Global genetic population structure of Bacillus anthracis. PLoS One, 2(5):e461. Epub 2007/05/24. doi: 10.1371/journal.pone.0000461. WIELINGA P.R., HAMIDJAJA R.A., AGREN J., KNUTSSON R., SEGERMAN B., FRICKER M., EHLING-SCHULZ M., DE GROOT A., BURTON J., BROOKS T., JANSE I. & VAN ROTTERDAM B. (2011). A multiplex real-time PCR for identifying and differentiating B. anthracis virulent types. Int. J. Food Microbiol., 145, Suppl. 1: S137-44. doi: 10.1016/j.ijfoodmicro.2010.07.039. Epub 2010 Aug 10. PMID: 20826037. WORLD HEALTH ORGANIZATION (WHO) (1965). General Requirements for Manufacturing Establishments and Control Laboratories: Distribution of aliquots of vaccine into single and multidose containers. Requirements for Biological Substances No. 1. WHO Technical Report No. 363. WHO, Geneva, Switzerland, 16–17. Chapter 3.1.1. – Anthrax 14 WOAH Terrestrial Manual 2023 WORLD HEALTH ORGANIZATION (WHO) (1967). World Health Organization Expert Committee on Biological Standardization Requirements for Anthrax Spore Vaccine (Live – for Veterinary Use). Requirements for Biological Substances No. 13. WHO Technical Report Series No. 361. WHO, Geneva, Switzerland. WORLD HEALTH ORGANIZATION (WHO) (2008). Anthrax in humans and animals, Fourth Edition. WHO Press, World Health Organization, Geneva, Switzerland. NB: There are WOAH Reference Laboratories for anthrax (please consult the WOAH Web site: Please contact the WOAH Reference Laboratories for any further information on anthrax diagnostic tests, reagents and vaccines NB: FIRST ADOPTED IN 1989. MOST RECENT UPDATES ADOPTED IN 2023.
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Section 3.6 : Derivatives of Exponential and Logarithm Functions
The next set of functions that we want to take a look at are exponential and logarithm functions. The most common exponential and logarithm functions in a calculus course are the natural exponential function, ex, and the natural logarithm function, ln(x). We will take a more general approach however and look at the general exponential and logarithm function.
Exponential Functions
We’ll start off by looking at the exponential function,
f(x)=ax
We want to differentiate this. The power rule that we looked at a couple of sections ago won’t work as that required the exponent to be a fixed number and the base to be a variable. That is exactly the opposite from what we’ve got with this function. So, we’re going to have to start with the definition of the derivative.
f′(x)=limh→0f(x+h)−f(x)h=limh→0ax+h−axh=limh→0axah−axh=limh→0ax(ah−1)h
Now, the axis not affected by the limit since it doesn’t have any h’s in it and so is a constant as far as the limit is concerned. We can therefore factor this out of the limit. This gives,
f′(x)=axlimh→0ah−1h
Now let’s notice that the limit we’ve got above is exactly the definition of the derivative of f(x)=ax at x=0, i.e. f′(0). Therefore, the derivative becomes,
f′(x)=f′(0)ax
So, we are kind of stuck. We need to know the derivative in order to get the derivative!
There is one value of a that we can deal with at this point. Back in the Exponential Functions section of the Review chapter we stated that e=2.71828182845905… What we didn’t do however is actually define where e comes from. There are in fact a variety of ways to define e. Here are three of them.
Some Definitions of e
e=limn→∞(1+1n)n
e is the unique positive number for which limh→0eh−1h=1
e=∞∑n=01n!
The second one is the important one for us because that limit is exactly the limit that we’re working with above. So, this definition leads to the following fact,
Fact 1
For the natural exponential function, f(x)=ex we have f′(0)=limh→0eh−1h=1.
So, provided we are using the natural exponential function we get the following.
f(x)=ex⇒f′(x)=ex
At this point we’re missing some knowledge that will allow us to easily get the derivative for a general function. Eventually we will be able to show that for a general exponential function we have,
f(x)=ax⇒f′(x)=axln(a)
Logarithm Functions
Let’s now briefly get the derivatives for logarithms. In this case we will need to start with the following fact about functions that are inverses of each other.
Fact 2
If f(x) and g(x) are inverses of each other then,
g′(x)=1f′(g(x))
So, how is this fact useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know what the derivative of the natural exponential function is!
So, if we have f(x)=ex and g(x)=lnx then,
g′(x)=1f′(g(x))=1eg(x)=1elnx=1x
The last step just uses the fact that the two functions are inverses of each other.
Putting this all together gives,
ddx(lnx)=1xx>0
Note that we need to require that x>0 since this is required for the logarithm and so must also be required for its derivative. It can also be shown that,
ddx(ln|x|)=1xx≠0
Using this all we need to avoid is x=0.
In this case, unlike the exponential function case, we can actually find the derivative of the general logarithm function. All that we need is the derivative of the natural logarithm, which we just found, and the change of base formula. Using the change of base formula we can write a general logarithm as,
logax=lnxlna
Differentiation is then fairly simple.
ddx(logax)=ddx(lnxlna)=1lnaddx(lnx)=1xlna
We took advantage of the fact that a was a constant and so lna is also a constant and can be factored out of the derivative. Putting all this together gives,
ddx(logax)=1xlna
Here is a summary of the derivatives in this section.
ddx(ex)=exddx(ax)=axlnaddx(lnx)=1xddx(logax)=1xlna
Okay, now that we have the derivations of the formulas out of the way let’s compute a couple of derivatives.
Example 1 Differentiate each of the following functions.
R(w)=4w−5log9w
f(x)=3ex+10x3lnx
y=5ex3ex+1
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a R(w)=4w−5log9w Show Solution
This will be the only example that doesn’t involve the natural exponential and natural logarithm functions.
R′(w)=4wln4−5wln9
b f(x)=3ex+10x3lnx Show Solution
Not much to this one. Just remember to use the product rule on the second term.
f′(x)=3ex+30x2lnx+10x3(1x)=3ex+30x2lnx+10x2
c y=5ex3ex+1 Show Solution
We’ll need to use the quotient rule on this one.
y′=5ex(3ex+1)−(5ex)(3ex)(3ex+1)2=15e2x+5ex−15e2x(3ex+1)2=5ex(3ex+1)2
There’s really not a lot to differentiating natural logarithms and natural exponential functions at this point as long as you remember the formulas. In later sections as we get more formulas under our belt they will become more complicated.
Next, we need to do our obligatory application/interpretation problem so we don’t forget about them.
Example 2 Suppose that the position of an object is given by
s(t)=tet
Does the object ever stop moving?
Show Solution
First, we will need the derivative. We need this to determine if the object ever stops moving since at that point (provided there is one) the velocity will be zero and recall that the derivative of the position function is the velocity of the object.
The derivative is,
s′(t)=et+tet=(1+t)et
So, we need to determine if the derivative is ever zero. To do this we will need to solve,
(1+t)et=0
Now, we know that exponential functions are never zero and so this will only be zero at t=−1. So, if we are going to allow negative values of t then the object will stop moving once at t=−1. If we aren’t going to allow negative values of t then the object will never stop moving.
Before moving on to the next section we need to go back over a couple of derivatives to make sure that we don’t confuse the two. The two derivatives are,
ddx(xn)=nxn−1Power Ruleddx(ax)=axlnaDerivative of an exponential function
It is important to note that with the Power rule the exponent MUST be a constant and the base MUST be a variable while we need exactly the opposite for the derivative of an exponential function. For an exponential function the exponent MUST be a variable and the base MUST be a constant.
It is easy to get locked into one of these formulas and just use it for both of these. We also haven’t even talked about what to do if both the exponent and the base involve variables. We’ll see this situation in a later section.
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Use of Desmopressin in Hyponatremia: Foe and Friend - PMC
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. 2019 Mar 14;1(2):65–70. doi: 10.1016/j.xkme.2019.02.002
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Use of Desmopressin in Hyponatremia: Foe and Friend
Steven G Achinger
Steven G Achinger
1 Department of Nephrology, Watson Clinic, LLP, Lakeland, FL
Find articles by Steven G Achinger
1, Juan Carlos Ayus
Juan Carlos Ayus
2 Renal Consultants of Houston, Houston, TX
3 Department of Nephrology, University of California, Irvine, Irvine, CA
Find articles by Juan Carlos Ayus
2,3,∗
Author information
Article notes
Copyright and License information
1 Department of Nephrology, Watson Clinic, LLP, Lakeland, FL
2 Renal Consultants of Houston, Houston, TX
3 Department of Nephrology, University of California, Irvine, Irvine, CA
∗
Address for Correspondence: Juan Carlos Ayus, MD, Renal Consultants of Houston, 2412 Westgate St, Houston, TX 77019. carlosayus@yahoo.com
Collection date 2019 Mar-Apr.
© 2019 The Authors
This is an open access article under the CC BY-NC-ND license (
PMC Copyright notice
PMCID: PMC7380358 PMID: 32734186
Abstract
Use of desmopressin (1-deamino-8-d-arginine vasopressin; DDAVP), a synthetic vasopressin receptor agonist, has expanded in recent years. Desmopressin leads to renal water retention, and iatrogenic hyponatremia may result if fluid intake is not appropriately restricted. It is common practice to stop a medication that is causing toxicity, and this advice is promulgated in Micromedex, which suggests withholding desmopressin if hyponatremia occurs. If intravenous saline solution is administered and desmopressin is withheld at the same time, rapid changes in serum sodium levels may result, which puts the patient at risk for demyelinating lesions. In the management of desmopressin-associated hyponatremia with neurologic symptoms, the drug should not be withheld despite the presence of hyponatremia. The medication should be continued while administering intravenous hypertonic saline solution. Desmopressin is also used to minimize water excretion during the correction of hyponatremia during water diuresis. When treating hyponatremia, clinicians should monitor closely to avoid free-water diuresis. To prevent ongoing water losses in urine and overly rapid “autocorrection” of serum sodium level, desmopressin can be given to reduce free-water losses. These treatment recommendations are the authors’ perspective from previously published work and personal clinical experience.
Index Words: Hyponatremia, DDAVP, central pontine myelinolysis, desmopressin
Introduction
Use of desmopressin (1-deamino-8-d-arginine vasopressin; DDAVP, Ferring Pharmaceuticals), a synthetic vasopressin receptor agonist, has expanded beyond the treatment of diabetes insipidus in recent years. Desmopressin leads to renal water retention, and iatrogenic hyponatremia may result, which is not an infrequent clinical occurrence.1, 2, 3 The increase in desmopressin use for causes unrelated to central diabetes insipidus, ease of administration of intranasal preparations, and recent approval (March 2007)4 of desmopressin for the treatment of enuresis in the elderly have all been factors favoring the increasing use of this medication in the outpatient setting.5
Desmopressin use in nursing home settings for enuresis can be particularly problematic because many patients are actively encouraged to drink liquids to avoid dehydration. Additionally, desmopressin is increasingly being used in the hospital setting specifically for limiting free-water excretion when managing complex hyponatremia cases in which overcorrection of serum sodium level is a concern. Therefore, desmopressin can either be a foe or a friend in the management of hyponatremia, on one hand leading to iatrogenic hyponatremia while on the other hand being used to limit free-water excretion and potential overcorrection of serum sodium level.
Desmopressin-Associated Hyponatremia: Desmopressin as a Foe
An illustrative case involves a 77-year-old woman who is a nursing home resident. She has severe cognitive impairment and has been taking desmopressin for the treatment of enuresis. She is receiving oral desmopressin each night before sleep. The nursing staff at this center encourages all patients to drink plenty of fluids, especially during the warmer summer months. She presents to the emergency department at 10:00 am with increased confusion and having had a tonic-clonic seizure. Urine osmolality is 550 mOsm/kg and urine sodium excretion is 85 mmol/L. Serum sodium level is 109 mmol/L.
Desmopressin therapy is discontinued and the patient is treated with 3% saline solution in the emergency department at a rate of 15 mL/h. Six hours after admission to the intensive care unit, serum sodium level is 122 mmol/L and urine output has increased substantially. Urine osmolality is now 80 mOsm/kg; 3% saline solution treatment is discontinued and the patient is placed on 125 mL/h of 5% dextrose and water intravenously. Six hours later, the patient’s serum sodium level is 133 mmol/L and her high urine output continues. Although the infusion rate of dextrose and water is increased to 200 mL/h, serum sodium level stabilizes at 141 mmol/L. Three to 4 days later, the patient starts to develop spasticity and becomes poorly responsive; brain magnetic resonance imaging findings are consistent with cerebral demyelination.
Desmopressin-associated hyponatremia occurs when failure to appropriately restrict water intake while taking desmopressin leads to hyponatremia.1 Because desmopressin will cause the kidneys to concentrate urine, fluid intake must be carefully monitored. Urine osmolality during therapy with desmopressin has been shown to increase to the 600-mOsm/kg range in the critically ill1 and up to 879 mOsm/kg in healthy volunteers,6 leading to abrupt decreases in urine output that are sustained for up to 8 to 12 hours.7 For this reason, improper fluid restriction during desmopressin administration can lead to hyponatremia. It is important that patients taking desmopressin be counseled about the proper degree of fluid intake and that hospitalized patients do not receive hypotonic intravenous fluids, unless during the correction of hypernatremia. Women are more sensitive to the effects of desmopressin and experience worse outcomes from hyponatremia. Therefore, doses may need moderation in this group.8, 9, 10
Hyponatremic encephalopathy, symptomatic cerebral edema secondary to hyponatremia, can have a fulminant presentation, but early signs are nonspecific, such as nausea, vomiting, and headaches, and can go unrecognized.8, 9, 11 Early manifestations are signs of cerebral edema; as the edema worsens, seizures may occur.7 If uncorrected, the ultimate manifestations (respiratory failure and death) are due to brainstem herniation.8
It is common practice to stop treatment with a medication that is causing toxicity, and this advice is promulgated in Micromedex (Thomson Reuters [Healthcare] Inc), which suggests withholding desmopressin if hyponatremia occurs; however, no suggestions regarding fluid therapy are offered.12 In the case of desmopressin-associated hyponatremia, withholding the drug may have untoward consequences. Therefore, treatment of symptomatic desmopressin-associated hyponatremia with neurologic symptoms can be a clinical challenge. Increasing serum sodium level with hypertonic saline solution acutely is indicated; however, if use of the medication is simply discontinued, a spontaneous free-water diuresis will occur due to the decrease in urine osmolality, and rapid autocorrection of serum sodium level is possible. If desmopressin is withheld while administering intravenous saline solution, there also is potential for rapid changes in serum sodium level.1, 2 These factors combined can contribute to overly rapid correction of serum sodium level, putting the patient at risk for neurologic injury due to osmotic demyelination syndrome. A patient with desmopressin-associated hyponatremia will initially be in an antidiuretic state and, if acutely symptomatic, will need early intervention with hypertonic saline solution to increase serum sodium concentration and avoid cerebral edema.
Our group recently reported 13 cases of neurologic injury and/or death due to inappropriate correction of serum sodium level when stopping desmopressin treatment during the management of symptomatic desmopressin-associated hyponatremia.1 This report highlights the potential dangers of desmopressin-induced hyponatremia and severe neurologic outcomes that can ensue.2 Discontinuing desmopressin treatment in the management of desmopressin-associated hyponatremia can lead to autocorrection of serum sodium level and significant overcorrection can occur, especially if intravenous saline solution (especially a hypertonic solution) is given at the same time (as was seen in our case series in which serum sodium level corrected on average by 37 mEq/L over 48 hours).1 If desmopressin treatment is stopped, free-water excretion can then occur unabated. Of additional concern, if desmopressin is withheld in patients with central diabetes insipidus, overcorrection past normonatremia is possible and is an additional risk factor for cerebral demyelination.13
In managing desmopressin-associated hyponatremia with neurologic symptoms, despite the presence of hyponatremia, use of the medication should be continued.1, 2, 7 Symptomatic patients should be given hypertonic saline solution while continuing to receive desmopressin. The approach to treatment of desmopressin-associated hyponatremia with hypertonic saline solution is summarized in Figure 1. The most critical point to emphasize is that in cases of desmopressin-associated hyponatremia, the medication should be continued despite symptomatic hyponatremia, and hypertonic saline solution should be given to treat the hyponatremia.
Figure 1.
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Treatment of desmopressin (DDAVP)-associated hyponatremic encephalopathy.
Desmopressin as a Therapeutic Tool in Managing Severe Hyponatremia: Desmopressin as a Friend
This next case demonstrates severe symptomatic hyponatremia due to thiazide use7 in which 3% saline solution was used initially and desmopressin was added to treat a water diuresis and prevent overcorrection (Fig 2).
Figure 2.
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Treatment of severe symptomatic hyponatremia with intravenous (IV) 3% saline solution and subcutaneous desmopressin (DDAVP). Abbreviation: Osm, osmolality. Achinger and Ayus7; reproduced with permission from Wolters Kluwer Health, Inc.
A 63-year-old woman with a history of alcohol abuse presents to the emergency department with 4 days of confusion followed by a tonic-clonic seizure, after which she is transported to the hospital by ambulance. She is found to be obtunded and her initial serum sodium level is 102 mEq/L. Following the initial correction of serum sodium level, the patient starts to undergo a water diuresis. Her sodium level initially corrects 13 mEq in the first 12 hours and she has no further seizures. With ongoing water losses, she would quickly exceed the safe limit of sodium level correction for the first 48 hours. Because the long-term consumption of alcohol raises the risk for cirrhosis, it is important that serum sodium level correction not be excessive in such cases. The decision is made to therapeutically slow the correction of serum sodium level with desmopressin. She is kept strictly without enteral fluid intake and is treated with a combination of subcutaneously administered desmopressin and intravenous 0.9% saline solution. When a patient is treated with hypertonic saline solution, ongoing water losses need to be continually assessed to be vigilant for water diuresis, as occurred in the case presented.
We have previously introduced the concept that desmopressin could be used to minimize water excretion during correction of hyponatremia in cases that involve the risk for large-volume water diuresis; for example, compulsive water drinking, cortisol deficiency, thyroid deficiency, and medication-induced hyponatremia (eg, thiazide diuretics).14 This concept has been validated in several subsequent articles and case reports, showing the clinical utility of this therapeutic maneuver.1, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27 This situation is termed a reversed urine osmolality.
Conceptually, reversed urine osmolality is a convenient nomenclature for categorizing conditions in which urine osmolality changes from high to low in the course of treatment of hyponatremia. Reversed urine osmolality is defined in contrast to fixed urine osmolality, which does not decrease significantly during the treatment of hyponatremia (Fig 3). Examples of hyponatremia causes for which reversed osmolality tend to be problematic are volume depletion, adrenal insufficiency, medication associated (especially desmopressin), and hypothyroidism.
Figure 3.
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Changes in urine osmolality during hyponatremia treatment.
When treating hyponatremia, clinicians need to monitor closely to avoid free-water diuresis. Desmopressin can be given to increase the urine concentration and reduce free-water loss, thereby limiting autocorrection of serum sodium level. It is vital that this be done carefully, and the patient must have strict fluid restrictions or have no enteral intake. Severe hyponatremia can develop if there is unrestricted fluid intake while desmopressin is being administered. The first sign that water diuresis is occurring is increased urine output; thus, hourly urine output should be monitored in all patients with hyponatremic encephalopathy, especially those with drug-induced hyponatremia. Desmopressin should only be used in situations in which free-water diuresis is either occurring or is imminent. Indiscriminate use of desmopressin in the treatment of hyponatremia can lead to iatrogenic hyponatremia and unnecessarily long durations of fluid restrictions.
Medication-induced hyponatremia is a common clinical syndrome.5 Thiazide diuretics, select antidepressants, and anticonvulsants are frequently associated with hyponatremia. In all cases of drug-induced hyponatremia, the potential exists for rapid correction of hyponatremia, especially when the offending agent is withdrawn because there is a risk for water diuresis when urinary concentration decreases and free-water excretion increases. We refer to this situation as reversed urine osmolality, in which initially there is antidiuretic hormone release and high urine osmolality due to a reversible condition, which reverses when this condition resolves and urine osmolality decreases. This situation can lead to unanticipated free-water losses. This can occur in several commonly seen clinical scenarios (Fig 4). In this regard, desmopressin-associated hyponatremia shares this basic similarity with all cases of drug-induced hyponatremia.
Figure 4.
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Recognizing clinical scenarios in which desmopressin (DDAVP) may be needed to correct hyponatremia safely. Abbreviation: SIADH, syndrome of inappropriate antidiuretic hormone. Achinger and Ayus7; reproduced with permission from Wolters Kluwer Health, Inc.
Because desmopressin reduces free-water excretion, it is very important that exogenous fluid intake (both enteral and parenteral) be restricted to prevent iatrogenic hyponatremia and possible treatment-related injury. Hospitalized patients often have conditions that can lead to increased thirst or increased fluid intake (such as dry mouth, delirium, or psychiatric illness) and iatrogenic hyponatremia can develop. We do not advocate the routine use of desmopressin in treating hyponatremia; rather, desmopressin use is advisable in patients with either: (1) desmopressin-associated hyponatremia accompanied by symptoms of encephalopathy in whom prompt correction of hyponatremia is warranted with hypertonic saline solution or (2) high risk for overcorrection of hyponatremia due to free-water diuresis.
We recommend using subcutaneous desmopressin at 2 μg every 12 hours or intranasal desmopressin, 20 μg, every 12 hours; however, other routes of administration also may be appropriate. The hourly urine output should decrease substantially in the first 1 to 2 hours after administration. To further verify therapeutic efficacy, urine osmolality should be checked to ensure that urine osmolality has increased; typically, in the 500- to 600-mOsm/kg range should ensure a significant antiaquaretic effect.
The majority of cases in the literature reported to date describe the use of desmopressin to prevent overcorrection when water diuresis has occurred or to use as an adjunct with hypotonic fluids to re-lower sodium level when overcorrection has taken place (Table 1). However, the use of desmopressin in cases of syndrome of inappropriate antidiuretic hormone (SIADH) or other states associated with an elevated vasopressin level has also been advocated in the literature.17 In states of elevated circulating vasopressin such as SIADH, we believe that it is of no benefit to administer desmopressin. This treatment is potentially deleterious if patients are not adequately fluid restricted and may lead to prolonged recovery times.26, 27 We advocate that patients treated for hyponatremic encephalopathy be monitored for the development of water diuresis with accurate and timely measurements of urine output and desmopressin should be administered only in cases in which free-water diuresis is occurring.
Table 1.
Studies Evaluating the Use of Desmopressin in Prevention of Overcorrection of Hyponatremia
| Study | Study Type | Timing of Desmopressin Administration | Outcome |
--- --- |
| Goldszmidt & Iliescu15 (2000) | Single case report | Desmopressin administered after sodium increased 19 mmol/L in first 19 h of therapy, in patient with polydipsia | No neurologic injury noted |
| Perianayagam et al16 (2008) | Retrospective review of 20 cases | Retrospective chart review of 6 patients with hyponatremia administered desmopressin after sodium corrected by >12 mEq/L in 24 h and 14 patients with hyponatremia given desmopressin concurrently with 3% sodium chloride solution | No neurologic injury noted in either group |
| Sterns et al17 (2010) | Single case report | Desmopressin administered concurrently with 3% sodium chloride solution in patient with alcoholism, using a thiazide diuretic and serum sodium of 96 mEq/L | No neurologic injury noted |
| Tomlin et al18 (2011) | Single case report | Desmopressin administered after sodium increased 12 mmol/L in first 18 h of therapy in patient with volume depletion and sodium of 109 mmol/L | No neurologic injury noted |
| Quinn et al19 (2012) | Single case report | Use of desmopressin and fluid restriction to treat patient with psychogenic polydipsia and seizures | No long-term sequelae |
| Sood et al20 (2013) | Retrospective chart review of 25 cases | Use of desmopressin and hypertonic saline solution at outset of therapy | No neurologic injury noted |
| Lum21 (2013) | Single case report | Desmopressin administrated to patient with schizophrenia and alcohol abuse; serum sodium increased from 106 to 126 mmol/L in first 2 hospital d, then stabilized | No neurologic injury noted |
| Gharaibeh et al22 (2013) | Single case report | Desmopressin administered with hypotonic fluids after serum sodium increased rapidly (19 mEq/L) in first 12 h of therapy | No neurologic injury noted |
| Rafat et al23 (2014) | Retrospective review of 20 cases | Desmopressin administered with hypotonic fluids after serum sodium increased rapidly (19 mEq/L) in first 12 h of therapy | 1 case of mild osmotic demyelination in patient with comorbid alcohol abuse |
| Changal et al24 (2014) | Single case report | Desmopressin administered with hypotonic fluids after development of central pontine myelinolysis | Survival with residual neurologic deficit |
| Achinger et al1 (2014) | Case series | 2 patients with hyponatremia and neurologic symptoms treated with desmopressin at onset of free-water diuresis; serum sodium had not yet increased beyond acceptable limits; overall correction 11 mEq/L over 48 h | No neurologic symptoms or injury |
| De Vecchis et al25 (2017) | Single case report | Desmopressin administered after rapid correction of sodium in patient with congestive heart failure | No neurologic injury noted |
| MacMillan & Cavalcanti26 (2017) | Retrospective observational study of 1,450 hospital admissions | Desmopressin administered in 254 patients as either a reactive or proactive strategy | 4/1,450 patients had suspected demyelination syndrome; hospital length of stay was longer in proactive strategy group |
| Ward et al27 (2018) | Retrospective observational comparison study | 16 patients received desmopressin, 5 patients received proactive strategy, 9 patients received reactive strategy, 2 received “reserve” therapy to reverse an overcorrection | No episodes of overcorrection or neurologic injury occurred; no difference in change in serum sodium, but desmopressin treatment group had longer length of hospital stay |
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Summary
Desmopressin use has been associated with iatrogenic hyponatremia, and treatment of this condition has potential pitfalls if the medication is withheld during therapy. Continuing desmopressin use during treatment of severe symptomatic desmopressin-associated hyponatremia appears to be a better strategy for managing this condition. This is because continuing the medication prevents free-water diuresis from occurring. Furthermore, in other causes of hyponatremia associated with reversed urine osmolality (such as drug induced or polydipsia), desmopressin can be used to slow the autocorrection of serum sodium level if free-water diuresis complicates therapy. The use of desmopressin in cases of SIADH or other states associated with an elevated vasopressin level, as has been advocated in the literature, is unnecessary in our view and subjects patients to unneeded therapy and may lead to prolonged recovery times.
Article Information
Authors’ Full Names and Academic Degrees
Steven G. Achinger, MD, and Juan Carlos Ayus, MD.
Support
None.
Financial Disclosure
The authors declare that they have no relevant financial interests.
Peer Review
Received November 5, 2018. Evaluated by 2 external peer reviewers, with direct editorial input from the Editor-in-Chief. Accepted in revised form February 25,2019.
Footnotes
Complete author and article information provided before references.
References
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Desmopressin-Associated Hyponatremia: Desmopressin as a Foe
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https://mntca.umn.edu/sites/mntca.umn.edu/files/files/media/murphy_dealing_with_urban_gingko_seeds_11.14.18_0.pdf
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Dealing with Urban Ginkgo Seeds: Turning a nuisance into a new local food product By: Ryan Murphy Department of Forest Resources University of Minnesota Ginkgo biloba can be a divisive tree in the urban environment. Many enthusiasts believe ginkgo to be one of the best urban tree species for the Northern Midwest region. It has seemingly no pest or pathogen issues, resists urban pollutants, puts up with tough planting locations and turns a vibrant golden yellow in the fall. On the flipside, many know of ginkgo simply for its obnoxious seed, which releases a strong odor as the outer seed coat decomposes. It is this odiferous aspect of ginkgo that accounts for the majority of removals in urban locations. But does it need to be this way? A small project housed between the Department of Forest Resources and Department of Horticultural Science is asking this question. Ginkgo is an important link to the past both botanically and culturally. Ginkgo biloba is the only living species of a once diverse division known as Ginkgophyta. These seed plants were common across the northern hemisphere as far back as 300 million years ago (Leistner & Drewke 2010). Now, it is believed there are only two small remaining populations of wild ginkgo trees located in China (Gong et al. 2008). G. biloba is currently on the endangered species list - a notion that is odd to comprehend due to the tree’s ubiquity in the landscape. The earliest modern ginkgo trees arrived in the United States in the late 18th century. Ginkgo is a dioecious species, and it became apparent to astute and non-astute observers alike that the seedless male trees are more desirable for the urban environment than the seed producing female trees. In spite of this knowledge, a number of seed producing trees were still planted into the landscape. The reason for the seeming mishap is unclear. One explanation is that selected stock plants were falsely believed to be male cultivars when in fact these trees were females that had yet to reach sexual maturity, a consequence of ginkgo taking multiple decades to reach seed baring age. There is also the not too uncommon theory that these seed producing trees resulted from a sex change of males converting to female. This theory is not totally unfounded as there are a number of accounts from around the world of very old male ginkgo trees spontaneously producing seed (Crane 2013). What is common in these accounts, however, is that there does not seem to be a complete canopy wide change of sex, rather isolated branches in the canopy producing some seeds. Interestingly, there are also accounts of monoecious trees. In one research trial of 638 seedling ginkgo, four trees were found to be monoecious, predominated by male catkins with a few female ovules (Santamour et al. 1983). Modern genetic tools have allowed researchers to identify a single marker which can be utilized in determining the sex of ginkgo seedlings (Jiang et al. 2003, Echenard et al. 2008). Simply stated, the ginkgo seeds for many are a public nuisance. The smell is produced by the release of butanoic and hexanoic acid from the outer-fleshy seed coat known as the sarcotesta (Tredici 2000). Compounding the matter is the fact that gingko is a prolific seed producer, with mature trees dropping hundreds of pounds of seed each year. The presence of seed producing trees in the near vicinity of houses and public sidewalks results in a number of requests for ginkgo tree removals each year. For other residents, however, the seed is seen as a healthful food and medicine. This is especially so for East Asian cultures. Cultivation of ginkgo trees and use of the seed as a food source dates back thousands of years in China. In the present day, commercial seed production is estimated at between 5,000-6,000 tons per year (He et al. 1997). In Western cultures, however, it was the leaf extract that gained popularity as a supplement with purported neurological benefits. The historic use of ginkgo seed as a healthful food, paired with the negative public perception of the seed in the urban environment, led university researchers in the departments of Forest Resources and Horticultural Science to undertake a project to assess the potential for urban produced ginkgo seed to be harvested commercially for human consumption. The benefits of collection and subsequent sale of urban ginkgo seed are two-fold; 1) collection of seeds from the landscape will reduce the negative impact on residents, resulting in fewer requests for removals and thereby retain the environmental benefits of these potentially long-lived trees, and 2) the sale of urban ginkgo seed will provide a new local food revenue source. Work on this project began with a grant from the Minnesota Department of Agriculture’s Specialty Crop Block Grant Program to ensure that the urban grown seeds do not pose any health risks to those consuming them. This is being accomplished by assessing seeds for the presence of any heavy metals as well as a compound called ginkgotoxin (4’-O-methylpyridoxine), a known neurotoxin. So far the results of the food safety research are promising. The challenge moving forward will be creating the mechanisms and supply chain to harvest and sell the seeds. Future research on harvest mechanisms will look at ways to efficiently collect seeds from urban trees. For example, mechanically shaking the trees in the autumn so all the seeds are removed at one time. And, of course, we can’t forget a good method for cleaning the seeds. The other critical piece of the puzzle is public education. It is understandable that citizens are annoyed by the presence of the seed each year. But where problems lie, so does opportunity. Hopefully someday folks will have more than just tree removal as an option for managing the seeds. A. raw ginkgo nuts, B. ginkgo nuts (no shell), C. honey ginkgo nuts, a dish to nourish the lungs and trachea, D. dry fried ginkgo nuts are used to relieve asthma, E. salty roasted ginkgo nuts are used as healthy snacks, F. ginkgo shrimp is a common Chinese dish, G. ginkgo asparagus, another famous Chinese dish For more information on the project, contact Ryan Murphy at murph523@umn.edu. Citations: Crane, P. (2013) Ginkgo: the tree that time forgot. Yale University Press. New Haven and London. Echenard, V., Lefort, F., Calmin, G., Perroulaz, R., Belhahri, L. (2008) A New and Improved Automated Technology for Early Sex Determination of Ginkgo biloba. Arboriculture & Urban Forestry 34(5):300-307 Gong, W., Chen, C., Dobes, C., Fu, C.X., Kock, M.A. (2008) Phylogeography of a living fossil: Pleistocene glaciations forced Ginkgo biloba L. (Ginkgoaceae) into two refuge areas in China with limited subsequent postglacial expansion. Molecular Phylogenetics and Evolution 48, 1094-1105. He, S.-A., Gu, Y., Pang, Z.-J. (1997) Resources and Prospects of Ginkgo biloba in China. Ginkgo – A Global Treasure: From Biology to Medicine. Springer-Verlag Tokyo. pp 373-383. Jiang, L., You, R.L., Li, M.X., Shi, C. (2003) Identification of a sex associated RAPD marker in Ginkgo biloba. Acta Botanica Sinica 45:742-747 Leistner, E. & Drewke, C. (2010) Ginkgo biloba and Ginkgotoxin. Journal of Natural Products, 73, 86-92 Santamour Jr., F.S., He, S., Ewert, T.E. (1983) Growth, Survival and Sex Expression in Ginkgo. Journal of Arboriculture 9(6): June. Tredici, P.D. (2000) The Evolution, Ecology, and Cultivation of Ginkgo biloba. Ginkgo biloba, Hardwood Academic Publishers p. 7-23
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https://www.exploratorium.edu/sites/default/files/pdfs/GPGFinalReport90.pdf
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SUMMATIVE EVALUATION OF GEOMETRY PLAYGROUND for The Exploratorium Selinda Research Associates, Inc. January 2012 Selinda Research Associates, Inc. ii Contributors Evaluation Team: Selinda Research Associates, Inc. Deborah Perry, Director, and project manager for this study. Eric D. Gyllenhaal, Senior Research Associate, and lead researcher for this study. Exploratorium Josh Gutwill, Acting Director of Visitor Research, and Co-PI for Geometry Playground Peggy Monahan, Project Director for Geometry Playground Toni Dancu, Lead Researcher for Geometry Playground Suggested Citation: Selinda Research Associates. (2012). Summative evaluation of Geometry Playground. Unpublished manuscript, Exploratorium, San Francisco, CA. Please address all communication to Deborah Perry, DLPerry@SelindaResearch.com All photographs in this report © 2011 Selinda Research Associates, Inc. Selinda Research Associates, Inc. iii ACKNOWLEDGEMENTS First and foremost we would like to thank the members of the Geometry Playground evaluation team for helping guide our work: Josh Gutwill, Toni Dancu, and Peggy Monahan provided invaluable background on the exhibit development process including front-end and formative studies; research on spatial reasoning; assistance with the evaluation plans and implementation; and input and feedback on numerous documents including the site visit and final reports. We are grateful for the Exploratorium’s continued commitment to research, visitor studies, and the incorporation of evaluation recommendations in their pursuit of rich and engaging visitor experiences. We would like to thank all the staff at the Exploratorium and two host museums who helped facilitate our site visits. At the Exploratorium, our thanks to the members of the Geometry Playground evaluation team and exhibit team, especially P.I. Thomas Rockwell and Co-P.I. Josh Gutwill. Special thanks also to Nina Hido and Toni Dancu, who introduced us to the exhibition and assisted with data collection, and to the exhibit developers, writers, project managers, and program developers who talked with us in meetings and individually. At the Science Museum of Minnesota: Sarah Cohn, Kirsten Ellenbogen, and J. Newlin; and at Don Harrington Discovery Center: Sarah Cook, Chip Lindsey, and Dick Falkard. Geometry Playground was funded in large part by the National Science Foundation’s Division of Research on Learning in Formal and Informal Settings (NSF/DRL #0610436). We are grateful to NSF for their continuing support of evaluation in informal science education projects. Additional funding was also provided by the Gordon and Betty Moore Foundation, grant #1282. We are also grateful to the many additional organizations and individuals who supported Geometry Playground at the host museums. Selinda Research Associates, Inc. iv TABLE OF CONTENTS Acknowledgements ...................................................................................................................... iii List of Tables ................................................................................................................................. vi Abstract ....................................................................................................................................... vii Executive Summary ................................................................................................................... viii Introduction ................................................................................................................................... 1 Project Goals and Audiences .................................................................................................................. 1 Project Components ................................................................................................................................ 2 Overview of Project Evaluations ............................................................................................................ 3 Methodology and Methods ........................................................................................................... 5 Methodology ............................................................................................................................................. 5 Overview of the Study ............................................................................................................................. 6 Methods .................................................................................................................................................... 9 Limitations ............................................................................................................................................. 14 Findings: Design and Layout ...................................................................................................... 15 Overview of the Exhibition ................................................................................................................... 15 Exploratorium ....................................................................................................................................... 17 Science Museum of Minnesota ............................................................................................................. 18 Don Harrington Discovery Center ....................................................................................................... 20 Findings: The Visitor Experience .............................................................................................. 23 Overall Experience ................................................................................................................................ 23 Visitor Flow ............................................................................................................................................ 23 Perceived Cohesion of the Exhibition .................................................................................................. 24 Perceived Focus of Exhibition .............................................................................................................. 25 Time at Exhibits ..................................................................................................................................... 25 Experiences with Immersive Exhibits ................................................................................................. 26 Caregiver-Child Interactions ............................................................................................................... 31 Looking at Different Subgroups .......................................................................................................... 37 The Role of Labels and Other Graphics ............................................................................................. 41 Effectiveness of Paired Exhibits ........................................................................................................... 47 Focus of Exhibition ................................................................................................................................ 48 Geometric Concepts & Shapes ............................................................................................................. 49 Selinda Research Associates, Inc. v Findings: Spatial Reasoning ....................................................................................................... 51 Findings: Travelability ................................................................................................................ 63 Findings: The Website ................................................................................................................. 65 Conclusions ................................................................................................................................... 69 References ..................................................................................................................................... 72 Appendixes ................................................................................................................................... 75 Appendix A. Description of Geometry Playground Exhibits .............................................................. 76 Appendix B. Photographs of Geometry Playground Exhibits ............................................................ 86 Appendix C. Annotated Bibliography of Reports & Presentations ................................................ 123 Appendix D. Topical Framework ...................................................................................................... 126 Appendix E. Sources of Data .............................................................................................................. 132 Appendix F. Description of Respondents .......................................................................................... 133 Appendix G. Installations at the Host Science Centers ................................................................... 134 Appendix H. Inventory of Engagements at the Large Climbing Structures ................................. 160 Appendix I. Taxonomy of Spatial Reasoning Skills ......................................................................... 164 Appendix J. Case Studies of Spatial Reasoning at Selected Exhibits ............................................. 169 Appendix K. Screenshots of the Website ........................................................................................... 182 Appendix L. Critical Review of the Website ..................................................................................... 193 Selinda Research Associates, Inc. vi LIST OF TABLES Table 1. page 7 Data collection site visits Table 2. page 16 Names and classification of Geometry Playground exhibits Table 3. page 52 Intended spatial reasoning skills in Geometry Playground Table 4. page 53 Inventory of spatial reasoning skills practiced by visitors to Geometry Playground Table 5. page 55 Hierarchy of spatial reasoning skills practiced by visitors to Geometry Playground Table 6. page 57 Knowledge hierarchy defining variables Table 7. page 57 Hierarchy of spatial reasoning skills practiced by visitors at Space-filling Blocks Table 8. page 58 Hierarchy of spatial reasoning skills practiced by visitors at Stack of Stars Table 9. page 59 Hierarchy of spatial reasoning skills practiced by visitors at the Gyroid Table E-1. page 129 Overview of sources of data Table F-1. page 130 Respondents observed and/or interviewed Table F-2. page 130 Staff interviewed for this study Table G-1. page 131 Geometry Playground exhibits at Explo Table G-2. page 140 Geometry Playground exhibits at SMM Table G-3. page 147 Geometry Playground exhibits at DHDC Table H-1. page 157 Inventory of engagements by children Table H-2. page 159 Inventory of engagements by adults Table I-1. page 161 Taxonomy of spatial reasoning skills Table J-1. page 172 Hierarchy of spatial reasoning skills for putting Space-filling Blocks into place Table J-2. page 174 Hierarchy of spatial reasoning skills for planning a structure built with Space-filling Blocks Table L-1. page 191 Critical review of the Geometry Playground website Selinda Research Associates, Inc. vii ABSTRACT The summative evaluation of the National Science Foundation (NSF)-funded Geometry Playground traveling exhibition was a two-year naturalistic study to examine (a) the ways and extent to which the exhibition promoted the practice of spatial reasoning skills, and appreciation for geometry, and (b) its influence on museum professionals’ thinking across three venues: the Exploratorium (San Francisco, CA), the Science Museum of Minnesota (St. Paul, MN), and the Don Harrington Discovery Center (Amarillo, TX). The study took place from December 2009 through November 2011 and included five site visits to the first three host museums, observations of, and interviews with, purposively selected casual museum visitors and museum staff, and reviews of documents including an extensive critical review of the project website. Data collection also included interviews with museum staff and a critical review of the project’s website. Data analysis was conducted using modified inductive constant comparison. A total of approximately 164 contact hours (including observations, interviews, and on-site debriefs) were spent with 354 respondent groups (1,620 individuals). Findings indicated many visitors of all ages engaged in a variety of types of spatial reasoning within the exhibition, and in some cases moved up a level or two on a knowledge hierarchy. Social interactions among visitors tended to be of three types: Far Out, Up Close, and Hybrid. Some social interactions and ability to engage with the exhibits were limited by the need for just one thing, e.g. a hint to notice or do something. Most visitors to the large climbing structures tended to engage in age-appropriate playground behaviors. At the smaller immersives and table-top units engagements were more typical of museum behaviors including teaching/learning behaviors. Most visitors tended to think primarily about geometry rather than spatial reasoning. When they did think about spatial reasoning, it helped them move up a level on the knowledge hierarchy. Engagement with spatial reasoning and geometry appeared to be equitable for girls and boys. Conceptual pairings of some of the exhibits worked well when the exhibits were in close proximity and/or when the pairings were pointed out to visitors. But most visitors were not aware of the pairings. The exhibits appeared to travel well. Challenges included the size, weight, and complexity of installation required for the large climbing structures, and some wear and tear of exhibit components. The flexible floor plan had both advantages and disadvantages. There was limited guidance and training for floor staff, ultimately resulting in uneven support and guidance for visitors. The exhibition contributed to many visitors’ positive feelings about, and appreciation for, the aesthetics of geometry. Host museum staff gained some appreciation for the exhibition itself, for the ways in which visitors learned about geometry, and for the appropriateness of large-scale climbing structures within a science museum setting. Selinda Research Associates, Inc. viii EXECUTIVE SUMMARY Introduction. The National Science Foundation (NSF)-funded Geometry Playground was a multi-faceted project whose major goals included developing a large geometry-focused traveling exhibition and raising the museum profession’s awareness of the roles that geometry, spatial reasoning, and immersion can play in science center exhibits and programming. The project’s major product was an approximately 5,000-square-foot traveling exhibition consisting of a mixture of large- and medium-sized exhibit units that invited and encouraged whole-body involvement with geometry, and smaller table-top units, some of which addressed the same and additional spatial reasoning and geometric concepts as the immersive structures. The project also developed a website and online Pathways educational materials and included a research component. The two major audiences for the project were small social groups of children 5-12 years old with their caregivers, and museum professionals. Overview of the evaluation. The summative evaluation took place from December 2009 through January 2012. The major focus of the evaluation was the traveling exhibition, hosted by three science centers during this study. The evaluation looked at the effectiveness of the exhibition as an opportunity for visitors to practice spatial reasoning. The summative evaluation also examined how well the exhibition traveled and the influences of the exhibition on host science center staff. Finally, the evaluation also included a critical review of the Geometry Playground website. Methodology and methods. The summative evaluation used a naturalistic methodology and primarily qualitative methods. The team included staff from both the Exploratorium and Selinda Research Associates (SRA). SRA conducted five site visits to the three venues: the Exploratorium (Explo), the Science Museum of Minnesota (SMM), and the Don Harrington Discovery Center (DHDC). Visitors and staff were observed and interviewed and project-related documents were reviewed, including websites and publicity signage. Most respondents for this study were purposively selected museum visitors in naturally occurring social groups. Data analysis was conducted using modified inductive constant comparison. A total of approximately164 contact hours (including observations, interviews, and on-site debriefs) were spent with 354 respondent groups (1,620 individuals). Exhibition design and layout. The 31 Geometry Playground exhibits were classified into six general types: (a) large immersives (two large climbing structures: the Gyroid and Stack of Stars); (b) medium immersives (smaller exhibits that involved the whole body, including a few smaller climbing structures); (c) table-top units (mostly interactives); two types of “contemplative” exhibits: (d) static objects behind glass and (e) photo panels; and (f) some exhibit furniture. A few of the exhibit units were “conceptually paired.” These conceptual pairs consisted of at least one large or medium immersive exhibit dealing with a certain geometric idea and one smaller table-top exhibit focusing on the same concept which engaged visitors with the concept in different ways. Selinda Research Associates, Inc. ix Most of the 31 exhibit units were installed in all three venues, and at all three, the layouts were open, with porous boundaries rather than solid walls separating them from adjacent exhibitions. However, the floor plans in each of the three venues were significantly different, and the conceptual pairs were installed differently at all three venues. Three important differences among the installations were: (a) at SMM, the exhibition was split and installed on two floors; (b) at DHDC, the Gyroid was reduced to two, rather than three, layers tall; and (c) labeling at the Gyroid and Stack of Stars differed. Visitor flow and perceived cohesion of the exhibition. Three distinct visitor flow patterns were identified: (a) walk through the exhibition on a linear traverse and return later; (b) explore in a less linear way, often moving in and out across the porous boundaries of the exhibition; and (c) settle and stay at a single exhibit for a long time, often one of the large climbing structures. One important consequence of these flow patterns was that visitors often did not perceive the exhibition as a coherent whole. This led to misunderstandings of what some of the exhibits were about and limited their engagements with spatial reasoning. Experiences with immersive exhibits. The two large immersive climbing structures inspired playground behaviors among children and adults alike. As at a playground, children tended to climb, whereas adults tended to sit. However, the data indicated that some caregivers and children displayed an interesting mix of playground and museum behaviors, such as thinking about geometric shapes, reading exhibit labels, and taking photographs. While most adult engagements at the two large climbing structures tended to be minimal, passive, and reminiscent of the ways adults engage in playgrounds, caregivers were more engaged at the smaller climbing structures like the Shape Cubes. Children’s play at the smaller immersives seemed less playground-like and more museum-like than with the large immersives. Transitioning between exhibits. Because the different types of exhibit units were mixed throughout the exhibition, visitors frequently transitioned between exhibits that required significantly different levels of energy, activity, and cognitive engagement. Most often when children left the large climbing structures, they (and their parents) resumed their usual science museum roles. The data indicated no significant negative spillover effect from their playground behavior, as they moved from the climbing activities to other exhibits. However, there was evidence of some positive spillover effects. For example, there were indications that spending physically active time on the climbing structures may have helped children focus on the surrounding exhibits. Safety issues. Because the exhibition included two large climbing structures, the development team and host museum staff were concerned about safety issues. Various attempts were made to increase safety at the large immersives, including warning signs, age limits, controlled access, and close staff supervision. Although caregiver visitors seemed aware of the safety dangers on the climbing structures, most did not express much concern about them. More parents tended to stand close to Stack of Stars and stayed more engaged with their children there than they did at the Gyroid. Findings about the effectiveness of the suggested age limits were not conclusive. Selinda Research Associates, Inc. x Caregiver-child interactions. Three broad categories of caregiver-child interactions emerged: Far Out, Up Close, and Hybrid, based on a combination of physical proximity and intimacy of interaction. Far Out interactions were: (a) On Your Own, (b) Observer, and (c) Safety and Supervision. Up Close interactions were: (a) Photo Op, (b) Games, and (c) Meaning Making. Hybrid interactions were: (a) Check In, and (b) Recognition and Pride. These categories defined certain types of engagements, not types of visitor groups, as members of any particular group moved frequently and fluidly between these different modes of social interactions. Variations of these engagements were observed at all the exhibits in Geometry Playground. In general however, Far Out and Hybrid interactions were characteristic of many engagements at the large immersive experiences, while Up Close interactions were more characteristic of the engagements at the medium immersives and table-top exhibits. Many social interactions, regardless of the type, were found to be limited by the need for just one thing to get something meaningful out of the exhibit, usually a hint or instruction to notice or do something. Equitable experiences for different subgroups. Although visitors of various ages tended to have different experiences—based on their physical size, level of physical activity, inclination towards reading, background knowledge, and other age-related factors—Geometry Playground provided appealing and effective experiences for visitors of all ages and genders. Indications were that girls and boys were equally attracted to, and engaged at, the exhibits. By enticing many visitors to go beyond their everyday experiences, Geometry Playground provided equitable experiences for most visitors. Labels. The labels in Geometry Playground were of six different (but often overlapping) types: (a) instructions, (b) warnings, (c) interpretive panels, (d) object labels, (e) quotations, and (f) titles. Many visitors used the labels to discover important information about the exhibit units, and geometry concepts in general. Sometimes the instruction labels left visitors in need of just one thing to be successful at the exhibit. The warning labels (located on the climbing structures) proved mostly effective at limiting excessive behaviors, but may also have unintentionally limited desirable engagements by visitors younger than the suggested ages. Bench-mounted interpretive labels at the climbing structures seemed most useful when they helped caregivers either discuss the label content with their children, or encourage their children to try suggested activities. Object labels helped visitors appreciate both the real-world applications and aesthetic aspects of geometry. Quotations included important ideas about geometry and spatial reasoning, but few visitors attended to them. The exhibition title was displayed in very few places and most visitors were unaware of it. When they did notice it, it communicated many important ideas: geometry is fun, something we can play with, and geometry goes beyond what we learned about in school. Some exhibit titles also communicated important ideas to visitors (for example Stack of Stars), but again, the titles often were not noticed. A few exhibits did not have labels, and in some installations of some other exhibits, the labels were not included. When labels were not included, fewer successful meaning-making interactions took place. Selinda Research Associates, Inc. xi Effectiveness of paired exhibits. Conceptual pairings, when visitors recognized them, contributed to both spatial reasoning about the exhibits, and increased understanding and thoughtfulness about geometric concepts. However, most visitors did not recognize most conceptual pairs. Because conceptual pairs were so effective when visitors did recognize them, the idea of pairing exhibits seems to be a sound one. Visitors’ relationships to geometry. Most respondents recognized that they were in a large exhibition about geometry or “shapes.” Some made connections to real-world geometry, and many expressed an aesthetic appreciation for geometry. Some visitors indicated they learned about, or expanded their understanding of geometric concepts and 2- and 3-D shapes. The few respondents who did not make the connection between the exhibits and geometry tended to be those who had spent a very limited amount of time in the exhibition, or visitors who had not yet formally studied geometry. Spatial reasoning. Visitors of all ages effectively engaged in practicing a wide range of spatial reasoning skills for extended periods of time. A six-level spatial reasoning knowledge hierarchy emerged: Level 0-No engagement, Level 1-Mindless engagement, Level 2-Trial and error, Level 3-Deliberateness, Level 4-Mental objects and/or complex rules, and Level 5-Absence of physical objects. Some visitors spent time practicing spatial reasoning skills within one level, while some others progressed a level or two. Most respondents were not meta-cognitive, or aware that they were doing spatial reasoning. This was not a goal of the exhibition, but when visitors thought more deliberately about their spatial reasoning engagements with the exhibits, they often moved up the knowledge hierarchy. Travelability. Most of the exhibit units were easily assembled, set up, and installed. The exhibition manual was a useful resource, and Explo staff were available via telephone when needed. Challenges included the size and weight of some of the exhibit units, and the assembly time for the large climbing structures and their rubberized platforms. The exhibition’s flexible floor plan proved to be a mixed blessing. The exhibition could be adapted to accommodate the specific needs of each venue, but this often resulted in missed opportunities for the goals of the exhibition to be achieved, or meant that some subtleties of the exhibit developers’ intentions went unrealized. Finally, although many host museum staff were pleased that the exhibit units required minimal maintenance, by the time the exhibition arrived at the third venue, some of the components were worn enough to need replacement. The exhibition website. The Geometry Playground website provided potential users with rich resources that could serve exhibition visitors in important ways. However, the website was not mentioned in the exhibition or in the publicity materials, so most visitors were not aware of it. Selinda Research Associates, Inc. xii The site’s navigation was clear and easy to use on most desktop or laptop computers, but it was too complex to be useful for mobile devices, and it did not work well with Apple products such as iPhones and iPads. In addition, there was no search function, which would have helped users locate materials and information. Many aspects of the website did a good job of supporting the exhibition’s goals by, for instance, helping users appreciate the beauty and aesthetic aspects of geometry and highlighting aspects of geometry in everyday life. However, there were few discussions of spatial reasoning, including the activities aimed at educators. Conclusions. The Geometry Playground exhibition provided engaging and enjoyable experiences for most visitors. Many visitors expanded their thinking about geometry in useful and important ways. The mix of large- and medium-sized immersive exhibits, hands-on table-top interactives, and contemplative exhibits provided a refreshing combination of experiences that visitors took advantage of and appreciated. Although the exhibition was targeted to 5- to 12-year-old children and their adult caregivers, visitors of all ages effectively practiced a wide range of spatial reasoning skills. The exhibition was also effective at raising visitors’ awareness about aesthetic aspects of geometry and about ways that geometry is encountered in everyday life. The exhibition had mixed results with its goal of encouraging social interactions between caregivers and children 5-12, both because the climbing structures seemed to encourage caregivers to disengage from their children, and because most labels were not designed to specifically support such interactions. While not an explicit project goal, the exhibition missed the opportunity to help visitors become more aware of, and deliberate about, their spatial reasoning activities. The Geometry Playground exhibition will likely contribute to the advancement of the field’s understanding of immersive visitor experiences, spatial reasoning, math, and geometry in informal settings. Selinda Research Associates, Inc. 1 INTRODUCTION Geometry Playground was a multi-faceted project whose major goals included developing a large, geometry-focused traveling exhibition and raising the museum profession’s awareness of the roles that geometry, spatial reasoning, and immersion can play in science center exhibits and programming. The project’s major product was an approximately 5,000-square-foot traveling exhibition with 31 exhibit units. The exhibition was designed to provide a variety of types of opportunities for visitors to engage with geometry, and practice a range of spatial reasoning skills. The project also developed a website and educational materials, and included a research component. These are described briefly below. Funded in part by the National Science Foundation (NSF), Geometry Playground was developed by the Exploratorium (Explo) with input from the Science Museum of Minnesota (SMM), which participated in exhibit development and remediation. Other institutions participated in workshops about large immersive exhibits and reviewed some of the processes and exhibits under development. The project kicked off in early 2007 and received a one-year, no-cost extension, with an end date of August 31, 2011. At the time of the writing of this report, the exhibition had traveled to the following four venues, and future venues were being planned. June 2010 – September 2010 Exploratorium, San Francisco, CA October 2010 – January 2011 Science Museum of Minnesota, St. Paul, MN February 2011 – May 2011 Don Harrington Discovery Center, Amarillo, TX June 2011 – June 2012 Reuben H. Fleet Science Center, San Diego, CA Project Goals and Audiences Following is a list of project goals developed early in the project.1 1. Foster spatial thinking among children 5-12 and their adult caregivers. 2. Increase engagement with geometry activities among children 5-12 and their adult caregivers. 3. Encourage social interactions between adults and children 5-12 during their visit to the exhibition. 4. Increase positive experiences and associations with geometry for children 5-12 and their adult caregivers. 5. Generate interest among museum professionals in immersion as an educational tool in informal settings. 6. Create and disseminate more examples of immersive experiences in museums. 7. Build understanding of the effects of immersive experiences in informal environments. The two major target audiences for the project were identified as (a) small groups with children 5-12 years old with their caregivers, and (b) museum professionals. 1 These goals are taken from a series of Impacts and Indicators worksheets developed by the Exploratorium for NSF. Selinda Research Associates, Inc. 2 Project Components The Geometry Playground project included a variety of components. The major ones are described briefly below. Exhibition The Geometry Playground traveling exhibition was designed to be installed entirely inside host museum’s exhibit galleries.2 It was composed of a mixture of large, immersive, and climbable exhibit units that invited and encouraged whole-body involvement with geometry, and smaller table-top units, some of which addressed the same and additional geometric concepts as the immersive structures. The traveling exhibition was composed of 31 exhibit units, although not all venues installed all units. The exhibit units included two large immersive climbing structures, 10 medium-sized immersive exhibits, 12 smaller table-top units, and a number of static objects and furniture. (See Table 2 under Findings: Design and Layout for a listing of the exhibits summarized by type.) For a more complete description of the exhibit units, see Appendix A, and to see photographs of each of the units, see Appendix B. Website The project also developed an accompanying Geometry Playground website.3 This website included information about the exhibits and their creators, as well as activities that website visitors could do within the exhibition or anywhere else, including playgrounds and other outdoor spaces. These activities related closely to key concepts and approaches used in the exhibition and could be either read online or printed out. In addition to the primary Geometry Playground website developed by the Exploratorium, each of the host museums mentioned the exhibition on its website home page or exhibition pages. Some of the host venues also developed more extensive web modules dedicated to Geometry Playground and included these on their institutions’ websites. However, not all of the host museums included links from their Geometry Playground website pages to the primary (Exploratorium) Geometry Playground website. Educational Activities An additional output of the project, with funding provided by the Moore Foundation, was a series of about 30 educational activities developed for use by teachers, homeschoolers, and families. Termed Pathways, these activities were developed for use at schools, homes, outdoors, and in the exhibition. The activities were accessible through the Resources section of Explo’s Geometry Playground website and at least one of the Geometry Playground web modules developed by the host museums. They were also made available to the local teacher community and on Explo’s Pinhole.4 2 This report uses the term exhibition to refer to the entire installation. Individual, stand-alone components, like the Gyroid or Build a Room, are referred to as exhibits or exhibit units. 3 As of the writing of this report, the URL for this website is: 4 Pinhole is an active listserv for alumni of the Exploratorium’s Teacher Institute, Selinda Research Associates, Inc. 3 Outdoor Exhibits In addition to the interior traveling exhibition, at the time of the writing of this report the Exploratorium was in the process of developing outdoor playground structures and accompanying signage and activities that encourage spatial reasoning. It is our understanding that these structures will be permanently installed outdoors at selected venues, perhaps including a science center, park, and school playground. While the original intent was to include some of these playground pieces in the summative study, the timing did not work out. Research Component In addition to products developed for public audiences, the Geometry Playground project also included a research component looking at spatial reasoning at the exhibits, as evidenced by visitors’ conversations about math and geometry. At the time of the writing of this report, the video data from this research were being analyzed. It is anticipated that this study will contribute significantly to the museum field’s understanding of how visitors engage with geometry and math in informal settings. Additional research undertaken as part of this project included a study focused on developing gender-equitable exhibits that promote use of spatial reasoning skills (Dancu, 2010) and a study comparing visitor experiences at immersive and table-top exhibits (Dancu & Gutwill, 2010). See Appendix C for a list of research presentations and papers published to date. Overview of Project Evaluations Evaluation for the Geometry Playground project included front-end, formative, and summative studies. Each of these is described briefly below. See Appendix C for a complete list of evaluation reports. Front-end Evaluation In search of effective approaches to developing exhibits and environments that foster intergenerational play involving adults and children, Hido & Gutwill (2007) undertook a front-end study at Adventure Playground in Berkeley, an outdoor facility at the marina where children and their adults are encouraged to play and build. The purpose of this study was to help identify “the design features of an environment that would engage adults and children ages 5-12 in whole-body play activities.” Formative Evaluation Prototype versions of the exhibits underwent extensive formative evaluation, including more than 48 studies of 21 exhibits. Formative evaluation took place at both the Exploratorium and the Science Museum of Minnesota, one of the partner museums. The exhibits and accompanying labels changed significantly based on the formative findings. Selinda Research Associates, Inc. 4 Summative Evaluation The summative evaluation of Geometry Playground began in December 2009, and concluded with the completion of this report in January 2012. Data collection took place from July 2010 through May 2011 and included site visits to three venues: Explo, SMM, and the Don Harrington Discovery Center (DHDC). The major focus of the summative evaluation was the exhibition as it toured the country. At each of the three venues, the exhibition footprints and floor plans were significantly different. Not all exhibit units were installed at each site, and sometimes, even when the same exhibit units were installed, the versions were slightly or significantly different. This summative study looked at both the effectiveness of the exhibition as a visitor experience, and at how well the exhibition traveled from one host to the next—what was referred to as the exhibition’s “travelability.” The summative also examined some aspects of the project’s contributions to the museum field, and included a detailed review of the Geometry Playground primary website. The evaluation did not look at the project’s educational programming or additional educational products or research. The remainder of this report covers the summative evaluation only. We hope it will provide useful feedback for both the Geometry Playground project team and the project’s funders. In addition, we hope it will be useful to other museums and science centers interested in incorporating math, geometry, and spatial reasoning in their exhibits and programs. Selinda Research Associates, Inc. 5 METHODOLOGY AND METHODS Methodology A research method is a technique for (or way of proceeding in) gathering evidence.… A methodology is a theory and analysis of how research does or should proceed.” (Harding, 1987, p. 2) People often confuse methodology and methods. In this study, we used a naturalistic methodology (Lincoln & Guba, 1985). Naturalistic inquiry is a field-based methodology that examines phenomenon as it naturally occurs—as opposed to an experimental or scientific methodology where hypotheses are generated and variables are controlled and manipulated. Relying to a large extent on naturalistic observations and respondent narratives, the goal of naturalistic methodology is to provide a holistic understanding of the phenomenon under investigation from a variety of perspectives. Naturalistic inquiry is a rigorous and disciplined approach to collecting, analyzing, and reporting data gathered in natural (as opposed to laboratory) settings and has proven to be particularly effective for conducting research in complex environments such as informal learning settings (Allen, Gutwill, Perry, Garibay, Ellenbogen, Heimlich, et al., 2007). Naturalistic methodologies emphasize collecting data using a mix of methods, primarily qualitative, and triangulating findings. Naturalistic evaluation reports tend to be narrative in format, enabling the researchers to capture and report the findings with rich descriptions of respondents’ behaviors and, when possible, in respondents’ own words. In a report written for the Smithsonian, a useful description of naturalistic evaluation is presented. The methods employed in this study are what we refer to as Naturalistic Evaluation. They are naturalistic because we observe and interact with persons as they actually engage in their museum visit. We approach them naturally; we engage with them in natural conversation; and we document their natural behavior. The methods are evaluative because we ask the visitor to assess the quality of his/her experience and to clarify the nature and meaning of such experience from their unique point of view. This strategy of combining observation and interview procedures has proven to be extremely helpful in revealing detailed aspects of the visitors’ experiences as well as the cognitive and affective impact that these experiences offer (Wolf & Tymitz, 1981, p. 2). Selinda Research Associates, Inc. 6 Overview of the Study Research Questions The primary evaluation question was: In what ways and to what extent did the Geometry Playground exhibition, stand-alone structures, and accompanying website facilitate meaning-making and promote the practice of spatial reasoning skills and appreciation for geometry? A secondary evaluation question was: In what ways and to what extent did various types of involvements with the Geometry Playground project contribute to selected museum staff members’ thinking about exhibition development, and especially their perceptions of and appreciation for the inclusion of (a) large-scale immersive structures, and (b) geometry and other math-related concepts in museum displays? Although these questions guided the research and are addressed extensively in this report, our investigations revealed a number of additional issues that we also discuss in the body of the report and its appendixes. The Evaluation Team A team comprised of staff from both the Exploratorium (Josh Gutwill, Peggy Monahan, and Toni Dancu) and Selinda Research Associates (Deborah Perry, and Eric Gyllenhaal) guided the summative evaluation. In this study the role of client was shared among the three Exploratorium (Explo) representatives. All three were present at planning meetings, and they shared client responsibilities. Josh Gutwill commissioned the study, and he and Peggy Monahan were primarily responsible for issues related to Explo goals, budget, and timelines. Toni Dancu was the primary communication point person, filtering feedback and emails between Explo and Selinda Research Associates (SRA). She also contributed to discussions about goals and timelines, and served as the liaison during SRA site visits to San Francisco and St. Paul. Deborah Perry served as project director for the evaluation study, and oversaw the timeline, schedule, and quality of the deliverable. She served as spokesperson for the study and took the lead in developing the evaluation plan and topical framework. She participated in the data collection, data analysis, and the writing of the final report. Eric Gyllenhaal was the lead researcher and was primarily responsible for conducting the study and implementing the evaluation plan. He took the lead in data collection, data analysis, the website critical review, and writing the first draft of the final report. Selinda Research Associates, Inc. 7 Research Design The summative evaluation study consisted of four phases. During the planning phase, December 2009 to June 2010, the evaluation plan was developed, research questions were identified, and a topical framework5 was developed. See Appendix D for the complete topical framework developed for this study. The data collection phase extended from July 2010 through May 2011, and included five site visits to the first three host museums (see Table 1). Table 1. Data collection site visits # Who? Where? When? 1 Eric & Deborah Exploratorium, San Francisco July 2010 2 Eric Science Museum of Minnesota, St. Paul December 2010 3 Eric Science Museum of Minnesota, St. Paul December 2010 4 Deborah Don Harrington Discovery Center, Amarillo, Texas May 2011 5 Eric Don Harrington Discovery Center, Amarillo, Texas May 2011 During these site visits, visitors and selected staff members were observed and interviewed, using methods described below. Additional phone interviews with staff were conducted from February through May 2011, and a critical review of the exhibition website took place during April and May 2011. See Appendix E for a complete list of data sources. Data analysis took place during and after data collection was completed (from July 2010 through July 2011). Brief site-visit summaries were submitted to the client after each of the first three site visits. Writing of the final report for this study began as data analysis wound down. Client reviews were built into the report-writing phase, including a multi-department team of reviewers from the Exploratorium, with feedback incorporated (as appropriate) into the final version of the report. Remediation While SRA did not conduct a formal remedial evaluation of the exhibition, remedial evaluations were conducted at Explo and SMM and revisions made to the exhibition. In addition, the initial installation at Explo served as a shake-down for the exhibition. Based on a preliminary analysis of findings from the first site visit, SRA researchers helped Explo staff identify challenges and determine necessary remediations before the exhibition traveled. Because SMM was a partner site, they continued to identify challenges and make necessary remediations before the exhibition traveled to non-partner sites, and SRA researchers tracked this process. After observing the installations at the first three host sites, SRA researchers made some preliminary recommendations for the installation at the fourth site in June 2011. Recommendations about installation at future sites are also included in this report. 5 A topical framework is a list of all the issues (or topics) that the summative evaluation was designed to explore. Selinda Research Associates, Inc. 8 Selecting Respondents Respondents for this study were selected using purposive—as opposed to the more familiar random—selection methods (Miles & Huberman, 1994). Purposive sampling is a technique where each respondent is selected based on the results of previous data sets in order to achieve the greatest variability in data. As data are gathered and preliminary analysis is conducted, new questions and areas of study emerge. Respondents were selected purposively to illuminate different types of experiences. Using purposive selection ensured that data was gathered from a variety of respondents with a maximum range of experiences. Children were often purposively selected for observation, but were not interviewed unless their parent’s permission was secured. Description of Respondents Respondents in this study were primarily casual museum visitors in naturally occurring social groups. Note that a “social group” was sometimes an individual visiting the museum alone. During the selection process, the researchers made a special effort to include as many different social group configurations as possible. Although we did not seek out intact school groups, scout troops, or birthday parties (or other organized groups), these groups were observed at times, especially when it was clear that doing so would contribute to our understanding of the effects and effectiveness of the exhibition. Also, when individuals or small groups who were sub-sets of these types of organized groups fit the selection criteria, they were included as respondents, for example when parent-chaperones were visiting with a group that included their own children. Because small family (and other) social groups are a typical audience for the host museums, they formed a large proportion of the respondents. We paid special attention to groups that matched the target audience for the exhibition: 5- to 12-year-old children and their caregivers, but did not limit ourselves to this population. It should be noted that many visitors to the exhibition were not part of the target audience (5- to 12-year-olds and their accompanying caregivers). In particular, visitors younger than 5-12 years old were a frequent—and in some situations, the most frequent—type of visitors. In addition, many groups that met the definition of the target audience also included preschool and younger siblings. Because these visitors were so prevalent, and because this study employed a naturalistic methodology—that is, one where the focus is on the naturally occurring world—we included findings on younger visitors’ experiences as well. Exact data about most respondents’ ages were not collected, unless it was mentioned as part of an interview. For this reason we refer to general descriptions of age (when appropriate) based on factors such as size and behavior, and perceived relationships within social groups. In general, the term “children” is used to describe children who appeared to be within the target age range of 5-12 years old, “younger children” to describe those younger than the target audience, and “older children” to describe those older than the target age range. Sometimes we refer specifically to those at the lower or upper ends of the targeted age range. One respondent was known to one of the researchers. This person visited the museum with their own family group, and agreed to a follow-up telephone interview. Selinda Research Associates, Inc. 9 While not a primary focus of the evaluation, it was important to the exhibition development team to understand how well the exhibition traveled to host museums and the ways in which the Geometry Playground exhibition contributed to museum staff members’ thinking about large-scale and immersive exhibit units. Therefore, respondents for this study also included 25 staff and volunteers from Explo, SMM, and DHDC. In some evaluation studies, the optimal number of respondents is determined during the planning phase. In naturalistic evaluations, the number of respondents is not pre-determined. Rather, data are collected until a state of redundancy is reached. Redundancy is the point at which trends have emerged, and no new information is found, in spite of repeated attempts to elicit alternative perspectives and new understandings. In this study, there was approximately 164 contact hours with respondents, with 354 respondent groups and 1,620 individuals. While the number of respondents is unusually high for a naturalistic study, it can be attributed to the large number of extended observations (see below), and observations of several school groups and camp groups. For complete tables summarizing respondents and contact hours, please see Appendix E and Appendix F. Ethical Treatment of Respondents SRA, Explo, and NSF are committed to the ethical treatment of respondents. Research adhered to standard professional practices for conducting research in informal settings. All respondents were guaranteed anonymity and gave informed consent. Signs were posted during all on-site evaluations informing visitors that data collection was taking place and explaining that they were under no obligation to participate, and that they could terminate the interview at any point. Disruption of visitors’ experiences was kept to a minimum. Methods The evaluation team used a variety of qualitative and quantitative methods to collect data for this study. Following is a brief description of each of the methods used. Observations A major method of gathering data for this study was observing visitors using the exhibition. As part of these observations, the researcher paid particular attention to the range of four types of visitor engagements: physical, social, emotional, and intellectual, as described below. These types of engagements are not—and are not meant to be—mutually exclusive. Note that all the examples of engagements listed below were not pre-determined, but emerged from the data. Physical engagements were all the physical things visitors did at the exhibits or activities, such as standing or sitting, reading labels, engaging in fantasy play, climbing onto or inside a structure, holding and rotating small loose pieces, assembling and reassembling structures, spending time, using an interactive, and so forth. Selinda Research Associates, Inc. 10 Intellectual engagements were all the ways in which visitors engaged cognitively and intellectually with the content of the exhibition. They included a respondent being thoughtful and reflective, making comparisons, practicing spatial reasoning skills, or relating something to their personal life. Intellectual engagements also included “aha” moments, or other indications that a respondent made a personally meaningful connection. Social engagements were the ways in which visitors engaged with each other to make meaning of their exhibition experiences. These included verbal exchanges and body language, such as storytelling, teaching-learning interactions, directing attention, asking and answering a question, explaining, and other social activities that contributed to active meaning-making. For the purposes of this study, it did not include social interactions that were not focused on the exhibition content, such as discussing the weather or where to go for lunch. Emotional engagements were all the ways that visitors engaged emotionally with the exhibition’s environments and interpretation. Researchers looked for a variety of emotional responses to both the content of the exhibition as well as the design and interpretation. These included appreciation, excitement, anger, surprise, satisfaction, frustration, or confusion. In looking at these four types of engagements, four different types of observations were conducted. During unobtrusive observations, a researcher would select a respondent group (see Selecting Respondents above), and then watch the visitor group as they explored the exhibition. The researchers tried to remain as unobtrusive as possible so they didn’t affect the visitor experience, but they did not try to disguise themselves. When the group moved around the exhibition, the researcher would follow them at a distance, taking notes about which exhibits and activities the group stopped at, how long they stayed, and what they did and said at each exhibit and activity. (Note that this aspect of unobtrusive observations was not a formal tracking and timing study.) Extended observations were a variation on unobtrusive observations. Similar to an unobtrusive observation, the researcher would try to minimize their presence as much as possible. Instead of following a single visitor group around the exhibition, the researcher would remain stationary, spending an extended period of time (often a half hour or more) at a single exhibit unit, or a small selection of related units. Similar to unobtrusive observations, the researcher observed and recorded how these groups engaged with the components, recorded segments of conversation, and sometimes timed how long individual groups stayed at a component. Incidental observations were also a variation on unobtrusive observations. They included short observations and snatches of conversation collected when the researchers were involved in other activities. For instance, researchers sometimes were observing one group when a member of a nearby group said or did something interesting or revealing. Rather than breaking off contact with the first group, the researcher made a few quick notes, adding context to the notes after the main activity was complete. This approach allowed researchers to record relatively rare events that otherwise would have been lost. Selinda Research Associates, Inc. 11 Finally, participant observations bridged the gap between observations and interviews. In some cases, opportunities arose for the researcher to become part of a visiting group and engage more directly with the respondents. In these situations, the researcher participated to some degree in the group experience, while clearly maintaining their role as researcher. This technique was useful at exhibits where there were opportunities to engage in a variety of ways and where outcomes were likely to be diverse, or where it was difficult to overhear what visitors were discussing. When feasible—and with respondents’ permission—participant observations were recorded, using a digital audio recorder and later transcribed for analysis. Most of the data collected during observations was qualitative. However, quantitative data was collected when needed to address particular questions. For instance, researchers regularly recorded the time a group spent engaged with an exhibit or series of exhibits, and the number of exhibits a group visited. Researchers sometimes recorded the numbers of respondents exhibiting a particular behavior, such as caregivers sitting or standing at the climbing structures, or the numbers of boys and girls present on the climbing structures, recorded at regular intervals. Details about sampling strategies are included in footnotes when these data are discussed later in the report. Depth Interviews Data were also collected for this study by conducting depth interviews with respondents. Depth interviews were open-ended and conversational in nature, allowing the respondents to set the direction of the conversation, and the researcher to gather data in the respondent’s own words. When feasible—and with respondents’ permission—depth interviews were recorded using a digital audio recorder and later transcribed for analysis. While researchers had loosely structured interview protocols to guide these conversations, the specific questions asked of any particular respondent were unique to that respondent. The researchers used their background and experience to ask non-threatening questions to elicit information that would enable them to answer the broader questions on the topical framework. They also recognized additional interesting threads that were not on the topical framework and often followed up on them. Visitor groups were not asked for any identifying information. Names and contact information for museum staff and volunteers have been held in confidence and are available only to selected SRA staff. Four types of depth interviews were conducted during this study. Intercept interviews. After a respondent group had been observed at one exhibit (or at a small group of related exhibits), researchers approached them for a conversation that focused on just that engagement. Intercept interviews typically lasted for 4-8 minutes, depending on the needs and interests of the group. Longer interviews often took place when respondents were willing to discuss exhibits they had recently left. Exit interviews. Conducted with visitor groups as they left the exhibition area, respondents were approached after they had been observed in at least four exhibit units, so that they had some familiarity with the range of exhibits available in Geometry Playground. Exit interviews with visitors lasted for 5-45 minutes depending on the needs and limitations of the group. Selinda Research Associates, Inc. 12 On-site staff and volunteer interviews. Depth interviews with staff were conducted face-to-face, often in the exhibition, but also in offices, hallways, and cafes, depending on the needs of the respondents. Interviews conducted in the exhibition often followed up on observations of the staff carrying out their duties, such as exhibit maintenance, monitoring safety at the climbing structures, or engaging with visitor groups at individual exhibits. On-site staff interviews lasted 5-15 minutes. Phone interviews. SRA researchers conducted a series of telephone interviews with selected museum staff, and with one visitor who agreed to participate in a follow-up phone interview. Respondents were invited to participate, and were assured of confidentiality. Using an open-ended depth interview protocol, phone interviews typically lasted 45-60 minutes. Document Reviews To understand what visitors might have seen or heard about Geometry Playground before encountering the exhibition, the researchers: (a) reviewed a number of handouts visitors received as they entered the host museums, (b) photographed and analyzed signage related to Geometry Playground, and (c) reviewed selected newspaper and social media websites related to Geometry Playground. Website Critical Review. The most extensive document review undertaken for this study was a critical review of the primary (Explo) website developed to accompany the exhibition. The website review was planned in collaboration with the evaluation and website development teams and was conducted by the SRA researchers, using a heuristic evaluation form adapted from earlier SRA studies.6 During a critical review, the evaluators use their experience to review the website with an eye towards the likely visitor experience. It is not an evaluation, in that it does not rely on direct visitor input, although if that is available, it will be incorporated. In the Geometry Playground website critical review, the two SRA evaluators relied on (a) their experience with informal learning, visitor studies, and instructional design, and (b) some limited visitor input gathered during site visits to host venues. The critical review took place primarily during April and May 2011. Data Analysis & Reporting Data analysis was conducted using a modified inductive constant comparison technique, whereby each unit of data was systematically compared with each previous unit to identify emergent trends (Lincoln & Guba, 1985). These themes were repeatedly checked against the topical framework to ensure the study remained on track. Data analysis took place in three stages: At the completion of each data collection “session” (i.e., interview and/or observation) the researcher found a quiet place to reflect and debrief. These debrief sessions were the first phase of data analysis. The process included fleshing out notes and writing a brief summary of what 6 See Jakob Nielsen’s website for a discussion of this approach: The heuristic started with a template developed by Nielsen but also incorporated questions from the summative evaluation’s topical framework. Selinda Research Associates, Inc. 13 was found out. It included identifying questions that were answered, new issues that were raised, and finally, developing a plan for the next observation/interview session. At the end of each site visit, a preliminary analysis of all the data was conducted and written into a Site Visit Summary. This summary documented the process used, and identified and discussed preliminary emergent themes. All themes and categories discussed in the Site Visit Summaries were preliminary, as they continued to evolve as more data were gathered. After all data from all site visits were gathered, a final analysis of the aggregated data was conducted, triangulating with other data as appropriate. It should be noted that although a variety of data collection strategies were employed, data gathered by different methods were not treated separately. For example, the results of document reviews or a summary of interview findings are not presented separately. In accordance with standards for naturalistic inquiry, data were synthesized and integrated to develop a comprehensive and multi-faceted understanding of different issues from a variety of angles. This report discusses findings on a range of topics as spelled out in the topical framework, integrating all the data relevant to each topic. All findings are a synthesis of the results that emerged from interviews, reviews of online sources and documents, and observations. It should also be noted that, for the most part, the reader will not find percentages or other statistics in this report. We have found that the inclusion of statistics in a naturalistic report that relies primarily on purposive sampling can create a false impression of specificity or strength of finding. Instead, in accordance with standards for naturalistic inquiry, we use the adjectives most, many, some, few, and none as a more accurate way to describe tendencies. In reporting aggregate information such as interview data, we made the decision not to tabulate percentages of different kinds of responses offered by visitors. It is our view that once numerical ratings are assigned to judgmental data, there is a great tendency on the part of the reader to engage in hyperbole and misinterpretation. Rather, we have intentionally used adjectives such as “most,” “many,” “some,” “few,” or “none” to help portray aggregate tendencies. If we simply say visitors, then it implies that almost everyone was in agreement (Wolf & Tymitz, 1981). Finally, SRA is committed to the use of non-sexist language whenever possible. In this report, we designated the gender of a person only when it seemed important to either the context of what was said, or when its inclusion would result in more interesting prose and it would not perpetuate a stereotype. Furthermore, we have chosen to replace the false generics “he” and “his,” and the awkwardness of “he or she” and “his or hers,” with “they” and “their.” Grammarians may object to this obvious lack of agreement in number. Given the choice between (a) agreement in number, (b) agreement in gender, and (c) less awkward prose, we have opted for the latter two (Miller & Swift, 1980, p. 35-47). Selinda Research Associates, Inc. 14 Limitations When conducting a research study using naturalistic methodologies, it is standard practice to continue collecting data until a state of redundancy is reached on as many of the issues under investigation as possible. In the real world, redundancy is difficult to achieve for all items on the topical framework, primarily due to limited resources. In this study, redundancy was achieved for many of the issues listed in the topical framework. However, in some areas, researchers were unable to explore the issue in enough depth to reach redundancy. Issues that could not be resolved satisfactorily were either not included in the final report or were identified where appropriate. As described previously, the exhibition as a whole—as well as many of the exhibit units individually—continued to evolve throughout the data collection process. This study attempts to capture some of that evolution, but ultimately, we describe a composite of visitor experiences across venues, and across time. This study does not compare installations across venues, except to the extent that such description helps paint a more complete picture of the visitor experience. It should be noted that this evaluation did not focus on the effectiveness of the individual exhibit units, but rather the exhibition as a whole. That said, there are detailed descriptions of the individual units in Appendixes A, B, and G. In studying the project’s contributions to the museum field, we portray a very narrow slice, limiting our findings to data we were able to gather through interviews with some museum professionals, as well as our first-hand observations. For example, our findings do not include an examination of professional development that took place during the development of the exhibition, at meetings with project partners, or at professional meetings earlier in the project, and, of course, they do not cover contributions made by this report nor any subsequent publications and presentations about Geometry Playground. As noted in the section on Document Reviews above, although an extensive critical review of the Geometry Playground’s website was conducted, this was not an assessment of actual visitor use or impact. Finally, this evaluation did not study the outdoor exhibits, nor the large immersive exhibits at other museums, as they were not installed during the time of this study. Selinda Research Associates, Inc. 15 FINDINGS: DESIGN AND LAYOUT Overview of the Exhibition A roughly 5,000 square foot traveling exhibition, Geometry Playground included 31 exhibit units, classified into six general types: (a) large immersives7 (the two climbing structures, the Gyroid and Stack of Stars); (b) medium immersives (e.g., Shape Cubes, Hopscotch, and Sphere Cube); (c) table-top units (most of the interactives, such as Drawing in Circles and Dividing Space); two types of “contemplative” exhibits: (e) static objects behind glass (Geometry Garden); and (e) photo panels (Geometry in the Wild); and (f) exhibit furniture (Entrance Structure and Shape Stools). A list of all the exhibits and their classifications is included on the next page in Table 2. The exhibits are described in detail in Appendix A, and illustrated by photographs in Appendix B. During the three years of exhibition development, three broad exhibit themes emerged. These themes were used to organize the Geometry in the Wild photo essay and the exhibition website: (a) Geometry of Seeing, (b) Geometry of Motion, and (c) Geometry of Fitting Things Together. These categories were used to organize Appendixes A and B of this report. However, these categories were not reflected in the exhibition floor plans as they were not a driving force of the project. There was no official floor plan for the exhibition, the installation at each site being determined by the host institution based on their specific needs and requirements. However, at all venues we visited, exhibit units of all types were installed near each other. In other words, just a few feet from the large climbing structures for example, there were mechanical interactives, medium climbing structures and immersives, display cases, furniture, and large pieces of art. Several sets of exhibits were developed so that the same—or similar—geometric concepts were presented as both immersive and non-immersive experiences allowing visitors to engage with the concept in different ways. Variously referred to as “conceptual pairs” and “content pairs,” these included the following units. (Note that immersive exhibits are listed first, followed by their smaller table-top conceptual pair(s).) - Stack of Stars :: Space-filling Blocks (both SRDs and TOs, see Table 2) - Hopscotch, Distorted Chair :: Distorted Drawing Six general types of labels accompanied exhibit units: (a) instructions (e.g., how to use the interactives); (b) warnings (e.g., listing age limits for the climbing structures); (c) interpretive panels; (d) object labels (for Geometry Garden); (e) quotations (mostly incorporated into the Entrance Structure); and (f) titles. 7 The NSF grant proposal states that, for this project, immersion is defined as any experience in which the exhibit is not only larger than the visitors, but at least partially surrounds them. Selinda Research Associates, Inc. 16 Table 2. Names and classification of Geometry Playground exhibits large immersive medium immersive table top static objects photo panels exhibit furniture 1 Hopscotch x 2 Distorted Chair x 3 Distorted Drawing x 4 c Projected Puzzle x 5 Conical Mirror x 6 Build a Room x 7 Spin a Shape x Shape Cubes 8 c square (large and small) x 9 c ellipse x 10 c trapezoid x 11 c triangle (large and small) x Shape Stools 12 c square x 13 c ellipse x 14 c trapezoid x 15 c triangle x 16 Geometry in Motion x 17 Drawing in Circles x 18 c Gyroid x 19 c Stack of Stars x Space-filling Blocks 20 stellated rhombic dodecahedrons (SRDs) x 21 truncated octahedrons (TOs) x 22 Stellator x 23 Tile Designer x 24 Geometron x 25 Dividing Space x 26 Polyhedra x 27 Gear Cube x 28 Geometry Garden x 29 Geometry in the Wild x 30 c Sphere Cube x 31 Entrance Structure x c = Exhibits that at least some visitors used as climbing structures. Selinda Research Associates, Inc. 17 As mentioned previously, during this evaluation study we saw the Geometry Playground exhibition in three different venues: the Exploratorium in San Francisco; the Science Museum of Minnesota in St. Paul; and the Don Harrington Discovery Center in Amarillo, Texas. Although most of the 31 exhibit units were installed in all three venues, some museums had units that others did not, and the layouts in each of the three venues varied significantly. This allowed us to study the effectiveness of different exhibit units under different conditions. The design and layout differences are described in this section, and differences in effectiveness are discussed later in the report. See Appendix G for a complete list of exhibit units at each host museum and photos of the different installations. Exploratorium The exhibition opened at the Exploratorium (Explo) in late June 2010, and stayed open through the summer. Although most of the exhibition was in place during our site visit, several exhibits had not yet been installed or were installed while we were there, and several additional exhibits were not entirely completed even though they were on the floor. Also, although almost all the labels were installed, some were trial versions still considered to be under development. At Explo, Geometry Playground was located in the center of the museum, set off from the rest of the museum’s exhibits and service areas on all but one side. The north side of the exhibition served as a corridor, with one geometry exhibit (Projected Puzzle) to the left as visitors approached the Entrance Structure (Fig. G-1). The exhibit was bounded on the east side by a wall (Fig. G-2). On the west side, a corridor separated the exhibition from the food service area (Fig. G-3). There were a few small exhibits on the west side of the corridor that looked like they could be part of the Geometry Playground, but were not. The south side of the exhibition was not separated from other exhibitions; for example, two of Explo’s permanent exhibits—the Distorted Room exhibit and the Studio exhibit—were immediately adjacent to the Gyroid and other Geometry Playground exhibits. This meant that the north, west, and south sides of the exhibition were open and permeable to visitors, who were able to move in and out of Geometry Playground without encountering any physical or visual boundaries. The two large climbing structures, Stack of Stars (Fig. G-4) and the Gyroid (Fig. G-9) were on separate, raised platforms, surrounded by benches, rails, and low walls. Both were accessible through step-up openings as well as low ramps (e.g., Fig. G-6). Warning labels at the entrances stated recommended age limits (Fig. G-9), and interpretative labels were mounted both on the low walls and on some of the benches (Fig. B-29). The interpretive labels on the benches were mounted so that as visitors sat and watched their children, they would also be able to read the labels (Fig. G-9). The vitrines and cases of Geometry Garden were located in the center of the exhibition (Fig. G-10 and G-11). During the time we were on site, Exploratorium Explainers were assigned to monitor visitor engagement at the Gyroid. At Explo, exhibits in one of the conceptual pairs were installed close to one another. The star-shaped Space-filling Blocks (SRDs) table-top interactive was located next to its conceptual pair, Selinda Research Associates, Inc. 18 the Stack of Stars large immersive climbing structure, separated only by a bench and low rail (Fig. G-5). The other Space-filling Blocks exhibit (TOs) was on the opposite side of the climbing structure, separated by a ramp and low wall (Fig. G-6). The Stellator, an exhibit that was loosely related although not part of the conceptual pairing, was located a short distance away (Fig. G-7). The other conceptual pairs, i.e., the three exhibits that used cylindrical mirrors (Hopscotch, Distorted Chair, and Distorted Drawing), were all widely separated, with the Distorted Chair (Fig. G-8) and Hopscotch (Fig. G-2) located on opposite sides of the exhibition, and Distorted Drawing in between. The Entrance Structure and Projected Puzzle were the first exhibits visitors encountered as they walked from the museum entrance towards the exhibition (Fig. G-1). Visitors walking through the museum’s main exhibit area could reach the geometry exhibits either by walking through the doorway, which framed the center of the exhibition, or around the Entrance Structure. The Entrance Structure was the only exhibit unit that indicated the exhibition’s title. In addition to the exhibition itself, Explo’s Geometry Playground website launched at about the same time as the exhibition opened to the public. (Screen shots of several pages are in Appendix K, and the website is described and reviewed later in this report.) Signage informing visitors of the Geometry Playground exhibition was located outside the Exploratorium entrance (Fig. G-13). Special events related to the exhibition included a Geometry in Motion film series8 and Geometric Threads demonstration and exhibit space. Both of these were advertised on a special events panel inside the entrance (although the exhibition itself was not shown on the adjacent visitor orientation map). The Geometry in Motion films were additionally advertised on a theater marquee and poster (Fig. G-14), and a label adjacent to the exhibition listed the Geometric Threads schedule (Fig. G-15). Science Museum of Minnesota Geometry Playground was installed at the Science Museum of Minnesota (SMM) from October 2010 to early January 2011. SMM had been a partner on the Geometry Playground project from the very beginning. They had played a role in initial exhibit development, even serving as a test site for prototype versions of many exhibits. SMM exhibits staff assisted with remediation and final development of some of the exhibit units, and SMM evaluators conducted formative testing of label texts for some exhibits. Although the Geometry Playground exhibition was designed as a single installation, circumstances at SMM required that it be installed on two separate floors (the 4th floor, Fig. G-17, and the 3rd floor, Fig. G-169). A list of exhibits on each floor is included as Table G-2 in Appendix G. Both of the large immersive climbing structures were installed on the 4th floor. 8 Note that this film series shared a title with the Geometry in Motion table-top exhibit that was part of the regular Geometry Playground exhibition, but this was a coincidence. 9 Note that most visitors enter SMM at an upper floor and descend to additional exhibit floors. Although the 4th floor is above the 3rd floor, most visitors to Geometry Playground encountered the exhibits on the 4th floor first. Selinda Research Associates, Inc. 19 Most of the medium-sized immersive climbing structures—like the crawl-through Shape Cubes and Sphere Cube—were installed on the 3rd floor. Visitors were able to move directly from one floor’s installation to the next by either the central staircase, elevators located adjacent to the exhibition, or by an additional staircase along the outer wall of the Museum. Several sides of Geometry Playground were adjacent to other permanent SMM exhibits (e.g., exhibits on light, blood circulation, and dinosaurs), and a T. rex skull with a movable jaw was located inside one edge of the 3rd floor installation (Fig. G-18). Although all of the 31 exhibit units were installed, a few labels were still under development. The installation of the Gyroid and Stack of Stars exhibits was similar at both Explo and SMM, i.e., they both were installed on their own platforms, some of the labels were the same, and the structures were full-size. However, at the Gyroid at SMM, there appeared to be a greater focus on monitoring and controlling visitor behavior than at Explo, especially when the exhibit was crowded. To limit access, the Gyroid had black metal rails and plastic-coated metal cables and an entrance sign that posted an age limit and encouraged safe behavior. The Gyroid was also well-staffed during peak times in the galleries. During our time on site, the floor staff enforced the age limits on the entrance signs for the Gyroid (Fig. B-26). At SMM, the bench-mounted labels were removed from the Gyroid because of concerns for visitor safety, and the bench-mounted labels for Stack of Stars were mounted on the other side of the benches, facing away from the climbing structure (so one had to turn away from the structure to read them, as seen in Figs. B-30 and B-37). Compared with the Gyroid, access to Stack of Stars at SMM was more open (see Fig. B-30), with no metal rails or cables, although a similar age-limit entrance sign encouraging safe behavior was in place (Fig. B-31). Floor staff at the Gyroid loosely monitored visitor behavior at Stack of Stars, but did not try to enforce the age-limit there. Similar to the installation at Explo, the conceptually paired Stack of Stars and Space-filling Blocks were installed immediately adjacent to one another at SMM, separated only by benches that young children could climb on to gain access to the blocks (Fig. B-33 and Fig. B-35). The Stellator, an exhibit loosely related to the conceptual pair, was also immediately adjacent to the Stack of Stars platform, located between the two Space-filling Blocks exhibits (Fig. B-37 and Fig. B-38). The three conceptually paired cylindrical mirror exhibits (Hopscotch, Distorted Chair, and Distorted Drawing) were all located on the 3rd floor, but they were not adjacent to one another. Hopscotch was in the atrium, where it could be clearly seen by visitors on the central staircase (Fig. B-1). Distorted Chair was on the opposite end of the exhibition, facing the elevators—the first thing visitors encountered as they exited the elevators. Distorted Drawing was roughly half way between the two immersives, surrounded by other exhibits. The first thing that visitors encountered as they walked down the staircase to the 3rd floor was the big orange dot for Projected Puzzle and Geometry Garden (middle foreground of Fig. G-17). Build a Room was located at the edge of the 3rd floor, facing away from the bright atrium and Selinda Research Associates, Inc. 20 shielded by partial walls (Fig. B-13). Geometron and Dividing Space were also located in a darker corner of the 3rd floor. The Entrance Structure was installed by the 4th floor elevators (Fig. G-19), and was the main access point where the exhibition’s title was displayed. Smaller signs near the stairs on each floor informed visitors that there was more of the exhibition on another floor (Fig. G-20). Geometry Playground was also featured prominently on the SMM website home page (Fig. G-23), which linked through to a small web module that SMM developed especially for the exhibition (Fig. G-24). The SMM Geometry Playground web module included links to PDF versions of many of the Pathways educational activities developed as part of the Geometry Playground project. The SMM site also linked to Explo’s Geometry Playground website (Fig. G-24, right menu, “Official Exhibition Site”). Additional signage for the exhibition was located behind the box office registers (Fig. G-21), and the exhibition’s two floors were also listed on the elevator placards (Fig. G-22). Don Harrington Discovery Center Geometry Playground was installed at Don Harrington Discovery Center (DHDC) from early February 2011 to the end of May 2011. Although DHDC was not a partner in the Geometry Playground project, it had other ties with the Exploratorium, such as an affiliation with the Exploratorium’s ExNet division,10 and the Executive Director and several other staff members who were former Explo employees. DHDC had several other traveling and temporary exhibitions on display concurrent with Geometry Playground, including several in the same wing. In addition, a traveling dinosaur exhibition was displayed in the other wing of the museum and attracted larger-than-normal crowds to the museum. Entering through a hallway (Fig. G-25), DHDC visitors walked past a Shape Cube to the Hopscotch exhibit, which was separated from the rest of Geometry Playground and surrounded by elements from other exhibitions (Fig. G-26). Although the floor plan of the exhibition was significantly different, the rest of the exhibition layout was somewhat similar to the Explo layout: the exhibition was all on one floor, in one large space, open on three sides, and with porous boundaries—visitors could, and did, move to and from Geometry Playground into surrounding exhibitions. In addition (and similar as well to SMM), the boundaries of Geometry Playground at DHDC were discontinuous, with several Geometry Playground exhibits set within, and surrounded by, other exhibitions. 10 As stated on the ExNet website, “The Exploratorium Network for Exhibit-Based Teaching (ExNET) is a hybrid exhibit and teaching program that shares the fruits of forty years of research by the Exploratorium and affiliated partners with a diverse group of science-rich institutions around the United States and the world.” Selinda Research Associates, Inc. 21 The Gyroid and Stack of Stars underwent the biggest changes to individual exhibits. Unlike at Explo and SMM, where each of these units was installed on its own platform, at DHDC both exhibits were installed on a single large, padded platform. The Gyroid installation at DHDC included only two of the three levels (Fig. G-27). Unlike at Explo and SMM, the two large climbing structures were unstaffed for the most part, and access to the structures was open, with no barriers. (Compare Figs. G-27 and G-28 at DHDC, with Fig. B-25 at SMM). The Gyroid included two wall-mounted interpretive panels and a warning sign at an entrance (Fig. G-28A). Circular activity labels (that had been piloted at SMM) were riveted onto the Gyroid surface (Fig. G-28B). The only Stack of Stars label was an entrance warning label mounted at the corner of the platform (Fig. G-29). The wall- and bench-mounted labels installed at both Explo and SMM were not installed at DHDC. As at Explo and SMM, both Shape-filling Blocks exhibits (SRDs and TOs) were next to Stack of Stars. At DHDC, however, they were even more closely conceptually paired, with no bench between them (Fig. G-29). The loosely related Stellator and Tile Designer exhibit units were a short distance away. Distorted Drawing and Distorted Chair (conceptual pairs) were adjacent in a far corner of the exhibition (Fig. G-30), although Hopscotch (part of the conceptual pairing) was at a distance, at the entrance to the exhibition. The Projected Puzzle was located near the far wall of the exhibit hall (Fig. G-31), with the large orange dot on the floor near an entrance to the Gyroid platform. The Geometry Garden was split into two sections of cases and installed on either side of the entrance to another exhibition on the north side of the hall (Fig. G-32), with one side of the larger section facing a dividing wall. Drawing in Circles and the small triangle Shape Cube were installed within another exhibition (Fig. G-33A). The Entrance Structure was not installed as such, although several parts of it—including the title of the exhibition—were installed near the back of the exhibition (Fig. G-29 and G-31). Also, relatively few of the Geometry in the Wild photo essay panels were included in the DHDC installation, and those that were, were installed around the large climbing structures. The enclosure around Geometry Garden was not installed at DHDC. Other differences noted from previous installations included higher levels of ambient lighting around the Geometron, Dividing Space, and Build a Room, compared to Explo and SMM. Also, at the time of our site visits to DHDC, the “light painter” at Build a Room had been replaced with a couple of flashlights (see Fig. G-33B). The original “light painter” (see Fig. B-14) had apparently been damaged when visitors used it as a hammer and pounded on the small table in this exhibit. The DHDC website had Geometry Playground listed on its home page on most days, along with the other temporary exhibitions and upcoming special events and programs (Fig. G-36). The “Learn more” link took users to a web page with a short description of the exhibition (Fig. G-37). This page did not include a link to Explo’s Geometry Playground website. Selinda Research Associates, Inc. 22 There were a few signs advertising Geometry Playground outside the exhibition hall, although none at the DHDC entrance (Fig. G-34). The only mention of the name of the exhibition near its entrance was on a donor panel (Fig. G-35A). Small posters inside the men’s rooms mirrored the Explo website’s home page by spelling out the thematic classification of the exhibition into the geometries of seeing, moving, and fitting things together (Fig. G-35B). Selinda Research Associates, Inc. 23 FINDINGS: THE VISITOR EXPERIENCE In this section of the report, we discuss the many ways in which visitors engaged with the exhibition at the three host museums, including discussion of some of the outcomes of these engagements. Most of our discussion of spatial reasoning is deferred to a later section of the report (Findings: Spatial Reasoning). Overall Experience The data from this study indicated that the Geometry Playground exhibition provided fun and educationally effective experiences for a wide range of visitors, including children in the target age range and their families. Most visitors to the exhibition enjoyed themselves, and engaged in a wide range of whole-body and smaller-scale physical engagements. We talked with many visitors who were fascinated by the Gyroid, Shape Cubes, Tessellating Blocks, and geometric artifacts on display, and seemed primed to follow up on that interest. For example, one visitor said she wanted to install a shape cube in her backyard and another said she was thinking about ways to adapt a section of the Gyroid as a sofa. These experiences helped raise many visitors’ awareness of aesthetic aspects of geometry and about ways that geometry is encountered in everyday life. Visitors worked hard to explore, navigate, and construct in both two- and three-dimensional spaces; they supported each other in figuring out how to climb to the top of a stack of stars, or to draw a room in three dimensions only to have it appear on a two-dimensional screen. When they engaged with the challenges presented at the exhibits, some visitors practiced increasingly sophisticated spatial reasoning as they discovered and applied rules and developed mental objects that helped them reach their goals. The mix of large- and medium-sized immersive exhibits, hands-on table-top interactives, and contemplative exhibits provided a refreshing combination of experiences that virtually all visitors took advantage of and appeared to appreciate. And although the exhibition was targeted to 5- to 12-year-old children and their adult caregivers, there was strong evidence that visitors of all ages effectively engaged in practicing a wide range of spatial reasoning skills. The most effective adult-child interactions took place at the table-top interactives, which were designed to provide opportunities for families to engage together as groups. At the large climbing structures and medium-sized immersives, most adults tended to stand back or sit down, sometimes watching and occasionally talking with their active children and sometimes doing other things with their time. The remainder of this report describes in detail many of the specific findings about visitors’ engagements with, and experiences in, the Geometry Playground exhibition. Visitor Flow As noted in the previous section, the floor plans for Geometry Playground at all three host museums were open and porous, and in some cases overlapped with surrounding exhibitions. There were indications that many visitors used only part of the exhibition—sometimes only two or three exhibits—before moving on to other exhibitions. Regardless of the site, we identified three distinct visitor flow patterns. Selinda Research Associates, Inc. 24 Linear traverse and return. Moving more or less linearly through Geometry Playground, some visitors started at one side of the exhibition, usually stopping at several geometry exhibits before continuing on to other nearby exhibitions. There were indications that many—maybe most—of these visitors stopped again at Geometry Playground later during their visit, perhaps a half hour or more after their first experience with it. Porous boundaries. Most visitor groups moved in a less linear way, passing back and forth across the boundaries of Geometry Playground several times within a 10- or 20-minute interval. For instance, at the Exploratorium, one family visited a geometry exhibit or two, and then went to the adjacent Distorted Room exhibit, and returned to interact with a few more geometry exhibits before moving off to see the (non-Geometry Playground) exhibits near the restrooms for 10 minutes or so. They then returned to Geometry Playground and spent several minutes on each of the climbing structures. At SMM, there were indications that many visitor groups sampled several of the exhibitions on one floor—including Geometry Playground—before moving to the next floor and sampling from Geometry Playground and other exhibitions on that level. Because most of the many entry points to the exhibition were not labeled as such, most visitors entered at least some parts of Geometry Playground without passing a sign identifying the exhibition. Settle and stay. After initial explorations, some groups found a single exhibit they really enjoyed and then stayed there for a long time, sometimes for 20 minutes or more. This often happened at the two large climbing structures, where adult caregivers sat as children played. Remaining at one exhibit sometimes happened at other exhibits. In particular, a few visitor groups were observed spending long periods of time carefully looking at and talking about the objects in the Geometry Garden. Perceived Cohesion of the Exhibition The data indicated that most visitors to Geometry Playground perceived the exhibition not as a coherent whole, but rather as a series of individual exhibits, similar to other exhibit units in many science centers. Because of its loosely organized floor plans and open boundaries, even visitors with a strong interest in geometry sometimes had a hard time finding and engaging with all the geometry exhibits. For instance, respondents who identified themselves as repeat visitors sometimes asked us if a given exhibit unit (such as Build a Room) was new, since they had visited Geometry Playground several weeks earlier and did not remember seeing it. One important consequence of these flow patterns was that visitors often formed their impressions of Geometry Playground and its themes based on incomplete and/or interrupted experiences. The lack of perceived cohesion was particularly evident at SMM, where the large and medium immersives were on separate floors. Visitors were rarely seen moving directly from one level of the exhibition to the other, and, especially on the 3rd floor, most visitors we observed used only a few Geometry Playground exhibits before moving to parts of an adjacent exhibition (e.g., dinosaurs, light, blood circulation, Science of a Sphere). Selinda Research Associates, Inc. 25 In addition, many respondents expressed differences in their perceptions of the two floors. Many respondents perceived the 4th floor of Geometry Playground as a coherent exhibit; adult respondents often called it “the Geometry Playground” or “Geometry Playspace.” However, respondents had a somewhat more difficult time seeing the 3rd floor as a coherent exhibit, separate from the surrounding exhibitions. For instance, one respondent said she thought the Distorted Chair might be about physics, part of the Light exhibition whose neon sign she saw behind it. Many respondents did not realize that the two floors were part of the same exhibition. Perceived Focus of Exhibition Another consequence of the open floor plans was that, for most visitor groups, Geometry Playground lacked a clearly defined entrance that identified and set the stage for the exhibition. Even when the Entrance Structure was installed (for example at Explo and SMM), there were indications that it played a small role in many visitors’ initial experience of the exhibition. At Explo, more than half the visitors we observed approaching the exhibition from the north walked through the door in the center of Entrance Structure, and more than half of those visitors stopped at one of the exhibits near the door. However, there were indications that few noticed the exhibition title, and unlike other exhibitions with a more prominent title, we heard no one state the name of the exhibition as they entered it. At SMM, most visitors’ first encounters with the exhibition were from the stairways or from the 3rd floor elevator, none of which had clear signage identifying the exhibition. At DHDC, most visitors walked through or past other exhibitions to reach Geometry Playground, and the only sign identifying the exhibition was on the back wall. For respondents who knew about Geometry Playground before they arrived at the museum, this was not as much of an issue. They were expecting to find geometry exhibits that they could play with, and they usually recognized them when they arrived at their destination. However, most respondents we talked with across all three host museums were not aware of Geometry Playground in advance. They discovered and made sense of the exhibition by exploring it. When we interviewed these respondents after they had engaged with one or two exhibits, some of them were aware that they were in a special place with exhibits about shapes or geometry; but often they were not. When we interviewed such visitors after they had visited five or six exhibits, they usually realized they were in a place with fun exhibits about shapes or geometry, but most were still unaware of the exhibition’s name. That seemed unfortunate, because when respondents could tell us they were visiting “Geometry Playground,” they often used the name as a starting point for describing both their understanding of the exhibition and their feelings about their engagements, saying things like, “This makes geometry fun!” Being aware of the title Geometry Playground also helped many of these visitors begin thinking about this as an exhibition about movement and other aspects of spatial reasoning, rather than one focused primarily on the geometry they learned in school. Time at Exhibits Many visitors to Geometry Playground spent several minutes engaged with individual exhibits. While a formal tracking and timing study was not part of the summative evaluation, we did time Selinda Research Associates, Inc. 26 some visitors as they engaged with exhibits, and documented engagements of 5, 10, 15, and even 20 minutes at specific exhibits, with the longest engagements at the climbing structures, and also at exhibits where each member of the group waited to take their turn, such as Drawing in Circles and Build a Room. Even when the exhibition was not very crowded, visitors sometimes had to wait several minutes or more to use the exhibits that could be used by only one visitor at a time. Experiences with Immersive Exhibits As noted earlier, many of the Geometry Playground exhibit units were categorized as either large or medium immersives (see Table 2 under Findings: Design and Layout). Immersive exhibits were defined as those exhibits where the phenomenon extends from in front of you to either behind your back, over your head, or under your feet. Immersion was also defined in the original grant proposal as “any experience in which the exhibit is not only larger than the visitors, but at least partially surrounds them.” Immersive exhibits were of particular interest in this project because the exhibit team hypothesized that these whole body experiences would contribute in important ways to visitors engaging in and practicing spatial reasoning. Large Immersives One area of interest in this study was the extent to which the two large immersive climbing structures—i.e., Stack of Stars and the Gyroid—would elicit playground behaviors and whether these behaviors contributed to, or detracted from, the goals of the exhibition: i.e., to engage visitors in spatial reasoning. Playground behaviors were loosely defined as behaviors one would see on a playground, for example, children engaging in whole body gross motor activities, and adults sitting on the sidelines watching their children play or engaged in other activities such as talking on a cell phone, The following sections will explore these issues in detail. The data indicated that the large climbing structures did inspire playground behaviors among children and adults alike. Children tended to climb, whereas adults tended to sit. Seeing adults resting as children played seemed to reinforce the playground ambiance of the exhibition. There was no other place in the museums where children were playing so physically in such large numbers. As at a playground, they climbed over, through, and when possible, under the structures, interacting with other children as they played. When the climbing structures were not closely supervised by museum staff, children also raced to the top of the structures and played games like tag, king of the hill, and play fighting. As often seen at a playground, many caregivers sat and rested, used cell phones, and gave their children snacks. Caregivers with children who needed a little more help—such as younger visitors, those unsure of themselves, or those with physical limitations—often stayed close to their children, standing beside the structures as their children climbed. The effects of the large climbing structures were particularly evident at SMM, where the large and medium immersives were on separate floors. When the two floors of Geometry Playground were crowded, the 4th floor with its climbing structures looked much more active and high energy than the 3rd floor, in part because the children on the climbing structures were so active. Selinda Research Associates, Inc. 27 The large immersives attracted long and active engagements by many children, but less engagement by most parents and other adults. But the Geometry Playground large climbing structures didn’t operate exclusively as a playground. Both caregivers and children displayed an interesting mix of playground and more museum-like behaviors. For example, as expected in a museum, a few of the children who were climbing the structures also seemed to be thinking about the shapes as they engaged with them. And unlike at a playground, some caregivers read labels. There also seemed to be more photos and videos taken to record the experience than at a neighborhood playground. We should note that we also observed a few adults climbing on the structures. Medium Immersives While the large immersives quickly and clearly elicited climbing and other playground behaviors among visitors, the medium immersives fell into two distinct camps: those used by visitors as a type of climbing structure—for example the Shape Cubes, the Sphere Cube, and the Projected Puzzle—and those that were virtually never used as climbing structures, i.e., Hopscotch, Build A Room, Distorted Drawing, and Polyhedra. While most adult engagements at the large climbing structures tended to be minimal, passive, and reminiscent of the ways adults engage in playgrounds, there were indications that many caregivers were more engaged at the medium immersives that inspired climbing. For example, whereas most adults tended not to climb in the Shape Cubes, they engaged by bending over to look inside, admiring the woodwork, taking photos of their kids climbing through them, and sometimes reading labels and talking about the internal shapes. At the Sphere Cube, however, although a few adults climbed inside (including a mom who laid side-by-side with her daughter, both on their backs, looking at something on a cell phone), most stood back and watched as their children climbed inside. With the non-climbing medium immersives, many adults were even more engaged, participating in ways that are typical in museum settings. For example, many adults entered the Build a Room space with their children, and caregivers often assumed a coaching role as their children used the light wand. We also observed many adults without children using the light wand and video screen. Adults and children often tried out the Distorted Chair together. At Hopscotch, however, caregivers more often watched as their children used the exhibit, sometimes reminding their children to look in the mirror as they hopped. Most adult visitors at the Projected Puzzle exhibit tended to stand back and observe the phenomenon at a distance, as it was designed. But many younger visitors walked through it, and toddlers often used it as a climbing structure. Many of these children did not seem to understand that this exhibit was something more, unless an adult called it to their attention and perhaps walked them back to the large orange dot where they could best see the phenomenon. Selinda Research Associates, Inc. 28 Children’s play at and around the medium immersives seemed less playground-like and more museum-like than with the large immersives, perhaps because of the exhibits’ smaller size and the shorter engagement spans they inspired,. For instance, the Hopscotch board was treated more as an optical illusion than a place to play hopscotch. The Sphere Cube and the Shape Cubes when they stood alone seemed most playground-like, as children climbed within them as caregivers watched; however, both exhibits were frequently treated as photo opportunities, which seemed more of a tourist-like or museum behavior than a playground one. Also, when the smaller interpretive Shape Cubes were mounted beside the larger matching cubes, caregivers often called them to children’s attention in a museum-like way. Transitioning Between Exhibits As mentioned previously, the different types of exhibit units (large immersives, medium immersives, table-top interactives, display cases, photo panels, and exhibit furniture) were mixed throughout the exhibition. This meant that visitors who were engaged in quiet contemplation were not far from others involved in active exploration and raucous play. It also meant that visitors would frequently transition between exhibit units requiring significantly different levels of energy, activity, and cognitive engagement. One of the concerns of the exhibit development team was that children might not be able to calm down after climbing on the Gyroid or Stack of Stars. If there was a spillover effect in children’s high-energy behaviors from the large immersives to the surrounding non-immersive exhibits, it might diminish the exhibition’s overall effectiveness. In spite of these concerns, the data indicated no significant negative spillover effect. For instance, we observed many children who went immediately from Stack of Stars to several-minute engagements with the Space-filling Blocks, and sometimes back again; or from the Gyroid to a similarly deep engagement with nearby table-top exhibits. Also, at SMM’s two-floor installation, although there was in general a higher energy level on the floors that had the large climbing structures, the energy level at the non-immersives on both floors was similar. This suggests that there was little spillover of the high-energy physicality observed at the large immersives to the smaller exhibits. One reason for this might be that most elementary school children have lots of experience transitioning from playground to classroom. Perhaps they put that experience to good use in Geometry Playground, as they transitioned from one type of playful engagement to another. While for the most part there appeared to be minimal spillover effect, we did observe a few exceptions. For example, chase games that began on the rubberized surfaces around the large climbing structures sometimes continued into the surrounding areas, and children who spotted Stack of Stars from atop the Gyroid sometimes ran from one exhibit to the other. And in one notable exception, we observed a birthday party with ten 6- and 7-year-old boys crowded into the Sphere Cube. Worked up by one of the dads going “booooo, boooo” through the cube’s outer grid, the boys started climbing the sides, sliding down, and wrestling. When they moved to Dividing Space, they then participated in a particularly aggressive episode of “smash and crash.” Selinda Research Associates, Inc. 29 In spite of these few exceptions, we found most often that when children left a large immersive, they (and their parents) resumed their usual science museum roles, and school groups resumed their school-group behaviors. While the data indicated little negative spillover effect in children’s high-energy behaviors from the large immersives to the surrounding non-immersive exhibits, there was evidence of some positive spillover effects. For example, many adults and children explained that spending physically active time on the climbing structures later helped the children focus on the surrounding exhibits. One dad said it was good to give his preschool son a break so he was refreshed for another set of regular exhibits. Another mother put it this way: For young kids it’s great, because it gets the energy out, and then you can go see the different areas and say, ‘Oh, look at this, this does this,’ versus just running around down here. A 9th grade boy related it to school. It’s kind of like in school after recess you’d be more focused afterwards…[You] are more into learning more things. Some parents also said their children’s climbing provided benefits for them. One mother talked about how it kept her active young son occupied so she could use the nearby exhibits. When we talked with another mom about the perceived value of her children climbing without her involvement, she explained that sitting here was fine with her because the kids were enjoying climbing. She explained how they sat through the IMAX movie for her, and now she is sitting through this for the kids. “It’s an exchange.” One dad explained in some detail that he would only be concerned if the exhibition was crowded, but that otherwise it provides a good balance of climbing, getting tired, resting, and climbing again. The activity between climbing and table-top activity would probably depend on the individual child and the supervision. And probably the dynamic of how many children were in here too. If you have a whole bunch of kids, and it gets a little wilder, maybe yes, [the energy level] would be problematic. As far as we were concerned, pretty much just the three of us, and it wasn’t really crowded in here, I didn’t really have a problem [with] the interchange between the climbing and the table-top activities. In a sense it may be a good transition because you can climb a little bit and when you get tired you can do something else, and when you get rested up you can climb some more. Given these types of statements, it was not surprising that most respondents supported the inclusion of playground-like structures in science museums. The few dissenters were adults who weren’t sure if children were getting anything out of all this effort and time. For instance, an aunt with two elementary school age boys said she didn’t mind her nephews playing with exhibits, as long as they got something out of it—but she didn’t think they were getting much out of the Selinda Research Associates, Inc. 30 Gyroid. Also, most of the parents we talked with had trouble explaining what their children might be getting out of their climbing experiences, but they felt the benefits came later as their children focused on other exhibits. Safety at the Large Climbing Structures As might be expected, the presence of two large climbing structures raised the question of visitor safety at these exhibits. The Explo team worked hard to make these structures as safe as possible including consulting with playground safety experts during exhibit development, controlling visitor access to the structures with cables and benches, promoting appropriate use via safety messages, and assigning museum staff to supervise visitor behavior, especially during busy times. At the Exploratorium—the first installation of the exhibition—access to the climbing structures was limited to three entrances, each marked by signage informing parents that their supervision was required, and that the structures were designed for children five years and older. Floor staffers were usually posted at the Gyroid and helped monitor and control children’s behavior. When the exhibition moved to SMM, staff were also assigned to the Gyroid. In addition, the age range posted on the signage on the Gyroid was increased to seven years and older, and all but one entrance to the Gyroid was sealed off during busy times, such as during school tour hours and winter break. Floor staff allowed a limited number of children on this climbing structure, and supervised them closely, making sure accompanying adults attended to the youngest children. When only a few family groups were present, all the Gyroid entrances were open, and the seven-and-up rule was loosely enforced, or in some cases enforced by caregivers rather than staff. Access to Stack of Stars was more open than at the Gyroid, although the five-and-up signs were posted at the entrances to the platform. At DHDC, only two levels of the Gyroid were installed, and access to both climbing structures was open at all times. The same warning signs used at SMM were posted near entrances to the platform: seven and up for the Gyroid; five and up for Stack of Stars. One floor staff member was responsible for an entire gallery with several exhibitions, so staff supervision at the climbing structures was minimal. Many children in the target age range seemed to recognize that there was some danger with the climbing structures. For most of them, it was part of the challenge and thus part of the fun. However, some children hung back because of their fears. In these cases, parents often stayed close to help them climb despite their fears. Some visitors told us that Stack of Stars looked more dangerous than the Gyroid (because of all those sharp-looking points on the stars). The data indicated that some accidents did occur on both structures and at all three venues, but the number and seriousness of these accidents was not tracked as part of this study. One question that emerged during data collection was the usefulness and effectiveness of the age limit signs. When the exhibition was at Explo, the Stack of Stars warning signs (Fig. B-31) indicated that the exhibit was “designed for children 5 and up. Adult supervision recommended.” A similar sign was posted at the Gyroid. Selinda Research Associates, Inc. 31 When the exhibit moved to SMM and DHDC the sign for Stack of Stars remained essentially the same, but at the Gyroid, the minimum age was raised to seven, and the wording was a little stronger: “Children under 7 aren’t ready for this climber. Adult supervision strongly recommended.” (Fig. B-26). While at times it felt that the signs helped caregivers make and enforce decisions (for example as something adults could use to limit or moderate children’s overly rambunctious or potentially unsafe behavior), we also observed instances where the people who needed the signs were the very ones who weren’t reading them. For example, at some times the structures became overly crowded. In these situations, having an age limit posted seemed an appropriate way to limit the numbers of children on the structure and also make sure the ones who were climbing were mature enough to take care of themselves. But some parents of younger or overly rambunctious children simply ignored the signs. On the other hand, visitors for whom the signs really didn’t apply, sometimes took them too literately. For example, we observed one little boy—he appeared to be about three-and-a-half years old—approach Stack of Stars. There was no one else on either of the climbing structures— the entire exhibition had very few visitors—and he was there with two attentive adults, presumably his parents. In this situation, it could have been a safe and enjoyable experience for him to climb Stack of Stars as he would have had the structure to himself, and he had two attentive adults to monitor his climbing. But the adults—seeing the sign saying it was for ages five and up—continued to try to steer him away from the structure. They grew increasingly frustrated by his persistence in trying to get onto it, finally threatening that they would leave the museum “if you misbehave.” In this example, the signs served to turn what could have otherwise been a safe and enjoyable experience into something tense and uncomfortable. While these (and similar situations) were interesting examples of some of the issues surrounding the posting of age limit and safety messages, further exploration is needed to come up with definitive conclusions or recommendations. Caregiver-Child Interactions As part of this summative evaluation we were particularly interested in looking at adult-child social interactions, given that this was an exhibition that incorporated playground pieces and elicited playground behaviors. Research continues to show the important role that social interaction plays in informal learning environments, and the Geometry Playground exhibition offered a unique opportunity to study this. Range of Social Interactions In studying social interactions, a range of types of engagements emerged. We identified eight different types, which we loosely classified into three broad categories, depending on physical proximity: Far Out, Up Close, and Hybrid. It should be stressed that these are very loose categories and define certain types of engagements, not types of visitor groups. The members of Selinda Research Associates, Inc. 32 any particular group moved frequently and fluidly between these different modes of social interactions, sometimes within very short time periods. Also, variations of all of these engagements were observed at all the exhibits in Geometry Playground. The purpose in presenting them here is to describe how visitors tended to use the exhibition, not to make judgments about their ultimate usefulness. Far Out Far Out social interactions included those characterized by visitors located at greater distances from one another. We identified three types of Far Out social interactions: On Your Own: On Your Own engagements tended to be the least engaged. This was when a child and their companion adult really did not socially interact to any great extent, at least not during the time of physical engagement with the exhibits. Sometimes this was an adult letting their child explore an exhibit—or sometimes the entire exhibition—while they pursued their own interests. It also included turning a child loose to explore the museum on their own. This approach seemed to happen because many parents—especially museum members and frequent visitors—felt their children were safe exploring on their own. This also gave parents a chance to engage with the exhibits they wanted to see in ways appropriate to their interest and prior knowledge. This was also the preferred style for some teachers and chaperones on school field trips. While not atypical of some science museum experiences, we speculate that On Your Own social engagements may have happened more frequently at the Geometry Playground exhibition than at other science museum exhibitions. It could be argued that On Your Own interactions are typical of adult-child playground interactions, and because so many of the Geometry Playground exhibits elicited playground behavior, that is perhaps why it emerged as an important category of social interaction at Geometry Playground. Observer: Observer social interactions consisted of one person engaging with an exhibit unit, usually a child, while the other, usually an adult, casually observed. It was characterized by a fairly independent exploration of a particular exhibit unit, and when the engaged person was finished, moving on together towards a new exhibit. This social engagement was characterized by a minimum of conversation and tended to be logistical (“Are we ready to move on?”) rather than content-focused. Although there were many exceptions, most Observer social interactions took place at table-top units, whereas Far Out interactions tended to occur at the large climbing structures. Although not ideal, this type of social engagement is typical at many museum exhibits including those in science museums. McManus (1987, 1988) conducted research that demonstrated a strong positive relationship between the amount of social interaction and the amount of (broadly defined) learning at exhibits. In a related vein, Diamond (1980, 1986) demonstrated that visitors tend to “learn” from one another rather than directly from the exhibit itself. Safety and Supervision: Safety and Supervision interaction was similar to Observer in that it was somewhat detached, but unlike Observer, it had a clear purpose, i.e., to monitor and oversee Selinda Research Associates, Inc. 33 behavior. It was characterized by adults remaining on the sidelines but periodically calling out to children to “be careful,” “don’t climb as high,” or “play nice.” Safety and Supervision was also typical of the social interactions when school groups were in the exhibition. Adults tended to stand back, try to keep an eye on all the children, and enforce discipline. They seemed more like playground monitors than teachers. While common at the large climbing structures, Safety and Supervision was not seen to the same extent at the other exhibits. Again, these interactions were likely a function of the playground behavior that the climbing structures elicited. Up Close Up Close social interactions tended to take place in closer physical proximity than Far Out interactions. We identified three types of Up Close interactions: Photo Op: These exchanges were focused on the end goal of taking a photograph and documenting a visit. They included requests by a visitor to have their photograph taken, as well as directions to someone to position themselves for a photograph, and suggestions for where and how to stand or sit. Many museum exhibits elicit photography, and in fact many have specific photo opportunities built into them. Although specific photo opportunities were not part of the Geometry Playground design, many visitors engaged in these types of interactions. Although photographs were taken throughout the exhibition, this was a particularly prevalent social interaction at the Shape Cubes, and at the two large immersives, Stack of Stars and the Gyroid. Games: We saw many spontaneous games erupt in Geometry Playground. These included tag and hide and seek (engaged in by groups comprised of both adults and children as well as children-only groups), and also large climbing structure games (engaged in by children only) such as king of the hill, and keep-my-sister-from-reaching-the-top. Spontaneous game-playing observed at Geometry Playground is atypical of many science museum exhibitions. We speculate that the large climbing structures naturally spawned some competitiveness as children raced to the top, and the Shape Cubes, furniture, and large cylindrical mirrors made intergenerational tag and hide-and-seek a natural. Meaning-Making: Meaning Making social interactions focused on trying to make sense from an exhibit experience. They included such teaching/learning exchanges as: (a) directing and focusing attention, (b) explaining and suggesting, (c) asking and answering questions, (d) goal setting, and (e) giving encouragement and feedback. This sort of engagement was most often displayed by parents, grandparents, and aunts and uncles in small family groups and happened much less often with school and camp groups. Although the exhibits were designed with children five and older in mind, this sort of close engagement by caregivers extended the effective age range of some of the exhibits down to much younger ages. While some visitors did engage in meaning-making with other members of their social group, only a few did so at the large climbing structures. At all the exhibits, the biggest inhibitor to engaging in this type of social meaning-making appeared to be just one thing (see below). Selinda Research Associates, Inc. 34 Hybrid Some visitor social interactions could only be classified as a type of hybrid between Far Out and Up Close. Check In: This type of social interaction was typical at the Gyroid and Stack of Stars when children would climb the structures for a short time while adults either supervised from afar, observed, or did their own thing. Then the child would come running back to check in, and the interaction would switch to Up Close with the parent pointing something out or helping the child make sense of what they were doing. This type of social interaction was observed almost exclusively at the Gyroid and Stack of Stars. Again, this tends to be somewhat typical of playground behavior, but less typical of museum exhibits. Recognition & Pride: Another type of social interaction took place as a mixture of Far Out and Up Close. Among many visitors, a prominent affective response to their engagements with the exhibits was pride in their accomplishments, which they enjoyed sharing with others in their social groups, both adults and children. Many visitors worked long and hard to construct things at exhibits like Space-filling Blocks and Geometry in Motion, and they often sought out and received others’ admiration. Not surprisingly, as is typical in many science museums, Recognition & Pride tended to happen at Geometry Playground table-top exhibits that actively involved visitors in creating something such as (for example) Space-filling Blocks, Distorted Drawing, Build a Room, and Tile Designer. But Recognition and Pride also took place at the large immersives when some children proudly proclaimed to their adults that they had reached the top. These categories of social interactions are not the definitive declaration of how all visitors engaged with one another, but have been included to describe the range of social engagements that emerged from our limited time in the exhibition. While some may be perceived as more or less important than others, it is up to the Explo team to decide if there are certain types of engagements they feel should be downplayed, and others they wish to see increased. That said, for many visitors to science museums in general, engaging in active Meaning-Making is an important component of their visit. As described immediately below, we found that many visitors to all of the Geometry Playground exhibits ran up against just one thing that prevented them from engaging in this type of social meaning-making. Just One Thing In observing the ways that social groups engaged with the exhibition, we noted that many visitors seemed to need just one thing—usually a hint or instruction to do or notice something— in order to get something meaningful out of the exhibit. Sometimes these just one things were for really basic things like getting the exhibit to do what it was supposed to do. In the best situations, these were provided by someone else in the group, for example, when an adult provided them for a child. We identified a few just one things that, when someone in the social group provided them, the meaningfulness of the visitor experience improved dramatically. Selinda Research Associates, Inc. 35 • Hopscotch and Distorted Drawing: “Look in the mirror.” • Projected Puzzle: “Stand on the orange dot.” • Spin a Shape: “Keep turning the base of the card until it fits into place.” • Space-filling Blocks: “Keep trying. If you turn it just the right way, it fits perfectly.” • Stack of Stars: “Look, it’s made of star shapes!” • Dividing Space: “Notice the color patterns around the pegs and pucks.” • Geometry in Motion: “Connect the pieces with the small metal magnets.” • Gear Cube: “Keep turning. Watch closely or you’ll miss the cube.” At many exhibits, these just one things were carefully incorporated into the exhibit labels or design, but the data indicated that many visitors did not use them. This tended to be either because (a) the group was operating in a Far Out mode, (b) the just one things were difficult to find in the labels, or (c) the hints didn’t stand out as something important to pay attention to. In some cases, the labels used straightforward language, but it was hard for visitors to recognize it as a just one thing or use it as part of their everyday conversation. Compare for example, the Stack of Stars label text below, with the just one thing we heard a few visitors say: “Look, it’s made of star shapes!” This climbing structure is made of stars that all have the same shape. In other situations, the just one thing was buried among other text on the label. An example of this was the direction to stand on the orange dot in order to use the Projected Puzzle. When visitors started at the second label located next to the structure (see Fig. B-9), most visitors did not find the just one thing (“walk over to the orange spot”) that was at the very end of the second paragraph of text. PROJECTED PUZZLE To see the words, you have to stand in the right spot. There is a message hidden here. To see the parts fit together to look like a flat image, walk over to the orange spot on the floor. Engagements at the Large Climbing Structures Engagements at all exhibit units included a wide range of behaviors, and virtually all types of engagements were observed at all exhibits (with obvious exceptions such as, for example, climbing was not in evidence at the Gear Cube or Geometry Garden). The data indicated however, that certain types of engagements were more typical at the two large climbing structures than at other smaller exhibits, where there tended to be more instances of typical museum behaviors such as reading labels and meaning-making, and fewer instances of typical playground behaviors such as the Far Out engagements described above. Because large climbing Selinda Research Associates, Inc. 36 structures in museum settings are still relatively rare, following is a detailed look at how children and adults tended to engage with these large immersive exhibits in Geometry Playground. Children. Looking over the full range of children’s behaviors at the climbing structures (see Appendix H, Table H-1, for a list of children’s behaviors identified during site visits), we identified general patterns in the ways that children engaged while climbing the structures. First, we observed many children who engaged with exploratory and playful behaviors on the structures. Children seemed to spend the first minute or so of their engagement exploring the structure. Then, especially when the climbing structures were loosely supervised by staff, children sometimes became more playful—even playing games like tag and race to the top. Both exploratory and playful engagements seemed to contribute to children’s spatial orientation and navigation through the spaces. As they explored and played, most children remained at least loosely connected to their caregivers, shouting to them from time to time and sometimes leaving the structure to check in with a caregiver sitting on a bench. Children also looked out for each other. Friends helped friends find the best way to climb, and older children sometimes helped and looked out for younger children. Adults. As their children engaged at the Gyroid and Stack of Stars, we found that many caregivers’ behaviors changed from ways they had been observed previously at other exhibits (see Appendix H, Table H-2, for a list of caregiver behaviors). As expected, adults rarely climbed on the Gyroid or Stack of Stars, even after signs were added encouraging them to do so. But even when adults had been consistently Up Close in their engagements previously, and even when they had engaged in active meaning-making in particular, many caregivers appeared to disengage intellectually from these exhibits as they sat on the benches, and their social and emotional engagements tended to be limited to occasional exchanges with their children. Other caregivers wandered off to look at and engage with other exhibits. Most caregivers stayed intermittently engaged with their children even as they also attended to other needs, such as nursing a baby, using their cell phones, and deciding which parts of the museum to visit next. We also noted that standing caregivers seemed to be more actively engaged with their children and the exhibits than sitting caregivers. Some caregivers stood close to the structures because they felt they needed to be there in case something happened to their child, while others wanted a closer look at what was going on with the exhibits. It’s also important to note that we observed some caregivers reading labels; however, this was a relatively uncommon engagement, and most read only one or two of the many labels available. Few caregivers used the term Gyroid when talking with their children and we noted that many children did not know what to call the Gyroid. We also noted that only a few caregivers called their children’s attention to the star shapes comprising the Stack of Stars structure, and how they fit together. When respondents mentioned the shapes in Stack of Stars most described it as being composed of triangles. When adults did read the labels, the data indicated that some of them used the information on the labels to engage in meaningful conversations with their children. Most parents talked with their children infrequently, and when they did it was often about what the children had been doing on the climbing structures and what they might do next, rather than helping them notice and pay attention to the shapes of the structure. Selinda Research Associates, Inc. 37 Looking at Different Subgroups One of the goals of this evaluation study was to look at ways in which the Geometry Playground experience might differ based on visitors’ age and gender. Were there ways in which physical, social, emotional, and intellectual engagements, spatial reasoning, and learning outcomes differed among children and adults, girls and boys, women and men throughout the exhibition? If so, to what extent might these differences be considered equitable? The data gathered suggested that Geometry Playground provided valuable experiences for a wide range of visitors with few exceptions. The differences among subgroups, and the factors that that seemed to be responsible for these differences, are discussed in this section of the report. Factors Related to Age There were many indications that Geometry Playground worked well for visitors across a broad spectrum of ages, from toddlers to seniors. Visitors of all ages engaged with the exhibits, practicing spatial reasoning skills and experiencing geometry in a range of ways, depending on the physical, emotional, social, and intellectual capabilities and tendencies they brought with them to the experience. Some of these differences were related to visitors’ physical size and agility, level of physical activity, and the emotional and social tendencies that seem to accompany these physical attributes. Most children in the target age range immersed themselves at every opportunity, climbing on, over, and through the Gyroid and Stack of Stars and medium immersives. However, most adults experienced the immersive exhibits from the outside, looking in. The climbing structures in particular provided very different experiences for these two groups. Adults climbed on the Gyroid or Stack of Stars much less often than children in the target age range or somewhat older teens and young adults. When adults did climb these structures, those we observed most often reached the top and then sat or stood for awhile and then climbed back down—most did not climb around and through, exploring every section, as children in the target age ranges did. In other words, adults seemed to climb in purposeful rather than playful ways. This held true even when the small signs were added within the Gyroid encouraging caregivers to climb with their children. While most adults at the large climbing structures stood or sat back and watched, it was encouraging to see that a few did climb. When we asked one adult respondent if he had climbed in the Gyroid with his son after reading the label that said: “Climb with your kids!” he responded emphatically “No!” and offered the following explanation. I’m still thinking there are probably some limits….I’m learning as I get older that I have limits. Another adult respondent indicated that he didn’t climb on the Gyroid because of “adult dignity” although he indicated that he might have climbed it if it was an adult-only night. Selinda Research Associates, Inc. 38 Visitors’ experiences at the climbing structures were markedly different than their experiences at the contemplative exhibits, in particular the object-filled vitrines and cases of Geometry Garden and the photo essays of Geometry in the Wild. At Explo and SMM, Geometry Garden was centrally located and appeared to be visited by more respondents. Some groups spent 10 to 15 minutes peering closely at case after case, looking closely at objects, reading labels, and discussing what they saw. Geometry Garden appeared to attract both adults and children, although not surprisingly, adults and children at the upper end of the target age range tended to spend longer amounts of time, whereas it was not unusual for children at the younger end to observe a case or two and then run off to explore another exhibit.11 Visitor experiences at the other contemplative exhibit—the photo essays of Geometry in the Wild—were more divided along age lines. We found that few respondents spent time at this exhibit, and almost all who did stop to study photos and read the labels were adults. So, although the immersive exhibits clearly catered to visitors who were more physically active—especially children—the contemplative exhibits provided a quieter and less physically active alternative for both adults and certain children. It was fascinating to see such different activity levels adjacent to each other. Geometry Playground demonstrated that geometry in informal settings can encompass these vastly different experiences. The exhibition appeared to successfully engage visitors of all ages, backgrounds, and needs, by offering a wide range of kinds of experiences. Adults and children also differed in reading behavior, especially when it focused on the interpretive panels rather than just the instruction labels. As is typical in many museum settings, the adults tended to read more labels than did the children. At the climbing structures, this meant that some children discovered the underlying geometric concepts through climbing and physically exploring the structures, and some adults discovered them from reading a label. Some of these adults then took on more of an interpretive role within their social group as they shared what they read with others in their group. For example, at the Gyroid, some children figured out that it was constructed of identical chips during their immersive explorations, some adults figured out its construction (i.e., that it was constructed of identical chips) through close inspection while standing outside the structure, and some others read about it on a label. In contrast, at Stack of Stars few respondents figured out that it was constructed of stars or that it illustrated the concept of tessellation by simply immersing themselves in it or examining it from the outside. When respondents did notice the stars and/or the tessellations inherent to Stack of Stars, they tended to have either read it on the label, or compared the smaller table-top SRDs with the larger structure. 11 At Explo and SMM, Geometry Garden was centrally located and many respondents visited it. At DHDC, the cases were split apart and assembled on either side of a doorway to another exhibit (Fig. G-32). Geometry Garden received less attention from visitors to DHDC, although it’s not clear whether this was because of its location or because most adult respondents were parents with children younger than the target audience. Selinda Research Associates, Inc. 39 Another age-related factor influencing visitors’ experiences in Geometry Playground was visitors’ level of prior background knowledge. Many children younger than eight or nine years old, for example, said they were not familiar with the term “geometry.” Middle school and older children and adults had more experience with this term, including understanding what it meant in at least a school setting. It was clear that the name of the exhibition had more meaning to older children and adults, and the ideas in many of the exhibits connected readily to a whole range of concepts they had learned about in school and elsewhere. For instance, many visitors to Space-filling Blocks, Stack of Stars, and Tile Designer made strong connections with their earlier experiences with the geometric concept of tessellation. Tessellation seemed to be a particularly popular idea among those who had experienced it before, and we saw some respondents’ faces light up as they realized the connection. On the other hand, learning about tessellation for the first time was certainly a valuable experience for those unfamiliar with it, but who were able to discuss the idea with their caregivers or others in their social group. While many respondents missed the concept of tessellation, for those who were able to explore it, experiences tended to be one of two equally valuable, but qualitatively different, experiences based on their prior knowledge: (a) recognizing and extending existing ideas about tessellation for those who were already familiar with the concept, or (b) learning about tessellation for the first time. Finally, we talked with many adult respondents who had degrees and work experience in math, science, or engineering. There were indications that many of these respondents were seeing things in the exhibits and understanding them in ways that went far beyond anything explained in the labels. In some cases, their thinking about the exhibits approached that of the developers who designed and built them. And there were indications that even “experts” could learn something knew. One university professor, who teaches statistics and decision analysis and visited the exhibition with his brother (a high school math teacher), explained that he had heard of and seen origami previously but didn’t realize it was made from a single sheet of paper. He also explained how he had been told about a Klein bottle when he was a child, and had never really understood it, but after spending time in the Geometry Garden, he now did. Some of these “expert” visitors shared their knowledge with others in their group, especially if they were accompanied by adults or teens. A few experts kept up a running commentary about what they saw and did, based partly on their exhibit experiences and the labels, and partly on their previous experiences and understanding of these and related phenomena. Expert visitors with children in the target age range and lower often held back, coaching their children through their experiences rather than explaining the exhibits in depth. However, these visitors’ expertise shown through during depth interviews, when they often expressed insightful views about both the exhibits and what their children were getting out of their Geometry Playground experiences. Most of these expert visitors expressed their enjoyment of, and enthusiasm for, the exhibits; only a few expressed their disappointment with the simplicity of the information presented in the labels. Selinda Research Associates, Inc. 40 Although different-aged visitors had very different experiences—based on their physical size, level of physical activity, inclination towards reading, background knowledge, and other age-related factors—Geometry Playground provided appealing and effective experiences for visitors of all ages. Gender-related Differences An important focus of the development of the exhibition was to ensure that spatial reasoning experiences for girls and boys would be equitable. This evaluation study focused on all visitors’ experiences with geometry and spatial reasoning, regardless of gender. However, our preliminary findings on this issue indicated that both girls and boys were highly engaged in both the immersive and non-immersive exhibits. Data indicated that both girls and boys practiced spatial reasoning skills and engaged in equally valuable experiences. We did find some preliminary indications at one of the venues during a focused 110-minute mid-day period, that slightly more boys used the Gyroid and Stack of Stars than did girls during that same period of time. During this time, we made a total of 76 counts and found that there were more boys than girls on the structures 50% of the time; more girls than boys 34% of the time; and equal numbers 16% of the time. Although we don’t know whether the sample consisted of equal numbers of boys and girls to begin with, because these were school groups we assume that the gender ratios within the visiting groups were approximately equal. These preliminary data suggest that during the time we observed, either (a) more boys climbed on the structures than girls, or (b) when boys were on the structures, they tended to stay longer. Limited resources prevented us from repeating this count at other times and at other venues, and from doing similar counts at other exhibits. At Dividing Space we saw the most incidents of aggressive behaviors, mostly in the form of smashing of the pucks, and while there were many examples of this from both genders, preliminary indications were that this happened more frequently with boys. Equitability When we looked at the differences of experiences between subgroups, we found the biggest differences between adults and children. If there were gender differences, these seemed to be slight, and typical of what one would find in any environment. We should note that, although we label these observations by age and gender, we could just as easily have compared physically active and less active respondents, readers and non-readers, or more attentive and less attentive children. The most important finding is that members of all of these subgroups found ways to engage profitably with the exhibits. The exhibition provided a range of challenges, so that visitors with different abilities could find one that felt comfortable, but then could push themselves to another level. Physically active visitors of both genders and, to a lesser extent, all ages, were provided with a range of immersive exhibits that allowed them to engage in physical ways. Less active visitors could find appropriate physical challenges if they chose to stretch themselves, or a range of less physically demanding interactive experiences if they chose not to climb in, over, or through the immersive exhibits. Small children could explore freely in the medium immersives, Selinda Research Associates, Inc. 41 climb the Stack of Stars with a parent standing a few feet away, and perhaps explore the lower reaches of the Gyroid with a parent right behind them. Those with shorter attention spans found experiences that paid off quickly, like Hopscotch or Distorted Chair. But they sometimes found their attention engaged in ways that might not have happened had they not just “burned off some energy” on the Gyroid and Stack of Stars. Visitors who liked to dig deep could spend 5-10 minutes testing many possibilities at Space-filling Blocks or Geometry in Motion, making an intricate artwork at Drawing in Circles, or studying the objects and graphics in Geometry Garden, Polyhedra, or Geometry in the Wild. In summary, the exhibition appeared to provide equitably a meaningful and satisfying range of experiences. Children and adults, girls and boys, and women and men could find exhibits that felt comfortable to them, and then push further if they were so motivated. Older children who no longer frequented playgrounds could climb the Gyroid, adults could play with blocks, girls could construct elaborate linkages at Geometry in Motion, and boys could design artworks on the giant spirograph (Drawing in Circles). Equitability in an exhibition seems partly about providing opportunities, and partly about inspiring the visitors to fulfill those opportunities in ways that go beyond what they might do in everyday life. By enticing many visitors to go beyond their everyday experiences, Geometry Playground provided equitable experiences for most visitors. The Role of Labels and Other Graphics Not surprisingly, labels contributed to visitors’ experiences in important ways. In this section, we discuss some of the various ways the labels contributed, and the roles they played. As noted earlier in this report, the labels in Geometry Playground were six different types: (a) instructions, (b) warnings, (c) interpretive panels, (d) object labels, (e) quotations, and (f) titles. Note that some label panels fulfilled more than one label function. For instance, many interpretive panels included instructions, warning labels included small bits of interpretation, and object labels sometimes included interpretation and object names. Instructions Instruction labels were either short, imperative statements incorporated as small stand-alone labels (e.g., the “Look in the mirror while you play hopscotch” label atop the cylindrical mirror in Fig. B-2) or text incorporated into larger panels (as at Distorted Chair, Fig. B-4). The instruction labels at Build a Room included a small label applied directly to the light wand (Fig. B-14) and a larger label that incorporated graphics showing visitors what to do (Fig. B-15). Short imperative instruction labels were mounted inside the Gyroid (Fig. B-27). And, although most instructions were stated as an imperative, some also provided hints, as at the Projected Puzzle (Fig. B-10). As is typical in exhibitions, most visitors engaged with interactives and immersives before looking at labels of any sort. When they had problems figuring out what to do, they looked for instructions to help. Caregivers often found the instructions as they stood back and watched their children’s initial engagement with the exhibits and sometimes intervened if their children were having problems or using the exhibit in ways that differed from the instructions. We found that Selinda Research Associates, Inc. 42 engagements were enhanced when instructions included text that caregivers could either read aloud or easily paraphrase for their children. In a previous section (under Caregiver-Child Interactions) we described how many visitors’ experiences were limited when they needed just one thing in order to engage meaningfully. In other words, visitors needed something to jumpstart their successful use of the exhibit, and without it, their experience was limited. Some of the labels incorporated this information. Examples included the “Look in the mirror” label at Hopscotch (Fig. B-2), the suggestion to stretch out on the Distorted Chair (Fig. B-4), and the labels included on the light wand at Build a Room (Fig. B-14). As noted previously, sometimes visitors noticed the embedded just one things, and sometimes they didn’t. In a few other exhibits however, the just one thing was missing. For example, the Gear Cube’s crank had to be rotated 10 times for visitors to see it progress from one shape to the next. Many respondents either gave up after too few rotations, or glanced away during the critical moment, resulting in a less meaningful experience. The just one thing was also missing at Spin a Shape, Geometry in Motion, and others. The challenge labels mounted inside the Gyroid exhibit12 (see Fig G28-b) were a special type of instruction label. We were able to observe visitors use the Gyroid both with and without these labels installed, and we talked with a few visitors who said they had read them. These preliminary data indicated that most respondents did not use these labels, but further study would be necessary to reach redundancy regarding their effectiveness. Warnings Warning labels were mounted at the entrances to the platform for the Gyroid (Fig. B-26) and Stack of Stars (Fig. B-31). These labels stated the age limits for the climbing structures and encouraged adults to supervise children’s behavior. At the Gyroid: Ages 7+ PLAY IT SAFE Children under 7 aren’t ready for this climber. Adult supervision strongly recommended. No aggressive or unsafe behavior. At Stack of Stars: Ages 5+ PLAY IT SAFE Adult supervision recommended. Designed for children 5 and up. 12 Note that these were installed only at the DHDC venue. Selinda Research Associates, Inc. 43 While most caregivers tended to heed the warning labels, the data indicated that some caregivers with preschoolers ignored the suggestions. At SMM, we talked with some parents who completely left the 4th floor when they saw the sign indicating that their children were too young to climb these structures. This seemed like a loss, because we observed children as young as 18 months safely climbing on the lower reaches of these structures with close adult supervision. This was the case in all three venues, but even more so at DHDC when the Gyroid was one-third lower in height. As discussed previously in the section under Safety, this would require more study to come to a firm conclusion. Interpretive Panels Most exhibits included at least one large label panel, which we refer to here as interpretive panels. Some examples include the Gyroid’s wall- and bench-mounted labels (Figs. B-28 and B-29), Stack of Stars bench-mounted labels (Fig. B-32), Spin a Shape (Fig. B-16, back panel), and Drawing in Circles (the entire back panel in Fig. B-22). Some examples of exhibits that did not include interpretive panels were Sphere Cube, Hopscotch, Gear Cube, and the large Shape Cubes (when they were not accompanied by small cubes). Several exhibits were designed by Explo artists in residence, and these included a special type of interpretive label with information about the artist (Fig. B-47). The Projected Puzzle’s stanchion label (Fig. B-10) provided an example of how instructions and interpretation sometimes blended together. Figuring out how to view the puzzle drove home the idea that, with a projection of this sort, there is one best place from which to view it. The stanchion label was effective at SMM because the large orange spot and the stanchion label were set up at the base of the central stairway where most visitors first entered the 3rd floor of the exhibition. When the sign was located where most visitors saw it before engaging with the structure, many respondents looked at the stanchion label, then the puzzle, then stepped to the center of the dot and looked again, and finally shared their discovery with others in their group, just the way the exhibit was intended to be used. Because the instructions were communicated with illustrations as well as text, children as young as six years old sometimes figured out how to view the puzzle on their own. There was a second Projected Puzzle interpretive panel, however, that was located right beside the structure (Fig. B-9). We found that when visitors started using the exhibit by reading this sign, they tended to bend over and maybe take a few steps away, but no visitor walked all the way from the puzzle to the orange spot, i.e., the location at which the puzzle “works.” In contrast to SMM, at DHDC the orange spot and stanchion label were located in the center of the exhibition, beside an entrance to the large climbing structure platform. Data indicated that very few visitors paid attention to the stanchion label, orange spot, or puzzle structure from this position. This suggested that the layout of the exhibition can make a big difference to the effectiveness of exhibit interpretation. With this set up, few respondents read either of the labels for this exhibit, and we saw no visitors who read the second label (the one next to the puzzle) and then walked over to the orange dot, per the label instructions. Selinda Research Associates, Inc. 44 The interpretive panels at several exhibits included examples of “real-world geometry” (Fig. B-28) and these labels were the primary way visitors discovered ways in which geometry could be seen and applied in everyday life. For example, the illustrations of the ballerina at Spin a Shape and the steam locomotive at Geometry in Motion were understood by some children in the target age range plus many adults. The data indicated that children sometimes noticed and talked about them on their own, and sometimes caregivers pointed them out to their children. At Dividing Space, the description of Voronoi diagrams used both text and graphics to explain this concept. Some adults—especially caregivers who stood back and watched their children— noticed and sometimes read this label for 20 seconds or more. Some also pointed it out to and discussed it with other adults in their groups. However, few caregivers tried to explain the real-world applications of the Voronoi concept to the children in their group. A few of the exhibits, such as the triangle Shape Cube and Build a Room, included small real-world geometry labels mounted on exhibit walls or sides, apart from other labels, although the data indicated that few respondents read these. Although Geometry Playground’s goals focused on experiencing geometry and practicing spatial reasoning skills, some interpretive panels also included explanations of what might be termed the “physics” behind the phenomena, while still emphasizing the geometrical aspects of the phenomena in question. At Distorted Chair (Fig. B-4), physics and geometry were inextricably linked, although a few respondents interpreted this exhibit as being about light (an idea that was reinforced for them when they tried to take flash photos of the reflection in the mirror). The panel at Spin a Shape named and defined the concept of rotational solids in geometric terms, but did not explain the perceptual aspects of this phenomenon. Many caregivers who tried to explain to their children why the spinning shape changed as it did struggled to find the right words; a few were successful, but most were not. The panel at Geometry in Motion included both geometry and physics in its explanation of mechanical linkages and—with its real-world example of a locomotive—seemed relatively easy for many children at the upper end of the target audience to understand on their own and for caregivers to explain to younger children. It appeared that the simple explanation of physics and perception along with geometry, accompanied by the incorporation of real-world examples, was an effective approach. At Explo, the Gyroid and Stack of Stars interpretative labels were displayed in two different ways: (a) panels mounted on the wall that enclosed some sections of the two exhibits (Fig. B-28), or (b) panels mounted on the benches where some caregivers sat as they watched their children (Fig. B-29). The wall-mounted labels were most often read by adults who were enjoying some “own time” as their children climbed. A few children, especially those at the upper end of the target audience and older, also stopped and read these labels. The bench-mounted labels were most often read by adult caregivers as they sat and watched their children, although some visitors stood outside the enclosures and read them. Most adult visitors who read them appeared to find some useful information for themselves, but they weren’t able to use them in meaningful ways with their children. There were indications that the labels could have supported adult-child interactions by (a) helping caregivers relate the information on the label to the climbing structure Selinda Research Associates, Inc. 45 in a concrete way so they could talk about it with their children, or (b) suggesting activities that parents could encourage their children to try while climbing in the Gyroid, like finding the shortest way to the top or figuring out how many “rooms” were inside. Object Labels Most of the artifacts and specimens displayed in the Geometry Garden included object labels that listed the object name, some interpretive information, and, when appropriate, the name of the artist (e.g., Figs. B-51 and B-52). These labels were read, at least in part, by most of the visitors who stopped to examine these objects, sometimes stimulating discussion among adults or conversations between caregivers and children. There were indications that both the objects and their labels played important roles in helping some visitors appreciate both the real-world applications and aesthetic aspects of geometry. Quotations and Stand-alone Texts Quotation labels and stand-alone texts were mostly displayed on the Entrance Structure (Fig. G-1 and Fig. G-19) or on the back wall of the exhibition space (Fig. G-31). They included some important thoughts about geometry and the role it plays in people’s lives. For instance, one quotation read as follows: Geometry is a skill of the eyes and the hands as well as the mind. Jean Pedersen, American mathematician, 1981 The term “spatial reasoning” was found as stand-alone text only on the Entrance Structure.13 This label read as follows: You’ve been doing geometry all your life. With every image you see, every move you make, and every object you hold. This geometry is more than what you study in school—it’s how you make sense of space. Here you’ll play with shapes and spaces. And while you’re using your body, you’ll also be using your mind. This natural mental geometry—called spatial reasoning— is what people use to knit a sweater, copy a dance move, or navigate across town. These labels—along with Geometry in the Wild (Figs. B-53, B-54, B-55)—included important ideas about geometry and spatial reasoning, but data indicated that few visitors attended to them. Titles The title of an exhibition is an important opportunity to succinctly inform visitors of the main focus of the exhibition. There were indications that when visitors were aware of it, the Geometry Playground title communicated many important ideas to visitors. It said that the exhibition was about geometry, that it was going to be fun, that you’d be using your whole body. 13 The Entrance Structure was installed as an entrance to the exhibition only at Explo and SMM. Selinda Research Associates, Inc. 46 For many of the visitors who were aware of, and thought about, the name, it was more than just a title; it was an idea worth contemplating. The title paired two things that most visitors didn’t naturally put together, ideas that made perfect sense once they were familiar with the exhibits, especially the large and medium immersives. The phrase “geometry playground” was both a call to action and a take-home message: geometry is fun, something we can play with, and geometry goes beyond what we learned about in school. However, data indicated that most visitors were unaware of the title of the exhibition. In analyzing why this might be so, we found that the exhibition’s title was not displayed in very many places, regardless of the venue. Although prominent on the Entrance Structure (Fig. G-1 and Fig. G-19), because of the porous nature of the exhibition, most visitors missed this structure and thus the exhibition’s title. In addition, at one venue the Entrance Structure was installed as separate panels at the back and sides of the exhibition (Fig. G-29 and Fig. G-31). At one venue, the title of the exhibition was included in a few additional places (Fig. G-20 and Fig. G-21) and frequently featured on the museum’s website homepage. At this venue more respondents were aware of the title. We found similar challenges with the titles of the two large climbing structures in particular. In both of these exhibits, the warning labels were also title labels, but while most paid attention to the safety messages, only a few visitors noticed the title portion. When visitors didn’t notice the names of these exhibits, they often had a difficult time talking about their experiences. A few visitors read the names on the bench- and wall-mounted signs, and when they did, some of these visitors could name the structures. But for the most part, few visitors knew what to call either exhibit. On Stack of Stars in particular, many visitors missed the idea that this exhibit was made up of stars. In fact, during interviews when respondents talked about shapes, they tended to describe the 2-D triangular faces of the stars rather than the 3-D star shapes. What Happened in the Absence of Labels As described above, there were many indications that labels played an important part in most visitors’ experiences. It was unfortunate when labels were minimally included, or not included at all on some exhibit units, for example: Sphere Cube; some of the Shape Cubes; the Gyroid; and Stack of Stars. Missing labels represented missed opportunities for engagement and learning. For example, caregivers were often stumped about what to say to their children in the Sphere Cube as they stood and watched them play. A few caregivers tried to describe the construction of this exhibit, pointing out the many circles that intersected at its walls, but they failed to come up with words that their children could understand. At the Shape Cubes, most respondents did not discover that it was more comfortable to slide through on their backs than crawl through on their knees. We also noticed that many young visitors focused more on the twisty and looping tunnel interiors and missed the simple shapes that had been cut out to make the tunnels. This happened even when the small interpretive cubes were included as part of the Triangle and Square Shape Cubes, perhaps because these were placed off to the side, where visitors were less apt to see them. Small Selinda Research Associates, Inc. 47 labels might be added right beside the tunnel entrances, for instance, printing the shape name at the tunnel mouth, and including a small photo of an adult visitor sliding through on their back. The lack of bench-mounted labels at the Gyroid and Stack of Stars climbing structures at some venues was particularly problematic, since these labels provided important opportunities for caregivers to read labels as their children climbed, and then pass on activity ideas or bits of information to their children. At the venues with bench labels installed, even when they were mounted facing away from the climbing structure (as at SMM, Fig. B-37), some caregivers used the information they were able to glean to help explain something to their children, direct their children’s attention, or suggest an activity. But when the labels were absent, fewer meaning-making conversations took place. Effectiveness of Paired Exhibits As noted previously, a few exhibits were designed as conceptual pairings, so that visitors would be able to engage with a geometric concept in different ways. The conceptual pairs in the final exhibition were as follows, with the immersive exhibits listed first: - Stack of Stars :: Space-filling Blocks (both SRDs and TOs) - Hopscotch, Distorted Chair :: Distorted Drawing In addition to the exhibits designed to be close conceptual pairs, the data indicated that some visitors made their own conceptual linkages, especially between the numerous tessellation exhibits. Tessellation was a familiar concept for some visitors, and they noticed that several exhibits dealt with this topic, including especially the Shape-filling Blocks and Tile Designer exhibits. A few visitors also talked about exhibits that reminded them of the repeated patterns they had seen in works by Escher, which included both the tessellation exhibits and, for at least one visitor, the Gyroid. For these visitors, the Stack of Stars :: Space-filling Blocks conceptual pair was expanded to also include the Tile Designer, and the Gyroid. An additional conceptual pairing was made by a few visitors who recognized the process of stellation in the Stellator and in the two Space-filling Blocks exhibits. For these visitors, the Stack of Stars :: Space-filling Blocks conceptual pair was expanded to include the Stellator. The data indicated that visitors could be successful and have rich and meaningful experiences at either or both of the intended conceptually paired exhibits without being aware of their conceptual link, but that when they did become aware of the relationship, it was a particularly powerful experience. For most visitors, to make the connection the exhibits had to be in close proximity to one another (so that visitors could stumble across the connection on their own), or the connection needed to be pointed out by a label or a person. At one museum in particular, floor staff sometimes engaged visitors with an activity at Hopscotch and then sent them over to do the Distorted Chair exhibit, thereby helping them make the connection. In the absence of a label or knowledgeable and motivated staff person, however, most visitors did not notice the intended conceptual pairs. This was not surprising considering that many times, the conceptual pairs were not located near one another. The one exception to this was the pairing of the Stack of Stars immersive exhibit, and the SRDs portion of the Space-filling Blocks Selinda Research Associates, Inc. 48 table top. This pairing was supported by the layout of the exhibition at all three host sites and by exhibit labels, and was recognized by many visitors and used to enhance their experiences with both exhibits. We also found that most of the visitors who indicated that they thought of Stack of Stars in terms of its construction (a spatial reasoning skill) and tessellation (an important geometric concept) did so primarily because they had engaged with the small SRDs. While not designed as conceptual pairs, another potentially powerful pairing opportunity arose between the Shape Cubes and the Shape Stools, as the Shape Stools were concrete evidence of the negative space within the Shape Cubes. The data indicated however, that the stools were rarely located close to their respective Shape Cubes. Two of the large Shape Cubes (the square and the triangle) had small table-top versions to facilitate deeper meaning-making experiences at the Cubes (see Fig. B-20). When the small versions were mounted adjacent to one another, some visitors noticed them and used them to enhance their understanding. But again, their placement was not consistent, and at other venues, they were either absent or separated. Finally, a few respondents independently suggested that the exhibition could benefit from an additional conceptual pair. These respondents pointed out how valuable it would have been to have both a large and small version of the Gyroid, perhaps with a rope that could be threaded through the smaller version to demonstrate the idea that a Gyroid is made of two “rooms” that never intersect. Conceptual pairings, when they were recognized, contributed to both spatial reasoning about the exhibits and increased understanding and thoughtfulness about geometric concepts. However, because this recognition depended mostly on what visitors brought to the engagements, rather than on the design and layouts of the exhibits, this did not happen very often. Given the success of the limited conceptual pairings that did take place, the idea of pairing exhibits seems a sound one that we hope will be pursued in future venues. Installing the intended conceptual pairs—as well as the other loosely conceptually related exhibits—in close proximity to one another, training floor staff to help visitors recognize and think more deeply about the various conceptual pairings, and using the labels to point them out could help create rich opportunities for visitors to evolve their geometric understandings in powerful ways. Focus of Exhibition Most respondents described the exhibition as being about geometry, or as some put it, about shapes. When respondents described a more nuanced theme, it was often that the exhibition made geometry more fun or easier to understand than the subject they remembered in school. For instance, a 9th grader said that the exhibition “kind of explained everything about it and made it easy to understand. It made geometry easy, if that makes sense.” One mother said she brought her primary-aged children to Geometry Playground because they were struggling with school math, as she did as a child. She said she wanted to help them see that math and geometry could be fun. Some respondents talked about what the exhibition’s labels termed real-world geometry, although they usually expressed this in their own terms. For instance, many respondents Selinda Research Associates, Inc. 49 recognized Drawing in Circles as a giant spirograph and the Geometron as something like a kaleidoscope.14 They often used these terms when talking about these exhibits with others in their groups. Some respondents talked about the unexpected geometric shapes and patterns they saw in commonplace objects, like the shells and other natural objects in the Geometry Garden. And some respondents saw graphics or read labels that helped them connect the exhibits to things they had heard about before visiting the exhibition. For instance, some respondents who created structures at Geometry in Motion saw the photo of the steam locomotive and related their constructions to train wheels, and a few respondents who read the real-world geometry label at Dividing Space talked about Voronoi diagrams with others in their group. A few respondents did not make the connection between the Geometry Playground exhibits and geometry. These visitors tended to be those who had spent a very limited amount of time in the exhibition, or thought the exhibition was about a related subject such as physics or light, especially after having spent time with the Distorted Chair, or Hopscotch. Also, children who had not yet studied geometry in a formal way did not appear to recognize the links to school math unless their parents pointed it out. In accordance with the project goals, many visitors also talked about an aesthetic appreciation for geometry, especially when they had spent time at the Geometry Garden, Geometron, Shape Cubes, Shape Stools, and Tile Designer. A few specific pieces from the Geometry Garden were mentioned in particular: the Klein bottle, origami, natural objects, and the structure built out of pencils. Geometric Concepts & Shapes While learning about geometric concepts and 2-D and 3-D shapes was not an explicit goal of the Geometry Playground exhibition, data indicated that some of this was in fact taking place. Examples of exhibits where this happened include: the Gyroid, Polyhedra, Stack of Stars, Space-filling Blocks, Stellator, Shape Cube, and Shape Stools. Data indicated that some respondents learned for the first time, and/or expanded their understanding of, one or more of the following words and concepts: tessellation, stellation, truncation, gyroid, octahedron, polyhedra (and all the related shapes inside this booth). However, there were also indications that the exhibit did not help visitors learn or expand their understanding of some of the more basic concepts, such as: square, sphere, ellipse, triangle, trapezoid, and cone, in spite of the many inherent learning opportunities (for example, seeing what it felt like to be inside a triangle). Some visitors younger than the target audience were able to engage meaningfully with the Geometry Playground exhibits, for example, by climbing Projected Puzzle, looking at the objects in the Geometry Garden, drawing at Distorted Drawing and Drawing in Circles, climbing the lower rungs of both the Gyroid and Stack of Stars, and wiggling through Shape Cubes. However, for most of these young respondents, the experience was exclusively “playground” rather than “geometry” when, with some small and unobtrusive additions, it could 14 These terms were not included in the exhibit labels. Selinda Research Associates, Inc. 50 have been both for many more. The data indicated that there were many missed opportunities to help them and older visitors with less geometry experience, start thinking geometrically, even if this meant simply noticing and naming various shapes throughout the exhibition. The addition of a few key elements (for example, names of simple shapes) could have a powerful impact on the meaningfulness of the experience for these mostly non-target, but otherwise frequent, visitors. Selinda Research Associates, Inc. 51 FINDINGS: SPATIAL REASONING One of the intended outcomes for visitors in Geometry Playground was that they would gain experience with two- and three-dimensional space, and practice spatial reasoning skills. This section of the summative evaluation documents and describes the many different goal-related types of spatial reasoning that took place in Geometry Playground, and some ways that the exhibition had the potential to—and in some cases did—go beyond its stated goals. Spatial Reasoning in Everyday Life and in the Exhibition It can be argued that spatial reasoning is so basic to our lives that we practice it almost everywhere we go. Any science museum is filled with exhibits that engage visitors in spatial reasoning skills of one sort or another. The following label on the Entrance Structure to the Geometry Playground exhibition explains it well. You’ve been doing geometry all your life. With every image you see, every move you make, and every object you hold. This geometry is more than what you study in school—it’s how you make sense of space. Here you’ll play with shapes and spaces. And while you’re using your body, you’ll also be using your mind. This natural mental geometry—called spatial reasoning—is what people use to knit a sweater, copy a dance move, or navigate across town. The question then is how was the spatial reasoning in Geometry Playground any different than just maneuvering around the museum? The data indicated that the Geometry Playground exhibition provided opportunities for visitors to engage in spatial reasoning: (a) in new ways, (b) for longer periods of time, and (c) to do so more deliberately than they typically would when just navigating life. Below is a brief description of some of the ways visitors engaged with geometry and spatial reasoning while in the exhibition. Practicing skills. Geometry Playground got many visitors practicing skills they may not use much anymore. For instance, the Gyroid got some teens—and even a few adults—climbing around like seven-year-olds, and Shape-filling Blocks got visitors of all ages playing with blocks. Unfamiliar uses. Geometry Playground got many visitors using familiar things, but in slightly different ways than they were used to using them. For example, the blocks they played with were odd shapes, the climbing structures were totally unique (the biggest Gyroid ever!), and when they drew with a “spray can of paint,” they did it in empty space, and it showed up on TV. In other words, many visitors saw the familiar in a new way. Prolonged engagement. Most visitors had fun at the exhibits, so they kept playing and kept trying different things. As they engaged for two, five, even 10 minutes or more, some got better at the basic physical spatial reasoning skills and then began to engage in more sophisticated and intellectually engaged ways. Selinda Research Associates, Inc. 52 Geometry is everywhere. When visitors encountered the title of the exhibition, they tended to attach the term “geometry” to what they did in the exhibition. Sometimes they did and saw things they didn’t necessarily associate with geometry, and so began to expand their definition of geometry, and to recognize it in other aspects of their daily lives. Geometry can fun. Some visitors who had not enjoyed geometry in school enjoyed themselves in this exhibition, developing positive feelings for a subject they had previously associated with negative feelings. Definition of Spatial Reasoning The definition of spatial reasoning in the Geometry Playground project included the two following descriptions. [Spatial reasoning skills are] those mental skills concerned with understanding, manipulating, reorganizing, or interpreting relationships visually (Tartre, 1990, p. 216, in Dancu, Gutwill, & Sindorf, 2009, p. 4) In practice, this means visualizing objects from multiple perspectives; rotating, flipping or inverting mental objects; recognizing spatial relationships among objects; perceiving spaces and their properties (Dancu, Gutwill, & Sindorf, 2009, p. 4). These definitions were further operationalized into eleven broad categories of skills (see Table 3) (Dancu & Gutwill, 2010; Gutwill, 2010; Gyllenhaal, 2010a; Gyllenhaal, 2011; Gyllenhaal & Perry, 2010). Table 3. Intended spatial reasoning skills in Geometry Playground. Categories Skills Shape Explorations of shapes (with your hands or body, also exploration of the geometry of an object, such as the edges of a box or table) Size Explorations of size Curvature Explorations of curvature Orientation Explorations of orientation (the direction an object or body is facing relative to the background or to another object) Location Explorations of locations of objects (including body) relative to the background or to another object) Congruence Explorations of congruence (comparing objects to other objects, or to other object pre- and post-transformation) Rotation Rotation (aligning shapes or spaces by physically or mentally rotating them) Movement Movement (moving in space, judging relative velocity/arrival time, distance/intercept) Selinda Research Associates, Inc. 53 Construction Construction (creating, building or developing shapes and forms). Drawing (perhaps as a type of Construction) Transformation Transformation (changing from 2-D to 3-D or from morphed to anamorphed) Prediction Spatial inference (e.g., predicting 3-D from 2-D) Using a grounded theory approach where findings emerge from the data, this classification scheme evolved into an expanded Table 4. This revised framework includes all of the original spatial reasoning skills listed in Table 3, and expands them to incorporate the full range of spatial reasoning that emerged from the data, especially in the larger, immersive spaces, as well as many of the mental skills that visitors practiced. This expanded framework groups spatial reasoning into those skills visitors practiced primarily physically, with their senses, hands, and bodies, and those they practiced primarily mentally, with their minds. Note that both of these broad categories (physically and mentally) involve some aspects of the other; in other words, when visitors were primarily using their senses, hands, and bodies, they were also mentally engaged, and vice versa. Table 4. Inventory of spatial reasoning skills practiced by visitors to Geometry Playground Ways visitors engaged primarily with their senses, hands, and body: Physical exploration of objects: shapes locations of objects size orientation of objects curvature congruence transformation Rotation of physical objects Ways visitors engaged primarily with their minds: Recognition of spatial relationships among objects Perception of spaces and their properties Spatial visualization and the creation of mental images or objects The manipulation of mental images or objects: rotated flipped or inverted transformed Selinda Research Associates, Inc. 54 Dynamic spatial reasoning: judging relative velocity/arrival time judging distance/intercept Recognition/classification of shapes based on mental objects Spatial orientation and navigation, including wayfinding Developing and applying rules to spatial problems Planning, constructing, and deconstructing objects and structures Spatial inference to solve spatial problems While the above table presents a quick snapshot of the range and types of spatial reasoning we observed in the Geometry Playground exhibition, Appendix I, Taxonomy of Spatial Reasoning Skills, contains additional descriptions and specific examples of exhibits. General Spatial Reasoning Knowledge Hierarchy In addition to examining specific evidence of spatial reasoning being practiced at various exhibit units, we found it useful to employ a knowledge hierarchy technique to fully understand visitors’ spatial reasoning in the exhibition. A knowledge hierarchy is a tool for understanding the range of ways visitors understand and experience an exhibit within the context of the exhibit developer’s intentions. Rather than assessing exclusively whether or not visitors achieved a pre-determined learning outcome, or engaged in a pre-defined way, a knowledge hierarchy describes visitors’ learning or engagement based on their experiences within the context of the intended outcomes. It should be noted that the knowledge hierarchy is not pre-determined, but emerges from the evaluation data. The knowledge hierarchy…is based on the assumption that inherent in each exhibit is an internal knowledge structure. This knowledge structure is at the intersection of the exhibit developer’s, and the visitor’s, organization and understanding of the topic. A knowledge hierarchy is simply a description of this range of understandings. It is not the range of knowledge visitors have about a topic, but the range of knowledge within the context of the exhibit. (Perry, 1993) The knowledge hierarchy technique has been used in many evaluations and some research studies over the years and has proven to be a useful framework for describing the range of visitor experiences. While usually used to describe visitor learning outcomes, it proved useful in this study to describe the range of ways visitors practiced spatial reasoning in Geometry Playground. The data indicated a knowledge hierarchy as described below in Table 5. This hierarchy has six levels, from Level 0, representing those visitors who had no engagement with spatial reasoning at an exhibit, through Level 5, which represents engagement on the most abstract level, based on mental objects of the exhibit rather than the physical exhibit itself. Selinda Research Associates, Inc. 55 Table 5. Hierarchy of spatial reasoning skills practiced by visitors to Geometry Playground Level 0 No engagement. These visitors were not engaged with the exhibition at all, for example they were reading a book, checking email, standing and talking about vacation plans, etc. Level 1 Mindless engagement. These visitors were minimally engaged mentally, although they were often physically active. They moved unconsciously and automatically. There was no intellectual engagement about what they were doing, in part because they were not trying to accomplish anything in particular. An example of this was the visitor at the SRDs who was absent-mindedly moving the shapes around while gazing elsewhere in the exhibition. Level 2 Trial and error. These visitors were aware and thoughtful about what they were doing, in part because they were trying to accomplish something. However, they did not have any particular ideas about how to accomplish their goal, so their engagement was characterized by trial and error, not based on a rule or plan. These visitors did not predict the results of their actions. Although they were sometimes successful, success was accidental. An example of this was the visitor who realized that they were supposed to fit the SRD shapes into the holes in the base, but who just spun and twisted them randomly until they fell into place. Level 3 Deliberateness. These visitors were deliberate and thoughtful, and were beginning to develop a hypothesis or plan to guide their efforts. These visitors might not have been successful all the time, but they developed a plan and predicted (at least implicitly) that their approach would be successful, although their plan was not always successful. They were also deliberate in how they executed their plan. And example at this level was the visitor at the TOs who said, “I’m going to make sure that the flat surface is always on the bottom.” Level 4 Mental objects and complex rules. These visitors had well-developed mental objects and had complex rules about how to apply those mental objects to achieve their goal. These visitors successfully predicted the results of their actions much of the time, but still needed to test their hypotheses with physical objects. Level 5 Absence of physical objects. These visitors operated successfully with mental objects and complex rules without the need to use physical objects. Examples included visitors who told us how they disassembled the Gyroid into its constituent “chips” in their heads. Selinda Research Associates, Inc. 56 This knowledge hierarchy (Table 5) describes a wide range of levels of engaging in spatial reasoning that emerged from the data. We observed and talked with visitors at all levels of the hierarchy at all the exhibits. Many visitors practiced spatial reasoning skills that were characterized by a lack of deliberateness. In other words, these engagements tended to be primarily mindless engagement or trial and error (Levels 1 and 2). Some visitors used more deliberate and thoughtful strategies (Level 3), and a few used mental objects and complex rules (Level 4), with a few of those operating in the absence of physical objects (Level 5). We also saw evidence, however, of some children in the target age range moving up a level (or sometimes two) when they engaged with an exhibit for several minutes. For example, starting with mindless engagement sometimes led to discoveries that sparked a desire to accomplish something, resulting in the visitor engaging in trial and error to try to achieve their goal. Similarly, some visitors who used primarily trial and error noticed that the results often led to the discovery of patterns and eventual rules, which led to more deliberate and thoughtful practicing. And for a few visitors, continued practice with deliberate, thoughtful and rule-based spatial reasoning fostered the development of simple mental models and moved them further along the hierarchy. When children in the target age range were able to reach Levels 4 or 5 however, it was usually because an adult played a role. The adults—usually parents or grandparents—gave instructions, directed attention to important parts of the exhibit, pointed out or read label texts, or talked with the children about their explorations. When these teams of parents and children worked together, the exhibits achieved their full potential as spatial reasoning experiences. With this knowledge hierarchy, eight variables worked together to define the various levels: the use of physical objects, the presence of a goal, being thoughtful, being deliberate, making predictions, developing and using rules, using mental objects, and the likelihood of success. These are charted in Table 6 on the next page. To give a clearer picture of how spatial reasoning was taking place at the exhibits in Geometry Playground, we have included detailed hierarchies for four specific exhibits: Space-filling Blocks, Stack of Stars, the Gyroid, and Dividing Space. Each of these hierarchies gives clear examples of how spatial reasoning was evidenced in slightly different ways. And in these hierarchies, we point out the various types of spatial reasoning skills from Table 4 that were in evidence. All of these hierarchy examples begin at Level 1, as Level 0 (no engagement) was the same at all of them, i.e., an absence of engagement. Selinda Research Associates, Inc. 57 Table 6. Knowledge hierarchy defining variables characterized by: manipulate objects have a goal be thoughtful be deliberate make predictions use rules use mental objects high likelihood of success Level 0 lack of engagement Level 1 mindless engagement X Level 2 trial & error X X X Level 3 thoughtful & rule-based X X X X X X X Level 4 mental objects & complex rules X X X X X X X X Level 5 absence of physical objects X X X X X X X Space-filling Blocks Knowledge Hierarchy This exhibit provided a clear example of how the physical exploration of objects, combined with thinking about what they were doing, led to increasingly sophisticated practice of spatial reasoning skills, even among children at the lower end of the target age range. Table 7. Hierarchy of spatial reasoning skills practiced by visitors at Space-filling Blocks characterized by: Examples of each level for Space-filling Blocks: Level 1 mindless engagement Some visitors (often but not always children in pre-school and lower elementary grades) dropped the blocks onto the platform without trying to get them to fit, or tossed the blocks up in the air and let them fall on the platform. They did not yet realize that the blocks were supposed to fit perfectly in the holes. Some aspects of physical exploration began to happen on this level, but it focused on individual blocks rather than how they fit together. If and when the focus shifted to fitting the blocks together, visitors moved to the next level of the hierarchy. Level 2 trial & error These respondents knew they wanted to fit the block into a depression in the surface, but to achieve that fit they placed the block in the hole and just rotated it repeatedly and randomly until it fit. Level 3 thoughtful & rule-based These respondents fit the blocks by aligning them as they moved them into place based on a rule or rules about the direction that a part of the Selinda Research Associates, Inc. 58 block should be oriented. They would pick up a block, look at it, reorient it, and then move it into place. For instance, one six-year-old girl told us that a square always had to be on the bottom of the Truncated Octahedron (TO) block. Other respondents had rules about how to orient the tops of the stars (SRDs). Level 4 mental objects & complex rules These respondents told us that they had a picture in their heads about how a block should be oriented even before they picked it up. That picture or mental object included both the block and the hole it was going to fill—and in at least one case, the respondent told us that he visualized his activity as filling space, which he referred to as three-dimensional tessellation. As visitors thought about tessellation, they sometimes moved to Level 5 of this hierarchy. Level 5 absence of physical objects We observed three ways in which visitors functioned on this level. (1) Some parents appeared to approach this level as they gave their children advice about placing the blocks without actually touching the blocks themselves. (2) Some adults and older teens developed plans for structures they wanted to build in advance. (3) Some visitors thought about 3-dimensional tessellation in abstract ways, recalling, for instance, other places where they encountered this phenomenon. Perhaps the most interesting feature of this hierarchy was how experiences with physical manipulation of objects led to both increased skill in placing blocks and, in some visitors, development of mental objects. The desire to build led to an increase in both physical and mental skills, at least within the narrow and short-term context of this exhibit. It was also noteworthy that—through physical manipulation of the blocks—even visitors at the younger end of the target age range could move up the hierarchy to Level 3 and, sometimes, approach Level 4. Stack of Stars Knowledge Hierarchy This exhibit provided an example of how labels and comparisons among exhibits helped move visitors’ experiences from a primarily physical exploration to one more intellectually engaging. Table 8. Hierarchy of spatial reasoning skills practiced by visitors at Stack of Stars characterized by: Examples of each level for Stack of Stars: Level 1 mindless engagement Some visitors who had lots of experience on playground structures approached Stack of Stars and quickly climbed right to the top without much conscious intellectual engagement. There was little challenge here for them, so they often labeled Stack of Stars “for little kids” and moved on—perhaps to the Gyroid, which they felt was challenging enough to be “for big kids.” Level 2 trial & error Some respondents found that climbing Stack of Stars took some thought. They had to discover the hand/foot-holds and figure out how to use them. They had to try out small challenges presented by the lower reaches of the structure and find out if they could master them. Physical explorations of size, shape, orientation, and congruence were important aspects of spatial reasoning on this level. Once they reached the top, Selinda Research Associates, Inc. 59 they had to figure out how to get back down, again using trial and error as they dangled feet to try to catch hold, then pull themselves back up if that did not work. Level 3 thoughtful & rule-based Once they were familiar with the structure, some respondents engaged with a number of thoughtful forms of navigation—a more mindful sort of spatial reasoning. For instance, they would purposefully try out different routes, or plan their next step in advance, or play with other children in ways that used their knowledge of the structure (i.e., wayfinding rather than random exploration). Or they would help other children, explaining to them how to climb or helping them find routes to the top that fit their physical limitations. Level 4 mental objects & complex rules Reaching this level for this exhibit involved recognizing that the climbing structure was built of star-shaped solids, and that they fit together in a certain way—they were tessellated in three dimensions. In our observations, this most often happened when a parent read a label to a child or when a visitor at the small SRDs recognized that the climbing structure was built of giant versions of the blocks they were holding in their hands. Level 5 absence of physical objects Reaching this level involved thinking about 3-D tessellation in the abstract, including thinking of other examples they had known. We encountered a few examples of this among parents who stood back and read the exhibit labels, then pondered the label concepts as they watched their children play. This exhibit was able to engage a wide range of visitors, even those who did not climb it. It’s interesting that, for climbers, it took the intervention of a label or a comparison with another exhibit to transform physical exploration into a Level 4 experience. It was also interesting that some label readers (and diagram viewers) could reach Level 5 without starting with the physical experience of climbing on Stack of Stars or playing with the star-shaped blocks. Gyroid Knowledge Hierarchy This exhibit provided examples of several interesting phenomena related to the practice of spatial reasoning, extending some of the ideas encountered on the Stack of Stars hierarchy. Table 9. Hierarchy of spatial reasoning skills practiced by visitors at the Gyroid characterized by: Examples of each level for the Gyroid: Level 1 mindless engagement Compared to Stack of Stars, the Gyroid presented a larger, less familiar, and more slippery challenge for climbers. It was harder to practice mindless engagement at the Gyroid, but we did observe an eight-year-old girl who quickly made it to the top to have her picture taken. When it came time to get down, she seemed somewhat surprised at where she was and hesitant about how to proceed. Level 2 trial & error Thinking about where and how to start your climb got most visitors to Level 2, including parents who helped their younger children get started. Climbers had to discover that the surface was slippery and hard Selinda Research Associates, Inc. 60 to grasp, decide exactly where to start, find the hand/foot-holds and try them out, and perhaps figure out if they could gain purchase at the edges of the “chips.” As at Stack of Stars, they had to figure out what they could do given the limits of their size and strength and courage— physical explorations of size, shape, orientation, and congruence were important aspects of spatial reasoning on this level, even for older children, teens, and adults. Level 3 thoughtful & rule-based Once they reached the top, climbers had to figure out how to get back down. Because an error could really hurt, most visitors became more thoughtful about how to get back down. Some reconceived the Gyroid as a curving slide, and others carefully studied the distance to the ground before they jumped, comparing it with whatever mental model they had of their own body and its limitations. At this level some respondents started to develop an interest in, and understanding of, the Gyroid’s structure. They noticed that the structure was built of pieces that seemed all the same size and shape, or they pondered the tube-like nature of its interior. The Gyroid was much too big to take in all at once, so these kinds of thoughts almost required development of a mental object of the Gyroid—which lifted them to the next level of the hierarchy. Level 4 mental objects & complex rules Once they were familiar with the structure, some children engaged in more thoughtful navigation—especially if they decided to race a friend or a parent’s stopwatch to the top. If they decided to climb the outside of the structure, they planned their route in advance or stage-by-stage. As they became more familiar with the structure, they built mental maps of it—which came in handy if a friend started chasing them and they had to escape. A few children in the target age range discovered on their own that it was impossible to reach some parts of the Gyroid from where they were, unless they climbed to the outside of the structure. One 10-year-old who had noticed that the Gyroid was made of identical chips described how he deconstructed the structure in his mind. Level 5 absence of physical objects To reach this level, visitors thought only with mental objects based on what they knew or remembered about the Gyroid structure. We observed three approaches to this level: (1) Climbers who worked their way up the hierarchy, reaching Level 5 based on intellectual engagements that were firmly based in physical explorations. For instance, we observed two 10-year-olds who turned their backs on the Gyroid and described their discoveries about it to their parents, telling them about the two separate tubes or how the Gyroid was put together. (2) A few non-climbers stood back and studied the exhibit, engaging in visual explorations of both the exhibit itself and the ways that other visitors were engaging with it. A few visitors told us how they deconstructed their mental version of the Gyroid into its constituent parts or mentally rotated a segment of the Gyroid to make a more comfortable couch for their living room. (3) Some visitors read the exhibit labels—their understanding of the Gyroid’s construction or non- Selinda Research Associates, Inc. 61 intersecting tubes came from the labels, rather than direct physical experience. We especially enjoyed talking with the 10-year-olds whose physical explorations of the Gyroid led to such accomplished understandings of its construction. It was also interesting that label readers—if also gifted with an appropriately spatial imagination—were able to reach Level 5, apparently without starting on the lower, more physical levels. We were also impressed that visual explorations seemed to be effective on their own for some visitors. We watched a few adults whose children were climbing, step back, study the Gyroid, and walk around for different views. When interviewed, they talked as if they had mentally projected themselves into their mental object of the Gyroid, perhaps aided by their observations of others climbing in and through the Gyroid. Being Aware of Spatial Reasoning As noted previously, most visitors who engaged with the Geometry Playground exhibits practiced a wide range of spatial reasoning skills. When we talked with respondents, however, the data indicated that most visitors were thinking about the exhibits and their engagement with them in terms of geometry, not spatial reasoning. In other words, many visitors practiced spatial reasoning skills, but most focused on shapes and geometry rather than the more sophisticated enterprise described by Tarte at the beginning of this section as “understanding, manipulating, reorganizing, or interpreting relationships visually.” It certainly was not necessary for visitors to be thinking deeply about spatial reasoning in order to practice spatial reasoning in the exhibition. Getting visitors to think about spatial reasoning was not a stated goal of the exhibition. However, there were some indications that when visitors did think actively about movement and transformation through space, as opposed to simply playing with shapes, they were more likely to evolve in how they engaged in spatial reasoning. For example, when adults were able to help children on the large climbing structures notice and pay attention to how they were moving through and among the various shapes, there were indications that children were more deliberate and thoughtful in their engagements. We know from another study that visitors can reflect about their mathematical experiences in exhibitions (Gyllenhaal, 2006). It is likely that helping visitors be more aware of, and articulate about, their spatial experiences and related geometric concepts would help many visitors have even richer experiences in the exhibition. There were also indications that all this practice of spatial reasoning led to other desirable outcomes. For example, a move up the hierarchy indicated at least short-term improvement at the skills that were being practiced. Visitors who reached higher levels evidenced new or better understandings of geometric concepts like tessellation and gyroids. And many visitors expressed positive feelings about their accomplishments, the exhibits, and the topic of geometry. It would be interesting to see how things might be different if adults had more opportunity to read about how playground equipment, blocks, construction skills, and sports help develop Selinda Research Associates, Inc. 62 children’s spatial reasoning skills15 or if host museum staff had resources that helped them learn more about spatial reasoning skills and how to encourage and talk about those skills with museum visitors. In conclusion, it was clear that visitors of all ages were practicing a wide range of different types of spatial reasoning skills in the Geometry Playground exhibition. Most used a wide range of physical skills as they began their explorations of an exhibit. Some of those stuck with it and both (a) practiced the more basic skills through repetition, and (b) became more intellectually engaged as they tried to meet the challenges posed at the exhibit. Some of those visitors developed mental objects that they used to help accomplish their goals, while some visitors’ practice of spatial reasoning grew more sophisticated as they engaged with the exhibition—they “moved up” a level or two on the knowledge hierarchy. 15 That opportunity was nearby, but it was on the outer wall of the Gyroid structure and went largely unnoticed by most visitors. Selinda Research Associates, Inc. 63 FINDINGS: TRAVELABILITY As part of the summative evaluation, the Explo team was interested in a number of questions related to how well the exhibition traveled to the Science Museum of Minnesota (SMM) and the Don Harrington Discovery Center (DHDC). The term “travelability” was defined quite broadly to include all aspects of exhibition planning, design, production, and implementation related to making an exhibition that could travel from one museum to another over several years. The aim of this part of the study was to: (a) come up with a list of lessons learned from the traveling of Geometry Playground to the first two hosts, (b) make recommendations that could improve Geometry Playground’s travelability for future hosts, and (c) contribute to the development of future Explo traveling exhibitions. The many findings and recommendations about travelability were summarized in an interim report (Gyllenhaal, 2010b). The remainder of this section summarizes the important findings. For the most part, the Geometry Playground exhibition appeared to travel well it’s first two off-site venues, in large part because the exhibition was developed with travelability in mind. It had a very flexible floor plan so it could be (and was) adapted to accommodate the specific needs of each venue. The exhibits were designed to use easy-to-obtain parts and to allow for relatively quick assembly and disassembly of most of the larger units. Explo staff developed a special crating system for the small and delicate pieces in Geometry Garden, and most of the other exhibit units were shipped as-is without damage. Most of the exhibition was quickly set up and installed at both venues, with the major exceptions being the large climbing structures and their platforms. Explo’s on-site installers worked closely with host staff so that they were familiar with each of the exhibit’s routine and special maintenance needs. The exhibition manual included detailed information about each exhibit, which proved to be a useful resource. Once the exhibition was installed, the exhibits allowed easy access to the inner workings for maintenance. After the break-in period at the Exploratorium and the first host museum, data indicated that host staff felt the exhibition was easy to maintain, with major breakdowns rarely occurring. When maintenance issues came up, Explo staff were readily available via telephone or e-mail. Also, upon request, Explo staff provided assistance with developing on-the-floor interpretation. As with all traveling exhibitions, the data indicated that some issues proved more challenging. One of the most obvious was the size and complexity of the large climbing structures, both in terms of their footprints, but also their height and the complexity of assembling them. While the other exhibit units were assembled relatively quickly, the two large climbing structures took the vast majority of the installation time. Setting up the rubberized surface around the structures was particularly delicate and difficult. There also appeared to be limited guidance and training for staff who worked with visitors on the floor, ultimately resulting in uneven support and guidance for visitors from these front line folks, especially with pointing out conceptual pairs, relating the exhibit units to everyday life, or providing the just one things. Selinda Research Associates, Inc. 64 Unlike many traveling exhibitions that we have seen, there appeared to be no consistently recommended (or required) floor plan. While this flexibility enabled each institution to adapt the exhibition to their facility, it also resulted in many missed opportunities for the goals of the exhibition to be achieved, and meant that some of the subtleties of the exhibit developers’ intentions went unrealized. The lack of deliberate conceptual pairings (i.e., by placing pairs in close proximity) was one such missed opportunity, but also things like providing a powerful entrance experience. Bleeding into other host exhibitions in particular meant that visitors were sometimes confused about the main theme of the exhibition, and the main point of many of the individual exhibits. There were also indications that because of the size and weight of some of the pieces, the exhibition required an unusually high number of staff and fork lift equipment. This, combined with the large size and height of the exhibition once it was installed, will probably preclude most smaller museums from hosting it. Finally, we noticed that by the time the exhibition arrived at the third venue, some of the components were getting quite worn. For example, the labels inside the Gyroid were peeling off (Fig. B-58), the spray can of light was broken and replaced by a flashlight that resulted in drawings with only intermittent lines (Fig. B-59), the table in Build a Room was missing some of its reflective tape, and it was deeply pocked from visitors hitting it with the spray can (Fig. B-60), the SRDs were dirty and starting to disintegrate (Fig. B-61), the Dividing Space exhibit no longer defined a line equidistant (or approximately equidistant) between two pegs (Fig. B-62), and the paddles in Spin a Shape were missing some (and in some cases all) of their reflective paint (Fig. B-63). With the exception of the Gyroid labels and worn SRDs, which were primarily aesthetic, these maintenance issues resulted in visitors not being able to experience the exhibits in their intended ways, and inhibited visitors from practicing spatial reasoning skills and becoming more deliberate and articulate about their practice. Selinda Research Associates, Inc. 65 FINDINGS: THE WEBSITE An important component of the Geometry Playground project was a website developed by the Exploratorium. As part of this summative study, we were interested in how the website was designed, and what role it played in visitors’ experiences. Description of the Website As of the writing of this report, the Geometry Playground website could be accessed at: As noted previously, this website included information about the exhibits and their creators, as well as activities that website visitors could do either within the exhibition or anywhere else, including classrooms, playgrounds, and other outdoor spaces. These activities related closely to key concepts and approaches used in the exhibition and could be either read online or printed out. (See Appendix K for screen shots of selected web pages.) The Geometry Playground website was designed with limited funding, primarily to support the traveling exhibition. The activities were designed to be used in conjunction with and to enhance a visit to the exhibition, but it was also important that they be able to stand alone. The expectation was that the majority of website users would come to the site because they already knew about the exhibition and were looking for further information. In addition, the project team asked us to consider ways to adapt the current website for mobile use, especially in relation to the anticipated installation of Geometry Playground-related structures on one or more San Francisco outdoor playgrounds. The intended audience for the website was described as primarily adults, and included (a) parents who were anticipating a family trip to the exhibition and were seeking information to prepare for their visit, (b) teachers and parents of homeschooled children who were planning a trip and looking for resources ahead of time, and (c) parents and teachers who had already been to the exhibition and were seeking follow up activities. It was anticipated that host venues would be able to provide a link to the Explo website from their websites, or that visitors could conduct an Internet search for the exhibition title. The website was also considered to be a way to disseminate the Geometry Playground project’s work to a larger audience, thereby extending the reach of materials developed for the exhibition, like the photo essays. The website also provided information about the artists who worked on the project, and briefly described some of the other elements of the project, like the research projects and other partnerships. In designing the site, the development team aspired to the same goals as those of the exhibition for inclusiveness and aesthetic appeal, providing examples of geometry in real life as well as showcasing the inherent beauty of geometry. Selinda Research Associates, Inc. 66 Design of the Website The following section of the report presents a brief summary of the findings about the design of the website from the critical review, as described in the Methods section above. The review took place in April and May 2011. It should be noted that the website may have changed since that time. It’s also important to note that the following comments are based on our critical review and include limited visitor data, except as noted. The critical review includes recommendations for redeveloping the website to improve usability and effectiveness on mobile devices and for developing websites for future Exploratorium exhibitions. Thus some of the recommendations go beyond what was possible for the current website, given the resources available for its initial development. The complete detailed critical review is in Appendix L. In spite of limited funding, the Geometry Playground website included rich resources that could serve exhibition visitors in important ways. Using the title of the exhibition as the search term, the website was easy to find through various Web search engines. However, as noted previously, most visitors were not aware of the exhibition’s name. In addition, data indicated that none of the venues informed exhibition visitors that there was a website, or included mention of it in any of its promotional materials. Only one of the two traveling host venues included a link to the Geometry Playground website on its own website. We found little evidence that the site would be able to attract its intended audience—i.e., people who were planning a visit to the exhibition, visitors within the exhibition, or those who had already visited. This was further confirmed by the many visitors with whom we spoke who had not visited the website, and were not aware of its existence. Web users who visited the site might not have recognized immediately that the website supported a major traveling exhibition, especially if they did not start their visit on the Geometry Playground homepage. If website visitors read some of the text and explored around a bit on the homepage, however, they might have found a range of interesting information and ideas about the exhibition. If website visitors first arrived at some of the internal pages, such as the Activities and Links page or one of the individual activities, they would have found no direct reference to the exhibition. The navigation was pretty clear and easy to use on most desktop or laptop computers (with a few notable exceptions that are described in detail in the full critical review in Appendix K). However, the design was too complex to be useful for mobile devices, and—because it was Flash based—did not work well with iPhones and iPads. There was no search function specific to the Geometry Playground website, which made it difficult to find or re-find information and photos about specific topics or activities. The text was, for the most part, short and clearly written. Most of the photographs were beautiful and gave a good feel for the exhibition, and the videos provided interesting perspectives on the artists in residence and other aspects of the exhibition. These videos worked well on desktop and laptop devices, but—again because of the use of Flash for navigation and the display of graphics—these were not usable when tested with the researchers’ Apple mobile devices. The activities seemed like they would be interesting and useful to teachers, and many were well designed. However, some had typos and confusing instructions. We also were concerned that Selinda Research Associates, Inc. 67 many teachers and homeschoolers would never find out about these activities, because they did not rank high on web searches for word combinations like “geometry activities” and “geometry educational activities.” The Geometry Playground exhibition included many amazing and unique exhibits, including pieces developed by artists in residence and by exhibit developers. For instance, we talked with many visitors who were fascinated by the Gyroid, Shape Cubes, and tessellating blocks and seemed primed to follow up on that interest. One visitor said she wanted to install a shape cube in her backyard and another said she was thinking about ways to adapt a section of the Gyroid as a sofa. Unfortunately, these and other website visitors looking for in-depth information about specific exhibit units or geometry topics would probably have been disappointed. Beyond some of the artist-in-residence pieces, there was little in-depth information about most exhibits—even unique and interesting exhibits like the Gyroid. Links to geometry topics were either non-existent or difficult to find because they were located near the bottom of the Activities and Links page, and users had to click on each individual category of links to see the full list. Many aspects of the website did a really good job of supporting the exhibition’s goals by, for instance, helping users appreciate the beauty and aesthetic aspects of geometry and highlighting aspects of geometry in everyday life. However, spatial reasoning—a major focus of the exhibition—was almost completely missing from the website, including the activities aimed at educators. The critical review noted that there were few online activities developed for use at specific exhibits, and that for most users of the website, their visit would likely have been a fairly passive—although interesting and aesthetically pleasing—experience. Overall, if visitors to the exhibition were able to find it, the website appeared likely to have been able to expand their museum visit in several ways, allowing them to gain interesting perspectives on the exhibition. The activities were likely to provide a broad range of intellectual, as well as physical, social, and emotional, engagements for visitors. Aside from the activities, however, a website visit was likely a fairly passive experience. The full critical review, found in Appendix L, describes in more detail many of the strengths of the Geometry Playground website, and includes specific recommendations for addressing the challenges presented by the version we reviewed. Use on Mobile Devices As stated earlier, the intended audience for the Geometry Playground website was primarily exhibition visitors, such as parents and teachers who would access the website before, during, or after their visit to the host museum. As part of the critical review, we were asked to also look at the efficacy of a Geometry Playground mobile presence, perhaps as a place where adult visitors to the exhibition and/or to the planned outdoor playground structures could access information about the exhibit structures and activities as their children and students played on the structures. As mentioned previously, the critical review indicated that the current website did not work very well as a mobile experience. In order to make the current website useful as a mobile presence, it Selinda Research Associates, Inc. 68 would require a significant investment of time beyond the scope of the Geometry Playground project or goals, and would likely require a total redesign of most of the website’s navigation, including, for example, simplifying the categories, conceptualizing how it links up with the on-the-floor exhibit experience, and redeveloping the videos and slideshows. The activities would also have to be vastly simplified into something someone would want to read and use on a mobile device. And, of course, adequate visitor field-testing would be an important component of the development of such a site. Jakob Nielsen (2011) strongly advocates for separate designs for mobile and desktop sites. Anybody wanting to launch a mobile intranet portal is advised to heed the strong finding from our research with mobile websites: Good mobile usability requires a separate design with a reduced feature set for mobile use cases, focusing on time- and location-dependent tasks. It's not enough to make an existing portal accessible through phones because the UI is optimized for desktop use. (Emphasis in original.) Social Media Social media can be an important tool and a powerful opportunity to involve visitors in rich and engaging ways. At Geometry Playground, social media was limited to a dedicated Flickr site: that was a venue for web users to post their geometry-themed photographs. As of the writing of this report, the site had 466 members, included over 3,500 photographs, and had four discussion topics. However, at the time of our review, the most recently posted comment was 8 months prior. Social media requires significant “care and feeding” in order to realize its potential. The Geometry Playground Flickr site was not an important part of the Geometry Playground project’s goals, but it still served as a solid initial social media presence, although it languished because of lack of attention.16 As this report is being written, both mobility and social media are hot topics at conferences and on museum-related blogs and e-mail lists. We hope that the Exploratorium will continue to play an active role in experimenting with exhibition-related websites, and exhibition-themed mobile and social media presences. 16 Note that as of the writing of this report, the home page still stated that the exhibition “is currently under development at the Exploratorium and will begin touring science museums around the country in 2010.” Selinda Research Associates, Inc. 69 CONCLUSIONS The Geometry Playground exhibition proved to be an engaging and enjoyable experience for most of its visitors. Few visitors expressed negative feelings about their experiences, and many expanded their thinking about geometry in useful and important ways. The mix of large and small climbing structures, medium-sized immersive interactive exhibits, hands-on table-top interactives, and contemplative exhibits, display cases, and even furniture, provided a refreshing combination of experiences that visitors took advantage of and appeared to appreciate. And although the exhibition was targeted to 5- to 12-year-old children and their adult caregivers, there was strong evidence that visitors of all ages effectively practiced a wide range of spatial reasoning skills. One limitation of the exhibition was that it did not help many visitors become meta-cognitive, deliberate, or aware that they were doing spatial reasoning. This was partly due to the fact that aspects of the exhibition that encouraged meta-cognition—i.e., the quotes on the Entrance Structure and Geometry in the Wild—were passive experiences, located in places where few visitors attended to them. The exhibition’s website also failed to support this sort of meta-cognition, even among educators. Meta-cognition and awareness among visitors were not stated goals of the exhibition, but by not focusing on this aspect of spatial reasoning, the exhibition missed out on an opportunity that would likely have contributed to even richer visitor experiences. Given the sorts of deep thinking we saw by some visitors to Geometry Playground, it seems like it could be an important goal of future math-related projects. Helping visitors be more aware of, and articulate about, their spatial experiences and related geometric concepts would help advance the public understanding of this critical aspect of mathematics, and perhaps inspire parents to include more spatial reasoning practice in their children’s lives outside of school. The exhibition also achieved many of its other goals, such as raising some visitors’ awareness about aesthetic aspects of geometry and about ways that geometry is encountered in everyday life. Their exhibition experiences also convinced many visitors that geometry can be fun, at least away from a school setting. We should also note, though, that more visitors reflected on having fun with geometry when they knew up front that the exhibition was going to be a playful experience with geometry, for example, when they were aware of the exhibition title. In the end, the exhibition increased positive experiences and associations with geometry for children 5-12 and their adult caregivers. The most effective social interactions took place at the interactives, which were, for the most part, designed in ways that provided opportunities for families to engage together as groups, or at least take turns in ways that kept them together as a group. At the interactives, some parents got ideas about how to help their children use the exhibits and how to talk with them about what they were doing. The exhibits that were less effective at encouraging social interactions were the climbing structures and some of the medium-sized immersives where adults and children tended to engage in more Far Out behaviors. The playground-like designs of the climbing structures seemed to lead many parents, especially those of older and more independent children, to disengage from the social experience, as they sat on the benches and did other things with their Selinda Research Associates, Inc. 70 time. This was not necessarily a bad thing in terms of family dynamics, and many caregivers seemed to appreciate having some time on their own. However, it did represent a missed opportunity for both children thinking and learning more about the structures they were climbing on, and parents learning about how playground experiences contribute to children’s skills that will eventually be needed for spatial reasoning about geometry and the many other disciplines where spatial reasoning is important. Parents look to labels for many reasons, some of which contribute to their engagements with their children. However, some immersives in Geometry Playground were not labeled at all (like the Sphere Cube and one installation of Stack of Stars) and some were labeled in ways that were not particularly useful to parents—like some of the Shape Cubes, and certain installations of the Gyroid. These exhibits in particular would likely have been more effective if they had had labels that were designed specifically to support family interactions. The idea of having conceptual pairs of exhibits was a powerful one. One of them—Stack of Stars and the small SRDs—was particularly effective and proved that this could be a valuable approach to designing visitor experiences. Having visitors recognize the conceptual pairs appeared to be an important component of the success of Stack of Stars :: small SRDs pairing. This was achieved by locating the two exhibits in close physical and visual proximity. Additional support might have been provided via labels and graphic panels that pointed out the pairings even more explicitly. As we followed the exhibition across three host museums, it became clear that many seemingly small things either enhanced or disrupted the visitor experience. The different layouts of the exhibition clearly affected which exhibits visitors noticed and engaged with. In addition, the proximity of certain exhibits, especially those that were conceptually related, made a difference in helping visitors move up various knowledge hierarchies. Consistent and effective exhibit maintenance was also a vital part of the visitor experience; when this was not possible, the exhibit’s goals were undercut. Data also indicated that a well-trained floor staff could greatly enhance the exhibition experience in lots of different ways. Publicity, and the design and placement of information on host museum websites, played a role as well. For an exhibition to maximize its effectiveness at meeting its educational goals, all these factors have to be taken into account. The website provided a strong, although mostly passive and non-social, desktop experience. The challenge here however was to make visitors aware of it so that they could use it as a helpful resource. The design of the site did not allow for effective visitor use as a mobile activity, and the inclusion of social media during the early stages of the project (i.e., the Flickr site) was a promising start, but was not kept current. The other major audience for the project was museum professionals. Many of the projects’ efforts with regard to professional audiences took place before the summative evaluation phase, and the project will continue to affect the field for several years to come through presentations and publications, particularly about the project’s research and evaluation efforts. The summative evaluation looked at only one small aspect of professional development, i.e., the affects of the hosting experience on staff at the various venues. We found that most host staff gained some Selinda Research Associates, Inc. 71 appreciation for the exhibition itself, for the ways in which visitors were learning about geometry, and about the appropriateness of large-scale climbing structures within a science museum setting. Their increased appreciation of immersion as an exhibit technique and spatial reasoning, based on their experiences hosting the exhibition, appeared more limited. We attribute this to the fact that there was limited communication about these topics with host staff through either written materials or training. Overall, the Geometry Playground exhibition provided visitors of all ages with a wide range of enjoyable, rich, and engaging experiences with geometry and spatial reasoning across a variety of locations. The exhibition will likely continue to contribute to the advancement of the field’s understanding of immersive visitor experiences, spatial reasoning, math, and geometry in informal settings. Selinda Research Associates, Inc. 72 REFERENCES Allen, S., Gutwill, J., Perry, D. L., Garibay, C., Ellenbogen, K. M., Heimlich, J. E., et al. (2007). Research in museums: Coping with complexity. In J. H. Falk, L. D. Dierking & S. Foutz (Eds.), In principle, in practice: Museums as learning institutions (pp. 17). Lanham, MD: Alta Mira Press. Boakes, N. J. (2009). Origami instruction in the middle school mathematics classroom: Its impact on spatial visualization and geometry knowledge of students. Research in Middle Level Education, 32(7), 1-12. Clements, D. H. (1998). Geometric and spatial thinking in young children. Retrieved March 2, 2007, from the World Wide Web: 436232 Dancu, T. N. (2010). Designing exhibits for gender equity. Unpublished doctoral dissertation, Portland State University, Portland, OR. Downloaded from the Exploratorium website July 5, 2011. Dissertation focused on developing gender equitable exhibits that promote use of spatial reasoning skills. Dancu, T. N., & Gutwill, J. (2010). Getting inside immersive exhibits: A study of Exploratorium's Geometry Playground. Downloaded from the Exploratorium website July 5, 2011. Poster session describing research that compared visitor experiences at immersive and table-top exhibits. Dancu, T. N., Gutwill, J., & Sindorf, L. (2009). Geometry Playground Pathways study. Downloaded from the Exploratorium website July 5, 2011. Diamond, J. (1980). The ethology of teaching: A perspective from the observations of families in science centers. Dissertation Abstracts International, (UMI No. 8113012). Diamond, J. (1986). The behavior of family groups in science museums. Curator, 29(2), 139-154. Gutwill, J. (2010). Fostering and assessing spatial reasoning. Downloaded from the Slideshare website July 5, 2011. Presentation to Visitor Studies Conference Gyllenhaal, E.D. (2011). Geometry Playground summative evaluation: Site Visit 3 Summary Science Museum of Minnesota. Unpublished manuscript, The Exploratorium, San Francisco, CA. Selinda Research Associates, Inc. 73 Gyllenhaal, E. D. (2010a). Geometry Playground summative evaluation: Site Visit 2 Summary Science Museum of Minnesota. Unpublished manuscript, The Exploratorium, San Francisco, CA. Gyllenhaal, E. D. (2010b). “Travelability” of Geometry Playground: Preliminary Findings. Unpublished manuscript, The Exploratorium, San Francisco, CA. Gyllenhaal, E. D. (2006). Memories of math: Visitors' experiences in an exhibition about calculus. Curator: The Museum Journal, 49(3), 345–364. Gyllenhaal, E.D. & Perry, D.L. (2011). Geometry Playground summative evaluation: Site Visit 4 Summary Don Harrington Discovery Center. Unpublished manuscript, Selinda Research Associates, Inc., Chicago. Gyllenhaal, E.D., & Perry, D.L. (2010). Site visit report: Geometry Playground at the Exploratorium. Unpublished manuscript, The Exploratorium, San Francisco, CA. Harding, S. (Ed.). (1987). Feminism & methodology. Bloomington: Indiana University Press. Hido, N., & Gutwill, J. (2007). Adventure playground – Front-end study. Unpublished manuscript, The Exploratorium, San Francisco, CA. Law, D.J., Pellegrino, J.W., & Hunt, E.B. (1993). Comparing the tortoise and the hare: Gender differences and experience in dynamic spatial reasoning tasks. Psychological Science, 4(1), 35-40. Lincoln, Y. S., & Guba, E. G. (1985). Naturalistic inquiry. Beverly Hills, CA: Sage Publications. McManus, P. M. (1987). It's the company you keep...:The social determination of learning-related behaviour in a science museum. The International Journal of Museum Management and Curatorship, 6, 263-270. McManus, P. M. (1988). Good companions: More on the social determination of learning-related behaviour in a science museum. The International Journal of Museum Management and Curatorship, 7, 37-44. Miles, M. B., & Huberman, A. M. (1994). Qualitative data analysis: An expanded sourcebook (2nd. ed.). Thousand Oaks, CA: Sage Publications. Miller, C., & Swift, K. (1980). The Handbook of Nonsexist Writing. New York: Harper and Row. National Council of Teachers of Mathematics (2000). Principles and standards for school mathematics. Reston, VA: National Council of Teachers of Mathematics. Selinda Research Associates, Inc. 74 Neilsen, J. (2011) Intranet portals: Personalization hot, mobile weak, governance essential. Downloaded August 17, 2011 from: Perry, D. L. (2012). What makes learning fun? Principles for the design of intrinsically motivating museum exhibits. Walnut Creek, CA: AltaMira Press Perry, D. L. (1993). Measuring learning with the knowledge hierarchy. Visitor studies: Theory, research and practice: Collected papers from the 1993 Visitor Studies Conference, 6, 73-77. Available online at: Tartre, L. (1990). Spatial orientation skill and mathematical problem solving. Journal for Research in Mathematics Education 21(3): 216-229. Wolf, R. L., & Tymitz, B. L. (1981). "Hey mom, that exhibit's alive": A study of visitor perceptions of the Coral Reef exhibit (Unpublished manuscript). Washington, DC: National Museum of Natural History, Smithsonian Institution. Selinda Research Associates, Inc. 75 APPENDIXES Appendix A. Description of Geometry Playground Exhibits Appendix B. Photographs of Geometry Playground Exhibits Appendix C. Annotated Bibliography of Reports & Presentations Appendix D. Topical Framework Appendix E. Sources of Data Appendix F. Description of Respondents Appendix G. Installations at the Host Science Centers Appendix H. Inventory of Engagements at the Large Climbing Structures Appendix I. Taxonomy of Spatial Reasoning Skills Appendix J. Case Studies of Spatial Reasoning at Selected Exhibits Appendix K. Screenshots of the Website Appendix L. Critical Review of the Website Selinda Research Associates, Inc. 76 Appendix A. Description of Geometry Playground Exhibits This appendix provides brief descriptions of the exhibits developed for Geometry Playground, along with the exhibits’ intended learning goals and the spatial reasoning skills each exhibit was designed to encourage. The exhibits are classified into three categories—Seeing, Moving, and Fitting Things Together—a classification that was developed early in the project, but one that was not reflected in the exhibition floor plans, as it was not a driving force of the project. All photos referred to in this appendix can be found in Appendix B. The information in this appendix was adapted from three documents supplied to the evaluation team early in the project: Exhibit Descriptions (Word file created 4/20/2010), GP Exhibits Learning Goals (Excel file dated 6/14/2010) and Exhibit Learning Goals (pdf file created 7/1/2010). For more information about spatial reasoning skills, see the section entitled “Findings: Spatial Reasoning” in the body of the report. The spatial skills discussed in that chapter are highlighted here with italics. SEEING Hopscotch This immersive exhibit consisted of a two-foot diameter, upright, cylindrical mirror surrounded by two different hopscotch-board graphics on the floor (Fig. B-1). It was one of four exhibits that included anamorphic mirrors.17 The hopscotch images appeared distorted when seen on the floor, but when viewed in the mirror looked normal or undistorted (as if they were made of straight lines, despite the curved surface of the mirror). A label atop the mirror reminded visitors to look in the mirror (Fig. B-2) where they could see a distorted version of their bodies hopping along the board. The developers’ goal was to provide a fun, whole-body exhibit that allowed visitors to experience how a large cylindrical mirror changes shapes in the reflection. The exhibit was designed to give visitors a chance to experience the same geometric concepts with their bodies as they experienced with their hands at the Distorted Drawing exhibit (described below). The exhibit allowed visitors to practice a range of spatial reasoning skills related to the transformations from the distorted images of the hopscotch boards on the floor to the undistorted reflections in the mirrors and the horizontally compressed reflections of their bodies. These observational skills included locating objects in space (e.g., the images relative to their reflections in the mirror) and exploring curvature and congruence (by comparing the distorted and undistorted boards and bodies). 17 An anamorphic mirror is a conical or cylindrical mirror that transforms a flat distorted image into a picture that, because it shifts position as the viewer moves, seems like a three-dimensional object. The exhibits using anamorphic mirrors were Hopscotch, Distorted Chair, Distorted Drawing, and Conical Mirror. Selinda Research Associates, Inc. 77 Distorted Chair Like the Hopscotch exhibit, Distorted Chair was an immersive experience using a cylindrical or anamorphic mirror (Fig. B-3). Visitors were encouraged to sit on a distorted plastic chair that looked normal or undistorted in the mirror, at the same time seeing their own horizontally squished reflections as they sat and moved about the chair. The team’s goal was to allow visitors to explore the geometry of a mirrored cylinder with their whole body. As with the Hopscotch exhibit, Distorted Chair was designed to allow visitors to practice a range of spatial reasoning skills. The spatial skills were related to the mirror’s transformation from distorted to normal (and vice versa), and included explorations of shapes, sizes, and curvature. The label (Fig. B-4) encouraged visitors to draw spatial inferences about the mirror’s effects. Distorted Drawing This interactive table-top exhibit complemented the whole-body Distorted Chair and Hopscotch exhibits by providing small-scale hands-on experiences with a cylindrical mirror (Fig. B-5). Visitors could view a series of flip cards with distorted images that transformed to normal in the mirror (Fig. B-6), including one card with a miniature distorted hopscotch board that explicitly linked this exhibit to the Hopscotch exhibit (Fig. B-7). Visitors could also use a chalkboard with a distorted grid to draw images and view them in the small cylindrical mirror (Fig. B-8). The developers’ goals were that, by experimenting with the chalkboard activities encouraging trial and error, visitors would experience the geometric effects of a cylindrical mirror and observe how the shape of the mirror affected the reflections. The flip-book activity was designed to allow visitors to observe images that have been distorted or “anamorphed” and then recognize their undistorted reflections in the curved mirror. Developers also intended that visitors experience artistic aspects of geometry at Distorted Drawing, as they created their own images and enjoyed those made by others. In addition to spatial skills related to transformations and exploration of curvature and congruence (described above), this exhibit was designed to encourage the spatial skill of construction by allowing visitors to create their own images. Label texts, the sequence of card transformations, and the hands-on chalkboard activities were designed to encourage visitors to make spatial inferences about the transformations. Projected Puzzle The puzzle in the name of this exhibit consists of the words, “LOOK HERE,” painted on a set of free-standing panels. What initially appear as almost randomly arranged patterns and splotches of color reveal their message once visitors find the perfect spot from which to view them—then they snap into focus (Figs. B-9 and B-10). Selinda Research Associates, Inc. 78 The developers’ goal was that visitors would experience how parts of an image spread out over three-dimensional space come together to create a seemingly flat or two-dimensional picture. The team also intended for visitors to think about how they experience objects in three-dimensional space and about how objects from different viewpoints can look different. In addition, this exhibit provided another instance where visitors could view geometry as art. The intended spatial reasoning skills for this exhibit included experiencing the transformation from three-dimensional to two-dimensional image and exploration of the locations and orientations of the objects and their own eyes that make this transformation possible. Conical Mirror Conical Mirror is a table-top interactive were visitors can transform images with a second type of anamorphic mirror, a mirrored cone (Figs. B-11 and B-12). It was designed so visitors could place a series of colorful discs around the mirrored cone and look down the point to see the discs’ images transformed (without poking their eyes out). The conical mirror more-or-less reflected every image inside out, so that a square was transformed into a clover and a sun into a flower, and hearts were flipped upside-down. The developers’ goal was to let visitors to explore the surprising transformations of images reflected by a conical mirror. This exhibit was also considered another opportunity for visitors to explore geometry as art. As with the other exhibits using anamorphic mirrors, Conical Mirror was designed so that visitors could practice spatial skills related to transformations of images and so they could explore curvature and congruence as they compared the original and transformed images. MOVING Build a Room This immersive exhibit was an open-sided room with grids on the walls and floor (Fig. B-13). It was designed so that visitors could use a light pen (Fig. B-14) to trace objects in the room and to draw in three dimensions, and then see the results on a monitor (as illustrated in the label, Fig. B-15). The whole-body immersive nature of this experience was intended to put visitors in the middle of their geometric artwork. The developers’ goal was to give visitors a chance to explore what drawing was like in three dimensions. As visitors drew, they could practice creativity and experience geometry as art. The design encouraged visitors to practice a range of spatial reasoning skills as they drew and examined their work, including explorations of shapes (especially the edges of things they drew). As they constructed their images, visitors could also explore the locations and orientations of objects and then could see their work in three dimensions transformed into two-dimensional images on the monitor screen. Selinda Research Associates, Inc. 79 Spin a Shape This table-top interactive allowed visitors to choose cards with simple two-dimensional shapes and fit them onto a spindle (Fig. B-16). At the push of a button the spindle spun the cards, transforming the two-dimensional image into a perceived three-dimensional shape (Fig. B-17). The stated goal for this exhibit was for visitors to experience how a flat shape, when rotated around an axis, transformed into a rotational solid. Developers also intended this exhibit to contribute to visitors’ aesthetic appreciation of geometry, and the label illustrated a real-world example: a spinning ballerina. The label challenging visitors to guess which card would make each illustrated shape, the exhibit encouraged explorations of congruence (comparing objects with one another). To meet this challenge, visitors had to mentally rotate the two-dimensional shapes and infer the transformation (an example of spatial inference). Shape Cubes and Shape Stools The Shape Cubes were sculptural tunnels made by rotating many identical simple shapes (triangle, square, etc). They were made by cutting simple shapes like squares and triangles out of a series of wooden squares—but first rotating the shape several degrees from one square to the next—and then gluing the squares together to make a cube or rectangle (Fig. B-18). The resulting tunnel was a twisting tube, with its inside surfaces more curved and complex than one might expect given the simple straight-sided shape used to generate it. Shape Cubes came in two sizes: a large size for visitors to crawl through (as in Fig. B-18) and a smaller size used to interpret the larger cubes (Fig. B-19). The cut-out shapes were not wasted. Instead, they were glued together, again rotating each successive panel, generating a curve-sided Shape Stool (Fig. B-20). The goal of this design was to allow visitors to experience with their body and hands how many flat surfaces can create a surprising curved surface inside the Shape Cubes. Additional goals were to inspire visitors to see geometry as art and to encourage their aesthetic appreciation of geometry. Spatial reasoning skills visitors could practice in the Shape Cubes included exploration of shapes, locations, and curvature of objects, along with experiencing rotation of those shapes, physically within the Shape Cubes, and mentally as visitors sought to understand how the tunnels were generated. Visitors could also practice congruence as they compared the solid shapes of the Shape Stools with the similarly shaped hollows within the Shape Cubes. Selinda Research Associates, Inc. 80 Geometry in Motion The Geometry in Motion exhibit allowed visitors to assemble combinations of wheels and rods, making mechanical linkages that transformed linear motion into circular motion and vice versa, (Fig. B-21). The loose parts snapped onto the board and to each other using cylindrical magnets (Fig. B-22). The label suggested a simple project for visitors to build—a waving hand—and showed a real-world example—the wheels on a steam locomotive. The developer’s goal was to give visitors’ first-hand experiences with mechanical linkages as they created their own moving sculptures. Visitors were also intended to practice a range of spatial reasoning skills through this construction activity, including explorations of the size and locations of objects, the rotations of objects, both physically and mentally, and making spatial inferences as constructions were planned. Drawing in Circles Exploratorium artist in residence, Tauba Auerbach,, created this large table-top spirograph (Fig. B-23). Visitors made full-arm motions as the gears swept around and created loop upon loop with colored pencils drawing on large sheets of white paper (Fig. B-24). The artist intended this piece to be as much about the experience as it was about the striking end product, which visitors could take home once it was finished. The large motions and oversize patterns generated by the gears revealed aspects of the overlapping loops that were harder to notice with small-scale classic spirographs. This drawing activity was designed to foster connections between math and art and aesthetic appreciation of geometry—to see geometry as art. It was also intended to help visitors build their art-making competency. Spatial reasoning skills intended to be practiced by visitors during construction of their drawings include explorations of shapes, sizes, and curvatures and making spatial inferences as visitors planned which gears to use for their drawing. Gyroid The Gyroid (Fig. B-25) was an immersive climbing structure based on a geometric form called a periodic minimal surface, discovered by Alan Schoen in 1970. The structure was assembled from many identical pieces that, when attached together, created two separate yet intertwined “rooms” that were mirror images of one another. The only way visitors could get from one room to another was to climb or jump out of one opening and then climb back in another. The Gyroid was mounted on a platform covered with rubberized tiles, with wooden benches set back at a safe distance (Fig. B-25). At some host museums rails, cables, and signage (Fig. B-26) were used to limit access to the structures, including limiting the overall numbers of children on the platform at any given time, and keeping children younger than seven years old from climbing on the structure. Activity labels were placed within the Gyroid at some host museums, suggesting things that visitors could do within the structure, including climb together (Fig. B-27). Selinda Research Associates, Inc. 81 Interpretive labels mounted on walls (Fig. B-28) or bolted to the surrounding benches (Fig. B-29) explained the origins and properties of the Gyroid and showed some real-world examples. The stated goals for this exhibit were to provide visitors with an immersive climbing experience as they explore a minimal surface structure, and to foster their aesthetic appreciation of geometry. The spatial reasoning skills encouraged by this exhibit were whole-body explorations of shapes, locations, and orientations of the structure, its components, and the bodies within it. The developers also intended visitors to explore congruence by comparing the two interlocking tunnels. FITTING THINGS TOGETHER Stack of Stars Stack of Stars (initially called Climb the Stars, at the first venue only) was the second large immersive climbing structure, built by tessellating three-dimensional star-like shapes (Fig. B-30), called stellated rhombic dodecahedrons (big SRDs). Tessellating meant the SRDs were packed together with no spaces between them—a concept that applies in both two and three dimensions. This large-scale, experience was a conceptual pair with a table-top set of small SRDs that visitors could use to build their own structures (see Space-filling Blocks, below). Like the Gyroid, this exhibit was mounted on a platform covered with rubberized tiles and surrounded by wooden benches (Fig. B-30). Labels at the entrances (between the benches) informed parents that this exhibit was designed for children five years and older (Fig. B-31). Interpretative labels were bolted to the benches (Fig. B-32) at some host museums. The exhibit developers intended that visitors would explore the tessellating shapes and be exposed to geometric terminology (e.g., stellated rhombic dodecahedrons) and to the concept of tessellation. Another goal was that this exhibit would contribute to visitors’ aesthetic appreciation of geometry. As with the Gyroid, Stack of Stars was intended to encourage spatial reasoning skills like immersive and whole-body explorations of shapes, locations, and orientations of the structure, including its components, and the bodies within it. The developers also intended visitors to explore congruence and size by comparing the big SRDs in the structure with the small SRDs in the adjacent Space-filling Blocks exhibits. Space-filling Blocks The table-top Space-filling Blocks exhibits included one unit with loose, hand-held versions of Stack of Stars stellated rhombic dodecahedrons (small SRDs, Figs. B-33 and B-34) and the other with hand-held truncated octahedrons (abbreviated as TOs, Figs. B-35 and B-36). The exhibits allowed visitors to design and build their own three-dimensional tessellating structures by fitting together the blocks with no spaces between them. Selinda Research Associates, Inc. 82 Developers intended that, by playing with, stacking, and building with the small SRD and TO blocks, visitors would experience three-dimensional tessellation and reason about the orientation of stacking required for their tessellation. This exhibit was also considered another opportunity for visitors to develop their appreciation of geometry. These construction exhibits were intended to encourage smaller-scale explorations of the shapes, locations, and orientations of the tessellating shapes. Visitors were also expected to rotate the blocks physically and mentally as they were moved into place. Developers also intended visitors to explore congruence and size by comparing the small SRDs with the big SRDs in the adjacent Stack of Stars. Stellator At this small exhibit (Fig. B-37), visitors could look into a window at a seemingly three-dimensional dodecahedron, with twelve flat sides (Fig. B-38). Then, as they turned a crank, the shape would be transformed as points grow smoothly from each side to reveal a stellated dodecahedron. When visitors reversed the crank, the points would disappear. Rather than using computer animation, the entire transformation was done with mirrors. This exhibit was designed to allow visitors to explore the differences and relationships between these two different shapes, exposing them to the concept of stellation. This exhibit was also intended as an opportunity for visitors to cultivate their aesthetic appreciation of geometry. Spatial reasoning skills that could be practiced here included the exploration of the two final shapes (and their intermediates) and of the transformation from the “regular” to stellated shape and back. In comparing and thinking about the two shapes, visitors could also practice congruence and spatial inference. Tile Designer This exhibit (Fig. B-39) allowed visitors to make simple manipulations to a line on a touch screen (Fig. B-40) that would immediately be applied to three wallpaper-like sections on a larger monitor (upper half of Fig. B-39). These visitor-generated patterns represented three types of tessellations. The stated goal of this exhibit was to help visitors understand that a single line segment can be multiplied, according to geometric rules, to create several different patterns and to engage visitors creatively with these phenomena. This was also an opportunity for visitors to see geometry as art and to cultivate their aesthetic appreciation of geometry. As they engaged with Tile Designer, visitors could engage with aspects of spatial reasoning such as exploration of shapes, orientation, and congruence. Selinda Research Associates, Inc. 83 Geometron Created by Geometry Playground’s first artist in residence, John Edmark, this exhibit was designed as a video kaleidoscope that put the visitors’ hands in the image—thus it could be both a table-top and an immersive exhibit (Fig. B-41). Visitors looked into a tapered kaleidoscope at a video screen showing them a live video capture of colorful stick-on plastic pieces and of their hands, multiplied many times into an icosahedral (20-sided) shape (Figs. B-42A and B). Visitors could explore the many symmetries in the image by moving glowing shapes in the camera’s view. The goal of this artistic piece was to let visitors practice making patterns on a three-dimensional shape. Visitors could create their own kaleidoscopic images and had the opportunity to manipulate and appreciate the symmetry of the patterns by the angled mirrors. This was also an opportunity for visitors to see geometry as art and to cultivate their aesthetic appreciation of geometry. Spatial reasoning skills practiced here could include explorations of the shapes, including translating (sliding them along), rotating, and aligning them on the clear plastic front of the exhibit. Visitors could also practice congruence, comparing the plastic pieces with their expressions in the video image, and spatial inference as they planned their next move, adding or removing pieces from the screen. Dividing Space With its sloping table-top and moveable pieces, this exhibit reminded some exhibit team members of the Plinko game from the television show, Price is Right. However, it was actually much more sophisticated, with a computer-vision system that tracked the positions of re-arrangeable red pegs on a table and two larger blue pucks that could slide across the table and bounce off the pegs (Fig. B-43). Based on these tracking data, a projector above the table drew polygons around each peg and the puck, and these polygons constantly changed as the pucks careened across the table. (See Fig. B-44 for a close up view of the pegs and pucks). These polygons were designed to bisect the distance between each object on the table, using the Voronoi algorithm. Thus Dividing Space was designed to allow visitors to experience how a Voronoi diagram responds in real time to the changing positions of objects on the sloping table. Developers’ intention was that, through play, visitors could observe how changes in one cell (or colored polygon) had an impact on adjacent cells. With repeated observations, some visitors might develop their abilities to predict those changes. In terms of spatial reasoning, visitors could practice their construction skills as they developed a course with the red pegs. Visitors could also explore the shapes, locations, and sizes of the polygons and their boundaries, and practice spatial inference as they tried to predict what would happen as red pegs were added or removed and as the blue pucks slid across the table. Selinda Research Associates, Inc. 84 Polyhedra Artist-in-Residence Stacy Speyer created a set of large, colorful polyhedra out of metal struts. These shapes were suspended within reach in a translucent silo-shaped structure (Fig. B-45), so that visitors could touch, rotate, and contemplate them (Fig. B-46). Ms. Speyer described her approach to the exhibit in a label attached outside the structure’s entrance (Fig. B-47) This artistic piece was developed to allow visitors to observe and experience the many colorful geometric polyhedra, and so they might appreciate their geometrical symmetries and relationships. This was also an opportunity for visitors to see geometry as art and to cultivate their aesthetic appreciation of geometry. Spatial reasoning skills intended to be practiced here included exploration of shapes as well as rotating the shapes both physically and mentally. Gear Cube The Gear Cube (Fig. B-48) was an oddly toothed creation that looked like it shouldn’t spin freely, but did. After up to ten revolutions, the gear segments rotated into a cubic form that disappeared with the next turn of the crank (Fig. B-49). The intention was to help visitors gain appreciation for the intricacy of gears and how these gears could create a surprising changing shape. This was also an opportunity for visitors to see geometry as art and to cultivate their aesthetic appreciation of geometry. Spatial reasoning skills intended to be practiced here included exploration of sizes, rotating the gear segments both physically and mentally, and both congruence and rotation. Geometry Garden The museum-like cases of Geometry Garden (Fig. B-50) displayed a range of geometric phenomena, including natural history objects and art pieces. Each object was labeled, often with fairly detailed descriptions (Figs. B-51 and B-52). The developer’s intention was to expose visitors to a gallery of objects in nature and art, including shapes and forms that illustrated a range of geometric concepts. This was also an opportunity for visitors to see geometry as art and to cultivate their aesthetic appreciation of geometry. Spatial reasoning skills intended to be practiced here included exploration of orientations, congruence, and transformation. Selinda Research Associates, Inc. 85 Geometry in the Wild Scattered throughout the exhibition were a series of low walls displaying photo essays—i.e., curated sets of images—of geometry from the real world (Fig. B-53). Some images were drawn from the photographs submitted to the project’s online Flickr pool. Texts described geometric concepts related to the images (Fig. B-54), including an explanation of the geometry that takes place on playgrounds (Fig. B-55). The intentions here were to help visitors gain appreciation for geometry in the everyday world and to cultivate visitors’ aesthetic appreciation of geometry. There were no specific goals related to spatial reasoning. Sphere Cube The Sphere Cube (Fig. B-56) was a medium-sized immersive exhibit that was at various stages of development during the summative evaluation. It was designed so visitors could climb inside a spherical hollow suspended in a grid-like cube (Fig. B-57). There were no labels on this during any of the data collection site visits, although our understanding was that these were being developed. The goal of this exhibit was to create a sense of enclosure and to foster groups hanging out and being immersed in geometric space. Selinda Research Associates, Inc. 86 Appendix B. Photographs of Geometry Playground Exhibits This appendix includes photographs taken of the Geometry Playground exhibits. The exhibits are classified into three categories—Seeing, Moving, and Fitting Things Together—a classification that was developed early in the project, but one that was not reflected in the exhibition floor plans, as it was not a driving force of the project. All exhibits and their accompanying spatial reasoning goals are described in detail in Appendix A. All photographs in this report © 2011 Selinda Research Associates, Inc. Selinda Research Associates, Inc. 87 SEEING Hopscotch Figure B-1. Hopscotch. (SMM) Figure B-2. Label on the Hopscotch cylindrical mirror. (SMM) Selinda Research Associates, Inc. 88 Distorted Chair Figure B-3. Distorted Chair. (SMM) Figure B-4. Label for Distorted Chair. (SMM). Selinda Research Associates, Inc. 89 Distorted Drawing Figure B-5. Distorted Drawing showing flip cards (foreground) and chalk board (background). (SMM) Figure B-6. Example of a flip card at Distorted Drawing. (DHDC) Selinda Research Associates, Inc. 90 Figure B-7. Flip card linking Distorted Drawing to Hopscotch immersive. (DHDC) Figure B-8. Straight lines drawn on chalk board at Distorted Drawing. (DHDC) Selinda Research Associates, Inc. 91 Projected Puzzle Figure B-9. Projected Puzzle viewed from the base of the central staircase. Note the orange dot on the floor indicating where you are supposed to stand, and the two signs, one by the dot and the other by the puzzle itself. (SMM) Figure B-10. Label on the stanchion located by the orange dot. (SMM) Selinda Research Associates, Inc. 92 Conical Mirror Figure B-11. Conical Mirror. (SMM) Figure B-12. Closer look at Conical Mirror. (DHDC) Selinda Research Associates, Inc. 93 MOVING Build a Room Figure B-13. Build a Room. (SMM) Figure B-14. Labels on light pen. (SMM) Selinda Research Associates, Inc. 94 Figure B-15. Revised exhibit label included a photo of a visitor using the light pen. (SMM) Selinda Research Associates, Inc. 95 Spin a Shape Figure B-16. Spin a Shape. (DHDC) Figure B-17. Spin a Shape card at rest and spinning. (DHDC) Selinda Research Associates, Inc. 96 Shape Cubes and Shape Stools Figure B-18. Large Shape Cube (Rotating through Trapezoids). (Explo) Figure B-19. Small Shape Cube with interpretation. (SMM) Selinda Research Associates, Inc. 97 Figure B-20. Shape Stool with large and small Shape Cubes in the background. (DHDC). Selinda Research Associates, Inc. 98 Geometry in Motion Figure B-21. Geometry in Motion. (SMM) Figure B-22. Closer view of Geometry in Motion. Note cylindrical magnets that hold the pieces together (uppermost right). (SMM) Selinda Research Associates, Inc. 99 Drawing in Circles Figure B-23. Drawing in Circles. (SMM) Figure B-24. Closer view of Drawing in Circles. Note blue pencil sharpener. (SMM) Selinda Research Associates, Inc. 100 The Gyroid Figure B-25. The Gyroid. Note the rubberized tiles under and around the structure, the benches set back on the platform, and the rails and cables restricting access. (SMM) Selinda Research Associates, Inc. 101 Figure B-26. Warning label used at the entrance to the Gyroid. (SMM) Figure B-27. Prototype version of the activity label mounted inside the Gyroid. (SMM) Selinda Research Associates, Inc. 102 Figure B-28. Wall-mounted interpretive label explaining the Gyroid structure and discussing real-world examples. (SMM) Figure B-29. Interpretive label mounted on a bench. (Explo) Selinda Research Associates, Inc. 103 Selinda Research Associates, Inc. 104 FITTING THINGS TOGETHER Stack of Stars Figure B-30. Stack of Stars. Note the rubberized tile-covered platform, labels, and surrounding benches. (SMM) Figure B-31. Warning label used at entrance to Stack of Stars. (SMM) Selinda Research Associates, Inc. 105 Figure B-32. Stack of Stars bench label. (SMM) Selinda Research Associates, Inc. 106 Space-filling Blocks (SRDs & TOs) Figure B-33. Space-filling Blocks with its star-shaped blocks (small SRDs) was adjacent to Stack of Stars. (SMM) Figure B-34. Space-filling Blocks (small SRDs) and label. (SMM) Selinda Research Associates, Inc. 107 Figure B-35. The Truncated Octahedron (TO) version of Space-filling Blocks also was adjacent to Stack of Stars. (SMM) Figure B-36. Closer view of the Truncated Octahedron (TO) version of Space-filling Blocks, and label. (SMM) Selinda Research Associates, Inc. 108 Selinda Research Associates, Inc. 109 Stellator Figure B-37. Stellator, located at one corner of Stack of Stars and between the two types of Space-filling Blocks. (SMM) Figure B-38. View inside the Stellator (non-stellated version). (DHDC) Selinda Research Associates, Inc. 110 Tile Designer Figure B-39. Tile Designer. (SMM) Figure B-40. Tile Designer control panel. (DHDC) Selinda Research Associates, Inc. 111 Geometron Figure B-41. Geometron. (SMM) Figures B-42A and B. Interior views of Geometron. (DHDC) Selinda Research Associates, Inc. 112 Dividing Space Figure B-43. Dividing Space. (SMM) Figure B-44. Details of red pegs and blue pucks. (SMM) Selinda Research Associates, Inc. 113 Polyhedra Figure B-45. Polyhedra. (SMM) Selinda Research Associates, Inc. 114 Figure B-46. Examples of objects and labels inside the Polyhedra structure. (SMM) Figure B-47. Information about the Polyhedra artist. (SMM) Selinda Research Associates, Inc. 115 Gear Cube Figure B-48. Gear Cube. (SMM) Figure B-49. Closer views of rotated and un-rotated Gear Cube. (DHDC) Selinda Research Associates, Inc. 116 Geometry Garden Figure B-50. Geometry Garden. (SMM) Figure B-51. Pencil Hexastix (SMM) Figure B-52. Klein bottle (SMM) Selinda Research Associates, Inc. 117 Geometry in the Wild Figure B-53. Geometry in the Wild on the outer side of the Gyroid wall. (SMM) Figure B-54. Closer view of Geometry in the Wild on the outer side of the Gyroid wall. (SMM) Selinda Research Associates, Inc. 118 Figure B-55. Closer view of Geometry in the Wild panels on the outer side of Stack of Stars wall. (SMM) Selinda Research Associates, Inc. 119 Sphere Cube Figure B-56. Sphere Cube. (SMM) Figure B-57. Closer view of Sphere Cube. (SMM) Selinda Research Associates, Inc. 120 EXHIBIT COMPONENT WEAR & TEAR Figure B-58. Worn challenge label inside the Gyroid. (DHDC) Figure B-59. Intermittent lines on the Figure B-60. Damaged table top and Build A Room screen. (DHDC) missing tape. (DHDC) Selinda Research Associates, Inc. 121 Figure B-61. Worn star blocks at Space-filling Blocks. (DHDC) Figure B-62. The dividing lines at Dividing Space were no longer equidistant between pegs. (DHDC) Selinda Research Associates, Inc. 122 Figure B-63. Paddles were missing reflective paint at Spin a Shape. (DHDC) Selinda Research Associates, Inc. 123 Appendix C. Annotated Bibliography of Reports & Presentations This appendix lists and briefly describes the evaluation and research studies undertaken as part of the Geometry Playground project. For more background, see the sections, Overview of Project Evaluations and Research Component, in the body of the report. For many of the following and additional Geometry Playground reports, see: Playground&BrowseSubmit=Browse&Authors=Any&Type=Any Front-end Study Hido, N., & Gutwill, J. (2007). Adventure Playground – Front-end study. Unpublished manuscript, The Exploratorium, San Francisco, CA. This study took place at Adventure Playground, “an outdoor facility at the Berkeley Marina, featuring staff and unique play experiences that encourage children to play and build creatively using simple tools.” The authors looked for approaches to developing exhibits and environments that foster intergenerational play involving adults and children. Formative Studies Nearly 50 formative evaluation studies were conducted on approximately 20 of the Geometry Playground exhibits. As of the writing of this report, several of these studies were to be posted to the Exploratorium’s website: As a partner institution, the Science Museum of Minnesota undertook four formative studies (with a focus on remediation) of labels for the Gyroid and Sphere Cube. Research Presentations and Reports Dancu, T. N. (2010). Designing exhibits for gender equity. Downloaded from the Exploratorium website July 5, 2011. Dissertation focused on developing gender equitable exhibits that promote use of spatial reasoning skills. Dancu, T., & Gutwill, J. (2010). Getting inside immersive exhibits: A study of Exploratorium's Geometry Playground. Downloaded from the Exploratorium website July 5, 2011. Poster session describing research that compared visitor experiences at immersive and table-top exhibits. Selinda Research Associates, Inc. 124 Dancu, T., Gutwill, J., & Sindorf, L. (2009). Geometry Playground Pathways study. Downloaded from the Exploratorium website July 5, 2011. Report on a study that explored the effects of in-school and museum-visit activities (termed Pathways) on San Francisco elementary school students’ attitudes toward geometry and spatial reasoning abilities. Gutwill, J. (2010). Fostering and assessing spatial reasoning. Presentation at the annual meeting of the Visitor Studies Association, Phoenix, AZ. Downloaded from the Slideshare website July 5, 2011. This presentation first described the Geometry Playground project and its learning goals. Then it defined and gave examples of spatial reasoning and discussed the initial stages of the research on spatial reasoning at Geometry Playground exhibits. Gutwill, J. (2011). Spatial Reasoning in the Geometry Playground. Presentation at the annual meeting of the Visitor Studies Association, Chicago, IL. Downloaded from: Presentation that described the measurement of verbal spatial reasoning and the results from a research study comparing visitors’ verbal spatial reasoning at immersive and table-top exhibits. Gutwill, J., Dimond, E., Boyd, K., Dancu, T., Hido, N. and Gilbert, H. (2011). Exploration of Immersive Exhibits: How, Why and What Impact? Presentation at the annual meeting of the Visitor Studies Association, Chicago, IL. Presentation of an overview of the Geometry Playground project with a focus on immersive exhibits: challenges for prototyping them, strategies for labeling them, and results from a research study of the visitor experience at immersive exhibits compared to table-top exhibits. Summative Evaluation Site Visit Summaries These brief summaries were written specifically for the Geometry Playground team and were not intended for wider distribution. Gyllenhaal, E.D., & Perry, D.L. (2010). Site visit report: Geometry Playground at the Exploratorium. Unpublished manuscript, The Exploratorium, San Francisco, CA. This summary of findings from the first site visit to the exhibition included recommendations for remediation of the exhibits as well as preliminary findings about parent-child interactions at the climbing structures, visitor flow through the exhibition, and visitors’ perceptions about the exhibition as a whole. Selinda Research Associates, Inc. 125 Gyllenhaal, E.D. (2010). Geometry Playground summative evaluation: Site Visit 2 Summary Science Museum of Minnesota. Unpublished manuscript, The Exploratorium, San Francisco, CA. This summary, from the second site visit (the first of two visits to SMM) compared visitor engagements on the two floors of this installation, looking especially at how visitor engagements at large and medium immersives affected their visits to the rest of the exhibition (termed “spillover effects”). It also analyzed preliminary findings about spatial reasoning at the exhibits. Gyllenhaal, E.D. (2011). Geometry Playground summative evaluation: Site Visit 3 Summary Science Museum of Minnesota. Unpublished manuscript, The Exploratorium, San Francisco, CA. The summary from the third site visit (the second of two site visits to SMM) expanded on themes discussed in the previous summary, including findings about possible positive spillover effects from engaging with the large immersives and looking more closely at roles played by labels and engagements with and spatial reasoning at table-top interactives. Gyllenhaal, E.D. & Perry, D.L. (2011). Geometry Playground summative evaluation: Site Visit 4 Summary Don Harrington Discovery Center. Unpublished manuscript, Selinda Research Associates, Inc., Chicago. This summary from the fourth and final site visit further expanded on previous themes. This summary was for internal use only to facilitate data analysis, and was not distributed to the Explo team but was incorporated into this final summative evaluation. Summative Evaluation Selinda Research Associates. (2012). Summative evaluation of Geometry Playground. Unpublished manuscript, Exploratorium, San Francisco, CA. This report is the complete final summative evaluation of the Geometry Playground exhibition at its first three venues. It can be found on www.informalscience.org. Selinda Research Associates, Inc. 126 Appendix D. Topical Framework Attached to the Geometry Playground Summative Evaluation Plan Selinda Research Associates, Inc. Dated June 17, 2010 Description and Context - How many exhibit units were there? - What were they, how were they positioned relative to one another, and how were some of them “paired”? - What were the layouts for the exhibition at the two museums, and how were they the same/different? - What were the intentions and hoped for outcomes for each unit? - What was the exhibit development process and how did the goals, intentions, and anticipated outcomes evolve over time? - In what ways did Geometry Playground differ from other Exploratorium exhibits in terms of intended audience, intellectual content, types of exhibits included, and overall design? - Which units were installed in the outdoor spaces: San Francisco playground, San Francisco park, and Ithaca Science Center? What was the layout and overall context at each location? Visitor Engagements Physical - What was the range of ways visitors used the exhibition as a whole? How did visitors move through the exhibition? How did they move between large and small exhibits? Between active, playful experiences and more passive engagements? Between exhibit content pairs? - Did visitors tend to stay in the exhibition and engage with most of the units, or did they move in and out, leave and return? - What was the range of ways that visitors engaged with individual exhibit units? In what ways and to what extent did visitors use the units to physically engage with shapes, spaces, and geometric processes and phenomena18? - In what ways were visitor physical engagements the same and different between large and small exhibit units? - What was the nature of the whole body experiences, and in what ways did these enhance or detract from the intentions of the exhibition? - What was the range of ways in which visitors immersed themselves in the larger units? In what ways was visitors’ immersion active and physical (like a playground space)? What other, less active forms of immersion took place? - In what ways were physical engagements for girls and boys, women and men the same and different throughout the exhibition? To what extent were these different engagements equitable? - In what ways were physical engagements for children and adults the same and different throughout the exhibition? 18 These may be some of the physical engagements that play a part in spatial reasoning. Selinda Research Associates, Inc. 127 - How did the physical engagements of the users of the playground and park installations compare and contrast with visitors’ use of the exhibit units? - In what ways and to what extent did visitors and users incorporate activities from the website into their engagements with the units and structures (both at the museum and in outdoor settings)? Did any visitors find and use the website before coming to the exhibition? If so, how did the website contribute to their exhibit experiences? Social - In what ways and to what extent were visitors engaged in teaching-learning (broadly defined) activities? - How were visitors jointly constructing meaning? - What forms of collaboration did visitors engage in, and how did this collaboration manifest itself, physically and intellectually? - In what ways were visitor social interactions the same and different between large and small exhibit units? - In what ways did social engagements vary when visitors’ immersive experiences were active and playground-like as opposed to more passive? - What was the nature of adult-child interactions, especially those between children in the target age range (5-12 years) and their adult caregivers? In what ways were these the same and different between different exhibit units? - In what ways were social engagements for children and adults the same and different throughout the exhibition? - In what ways and to what extent did adults behave differently when they perceive certain exhibits (and certain behaviors) to put their children at risk? Did they stay closer, exert more control, or behave in some other way? - In what ways were social engagements for girls and boys, women and men the same and different throughout the exhibition? To what extent were these different engagements equitable? - How did the social engagements of the users of the playground and park installations compare and contrast with visitors’ use of the exhibit units and structures? Intellectual - In what ways and to what extent were visitors’ minds engaged? - What intellectual processes were they using? Comparing and contrasting? Observing? Making connections? Spatial reasoning skills? - In what ways and to what extent were visitors’ intellectual engagements grounded in their physical engagements with the exhibits? - In what ways were visitor intellectual engagements the same and different between large and small exhibit units? - In what ways did intellectual engagement vary when visitors’ immersive experiences were active and playground-like as opposed to more passive? - In what ways and to what extent did visitors make intellectual connections between the designed exhibit content pairs? - In what ways were intellectual engagements for girls and boys, women and men the same and different throughout the exhibition? To what extent were these different engagements equitable? Selinda Research Associates, Inc. 128 - In what ways were intellectual engagements for children and adults the same and different throughout the exhibition? - How did the intellectual engagements of the users of the playground and park installations compare and contrast with visitors’ engagements with the exhibit units? Emotional - In what ways and to what extent were visitors emotionally engaged? - In what ways did visitors emotionally engage with the beauty and aesthetic aspects of geometry as portrayed in the exhibits? - In what ways were visitor emotional engagements the same and different between large and small exhibit units? - In what ways did emotional engagement vary when visitors’ immersive experiences were active and playground-like as opposed to more passive? - In what ways were emotional engagements for girls and boys, women and men the same and different throughout the exhibition? To what extent were these different engagements equitable? - In what ways were emotional engagements for children and adults the same and different throughout the exhibition? - How did parents and caregivers feel about exhibits that appeared to put their children at risk? How did children feel about the same exhibits? - In what ways and to what extent were visitors’ emotional engagements colored by their earlier experiences with school geometry and mathematics? - How did the emotional engagements of the users of the playground and park installations compare and contrast with visitors’ use of the exhibit units? Visitor Learning Outcomes - In what ways and to what extent were visitors “learning” geometry and math-related ideas and skills? - In what ways and to what extent did visitors make connections between geometry and the real world (e.g., in nature, architecture, and everyday life)? - What else were they “learning?” - In what ways and to what extent was there evidence of visitors developing and practicing spatial reasoning skills? What did those skills look like? In what ways did physical, social, intellectual, and emotional engagements interact to realize each skill? (As the study proceeds, we will probably develop a taxonomy of spatial reasoning skills as practiced in the exhibits.) - For each exhibit, which of the intended learning outcomes appeared to be achieved, and which were not?19 - In what ways did visitors perceive this as an exhibition (or exhibits) about experiencing and investigating shapes and spaces?20 About the kinds of geometry they may have been exposed to in school settings? About mathematics in a larger sense? - How did these perceptions evolve during their time in the exhibition? 19 What does this mean? These refer to outcomes stated or implied in labels and Exploratorium documentation. 20 We’re trying to think about how a visitor might talk about “spatial reasoning.” Selinda Research Associates, Inc. 129 - How did visitors’ connections with geometry and mathematics evolve as a consequence of engaging with this exhibition? - How did visitor interest in and attitudes towards geometry and mathematics change or evolve as a consequence of engaging with the exhibition or some part of it? - What was the role of the whole body immersive experiences in helping visitors change the ways they think and feel about geometry, spatial reasoning, and mathematics? How did immersive experiences differ from what’s experienced by those who stand back and watch? - In what ways did learning outcomes vary when visitors’ immersive experiences were active and playground-like as opposed to more passive? - When visitors/users made multiple visits and used the exhibition/structures repeatedly, how did their thinking evolve over time? - In what ways were learning outcomes for girls and boys, women and men the same and different throughout the exhibition? To what extent were these different engagements equitable? - In what ways were learning outcomes for children and adults the same and different throughout the exhibition? - What new ways of thinking or understanding did visitors/users leave the exhibition/structures with? How did visitors'/users' mathematical reasoning skills evolve as a result of interacting with the exhibition? How did their perceptions of and understandings of what geometry is and does evolve?21 Visitor Motivations - In what ways and to what extent were the six visitor motivations engaged: communication; curiosity; confidence; challenge; control; and play?22 - In which sections or at which exhibit units were different motivations more or less engaged? - In what ways were visitor motivations for girls and boys, women and men the same and different throughout the exhibition? To what extent were these different motivations equitable? Design & Installation - What aspects of the design of different units and sections contributed in what ways to meaningful visitor experiences? - How did the design of the large units encourage—or discourage—physical immersion? - What was the nature of the visitor experience when visitors—or some visitors—did not climb inside and immerse themselves in the large exhibit units? 21 Can you say more about what new ways of understanding might look like? Based on what we’ve seen of the labels and exhibit goals, visitors may come away understanding geometric terms that were unfamiliar to them, with new or revised understandings about geometric concepts, or with new ways of thinking about the subject of geometry. For instance, they may come away thinking of geometry as beauty or art, or talking about geometry as something that surrounds them in their lives or as something that is "apart" from their lives, something special. 22 Is this based on someone’s work that I could read? This is Deborah’s model of intrinsically motivating museum experiences, also known as, “What makes learning fun?” See for references and links. Selinda Research Associates, Inc. 130 - In what ways and to what extent did the design of “paired” exhibit units contribute to visitor experiences? - What about the design of the website? What aspects will likely contribute in what ways to meaningful user experiences? - What was the nature and likely usability of the user interface and overall design of the website?23 - How did the layout of the exhibition differ between locations, and how did this contribute to (or detract from) visitor experiences? - In what ways did the design and layout contribute to maintenance challenges? - In what ways and to what extent did different design elements work the same and differently for girls, boys, women, and men? For children and for adults? - In what ways and to what extent did the differences between Geometry Playground and other Exploratorium exhibitions manifest themselves in visitors’ experiences? Did it seem to attract a younger audience, and did that audience behave in different ways in the big, high energy exhibits? Did the cohesiveness and visual identity of Geometry Playground seem to have an impact of visitors’ experiences? In what other ways did Geometry Playground seem to differ from a more traditional Exploratorium experience? Influence on the field - What evidence was there that involvement with the Geometry Playground project advanced museum staff thinking about immersive exhibits and the ways in which they can engage museum visitors? - What evidence was there that involvement with the Geometry Playground project advanced museum staff thinking about exhibits that stimulate whole-body, active play—more like a children’s playground than a traditional science exhibit? For instance, what were perceptions and levels of comfort about the risks involved with playground-like structures? - What evidence was there that involvement with the Geometry Playground project advanced museum staff thinking about large-scale exhibits (like the large playground structures)? - What evidence was there that involvement with the Geometry Playground project advanced museum staff thinking about geometry and/or math exhibits? - In what ways did the pairing of large and small exhibit units influence museum staff thinking about the design of museum exhibitions? Travel-ability - How well did the exhibition travel to SMM? - What challenges did SMM staff face as they moved and installed the exhibition? What aspects of the traveling exhibition helped transport and installation go smoothly? - In what ways and to what extent did the layout of the exhibition at SMM differ from the layout at the Exploratorium and/or the recommended layout, and why? - What unanticipated challenges related to travel-ability were there at SMM? Lighting? Floor treatments? Noise and sound-proofing? Other? - Were there any maintenance issues specific to traveling the large-scale immersive structures? 23 Can you describe how you assess likely usability? See the description of heuristic evaluation/critical review of the website in the Evaluation Plan. Selinda Research Associates, Inc. 131 Selinda Research Associates, Inc. 132 Appendix E. Sources of Data Table E-1. Overview of sources of data Data type Groups Description Contact hours Unobtrusive observations 104 Unobtrusive observations of visitors in Geometry Playground. 59 Participant observations 3 Observations where the researcher took part in the engagement as a member of the group. 1.5 Extended observations 26 Observations of a series of visitor groups engaged at a single exhibit. 27.5 Incidental observations 17 Brief, informal observations to capture fleeting but revealing engagements in the exhibition. 4.25 Census on climbers 2 Periodic counts of children on climbers, identifying genders and classifying behaviors. 2.5 Depth interviews with visitors 43 Interviews of visitors after they had engaged with some or all of Geometry Playground. 27 195 Subtotal (visitors): 121.75 On-site interviews with staff 21 Interviews and less formal conversations with staff and volunteers during site visits to host museums. 20 Phone interviews with staff 10 Phone interviews with staff about the "travelability" of Geometry Playground. 23 31 Subtotal (staff): 43 226 Subtotal (all respondents): 164.75 Document reviews 40 documents Reviews of marketing signage, Web pages, and publications related to Geometry Playground. 20 40 Subtotal (documents) 20 TOTAL 184.75 Contact hours include both time spent on the floor with visitors or with staff, and time spent on the written debriefs of the encounters (including time back at the hotel after hours). It does not include time completing debriefs and analyzing recordings after the site visit. Selinda Research Associates, Inc. 133 Appendix F. Description of Respondents Table F-1. Respondents observed and/or interviewed Group Type Number of Groups Number of Individuals Groups with children 287 748 All-adult groups 20 42 Individual visitor (alone) 16 16 School/camp/birthday groups 31 789 Museum staff na 25 TOTALS: 354 1,620 Table F-2. Staff interviewed for this study Institution # Phone interviews # On-site interviews # Individual Staff interviewed Staff Roles Exploratorium 5 6 11 Exhibition Services staff, exhibit developers, installation supervisors, evaluators, administrative staff Science Museum of Minnesota 2 7 6 Special Exhibits staff, curatorial staff, evaluators, floor staff and volunteers Don Harrington Discovery Center 3 8 8 Exhibit and maintenance staff, program and visitor experience staff, administrative staff, floor staff Total: 10 21 25 Selinda Research Associates, Inc. 134 Appendix G. Installations at the Host Science Centers This appendix lists and has photographs of exhibits that were installed at each of the three host museums during our site visits. Exploratorium Table G-1. Geometry Playground exhibits at Explo – more or less from northeast to southwest Projected Puzzle Build a Room Stellator Distorted Drawing Drawing in Circles Hopscotch Shape Cubes Geometry Garden Stack of Stars Space-filling Blocks Stellated Rhombic Dodecahedrons (small SRDs) Truncated Octahedrons (TOs) Tile Designer Dividing Space Polyhedra Geometron Gyroid Distorted Chair Conical Mirror Entrance Structure Geometry in the Wild Shape Stools Gear Cube (was installed during our site visit) These exhibits were not on the floor during our site visit: Spin a Shape Geometry in Motion Small Shape Cubes Selinda Research Associates, Inc. 135 Figure G-1. Entrance area at Explo, including the Entrance Structure (on the right) and Projected Puzzle (on the left). Figure G-2. Hopscotch was on a raised platform at Explo, adjacent to the east wall of the museum. Selinda Research Associates, Inc. 136 Figure G-3. Shape Cubes along the edge of the exhibition, separated from the Café (in the background) by a loosely configured pathway. Figure G-4. Stack of Stars at Explo. Note benches and rails around the exhibit. Selinda Research Associates, Inc. 137 Figure G-5. The star-shaped Space-filling Blocks (SRDs) were immediately adjacent to Stack of Stars, separated from the climbing structure by a low rail and bench. Figure G-6. Space-filling Blocks (TOs) were adjacent to the ramp for Stack of Stars, separated from the climbing structure by the ramp and a low wall. (Stack of Stars was off to the left.) Selinda Research Associates, Inc. 138 Figure G-7. Stellator was adjacent to the Hopscotch exhibit, with several exhibits between it and Stack of Stars. Figure G-8. Distorted Chair with the Gyroid in the background. It was on the opposite side of the exhibition from the other units with cylindrical mirrors. Selinda Research Associates, Inc. 139 Figure G-9. The Gyroid at the Exploratorium. Note labels at entrance to platform and labels at benches. Figure G-10. Geometry Garden, with Gear Cube in the foreground. Selinda Research Associates, Inc. 140 Figure G-11. Geometry Garden from the other side, with Distorted Chair and a Shape Cube in the background. Figure G-12. Poster for Geometry Playground outside the main entrance. Selinda Research Associates, Inc. 141 Figure G-13. Visitor Orientation Map near the Exploratorium entrance. Geometry Playground was located in the skylight area, adjacent to the Café. Note also Geometry in Motion film program (left) and Geometric Threads exhibition (right) advertised on the lower panel. Figure G-14. Marquee for the Geometry in Motion Cinema Arts Series. Selinda Research Associates, Inc. 142 Figure G-15. Sign for Geometric Threads demonstrations and temporary exhibits. Selinda Research Associates, Inc. 143 Science Museum of Minnesota Table G-2. Geometry Playground exhibits at SMM – by floor 4th Floor Exhibits (upper level): Entrance [by elevators] Polyhedra Gyroid Stack of Stars Space-filling Blocks Stellated Rhombic Dodecahedrons (small SRDs) Truncated Octahedrons (TOs) Stellator Geometry in Motion Tile Designer Spin a Shape Gear Cube Geometry in the Wild (on both floors) 3rd Floor Exhibits (lower level): Hopscotch Projected Puzzle Build a Room Conical Mirror Shape Cubes Small Shape Cubes Shape Stools Geometry Garden Distorted Chair Dividing Space Sphere Cube Distorted Drawing Geometron Drawing in Circles Geometry in the Wild (on both floors) Selinda Research Associates, Inc. 144 Figure G-16. View of the 4th floor installation from SMM’s central staircase. The Stack of Stars exhibit is in the far right corner, partially obscured by an SMM exhibit about blood circulation. Figure G-17. View of the 3rd floor installation from the bottom of SMM’s central staircase. A T. rex skull (part of a different exhibition) is behind the red support column. Selinda Research Associates, Inc. 145 Figure G-18. View of the 3rd floor installation from a corridor beside the elevators. The T. rex skull separates several geometry exhibits from the rest of this floor’s Geometry Playground. Figure G-19. Entrance Structure at SMM, seen as visitors stepped off the elevators on the 4th floor. This was one of the few places where Geometry Playground was identified as a separate exhibition. Selinda Research Associates, Inc. 146 Figure G-20. “Explore More Geometry Playground” signs on 4th and 3rd floors. These were the only signs that announced the exhibition’s name as one entered from the central staircase, and one of the places that called out the two levels of the exhibition. Figure G-21. Geometry Playground posters behind box office registers. This was one of the additional places where the exhibition was identified as such for museum visitors. Selinda Research Associates, Inc. 147 Figure G-22. Placards inside the elevators called attention to the two levels of Geometry Playground. Selinda Research Associates, Inc. 148 Figure G-23. Screen capture of the SMM’s homepage at the time of the 3rd site visit. Geometry Playground is prominently featured. Selinda Research Associates, Inc. 149 Figure G-24. Screen capture of SMM’s Geometry Playground web page at the time of the 3rd site visit. The information that Geometry Playground is on two levels is near the bottom of the page. Selinda Research Associates, Inc. 150 Don Harrington Discovery Center Table G-3. Geometry Playground exhibits at DHDC – more or less from front to back Hopscotch – just inside the entrance Polyhedra Geometry in Motion Shape Cubes – one Shape Cube was in the hall outside the entrance to the exhibition and one was within the adjacent Joshua’s Journey exhibition Small Shape Cubes – one was installed separately in an exhibit area adjacent to the exhibition, and one was installed as part of its accompanying Shape Cube. Shape Stools Conical Mirror Sphere Cube Geometry Garden Drawing in Circles – in an exhibit area adjacent to the exhibition Gear Cube Projected Puzzle Distorted Drawing Distorted Chair Geometron Stellator Gyroid – only two of the three levels were installed. Stack of Stars Space-filling Blocks Stellated Rhombic Dodecahedrons (small SRDs) Truncated Octahedrons (TOs) Tile Designer Spin a Shape Build a Room Dividing Space Geometry in the Wild (some sections were on the backs of several Distorted Chair and Gyroid walls, but other sections were not installed) The Entrance Structure was not included, but quotation label panels were placed in gallery walls in several places. An additional anamorphic mirror was installed in the middle of the exhibition, near the large climbing structures. Selinda Research Associates, Inc. 151 Figure G-25. Entrance to the exhibition hall that included Geometry Playground. Figure G-26. Hopscotch was right inside the entrance, separated from other Geometry Playground exhibits. Selinda Research Associates, Inc. 152 Figure G-27. The Gyroid was two levels tall at DHDC, and the Gyroid and Stack of Stars were installed on the same platform. Figure G-28A. Labeling for the climbers was minimal. There were no bench labels and only two wall labels (for the Gyroid, foreground). There were warning labels for both climbing structures. Selinda Research Associates, Inc. 153 Figure G-28B. At DHDC, the circular activity labels that had been piloted at SMM were riveted onto the Gyroid surface. Figure G-29. Both Shape-filling Blocks exhibits (SRDs and TOs) were immediately adjacent to the Stack of Stars platform. (The Entrance Structure panels and the Dividing Space exhibit are in the background, against the wall.) Selinda Research Associates, Inc. 154 Figure G-30. Northeast corner of the exhibition, showing Distorted Drawing and Distorted Chair, Geometron (far left), and Geometry Garden (far right). Figure G-31. Projected Puzzle was near the back of the exhibition, and sections from the Entrance Structure (including the exhibition title) were mounted on the back wall. Selinda Research Associates, Inc. 155 Figure G-32. Geometry Garden was split in two and installed on either side of a doorway into another exhibition. One side of some of the Geometry Garden was close to a wall, with only a few feet to spare. Figure G-33A. Drawing in Circles (on the back wall towards the left) and the small triangle Shape Cube (against the glass wall to the right) were installed in an adjacent exhibit area. Selinda Research Associates, Inc. 156 Figure G-33B. One of the flashlights used as a light source for Build a Room during our site visits to DHDC. (Compare with Fig. B-14, the light wand that had been damaged by visitors.) Figure G-34. Signs at the entrance to DHDC announced other traveling exhibitions at DHDC (The T. rex Experience, Joshua’s Journey, Risk, and Bounce). Selinda Research Associates, Inc. 157 Figures G-35A and B. Signage for Geometry Playground at the museum included a donor panel at the gallery entrance and small posters in the men’s room. Selinda Research Associates, Inc. 158 Figure G-36. Home page of the DHDC website during March 2011. This page, and the location of the Geometry Playground information, changed frequently during the run of the show. Note that there were four other temporary exhibitions at DHDC during Geometry Playground’s run. Selinda Research Associates, Inc. 159 Figure G-37. DHDC website’s Geometry Playground page. Note that there was no link to the Exploratorium’s Geometry Playground website. Selinda Research Associates, Inc. 160 Appendix H. Inventory of Engagements at the Large Climbing Structures Table H-1. Inventory of engagements by children We observed these engagements and behaviors by children at the Gyroid and Stack of Stars. On the platform and on or within the climbers: Exploratory engagements (and behaviors): Glance at the label at the entrance. Walk around the structure and look for a place to start climbing. Climb up the outside of the structure. Climb through the inside of the structure. Try multiple entrances to the structure (sometimes small children searched for one they could enter without help from parents). Try multiple routes through the structure. Read a wall mounted label. Playful engagements (and behaviors): Sensory: Slide down (the Gyroid). Jump off the top of the structure. Pretend to be injured by the fall. Games/Challenge: Play tag. Race to the top. Chase another child through the structure. Try to avoid being caught by a child who is chasing you. Challenge another child to do what you just did. Fantasy: Play fight (often involved chasing as well as martial arts). Other engagements (and behaviors): Narrate your own experience: Talk about you are doing and what you are going to do next. (“I’m gonna go there.”) Often there was no obvious audience for this. Seek recognition: Shout out your accomplishments to your caregiver (“Look at me!” “Take my picture!”) or to everyone else (“I’m king of the world!”) Other: Take off your shoes and leave them at the entrance. (We saw this on one day at Explo and several days at DHDC.) (Were they thinking about the rules enforced at McDonald’s Playlands, which are also indoor playgrounds?) Verbal assistance: Help another child climb with verbal instructions. (“You gotta hold on…”) Physical assistance: Help another child who has a problem or gets stuck. Look below to try to avoid stepping on the other kids Selinda Research Associates, Inc. 161 Sit on or stand beside a bench and: Engage with their caregivers about their climbing experience: Describe what they had done, especially their accomplishments Tell parent something they discovered. Ask a question related to the exhibit. Tell parent what they wanted to do next. Ask parent to take a picture. Ask parent to help them climb. Complain about an injury. Complain about another child’s behavior. Read a label when parent points it out. Lean over the rail and use the small SRDs. Enjoy some own-time (not paying attention to the exhibits): Just rest – look into space. Nap. Look around the area to see what else and who else is there. Use cell phone. Listen to iPod. Eat or drink something. Look at parent’s photos taken earlier during the visit. Talk with people on the bench about other topics, like where to go next and when to eat lunch. Wait for the rest of your group to meet you there. Selinda Research Associates, Inc. 162 Table H-2 Inventory of engagements by adults We observed these engagements and behaviors by adult caregivers at the Gyroid and Stack of Stars. Stand on the platform and: Engage in active parenting: Teaching/learning: Read aloud/translate label at entrance. (“That’s called a JI-roid.”) Talk to child about what they are doing. Give verbal instructions to a child, such as where to put their feet. Reply to questions from a child. Provide an intellectually scaffolding experience. Care giving/nurturing: Spot for child – stand behind, ready to catch them if they fall – a kind of physical scaffolding. (Usually done with younger child; more often seen at Stack of Stars, where children tended to be younger on average). Climb inside (the Gyroid) to boost a small child to the next level. Rescue – pick a child off the structure when they seem insecure. Help a small child find the best place to begin exploring (such as a low opening into the structure). Search for things a child dropped inside the structure. Choose not to help a child (e.g., a really small child who wants to climb but can’t do it on their own) to prevent them from getting hurt. Note the age limits on the entrance signage and forbid a child from climbing— usually moving to a different part of the exhibition. Recognition: Admire/affirm a child’s accomplishments. (“Woo hoo, you did it!” from a parent.) Take photos of a child on the structure. Make videos of a child on the structure. Enjoy some own-time (not paying attention to children on the structure, but sometimes engaging with the exhibit): Walk around the structure and examine it from several perspectives. Read label on short wall section. Try climbing the structure themselves (although sometimes children asked adults not to climb on Stack of Stars). Use cell phone. Selinda Research Associates, Inc. 163 Sit on a bench and: Engage in active parenting: Watch child on the structure – just curious, or just in case. Respond to child’s queries and requests (calling from the climbing structure). Admire child’s accomplishments (“Look at me, I’m at the top.”). Take photos of child (usually when they are at the top, part of admiration?). Take videos of child climbing. Call out instructions, often safety related. Provide a challenge (“Let’s see you climb that!”). Be there for periodic check-ins by child who leaves the structure – talk to them about what they did or ask them questions (“Was it hard?”). Provide a knee for child’s head as they rest (one child fell asleep). Feed or nurse a child. Direct child’s visit when they seem ready to leave the structure – “There’s another one over there” (At Stack of Stars, referring to Gyroid). Enjoy some own-time (not paying attention to children on the structure): Read a label that’s right in front of them. (Only possible at Explo; this sometimes led to active parenting.). Lean over the rail and use the small SRDs (this often led to active parenting). Read museum brochure (this sometimes led to active parenting). Talk with other adults – usually in their own group. This was often about planning the visit – when and where to eat, where to go next. (So this may really be parenting). Just rest – look into space or close eyes. Nap. Look around the area to see what else and who else is there (this sometimes led to active parenting). Use cell phone/iPad. Listen to iPod/music player. Read a newspaper or book. Eat or drink something. Put on lipstick. Check photos taken earlier during the visit. Wait for the rest of your group to meet you there. Leave your child on the climbing structure and go elsewhere: Explore other exhibits in Geometry Playground. Explore elsewhere in the museum. Take another child to the restroom. Get some food or drinks. Selinda Research Associates, Inc. 164 Appendix I. Taxonomy of Spatial Reasoning Skills This taxonomy includes skills described in documents related to the project including the original grant proposal, and the exhibit development document GP Exhibits Learning Goals (revised after the exhibition opened). The descriptions of the skills and the structure of the taxonomy have been revised numerous times based on findings that emerged during data collection and analysis. Table I-1. Taxonomy of Spatial Reasoning Skills Skill Description/explanation (and source) Examples from the Geometry Playground exhibits Things you do primarily with your senses, hands, body (while mentally engaged): Physical and perceptual explorations of… ...shapes Perceiving or exploring shapes “with your hands or your body” (GP Exhibits Learning Goals). Respondents explored the Gyroid with their bodies; 3-dimensional shapes with their hands/arms at Space-filling Blocks and Build a Room; and 2-dimensional shapes with their hands at Distorted Drawing. ...locations of objects Perceiving or exploring positions of objects “including body, relative to background or another object” (GP Exhibits Learning Goals). With Dividing Space, respondents noticed locations of pegs and sliders relative to edges of colored shapes. ...size Perceiving or exploring sizes of similar or dissimilar objects; perceiving the size of something compared to a figurative “yardstick” (GP Exhibits Learning Goals). Some respondents noticed the size differences between the small SRDs and Stack of Stars structure. As they prepared to climb the Gyroid and Stack of Stars, children compared the size and reach of their body/limbs with the spaces they had to traverse. ...orientations of objects Perceiving or exploring “the direction of an object or body is facing relative to the background or another object” (GP Exhibits Learning Goals). Manipulating the orientation of blocks was a big part of the Space-filling Blocks experience. Respondents also oriented their bodies to climb comfortably through Shape Cubes and safely/quickly through the Gyroid. Selinda Research Associates, Inc. 165 ...curvature Perceiving, exploring, or constructing curved lines or surfaces (GP Exhibits Learning Goals) Respondents watched what happened to objects viewed in the cylindrical mirrors (curved lines became straight, straight lines became curved). Respondents generated their own curved lines at Drawing in Circles. ...congruence “Comparing objects to other objects” (GP Exhibits Learning Goals). Respondents compared small SRDs and the Stack of Stars structure to note the similarities. One group even built a miniature version of Stack of Stars using small SRDs. ...transformation Perceiving or exploring how things look before and after they have been changed in appearance. (This is, in some senses, a special case of congruence.) Respondents perceived the transformations at Spin a Shape and the cylindrical mirror exhibits. Rotation of physical objects “Aligning shapes or spaces by physically … rotating them” (GP Exhibits Learning Goals). Respondents rotated the TO and SRD blocks as they figuring out how to layer, stack, and build at Space-filling Blocks. Things you do primarily with your brain (with senses engaged): Recognition of spatial relationships among objects We define this as the intellectual correlate of the “physical and perceptual explorations of locations of objects” described above. This set of skills was practiced extensively as respondents drew with the light wand at Build a Room and manipulated pegs and slides (and thus colored polygons) at Dividing Space. Perception of spaces and their properties "Spatial perception [is] the ability to identify spatial relations with respect to participants’ own physical position in the presence of distracting information. The authors suggest that spatial perception tasks require participants to rely, at least partially, on kinesthetic cues." (Dancu, 2010, p.10) One interesting example was the young respondent at the Gyroid who figured out that, “when you climb in one side, you have to climb out at the top to get in the other side.” Selinda Research Associates, Inc. 166 Spatial visualization/ Creation of mental images or mental objects "Spatial visualization is understanding and performing imagined movements of two- and three-dimensional objects." "You need to be able to create a mental image and manipulate it. An image is not a ‘picture in the head.’ It is more abstract, more malleable, and less crisp than a picture.” (Clements, 1998, p. 18-19). Spatial visualization is "building and manipulating mental representations of two- and three-dimensional objects and perceiving an object from different perspectives" (NCTM 2000, p. 41 from Boakes 2009). Many respondents developed mental images/models of Space-filling Blocks (and of the holes they filled) as they honed their skills at moving the blocks into place. Some respondents developed mental images/models of the structures they hoped to build at Space-filling Blocks and Geometry in Motion. Respondents who drew in three dimensions at Build a Room seemed to be working from mental models of a chair. Some respondents developed mental images/models of the Gyroid, which they used for navigation or for de-constructing the structure in their minds. Like physical objects, mental objects can be…. …rotated “Aligning shapes or space by … mentally rotating them” (GP Exhibits Learning Goals). Mental rotation was an important component of the highly skilled placement of blocks that some respondents developed at Space-filling Blocks. One respondent said she wanted to make the Gyroid into a piece of furniture for her home, and she would “turn” it so the smooth, curved places would be better to lay or sit on. …flipped or inverted We define this as mentally making the equivalent of a 180 degree rotation of a [mental] object. That was one way to manipulate Space-filling Blocks as visitors tried to move fit them into place. …transformed Changed “from 2-D to 3-D or from morphed to anamorphed” (GP Exhibits Learning Goals). Respondents seemed to be mentally exploring these sorts of transformations at, for instance, Distorted Chair and especially Distorted Drawing (because of the multiple iterations this exhibit encouraged). Selinda Research Associates, Inc. 167 Dynamic spatial reasoning (objects in motion) Judging attributes where one or more objects are in motion, such as estimating relative velocity and/or arrival time as well as distance and intercept points (Law et al., 1993) Many of the things respondents did as they mis-used the Dividing Space exhibit involved the practice of dynamic spatial reasoning, including “estimating relative velocity and/or arrival time as well as distance and intercept points.” Judging distance and intercept points were also important for the playful activities that took place when the Gyroid was unsupervised, like jumping off and chase games. Recognition/ classification of shapes (based on mental objects) Children develop the ability to recognize, describe, and analyze shapes – including forming mental images or prototypes of them at more advanced levels of development (Clements, 1998). Many respondents identified at least a few shapes, such as the triangular faces on Stack of Stars big SRDs. However, there were also many missed opportunities for practicing this skill (e.g., relatively few respondents identified the shapes used to generate Shape Cubes). Spatial orientation/ Navigation "Spatial orientation is knowing where you are and how to get around in the world; that is, understanding and operating on relationships between different positions in space, especially with respect to your own position" (Clements, 1998). Respondents seemed to be practicing spatial orientation and/or navigation within the Gyroid as, for instance, they figured out the shortest way to get to the top and the best ways to avoid being caught when someone was chasing them. Noticing patterns, developing and applying rules (to spatial sorts of problems) Perceiving regularities in a system that allowed one to predict the outcomes of certain actions. (In some ways this is a special case of spatial inference.) (from the NSF grant proposal) Respondents seemed to be developing their own rules for working with Space-filling Blocks. For one young child it was a simple as putting a TO in a hole and then rotating it until it snapped into place. Older respondents seemed to develop more sophisticated rules involving recognizing which shapes or faces should be facing up or down. Selinda Research Associates, Inc. 168 Planning/ Constructing De-constructing Constructing is creating, building, and assembling parts to make a new whole. Planning is thinking out a construction before starting to build. De-constructing, as used here, is dis-assembling, physically or mentally, in order to better understand a structure. Space-filling Blocks and Geometry in Motion provided many examples of constructing, with or without much advance planning. Several visitors mentally de-constructed the Gyroid into its many identical parts. Spatial inference “Exploring cause and effect, and temporal sequence, relationships” (GP Exhibits Learning Goals). Geometry in Motion was all about causal relationships for many visitors. Selinda Research Associates, Inc. 169 Appendix J. Case Studies of Spatial Reasoning at Selected Exhibits As we gathered data about visitors’ experiences, we were struck by the range and intensity of engagements with spatial reasoning at the different exhibits. To give a richer and more nuanced picture of what these engagements looked like, this section presents case studies of selected exhibits, as a supplement to the findings in the body of the report. These case studies were chosen to illustrate important aspects of spatial reasoning as practiced by visitors at: Hopscotch, Distorted Chair, Distorted Drawing, Space-filling Blocks, and Stellator. Note that in these case studies when we refer to all, most, many, some, and few visitors, we are referring only to those visitors at the particular exhibit, not in the Geometry Playground exhibition as a whole. SEEING Hopscotch Exhibit description here. Photographs here. We observed this exhibit in all three venues, and the placement of it seemed to affect visitors’ use of it. (For example, it was more visible at certain institutions and thus was used by more visitors, see Fig. B-1). Hopscotch seemed to appeal somewhat more to children than adults; some caregivers watched and encouraged their children without immersing themselves in the experience. Although many visitors at the lower end of the target age range and younger just looked at themselves in the mirror and then walked on, most respondents noticed the hopscotch boards and many of those started hopping almost immediately. Few studied the images in the mirror before playing hopscotch. Hopping on the distorted board seemed to require most of respondents’ attention, especially on the first attempt, so many respondents watched their feet more than the mirror. Some needed to be reminded to look in the mirror as they hopped, a role that caregivers often assumed once they noticed the “Look in the mirror” sign (Fig. B-2). Many respondents only hopped on the board once or twice before moving on, and only a few tried both boards before leaving the exhibit. Respondents who hopped on the board where they faced the cylindrical mirror seemed to pay more attention to their reflection as they hopped, since it was harder to maintain balance when hopping in one direction while looking in another. Children at the lower end of the target age range and younger, especially showed a range of understandings about how to hop on the board. Some hopped on only one foot through the whole board; others landed on both feet on the side-by-side squares, which seemed to highlight the distortion of their bodies. As with Distorted Chair (below), respondents’ engagements began as mostly physical and emotional, but here the focus was as much on the physical challenge of hopping on the distorted board as on experiencing the reflections in the mirror. These physical engagements sometimes led to social engagements, such as pointing at the mirror and exclaiming about what happened there, but not as often as at Distorted Chair, perhaps because other members of the group stood back farther from the participants. There also seemed to be less photography here than at Selinda Research Associates, Inc. 170 Distorted Chair. There were few indications of deeper intellectual engagement at this exhibit; for instance, we saw few respondents stop their hopping and study their reflections in the mirror, talk about why the changes were happening with others in their group, or compare the two hopscotch boards with each other. On the Knowledge Hierarchy of Spatial Reasoning Skills (Table 5 in the body of the report), few of the visitors we talked with seemed to get beyond Level 2, Trial-and-Error. As with Distorted Chair, the Hopscotch cylindrical mirror generated two phenomena, one making the distorted hopscotch boards appear normal in the mirror and the other making the upright human body look squished, side to side. Once their attention was called to the mirror, most respondents seemed to pay more attention to the reflections of the boards than to the distortions of their own bodies. To call attention to the second phenomenon, some floor staff at SMM approached visitor groups and recruited a member to sit on the floor, legs out straight, so that everyone could see how his or her legs were distorted in the mirror. Group members often reacted with surprise and laughter, as if this was an aspect of the mirror that they had not attended to before. The developers’ goal for this experience was “to provide a fun, whole-body exhibit that allowed visitors to experience how a large cylindrical mirror changes shapes in the reflection,” and our observations suggested that this happened quite often with the visitors we observed. With the addition of the “Look in the mirror” label, most respondents noticed the transformation of the distorted image of the hopscotch board to its undistorted reflection in the mirror, although fewer attended closely to the horizontally compressed reflection of their own bodies. As intended by the designers, additional spatial reasoning skills practiced here included locating objects in space (finding your body and the board in the mirror); exploring curvature (noticing that what was curved on the floor looked straight in the mirror); and congruence (comparing the distorted and undistorted boards). There were indications that the skills practiced here were mostly perceptual, with few respondents making spatial inferences about what was happening in the mirrors. In other words, few respondents seemed to think about why the mirror did the things that it did—the experience was mostly about noticing the effects, not about explaining them. Although Hopscotch clearly fit the “playground” theme of the exhibition, there were indications that hopping on a distorted hopscotch board was not the best way to get visitors thinking deeply about the transformations that take place in a cylindrical mirror. When moving across the board, it was hard for participants to see how reflections changed in the mirror, and because other members of the group stood back and watched, there were fewer social engagements than at Distorted Chair. Also, although caregivers did some instructing (saying “look in the mirror”), there was less facilitating the experience than at Distorted Drawing (as in telling children what to look for or explaining what they saw). That may be, in part, because this exhibit did not include an interpretive panel, so that parents were on their own when it came to calling attention to and explaining what was going on. Selinda Research Associates, Inc. 171 Distorted Chair Exhibit description here. Photographs here. We observed this exhibit in all three venues, and its location seemed to affect visitors’ use of it. (When it was more visible it was used more often, especially when it was located near a major entrance to the exhibition, see Fig. B-3). This exhibit appealed to both adults and children; all ages engaged with it, with adults immersing themselves in the experience to an extent that exceeded most of the other medium immersives. As they engaged with Distorted Chair, respondents often stood back and looked at the mirror and the chair for a few seconds, usually engaging with the exhibit before looking at the label. Then one or more members of the group sat in the chair, looked at themselves and the chair in the cylindrical mirror. Both those in the chair and other members of the group noticed and often commented on the transformations to both the chair and the respondents’ bodies. Sometimes one member of the group noticed the label and, based on the photo or text, suggested to the person in the chair that they should lie on the chair and stretch out their leg. The resulting transformation often produced positive exclamations (“Cool!”), and then other members of the group said they wanted to take their turn in the chair. Because the label usually faced away from the chair, it could not be read by visitors sitting in the chair; therefore, many visitors who were on their own (or in couples where both sat in the chair) did not experience transformations to their bodies to the same extent as those in larger groups. As with Hopscotch , the engagements began as mostly physical and emotional—it was all about experiencing and enjoying the phenomenon at first. The physical engagements usually led to social engagements, such as pointing at the mirror and exclaiming about what happened there, sharing space on the chair, taking turns, and recording the experience with photos or videos. Respondents who tried to take flash photos were sometimes confronted with a bit of a dilemma, since most found it very difficult to take a photo that included the mirror but not the flash. Some respondents did not respond to this challenge (other than turning off the flash), but others engaged intellectually as they tried to take a flash-free photo of the mirror. The cylindrical mirror at Distorted Chair generated two phenomena, one kind of obvious and funny (taking something normal—a human body—and making it look smashed flat or very short) and the other amazing and thought-provoking (making something distorted—the chair— look normal). The first phenomenon was funhouse-like and immersive, but the second seemed to have the potential to stimulate deeper thinking. That said, most visitors’ experiences with this exhibit seemed to be just a bit beyond the fun house level. They noticed that chair changed appearance in the mirror, but they couldn’t do much beyond that with the chair (unlike their bodies, which they could experiment with in numerous ways). Does noticing that transformation count as spatial reasoning? We think it does, although on a basic level, more about spatial perceptions than spatial inferences. The nice thing about cylindrical mirrors, as opposed to fun-house mirrors, is that they are more predictable and easier to figure out. However, the experience of Distorted Chair did not seem to push most visitors towards deep levels of intellectual engagements, although more visitors seemed to think more deeply here than at Hopscotch . On the Knowledge Hierarchy of Spatial Reasoning Skills (Table 5 in the main report), some visitors Selinda Research Associates, Inc. 172 seemed to be operating on Level 3 as they developed an understanding of how the mirror was deforming their limbs, and then tried new positions or repeated positions they had seen others try while waiting to sit in the chair. One wonders what the experience would have been like if visitors had been able to experiment by moving the chair, as well as their bodies. Would they discover that the distortion was corrected only when the chair was oriented in a particular direction? And what would they make of that? So, the exhibit largely achieved the team’s goal of having visitors to explore geometric aspects of a mirrored cylinder using their whole bodies. Most respondents both perceived the mirror’s transformation of the chair from distorted to normal, and explored the transformations of their own bodies from normal to horizontally compressed as they moved them relative to the mirror. In exploring the transformations, visitors seemed to focus mostly on changes in the shape and relative lengths of their own limbs and the transformation of the curved edges of the chair to the straightened image in the mirror. Some respondents also experimented with various body positions to see what happened, but usually in ways that suggested trial-and-error explorations rather than well thought out experiments to test their ideas about the mirror’s effects. Most visitors did not seem to think much about how the transformation happened—they enjoyed the experience without drawing spatial inferences about the effects of the mirror’s curvature. There were indications that most visitors did not make the conceptual link between this exhibit and the Hopscotch or Distorted Drawing exhibits, unless the links were called to their attention by floor staff or researchers. That seems unfortunate, because thinking about the similarities and differences in these exhibits could have stimulated deeper thinking about the geometric transformations that took place within the cylindrical mirrors. Distorted Drawing Exhibit description here. Photographs here. We observed this exhibit in all three venues. This exhibit appealed to both adults and children; all ages engaged with it, although not equally, since caregivers sometimes stood behind their children and provided support and encouragement. A few caregivers also stayed and tried the activities on their own after their children ran off to other exhibits. Most respondents began their engagements with Distorted Drawing by sitting down in one of the three seats and checking out their options. Those who sat at the flip cards usually looked at three or more cards, often tracing their fingers over the edges of a distorted illustration while looking in the mirror. We sometimes overheard comments about the card images, such as, “The car is backwards.” Those who sat at one of the chalkboards usually attempted one or more drawings, sometimes, but not always, looking to the nearby labels for inspiration. Some caregivers reminded children to look in the mirror and sometimes talked with them about what they were seeing. In contrast to Hopscotch and Distorted Chair, we saw a full range of engagements at Distorted Drawing, including higher levels of intellectual engagement than at the two immersive versions Selinda Research Associates, Inc. 173 of the cylindrical mirror. Social engagements seemed stronger than at Hopscotch, and, although the overall level of social engagement was about the same as at Distorted Chair, we saw more teaching-learning at Distorted Drawing. On the Knowledge Hierarchy of Spatial Reasoning Skills (Table 5 in the body of the report), many of the visitors who stuck with it seemed to be functioning on Level 3, as demonstrated by their drawings on the chalk board. We noted a few instances where respondents noticed the link between this exhibit and Hopscotch. For instance, a girl about 10 years old flipped through several cards with her caregiver, eventually coming to the one with the distorted hopscotch board. The girl started hopping on the distorted board with her fingers. The caregiver recognized the image and said, “It’s like the one over there,” pointing towards the large Hopscotch exhibit. However, we also noted similar situations where respondents encountered the hopscotch card but did not talk about it with others in their group. For instance another 10-year-old and his caregiver found the hopscotch image, and the caregiver said, “You can play hopscotch with your fingers,” but did not point out a link to the larger Hopscotch exhibit. Also, we did not notice any instances where respondents linked the small mirror at Distorted Drawing to the large mirror at Distorted Chair. Perhaps it would have been effective to include a photo of the Distorted Chair, along with some text designed to help visitors notice the link between the two exhibits. Visitors seemed to be achieving the developers’ goal that visitors “experience the geometric effects of a cylindrical mirror and observe how the shape of the mirror affected the reflections.” Visitors also practiced a variety of perceptual spatial skills related to the transformations of images—both the cards and their own work—through reflection in the cylindrical mirror. Both the construction activity using the chalkboard and the finger-tracing of the card images aided visitors’ explorations of curvature and congruence. And, in contrast to the two immersive versions of the cylindrical mirror, more visitors seemed to make spatial inferences about the transformations, especially when they viewed a sequence of cards or created the images themselves. Indeed, the challenge to make a letter T or to write your name in chalk pretty much demanded that visitors make a prediction about what would happen as they drew with chalk, and then test that prediction by looking in the mirror. Selinda Research Associates, Inc. 174 FITTING THINGS TOGETHER Space-filling Blocks Exhibit description here. Photographs here. We watched a lot of really wonderful engagements at these two table-top exhibits. Among other things, these were very motivating experiences. Looking at visitor engagements using the Selinda model,24 we noted that respondents seemed to feel in control, even powerful, and when their control was channeled in the right direction it was wonderful (e.g., building a tall sculpture with the TOs). Curiosity was engaged when respondents noticed the similarity between the small SRD blocks and the big SRDs in the Stack of Stars structure, or when they noticed the term “tessellation” in the labels and thought about other places they had encountered that idea. Challenge was an interesting motivation here; these were not easy activities, but visitors seemed to stay engaged, sticking with them and working things through. There also seemed to be a nice balance of challenge and confidence. Many respondents approached these exhibits in a playful way, although it was often a focused and intense sort of play—more akin to playing a challenging computer game than to climbing around on Stack of Stars. However, some visitors found alternative ways to play with the blocks. For instance, one young boy picked up a small SRD, called it “a bouncy ball,” and threw it hard on the Stack of Stars rubberized surface. (It really did bounce high, at least 4 feet.) That said, there were many indications that Spacing-filling Blocks were fun exhibits when used as intended. As one young respondent said, “These are better than Legos; I’m serious!” There were also indications that developers’ expectations about spatial reasoning at these exhibits were fulfilled. Especially at SMM and DHDC, we saw many groups whose members both built with the small SRDs, and climbed on the structure built of large SRDs. Of this subgroup, many noticed that both exhibits used the same star-like shapes. Some visitors really did explore congruence and size by holding up a small SRD and comparing it with the big SRDs, often calling out the similarities to others in their groups. Constructing with the blocks engaged respondents in explorations of the shapes, locations, and orientations of the blocks and of the spaces in which they fit. Respondents rotated the blocks physically as they moved them into place—often randomly, but sometimes systematically—and with practice some visitors began rotating the blocks mentally before or as they moved them into place. This last observation was particularly interesting to us. Respondents who stuck with this exhibit really did get better at packing the blocks together, and there were indications that some respondents used mental objects to guide their placement of the blocks. For instance, some respondents told us that they could envision how the block should be oriented within the structure before they moved it into place, and some also described a mental image of the hole in which the block would fit. 24 The Selinda model of intrinsically motivating museum experiences includes six visitor motivations: communication, curiosity, confidence, challenge, control, and play (Perry, 2012). For additional references and links, see Selinda Research Associates, Inc. 175 Building from these findings, we developed a hierarchy of spatial reasoning skills that applied specifically to this exhibit (Table J-1). Note that each of these hierarchies stands alone, i.e. the levels don’t necessarily line up directly with the more general Knowledge Hierarchy of Spatial Reasoning Skills (Table 5) or with each other. Knowledge hierarchies by definition vary greatly in the number of levels, as well as the type of engagements at each level. Table J-1. Hierarchy of spatial reasoning skills for putting Space-filling Blocks into place Level 0 – Couldn’t figure out what to do with the blocks. Some young children were at this level. They needed help from their siblings or caregivers to figure out that the blocks were supposed to go in the holes, not on a flat surface. Level 1 – Trial and error, with imperfect fit. These respondents knew they had to fit the block into a depression in the surface, but to achieve that fit they placed the block in the hole and just rotated it repeatedly and randomly until it fit close enough for their standards, although it failed to nest perfectly. We saw a few younger children at this stage; one left the surface with five or six imperfectly placed stars. Note that this was a rule-based system; it was just an unsophisticated one. Here are some examples of imperfect fit: Level 2 – Trial and error, with perfect fit. These respondents took Level 1 to the next step, placing blacks into the hole where they should fit and rotating them more-or-less randomly until they fell into place and nested perfectly. Most children and many adults seemed to be working at this level most of the time. In some ways this was a simple rule-based system, but the rule seemed to be, “rotate the block every which way until it fits.” Mental objects did not influence the strategy for placing the blocks, but mental objects may have played a role in judging success—i.e., it looked and felt right when the blocks fit. Selinda Research Associates, Inc. 176 Level 3 – Rule-based fitting. These respondents fit the blocks by aligning them as they moved them into place, based on a rule or rules about the direction that a part of the block should be oriented. They would pick up a block, look at it, reorient it, and then move it into place. For instance, a six-year-old girl told us that a square always had to be on the bottom of the Truncated Octahedron (TO) block. Other respondents had rules about how to orient the tops of the stars. Although this was mostly a rule-based system, some of these respondents seemed to have an incipient mental image of the way a block should look that they referenced as they moved the block into place. However, some of the rule-based systems were incomplete. For instance, the six-year-old did not have a rule for orienting the sides of the TO blocks, so she had to spin them a bit to achieve perfect fit. Level 4 – Mental object-based placement. There were indications that these respondents had a picture in their heads about how a block should be oriented either before they picked it up or as they moved it into place. That picture or mental object included both the block and the hole it was going to fill—and in at least one case, the respondent told us that he visualized his activity as filling space, which he referred to as three-dimensional tessellation. (He had been reading about tessellation before his visit, in addition to reading the exhibit labels.) There were indications that most respondents who stayed with the blocks for several minutes moved up a level or two on the Table J-1 hierarchy as they played. Children at the lower end of the target age range who stayed with it usually reached Level 2, although they sometimes needed some adult guidance to get there.25 Some of these younger children also reached Level 3, although when asked to talk about the rules they were following, they sometimes had trouble doing so with just words; they often resorting to gestures or demonstrations to complete their explanations. Level 4 was more often reached by children at the upper end of the target age range and older, and adults, who also had trouble explaining how they placed the blocks, since their mental objects were, by definition, based on images rather than words. When respondents could talk articulately about block placement, their deeds sometimes lagged behind their words. For instance, one respondent described the differences between the two sides of the TO exhibit, but when he tried to show us how they were different, he fumbled the block placement. Either his rule for placing blocks was incorrect or he didn’t have enough practice doing it. Placing the blocks was only part of the fun at these exhibits. The second and perhaps more enjoyable challenge was to build structures with the blocks. Indeed that desire to build often drove most visitors’ engagements here. We were able to construct a second hierarchy based on how much planning visitors put into the constructions, especially how much of the structure was based on a mental image or model developed before construction began (Table J-2). 25 See Jeanne Vergeront’s blog post for an example of this: Selinda Research Associates, Inc. 177 Table J-2. Hierarchy of spatial reasoning skills for planning a structure built with Space-filling Blocks Level 0 – No plan. Many visitors started at this level because they first focused on placing individual blocks. Some just placed blocks more or less randomly when the board was wide open, without filling the space between them. Some placed blocks into whatever spaces were available on the board. This level was about fitting shapes rather than making a structure. Level 1 – Fill one layer (or part of one layer) with close-fitting blocks. This is about filling a horizontal planar space rather than fitting shapes into holes. Level 2 – Fill multiple layers, with no larger plan than that. This is still about filling a second level with the idea of building up into the next dimension. Level 3 – Copy an existing structure made by someone else. This was about building a structure, but visitors on this level did not imagine their own structures, they just copied something that they could already see. These visitors worked from a physical image or thing that they were trying to duplicate, not from a mental image or model. The most sophisticated example of this approach that we saw was completed by a team of two fathers and their daughters (about 10 years old). They built an exact copy of the Stack of Stars, block by block, by looking closely at the climbing structure and then placing the small SRDs based on what they saw. Although this was a difficult task and required sophisticated block-placement skills, they kept referring back to the original structure, walking up to it and around it to study it closely. They Selinda Research Associates, Inc. 178 seemed to hold mental images of small parts of the structure in their heads as they constructed them, but they did not seem to hold a mental model of the entire structures. Level 4 – Imagine and plan a simple structure based on a mental image or model. This level was about building something based on an image in the visitor’s head, although the image was a relatively simple one, like a single-row wall, one-level shape, a simple tower (four or five blocks at least), or simple pyramid (with each level of the pyramid smaller than the one below). Level 5 – Imagine and plan a complex structure based on a mental image or model. This level was also about building something based on an image in the visitor’s head, but the image was a more complex one, like a straight-walled tower. Here are examples of some Level 5 structures, built based on complex mental plans: The important thing with the planning hierarchy was not the final product but the way the visitor reached that product. For instance, a pyramid of TO blocks could be Level 2 structure if it emerged more-or-less unplanned as a visitor just kept adding more layers, each less extensive than the one below; a Level 3 plan if the visitor copied someone else’s pyramid; a Level 4 plan if it was invented (or reinvented) by the visitor—i.e., planned in advance based on an idea of what a pyramid should look like. Then there’s the case of the boy, about 10 years old, who told us he was building a Ziggurat (a Mesopotamian-stepped pyramid). He seemed to be starting with an image of a real pyramid, adapting it to the materials at hand and taking into account their limitations. As he explained, “I was thinking about how stable it would be, and how tall I could get it.” Perhaps we should consider that Pyramid (shown below) to be a Level 5 structure. Selinda Research Associates, Inc. 179 Since this is an evaluation study, it’s important to consider if and how the two Space-filling Block units supported visitors on various levels, and if there were missed opportunities that could be included in future manifestations of the exhibit. The following attributes of these exhibits seemed to support visitors moving up the hierarchies: - The two types of blocks—TOs and small SRDs—presented different challenges, with one easier than the other (TOs easier than SRDs). - Placing the blocks was a challenge at first, but even children at the lower end of the target age range were able to get better at it if they applied themselves. It was a “just right” challenge for most children in the target age range. And the challenge naturally evolved as visitors’ became increasingly experienced. - The congruence between the small and large SRDs got visitors thinking and talking about these interesting shapes. - There were lots of blocks to play with most of the time. - The two sides to each exhibit each had a different template (with different shaped holes to put the blocks in). Visitors who figured out one side had to reconsider their rules or mental objects if they started building on the other side. - The idea of tessellation in three dimensions built on a concept that some visitors were already familiar with (tessellation in two dimensions). Are there ways this exhibit could have been better? We can think of a few revisions to the interpretive panel that might be helpful. First, the instructions might be changed so they are more encouraging and conversational. They currently read: Pile up the blocks so they fit together with no space between them. (It may be harder than it looks!) We suggest testing the following revision: Keep rotating the blocks until you get a perfect fit. Fit them together so there’s no space between them. Selinda Research Associates, Inc. 180 The labels also included hints, for example: Imagine what shape would fit into a space on the table, then find that shape on a block. These hints seemed particularly useful since they matched the sorts of spatial reasoning we observed at this exhibit. We suggest highlighting the hints using larger font size and by placing them in the center of the panel (moving the interpretive explanatory text to the right side). Also, the second hint, Turn the block so it points in the same direction as its neighbor, seems like a rule-based approach to block placement. Since rule-based placement came earlier in the block-placement hierarchy, we suggest listing this rule first. The other hint (mentioned above) is a mental object-based approach. That approach comes later in the hierarchy and so should be listed second on the label. Overall, the exhibit teams’ intentions for these exhibits seemed to be fulfilled for most groups of visitors who engaged with this exhibit. By playing with, stacking, and building with the SRD and TO blocks, visitors gained first-hand experience with three-dimensional tessellation. They reasoned in a very hands-on way about the orientation of stacking required to build a structure, and building a required tessellation even when they were not familiar with that concept. And visitors who were successful at these exhibits seemed to leave with good feelings about the experience, including to varying degrees about the exhibit itself, the geometric ideas behind it, and their own abilities at building with novel-shaped blocks. Stellator Exhibit description here. Photographs here. This small, relatively inconspicuous exhibit seemed to attract more attention when it was placed in the flow of traffic (as at DHDC) than when was it was tucked against a pillar (as at Explo or SMM). When visitors noticed the circular window, looked inside, turned the crank, and watched what happened, developer’s intentions were often met. Visitors enjoyed watching the transformation, experienced something they may have never imagined could happen, and often shared their newly developed aesthetic appreciation for this exhibit with others in their group, calling them over and saying, “Look here!” Few visitors used the term, stellation, for what they saw, but they understood that there was a way to grow stars inside that box. In terms of spatial reasoning skills, there were indications that respondents enjoyed controlling and exploring the transformation with the crank, magically moving from the “regular” to stellated shape and back. The congruence explored by visitors was a fairly continuous one— about the process and intermediate steps as much as the end products. We saw relatively little evidence of spatial inference at this exhibit, perhaps because respondents could get the “point” pretty quickly. For most respondents, this was a stand-alone experience, enjoyed but not linked to anything that made them think about it. Few visitors seemed to get beyond Level 2, Trial and Error, on the Knowledge Hierarchy of Spatial Reasoning Skills (Table 5 in the body of the report). However, we did see one case that got us thinking about the possibilities. This took place at SMM, where the exhibit was pushed up against a pillar but more-or-less between the two tables with Space-filling Blocks (Fig. B-37). An older woman was standing near the Space-filling Blocks exhibit, watching the exhibits and the younger adults in Selinda Research Associates, Inc. 181 her group, occasionally informing them of her discoveries (“Here are some more space-filling blocks”) but not building anything herself. She noticed the Stellator case and walked over to give it a try. She looked up and said, “That’s the two different tables—one with the octahedrons and one with the stars.” She wasn’t exactly right—the small SRDs weren’t really formed from octahedrons—but that seemed like a really good and useful connection (and another example of congruence). The exhibit team might consider ways to arrange these exhibits so that more visitors make this connection. Selinda Research Associates, Inc. 182 Appendix K. Screenshots of the Website Figure K-1. Home page for the Geometry Playground website. Selinda Research Associates, Inc. 183 Figure K-2. Example of one of the pages about one of the exhibit themes: Geometry of Seeing. Selinda Research Associates, Inc. 184 Figure K-3. Geometry Garden page. Selinda Research Associates, Inc. 185 Figure K-4. Artists in residence page. Selinda Research Associates, Inc. 186 Figure K-5A. Activities and Links page, upper screens. Selinda Research Associates, Inc. 187 Figure K-5B. Activities and Links page, lower screens. The Links section has not been expanded, thus shows only categories of links. Selinda Research Associates, Inc. 188 Figure K-6A. Fully-expanded Links section on Activities and Links page, upper screens of links. Selinda Research Associates, Inc. 189 Figure K-6B. Fully-expanded Links section on Activities and Links page, lower screen of links. Selinda Research Associates, Inc. 190 Figure K-7. Geometry Playground Flickr page. Selinda Research Associates, Inc. 191 Figure K-8. Video of visitors on Stack of Stars (when it was called Climb the Stars). Figure K-9. Part of Geometry Playground home page as it displayed on an iPad screen. Selinda Research Associates, Inc. 192 Figure K-10. Geometry of Moving page displayed on an iPad screen Figure K-11. Example of a QR code cell-phone-readable webpage link. Selinda Research Associates, Inc. 193 Appendix L. Critical Review of the Website Introduction This critical review of the Geometry Playground website was intended to (a) provide the project’s web development team with feedback on the design and likely effectiveness current website at meetings its goals for its intended audience; (b) make recommendations for the development of a mobile version of the project’s website; and (c) help the website team identify lessons learned from this website that can be applied to the development of future exhibition websites. The intended audience, purpose, and viewer experience for the website are outlined in the body of the report, Description of the Website. The review was conducted by two experienced website reviewers who completed a heuristic review using a format that Selinda Research Associates developed over multiple projects. (The heuristic was adapted from a template developed by Jakob Nielsen: Findings and Recommendations A short summary of the overall findings of the critical review are in the section Design of the Website. This appendix includes detailed discussions of the review findings, along with specific recommendations associated with the findings. Recommendations are written as imperatives, although we understand that funding for revisions are limited at this time. To help the web team prioritize revisions, recommendations are prioritized using three levels. Level 1 recommendations are the things that probably are not going to “cost” a lot, will be relatively straightforward to implement, and will have a big payoff. In other words, our recommendation is that Level 1 recommendations should be done, as soon as possible, even if there is no money and little staff time. Level 2 recommendations are things that should be completed if they possibly can. These revisions will “cost” a bit more, but will also make a big difference in web visitors’ experiences. Level 2 changes might also include those things that might not take much effort to implement, but won’t make that much difference. Level 3 recommendations are more along the lines of “lessons learned.” These are things to keep in mind for next time, or if additional money and staff time becomes available. Selinda Research Associates, Inc. 194 Table L-1. Critical review of the Geometry Playground website Feature Strengths Challenges Recommendations Overall design of this review This was a cool website with rich resources that could serve visitors in important ways. We will point out many beautiful, interesting, and useful features of the website in this “strengths” column (arranged by topic, below) However, there were some usability issues that arose as we explored the website, as well as other issues that might limit the effectiveness of the website for some users. We discuss those issues in this “challenges” column (by topic, below) In addition, there were indications that many exhibition visitors wouldn’t even know this website exists. We also discuss these sorts of “findability” issues in this challenges column. Finally, compared with the exhibition, the website engaged visitors in a more limited range of ways. In particular, it seemed like a pre-social web experience, with interesting information about some aspects of the exhibition, but with no chance to, say, interact with the exhibits’ developers or with other visitors. These sorts of issues are also discussed in the challenges column. To meet the challenges presented in column two, we make specific recommendations in this column. Each recommendation responds to one or more of the challenges posed in the cell immediately to the left. Most recommendations are prioritized into three levels, as discussed above. A few recommendations are aimed at shaping the development of future exhibition websites rather than remediating the Geometry Playground website. Such recommendations are not prioritized. Finding out about the website Googling “Geometry Playground,” the exhibition website was the first one listed on the first page of results (as Geometry Playground Exhibition | Exploratorium) This website was not called out in the exhibition—there were no links or hints to its existence on the exhibit labels. On the SMM’s website, the link to this website was kind of buried within a menu on the left side on an interior page (of SMM’s own Add links to the website home page on exhibit labels. (Level 1) Use a shortened version of the URL in labels (e.g. www.gp.edu or something so it’s really easy to type in). www.geometryplaygrou nd.edu should work, too Selinda Research Associates, Inc. 195 Feature Strengths Challenges Recommendations The same result was obtained with a Yahoo search from an iPhone. There were links to the website on at least one host’s website (SMM). Geometry Playground website). Although two other hosts had Geometry Playground web pages (DHDC) or small websites (Reuban Fleet), neither of these sites linked to the Exploratorium’s Geometry Playground website. There were no links on other publicity signage that we saw at host museums (leave off the “exploratorium” part of the address if at all possible.) (Level 1) Also link to pages within the website where appropriate. For instance, labels on the Artists in Residence exhibits should inform visitors where they can go with their mobile phones to see the artists’ videos. (Level 1) Include website links in the signage/poster/publicity templates provided to host museums. (Level 1) Negotiate with host museums to feature prominent links on their own website (as a service to their visitors and to the educational goals of the exhibition). (Level 2) Include a dedicated computer in the exhibition that connects directly to the website and the sites it directly links to. This could be placed near the climbing structures, so parents could look at it while their children play. This would be particularly wonderful because visitors could see the Artist in Residence Selinda Research Associates, Inc. 196 Feature Strengths Challenges Recommendations videos immediately before or after they see and play with examples of their work. (Even when visitors can connect with these using their mobile devices, it would be nice to see them also on a larger screen.) (Level 3) Consider adopting technologies like QR cell-phone readable webpage links used at other informal education institutions (e.g., Fig. K-11). (Level 2) As these changes are implemented, keep track of how visitors arrive at the website, and what pages they visit on it. Use these results to plan publicity for future exhibition websites. (Level 2) Identifying the purpose of website The homepage graphics and text clearly communicated the geometric aspects of the website and the experiential nature of the geometry playground experience. The first three sentences below the name Geometry Playground (upper left of screen) help At first glance it was hard to tell from the home page that Geometry Playground was a physical exhibition that website users might be able to visit one day. Site users needed to read carefully and scan (and mouse over) the menu choices to see that all the choices related to a traveling exhibition. In the uppermost left of the screen, one could see the name “Exploratorium,” but one had to infer that this is Add a tagline below the homepage title: Geometry Playground, then something like “A traveling exhibition by the Exploratorium.” (Level 1) Switch the first and second sentences on the home page – first tell what it is (an exhibition), then tell how it will “change visitors’ lives.” (Level 1) Selinda Research Associates, Inc. 197 Feature Strengths Challenges Recommendations connect the website to the exhibition (see Fig. K-1 in Appendix K). At the next level of pages within the website (e.g., the exhibit pages, Fig. K-2, and Geometry Garden, Fig. K-3), the purpose of the website (to support the exhibition) becomes clear. their exhibition. The first two sentences below “Geometry Playground” were in the wrong order—the exhibition should be mentioned in the first sentence. The use of font sizes on the upper left part of the homepage was inconsistent. Standardize the font size: The third sentence (i.e., the second paragraph) was larger text on our computers. (Level 1) Relationshi p between website and exhibition The website provided a good overview of the exhibition, especially the pages with slideshows of selected exhibits and exhibit themes (e.g., Fig. K-2) The Geometry Garden page (Fig. K-3) allowed exhibit visitors to revisit some of the striking pieces from this exhibit. The artist-in-residence pages (e.g., Fig. K-4) provided an in-depth experience with the artists who designed several of the exhibits. If exhibition visitors went to the website seeking more information about cool exhibits like the Gyroid or how an anamorphic mirror works might feel frustrated that there is no in-depth information about this exhibits and no links to such information. If a user wanted to locate photos or information about a particular exhibit, like Build a Room or Hopscotch , they would have trouble finding it, because there was no search function specific to this website and no list or catalog of exhibits. (The seeing-moving-fitting together classification was not an intuitive way to search for these exhibits.) Develop an alternative way, beyond the slideshows, to access photos, information, and links about the exhibits. This might be a menu of exhibit photos rather than a list of exhibit names (which visitors might not remember anyway). (Level 3) Track use of this page to see how often website users access it and the individual exhibits it portrays. (Level 3) Selinda Research Associates, Inc. 198 Home page navigation As shown in Figure K-1 (Appendix K), there was not a standard menu bar on the home page. Instead, three major sections of the website (related to three aspects of the exhibition) could be accessed by clicking on segments of a menu graphic on the left/center of the screen. The organization of the three themes was described briefly on mousing over and illustrated by images of people at an exhibit from that section. This was an attractive way of presenting a useful perspective on the exhibits—that geometry is about doing things. The column on the right of the home page gave access to several other sorts of features, with names, photo, and short description (Geometry Garden, Artists in Residence, Activities and Links, and About the Project) in an intuitive kind of way. Following web conventions, a The major challenge presented by the home page navigation was that the vertical left and right menus from the home page (Fig. K-1) were reorganized into horizontal tabs on the internal pages of the website (e.g. Fig. K-2, above and below the title, “Geometry Playground”). There were indications that some users will find this confusing. (See recommendations in the cell below this one.) Selinda Research Associates, Inc. 199 horizontal menu at the bottom of the screen provided links to major parts of the Exploratorium’s main website. Navigating the rest of the website As seen in Figure K-2, the use of tabs at the top of the page for major sections of the website followed standard web conventions and should be familiar to most users. Clicking through these menu choices, the color pallets for each page helped link it to the tab used to access it. Following web conventions, if one clicked on the Exploratorium name/symbol in the uppermost left, they are taken to the Exploratorium’s home page. If one clicked on “Geometry Playground” graphic (between the rows of tabs) you go to the Geometry Playground website home page. As a backup (and for text readers) there was a small “Home” link right under the graphic, so there was the best of both worlds, as it were. The menus that were vertical on the home page (Fig. K-1) become horizontal tabs on the internal pages (Fig. K-2). This lack of consistency was disconcerting and some website users probably won’t immediately make the connection. The Activities and links were on the same page (Figs. K-5 and K-6). Reviewers had to scroll down and click to open “hidden” menus to discover the full palate of links. That meant that most website users will probably miss most links. On many pages the tabs for Geometry Garden and Activities and Links are not linked back to the main page for that section. For instance, when one is looking at photos of artifacts in Geometry Garden, one can’t get back to the Geometry Garden “home” page, other than via back arrows. Note that this type of navigation does work on the three Moving, Seeing, Fitting sections. With the activities, some opened on the same page (of tab), and others (the pdf files) opened a new page (or Do not change the menu/tabs navigation for this website, but keep the horizontal-to-vertical issue in mind for future websites. Develop separate tabs for activities and links— they are really very different functions for the website, and would likely be of interest to different sorts of users. (Level 3) The tabs for every section of the website should always link back to the main or home page for that section. (Level 1) Do not change the conventions for opening and closing the Activity pages for this website, but keep the issue in mind for future websites. Selinda Research Associates, Inc. 200 tab). This got a bit confusing when reviewers tried to get back to the Activities page— should they use the back arrow or close the page/tab? Design details and web conventions The website used well-designed and consistent standards for fonts, navigation, color pallets, and tone and style of writing. The website looked really good! Reviewers found it kind of annoying that the website was not centered on the screen, particularly with rather wide laptop screens. Also, to close the window of a few of the video clips, one had to click in the lower right, rather than the upper right as is standard (see screen shot of Stack of Stars video Fig. K-8.) Center the website on the screen. (Level 1) Standardize the closing of windows for video clips to upper right. (Level 2) Perceived richness and depth of website The website seemed to be rich and deep in several important areas, such as activities, art/aesthetics, and everyday geometry. There was not much depth for most of the individual Geometry Playground exhibits, and there were no links to more information about most of the exhibits. As mentioned earlier, develop an alternative way, beyond the slideshows, to access photos, information, and links about the exhibits, then track use of these pages to see how often website users access them. (Level 3) Graphics Most of the graphics were photos, and these were great and gave a good feel for the exhibits. The inclusion of people in many of the photographs was great. There appeared to be a good balance of still photos and video clips. The slide shows and other aspects of the graphics either did not display well on Apple products like the iPhone and iPad (e.g., Fig. K-9) or could not be displayed at all (e.g., Fig. K-10). The error messages that appeared when this happened were misleading. Some of the photos in the Exhibits photo essays were of prototype versions of the exhibits (e.g., Geometry in For future exhibition websites (and for the mobile website), include an Apple friendly graphic and slide-show format. (Level 3) Update the Exhibits section with exhibit photos from the completed/installed exhibition. (Level 1) Selinda Research Associates, Inc. 201 The line drawings with various activities also seemed clear and very useful. Motion) or otherwise out of date (e.g., the Gyroid at DHDC was two levels high and apparently will not be installed in San Diego). Videos The video clips were interesting and enjoyable, adding useful perspectives to the exhibition experience. Most video clips included the duration of the clip. This is helpful for viewers. A few video clips did not include durations, e.g. the Gyroid and Star Climber loops (see photo of Star Climber at end). The video clips (and the slide-show photographs) could not be accessed via iPhone or iPad. The reviewers would have liked to see videos of the Exploratorium developers who created the Gyroid, Build a Room, and so forth. These folks are educators/artists/engineers in their own right, and telling their stories also would provide visitors with additional information on some of the most popular exhibits in Geometry Playground. (Perhaps they could have even talked about the spatial reasoning skills that visitors practiced at these exhibits.) The time-lapse video of the Stack of Stars (then called Climb the Stars) missed an opportunity to add perspective—it could have shown the stars being assembled, thereby reinforcing the tessellation concept (which some visitors missed in the exhibition). Add durations to the descriptions on all video clips. (Level 1) For future exhibition websites (and for the mobile website), include an Apple friendly video format. (Level 3) For future exhibition websites, include videos about the Exploratorium’s own exhibit developers, too! (Level 3) Change the name on the “Climb the Stars” video to “Stack of Stars” so it is consistent with the exhibition. (Level 1) Make a video of Stack of Stars being assembled at one of the host museums and put it on the website, with appropriate commentary. (Level 3) Website The text on the web Good job! Selinda Research Associates, Inc. 202 text pages was short and to the point and use of geometry terms was pretty minimal and explained in context. Terms used on the activity pages were explained in context, and there was even a sheet identifying basic geometric shapes. Finding appropriate activities There was a link to the “Activities and links” page from the homepage (lower right) and every other page (Activities and links tab). When we searched for the activities from the Exploratorium’s home page, there was a search box in the upper right (good!). Searching “geometry activities,” the first link was the Activities page of the exhibition website. Searching “geometry,” the first activity was the sixth entry on the list of returns (gift shop was first, exhibition website was fifth). So, the Activities page was relatively easy to find from the Explo home page. But what about a teacher just browsing through resources (without knowing the there are geometry resources available)? Starting from the Exploratorium home page, here’s what happened: Ø Clicked top left, “Who are you?” tab Ø Clicked Educators (stayed on default For elementary educators tab at bottom) Ø Looked at For Educators menu on lower right. Nothing obvious, so clicked top choice: Digital library. Ø This page added a new choice to the lower right menu: Activities, so clicked this. Ø This took us to a page labeled EXPLORE. There was a menu of activity thumbnails, and the first five were all Geometry Playground activities (on the exhibition website). We went back to the Because you really have to fight for attention on the Web – and even on your own organization’s website, we recommend: • Add a link to Math and Geometry activities under the Exploratorium home page EXPLORE tab. (Level 1) • Ask a bunch of teacher bloggers and teacher websites to add links to the exhibition’s Activities page, which should improve its Google ranking (plus get some attention directly from these sites’ readers). (Level 2) Fix the broken links. (Level 1) Selinda Research Associates, Inc. 203 Exploratorium home page and noticed that the EXPLORE tab was the second tab on the top, right next to “Who are you?” So we could have gotten to this page much faster if we had used a different route. However, when mousing over the EXPLORE tab on the home page, a list of choices popped up right below it. None of those choices was for math or geometry – but if one clicked on Everyday Science, the first five activities were Geometry Playground. (This same list of choices is available as a menu bar at the bottom of the first screen for the EXPLORE page.) We decided to enter the Exploratorium website as a parent: Ø Clicked top left, “Who are you?” tab Ø Clicked Parent Ø At bottom of page were two choices: explore websites we’ve made and explore things to make and do. The second choice took us to the EXPLORE page with geometry activities as the top five thumbnails/choices. The first choice took us to the same page, but we noticed that there were links to a bunch of websites at the bottom of the page: including the Geometry Selinda Research Associates, Inc. 204 Playground website. Just to check, we went to the Google homepage and searched for “geometry activities” (during April, 2011). The link to the exhibition’s Activities and Links page was on page 4 of the results (10 per page), seventh down. That suggests relatively few teachers will get there by this route. During our explorations, we found one “dead” link. Geometry Playground Home > Geometry of Moving > Walk A Math Trail Activity > [The menu at the top has an additional tab that did not work: “Events”] When a user with a Mac clicked on this link: yground/ Activities/GeometricShapesChart.pdf ….they got a page of gibberish sometimes, even from within the Geometry Playground website. But then other times, it was fine. Design of activities The activities were mostly great ideas, and many were well carried out. We were glad to see that some activities were called out from the exhibit pages and shown as web pages (rather than pdfs). It was good to have The printer-friendly versions of the web-page activities lacked the nice formatting of the pdfs. At least some of the activities did not seem particularly well-designed or tested. For example, the handout to make your own pocket protector did not specify that you need to start with a square piece of paper Make the web-page activities available as pdfs (as the printer-friendly version). (Level 2) We didn’t go through every activity in detail, but we assume that if there are problems on these few that we did look at, there will be problems on others. All Selinda Research Associates, Inc. 205 the printer-friendly versions of the web-page activities in addition to the printer-friendly pdfs for the other activities. within the Try This narrative (only under Materials needed, but many folks will not start there.) The Traveling Networks activity was also confusing in places. If someone didn’t quite get what a point was, or for example that an edge can be curved, it could be confusing. And we were not sure what this meant “If you found that one of these networks can’t be travelled…” We feel like we’re missing something (and we’re not going to take the time to figure it out right now…..similar to a classroom teacher, or parent we’d imagine.) the activities should be reviewed by an editor and tested by educators and then revised accordingly. (Level 3) Searching within the website The Exploratorium’s home page has a search box in the upper right corner that proved very useful. The lack of a search function for the Geometry Playground website was problematic. For example, during initial explorations we came across an activity about making a pocket protractor. We had problems locating it again when we wanted to re-visit the activity. We tried to guess the grade level (Fig. K-5A, but missed the Outdoor Activities on the bottom (Fig. K-5B), possibly because we assumed those would be cross-listed by grade. That would have been much easier if there had been a search box on every Geometry Playground page. (When we went to the Exploratorium home page and searched from there, we found the Pocket Protractor Add a search box in the upper right corner of every page. When there are multiple categorization schemes (e.g. grade level; indoor/outdoor; activities for educators; etc.), a search box is essential. (Level 2) Alternatively create a type of index so each activity can be listed in multiple categories. (Level 3) Selinda Research Associates, Inc. 206 activity right away.) Links to other websites The links on the lower left side of the Activities and Links page (Fig. K-5B) seemed really interesting and useful to anyone with an interest in those topics. The links to geometry artists seemed particularly special— we’re not sure how else visitors could find out that these artists exist. Each of the activity pdfs has a link to the Geometry Playground website activity page at the top of the first page (good!) Because most of these external links are kind buried on the Activities and Links page, and because you have to click on a topic to see the actual links (Fig. K-6A and B), many users will never discover these wonderful resources. There was no section with links to further information about popular exhibits, like the Gyroid, or popular topics, like tessellation. If visitors were interested in learning more about these topics, they had to do so without guidance from the Exploratorium. If these links were included, some users might realize that these are topics they could explore on the web (something they might not have thought of on their own). We did not find a link to the Exhibit Services webpage about Geometry Playground < services/ rentals/geometry-playground/ > or vice versa, and the vice versa may be at least as, or more important. (There was, however, an e-mail address for museums that wanted to rent the exhibit.) Develop a separate page for links, so these are not “buried” under all the activity links. Add this page to the navigation structure of the website. Include links to specific sections on the Link page on various pages of the website. For instance, the Artists in Residence page should include links to not only those three artists’ web pages, but to the other artists listed on the current Activities and Links page. (Otherwise, how are users going to know who to search for on the web?) Include a section with links specifically related to exhibits and the topics they cover. Maybe even include the photos of the exhibits beside the links, for visitors who don’t remember the exhibit names. (Note that the texts accompanied these links could bring some web users to this website as they search for those topics on Google, etc.) Selinda Research Associates, Inc. 207 Serving subgroups of website visitors The website seemed relatively educator friendly. For teachers of various grade levels, the activities on the Activities and Links page were arranged on a menu (on the right side of the screen) by grade level. When you go to a specific activity, the grade level was repeated at the top and bottom of every activity. National Council of Teachers of Mathematics national education standards were at the bottom of some activities. National Council of Teachers of Mathematics national education standards were not included on some activities. Groups such as students, parents, artists, and geeks were not called out on the Geometry Playground website (as some of they were in the Who Are You? section on the Exploratorium home page). Include national education standards with all activities (or say why they cannot be included). Find ways to reach out to various subgroups of visitors and entice them to the Geometry Playground website. For instance, ad a link to the Geometry Playground Artists in residence page to the Who Are You: Artists page on the main website (and also make sure they can easily find the list Geometry artists links). Use on mobile devices The summative evaluation revealed great potential for integrating mobile technologies into this “playground” exhibit. For instance, the evaluators observed many adults sitting at the climbing structures using cell phones while their children played, and similar behaviors can be expected once structures are added to outdoor playground. There should be ways to connect with many of these adults (at least after they have The website did not function well on iPhones, iPads, and other Apple devices, mostly because Apple does not support Adobe Flash Player. For instance, the home page text overlapped on a iPad screen (Fig. K-9), and the videos and photo essays did not play on the iPad (Fig. K-10). Develop a separate mobile homepage that will work well on any platform. (Level 3) Make videos accessed through the mobile homepage available in two formats, including a format that works on iPhones and other Apple devices. (Level 3) Use QR codes (Fig. K-11), shortened URLs, and other approaches to help mobile users quickly connect to the website or directly to relevant pages, videos, and activities. (Level 3) Selinda Research Associates, Inc. 208 checked their e-mail and Facebook pages). The activities displayed pretty well on an iPhone or iPad, including both the screen-friendly and printer-friendly versions. Use of social media Social media— specifically the project’s Flickr site— played an important role early in the project, and was still available to website visitors. Many of the activities required or encouraged and benefited from social interactions. The website, as currently designed, could be a shared experience, but little about its design seemed to push users in that direction (with the exception of some of the activities). The Flickr site, as this report was written, was largely inactive and presented out-of-date information about the exhibition. It said: The Geometry Playground is a combination of interactive exhibition, artful playground, and research project in the service of creating and evaluating a strong experiential foundation for geometrical learning. This exhibition is currently under development at the Exploratorium and will begin touring science museums around the country in 2010. Inspire us! Send us your pictures of shapes, architecture, lines, volumes, macro and microscopic imagery. Selected images and activities from this Flickr group will be featured on an accompanying exhibit website. This outdated information told visitors that no one from the Exploratorium was paying attention to this pool. Future exhibit websites should be designed as interactive parts of the exhibition—especially the mobile websites. Part of the exhibition experience should overlap with the social web (e.g., by keeping the Flickr site as an active part of the web experience). (Level 3) Encourage and facilitate meaningful social interactions among visitors when they are using the mobile website, especially when it’s used in playgrounds and in the exhibition. (Level 3) Highlight the link to the Flickr photo pool, so more website visitors will find it. (Level 2) Update the text information on the Flickr page. (Level 1) Highlight the Flickr page in the exhibition, Selinda Research Associates, Inc. 209 perhaps at the Geometry in the Wild photo exhibits and at Geometry Garden. Encourage exhibition visitors to add their own photos to the site. (Level 1) Add photos of the completed exhibition and a link to the Geometry Playground website to the Flickr pool. (Level 1) Privacy policy The Exploratorium’s privacy policy had a link at the very bottom of the page, which followed accepted conventions for this sort of thing. The privacy policy was reproduced in small font, and it included information that would be extraneous for most website users. There were actually two separate policies, one was a privacy policy and the other was a use policy. The use policy was especially densely worded and reproduced. Revise the privacy and use policies to put the important information at the top and in easy to read language and fonts. (Level 2) Both policies should present a short bulleted list up front as a preface (or a fleshed out table of contents) with the more detailed information following. These bulleted points might include, “We collect cookies.” “We don’t sell your information.” And so forth. Recovering for users’ errors The website did very well at helping users recover from their errors. For instance: 1. When we misspelled geometry as “geometry” in the search box on Exploratorium’s When iPhone and iPad users tried to play Flash-based slide shows and videos, they received a message to upload a newer version of Adobe Flash Player (which Apple does not support). Develop alternative formats for slide shows and videos that can be accessed by iPhone and iPad users. (Level 3) For future websites, keep in mind the disadvantages of developing Flash-based Selinda Research Associates, Inc. 210 home , the response suggested the correct spelling. 2. When we misspelled the name of the exhibit as “gemetryplaygro und” in the URL on my browser’s address bar, we got the following message: “We’re sorry ...the page you requested is not available. “The page you were looking for may not be available anymore or it may have been moved. Please make a selection using the top menu or use the search function below. If you're still having trouble finding a page feel free to send an email to the webmaster.” and gave his real name and email address. We also got the same error message on a “dead” link. These seemed like good responses. The option to email a real person with a real name is friendly and respectful. applications. Selinda Research Associates, Inc. 211 Likely outcomes for website visitors The website will probably stimulate some visitors to seek more information about certain topics—most likely geometry artists, because those videos are pretty interesting—although possibly other topics as well. However, because most of the external links were kind of buried on the Activities and Links page (Fig. K-5B and K-6A-B), many visitors will miss out on opportunities to follow up on their interests. As mentioned above, split the Activities and Links page into two separate pages. (Level 3) Motivating and facilitating visits to the exhibition The website made the exhibition seem pretty cool and will likely make users want to visit if it’s at a nearby host museum. There was a link to a list of host museums and scheduled dates for the traveling exhibition. The information could be accessed from a text link near the top of the home page, which led to the list at the bottom of the About page. Although motivating visits to the exhibition was not a goal expressed by the project web team, it probably should be. We expect that most people who find the site will not have visited the exhibition The exhibition schedule could not be directly accessed from pages other than the home page. The exhibition schedule did not include links to the host museums’ websites or to their Geometry Playground web pages. There was no mention of things that parents and/or educators (and other visitors) could do to prepare for and then follow up on a visit to Geometry Playground. Move the information about host sites for the traveling exhibition to the top of the About page (above visitor research and geometry on playgrounds, which more narrowly focused on a professional audience than at exhibition visitors). (Level 1) Include links to host museums and to their Geometry Playground pages. (Level 1) Add a page specifically about visiting the exhibition and include a link to this page on every website page. This could include: • Info and links for current, future, and past hosts, with links to their home pages and Geometry Playground pages. • Information about preparing for and Selinda Research Associates, Inc. 212 following up on a visit to the exhibition, aimed at parents and educators. • Perhaps called something like Visiting Geometry Playground, just like people have come to expect to find for Visiting the Museum. (Level 3) Contributio ns to project goals (e.g., spatial reasoning, aesthetics and everyday geometry) Many aspects of the website did a really good job of helping users appreciate the beauty and aesthetic aspects of geometry. For instance, the three photo essays associated with the three sections of the exhibition included many beautiful photos illustrating beautiful phenomena. The Geometry Garden section included many photos of beautiful artworks, artifacts, and natural history specimens. The Artists in Residence videos put a human face on the pursuit of geometric beauty. There was also a link to even more beautiful images in the Geometry Playground Flickr photo pool. It was disappointing that spatial reasoning—a major focus of the exhibition—was almost completely missing from the website. Providing information and examples of spatial reasoning could have been a very important way to extend and expand visitors’ engagements with the exhibits, especially for educators and parents. The language of spatial reasoning was largely missing from the activities, and the range of spatial reasoning skills practiced in the activities seemed rather minimal compared with the exhibits. It seemed like the activity developers were thinking geometry, but not spatial reasoning. The link to the Geometry Playground Flickr pool was small and farther down on the home page (on the second screen down on some computers). Create an engaging website experience that is specifically about spatial reasoning, such as a video or slide show. Something like the “Math Trails: Making Math Concrete” video could be a good inspiration, although it should be possible to develop a more interactive approach to the topic. (Level 3) Develop an activity or two specifically about spatial reasoning skills—calling them out for both students and teachers. (Level 3) Highlight the link to the Flickr photo pool on the home page and elsewhere, so more website visitors will find it. (Level 2) Tag or otherwise call out real-world and Selinda Research Associates, Inc. 213 Also, various aspects of the website did a really good job of highlighting geometry in everyday life. The three photo essays associated with the three sections of the exhibition included real-world applications. The Geometry Garden section included photos of real-world artifacts and natural history specimens, as did the Geometry Playground Flickr photo pool. aesthetic images in the Flickr pool. (Level 2) Expanding users’ engagement with exhibition themes and goals The website seemed likely to expand visitors’ engagements with the exhibition in several ways. The strongest aspects of the site were: • Giving visitors a way of thinking about geometry that goes beyond just shapes, e.g., Geometry of Moving, etc., Geometry Garden, and more. • The videos of the artists who created some of these exhibits. • The activities. Website users could Aside from the activities, the website was a fairly passive experience. Users looked at photos and videos and did a bit of reading. That didn’t really match the high levels of physical, social, emotional, and intellectual engagement that were common in the exhibition. There were few online activities developed for use at specific exhibits. (However, there were links to other website’s online activities on the Activities and Links page.) Develop activities and information specifically for use at particular exhibits, especially for the mobile website. There are lots of great opportunities here, like activity suggestions and extra information that parents can access with cell phones and tablets while sitting at the climbing structures (perhaps by scanning QR codes). (Level 3) Selinda Research Associates, Inc. 214 gain interesting perspectives on some of the exhibits, including perspectives that were not highlighted in the exhibition (or not shown in the same way). Example: Video of ball rolling through computer animated Gyroid. The activities provided a broad range of intellectual engagements for visitors, as well as physical, social, and emotional engagements.
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Factoring Cubic Polynomials Robert G. Underwood 1. Introduction There are at least two ways in which using the famous Cardano formulas (1545) to factor cubic polynomials present more difficul-ties than the quadratic formula poses when factoring quadratic polynomials. First and obviously, with its cube roots and roots of roots, the Cardano formulas involve computations that are more complicated. Second, we get less in return; in particular, with the Cardano formulas we only obtain one out of three factors, whereas the quadratic formula yields both factors of any given quadratic polynomial. In this paper, we will describe a natural procedure that will lead us to all three factors of an arbitrary cubic polyno-mial with real number coefficients.
Let R denote the real field, and let R[X] denote the ring of polynomials over R. Consider the cubic polynomial p(X) = X3 − k ∈R[X]. Let C denote the field of complex numbers, and let ζ be a primitive 3rd root of unity, that is, ζ is one of the two complex solutions of the equation X3 = 1. One calculates ζ = −1 + i √ 3 2 , where i = √−1.
The cubic p(X) factors completely over C. Indeed, if 3 √ k is the real 3rd root of k, then we have the factorization X3 −k = (X − 3 √ k)(X −ζ 3 √ k)(X −ζ2 3 √ k), and the roots of X3 −k are 3 √ k, ζ 3 √ k, ζ2 3 √ k.
26 Alabama Journal of Mathematics In this paper we show how to generate the roots of any cubic polynomial over R in an analogous manner, using ζ, and a solution (a, b) of the equation X3 + Y 3 = 1 over C.
2. Generating the roots of a cubic polynomial The problem of finding all the zeros of an arbitrary monic cubic polynomial f(X) = X3 + αX2 + βX + γ is equivalent to the problem of finding the zeros of the reduced cubic g(X) = X3 + δX + ².
We see that if c is a zero of g(X), then c −(1/3)α will be a zero of f(X), cf. [3, p. 568]. It is not difficult to show that any reduced cubic can be written in the form p(X) = X3 −3k2abX + k3 where (a, b) ∈C × C is a solution to the equation X3 + Y 3 = 1.
For example, if p(X) = X3 −12X + 8, we choose k = 2 and a and b so that a3 + b3 = 1 with ab = 1. Hence the required solution is (a, 1/a), where a is the root a = 3 s 1 2 + i √ 3 2 of the quadratic equation in X3: X6 −X3 + 1 = 0.
With this in mind, we prove the following: Theorem.
Suppose p(X) is a reduced cubic in the form p(X) = X3 −3k2abX + k3 where (a, b) is a solution of the equation X3 + Y 3 = 1, and k is a real number. Then the roots of p(X) are Spring 2002 27 c1 = −ka −kb c2 = −kaζ −kbζ2 c3 = −kaζ2 −kbζ.
Proof.
One could show directly that p(ci) = 0 for i = 1, 2, 3, but we present a different proof.
Consider C3, the vector space generated by ordered triples (r, s, t) of complex numbers. Set r = ka, s = kb, and t = c1 = −ka −kb. Then the set S = {(ka, kb, t), (t, ka, kb), (kb, t, ka)} is linearly dependent over C. To see this, observe that: (−1)(ka, kb, c1) + (−1)(c1, ka, kb) = (kb, c1, ka) is a dependence relation. It follows that the matrix A = ⎡ ⎣ ka kb c1 c1 ka kb kb c1 ka ⎤ ⎦ is singular, since the rows of A are linearly dependent. Thus det(A) = k3a3 + k3b3 + c3 1 −3k2abc1 = 0, which we may rewrite as k3a3 + k3b3 −k3 + (c3 1 −3k2abc1 + k3) = 0, or c3 1 −3k2abc1 + k3 = 0, since (a, b) is a solution of X3 + Y 3 = 1. We conclude that c1 = −ka−kb is a root of our reduced cubic p(X) = X3 −3k2abX + k3.
28 Alabama Journal of Mathematics To find the other roots of the cubic p(X), we find other values of t so that the set S = {(ka, kb, t), (t, ka, kb), (kb, t, ka)} is linearly dependent. Another such value is t = c2 = −kaζ −kbζ2.
To see this, observe that the set S is linearly dependent via the relation (−ζ2)(ka, kb, c2) + (−ζ)(c2, ka, kb) = (kb, c2, ka).
The matrix A = ⎡ ⎣ ka kb c2 c2 ka kb kb c2 ka ⎤ ⎦ is singular, and det(A) = k3a3 + k3b3 + c3 2 −3k2abc2 = 0, which, in turn, implies that c3 2 −3k2abc2 + k3 = 0.
Thus c2 is the second root of p(X).
Finally, if we put t = c3 = −kaζ2 −kbζ the set S = {(ka, kb, t), (t, ka, kb), (kb, t, ka)} is linearly dependent via the relation (−ζ)(ka, kb, c3) + (−ζ2)(c3, ka, kb) = (kb, c3, ka).
The matrix A = ⎡ ⎣ ka kb c3 c3 ka kb kb c3 ka ⎤ ⎦ is singular, and det(A) = k3a3 + k3b3 + c3 3 −3k2abc3 = 0, Spring 2002 29 which, in turn, implies that c3 3 −3k2abc3 + k3 = 0.
Thus c3 is the third and last root of p(X).
¦ Returning to our example p(X) = X3 −12X + 8, we obtain the factorization p(X) = (X + 2a + 2a−1)(X + 2ζa + 2ζ2a−1)(X + 2ζ2a + 2ζa−1), where a = 3 s 1 2 + i √ 3 2 .
We close with a final thought. Let Q denote the rational num-bers. Suppose p(X) = X3 −3k2abX + k3 is a reduced cubic over Q with exactly one rational root with k2ab 6= 0. Because Fermat’s Last Theorem is true for the case n = 3, there are no non-trivial solutions to the equation X3 + Y 3 = 1 in Q × Q. Hence if we use our cubic formula to calculate this rational zero, we must involve non-rational reals or non-real complex numbers. We wonder if this is the case using the standard cubic formula, see [3, Theorem 51.3].
References K. Hoffman, R. Kunze, Linear Algebra, Second Edition, Prentice-Hall, New Jersey, 1971.
K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, Second Edition, Springer-Verlag, New York, 1990.
S. Warner, Modern Algebra, Vol. II, Prentice-Hall, New Jersey, 1965.
Department of Mathematics Auburn University Montgomery Montgomery, AL 36124 underw@strudel.aum.edu 30 Alabama Journal of Mathematics
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calculus - condition for two curves to have a common tangent - Mathematics Stack Exchange
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condition for two curves to have a common tangent
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Consider the curves b y=x 2+x+b 24 b y=x 2+x+b 24 and b x=y 2+y+b 24 b x=y 2+y+b 24 if both the curves have a common tangent then the what values b can take ?
i observed that both curves are inverse of each other or they are symmetrical w.r.t to y=x y=x . My intuition is that this is the key to the problem But i am not able to go further.
So i tried using the general formula for tangent to a parabola of the form (y−f)2=4 a(x−g)(y−f)2=4 a(x−g) which is y−f=m(x−g)+a m y−f=m(x−g)+a m but the expression becomes too complicated.Finally i tried drawing a rough sketch to get some idea but i have to consider too many sub cases.
I came across this problem in a mock test that expects you to solve the problem within 2 minutes.So i guess there is a relatively simple method to approach this problem. Any help will be appreciated .
Edit I agree with all the answers given below.But when i looked at the solution i have no clue of what they are trying to do.i understand how they got b=3 2 a n d 2 3 b=3 2 a n d 2 3.I dont understand the rest of the part it is as follows.
let a=1 b a=1 b if the 2 curves intresect at P1 and P2 but at P1 tangent to first curve is perpendicular to y=x so it is tangent to second curve at P1.
slope of tangent=2 a x+a 2 a x+a .As (a,x) satisfies this
2 a x+a=−1 2 a x+a=−1 and solving with x=a x 2+a x+1 24 x=a x 2+a x+1 24
a=−13+601√12=1 b a=−13+601 12=1 b
i have written the answer as given .It looks quite absurd .
calculus
algebra-precalculus
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edited Aug 20, 2020 at 13:07
Hari Ramakrishnan SudhakarHari Ramakrishnan Sudhakar
asked Aug 20, 2020 at 7:19
Hari Ramakrishnan SudhakarHari Ramakrishnan Sudhakar
12.1k 2 2 gold badges 16 16 silver badges 53 53 bronze badges
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The answer can be given so : for any b b, these curves have (at least) a common tangent. It can be more than one, but this is another question :))Jean Marie –Jean Marie 2020-08-20 10:33:12 +00:00 Commented Aug 20, 2020 at 10:33
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The parabolas are about the line y=x y=x (they are mirror image of each other about the line y=x y=x). So their common tangents will also be symmetric aboy y=x y=x. Two possibilities arise:
Case 1: x=y x=y is a common tangent:
If they have to have one common tangent then by putting y=x y=x in one of them gives
x 2+(1−b)x+b/24=0 x 2+(1−b)x+b/24=0
, this quadratic needs to have only one real root, so the condition B 2=4 A C B 2=4 A C needs to be satisfied:
(1−b)2=b 6⟹6 b 2−12 b+6=0⟹b=13±5 12⟹b=3 2,2 3.(1−b)2=b 6⟹6 b 2−12 b+6=0⟹b=13±5 12⟹b=3 2,2 3.
Only when b=3/2,2/3 b=3/2,2/3 y=x y=x is the common tangent.
Case 2: when x+y=−k x+y=−k is the common tangent (more general)
Then we put y=−k−x y=−k−x in the first parabola to get
x 2−(1+b)x+b k+b/24=0 x 2−(1+b)x+b k+b/24=0
For tangency we demand B=4 A C B=4 A C, we get
(1+b)2=4 b k+b/24⟹k=(1+b)2 4 b−1 24(∗)(1+b)2=4 b k+b/24⟹k=(1+b)2 4 b−1 24(∗)
Therefor for any real value of b b x+y=−k x+y=−k, will a common tangent to these parabolas whe k k comes from (∗)(∗).
Case 3: two common tangents
Interestingly, b=3/2,2/3 b=3/2,2/3 gives k=1.k=1. So x+y=1 x+y=1 and y=x y=x are two common tangents to the given two parabolas.
See the figures below for b=4 b=4 (one common tangent, x+y=73/48.x+y=73/48.) and for b=3/2 b=3/2 (two common tangents y=x,x+y=1 y=x,x+y=1).
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edited Aug 20, 2020 at 8:48
answered Aug 20, 2020 at 7:42
Z AhmedZ Ahmed
46.8k 2 2 gold badges 17 17 silver badges 56 56 bronze badges
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2 There are two other solutions for b=1 36(−13±601−−−√)b=1 36(−13±601). I got them solving the 4 4-th degree system and looking for b b values that gave double roots. I got b=3 2;b=2 3 b=3 2;b=2 3, too Raffaele –Raffaele 2020-08-20 08:33:44 +00:00 Commented Aug 20, 2020 at 8:33
1 Look here imgur.com/16pnSeJRaffaele –Raffaele 2020-08-20 08:39:23 +00:00 Commented Aug 20, 2020 at 8:39
@Raffaele, interesting but see the case (2) of mine, any real value of b b will give a common tangent x+y=−k x+y=−k, where k k is an interesting function of b b. I mean to say that any value of b b around your two critical irrational values will give a common tangent . On e has two bring out the critical geometric feature of there two interesting irrational critical values of b of yours.Z Ahmed –Z Ahmed 2020-08-20 09:01:38 +00:00 Commented Aug 20, 2020 at 9:01
@HariRamakrishnanSudhakar I used Wolfram Mathematica to solve the system and got the quartic (576 b 3−624 b)x−24 b 3−25 b 2+(−624 b−576)x 2−576 x 4−1152 x 3=0(576 b 3−624 b)x−24 b 3−25 b 2+(−624 b−576)x 2−576 x 4−1152 x 3=0 to be factored as (24 b x−b−24 x 2−24 x)(24 b 2+24 b x+25 b+24 x 2+24 x)=0(24 b x−b−24 x 2−24 x)(24 b 2+24 b x+25 b+24 x 2+24 x)=0 the discriminants of the two factors are zero for the values I gave in the other comment. As it is not my work, I did not answer directly.Raffaele –Raffaele 2020-08-20 11:32:53 +00:00 Commented Aug 20, 2020 at 11:32
@DrZafarAhmedDSc i am able to follow your solution Thank you. But i am not able to comprehend the solution given .i think they are ignoring your second case .Could you also explain what they are trying to do ?Hari Ramakrishnan Sudhakar –Hari Ramakrishnan Sudhakar 2020-08-20 13:11:45 +00:00 Commented Aug 20, 2020 at 13:11
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Consider the straight line of equation
x+y=c.x+y=c.
We intersect it with the parabola
b y=x 2+x+b 24 b y=x 2+x+b 24
and by eliminating y y we get a quadratic equation. The discriminant is
(b+1)2−b 6+4 b c(b+1)2−b 6+4 b c
and it cancels (double root) when
c=1 24−(b+1)2 4 b.c=1 24−(b+1)2 4 b.
This correspond to a tangent to the first parabola, and by exchange of x,y x,y, it is also tangent to the second one. Hence there is a common tangent for all b≠0 b≠0.
F.i., with b=4 b=4,
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answered Aug 20, 2020 at 8:18
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15.401 15.401 Finance Theory 15.401 Finance Theory MIT Sloan MBA Program Andrew W. Lo Andrew W. Lo Harris & Harris Group Professor, MIT Sloan School Harris & Harris Group Professor, MIT Sloan School Lectures 2 Lectures 2– –3 3: Present Value Relations : Present Value Relations © 2007–2008 by Andrew W. Lo © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 2 Critical Concepts Critical Concepts Cashflows and Assets The Present Value Operator The Time Value of Money Special Cashflows: The Perpetuity Special Cashflows: The Annuity Compounding Inflation Extensions and Qualifications Readings: Brealey, Myers, and Allen Chapters 2–3 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 3 Cashflows and Assets Cashflows and Assets Key Question: What Is An “Asset”?
Business entity Property, plant, and equipment Patents, R&D Stocks, bonds, options, … Knowledge, reputation, opportunities, etc.
From A Business Perspective, An Asset Is A Sequence of Cashflows © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 4 Cashflows and Assets Cashflows and Assets Examples of Assets as Cashflows Boeing is evaluating whether to proceed with development of a new regional jet. You expect development to take 3 years, cost roughly $850 million, and you hope to get unit costs down to $33 million. You forecast that Boeing can sell 30 planes every year at an average price of $41 million.
Firms in the S&P 500 are expected to earn, collectively, $66 this year and to pay dividends of $24 per share, adjusted to index. Dividends and earnings have grown 6.6% annually (or about 3.2% in real terms) since 1926.
You were just hired by HP. Your initial pay package includes a grant of 50,000 stock options with a strike price of $24.92 and an expiration date of 10 years. HP’s stock price has varied between $16.08 and $26.03 during the past two years.
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 5 Cashflows and Assets Cashflows and Assets Valuing An Asset Requires Valuing A Sequence of Cashflows Sequences of cashflows are the “basic building blocks” of finance Always Draw A Timeline To Visualize The Timing of Cashflows © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 6 The Present Value Operator The Present Value Operator What is Vt?
What factors are involved in determining the value of any object?
– Subjective?
– Objective?
How is value determined?
There Are Two Distinct Cases No Uncertainty – We have a complete solution Uncertainty – We have a partial solution (approximation) – The reason: synergies and other interaction effects Value is determined the same way, but we want to understand how © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 7 The Present Value Operator The Present Value Operator Key Insight: Cashflows At Different Dates Are Different “Currencies” Consider manipulating foreign currencies ¥150 + £300 = ??
?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 8 The Present Value Operator The Present Value Operator Key Insight: Cashflows At Different Dates Are Different “Currencies” Consider manipulating foreign currencies ¥150 + £300 = ??450 Cannot add currencies without first converting into common currency ¥150 + (£300) × (153 ¥ / £) = ¥46,050.00 (¥150) × (0.0065 £ / ¥ ) + £300 = £ 300.98 Given exchange rates, either currency can be used as “numeraire” Same idea for cashflows of different dates ?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 9 The Present Value Operator The Present Value Operator Key Insight: Cashflows At Different Dates Are Different “Currencies” Past and future cannot be combined without first converting them Once “exchange rates” are given, combining cashflows is trivial A numeraire date should be picked, typically t=0 or “today” Cashflows can then be converted to present value © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 10 The Present Value Operator The Present Value Operator Net Present Value: “Net” of Initial Cost or Investment Can be captured by date-0 cashflow CF0 If there is an initial investment, then CF0 < 0 Note that any CFt can be negative (future costs) V0 is a completely general expression for net present value How Can We Decompose V0 Into Present Value of Revenues and Costs?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 11 The Present Value Operator The Present Value Operator Example: Suppose we have the following “exchange rates”: What is the net present value of a project requiring a current investment of $10MM with cashflows of $5MM in Year 1 and $7MM in Year 2?
Suppose a buyer wishes to purchase this project but pay for it two years from now. How much should you ask for?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 12 The Present Value Operator The Present Value Operator Example: Suppose we have the following “exchange rates”: What is the net present value of a project requiring an investment of $8MM in Year 2, with a cashflow of $2MM immediately and a cashflow of $5 in Year 1?
Suppose a buyer wishes to purchase this project but pay for it two years from now. How much should you ask for?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 13 The Time Value of Money The Time Value of Money Implicit Assumptions/Requirements For NPV Calculations Cashflows are known (magnitudes, signs, timing) Exchange rates are known No frictions in currency conversions Do These Assumptions Hold in Practice?
Which assumptions are most often violated?
Which assumptions are most plausible?
Until Lecture 12, We Will Take These Assumptions As Truth Focus now on exchange rates Where do they come from, how are they determined?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 14 The Time Value of Money The Time Value of Money What Determines The Growth of $1 Over T Years?
$1 today should be worth more than $1 in the future (why?) Supply and demand Opportunity cost of capital r Equivalence of $1 today and any other single choice above Other choices are future values of $1 today $1 in Year 0 = $1 × (1 + r) in Year 1 $1 in Year 0 = $1 × (1 + r)2 in Year 2 .
.
.
$1 in Year 0 = $1 × (1 + r)T in Year T © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 15 The Time Value of Money The Time Value of Money What Determines The Value Today of $1 In Year-T?
$1 in Year-T should be worth less than $1 today (why?) Supply and demand Opportunity cost of capital r These are our “exchange rates” ($t/$0) or discount factors $1/(1 + r) in Year 0 = $1 in Year 1 $1/(1 + r)2 in Year 0 = $1 in Year 2 .
.
.
$1/(1 + r)T in Year 0 = $1 in Year T © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 16 The Time Value of Money The Time Value of Money We Now Have An Explicit Expression for V0: Using this expression, any cashflow can be valued!
Take positive-NPV projects, reject negative NPV-projects Projects ranked by magnitudes of NPV All capital budgeting and corporate finance reduces to this expression However, we still require many assumptions (perfect markets) V0 = CF0 + 1 (1 + r) × CF1 + 1 (1 + r)2 × CF2 + · · · V0 = CF0 + CF1 (1 + r) + CF2 (1 + r)2 + · · · © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 17 The Time Value of Money The Time Value of Money Example: Suppose you have $1 today and the interest rate is 5%. How much will you have in … 1 year … $1 × 1.05 = $1.05 2 years … $1 × 1.05 × 1.05 = $1.103 3 years … $1 × 1.05 × 1.05 × 1.05 = $1.158 $1 today is equivalent to $ in t years $1 in t years is equivalent to $ today © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 18 The Time Value of Money The Time Value of Money PV of $1 Received In Year t PV of $1 Received In Year t $0.0 $0.2 $0.4 $0.6 $0.8 $1.0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Year when $1 is received r = 0.04 r = 0.08 r = 0.12 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 19 The Time Value of Money The Time Value of Money Example: Your firm spends $800,000 annually for electricity at its Boston headquarters. Johnson Controls offers to install a new computer-controlled lighting system that will reduce electric bills by $90,000 in each of the next three years. If the system costs $230,000 fully installed, is this a good investment?
Lighting System Assume the cost savings are known with certainty and the interest rate is 4% Year 0 1 2 3 Cashflow -230,000 90,000 90,000 90,000 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 20 The Time Value of Money The Time Value of Money Example: Lighting System Go ahead – project looks good!
Year 0 1 2 3 Cashflow -230,000 90,000 90,000 90,000 ÷ 1.04 (1.04)2 (1.04)3 PV -230,000 86,538 83,210 80,010 NPV = -230,000 + 86,538 + 83,210 + 80,010 = $19,758 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 21 The Time Value of Money The Time Value of Money Example: CNOOC recently made an offer of $67 per share for Unocal. As part of the takeover, CNOOC will receive $7 billion in ‘cheap’ loans from its parent company: a zero-interest, 2-year loan of $2.5 billion and a 3.5%, 30-year loan of $4.5 billion. If CNOOC normal borrowing rate is 8%, how much is the interest subsidy worth?
Interest Savings, Loan 1: 2.5 × (0.08 – 0.000) = $0.2 billion Interest Savings, Loan 2: 4.5 × (0.08 – 0.035) = $0.2 billion © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 22 Special Special Cashflows Cashflows: The Perpetuity : The Perpetuity Perpetuity Pays Constant Cashflow C Forever How much is an infinite cashflow of C each year worth?
How can we value it?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 23 Special Special Cashflows Cashflows: The Perpetuity : The Perpetuity Growing Perpetuity Pays Growing Cashflow C(1+g)t Forever How much is an infinite growing cashflow of C each year worth?
How can we value it?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 24 Special Special Cashflows Cashflows: The Annuity : The Annuity Annuity Pays Constant Cashflow C For T Periods Simple application of V0 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 25 Special Special Cashflows Cashflows: The Annuity : The Annuity Annuity Pays Constant Cashflow C For T Periods Sometimes written as a product: © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 26 Special Special Cashflows Cashflows: The Annuity : The Annuity Annuity Pays Constant Cashflow C For T Periods Related to perpetuity formula Perpetuity Minus Date-T Perpetuity Equals T-Period Annuity © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 27 Special Special Cashflows Cashflows: The Annuity : The Annuity Example: You just won the lottery and it pays $100,000 a year for 20 years. Are you a millionaire? Suppose that r = 10%.
What if the payments last for 50 years?
How about forever (a perpetuity)?
© 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 28 Compounding Compounding Interest May Be Credited/Charged More Often Than Annually Bank accounts: daily Mortgages and leases: monthly Bonds: semiannually Effective annual rate may differ from annual percentage rate Why?
Typical Compounding Conventions: Let r denote APR, n periods of compounding r/n is per-period rate for each period Effective annual rate (EAR) is 10% Compounded Annually, Semi-Annually, Quarterly, and Monthly rEAR ≡ (1 + r/n)n −1 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 29 Compounding Compounding Example: Car loan—‘Finance charge on the unpaid balance, computed daily, at the rate of 6.75% per year.’ If you borrow $10,000, how much would you owe in a year?
Daily interest rate = 6.75 / 365 = 0.0185% Day 1: Balance = 10,000.00 × 1.000185 = 10,001.85 Day 2: Balance = 10,001.85 × 1.000185 = 10,003.70 … … Day 365: Balance = 10,696.26 × 1.000185 = 10,698.24 EAR = 6.982% > 6.750% © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 30 Inflation Inflation What Is Inflation?
Change in real purchasing power of $1 over time Different from time-value of money (how?) For some countries, inflation is extremely problematic How to quantify its effects?
Wealth Wt ⇔Price Index It Wealth Wt+k ⇔Price Index It+k Increase in Cost of Living ≡ It+k/It = (1 + π)k “Real Wealth” f Wt+k ≡ Wt+k/(1 + π)k © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 31 Inflation Inflation © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 32 Inflation Inflation For NPV Calculations, Treat Inflation Consistently Discount real cashflows using real interest rates Discount nominal cashflows using nominal interest rates – Nominal cashflows ⇒expressed in actual-dollar cashflows – Real cashflows ⇒expressed in constant purchasing power – Nominal rate ⇒actual prevailing interest rate – Real rate ⇒interest rate adjusted for inflation © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 33 Inflation Inflation Example: This year you earned $100,000. You expect your earnings to grow 2% annually, in real terms, for the remaining 20 years of your career. Interest rates are currently 5% and inflation is 2%. What is the present value of your income?
Real Interest Rate = 1.05 / 1.02 – 1 = 2.94% Real Cashflows Present Value = $1,818,674 Year 1 2 … 20 Cashflow 102,000 104,040 … 148,595 ÷ 1.0294 1.029422 … 1.0294220 PV 99,086 98,180 … 83,219 © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 34 Extensions and Qualifications Extensions and Qualifications Taxes Currencies Term structure of interest rates Forecasting cashflows Choosing the right discount rate (risk adjustments) © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 35 Key Points Key Points Assets are sequences of cashflows Date-t cashflows are different from date-(t+k) cashflows Use “exchange rates” to convert one type of cashflow into another PV and FV related by “exchange rates” Exchange rates are determined by supply/demand Opportunity cost of capital: expected return on equivalent investments in financial markets For NPV calculations, visualize cashflows first Decision rule: accept positive NPV projects, reject negative ones Special cashflows: perpetuities and annuities Compounding Inflation Extensions and Qualifications © 2007–2008 by Andrew W. Lo Lecture 2-3: Present Value Relations 15.401 Slide 36 Additional References Additional References Bodie, Z. and R. Merton, 2000, Finance. New Jersey: Prentice Hall.
Brealey, R., Myers, S., and F. Allen, 2006, Principles of Corporate Finance. New York: McGraw-Hill Irwin.
Copeland, T., Weston, F. and K. Shastri, 2003, Financial Theory and Corporate Policy, (4th Edition). Reading, MA: Addison-Wesley.
MIT OpenCourseWare 15.401 Finance Theory I Fall 2008 For information about citing these materials or our Terms of Use, visit:
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Which Condition Is A Cause Of Reversible Dementia (Causes You Can Treat)
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Home>Alzheimer’s & Dementia Care / Symptoms>Which Condition Is A Cause Of Reversible Dementia: Reversal Paths
Which Condition Is A Cause Of Reversible Dementia: Reversal Paths
Written byChristopher Ravn, on July 23, 2025 and updated on August 12, 2025
Scientifically Reviewed by Mai Nguyen
While dementia is often associated with irreversible conditions like Alzheimer’s, it’s crucial to understand that certain types of cognitive decline can be reversed. These reversible dementias stem from treatable medical conditions that, with timely diagnosis and intervention, can lead to significant improvement or even full resolution of symptoms. Early detection is paramount for the best possible cognitive outcomes.
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Christopher Ravn
Christopher Ravn is head of Revenue Operations at OptoCeutics and holds a Master's in Biomedical Engineering.
Key Takeaways
Unlike progressive dementias such as Alzheimer's, reversible dementia results from treatable underlying medical conditions, offering hope for improved cognitive function upon diagnosis and treatment.
Common causes of reversible dementia include vitamin deficiencies (especially B12), thyroid disorders, medication side effects, infections (like UTIs), depression (pseudodementia), and Normal Pressure Hydrocephalus (NPH).
Early and accurate diagnosis is critical; distinguishing reversible causes from irreversible dementias allows for prompt, targeted treatment that can stabilize or restore cognitive health.
Table of Contents
1. Which Condition Is A Cause Of Reversible Dementia
2. What Reversible Dementia Conditions Are Covered In Relias Training?
3. What Is A Reversible Type Of Dementia?
4. Can Dementia Caused By Medication Be Reversed?
5. Is Syphilis A Reversible Dementia?
6. Which One Is Reversible, Delirium Or Dementia?
7. What Dementia Causes Are Reversible?
8. Can Depression Cause Reversible Dementia Symptoms?
9. Can Infections Cause Dementia Symptoms That Are Reversible?
10. Is Vitamin B12 Deficiency A Cause Of Reversible Dementia?
11. Does Normal Pressure Hydrocephalus Cause Reversible Dementia Symptoms?
12. How Long Does Dementia Take To Reverse After Treatment?
13. Frequently Asked Questions About Which Condition Is A Cause Of Reversible Dementia
Which Condition Is A Cause Of Reversible Dementia
It is important to note that certain types of dementia can be reversible. Unlike Alzheimer’s Disease, reversible dementia results from other medical issues that, once diagnosed and treated, may lead to improvement or resolution of cognitive symptoms. Early detection is crucial, as it significantly enhances cognitive outcomes.
Various conditions can mimic dementia. Common reversible causes include vitamin deficiencies (such as B12), thyroid disorders (like hypothyroidism), normal pressure hydrocephalus (NPH), side effects from medications, infections, and depression. Addressing a thyroid imbalance might enhance memory, but it cannot reverse dementia. Educating older adults about correctable memory problems offers hope and motivates them to take control of their health.
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What Reversible Dementia Conditions Are Covered In Relias Training?
Healthcare certification tests, like Relias’, often feature reversible dementia to help dementia caregivers spot cognitive deterioration. These platforms stress the importance of a complete medical evaluation by showing caregivers how to find and treat possible reversible reasons using standardized examples.
Some common medically approved reversible reasons for this kind of training are:
Taking multiple drugs can cause confusion or forgetfulness.
Vitamin B12 and thiamine are two types of B vitamins, of which some individuals have inadequate levels.
Hypothyroidism and hyperthyroidism can both impair clear thinking.
Infections like UTIs and pneumonia can cause delirium that looks like dementia, especially in older people.
Normal Pressure Hydrocephalus (NPH) is a condition in which fluid builds up in the brain.
When cognitive symptoms occur, sadness is referred to as pseudodementia.
Subdural hematomas that occur after a fall are caused by bleeding in the brain.
Depending on the depth of the course and the intended audience, the list may change. The primary purpose is to train doctors to identify and treat treatable diseases. (Jandu et al., 2025; Alšauskė et al., 2024)
What Is A Reversible Type Of Dementia?
Reversible dementia can be treated or even cured with medical care, unlike Alzheimer’s Disease and other progressive and irreversible dementias. Reversible dementias, on the other hand, have underlying disorders that can be treated to bring back normal brain function.
Reversible dementia, similar to early-stage progressive dementia, can cause forgetfulness, disorientation, and difficulties with executive functions.The outcome is significant. With the proper treatment, like vitamin supplements for Wernicke’s encephalopathy (which is caused by a lack of thiamine) or shunt insertion for Normal Pressure Hydrocephalus (NPH), cognitive functions can stabilize or return to normal. This shows how important it is to make the right diagnosis to distinguish between treatable and non-treatable diseases and to restore cognitive health. It is also essential to incorporate a balanced diet with cognitive training, such as brain exercises to prevent Alzheimer’s, brain exercises for seniors, mental exercises for mild cognitive impairment, and cognitive rehabilitation exercises.
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Can Dementia Caused By Medication Be Reversed?
Dementia-like symptoms caused by drugs can be reversed, which makes them a key sign of cognitive decline. Drugs can damage the brain, affecting memory, attention, and thinking skills. Some common medications that cause problems are anticholinergics, benzodiazepines, opioids, antidepressants, and statins.
These medications can damage the brain by blocking neurotransmitters such as acetylcholine, which can slow brain activity or interfere with neuron communication. Benzodiazepines, for example, can make it harder for short-term memories to become long-term memories. When someone is diagnosed with drug-induced cognitive impairment, changing or quitting the drug that caused the problem under medical supervision can help their cognitive performance. A complete medication examination is necessary when looking at new dementia-like symptoms, especially in older adults who take a lot of medicines.
Is Syphilis A Reversible Dementia?
Neurosyphilis is a late-stage syphilis infection that affects the brain and nervous system and can lead to dementia. It is rare, yet it can cause memory loss, personality changes, confusion, and poor judgment, just like other types of dementia. (Fadel et al., 2024; Funayama et al., 2023)
Cognitive impairment caused by syphilis can happen years or even decades after the person gets the disease. To find Treponema pallidum, doctors look at blood and cerebrospinal fluid. High-dose penicillin is very important because it can stop the progression of neurosyphilis and improve or even restore cognitive abnormalities, especially if it is diagnosed early. The stage of neurosyphilis and the extent of neurological damage determine its reversibility, highlighting the importance of early diagnosis and treatment.
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Which One Is Reversible, Delirium Or Dementia?
Delirium is an abrupt change in awareness and thinking that lasts for hours or days. Infections, adverse effects of drugs, dehydration, or abnormalities in electrolytes are the most common causes. Once the cause of delirium is found and treated, cognitive function can be restored and the condition can be reversed. (Echeverría et al., 2022)
At the same time, brain illnesses like Alzheimer’s lead to dementia, which is a slow, persistent decline in cognitive functioning over months or years. Some dementias can be reversed, but most can’t. People commonly mix up delirium and dementia, especially older people. This might delay medical exams and treatment for delirium that could be reversed, which can make things worse. So, to give timely and efficient care, one needs to know that delirium can come on quickly and change.
What Dementia Causes Are Reversible?
Several medically recognized conditions can cause dementia-like symptoms that are reversible with proper treatment. These are often mistaken for permanent dementia, highlighting the importance of thorough diagnostic workups.
| Reversible Cause | How it Leads to Cognitive Changes | Treatment |
---
| Depression | Can cause pseudodementia, impacting concentration and memory. | Antidepressants, therapy |
| Vitamin B12 Deficiency | Essential for nerve health; deficiency impairs brain function. | B12 supplements (injections or oral) |
| Thyroid Disorders | Both hypo- and hyperthyroidism affect metabolism and brain activity. | Thyroid hormone medication (levothyroxine) or anti-thyroid drugs |
| Normal Pressure Hydrocephalus (NPH) | Excess cerebrospinal fluid compresses the brain, affecting gait, cognition, and continence. | Surgical shunt placement |
| Medication Side Effects | Various drugs (e.g., anticholinergics, sedatives) can confuse. | Adjusting or discontinuing offending medications |
| Infections (e.g., UTIs) | Systemic infections can cause delirium mimicking dementia. | Antibiotics, supportive care |
| Alcohol Abuse | Chronic abuse leads to brain damage and nutrient deficiencies. | Abstinence, nutritional support |
| Brain Tumors | It can exert pressure on brain tissue, disrupting function. | Surgery, radiation, chemotherapy |
Can Depression Cause Reversible Dementia Symptoms?
Pseudodementia is a term for reversible dementia symptoms that can happen with depression. Severe depression can make one forget things, have trouble concentrating, think slowly, and feel lost, which could be mistaken for dementia. Depression, not brain atrophy, is what causes these cognitive problems. (Mouta et al., 2023)
To tell the difference between pseudodementia and dementia, one needs a complete medical history, cognitive tests, and the ruling out of other medical conditions. People with pseudodementia often feel distressed about their memory issues, whereas those with dementia might not experience such feelings. It’s important to know that treating depression with antidepressants, therapy, or both can help with cognitive issues and bring back mental sharpness and recall. Mental health is essential when cognitive impairment occurs.
Can Infections Cause Dementia Symptoms That Are Reversible?
Yes, infections can cause dementia-like symptoms that usually go away on their own, especially in older adults.Delirium can be caused by UTIs, pneumonia, and even worse diseases, including encephalitis and meningitis. This delirium, similar to dementia, can cause confusion, forgetfulness, difficulty concentrating, and visual hallucinations.
When the body gets sick, it can cause inflammation that affects how the brain works and the balance of neurotransmitters. Unlike genuine dementia, infection-related delirium produces cognitive problems that start fast and change throughout the day. After the infection is diagnosed and treated with antibiotics or antivirals, delirium usually goes away, and cognitive function usually returns to normal. People who are vulnerable and have sudden changes in their thinking need to see a doctor right away. (Sankowski et al., 2015)
Is Vitamin B12 Deficiency A Cause Of Reversible Dementia?
It is understood that a deficiency in vitamin B12 can cause dementia, which might be reversible. This vitamin supports neuron health, helps produce red blood cells, and is essential for DNA creation, all of which are vital for brain function. Not getting enough B12 can cause neurological problems like memory loss, brain fog, confusion, disorientation, and personality changes that seem like irreversible dementia. (Wong et al., 2022)
Cognitive decline caused by a lack of B12 can be restored with therapy. Most of the time, treatment includes B12 shots, pills, and dietary changes to include more foods that are high in B12. Vegan individuals (since B12 is mainly in animal products), older adults (due to reduced absorption), and those with Crohn’s disease or pernicious anemia are at higher risk of deficiency. People with cognitive problems need to have their B12 levels checked regularly to find and address this reversible cause.
Does Normal Pressure Hydrocephalus Cause Reversible Dementia Symptoms?
Normal Pressure Hydrocephalus (NPH) can cause dementia symptoms that may resolve. NPH happens when CSF builds up in the brain’s ventricles in an unusual way, placing pressure on brain tissue. Fluid buildup uniquely affects cognitive function, sometimes leading to symptoms like memory loss and slower thinking that resemble dementia. It also causes problems with walking and shuffling, as well as urine incontinence.
To find out if someone has NPH, doctors do a neurological exam and an MRI or CT scan to look at the enlarged ventricles. Because the symptoms are similar, it is commonly mistaken for Alzheimer’s or Parkinson’s. The main difference is how easily it can be treated. The best way to treat NPH is via shunt surgery, which involves putting in a small tube to drain extra CSF from the brain so that it may be absorbed. NPH is an integral part of evaluating reversible dementia because it can significantly improve symptoms.
How Long Does Dementia Take To Reverse After Treatment?
The improvement of dementia symptoms after therapy can range from days to months, depending on the source and the speed of identification addressed. After an illness goes away, cognitive improvements may be noticeable within days for short-term issues like delirium. Adjusting or discontinuing the prescription may improve medication-related cognitive impairment within a few weeks.
However, if someone has a severe vitamin deficiency or Normal Pressure Hydrocephalus (NPH), it could take weeks or months for their brain to heal, and they will need ongoing monitoring and rehabilitation. Recovery is affected by a person’s health, age, and how long and severe their cognitive impairment was before treatment. Regular follow-up is important because some diseases need long-term care to prevent recurrence.
Are you noticing dementia-like symptoms in your loved one, such as memory loss or confusion?
Yes, I’m noticing dementia-like symptoms.
No, I’m not noticing dementia-like symptoms.
Have you consulted a doctor to evaluate for reversible causes like vitamin deficiencies or infections?
Yes, we’ve consulted a doctor.
No, we haven’t consulted a doctor.
← Back
SOLUTION:
Consulting a doctor is proactive. Follow medical advice for treatments like B12 supplements or antibiotics. Visit alz.org for reversible dementia resources and monitor progress with a neurologist.
← Back
SOLUTION:
Unaddressed symptoms need action. Schedule a doctor’s visit to check for reversible causes like infections. Visit alz.org for guidance and consult a neurologist promptly.
← Back
Are you monitoring for early cognitive changes or discussing preventive health with a doctor?
Yes, I’m monitoring or discussing preventive health.
No, I’m not monitoring or discussing preventive health.
← Back
SOLUTION:
Proactive monitoring is ideal. Continue tracking cognitive health and consult a doctor for preventive care. Access alz.org for dementia prevention resources.
← Back
SOLUTION:
Lack of monitoring risks delayed detection. Start observing cognitive changes and consult a doctor. Visit alz.org for early detection resources and preventive strategies.
← Back
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Frequently Asked Questions About Which Condition Is A Cause Of Reversible Dementia
Which Condition Is A Cause Of Reversible Dementia?
It is important to note that certain types of dementia can be reversible. Unlike Alzheimer’s Disease, reversible dementia results from other medical issues that, once diagnosed and treated, may lead to improvement or resolution of cognitive symptoms. Early detection is crucial, as it significantly enhances cognitive outcomes. Various conditions can mimic dementia. Common reversible causes include vitamin deficiencies (such as B12), thyroid disorders (like hypothyroidism), normal pressure hydrocephalus (NPH), side effects from medications, infections, and depression.
Can Vitamin B12 Deficiency Really Cause Reversible Dementia Symptoms?
It is understood that a deficiency in vitamin B12 can cause dementia, which might be reversible. This vitamin supports neuron health, helps produce red blood cells, and is essential for DNA creation, all of which are vital for brain function. Not getting enough B12 can cause neurological problems like memory loss, brain fog, confusion, disorientation, and personality changes that seem like irreversible dementia.
1000 667 Christopher Ravn Christopher Ravn July 23, 2025 August 12, 2025
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Sexual differentiation in humans - Wikipedia
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1 Sex determination
2 Reproductive systemToggle Reproductive system subsection
2.1 Internal genital differentiation
2.2 External genital differentiation
3 Secondary sexual characteristicsToggle Secondary sexual characteristics subsection
3.1 Breast development
4 Psychological and behavioral differentiation
5 Disorders of sex development
6 Timeline
7 See also
8 References
9 Further reading
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From Wikipedia, the free encyclopedia
Process of development of sex differences in humans
This scientific article needs additional citations to secondary or tertiary sources. Help add sources such as review articles, monographs, or textbooks. Please also establish the relevance for any primary research articles cited. Unsourced or poorly sourced material may be challenged and removed.(March 2023) (Learn how and when to remove this message)
The human Y chromosome showing the SRY gene which codes for a protein regulating sexual differentiation.
Sexual differentiation in humans is the process of development of sex differences in humans. It is defined as the development of phenotypic structures consequent to the action of hormones produced following gonadal determination.Sexual differentiation includes development of different genitalia and the internal genital tracts and body hair plays a role in sex identification.
The development of sexual differences begins with the XY sex-determination system that is present in humans, and complex mechanisms are responsible for the development of the phenotypic differences between male and femalehumans from an undifferentiated zygote. Females typically have two X chromosomes, and males typically have a Y chromosome and an X chromosome. At an early stage in embryonic development, both sexes possess equivalent internal structures. These are the mesonephric ducts and paramesonephric ducts. The presence of the SRY gene on the Y chromosome causes the development of the testes in males, and the subsequent release of hormones which cause the paramesonephric ducts to regress. In females, the mesonephric ducts regress.
Disorders of sexual development (DSD), encompassing conditions characterized by the appearance of undeveloped genitals that may be ambiguous, or look like those typical for the opposite sex, sometimes known as intersex, can be a result of genetic and hormonal factors.
Sex determination
[edit]
Further information: XY sex-determination system
Most mammals, including humans, have an XY sex-determination system: the Y chromosome carries factors responsible for triggering male development. In the absence of a Y chromosome, the fetus will undergo female development. This is because of the presence of the sex-determining region of the Y chromosome, also known as the SRY gene. Thus, male mammals typically have an X and a Y chromosome (XY), while female mammals typically have two X chromosomes (XX).
Chromosomal sex is determined at the time of fertilization; a chromosome from the sperm cell, either X or Y, fuses with the X chromosome in the egg cell. Gonadal sex refers to the gonads, that is the testicles or ovaries, depending on which genes are expressed. Phenotypic sex refers to the structures of the external and internal genitalia.
Six weeks elapse after fertilization before the first signs of sex differentiation can be observed in human embryos. The embryo and subsequent early fetus appear to be sexually indifferent, looking neither like a male or a female. Over the next several weeks, hormones are produced that cause undifferentiated tissue to transform into either male or female reproductive organs. This process is called sexual differentiation. The precursor of the internal female sex organs is called the Müllerian system.
Reproductive system
[edit]
Further information: Development of the reproductive system
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Figure One: The mesonephric System Pathway
Differentiation between the sexes of the sex organs occurs throughout embryological, fetal and later life. In both males and females, the sex organs consist of two structures: the internal genitalia and the external genitalia. In males, the gonads are the testicles and in females, they are the ovaries. These are the organs that produce gametes (egg and sperm), the reproductive cells that will eventually meet to form the fertilized egg (zygote).
As the zygote divides, it first becomes the embryo (which means 'growing within'), typically between zero and eight weeks, then from the eighth week until birth, it is considered the fetus (which means 'unborn offspring'). The internal genitalia are all the accessory glands and ducts that connect the gonads to the outside environment. The external genitalia consist of all the external reproductive structures. The sex of an early embryo cannot be determined because the reproductive structures do not differentiate until the seventh week. Prior to this, the child is considered bipotential because it cannot be identified as male or female.
Internal genital differentiation
[edit]
The internal genitalia consist of two accessory ducts: mesonephric ducts (Woffian duct) and paramesonephric ducts (Müllerian ducts). The mesonephric system is the precursor to the male genitalia and the paramesonephric to the female reproductive system. As development proceeds, one of the pairs of ducts develops while the other regresses. This depends on the presence or absence of the sex determining region of the Y chromosome, also known as the SRY gene. In the presence of a functional SRY gene, the bipotential gonads develop into testes. Gonads are histologically distinguishable by 6–8 weeks of gestation.
Subsequent development of one set and degeneration of the other depends on the presence or absence of two testicular hormones: testosterone and anti-Müllerian hormone (AMH). Disruption of typical development may result in the development of both, or neither, duct system, which may produce morphologically intersex individuals.
Males: The SRY gene when transcribed and processed produces SRY protein that binds to DNA and directs the development of the gonad into testes. Male development can only occur when the fetal testis secretes key hormones at a critical period in early gestation. The testes begin to secrete three hormones that influence the male internal and external genitalia: they secrete anti-Müllerian hormone (AMH), testosterone, and dihydrotestosterone (DHT). Anti-Müllerian hormone causes the paramesonephric ducts to regress. Testosterone converts the mesonephric ducts into male accessory structures, including the epididymides, vasa deferentia, and seminal vesicles. Testosterone will also control the descending of the testes from the abdomen. Many other genes found on other autosomes, including WT1, SOX9 and SF1 also play a role in gonadal development.
Females: Without testosterone and AMH, the mesonephric ducts degenerate and disappear. The paramesonephric ducts develop into the uterus, fallopian tubes, and upper vagina (the lower vagina develops from the urogenital sinus). There still remains a broad lack of information about the genetic controls of female development (as of 1992), and much remains unknown about the female embryonic process. The mesonephric ducts are not completely useless in the female case: they secrete WNT9B, which is necessary for the elongation of the paramesonephric ducts. Elongation also happens through the active migration of the paramesonephric epithelium, which happens through a phosphoinositide 3-kinase pathway.
External genital differentiation
[edit]
Development of external genitalia
By 7 weeks, a fetus has a genital tubercle, urogenital sinus, urogenital folds and labioscrotal swellings. In females, without excess androgens, these become the vulva (clitoris, vestibule, labia minora and labia majora respectively). Males become externally distinct between 8 and 12 weeks, as androgens enlarge the genital tubercle and cause the urogenital groove and sinus to fuse in the midline, producing an unambiguous penis with a phallic urethra, and the labioscrotal swellings become a thinned, rugate scrotum where the testicles are situated. Dihydrotestosterone will differentiate the remaining male characteristics of the external genitalia.
A sufficient amount of any androgen can cause external masculinization. The most potent is dihydrotestosterone (DHT), generated from testosterone in skin and genital tissue by the action of 5α-reductase. A male fetus may be incompletely masculinized if this enzyme is deficient. In some diseases and circumstances, other androgens may be present in high enough concentrations to cause partial or (rarely) complete masculinization of the external genitalia of a genetically female fetus. The testes begin to secrete three hormones that influence the male internal and external genitalia. They secrete anti-Müllerian hormone, testosterone, and Dihydrotestosterone. Anti-Müllerian hormone (AMH) causes the paramesonephric ducts to regress. Testosterone, which is secreted and converts the mesonephric ducts into male accessory structures, such as epididymis, vas deferens and seminal vesicle. Testosterone will also control the descending of the testes from the abdomen into the scrotum. Dihydrotestosterone, also known as (DHT) will differentiate the remaining male characteristics of the external genitalia.
Further sex differentiation of the external genitalia occurs at puberty, when androgen levels again become disparate. Male levels of testosterone directly induce growth of the penis, and indirectly (via DHT) the prostate.
Alfred Jost observed that while testosterone was required for mesonephric duct development, the regression of the paramesonephric duct was due to another substance. This was later determined to be paramesonephric inhibiting substance (MIS), a 140 kD dimeric glycoprotein that is produced by Sertoli cells. MIS blocks the development of paramesonephric ducts, promoting their regression. Today it’s better known as the anti-Müllerian hormone (AMH).
Secondary sexual characteristics
[edit]
Further information: Sexual characteristics
This section needs expansion. You can help by adding to it. (January 2020)
Breast development
[edit]
Visible differentiation occurs at puberty, when estradiol and other hormones cause breasts to develop in typical females.
Psychological and behavioral differentiation
[edit]
Human adults and children show many psychological and behavioral sex differences. Some (e.g. dress) are learned and cultural. Others are demonstrable across cultures and have both biological and learned determinants. For example, some studies claim girls are, on average, more verbally fluent than boys, but boys are, on average, better at spatial calculation. It seems likely that this is due to males generally having a greater area allocated to the space-specialized parietal cortex, while females generally have relatively more brain area allocated to the verbal-associative-specialized temporal cortex.
Disorders of sex development
[edit]
Further information: Disorders of sex development and intersex
Disorders of sex determination (DSD) are classified into a multitude of categories. These categories consists of different types of female disorders along with categories specifically for male DSDs. There are also sex chromosomal DSDs such as, the later mentioned, Klinefelter and Turner syndrome
The following are some of the conditions associated with atypical determination and differentiation process:
A zygote with only X chromosome (XO) results in Turner syndrome and will develop with female characteristics.
Congenital adrenal hyperplasia –Inability of adrenal to produce sufficient cortisol, leading to increased production of testosterone resulting in severe masculinization of 46 XX females. The condition also occurs in XY males, as they suffer from the effects of low cortisol and salt-wasting, not virilization.
Persistent Müllerian duct syndrome – A rare type of pseudohermaphroditism that occurs in 46 XY males, caused by either a mutation in the Müllerian inhibiting substance (MIS) gene, on 19p13, or its type II receptor, 12q13. Results in a retention of Müllerian ducts (persistence of rudimentary uterus and fallopian tubes in otherwise normally virilized males), unilateral or bilateral undescended testes, and sometimes causes infertility.
XY differences of sex development – Atypical androgen production or inadequate androgen response, which can cause incomplete masculinization in XY males. Varies from mild failure of masculinization with undescended testes to complete sex reversal and female phenotype (Androgen insensitivity syndrome)
Swyer syndrome. A form of complete gonadal dysgenesis, mostly due to mutations in the first step of sex determination; the SRY genes.
A 5-alpha-reductase deficiency results in atypical development characterized by female phenotype or under virilized male phenotype with development of the epididymis, vas deferens, seminal vesicle, and ejaculatory duct, but also a pseudovagina. This is because testosterone is converted to the more potent DHT by 5-alpha reductase. DHT is necessary to exert androgenic effects farther from the site of testosterone production, where the concentrations of testosterone are too low to have any potency.
Klinefelter syndrome (47,XXY)- A chromosomal disorder that results in an extra X chromosome in males. This leads to hormonal problems later on. However, this disorder while in some cases easily identified, sometimes is not extreme and can not be determined until after puberty if at all.
Timeline
[edit]
Human prenatal sexual differentiation| Fetal age (weeks) | Crown-rump length (mm) | Sex differentiating events |
---
| 1 | blastocyst | X-inactivation, if more than one X chromosome present |
| 4 | 2–3 | Development of Wolffian ducts |
| 5 | 7 | Migration of primordial germ cells in the undifferentiated gonad |
| 6 | 10–15 | Development of Müllerian ducts |
| 7 | 13–20 | Male: Differentiation of seminiferous tubules |
| 8 | 30 | Male: Regression of Müllerian ducts |
| 8 | 32–35 | Male: Appearance of Leydig cells. First synthesis of testosterone |
| 9 | 43 | Male: Total regression of Müllerian ducts. Female: Loss of sensitivity of Müllerian ducts in the female fetus |
| 9 | 43 | Female: First meiotic prophase in oogonia |
| 10 | 43–45 | Male: Beginning of masculinization of external genitalia |
| 10 | 50 | Female: Beginning of regression of Wolffian ducts |
| 12 | 70 | Male: Fetal testis is in the internal inguinal ring |
| 12–14 | 70–90 | Male: Penile urethra is completed |
| 14 | 90 | Male: Appearance of first spermatogonia |
| 16 | 100 | Female: Appearance of first ovarian follicles |
| 17 | 120 | Male: Numerous Leydig cells. Peak of testosterone secretion |
| 20 | 150 | Male: Regression of Leydig cells. Diminished testosterone secretion |
| 24 | 200 | Female: First multilayered ovarian follicles. Canalisation of the vagina |
| 28 | 230 | Female: Cessation of oogonia multiplication |
| 28 | 230 | Male: Descent of testis |
See also
[edit]
Neuroscience of sex differences
Sex differences in humans
Determination of sex
References
[edit]
^ abcHughes, Ieuan A. (2001). "Minireview: Sex Differentiation". Endocrinology. 142 (8): 3281–3287. doi:10.1210/endo.142.8.8406. PMID11459768.
^Sizonenko, P. C. (n.d.). "Human sexual differentiation". Reproductive health – via Geneva Foundation for Medical Education and Research.
^Mukherjee, Asit B.; Parsa, Nasser Z. (1990). "Determination of sex chromosomal constitution and chromosomal origin of drumsticks, drumstick-like structures, and other nuclear bodies in human blood cells at interphase by fluorescence in situ hybridization". Chromosoma. 99 (6): 432–435. doi:10.1007/BF01726695. PMID2176962. S2CID25732504.
^Kučinskas, Laimutis; Just, Walter (2005). "Human male sex determination and sexual differentiation: Pathways, molecular interactions and genetic disorders"(PDF). Medicina. 41 (8): 633–640. ISSN1010-660X. PMID16160410.
^ abcdRey, Rodolfo; Josso, Nathalie; Racine, Chrystèle (27 May 2020) [first published 2000]. "Sexual Differentiation". In Feingold, Kenneth R.; Anawalt, Bradley; Blackman Marc R.; et al. (eds.). Endotext [Internet]. South Dartmouth, Mass.: MDText.com, Inc. PMID25905232. Retrieved 28 March 2023 – via National Institutes of Health.
^Achermann, John; Jameson, Larry (2012). Fauci, Anthony S. (ed.). Harrison's principles of internal medicine (18th ed.). New York: McGraw-Hill Medical. pp.3046–3048. ISBN978-0-07-147693-5.
^Silverthorn, Dee, U.. (2010). Reproduction and Development. In: Human Physiology: an integrated approach. 5th ed. san francisco: Pearson education. pp. 828–831.
^"Learning Objectives". Albany.edu. Archived from the original on 3 May 2001. Retrieved 2 October 2017.
^ abFauci, Anthony S.; Harrison, T. R., eds. (2008). Harrison's principles of internal medicine (17th ed.). New York: McGraw-Hill Medical. pp.2339–2346. ISBN978-0-07-147693-5 – via Internet Archive.
^Fausto-Sterling, Anne (1992). Myths Of Gender: Biological Theories About Women And Men (revised ed.). New York: Basic Books. pp.81–82. ISBN978-0-4650-4792-5.
^Wilson, Danielle; Bordoni, Bruno (2025). "Embryology, Mullerian Ducts (Paramesonephric Ducts)". StatPearls. StatPearls Publishing. PMID32491659.
^Hughes, Ieuan A. . (June 12, 2011).[full citation needed]
^Jost, A.; Price, D.; Edwards, R. G. (1970). "Hormonal Factors in the Sex Differentiation of the Mammalian Foetus [and Discussion]". Philosophical Transactions of the Royal Society B: Biological Sciences. 259 (828): 119–131. Bibcode:1970RSPTB.259..119J. doi:10.1098/rstb.1970.0052. JSTOR2417046. PMID4399057.
^Halpern, Diane F. (2012). Sex Differences in Cognitive Abilities (4th ed.). New York: Psychology Press. ISBN978-1-8487-2940-7 – via Internet Archive.[page needed]
^Hänggi, Jürgen (2010). "Sexual Dimorphism in the Parietal Substrate Associated with Visuospatial Cognition Independent of General Intelligence". Journal of Cognitive Neuroscience. 22 (1): 139–155. doi:10.1162/jocn.2008.21175. PMID19199407.[page needed]
^ abFeldman Witchel, Selma (April 2018). "Disorders of sex development". Best Practice & Research. Clinical Obstetrics & Gynaecology. 48: 90–102. doi:10.1016/j.bpobgyn.2017.11.005. ISSN1532-1932. PMC5866176. PMID29503125.
^MacLaughlin, David T.; Donahoe, Patricia K. (2004). "Sex Determination and Differentiation". New England Journal of Medicine. 350 (4): 367–378. doi:10.1056/NEJMra022784. PMID14736929.
^"Klinefelter syndrome: MedlinePlus Genetics". medlineplus.gov. Retrieved 17 April 2025.
^PC Sizonenko in Pediatric Endocrinology, edited by J. Bertrand, R. Rappaport, and PC Sizonenko, (Baltimore: Williams & Wilkins, 1993), pp.88–99
^Tallaksen HB, Johannsen EB, Just J, Viuff MH, Gravholt CH, Skakkebæk A (August 2023). "The multi-omic landscape of sex chromosome abnormalities: current status and future directions". Endocrine Connections. 12 (9). doi:10.1530/EC-23-0011. PMC10448593. PMID37399516.
Further reading
[edit]
De Felici, M. (2010). "Germ stem cells in the mammalian adult ovary: Considerations by a fan of the primordial germ cells". Molecular Human Reproduction. 16 (9): 632–636. doi:10.1093/molehr/gaq006. PMID20086005.
Sharman, GB; Hughes, RL; Cooper, DW (1989). "The Chromosomal Basis of Sex-Differentiation in Marsupials". Australian Journal of Zoology. 37 (3): 451. doi:10.1071/ZO9890451.
Watson, CM; Margan, SH; Johnston, PG (1998). "Sex-chromosome elimination in the bandicoot Isoodon macrourus using Y-linked markers". Cytogenetics and Cell Genetics. 81 (1): 54–59. doi:10.1159/000015008. PMID9691176. S2CID20042866.
| v t e Sex differences in humans |
| Biology | Sexual differentiation Disorders In research Physiology Dimorphism Scientific measures | |
| Medicine and Health | Autoimmunity Life expectancy Health survival paradox Suicide Mental disorders Autism Depression Schizophrenia Substance abuse Stroke care |
| Neuroscience and Psychology | Cognition Coping Emotional expression Aggression Emotional intelligence Empathy Gender empathy gap Intelligence Memory Narcissism Neurosexism Sexuality Age disparity in relationships Attraction Desire Fantasy Jealousy |
| Sociology and Society | Crime Education in mathematics and reading in the U.S. Gender inequality Greater male expendability Greater male variability Leadership Religion Social capital Social support Sociolinguistics Gender-equality paradox |
| v t e Sex determination and differentiation |
| Overview | Sexual differentiation humans Development of the reproductive system gonads Mesonephric duct Paramesonephric duct |
| Genetic basis | Sex-determination system XY X0 ZW ZO Temperature-dependent Haplodiploidy Sex chromosome X chromosome Y chromosome Sex determining gene: SRY (mammal) DMRT1 (birds) |
| See also | Hermaphrodite Intersex Disorders of sex development Sex reversal |
| Category |
| v t e Sex |
| Biological terms | Sexual dimorphism Male Female Sexual differentiation Feminization Virilization Sex-determination system XY ZW XO ZO Temperature-dependent Haplodiploidy Sex chromosome Testis-determining factor Hermaphrodite Sequential hermaphroditism Simultaneous hermaphroditism Intersex parasexuality Sex as a biological variable |
| Sexual reproduction | Evolution of sexual reproduction Anisogamy Isogamy Germ cell Reproductive system Sex organ Mating Meiosis Gametogenesis Spermatogenesis Oogenesis Gamete spermatozoon ovum Fertilization External Internal Sexual selection Plant reproduction Fungal reproduction Sexual reproduction in animals Sexual intercourse Penile–vaginal intercourse Copulation Hormonal motivation Human reproduction Lordosis behavior Pelvic thrust |
| Sexuality | Plant sexuality Animal sexuality Human sexuality Mechanics Differentiation Activity |
| Sexual system | Dioecy Gonochorism Hermaphrodite Intersex |
| Category |
| v t e Human physiology of sexual reproduction |
| Menstrual cycle | Menarche Menstruation Follicular phase Ovulation Luteal phase |
| Gametogenesis | Spermatogenesis spermatogonium spermatocyte spermatid sperm Oogenesis oogonium oocyte ootid ovum Germ cell gonocyte gamete |
| Human sexual activity | Sexual arousal Sexual intercourse Masturbation Erection Penile Clitoral Orgasm Ejaculation Male Female Insemination Fertilization / Fertility Implantation Pregnancy Postpartum period Mechanics of sex Vaginal lubrication |
| Development of the reproductive system | Sexual differentiation Sexual dimorphism Feminization Virilization Puberty Gonadarche Tanner scale Pubarche Menarche Spermarche Adrenarche Maternal age/ Paternal age Menopause |
| Egg | Ovum Vivipary |
| Reproductive endocrinology and infertility | Hypothalamic–pituitary–gonadal axis Hypothalamic–pituitary–prolactin axis Andrology Hormone |
| Breast | Thelarche Development Lactation Breastfeeding |
| Human reproductive system | Male Female |
| Diseases and disorders | Anorgasmia Dysorgasmia Dyspareunia Erectile dysfunction Hard flaccid syndrome Premature ejaculation Vaginismus |
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Sexual differentiation in humans
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Enzyme cofactors and coenzymes | Biology | Khan Academy
Khan Academy
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Posted: 19 Aug 2015
The cofactors and coenzymes (organic cofactors) that help enzymes catalyze reactions
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Transcript:
we've already spent a couple of videos talking about enzymes and what i want to do in this video is dig a little bit deeper and focus on some actors that actually help enzymes and just as a reminder enzymes are around to help reactions to proceed to lower their activation energies to make the the reactions happen more frequently or to happen faster now we've already seen examples of enzymes and just to frame things in our brain properly sometimes in a textbook you'll see an enzyme like this you'll see a drawing like this and you'll say and people will call this the enzyme they'll call this the enzyme and then they'll call they'll call this right they'll say okay and it's acting on some type of a substrate right over here it's going to do something to that and this is nice for a very abstract textbook idea of a substrate locking into an enzyme like this but this isn't actually what it looks like in a biological system we have to remind ourselves when people talk about enzymes they're talking about proteins now there are these kind of rna enzymes called ribozymes but the great majority when we're talking about enzymes we tend to be talking about proteins and we spend a lot of time talking about how proteins are these structures there's polypeptides and all the side chains of the various amino acids fold the proteins in all sorts of different ways so uh if i were a better drawing for something like this you know it would be this this protein it's all folded in different ways maybe it has some alpha helices here maybe it has some beta sheets right over here it's all this kind of crazy stuff right over there and then the substrate might be some type of a molecule that is it gets embedded it gets embedded in the protein and you see some examples right over here this is actually a hexokinase model and you see the at least you can see a little bit of the atp right over there and it's a little harder to see the glucose that's going to be phosphorylated and this reaction is being facilitated by this big protein structure the hexokinase now what we're going to focus on in this video is that when we talk about an enzyme we're talking about proteins we're talking about a chain of amino acids but there's often other parts of the enzyme that aren't officially proteins and we even saw that when we talked about hexokinases when we talked about the phosphorylation of glucose we said hey the rate the way that it lowers the activation energy is you have these positive magnesium ions these positive magnesium ions that can keep the electrons and the phosphate groups a little bit busy draw them away so that the so that this hydroxyl group right over here can bond with this phosphate and not be interfered with these electrons well these these magnesium ions right over here they aren't officially part of the protein these are what we call cofactors so you might have a cofactor right over there that gets that latches on to the broader protein to become part of the enzyme and you actually need that for the reaction to proceed it plays a crucial role here so another drawing in the textbook you'll see something like this or even you know they'll draw they'll say okay in order for this reaction to proceed yes you need the substrate but you also need the cofactor the cofactor and once again it sounds like a fancy word but all it means is the non-protein part of an enzyme it's another molecule or or ion or atom that is involved in letting the enzyme perform its function that it's not formally it's not formally in a part of an amino acid or part of a side chain or part of the protein but it's another thing that needs to be there to help catalyze the reaction we saw that with hexokinase you had magnesium ions that the the the that the complex picks up and this is why you know when people talk about your vitamins and minerals a lot of the vitamins and minerals that you need they actually act as cofactors for four enzymes and so you can even see it in this drawing over here at least based on what i read these are the magnesium ions in green right over here and these are cofactors these are cofactors so cofactor non-protein part of your actual enzyme now we can subdivide cofactors even more we can divide them into organic cofactors and non or inorganic cofactors so if you have cofactors you have cofactors we've seen an inorganic cofactor a lot of these ions you'll see magnesium ions you'll see sodium ions you'll see calcium ions you'll see all sorts of things acting as cofactors oftentimes to distract electrons so that or to keep them busy so that electrons can proceed but you can also have organic ones you can also have organic molecules remember organic molecules these are just they'll involve carbon they have chains of carbons and other things and cofactors that are organic organic molecules we call them coenzymes coenzymes and there's a bunch of examples of coenzymes this right over here is the enzyme lactate dehydrogenase and it has the coenzyme and this coenzyme you are going to see a lot in your biological careers and a nad right over here notice this isn't just an ion it is an entire molecule it has carbon in it that's why we call it organic and it is not formally protein it's not it's not part of the the amino acids that make up the protein so that's what makes it a cofactor and since it's since it's an entire organic molecule we call this a coenzyme co-coenzyme but like any cofactor it plays a role in actually allowing the enzyme to do its function to facilitate reaction and this particular coenzyme nad which you're going to see a lot it helps facilitate the transfer of hydride ions hydride ions never or very seldom exist by themselves but it's a hydrogen with an extra electron so it has a negative so it has a negative charge so it allows to the transfer of this group or from from a from a substrate or to a substrate and that's because nad nad can accept a hydride anion right over here and become nadh and if you want to see its broader structure it's actually quite fascinating i'll probably do a whole video on nad because uh in so many textbooks growing up i just saw you know nad and nadh and i'm like what is this thing it's a fascinating molecule so what it can do is it can actually pick up the hydride anion right over here at this carbon you can actually form another bond with the hydrogen i'll do that in a future video i'll show the mechanism for it but it's a pretty cool molecule and i like to actually look at this molecule remember the whole focus of this is coenzymes but we see these patterns throughout biology because the name nicotinamide adenine dinucleotide exactly describes what it is nicotinamide nicotinamide right down here that is that is this piece of the molecule and this is the part that can accept a hydride or let go of a hydride so you could say this is the active part of the molecule adenine our good old friends we've we've seen adenine and dna and rna in atp so this is our good old friend adenine right over here and it says dinucleotide because we actually have two nucleotides paired together their phosphate groups are tied together and there's a couple cool ways to think about this you have an adenine right over here you have a ribose you have a phosphate group if you just looked at if you just looked at this piece right over here if you looked at this right over here this is your building block this this is your building block of r or this could be a building block of rna if you have an adenine right over there and if you include if you include let me undo this if you include all of this this right over here this is adp but the reason why it's called dinucleotides you can also divide it the other way you can say all right you have one nucleotide that has nicotinamide right over here so that's one of the nucleotides and then the new other nucleotide is right over here the one that involves the one that involves adenine and that's why it's called dinucleotide so hopefully this makes nad less of a mysterious molecule we'll see it in the future but i i like to look at it because it's it's got all these patterns it's got all these components that you see over and over again you see an atp you see it in rna over and over again but the this isn't the only this isn't the only cofactor or coenzyme there are many many others in fact when people say take your vitamins and your minerals that tends to be because they are cofactors vitamin c is a very important cofactor to be involved in enzymes that well i won't go into all of the different things that it can do these are two different views of vitamin c a space filling model and this is a ball and stick model right over here vitamin c folic acid once again two different views but these are all coenzymes they all work they all work you know if you have a protein right over here that you know it's all this really complex structure maybe you have some substrates but to help facilitate let me do the substrates in a different color in a different color so maybe you have some substrates so these are the things that the enzymes trying to catalyze the reactions for but then you could have some ions which would you know you could kind of view these as you would view these you would view the ions as cofactors and you could have organic cofactors like your the vitamin c or other things that we talked about that are also involved in help facilitating the mechanism or help facilitate the reaction and once again sometimes it might be to help stabilize some charge sometimes it might be to be an electron acceptor or a donor or a whole series of different things they can they can actually act as part of the reaction mechanism
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Pendiente y ángulo de inclinación de la recta conociendo dos puntos
Matemáticas profe Alex
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Posted: 15 Sep 2020
Ejemplo de cómo encontrar la pendiente de una recta cuando conocemos dos puntos, en este tenemos que encontrar la pendiente y también el ángulo de inclinación de la recta.
Curso completo de Ecuación de la recta:
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0:00 Saludo
0:22 Conceptos que debes saber
1:01 Solución del ejemplo
6:55 Explicación gráfica
8:42 Ejercicios de práctica
10:25 Despedida y videos recomendados
282 comments
Transcript:
Saludo qué tal amigos espero que estén muy bien soy el profe de álex y en este vídeo que está dentro del curso de ecuación de la recta vamos a aprender cómo encontrar la pendiente de una recta cuando conocemos dos puntos [Música] cuándo y en este vídeo vamos a hacer este Conceptos que debes saber ejercicio que bueno ya tenemos que practicar un poquito más cositas de la pendiente en este caso vamos a determinar la pendiente de la recta que pasa por dos puntos que bueno esto ya lo hemos hecho en todos los vídeos anteriores y ya les aclare de dónde es que sale esta fórmula y cómo es que funciona no simplemente aquí vamos a practicar aplicándola y nada más y luego otra cosa para la que sirve la pendiente es para encontrar el ángulo de inclinación de la recta porque porque hay una fórmula dado porque ya se sabe que la pendiente es igual a la tangente del ángulo si el número de la pendiente es exactamente lo mismo que la tangente Solución del ejemplo del ángulo que al final voy a hacer el dibujo de esto para aclararles un poquito más esto bueno entonces los invito a que se queden hasta el final primero que todo que voy a hacer pues encontrar la pendiente no acordémonos que bueno ya eso lo voy a hacer rápido porque ya lo hemos visto mucho en vídeos anteriores que pues aquí tenemos que mirar las coordenadas porque pues con esos que trabajan la pendiente no aquí sería ésta la coordenada x o más bien es la coordenada xy la coordenada del punto a y esta es la coordenada x la coordenada y del punto b que pues como este punto está primero escribo que esta es la x del primer punto y la del primer punto y pues esto sería la x del segundo punto y la jr el segundo punto si para aplicar la fórmula no entonces aquí escribo que la pendiente y aplicamos la formulita la aie del segundo punto menos la del primer punto entonces la ya del segundo punto es menos 1 y en el segundo punto menos la jr el primer punto que es 5 aquí dice que hay una división y en el denominador que escribimos pues lo que dice acá no la equis del segundo punto menos la equis del primer punto entonces la equis del segundo punto que es el número 2 - cuidado aquí la equis del primer punto que en este caso la equis del primer punto es negativa tenemos que escribir la negativa no esté menos será de la fórmula y ahora escribimos el segundo la equis del primer punto que es menos 4 si generalmente uno le hace un paréntesis para separar estos signos pero pues ya les digo a mis estudiantes que de una vez se salten ese paso como quedaron dos signos seguidos menos por menos eso es más y simplemente tenemos que hacer la operación no entonces aquí que nos queda bueno voy a hacerlo aquí al frente menos 15 eso es menos 6 acordemos que ahí no se multiplican signos porque es una revista o suma como lo quieran decir y abajo dice 2 + 4 que es en este caso se puede hacer la operación o más bien no está entero entonces voy a hacer la operación aquí nos queda menos por más que sería menos y 6 dividido en 6 que eso es 1 o sea que ya sabemos que la pendiente de la recta que pasa por estos dos puntos es menos 1 acordémonos lo que vimos en vídeos anteriores primero algo que podemos analizar aquí como es negativa ya sabe que es una recta que va bajando y bueno la pendiente es 1 que quiere decir ver espera que ya recuerden la pendiente no listo ya encontramos la pendiente que la primera parte ahora dice que después de encontrar la pendiente o bueno no importa si era después en este caso pues teníamos que encontrar la pendiente encontremos el ángulo de inclinación para eso nos acordamos de esta formulita no que nos dice que la pendiente es igual a la tangente del ángulo de inclinación de la recta como les digo ahorita hago el dibujo para aclararles un poquito más esto no entonces simplemente reemplazamos en esta fórmula miren que aquí dice pendiente pero la pendiente ya bueno voy a escribir aquí pendiente es igual o más bien voy a borrar para que llame que para el ejercicio así bien explicado no entonces voy a escribir esto que la pendiente es igual a la tangente del ángulo que es lo que tenemos que hacer aquí pues reemplazar la pendiente porque ya sabemos que la pendiente o al menos sólo aquí me queda fácil entonces al borrar la pendiente porque yo ya sé que vale menos 1 que es lo que tenemos que encontrar tenemos que encontrar el ángulo para encontrar el ángulo pues tenemos que despejar lo no y para despejar lo pues tenemos que quitar esta palabra tangente sí entonces como se quita la tangente aplicando la inversa de la tangente entonces voy a escribir esto nuevamente menos 1 igual sólo que voy a correr esto un poquito para explicarles bien qué es lo que vamos a hacer como vamos a quitar la tangente tenemos que aplicar la inversa de la tangente que la arco tangente sí que es la que me permite eliminar esta tangentes y entonces escribo aquí arco tangente que se puede escribir de dos formas tangente a la menos 1 como la escribimos a la derecha tenemos que escribirle a la izquierda tangente a la menos 1 de ese no y aquí podemos hacer el paréntesis también si aquí lo aclaro como para saber que es la tangente a la menos 1 que eso es arco tangente si no podemos escribir arq tangente como les decía porque escribimos esto porque está arco tangente me permite simplificar la tangente si por ser inversas el arco tangente de la tangente del ángulo es simplemente el ángulo entonces qué nos queda simplemente hacer esta operación en nuestra calculadora que debemos tener cuidado en dos cositas una que generalmente pues aún no le gusta que el ángulo lo dé en grados sí o sea 30 grados 50 grados entonces la calculadora para hacer esta operación que vamos a hacer aquí debe estar en grados no que en la parte superior de la pantalla debe decir la letra y no la letra g uno muchas veces piensa que la g de grados es la de no quede en inglés es la de degree aunque se escribe así degree que es grados en inglés bueno entonces la calculadora de estar en grados sí ahora sí ya cuando la calculadora está en grados entonces hacemos la operación tangente a la menos uno de menos uno entonces como se escribe tangente a la menos uno oprimimos la tecla shift sí porque generalmente tangente a la menos uno está ahí arriba de la tecla tangente shift tangente ahí nos aparece en la pantalla tangente a la menos uno abrimos paréntesis menos uno cerramos paréntesis y eso nos da menos 45 grados entonces aquí escribimos menos 45 grados igual aquí como quedado eliminado solamente nos quedó el ángulo o sea que ya sabemos el ángulo del ángulo es de menos 45 grados que quiere decir menos 45 grados quiere decir bueno desde aquí por el concepto de tangente y por lo que ya hemos visto en sobretodo en los primeros vídeos en el que les expliqué la tangente pues ya sabemos que cuando la tangente es negativa como les decía es porque la recta va bajando siempre sucede lo mismo aquí cuando el ángulo es negativo simplemente quiere decir que la recta va bajando entonces ahora Explicación gráfica sí observemos el gráfico para comprender un poquito mejor el tema no primero la tangente menos uno como era negativa ya se sabía que la recta iba bajando bueno aquí está el punto que era el punto menos 45 y aquí está el punto b que era el punto 2 coma menos 1 entonces menos 1 la pendiente va bajando así que si escogemos cualquier punto este y este por ejemplo cada unidad que corramos a la derecha va a bajar 1 eso quiere decir s menos 1 si eso era lo que quería ver pero bueno eso lo expliqué mejor en los vídeos anteriores lo importante es esto qué quiere decir este menos 45 acordemos que el ángulo siempre se observa aquí saliendo del eje x xi a la derecha saliendo del eje x hacia arriba o hacia abajo siempre que sea negativo nos está diciendo el ángulo que le estoy diciendo yo es el ángulo que se cuenta desde el eje x hacia abajo así mírelo sí o sea menos 45 grados quiere decir cuente el ángulo 45 pero hacia abajo este es el ángulo que me dio mi ejercicio menos 45 grados aquí escribo 45 porque ya lo marque abajo si si nos da positivo que quería decir primero que todo que la recta iría subiendo si como acaso aquí se otra recta y cualquiera si solamente por esto aplicarles esta recta va subiendo entonces ya se sabe que la pendiente nos va a dar positiva y en ese caso el resultado de la pendiente el perdón del ángulo siempre también nos va a dar positivo bueno si la pendiente es positiva el ángulo es positivo si la pendiente es negativa del ángulos negativos en este caso como la recta va subiendo entonces el ángulo nos daría positivo que querría decir que se va a medir aquí por la parte de arriba pero Ejercicios de práctica bueno ya con esto termino mi explicación como siempre por último les voy a dejar un ejercicio para que ustedes practiquen ustedes van a hacer lo mismo que yo hice no van a encontrar la pendiente de la recta que pasa por estos dos puntos sin aquí está la formulita por si se les olvidó y también van a encontrar el ángulo listos y la respuesta va a aparecer en 321 bueno todos los pasos son exactamente lo mismo no lo primero pues a mí me gusta siempre marcar las coordenadas no coordenada xy coordenadas del primer punto y coordenada xy coordenadas del segundo punto entonces la aie del segundo menos la jr el primero y el segundo que es 3 - la sie del primero que es menos 2 por eso es que menos por menos queda más ahora abajo la equis del segundo que es 4 menos la equis del primero que como es negativa también queda positivo más 13 más 12 541 es 5 y 5 sobre 5 es 1 en este caso medio positivo no es que siempre de 1 no en este caso medio positivo que quiere decir que la recta va subiendo como aquí da positivo ya se sabe que el ángulo también va a dar positivo entonces simplemente reemplazamos en la formulita que es la pendiente de es la tangente del ángulo como la pendiente es 1 pues reemplazamos y la tangente del ángulo la corrí para la derecha para que para poder escribir arco tangente este es el símbolo del arco tangente pero no quiere decir que sea la inversa multiplicativo de la tangente cuidado con eso no esto es arco tangente aquí nos queda arco tangente de uno y aquí la tangente se puede eliminar con la arco tangente y nos queda la tangente el arco tangente de uno que es 45° igual t Despedida y videos recomendados bueno amigos espero que les haya gustado esta clase ya saben aquí y en la descripción del vídeo les dejo los links del curso completo para que aprendan mucho más acerca de este tema o aquí les dejo algunos vídeos recomendados que sé que les van a servir muchísimo no olviden comentar suscribirse y darle un like a este vídeo y no siendo más bye bye
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Homework 9 MTH 829 Complex Analysis Joshua Ruiter February 13, 2018 Proposition 0.1 (Exercise X.12.2). Let g(z) = z7 + 4z4 + z3 + 1. In the region |z| < 1, g has four zeroes and in the region 1 < |z| < 2, f has three zeroes, counting multiplicity.
Proof. Let f(z) = z7. On the circle |z| = 2, we have |f(z) −g(z)| = |4z4 + z3 + 1| ≤|4z4| + |z3| + 1 = 4|z|4 + |z|3 + 1 = 73 |f(z)| = |2|7 = 128 In particular, |f(z) −g(z)| < |f(z)|, so the hypotheses of Rouche’s Theorem are satisfied, and we conclude that f and g have the same number of zeroes in the interior |z| < 2. Since we can see immediately that f has a zero of order 7 at zero and no others, this implies that g has seven zeroes in |z| < 2. Now let h(z) = 4z4. On the circle |z| = 1, we have |h(z) −g(z)| = |z7 + z3 + 1| ≤|z7| + |z3| + 1 = |z|7 + |z|3 + 1 = 3 |h(z)| = 4|z|4 = 4 Again we can apply Rouche’s Theorem to conclude that g and h have the same number of zeroes in |z| < 1. Since h has a zero of order four at zero and no other zeroes, g has four zeroes in |z| < 1. Since g is degree seven, it has seven zeroes, counting multiplicity. Since four of the zeroes lie in |z| < 1, and all seven lie in |z| < 2, the other three must be in 1 < |z| < 2.
Proposition 0.2 (Exercise X.12.3). For any ϵ > 0, the function g(z) = sin z + 1 z+i has infinitely many zeroes in the strip | Im z| < ϵ.
Proof. Let Kn = B(2πn, ϵ/2) be the closed ball of radius ϵ centered at 2πn, for n ∈Z.
Note that Kn ⊂{z : | Im z| < ϵ}. Define f(z) = sin z. Since f is 2π-periodic, we can compute the values of f on the boundary of Kn by considering f on the circle centered at zero. We may assume ϵ is small enough that the sets Kn are disjoint, so f does not vanish on ∂Kn = { ϵ 2eit : t ∈[0, 2π] since the only zeroes of sin z are the centers of the Kn. Define δ = min t∈[0,2π] n f ϵ 2eit o 1 We know δ exists since ∂Kn is compact, and δ > 0 since f does not vanish on ∂Kn. Also on ∂Kn, we have |f(z) −g(z)| = 1 z + i This goes to zero as |z| →∞, so we can choose N so that |z| > N = ⇒ 1 |z + i| < δ Then on ∂Kn we have |f(z) −g(z)| = 1 |z + i| < δ ≤|f(z)| Note that f, g are holomorphic away from −i, so we can apply Rouche’s Theorem to conclude that f, g have the same number of zeroes on the interior of Kn for n ≥N. Since f has one zero on the interior of every Kn, g has a zero on the interior of Kn for n ≥N. That is, g has infinitely many zeroes in S n∈Z Kn, which is a subset of | Im z| < ϵ.
Proposition 0.3 (Exercise X.12.7). Let 1 < a < ∞and define g(z) = z + a −ez. Then g has exactly one zero in the left half plane Re z < 0 and it is on the real axis.
Proof. Define f(z) = z + a. Let Kn be the rectangle with vertices ±nai, −2na ± nai for n ∈N with n > a. In the half plane Re z < 0, we have |ez| ≤1 (with equality only on the imaginary axis). Thus on the boundary of Kn, we have |f(z) −g(z)| = |ez| ≤1 |f(z)| = |z + a| ≥a > 1 The second inquality holds for n ≥a > 1, since then Kn contains the disk |z −(−a)| < a.
Thus we can apply Rouche’s Theorem, so f, g have the same number of zeroes in Kn. Clearly, f has one zero at −a, so g has one zero in Kn. Since n was arbitrary, g has only one zero in Re z < 0. Finally, note that g(−a) = −e−a < 0 g(0) = a −1 > 0 so by the intermediate value theorem, g has a zero in (−a, 0) so its single zero is real.
Proposition 0.4 (Exercise X.12.8). Let 0 < |a| < 1 and define g(z) = (z −1)nez −a. Then g has exactly n roots, each of multiplicity one, in the half plane Re z > 0. Furthermore, if |a| ≤2−n, then the roots all lie in the disk |z −1| < 1 2.
Proof. Let f(z) = (z −1)nez and for m ≥2 let Km be the right half of the disk of radius m centered at zero. Then the closed disk of radius one centered at one is contained in Km.
Note that in the right half plane, |ez| ≥1, so on the boundary of Km, we have |f(z)| = |z −1|n|ez| ≥|z −1|n ≥1 > |a| = |f(z) −g(z)| Thus by Rouche’s Theorem, f, g have the same number of zeroes in Km. Clearly f has a zero of multiplicity n at z = 1 and no other zeroes (ez never vanishes). Thus g also has n 2 zeroes, and since m was arbitrary, g has n zeroes in the right half plane. We claim these zeroes all have multiplicity one. The derivative of g is g′(z) = n(z −1)nez + (z −1)nez = (z −1)n−1ez(z −(1 −n)) The only zeroes of g′(z) occur at z = 1 and z = 1 −n. Pluggin in 1, we know it is not a zero of g, and 1 −n is not a zero of g in the right half plane. Since any zero of order 2 or more would have derivative zero at that point, all of g’s zeroes in the right half plane have order one. Now suppose |a| ≤2−n, and let K = {z : |z −1| < 1/2}. Then on ∂K, |f(z) −g(z)| = |a| ≤2−n ≤2−n|ez| = |f(z)| so by Rouche’s Theorem, the n zeroes of g are in K, since f has n zeroes in K.
Proposition 0.5 (Exercise X.16.1). The following domains are all conformally equivalent to the open unit disk. In each case, we construct an explicit univalent holomorphic map from the open unit disk to the domain.
1. The upper half plane Im z > 0 2. The whole plane with a slit along (−∞, 0] 3. The strip 0 < Im z < 1 4. The first quadrant, Im z > 0, Re z > 0 5. The intersection of the unit disk with the upper half plane 6. The unit disk minus the segment [0, 1) Proof. Throughout, we use the fact that a composition of univalent functions is univalent (when the domain and image line up appropriately). (1) The map φ(z) = z−i iz−1 satisfies φ(i) = 0 φ(1) = −1 φ(−1) = 1 φ(0) = i The first three say that φ maps the unit circle to the real axis. Since φ(0) = i, φ maps the interior of the unit circle (the open unit disk) to the side of the real axis containing i (the upper half plane). Since φ is a linear fractional transformation, it is univalent.
(2) Let φ be as above. We claim that the map ψ(z) = (−iφ(z))2 maps the open unit disk to the plane with slit (−∞, 0]. As already noted, φ maps the open unit disk bijectively to the upper half plane. Multiplying by −i rotates by an angle −π/2, so the upper half plane becomes the right half plane. The map z 7→z2 takes the right half plane, which we can write as {reiθ : r > 0, θ ∈(−π/2, π/2)} and doubles the angle and squares the modulus to send it to {r2ei2θ : r > 0, θ ∈(−π/2, π/2)} = {reiθ : r > 0, θ ∈(−π, π)} = C \ {−∞, 0] 3 which is precisely the slit plane we wanted. Note that z →z2 is univalent on the right half plane, since everything has angle in (−π/2, π/2), so doubling the angle doesn’t introduce any overlaps in the image.
(3) Let φ be as above. We claim that the map α(z) = 1 π Log φ(z) maps the open unit disk to the strip 0 < Im z < 1 univalently. As already noted, φ takes the open unit disk to the upper half plane. This region is simply connected, so there is a branch of log φ, and we just choose the principal branch. Note that Log(reiθ) = ln |r| + iθ so applying Log to the region {reiθ : r > 0, θ ∈(0, π)} gives the strip 0 < Im z < π, since r 7→ln |r| is onto all of R. Finally, the dilation z 7→1 πz shrinks the strip 0 < Im z < π to the strip 0 < Im z < 1. These maps are all univalent, so the composition is univalent.
(4) Let φ be as above. We claim that the map β(z) = p φ(z) maps the open unit disk univalently to the first quadrant. We know that φ maps the open unit disk to the upper half plane, which is simply connected, so there is a branch of √φ. Taking the principal branch, √z takes the upper half plane and halves angles (and shrinks radii), so θ ∈(0, π) gets shrunk to θ ∈(0, π/2). Thus √z takes the upper half plane to the first quadrant.
(5) Define the linear fractional transformation f(z) = −z−1 z−1 . It satisfies f(1) = ∞ f(−i) = −i f(i) = i f(−1) = 0 f(−1 + i) = −1 5 + i2 5 so f maps the real axis onto the real axis and the imaginary axis onto the unit circle. Thus f maps each quadrant to one of four regions: the upper half of the unit disk, the lower half of the unit disk, the upper half plane minus the unit disk, or the lower half plane minus the unit disk. Since f(−1 + i) = −1 5 + i 2 5, f maps the second quadrant to the upper half of the unit disk.
Then define η(z) = f(iβ(z)), with β as above. We know that β maps the open unit disk to the first quadrant. Multiplication by i rotates the first quadrant to the second quadrant, and then η maps the second quadrant to the upper half of the unit disk. Thus η maps the open unit disk to the upper half of the unit disk, and η is a composition of univalent functions, so it is univalent.
(6) Let η be as above. Define χ(z) = η(z)2. We claim χ maps the open unit disk to the open unit disk minus the slit [0, 1). The image of η is {reiθ : 0 < r < 1, 0 < θ < π} so after applying the square map, this region becomes {r2e2iθ : 0 < r < 1, 0 < θ < π} = {reiθ : 0 < r < 1, 0 < θ < 2π} which is precisely the open unit disk minus the segment [0, 1).
For the sake of concreteness, here are explicit formulas for φ, ψ, α, β, η, χ.
φ(z) = z −1 iz −1 ψ(z) = (−i) z −1 iz −1 2 α(z) = 1 π Log z −1 iz −1 β(z) = r z −i iz −1 η(z) = −i q z−i iz−1 −1 i q z−i iz−1 −1 χ(z) = −i q z−i iz−1 −1 i q z−i iz−1 −1 2 4 Proposition 0.6 (Exercise X.16.5). The Koebe function, defined by f(z) = z (1 −z)2 is univalent on the open unit disk. Its image is C \ (∞, 1 4].
Proof. Let φ be the linear fractional transformation φ(z) = 1 + z 1 −z Then φ(−1) = 0, φ(i) = i, and φ(−i) = −i, so φ maps the unit circle to the imaginary axis.
Also, φ(0) = 1, so φ maps the open unit disk to the right half plane Re z > 0. It is univalent since all LFTs are univalent. Now note that 1 4(φ(z))2 −1 4 = 1 4 1 + z 1 −z 2 −1 4 = (1 + z)2 −(z −1)2 4(z −1)2 = (1 + 2z + z2) −(1 −2z + z2) 4(z −1)2 = 4z 4(z −1)2 = z (1 −z)2 = f(z) So we have rewritten f as a composition of several simpler maps: φ, z 7→z2, z 7→1 4z, and z 7→z −1 4. First, as we already showed, φ maps the unit disk univalently onto the right half plane. The map z 7→z2 is univalent on the right half plane, since it doubles the argument.
Its image is the slit plane C \ (−∞, 0]. The map z 7→1 4z is a simple dilation, which is a bijection of the slit plane onto itself. Then the final map is a translation, which is univalent, and maps the slit plane C \ (−∞, 0] to the shifted slit plane C \ (−∞, 1 4]. A composition of univalent functions is univalent, so f is univalent on the open unit disk.
Proposition 0.7 (Exercise X.19.3). Let G be a domain, and let A = f : G →C holomorphic ZZ G |f(x + iy)|dx dy ≤1 Then A is a normal family.
Proof. We claim that A is locally uniformly bounded. If we can show this, then the Stieltjes-Osgood Theorem says that it is a normal family. Let z0 ∈G, and choose r > 0 so that B(z0, r) ⊂G. Let f ∈A. Then f is holomorphic, so by the Mean Value Property (Exercise VII.6.1 of Sarason), f(z0) = 1 π(r/2)2 ZZ |z−z0| 0 such that |z −z0| < δ = ⇒|f(z) −f(z0)| < 4 πr2 Now define r′ = min(δ, r). Then in B(z0, r′), we have |f(z)| ≤ 8 πr2 (for all z in the ball). This process gives us a rule to assign to each w ∈G some rw > 0 so that in B(w, rw) we have the inequality |f(w)| ≤ 8 πr2 w . Let K ⊂G be compact. We can cover K by balls K ⊂ [ w∈K B(w, rw) Since K is compact, there is a finite collection w1, . . . , wn so that K ⊂ n [ i=1 B(wi, rwi) Now let r = min{rwi}. Since all the rw > 0, we also have r > 0. Then on all of K, we have |f(z)| ≤ 8 πr2 That is, 8 πr2 is a bound for all of A on K. Thus A is locally uniformly bounded, so by the Stieljes-Osgood theorem it is a normal family. (Note: Sarason includes an additional hypothesis that G is bounded, which this proof shows to be unnecessary. However, if G is unbounded, then one can use the mean value property to show that A includes only the zero function, so the result is trivial in that case regardless.) Proposition 0.8 (Exercise X.19.4). Let (fn)∞ n=1 be a locally uniformly bounded sequence of holomorphic functions on a domain G. Suppose that fn converges pointwise on A ⊂G where A has a limit point in G. Then fn converges locally uniformly on G.
Proof. First, we claim that any two locally uniformly convergent subsequences of fn must converge to the same limit.
Suppose fnk and fmk are are locally uniformly convergent subsequences of fn, with limits f, g respectively. By the Weierstrass convergence thoerem (VIII.15 of Sarason), f and g are holomorphic. Since fn converges to f and g pointwise on A, f, g agree on A. Then by the identity principle, f = g. This proves the claim.
Now we prove by contradiction that fn is locally uniformly Cauchy in G. Suppose it is not locally uniformly Cauchy. Then there exists a compact set K ⊂G and ϵ > 0 and z0 ∈K such that for every N ∈N, there exist mN, kN > N so that |fmN(z0) −fkN(z0)| > ϵ Then we can choose subsequences fmn and fkn of fn so that |fmn(z0) −fkn(z0)| > ϵ ∀n ∈N 6 By the Stieljes-Osgood Theorem, there exist locally uniformly convergent subsequences of fmn and fkn, which we denote by f ℓ mn and f p kn. By the previous claim, f ℓ mn and f p kn converge to the same limit. But this contradicts the fact that they can’t converge pointwise to the same value at z0, because there exists ϵ > 0 so that |f ℓ mn(z0) −f p kn(z0)| > ϵ ∀ℓ, p Thus fn is locally uniformly Cauchy. We showed in class that being locally uinformly Cauchy is equivalent to being locally uniformly convergent.
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Derivatives of Trig and Hyperbolic Functions
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Derivatives of Trig and Hyperbolic Functions
n0 f u n c t i o n s s i n(x) c o l o r s d e r i v a t i v e b a c k g r o u n d T h e z e r o t h d e r i v a t i v e o f s i n(x)i s s i n(x).
This Demonstration plots the selected function and its
th
n
derivative.
External Links
Cosine (Wolfram MathWorld)
Derivative (Wolfram MathWorld)
Derivatives of Sine
Hyperbolic Cosine (Wolfram MathWorld)
Hyperbolic Sine (Wolfram MathWorld)
Hyperbolic Tangent (Wolfram MathWorld)
Sine (Wolfram MathWorld)
Tangent (Wolfram MathWorld)
Permanent Citation
Itsaso Aranzabal
"Derivatives of Trig and Hyperbolic Functions"
Wolfram Demonstrations Project
Published: June 8, 2011
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Pentagon in Maths | Shape, Parts, Types, Examples
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The pentagon is a two-dimensional polygon with five straight sides and five interior angles. There are two main types of pentagons:
Regular Pentagon: When all sides and angles are equal, it is called a regular pentagon. In a regular pentagon, each interior angle measures 108°, and the sum of all interior angles is always 540°.
Irregular Pentagon: If the sides and angles are not equal, the pentagon is classified as an irregular pentagon.
Below is the figure of a regular Pentagon with sides s:
Note: The term "Pentagon" originates from the Greek words "Penta," which means "five," and "gonia," which means "angles." Thus, a Pentagon is a geometric figure defined by having five sides and five interior angles.
In the case of a regular pentagon, all five sides are equal in length, all five interior angles measure 108 degrees, and the shape possesses both reflectional and rotational symmetry about its center, resulting in five lines of symmetry.
Table of Content
Parts of Pentagon
Types of Pentagons
Angles in a Pentagon
Real-Life Examples of Pentagon
Properties of Pentagon
Formulas Related to the Pentagon
Parts of Pentagon
The parts of a pentagon are shown in the image below:
Let's learn about these parts in detail:
| Term | Definition |
--- |
| Side | One of the five line segments that together form the boundary of a pentagon. |
| Vertex | A point where two sides of the shape meet. It's also referred to as a corner. |
| Diagonal | A line segment connecting two non-adjacent vertices of the polygon. |
| Exterior Angle | The angle between one side of the polygon and the extension of an adjacent side. |
| Interior Angle | The angle formed inside the polygon between two adjacent sides. |
| Apothem | A line from the center of the polygon perpendicular to one of its sides. It also represents the shortest distance from the center to the side. |
Types of Pentagons
Pentagons can be classified into four types depending on their sides, angles, and vertices.
Based on the Length of the Side: Regular Pentagon, Irregular Pentagon.
Based on the Measure of Angle: Convex Pentagon, Concave Pentagon.
Other types: Equilateral Pentagon, Cyclic Pentagon.
Let's discuss them in detail.
Regular and Irregular Pentagons
A regular pentagon has all its sides of equal length, and all its angles have the same measure. This symmetry ensures that the polygon looks the same from any angle or side. In the case of a regular pentagon, it always appears identical.
An irregular pentagon lacks this symmetry because it has varying side lengths and angles. As a result, the shape may look different when observed from different angles or sides.
Convex and Concave Pentagon
A convex pentagon is a polygon in which all its vertices point outward, creating a shape that does not point inward. In a convex pentagon, no internal angle is greater than 180°.
A concave pentagon has a bowl-like structure between some sides and at least one vertex that points inward. In a concave pentagon, at least one internal angle is greater than 180°.
Equilateral Pentagon
An Equilateral Pentagon is a geometric shape where all five sides are of the same length. While the angles within this type of pentagon can vary within a specific range, it is referred to as "equilateral and equiangular" when all sides and angles are equal.
Cyclic Pentagon
A Cyclic Pentagon is a polygon in geometry where all of its vertices are positioned on the circumference of a circle. This characteristic of having its vertices on the circle's boundary is what defines it as a "cyclic" pentagon. A classic example of a cyclic pentagon is a regular pentagon.
Angles in a Pentagon
An angle is created when two sides of the Pentagon intersect at a common point called the Vertex. A pentagon has two types of angles.
Interior Angle
Exterior Angle
Let's discuss both these angles in detail.
Interior Angle of a Pentagon
An Interior Angle is the angle formed by two adjacent sides of the shape on the inside. When two straight lines intersect within the shape, they create Interior Angles.
Sum of Interior Angle of Pentagon = 540°
A Pentagon can be thought of as composed of three triangles. Therefore, the Total Sum of Angles in a Pentagon is equivalent to the sum of angles in three triangles, which is 3 times the sum of angles in one triangle (180 degrees). This results in a sum of 540 degrees for the Interior Angles of a Pentagon.
Sum of Interior Angles in any Polygon = 180° × (n − 2)
Where 'n' represents the number of sides. In the case of a Pentagon with 5 sides, this formula will be:
Sum of Interior Angles of a Pentagon = 180° × (5 − 2) = 3 × 180° = 540°.
Note: Each Interior Angle of a Regular Pentagon is equal to 540° ÷ n = 540° ÷ 5 = 108°.
Exterior Angle of a Pentagon
An Exterior Angle is the angle formed by two adjacent sides of the shape on the outside. It measures the angle at a specific vertex, but on the exterior of the shape.
Sum of Exterior Angles of a Pentagon = 360°.
To prove that the sum of exterior angles of a polygon is 360°, we can follow these steps:
We know the formula for the sum of interior angles of a regular polygon with 'n' sides, which is 180° × (n − 2).
Each interior angle in the polygon can be calculated as: 180° × (n-2)/n .
It's a known fact that each exterior angle in a polygon is supplementary to its corresponding interior angle.
So, Each exterior angle can be expressed as: [180°n - 180°n + 360°]/n, which simplifies to 360°/n.
To find the Total Sum of Exterior Angles for the polygon, we multiply the number of sides 'n' by the measure of each exterior angle (360°/n).
Applying this to a pentagon with 5 sides (n = 5), we observe that the sum of exterior angles for the pentagon is 5 x (360°/5) = 360°
Note: Each exterior angle of a regular pentagon is equal to 360° ÷ n = 360° ÷ 5 = 72°.
Real-Life Examples of Pentagon
A diamond can resemble a pentagon with its five sides and five corners.
The headquarters of the United States Department of Defense is famously known as the Pentagon due to its architectural resemblance to a pentagon shape.
A football is constructed from several black and white pentagonal patches of five-sided shape.
Echinoderms like sea stars exhibit a pentagonal symmetry in their body structure.
Properties of Pentagon
A Pentagon is a 2D shape featuring five sides and five interior angles. Its key properties include:
The sum of the interior angles in a Pentagon is always 540°.
For a Regular Pentagon:
All five sides are of equal length.
All interior angles are congruent, measuring 108° each.
All exterior angles are also congruent, with a measurement of 72°.
Regular pentagons have five lines of symmetry, dividing the shape into congruent parts.
They also possess five rotational symmetries.
Five diagonals intersect at a common point within the pentagon.
The ratio of the diagonal's length to the side's length in a regular pentagon is the golden ratio, (1 + √5)/2.
Line of Symmetry
The number of lines of symmetry in a regular polygon is equivalent to the number of its sides. These symmetrical lines extend from a vertex to the midpoint of the opposite side, creating a total of 5 lines that divide the pentagon into congruent halves. A regular pentagon has five lines of symmetry: one horizontal, one vertical, and three diagonals.
Formulas Related to the Pentagon
The following table shows the list of all the formulas necessary to calculate the various pentagon properties:
| Property | Formula |
--- |
| Perimeter | For Regular Pentagon: 5(a). For Irregular Pentagon: The Sum of all the sides. |
| Area | For Regular Pentagon: 415(5+25a2) If Apothem is given: ½ ⨉ P ⨉ a If only the radius is given: (25)r2(sin 72°) If only the side length is given: (4 tan36°)5s2 For Irregular Pentagon: Divide the pentagon into smaller polygons, then calculate the area of each polygon and sum them up. |
| Diagonal | Number of diagonals: 2n(n−3) Diagonal length: 21+5 a |
| Side | P/5 {where P is perimeter} |
People Also Read:
Triangle
Quadrilateral
Diagonal Formula
Pentagonal Pyramid
Pentagonal Prism
Polygon
Types of Polygons
Solved Examples of Pentagon
Example 1: Determine the area of a Regular Pentagon if it has a side length of 20 cm and an apothem of 15 cm.
Solution:
Given Data,
Side Length = 20 cm
Length of apothem = 15 cm
Area = ½ × perimeter × apothem.
In this case, the perimeter is 5 times the length of one side, which is 20 cm. So, the formula becomes:
Area = ½ × 5 × 20 × 15.
Solving this equation:
Area = ½ × 5 × 20 × 15 = ½ × 1500 = 750 square cm.
Hence, the area of the Regular Pentagon is 750 square cm.
Example 2: If the perimeter of a Regular Pentagon is 400 cm, find the length of each side.
Solution:
The perimeter of the Regular Pentagon is 400 cm.
The perimeter of a Regular Pentagon is equal to the product of the number of sides and the length of each side. In this case, there are 5 sides, so:
Perimeter = 5 × Side
Now, we can solve for the length of each side:
400 cm = 5 × Side
To find the length of each side, divide both sides of the equation by 5:
Side = 400 cm / 5 = 80 cm
So, the length of each side of the Regular Pentagon is 80 cm.
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Protein G genes: structure and distribution of IgG-binding and albumin-binding domains - PubMed
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. 1989 Mar;3(3):319-27.
doi: 10.1111/j.1365-2958.1989.tb00177.x.
Protein G genes: structure and distribution of IgG-binding and albumin-binding domains
U Sjöbring1,L Björck,W Kastern
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1 Department of Medical Microbiology, University of Lund, Sweden.
PMID: 2473373
DOI: 10.1111/j.1365-2958.1989.tb00177.x
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Protein G genes: structure and distribution of IgG-binding and albumin-binding domains
U Sjöbring et al. Mol Microbiol.1989 Mar.
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. 1989 Mar;3(3):319-27.
doi: 10.1111/j.1365-2958.1989.tb00177.x.
Authors
U Sjöbring1,L Björck,W Kastern
Affiliation
1 Department of Medical Microbiology, University of Lund, Sweden.
PMID: 2473373
DOI: 10.1111/j.1365-2958.1989.tb00177.x
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Abstract
Protein G (also designated Fc receptor type III) is the IgG-binding protein of group C and G streptococci. Protein G has also been shown to bind human serum albumin but at a site that is structurally separated from the IgG-binding region. From the known gene sequence of protein G, two synthetic oligonucleotides were constructed for use as probes in DNA-hybridization experiments to study the structure and distribution of the albumin- and IgG-binding regions in bacterial strains belonging to different species. Thus, one of the probes corresponded to repeats within the IgG-binding region (I) and the other corresponded to repeats in the albumin-binding encoding region (II). Probe I showed strong hybridization to DNA isolated from 31 human group C and G strains, whereas hybridization to probe II was variable. With the three restriction endonucleases used, three restriction patterns were found in Southern blot experiments. No fundamental difference could be detected in hybridization experiments, either between strains of group C and G streptococci, or between isolates of different clinical origin. No hybridization to DNA from other bacterial species was found.
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https://www.math.purdue.edu/files/academic/courses/2010spring/MA26200/4-2.pdf
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240 CHAPTER 4 Vector Spaces 5. Verify the commutative law of addition for vectors in R4.
6. Verify the associative law of addition for vectors in R4.
7. Verify properties (4.1.5)–(4.1.8) for vectors in R3.
8. Show with examples that if x is a vector in the first quadrant of R2 (i.e., both coordinates of x are posi-tive) and y is a vector in the third quadrant of R2 (i.e., both coordinates of y are negative), then the sum x+y could occur in any of the four quadrants.
4.2 Definition of a Vector Space In the previous section, we showed how the set Rn of all ordered n-tuples of real num-bers, together with the addition and scalar multiplication operations defined on it, has the same algebraic properties as the familiar algebra of geometric vectors. We now push this abstraction one step further and introduce the idea of a vector space. Such an ab-straction will enable us to develop a mathematical framework for studying a broad class of linear problems, such as systems of linear equations, linear differential equations, and systems of linear differential equations, which have far-reaching applications in all areas of applied mathematics, science, and engineering.
Let V be a nonempty set. For our purposes, it is useful to call the elements of V vectors and use the usual vector notation u, v, . . . , to denote these elements. For example, if V is the set of all 2 × 2 matrices, then the vectors in V are 2 × 2 matrices, whereas if V is the set of all positive integers, then the vectors in V are positive integers. We will be interested only in the case when the set V has an addition operation and a scalar multiplication operation defined on its elements in the following senses: Vector Addition: A rule for combining any two vectors in V . We will use the usual + sign to denote an addition operation, and the result of adding the vectors u and v will be denoted u + v.
Real (or Complex) Scalar Multiplication: A rule for combining each vector in V with any real (or complex) number. We will use the usual notation kv to denote the result of scalar multiplying the vector v by the real (or complex) number k.
To combine the two types of scalar multiplication, we let F denote the set of scalars for which the operation is defined. Thus, for us, F is either the set of all real numbers or the set of all complex numbers. For example, if V is the set of all 2 × 2 matrices with complex elements and F denotes the set of all complex numbers, then the usual operation of matrix addition is an addition operation on V , and the usual method of multiplying a matrix by a scalar is a scalar multiplication operation on V . Notice that the result of applying either of these operations is always another vector (2 × 2 matrix) in V .
As a further example, let V be the set of positive integers, and let F be the set of all real numbers. Then the usual operations of addition and multiplication within the real numbers define addition and scalar multiplication operations on V . Note in this case, however, that the scalar multiplication operation, in general, will not yield another vector in V , since when we multiply a positive integer by a real number, the result is not, in general, a positive integer.
We are now in a position to give a precise definition of a vector space.
4.2 Definition of a Vector Space 241 DEFINITION 4.2.1 Let V be a nonempty set (whose elements are called vectors) on which are defined an addition operation and a scalar multiplication operation with scalars in F. We call V a vector space over F, provided the following ten conditions are satisfied: A1. Closure under addition: For each pair of vectors u and v in V , the sum u + v is also in V . We say that V is closed under addition.
A2. Closure under scalar multiplication: For each vector v in V and each scalar k in F, the scalar multiple kv is also in V . We say that V is closed under scalar multiplication.
A3. Commutativity of addition: For all u, v ∈V , we have u + v = v + u.
A4. Associativity of addition: For all u, v, w ∈V , we have (u + v) + w = u + (v + w).
A5. Existence of a zero vector in V : In V there is a vector, denoted 0, satisfying v + 0 = v, for all v ∈V.
A6. Existence of additive inverses in V : For each vector v In V , there is a vector, denoted −v, in V such that v + (−v) = 0.
A7. Unit property: For all v ∈V , 1v = v.
A8. Associativity of scalar multiplication: For all v ∈V and all scalars r, s ∈F, (rs)v = r(sv).
A9. Distributive property of scalar multiplication over vector addition: For all u, v ∈V and all scalars r ∈F, r(u + v) = ru + rv.
A10. Distributive property of scalar multiplication over scalar addition: For all v ∈V and all scalars r, s ∈F, (r + s)v = rv + sv.
Remarks 1. A key point to note is that in order to define a vector space, we must start with all of the following: (a) A nonempty set of vectors V .
(b) A set of scalars F (either R or C).
242 CHAPTER 4 Vector Spaces (c) An addition operation defined on V .
(d) A scalar multiplication operation defined on V .
Then we must check that the axioms A1–A10 are satisfied.
2. Terminology: A vector space over the real numbers will be referred to as a real vector space, whereas a vector space over the complex numbers will be called a complex vector space.
3. As indicated in Definition 4.2.1, we will use boldface to denote vectors in a general vector space. In handwriting, it is strongly advised that vectors be denoted either as → v or as v ∼. This will avoid any confusion between vectors in V and scalars in F.
4. When we deal with a familiar vector space, we will use the usual notation for vectors in the space. For example, as seen below, the set Rn of ordered n-tuples is a vector space, and we will denote vectors here in the form (x1, x2, . . . , xn), as in the previous section. As another illustration, it is shown below that the set of all real-valued functions defined on an interval is a vector space, and we will denote the vectors in this vector space by f, g, . . . .
Examples of Vector Spaces 1. The set of all real numbers, together with the usual operations of addition and multiplication, is a real vector space.
2. The set of all complex numbers is a complex vector space when we use the usual operations of addition and multiplication by a complex number. It is also possible to restrict the set of scalars to R, in which case the set of complex numbers becomes a real vector space.
3. The set Rn, together with the operations of addition and scalar multiplication defined in (4.1.13) and (4.1.14), is a real vector space. As we saw in the previous section, the zero vector in Rn is the n-tuple of zeros (0, 0, . . . , 0), and the additive inverse of the vector v = (x1, x2, . . . , xn) is −v = (−x1, −x2, . . . , −xn).
Strictly speaking, for each of the examples above it is necessary to verify all of the axioms A1–A10 of a vector space. However, in these examples, the axioms hold immediately as well-known properties of real and complex numbers and n-tuples.
Example 4.2.2 Let V be the set of all 2 × 2 matrices with real elements. Show that V , together with the usual operations of matrix addition and multiplication of a matrix by a real number, is a real vector space.
Solution: We must verify the axioms A1–A10. If A and B are in V (that is, A and B are 2 × 2 matrices with real entries), then A + B and kA are in V for all real numbers k. Consequently, V is closed under addition and scalar multiplication, and therefore Axioms A1 and A2 of the vector space definition hold.
A3. Given two 2 × 2 matrices A = a1 a2 a3 a4 and B = b1 b2 b3 b4 , we have A + B = a1 a2 a3 a4 + b1 b2 b3 b4 = a1 + b1 a2 + b2 a3 + b3 a4 + b4 = b1 + a1 b2 + a2 b3 + a3 b4 + a4 = b1 b2 b3 b4 + a1 a2 a3 a4 = B + A.
4.2 Definition of a Vector Space 243 A4. Given three 2 × 2 matrices A = a1 a2 a3 a4 , B = b1 b2 b3 b4 , C = c1 c2 c3 c4 , we have (A + B) + C = a1 a2 a3 a4 + b1 b2 b3 b4 + c1 c2 c3 c4 = a1 + b1 a2 + b2 a3 + b3 a4 + b4 + c1 c2 c3 c4 = (a1 + b1) + c1 (a2 + b2) + c2 (a3 + b3) + c3 (a4 + b4) + c4 = a1 + (b1 + c1) a2 + (b2 + c2) a3 + (b3 + c3) a4 + (b4 + c4) = a1 a2 a3 a4 + b1 + c1 b2 + c2 b3 + c3 b4 + c4 = a1 a2 a3 a4 + b1 b2 b3 b4 + c1 c2 c3 c4 = A + (B + C).
A5. If A is any matrix in V , then A + 0 0 0 0 = A.
Thus, 02 is the zero vector in V .
A6. The additive inverse of A = a b c d is −A = −a −b −c −d , since A + (−A) = a + (−a) b + (−b) c + (−c) d + (−d) = 0 0 0 0 = 02.
A7. If A is any matrix in V , then 1A = A, thus verifying the unit property.
A8. Given a matrix A = a b c d and scalars r and s, we have (rs)A = (rs)a (rs)b (rs)c (rs)d = r(sa) r(sb) r(sc) r(sd) = r sa sb sc sd = r(sA), as required.
A9. Given matrices A = a1 a2 a3 a4 and B = b1 b2 b3 b4 and a scalar r, we have r(A + B) = r a1 a2 a3 a4 + b1 b2 b3 b4 = r a1 + b1 a2 + b2 a3 + b3 a4 + b4 = r(a1 + b1) r(a2 + b2) r(a3 + b3) r(a4 + b4) = ra1 + rb1 ra2 + rb2 ra3 + rb3 ra4 + rb4 = ra1 ra2 ra3 ra4 + rb1 rb2 rb3 rb4 = rA + rB.
244 CHAPTER 4 Vector Spaces A10. Given A, r, and s as in A8 above, we have (r + s)A = (r + s)a (r + s)b (r + s)c (r + s)d = ra + sa rb + sb rc + sc rd + sd = ra rb rc rd + sa sb sc sd = rA + sA, as required.
Thus V , together with the given operations, is a real vector space.
□ Remark In a manner similar to the previous example, it is easily established that the set of all m × n matrices with real entries is a real vector space when we use the usual operations of addition of matrices and multiplication of matrices by a real number. We will denote the vector space of all m × n matrices with real elements by Mm×n(R), and we denote the vector space of all n × n matrices with real elements by Mn(R).
Example 4.2.3 Let V be the set of all real-valued functions defined on an interval I. Define addition and scalar multiplication in V as follows. If f and g are in V and k is any real number, then f + g and kf are defined by (f + g)(x) = f (x) + g(x) for all x ∈I, (kf )(x) = kf (x) for all x ∈I.
Show that V , together with the given operations of addition and scalar multiplication, is a real vector space.
Solution: It follows from the given definitions of addition and scalar multiplication that if f and g are in V , and k is any real number, then f +g and kf are both real-valued functions on I and are therefore in V . Consequently, the closure axioms A1 and A2 hold.
We now check the remaining axioms.
A3. Let f and g be arbitrary functions in V . From the definition of function addition, we have (f + g)(x) = f (x) + g(x) = g(x) + f (x) = (g + f )(x), for all x ∈I. (The middle step here follows from the fact that f (x) and g(x) are real numbers associated with evaluating f and g at the input x, and real number addition commutes.) Consequently, f + g = g + f (since the values of f + g and g + f agree for every x ∈I), and so addition in V is commutative.
A4. Let f, g, h ∈V . Then for all x ∈I, we have (f + g) + h = (f + g)(x) + h(x) = [f (x) + g(x)] + h(x) = f (x) + [g(x) + h(x)] = f (x) + (g + h)(x) = f + (g + h).
Consequently, (f +g)+h = f +(g+h), so that addition in V is indeed associative.
A5. If we define the zero function, O, by O(x) = 0, for all x ∈I, then (f + O)(x) = f (x) + O(x) = f (x) + 0 = f (x), for all f ∈V and all x ∈I, which implies that f + O = f . Hence, O is the zero vector in V . (See Figure 4.2.1.) y f(x) x y I y —f(x) y O(x) Figure 4.2.1: In the vector space of all functions defined on an interval I, the additive inverse of a function f is obtained by reflecting the graph of f about the x-axis. The zero vector is the zero function O(x).
4.2 Definition of a Vector Space 245 A6. If f ∈V , then −f is defined by (−f )(x) = −f (x) for all x ∈I, since f + (−f ) = f (x) + (−f )(x) = f (x) −f (x) = 0 for all x ∈I. This implies that f + (−f ) = O.
A7. Let f ∈V . Then, by definition of the scalar multiplication operation, for all x ∈I, we have (1f )(x) = 1f (x) = f (x).
Consequently, 1f = f .
A8. Let f ∈V , and let r, s ∈R. Then, for all x ∈I, (rs)f = (rs)f (x) = r[sf (x)] = r[(sf )(x)].
Hence, the functions (rs)f and r(sf ) agree on every x ∈I, and hence (rs)f = r(sf ), as required.
A9. Let f, g ∈V and let r ∈R. Then, for all x ∈I, [r(f + g)] (x) = r[(f + g)(x)] = r[f (x) + g(x)] = rf (x) + rg(x) = (rf )(x) + (rg)(x) = (rf + rg)(x).
Hence, r(f + g) = rf + rg.
A10. Let f ∈V , and let r, s ∈R. Then for all x ∈I, (r+s)f = (r+s)f (x) = rf (x)+sf (x) = (rf )(x)+(sf )(x) = (rf +sf )(x), which proves that (r + s)f = rf + sf .
Since all parts of Definition 4.2.1 are satisfied, it follows that V , together with the given operations of addition and scalar multiplication, is a real vector space.
□ Remark As the previous two examples indicate, a full verification of the vector space definition can be somewhat tedious and lengthy, although it is usually straightforward.
Be careful to not leave out any important steps in such a verification.
The Vector Space Cn We now introduce the most important complex vector space. Let Cn denote the set of all ordered n-tuples of complex numbers. Thus, Cn = {(z1, z2, . . . , zn) : z1, z2, . . . , zn ∈C}.
We refer to the elements of Cn as vectors in Cn. A typical vector in Cn is (z1, z2, . . . , zn), where each zk is a complex number.
Example 4.2.4 The following are examples of vectors in C2 and C4, respectively: u = (2.1 −3i, −1.5 + 3.9i), v = (5 + 7i, 2 −i, 3 + 4i, −9 −17i).
□ In order to obtain a vector space, we must define appropriate operations of “vector addition” and “multiplication by a scalar” on the set of vectors in question. In the case of Cn, we are motivated by the corresponding operations in Rn and thus define the addition 246 CHAPTER 4 Vector Spaces and scalar multiplication operations componentwise. Thus, if u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) are vectors in Cn and k is an arbitrary complex number, then u + v = (u1 + v1, u2 + v2, . . . , un + vn), ku = (ku1, ku2, . . . , kun).
Example 4.2.5 If u = (1 −3i, 2 + 4i), v = (−2 + 4i, 5 −6i), and k = 2 + i, find u + kv.
Solution: We have u + kv = (1 −3i, 2 + 4i) + (2 + i)(−2 + 4i, 5 −6i) = (1 −3i, 2 + 4i) + (−8 + 6i, 16 −7i) = (−7 + 3i, 18 −3i).
□ It is straightforward to show that Cn, together with the given operations of addition and scalar multiplication, is a complex vector space.
Further Properties of Vector Spaces The main reason for formalizing the definition of an abstract vector space is that any results that we can prove based solely on the definition will then apply to all vector spaces we care to examine; that is, we do not have to prove separate results for geometric vectors, m × n matrices, vectors in Rn or Cn, or real-valued functions, and so on. The next theorem lists some results that can be proved using the vector space axioms.
Theorem 4.2.6 Let V be a vector space over F.
1. The zero vector is unique.
2. 0u = 0 for all u ∈V .
3. k0 = 0 for all scalars k ∈F.
4. The additive inverse of each element in V is unique.
5. For all u ∈V , −u = (−1)u.
6. If k is a scalar and u ∈V such that ku = 0, then either k = 0 or u = 0.
Proof 1. Suppose there were two zero vectors in V , denoted 01 and 02. Then, for any v ∈V , we would have v + 01 = v (4.2.1) and v + 02 = v.
(4.2.2) We must prove that 01 = 02. But, applying (4.2.1) with v = 02, we have 02 = 02 + 01 = 01 + 02 (Axiom A3) = 01 (from (4.2.2) with v = 01).
Consequently, 01 = 02, so the zero vector is unique in a vector space.
4.2 Definition of a Vector Space 247 2. Let u be an arbitrary element in a vector space V . Since 0 = 0 + 0, we have 0u = (0 + 0)u = 0u + 0u, by Axiom A10. Now Axiom A6 implies that the vector −(0u) exists, and adding it to both sides of the previous equation yields 0u + [−(0u)] = (0u + 0u) + [−(0u)].
Thus, since addition in a vector space is associative (Axiom A4), 0u + [−(0u)] = 0u + (0u + [−(0u)]).
Applying Axiom A6 on both sides and then using Axiom A5, this becomes 0 = 0u + 0 = 0u, and this completes the verification of (2).
3. Using the fact that 0 = 0 + 0 (by Axiom A5), the proof here proceeds along the same lines as the proof of result 2. We leave the verification to the reader as an exercise (Problem 21 ).
4. Let u ∈V be an arbitrary vector, and suppose that there were two additive inverses, say v and w, for u. According to Axiom A6, this implies that u + v = 0 (4.2.3) and u + w = 0.
(4.2.4) We wish to show that v = w. Now, Axiom A6 implies that a vector −v exists, so adding it on the right to both sides of (4.2.3) yields (u + v) + (−v) = 0 + (−v) = −v.
Applying Axioms A4 and A6 on the left side, we simplify this to u = −v.
Substituting this into (4.2.4) yields −v + w = 0.
Adding v to the left of both sides and applying Axioms A4 and A6 once more yields v = w, as desired.
5. To verify that −u = (−1)u for all u ∈V , we note that 0 = 0u = (1 + (−1))u = 1u + (−1)u = u + (−1)u, where we have used property 2 and Axioms A10 and A7. The equation above proves that (−1)u is an additive inverse of u, and by the uniqueness of additive inverses that we just proved, we conclude that (−1)u = −u, as desired.
Finally, we leave the proof of result 6 in Theorem 4.2.6 as an exercise (Problem 22).
248 CHAPTER 4 Vector Spaces Remark The proof of Theorem 4.2.6 involved a number of tedious and seemingly obvioussteps.Itisimportanttoremember,however,thatinanabstractvectorspaceweare not allowed to rely on past experience in deriving results for the first time. For instance, the statement “0 + 0 = 0” may seem intuitively clear, but in our newly developed mathematical structure, we must appeal specifically to the rules A1–A10 given for a vector space. Hence, the statement “0 + 0 = 0” should be viewed as a consequence of Axiom A5 and nothing else. Once we have proved these basic results, of course, then we are free to use them in any vector space context where they are needed. This is the whole advantage to working in the general vector space setting.
We end this section with a list of the most important vector spaces that will be required throughout the remainder of the text. In each case the addition and scalar mul-tiplication operations are the usual ones associated with the set of vectors.
• Rn, the (real) vector space of all ordered n-tuples of real numbers.
• Cn, the (complex) vector space of all ordered n-tuples of complex numbers.
• Mm×n(R), the (real) vector space of all m × n matrices with real elements.
• Mn(R), the (real) vector space of all n × n matrices with real elements.
• Ck(I), the vector space of all real-valued functions that are continuous and have (at least) k continuous derivatives on I. We will show that this set of vectors is a (real) vector space in the next section.
• Pn, the (real) vector space of all real-valued polynomials of degree ≤n with real coefficients. That is, Pn = {a0 + a1x + a2x2 + · · · + anxn : a0, a1, . . . , an ∈R}.
Weleavetheverification that Pn is a(real) vectorspaceasanexercise(Problem23).
Exercises for 4.2 Key Terms Vector space (real or complex), Closure under addition, Clo-sure under scalar multiplication, Commutativity of addition, Associativity of addition, Existence of zero vector, Existence of additive inverses, Unit property, Associativity of scalar multiplication, Distributive properties, Examples: Rn, Cn, Mn(R), Ck(I), Pn.
Skills • Be able to define a vector space. Specifically, be able to identify and list the ten axioms A1–A10 governing the vector space operations.
• Know each of the standard examples of vector spaces given at the end of the section, and know how to per-form the vector operations in these vector spaces.
• BeabletocheckwhetherornoteachoftheaxiomsA1– A10 holds for specific examples V . This includes, if possible, closure of V under vector addition and scalar multiplication, as well as identification of the zero vec-tor and the additive inverse of each vector in the set V .
• Be able to prove basic properties that hold generally for vector spaces V (see Theorem 4.2.6).
4.2 Definition of a Vector Space 249 True-False Review For Questions 1–8, decide if the given statement is true or false, and give a brief justification for your answer. If true, you can quote a relevant definition or theorem from the text.
If false, provide an example, illustration, or brief explanation of why the statement is false.
1. The zero vector in a vector space V is unique.
2. If v is a vector in a vector space V, and r and s are scalars such that rv = sv, then r = s.
3. The set Z of integers, together with the usual oper-ations of addition and scalar multiplication, forms a vector space.
4. If x and y are vectors in a vector space V , then the additive inverse of x + y is (−x) + (−y).
5. The additive inverse of a vector v in a vector space V is unique.
6. The set {0}, with the usual operations of addition and scalar multiplication, forms a vector space.
7. The set {0, 1}, with the usual operations of addition and scalar multiplication, forms a vector space.
8. The set of positive real numbers, with the usual op-erations of addition and scalar multiplication, forms a vector space.
Problems For Problems 1–5, determine whether the given set of vectors is closed under addition and closed under scalar multiplica-tion. In each case, take the set of scalars to be the set of all real numbers.
1. The set of all rational numbers.
2. The set of all upper triangular n×n matrices with real elements.
3. The set of all solutions to the differential equation y′+9y = 4x2. (Do not solve the differential equation.) 4. The set of all solutions to the differential equation y′ + 9y = 0. (Do not solve the differential equation.) 5. The set of all solutions to the homogeneous linear sys-tem Ax = 0.
6. Let S = {A ∈M2(R) : det(A) = 0}.
(a) Is the zero vector from M2(R) in S?
(b) Give an explicit example illustrating that S is not closed under matrix addition.
(c) Is S closed under scalar multiplication? Justify your answer.
7. Let N = {1, 2, . . . } denote the set of all positive in-tegers. Give three reasons why N, together with the usual operations of addition and scalar multiplication, is not a real vector space.
8. We have defined the set R2 = {(x, y) : x, y ∈R}, together with the addition and scalar multiplication operations as follows: (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2), k(x1, y1) = (kx1, ky1).
Give a complete verification that each of the vector space axioms is satisfied.
9. Determine the zero vector in the vector space M2×3(R), and the additive inverse of a general ele-ment. (Note that the vector space axioms A1–A4 and A7–A10 follow directly from matrix algebra.) 10. Generalize the previous exercise to find the zero vec-tor and the additive inverse of a general element of Mm×n(R).
11. Let P denote the set of all polynomials whose degree is exactly 2. Is P a vector space? Justify your answer.
12. On R+, the set of positive real numbers, define the operations of addition and scalar multiplication as follows: x + y = xy, c · x = xc.
Note that the multiplication and exponentiation ap-pearing on the right side of these formulas refer to the ordinary operations on real numbers. Determine whether R+, together with these algebraic operations, is a vector space.
13. On R2, define the operation of addition and multipli-cation by a real number as follows: (x1, y1) + (x2, y2) = (x1 −x2, y1 −y2), k(x1, y1) = (−kx1, −ky1).
Which of the axioms for a vector space are satisfied by R2 with these algebraic operations?
250 CHAPTER 4 Vector Spaces 14. On R2, define the operation of addition by (x1, y1) + (x2, y2) = (x1x2, y1y2).
Do axioms A5 and A6 in the definition of a vector space hold? Justify your answer.
15. On M2(R), define the operation of addition by A + B = AB, and use the usual scalar multiplication operation. De-termine which axioms for a vector space are satisfied by M2(R) with the above operations.
16. On M2(R), define the operations of addition and mul-tiplication by a real number (⊕and · , respectively) as follows: A ⊕B = −(A + B), k · A = −kA, where the operations on the right-hand sides of these equations are the usual ones associated with M2(R).
Determine which of the axioms for a vector space are satisfied by M2(R) with the operations ⊕and ·.
For Problems 17–18, verify that the given set of objects to-gether with the usual operations of addition and scalar mul-tiplication is a complex vector space.
17. C2.
18. M2(C), the set of all 2 × 2 matrices with complex entries.
19. Is C3 a real vector space? Explain.
20. Is R3 a complex vector space? Explain.
21. Prove part 3 of Theorem 4.2.6.
22. Prove part 6 of Theorem 4.2.6.
23. Prove that Pn is a vector space.
4.3 Subspaces Let us try to make contact between the abstract vector space idea and the solution of an applied problem. Vector spaces generally arise as the sets containing the unknowns in a given problem. For example, if we are solving a differential equation, then the basic unknown is a function, and therefore any solution to the differential equation will be an element of the vector space V of all functions defined on an appropriate interval. Con-sequently, the solution set of a differential equation is a subset of V . Similarly, consider the system of linear equations Ax = b, where A is an m × n matrix with real elements.
The basic unknown in this system, x, is a column n-vector, or equivalently a vector in Rn. Consequently, the solution set to the system is a subset of the vector space Rn. As these examples illustrate, the solution set of an applied problem is generally a subset of vectors from an appropriate vector space (schematically represented in Figure 4.3.1).
The question we will need to answer in the future is whether this subset of vectors is a vector space in its own right. The following definition introduces the terminology we will use: Vector space of unknowns Solution set of applied problem: Is S a vector space?
S S V Figure 4.3.1: The solution set S of an applied problem is a subset of the vector space V of unknowns in the problem.
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How to Calculate the Volume of a Rectangle?
ElevatEd
Math
November 12, 2024
We'll go over how to calculate a rectangular prism's volume. The ability to compute the volume of various forms is a useful one in the field of geometry. Although you've probably heard of figuring out a rectangle's area, the phrase "volume of a rectangle" is a little deceptive. More precisely, we speak of the volume of a cuboid, which is a three-dimensional form with rectangular faces on all sides. Finding the volume within a box and figuring out the capacity of storage containers are only two real-world uses for knowing how to compute this.
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This blog article provides an explanation of the formula and several practical uses.
What is Volume?
Let's define volume first before moving on to the computation. To put it simply, volume is the measurement of the three-dimensional area that an object takes up. Consider volume as a shape's capacity, or the amount of liquid, air, or solid that it can hold. Typically, cubic units—such as cubic meters or cubic centimeters—are used to measure it.
To find the volume of a rectangular prism, multiply the length, width, and height of the prism.
The formula for the Volume of a Rectangular Prism
The formula for calculating the volume of a rectangular prism is straightforward:
Using this formula:
. One of the rectangular base's longer sides is measured as length (L).
. One of the rectangular base's shorter sides is measured as width (W).
. The measurement of the distance between the rectangular prism's top and bottom
sides is called its height (H).
Step-by-Step Process
Step 1: Identify the dimensions
Determine the rectangular prism's length, width, and height first. Consider, for illustration, that you have a rectangular box with the following measurements:
. Length= 5 meters
. Width= 3 meters
. Height= 2meters
Step 2: Apply the Formula
Now, using the formula for volume:
Substitute the values:
Step 3: Calculate the Result
Next, perform the multiplication:
So, the volume of the rectangular prism is 30 cubic meters.
Units Matter!
Recall that the calculation of volume depends heavily on the units of measurement. The result will be expressed in cubic meters (m³) if the measurement is made in meters. Comparably, the result in cubic centimeters (cm³) will be obtained if the measurement is made in centimeters. Before calculating, always make sure that your dimensions are in the same unit; otherwise, your findings could be off.
Real-World Applications of Volume Calculation
It is useful to know how to compute a rectangular prism's volume in a variety of contexts, such as building, shipping, and housekeeping. Among the common uses are:
Storage and Packaging: Finding the volume of a box will tell you how much space it can have. Knowing volume helps you make the most of available space while moving or packing for storage.
Building & Construction: Architects and engineers can assess how much space is available or how much material (such as concrete) is needed by using the volume calculation while planning a room or building a structure.
Science and Engineering: Understanding the behavior of gases, fluids, or solid materials under various conditions is crucial in subjects like physics, chemistry, and mechanical engineering. Volume calculations play a major role in this process.
Common Mistakes to Avoid
Mixing Units: Before completing the computation, be sure you convert all units to the same system if your length is in meters and your breadth or height is in centimeters.
Incorrect Order of Operations: While multiplication is commutative (i.e., order doesn’t matter), it’s crucial to multiply all three dimensions and not mix them with surface area computations.
Forgetting Cubic Units: Since volume relates to three-dimensional space, it is important to always state the final result in cubic units, such as cubic meters or cubic centimeters.
The ability to calculate the volume of a rectangular prism is a fundamental one that has many real-world applications. Utilizing the straightforward formula Volume = Length × Width × Height
By using the formula Volume=Length×Width×Height, you may find the volume of an object in a hurry. Understanding how to calculate volume will help you better grasp the capacity of the objects around you, whether you're packing boxes or organizing a building project.
FAQs (Frequently Asked Questions)
Q1: What is a rectangular prism?
Ans: A rectangular prism is a three-dimensional shape with six faces, all of which are rectangles. It has length, width, and height.
Q2: What units should I use for calculating volume?
Ans: You can use any unit of measurement, such as centimeters, meters, or inches. Just ensure that all dimensions are in the same unit for accurate results. The volume will be in cubic units (e.g., cubic centimeters, cubic meters).
Q3: Can the volume be calculated if I only have the surface area?
Ans: Not directly. You would need at least two dimensions (length, width, or height) to calculate the volume. The surface area can provide some clues, but additional information is necessary.
Q4: What if the rectangular prism is not a perfect rectangle?
Ans: If the shape is not a perfect rectangular prism (for example, if it has rounded edges), you may need to approximate the volume using other methods, such as breaking it down into smaller rectangular prisms or using calculus for more complex shapes.
Q5: Are there any real-world applications for calculating volume?
Ans: Yes! Volume calculations are essential in various fields, including construction (to determine material needs), shipping (to calculate space in containers), and even cooking (to measure ingredients).
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Juvenile Idiopathic Arthritis
Akaluck Thatayatikom; Renee Modica; Alessandro De Leucio.
Author Information and Affiliations
Authors
Akaluck Thatayatikom; Renee Modica1; Alessandro De Leucio2.
Affiliations
1 University of Florida
2 Hopital universitaire des enfants Reine Fabiola
Last Update: January 16, 2023.
Continuing Education Activity
Juvenile idiopathic arthritis (JIA), a heterogeneous group of chronic arthritis, is the most common chronic rheumatological condition in children. There are 7 different JIA subtypes with distinct phenotypes, genetic predispositions, pathophysiology, laboratory findings, disease course, and prognosis. This activity reviews the evaluation and management of JIA and reviews the role of the healthcare team in evaluating and treating patients with this condition.
Objectives:
Describe the classic presentation and typical physical exam findings associated with juvenile idiopathic arthritis.
Explain the imaging findings associated with JIA and the most common tests for the evaluation of juvenile idiopathic arthritis.
Outline the imaging findings associated with JIA and the most common tests for the evaluation of juvenile idiopathic arthritis.
Review the importance of collaboration amongst the interdisciplinary team to improve outcomes for patients affected by juvenile idiopathic arthritis.
Access free multiple choice questions on this topic.
Introduction
Juvenile idiopathic arthritis (JIA) is a heterogeneous group of idiopathic inflammatory arthritis affecting children younger than 16 and lasting 6 weeks or longer. The terminology of chronic arthritis in children has evolved from juvenile chronic arthritis and juvenile rheumatoid arthritis to JIA since 1995. According to the consensus conference of the International League of Associations for Rheumatology in 2001, there are 7 JIA categories: a) oligoarthritis; b) rheumatoid factor (RF) positive polyarthritis; c) RF negative polyarthritis; d) systemic arthritis; e) psoriatic arthritis; f) enthesitis-related arthritis; g) undifferentiated arthritis. These subtypes have distinct phenotypes, genetic predispositions, pathophysiology, laboratory findings, disease course, and prognosis. Although chronic arthritis is mandatory for all subtypes, the extraarticular and the systemic manifestations characterized every specific subtype. Recently, a new preliminary data-driven classification for JIA is proposed and being formally validated by the Pediatric Rheumatology International Trial Organization.
Etiology
The cause and trigger of chronic arthritis in JIA remain unclear. Abnormal immune responses triggered by the interactions between environmental factors in a genetically susceptible individual are speculative. Some environmental factors such as antibiotic exposure and C-section deliveries are potential risks; however, breastfeeding and household siblings are possible protectives. The roles of microorganisms such as Parvovirus B19, Epstein-Barr virus, enteric bacteria, Chlamydophila pneumoniae, and streptococcal infections are still inconclusive.
Based on familial aggregation studies and the concordance rate of 25% to 40% in monozygotic twins, genetic factors play a significant role. Specific HLA alleles and non-HLA genes may be vulnerable to particular JIA subtypes and uveitis. Human leukocyte antigen(HLA)-A2, HLA-DRB1:11, and HLA-DRB1:08 are associated with oligoarticular JIA and RF negative polyarticular JIA. HLADRB1:01 and HLADRB1:04 are associated with RF-positive polyarthritis. HLADRB1:04 and DRB1:11 are associated with systemic JIA. HLA-B27:05 and HLA-B27:04 are the most common HLA-B27 subtypes associated with estrogen receptor alpha. HLADRB1:01 and DQA1:01 are related to psoriatic JIA. HLADRB1:11 and HLADRB1:13 are associated with uveitis. The higher number of the HLA-DR risk alleles predisposes to the earlier development of JIA.
Epidemiology
JIA is the most common rheumatic disease reported in children of the Western world. The incidence and prevalence are varied among 1.6 to 23 new cases for 100,000 children, and 3.8 to 400 cases per 100,000 children depending upon study designs, disease categories, and geographical areas. In a US and Canada study, the incidence of JIA is 0.041 to 0.061 per 1000 children. The Utah Population Database provides the prevalence of 1.2 per 1000 in white populations. The relative risk of JIA in siblings varies from 15 to 30, similar to the relative risk of type 1 diabetes.
The frequencies of different subtypes are 50% to 60% for oligoarthritis, 11% to 28% for RF negative polyarthritis, 2% to 7% for RF positive polyarthritis, 10% to 20% for systemic arthritis, 2% to 15% for psoriatic arthritis, 1% to 7% for enthesitis-related arthritis. Specific subtypes are more common in some geographical regions. RF-negative polyarthritis is more common in North America; oligoarthritis is more common in southern Europe. Systemic arthritis and enthesitis-related arthritis are more common in southeast Asia. Uveitis is highest in northern Europe and southern Europe, but it is lowest in Latin America, Africa, the Middle East, and Southeast Asia. Most JIA subtypes occur predominantly in females except enthesitis-related arthritis mainly affects males, and systemic JIA affects men and women equally.
Pathophysiology
The imbalance of regulatory T cells, Th1 (interferon-gamma secreting T cells), and Th17 (interleukin -17 secreting T cells) of adaptive immunity is the feature of most subtypes of JIA. IL-17 induces proinflammatory cytokines and matrix metalloproteinases, leading to joint damage in oligoarthritis, polyarthritis, and psoriatic arthritis. In early stages of erheumatoid arthritis, IL-23 is a crucial cytokine that leads to inflammation through IL-17 and tumor necrosis factor (TNF), and new bone formation through interleukin (IL)-22. In contrast, the critical immunopathophysiology of systemic arthritis is persistent activation of innate immunity, including monocytes, macrophages, and neutrophils. As a result, innate proinflammatory cytokines such as IL-1 beta, IL-6, and IL-18 contribute to symptoms and signs of systemic arthritis.
Histopathology
The synovial membrane of JIA shows a high degree of infiltrating inflammatory cells, including T and B lymphocytes, plasma cells, macrophages, and dendritic cells, villous hypertrophy, and hyperplasia, hyperplasia of synoviocytes, endothelial hyperplasia, and activation, and increased vascularisation with hyperemia. The findings are non-specific and similar to rheumatoid arthritis in adults.
History and Physical
The disease course of JIA is highly unpredictable: in some patients is observed a self-limiting disease while in others, there is an unremitting disease with a high risk of joint destruction. The JIA has the general pattern of inflammatory joint disease (synovitis, joint effusion, soft tissue swelling, osteopenia, bone edema, and erosions) with some additional elements related to developmental age, such as epiphyseal growth disturbances, premature physeal fusion, and limb length inequality.
The thorough history taking, including the age of onset, the affected joints, the duration of arthritis, the associated symptoms or diseases, and physical and MSK examinations are essential for diagnosis and classification of JIA. A diagnosis of JIA is considered in any children younger than 16 years with arthritis for at least six weeks and exclusion of other causes of chronic arthritis.
Evaluation
There is no specific test for diagnosis and predicting disease activity in JIA.
Laboratory
Initial laboratory tests should include complete blood count, erythrocyte sendimentation rate, C-reactive protein, antinuclear antibody, rheumatoid factor (RF), anti-cyclic citrullinated peptide antibodies (anti-CCP), and HLA-B27. The typical inflammatory markers are common, especially in oligoarthritis. A positive RF or anti-CCP provides little value for the diagnosis but may indicate a poorer disease course and outcome. Ferritin, fibrinogen, AST, triglyceride are recommended when there is a concern of macrophage activation syndrome (MAS). Other tests for excluding other diseases may be considerations depending on the differential diagnosis.
Imaging
Imaging serves to improve the certainty of a diagnosis of JIA, narrow the differential diagnosis, and evaluate joint damage. The heterogeneity of clinical manifestations usually requires a multimodal imaging approach. Radiography remains initial imaging used for symptomatic joints; however, the radiographic changes are undetectable in an early stage of JIA. The indirect signs of arthritis in radiography are soft tissue swelling, increased density of soft tissue as well as dislocation of fat folds. Other features are periarticular osteoporosis, joint space narrowing, bone erosion and deformity, and joint subluxation or ankyloses in an advanced stage.
Ultrasound (US) is an imaging modality easily accessible and non-irradiating that plays a significant role in the imaging of JIA. Also, the US allows the comparison with the contralateral side and the dynamic evaluation of joints. The US is capable of assessing synovial thickening, joint effusion, tenosynovitis, enthesitis, and bone erosions. US evaluation of the synovial thickening and the synovitis is particularly important for the diagnosis. They appear as abnormally hypoechoic tissue associated with joint lines or surrounding tendons. In addition to detecting synovitis, the US accurately guides for intra-articular corticosteroid injections. Ultrasound allows evaluation without sedating the patient.
Magnetic resonance imaging (MRI) is the modality gold standard for the study of JIA. All joints affected by pathological inflammatory phenomena can be easily examined in all possible plans and with an excellent contrast resolution of bone and soft tissues. It is the most sensitive imaging technique detecting synovitis. The standard MR imaging protocol needs to include: T1 spin-echo (SE) sequence, fat-suppressed sequence (classic T2 fat-sat; short tau inversion recovery, STIR; DIXON fat-suppression sequence); T1 fat-suppressed sequence precontrast and postcontrast. MRI is the only modality able to objective bone marrow edema and the most sensitive to detect bone erosions.
Clinical-Laboratory Classification
The clinical manifestations and the test results of RF and HLA-B27 will be used to categorized JIA subtypes based upon the ILAR classification. Oligoarthritis is defined as chronic arthritis affecting four joints or less during the first six months of disease. Persistent oligoarthritis is defined as the affected joints being 4 or less after the first 6 months, and extended oligoarthritis is defined as more than four affected joints after the first 6 months.
RF negative polyarthritis is defined as arthritis affecting 5 joints or more during the first 6 months of disease with a negative immunoglobulin (Ig)M RF. RF positive polyarthritis is arthritis affecting 5joints or more during the first 6 months of disease with a positive IgM RF on at least 2 tests three months apart.Systemic arthritis is arthritis with or preceded by a fever of at least a 2-week duration and accompanied by at least 1 of the following: evanescent erythematous rash, generalized lymph node enlargement, hepatomegaly, and/or splenomegaly, or serositis (pericarditis and/or pleuritis and/or peritonitis). See Image.Systemic Onset Juvenile Idiopathic Arthritis.
Psoriatic arthritis is defined as chronic arthritis with psoriasis or chronic arthritis with at least 2 of the following: dactylitis, nail pitting, onycholysis, or psoriasis in a first-degree relative. Enthesitis related arthritis (ERA) is defined as arthritis with enthesitis, or arthritis or enthesitis with at least 2 of SI joint tenderness and/or inflammatory lumbosacral pain, a positive HLA-B27, the onset of arthritis in a boy older than 6, acute anterior uveitis, history of ankylosing spondylitis, ERA, sacroiliitis with inflammatory bowel disease, reactive arthritis or acute anterior uveitis in a first-degree relative. Undifferentiated arthritis is defined as chronic arthritis, which does not fulfill criteria in any subtype or fulfills two or more subtypes.
Treatment / Management
Treatment of JIA requires anti-inflammatory and immunomodulatory drugs and physical therapy, and eventually, surgery, nutritional support, and psychosocial support may be needed. The choice of pharmacological treatment depends on the disease subtypes, disease severity and damage, associated disease, and family acceptance. Nonsteroidal anti-inflammatory drugs (NSAIDs) are the mainstay of initial symptomatic treatment for all subtypes. The NSAID use in JIA has decreased over time with modern aggressive treatment, including methotrexate and biologics. Physical therapy emphasizes range of motion with minimal stress on joints. Swimming is often a good option. Patients should participate in moderate fitness, flexibility, and strengthening exercises.
Differential Diagnosis
The differential diagnosis of chronic arthritis is broad, depending on the clinical presentation and the JIA subtypes. Since JIA is a diagnosis of exclusion, any patients with a positive review of the system should be considered for possible diseases. The following differential diagnosis should be considered:
A) Oligoarthritis need to exclude post-streptococcal reactive arthritis, Lyme arthritis, acute rheumatic fever, reactive arthritis, toxic synovitis, septic arthritis, pyomyositis, steroid-induced osteonecrosis, sickle cell disease, hemophilia, scurvy, osteomyelitis, chronic nonbacterial osteomyelitis (CNO), sports injury, non-accidental injury, pigmented villonodular synovitis, osteoid osteoma, bone tumors, neuroblastoma, leukemia, and lymphoma.
B) Polyarthritis need to exclude post-streptococcal reactive arthritis, Lyme arthritis, acute rheumatic fever, reactive arthritis, scurvy, CNO or chronic recurrent multifocal osteomyelitis, non-accidental injury, systemic lupus erythematosus (SLE), Mixed connective tissue disease, Sjögren syndrome, scleroderma, sarcoidosis, Blau syndrome, arthritis associated with inflammatory bowel diseases, Farber disease, benign hypermobility joint syndrome and amplified musculoskeletal pain syndrome
C) Systemic arthritis need to exclude infections (mycoplasma, cat scratch disease, bacterial endocarditis, Lyme disease), acute rheumatic fever, syndrome of periodic fever, aphthous stomatitis, pharyngitis and cervical adenitis (PFAPA syndrome), autoinflammatory syndromes, systemic vasculitis (polyarteritis nodosa, Kawasaki disease), inflammatory bowel disease, malignancy (leukemia, lymphoma, neuroblastoma), Castleman disease
D) Enthesitis-related arthritis. Apophysitis (especially Osgood-Schlatter, Sever disease), inflammatory bowel diseases, chronic recurrent multifocal osteomyelitis, amplified musculoskeletal pain syndrome.
Arthralgias are common early in the course of sJIA, but arthritis is not always prominent. Any number of joints may be involved when arthritis becomes apparent. Disease in the wrists, knees, and ankles is most typical. Hands, hips, cervical spine, and temporomandibular joints are also sometimes affected, unlike the oligoarticular and polyarticular subtypes of JIA Arthralgias are common early in the course of sJIA, but arthritis is not always prominent. Any number of joints may be involved when arthritis becomes apparent. Disease in the wrists, knees, and ankles is most typical, but the hands, hips, cervical spine, and temporomandibular joints are also sometimes affected. Unlike the oligoarticular and polyarticular subtypes of JIA. Musculoskeletal complaints in children with joint pain are common in general practice and increase significantly after ten years of age in general practice. The most common presentation of musculoskeletal complaints is joint pain and soft tissue pain (65%), and the most common cause of all ages of children is trauma (45%).
Toxicity and Adverse Effect Management
NSAIDs inhibit cyclooxygenases, a necessary enzyme for prostaglandin synthesis. Physiologic roles of prostaglandins are initiating and maintenance of the protective mucosal barrier of the stomach and promoting intrarenal plasma flow and electrolyte balance. The potential gastrointestinal (GI) toxicities and nephrotoxicity should be addressed with the family. A survey of pediatric rheumatologists’ experience reveals significantly more abdominal pain, easy bruising, epistaxis, headaches, and fatigue for nonselective NSAIDs than for selective COX-2 NSAIDs. However, a study in safety profile, including GI toxicity between nonselective NSAIDs (COX1 and COX2 inhibitors) and selective COX-2 inhibitor (celecoxib) is no different. Since over 50% of children with JIA developed of GI symptoms related to NSAID, corticosteroid, and methotrexate use, reviewing the medical history and previous or concurrent use of NSAIDs, corticosteroid or methotrexate is necessary to assess and minimize the potential risk of gastrointestinal toxicity. The potential nephropathy, such as acute interstitial nephritis or acute papillary necrosis, was rarely reported. The prevalence of nephrotoxicity in children is 0.4%, which is five times less than the prevalence in adults. The benefit of routine monitoring (blood and urine) of asymptomatic children with JIA receiving only NSAIDs is unclear.
Prognosis
The prognosis of JIA has changed dramatically in recent years thanks to the availability of novel drugs, which can inhibit the biological mechanisms responsible for persistent inflammation selectively. Prompt and accurate diagnosis and treatment are essential to prevent permanent joint damage and preserve joint functionality. Some studies support the possibility of the existence of a “window of opportunity” in early disease, during which prompt treatment induces higher rates of remission and improves long-term outcomes.
A recent study on 168 patients showed the remission off medication in 48.8% of cases, the remission on medication (or minimal disease activity) in 49.9% of cases, and only 1.3% of subjects were no-responders. No association was found between the state and duration of remission and age of patients, clinical features, disease course, or laboratory findings.
Complications
The most common complications of JIA are leg-length discrepancy and joint contracture. An extremely fearsome complication is macrophage activation syndrome, due to the uncontrolled activation and proliferation of T lymphocytes and macrophages. The frequency of this syndrome in patients with JIA is unknown, but some studies report that it occurs in up to 10% of cases.
Other important complications are growth retardation, low bone mineral density for chronologic age, severe hip involvement with the need oh hip prosthesis, and amyloidosis.
Deterrence and Patient Education
A chronic disabling pathology such as JIA requires proper child education and family support. Numerous experiences have been made on this topic. Among the most recent and interesting new studies, there is the use of comics for the education of children and artificial intelligence.
Pearls and Other Issues
Any child with a history of persisting limp after a minor trauma should look for other diseases, including JIA and leukemia. The JIA has to be suspected in any children younger than 16 years with arthritis for at least six weeks and exclusion of other causes of chronic arthritis.
Enhancing Healthcare Team Outcomes
A recent study was realized by the collaboration between physicians and patient associations. A questionnaire was proposed to 622 parents in 23 European countries. The questions covered various domains of JIA care, including demographics, diagnosis, referrals to different health care professionals, access to pain and fatigue management and support groups, the information they received about the disease, and awareness of and participation in research.
The data furnished a view on JIA patients’ and their parents’ perspectives regarding the care they receive. It’s essential to increase the awareness of the existence of supportive care, such as pain and fatigue management, as well as support groups, or to start new initiatives where these are lacking. [Level 4].
Pediatric pharmacists review medications and provide patient and family education about the importance of compliance and potential side effects, and report any concerns to the clinician staff. Pediatric and rehabilitation nurses provide direct care and facilitate communication between the interprofessional health team members. This type of collaboration among team members will lead to better outcomes for these patients. [Level 5]
Review Questions
Access free multiple choice questions on this topic.
Click here for a simplified version.
Comment on this article.
Figure
Systemic Onset Juvenile Idiopathic Arthritis. Juvenile idiopathic arthritis is usually accompanied by a salmon-colored maculopapular rash. Contributed by A Thatayatikom, MD
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: Disclosure: Akaluck Thatayatikom declares no relevant financial relationships with ineligible companies.
: Disclosure: Renee Modica declares no relevant financial relationships with ineligible companies.
: Disclosure: Alessandro De Leucio declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
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Thatayatikom A, Modica R, De Leucio A. Juvenile Idiopathic Arthritis. [Updated 2023 Jan 16]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Continuing Education Activity
Introduction
Etiology
Epidemiology
Pathophysiology
Histopathology
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Toxicity and Adverse Effect Management
Prognosis
Complications
Deterrence and Patient Education
Pearls and Other Issues
Enhancing Healthcare Team Outcomes
Review Questions
References
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NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 1. σ-algebras Let X be a set, and let 2X denote the set of all subsets of X. Let Ec denote the complement of E in X, and for E, F ⊂X, write E \ F = E ∩F c. Let E∆F denote the symmetric difference of E and F: E∆F := (E \ F) ∪(F \ E) = (E ∪F) \ (E ∩F).
Definition 1.1. Let X be a set. A Boolean algebra is a nonempty collection A ⊂2X which is closed under finite unions and complements. A σ-algebra is a Boolean algebra which is also closed under countable unions.
If M ⊂N ⊂2X are σ-algebras, then M is coarser than N ; likewise N is finer than M .
◁ Remark 1.2. If Eα is any collection of sets in X, then [ α Ec α !c = \ α Eα.
(1) Hence a Boolean algebra (resp. σ-algebra) is automatically closed under finite (resp.
countable) intersections.
It follows that a Boolean algebra (and a σ-algebra) on X always contains ∅and X. (Proof: X = E ∪Ec and ∅= E ∩Ec.) ⋄ Definition 1.3. A measurable space is a pair (X, M ) where M ⊂2X is a σ-algebra. A function f : X →Y from one measurable spaces (X, M ) to another (Y, N ) is called measurable if f −1(E) ∈M whenever E ∈N .
◁ Definition 1.4. A topological space X = (X, τ) consists of a set X and a subset τ of 2X such that (i) ∅, X ∈τ; (ii) τ is closed under finite intersections; (iii) τ is closed under arbitrary unions.
The set τ is a topology on X.
(a) Elements of τ are called open sets; (b) A subset S of X is closed if X \ S is open; (c) S is a Gδ if S = ∩∞ j=1Oj for open sets Oj; (d) S is an Fσ if it is an (at most) countable union of closed sets; Date: February 24, 2015.
1 2 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 (e) A subset C of X is compact, if for any collection F ⊂τ such that C ⊂∪{T : T ∈F} there exist a finite subset G ⊂F such that C ⊂∪{T : T ∈G}; (f) If τ and σ are both topologies on X, then τ is finer than σ (and σ is coarser than τ) if σ ⊂τ; and (g) If (X, τ) and (Y, σ) are topological spaces, a function f : X →Y is continuous if S ∈σ implies f −1(S) ∈τ.
◁ Example 1.5. If (X, d) is a metric space, then the collection τ of open sets (in the metric space sense) is a topology on X. There are important topologies in analysis that are not metrizable (do not come from a metric).
△ Example 1.6. Let X be a nonempty set.
(a) The power set 2X is the finest σ-algebra on X.
(b) At the other extreme, the set {∅, X} is the coarsest σ-algebra on X.
(c) Let X be an uncountable set. The collection M = {E ⊂X : E is at most countable or X \ E is at most countable } (2) is a σ-algebra (the proof is left as an exercise).
(d) If M ⊂2X a σ-algebra, and E is any nonempty subset of X, then ME := {A ∩E : A ∈M } ⊂2E is a σ-algebra on E (exercise).
(e) If {Mα : α ∈A} is a collection of σ-algebras on X, then their intersection ∩α∈AMα is also a σ-algebra (checking this statement is a simple exercise). Hence given any set E ⊂2X, we can define the σ-algebra M (E ) = \ {M : M is a σ-algebra andE ⊂M }.
(3) Note that the intersection is over a nonempty collection since E is a subset of the σ-algebra 2X. We call M (E ) the σ-algebra generated by E .
(f) An important instance of the construction in (d) is when X is a topological space and E is the collection of open sets of X. In this case the σ-algebra generated by E is called the Borel σ-algebra and is denoted BX. The Borel σ-algebra over R is studied more closely later.
△ The following proposition is trivial but useful.
Proposition 1.7. If M ⊂2X is a σ-algebra and E ⊂M , then M (E ) ⊂M .
† The proposition is used in the following way: suppose we want to prove that a particular statement is true for every set in some σ-algebra M (say, the Borel σ-algebra BX), which we know is generated by a collection of sets E (say, the open sets of X).
Then it suffices to prove that 1) the statement is true for every set in E , and 2) the collection of sets for which the statement is true forms a σ-algebra.
NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 3 A function f : X →Y between topological spaces is said to be Borel measurable if it is measurable when X and Y are equipped with their respective Borel σ-algebras.
Proposition 1.8. If X and Y are topological spaces, then every continuous function f : X →Y is Borel measurable.
† Proof. The proof is left as an exercise. (Hint: follow the strategy described after Propo-sition 1.7.) □ We will defer further discussion of measurable functions.
1.1. The Borel σ-algebra over R. Before going further, we take a closer look at the Borel σ-algebra over R. We begin with a useful lemma on the structure of open subsets of R: Lemma 1.9. Every nonempty open subset U ⊂R is an (at most countable) disjoint union of open intervals.
† Here we allow the “degenerate” intervals (−∞, a), (a, +∞), (−∞, +∞).
Proof outline. First verify that if I and J are intervals and I ∩J ̸= ∅, then I ∪J is an interval. Given x ∈U, let αx = sup{a : [x, a) ⊂U} βx = inf{b : (b, x] ⊂U} and let Ix = (αx, βx). Verify that, for x, y ∈U either Ix = Iy or Ix ∩Iy = ∅. Indeed, x ∼y if Ix = Iy is an equivalence relation on U. Hence, U = ∪x∈UIx expresses U as a disjoint union of nonempty intervals, say U = ∪p∈PIp where P is an index set and the Ip are nonempty intervals. For each q ∈Q ∩U there exists a unique pq such that q ∈Ipq. On the other hand, for each p ∈P there is a q ∈Q ∩U such that q ∈Ip. Thus, the mapping from Q ∩U to P defined by q 7→pq is onto. It follows that P is at most countable.
□ Proposition 1.10 (Generators of BR). Each of the following collections of sets E ⊂2R generates the Borel σ-algebra BR: (i) the open intervals E1 = {(a, b) : a, b ∈R} (ii) the closed intervals E2 = {[a, b] : a, b ∈R} (iii) the (left or right) half-open intervals E3 = {[a, b) : a, b ∈R} or E4 = {[a, b) : a, b ∈ R} (iv) the (left or right) open rays E5 = {(−∞, a) : a ∈R} or E6 = {(a, +∞) : a ∈R} (v) the (left or right) closed rays E7 = {(−∞, a] : a ∈R} or E8 = {[a, +∞) : a ∈R} † Proof. We prove only the open and closed interval cases, the rest are similar and left as exercises. The proof makes repeated use of Proposition 1.7. To prove M (E1) = BR, first 4 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 note that since each interval (a, b) is open, M (E1) ⊂BR by Proposition 1.7. Conversely, each open set U ⊂R is a countable union of open intervals, so M (E1) contains all the open sets of R, and since the open sets generate BR by definition, Proposition 1.7 implies BR ⊂M (E1). Thus M (E1) = BR.
For the closed intervals E2, first note that each closed set is a Borel set, since it is the complement of an open set; thus E2 ⊂BR so M (E2) ⊂BR by Proposition 1.7.
Conversely, each open interval (a, b) is a countable union of closed intervals [a+ 1 n, b−1 n].
More precisely, since a < b we can choose an integer N such that a + 1 N < b −1 N ; then (a, b) = ∞ [ n=N [a + 1 n, b −1 n].
It follows that E1 ⊂M (E2), so by Proposition 1.7 and the first part of the proof, BR = M (E1) ⊂M (E2).
□ 2. Measures Definition 2.1. Let X be a set and M a σ-algebra on X. A measure on M is a function µ : M →[0, +∞] such that (i) µ(∅) = 0, (ii) If {Ej}∞ j=1 is a sequence of disjoints sets in M , then µ ∞ [ j=1 Ej !
= ∞ X j=1 µ(Ej).
If µ(X) < ∞, then µ is called finite; if X = ∪∞ j=1Xj with µ(Xj) < ∞for each j then µ is called σ-finite.
Almost all of the measures of importance in analysis (and certainly all of the measure we will work with) are σ-finite.
A triple (X, M , µ) where X is a set, M is a σ-algebra and µ a measure on M , is called a measure space.
◁ Here are some simple measures and some procedures for producing new measures from old. Non-trivial examples of measures will have to wait for the Caratheodory and Hahn-Kolmogorov theorems in the following sections.
Example 2.2. (a) Let X be any set, and for E ⊂X let |E| denote the cardinality of E, in the sense of a finite number or ∞. The function µ : 2X →[0, +∞] defined by µ(E) = ( |E| if E is finite ∞ if E is infinite NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 5 is a measure on (X, 2X), called counting measure. It is finite if and only if X is finite, and σ-finite if and only if X is countable.
(b) Let X be an uncountable set and M the σ-algebra of (at most) countable and co-countable sets (Example 1.6(b)). For E ∈M define µ(E) = 0 if E is countable and µ(E) = +∞is E is co-countable. Then µ is a measure.
(c) Let (X, M , µ) be a measure space and E ∈M . Recall ME from Example 1.6(c).
The function µE(A) := µ(A ∩E) is a measure on (E, ME). (Why did we assume E ∈M ?) (d) (Linear combinations) If µ is a measure on M and c > 0, then (cµ)(E) : c · mu(E) is a measure, and if µ1, . . . µn are measures on the same M , then (µ1 + · · · µn)(E) := µ1(E) + · · · µn(E) is a measure. Likewise an infinite sum of measures P∞ n=1 µn is a measure. (The proof of this last fact requires a small amount of care.) (e) (Point masses) Let X = R and M = BR (or M = 2R). Fix a point x ∈R and Define for each E ∈M δx(E) = ( 1 if x ∈E 0 otherwise Then δx is a measure, called the point mass at x (or Dirac measure, or, much worse, Dirac delta function).
△ One can also define products and pull-backs of measures, compatible with the con-structions of product and pull-back σ-algebras; these examples will be postponed until we have built up some more machinery of measurable functions.
The disjointification trick in the next Proposition is often useful.
Proposition 2.3 (Disjointification). If M ⊂2X is a σ-algebra if and G1, G2, . . . is a sequence of sets from M , then there exists a sequence F1, F2, . . . of pairwise disjoint sets from M such that n [ j=1 Fj = n [ j=1 Gj for n either a positive integer or ∞.
If ∅̸= M ⊂2X is closed with respect to complements, finite intersections and countable disjoint unions, then M is a σ-algebra.
† Proof. The proof amounts to the observation that if {Gn} is a sequence of subsets of X, then the sets Fn = Gn \ n−1 [ k=1 Gk !
(4) are disjoint and in M , and Sn j=1 Fj = Sn j=1 Gj for all N (and thus S∞ n=1 Fn = S∞ n=1 Gn).
6 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 To prove the second part of the Proposition, given a sequence (Gn) from M use the disjointification trick to obtain a sequence of disjoint sets Fn ∈M such that ∪Gn = ∪Fn.
△ Theorem 2.4 (Basic properties of measures). Let (X, M , µ) be a measure space.
(a) (Monotonicity) If E, F ∈M and E ⊂F, then µ(F) = µ(F \ E) + µ(E).
In particular, µ(E) ≤µ(F) and if µ(E) < ∞, then µ(F \ E) = µ(F) −µ(E).
(b) (Subadditivity) If {Ej}∞ j=1 ⊂M , then µ(S∞ j=1 Ej) ≤P∞ j=1 µ(Ej) (c) (Monotone convergence for sets) If {Ej}∞ j=1 ⊂M and Ej ⊂Ej+1 ∀j, then lim µ(Ej) exists and moreover µ(∪Ej) = lim µ(Ej).
(d) (Dominated convergence for sets) If {Ej}∞ j=1 ⊂M and Ej ⊃Ej+1 ∀j, and µ(E1) < ∞, then lim µ(Ej) exists and moreover µ(∩Ej) = lim µ(Ej).
Proof. a) By additivity, µ(F) = µ(F \ E) + µ(E) ≥µ(E).
b) We use the disjointification trick (see Proposition 2.3): for each j ≥1 let Fj = Ej \ j−1 [ k=1 Ek !
.
Then the Fj are disjoint, and Fj ⊂Ej for all j, so by countable additivity and (a) µ ∞ [ j=1 Ej !
= µ ∞ [ j=1 Fj !
= ∞ X j=1 µ(Fj) ≤ ∞ X j=1 µ(Ej).
c) The sets Fj = Ej \ Ej−1 are disjoint sets whose union is S∞ j=1 Ej, and for each j, Sj k=1 Fk = Ej. So by countable additivity, µ ∞ [ j=1 Ej !
= µ ∞ [ j=1 Fj !
= ∞ X k=1 µ(Fk) = lim j→∞ j X k=1 µ(Fk) = lim j→∞µ j [ k=1 Fk !
= lim j→∞µ(Ej).
d) The sequence µ(Ej) is decreasing (by (a)) and bounded below, so lim µ(Ej) exists.
Let Fj = E1 \ Ej. Then Fj ⊂Fj+1 for all j, and S∞ j=1 Fj = E1 \ T∞ j=1 Ej. So by (c) NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 7 applied to the Fj, and since µ(E1) < ∞, µ(E1) −µ( ∞ \ j=1 Ej) = µ(E1 \ ∞ \ j=1 Ej) = lim µ(Fj) = lim(µ(E1) −µ(Ej)) = µ(E1) −lim µ(Ej).
Again since µ(E1) < ∞, it can be subtracted from both sides.
△ Remark 2.5. Note that in item (d) of Theorem 2.4, the hypothesis “µ(E1) < ∞” can be replaced by “µ(Ej) < ∞for some j”. However the finiteness hypothesis cannot be removed entirely. For instance, consider (N, 2N) equipped with counting measure, and let Ej = {k : k ≥j}. Then µ(Ej) = ∞for all j but µ(T∞ j=1 Ej) = µ(∅) = 0.
⋄ For any set X and subset E ⊂X, there is a function 1E : X →{0, 1} defined by 1E(x) = ( 1 if x ∈E 0 if x ̸∈E , called the characteristic function or indicator function of E. For a sequence of subsets (En) of X, say (En) converges to E pointwise if 1En →1E pointwise1. This notion allows the formulation of a more refined version of the dominated convergence theorem for sets, which foreshadows (and is a special case of) the dominated convergence theorem for the Lebesgue integral.
Definition 2.6. Let (X, M , µ) be a measure space. A null set is a set E ∈M with µ(E) = 0.
◁ It follows immediately from countable subadditivity that a countable union of null sets is null. The contrapositive of this statement is a measure-theoretic version of the pigeonhole principle: Proposition 2.7 (Pigeonhole principle for measures). If (En)∞ n=1 is a sequence of sets in M and µ(∪En) > 0, then µ(En) > 0 for some n.
† It will often be tempting to assert that if µ(E) = 0 and F ⊂E, then µ(F) = 0, but one must be careful: F need not be a measurable set. This caveat is not a big deal in practice, however, because we can always enlarge the σ-algebra on which a measure is defined so as to contain all subsets of null sets, and it will usually be convenient to do so.
Definition 2.8. If (X, M , µ) has the property that F ∈M whenever E ∈M , µ(E) = 0, and F ⊂E, then µ is complete.
◁ 1What would happen if we asked for uniform convergence?
8 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 Theorem 2.9. Let (X, M , µ) be a measure space. Let N := {N ∈M |µ(N) = 0}.
Define M := {E ∪F|E ∈M , F ⊂N for some N ∈N } Then M is a σ-algebra, and µ(E ∪F) := µ(E) is a well-defined function from M to [0, ∞] and is a complete measure on M such that µ|M = µ.
Proof. First note that M and N are both closed under countable unions, so M is as well. To see that M is closed under complements, consider E ∪F ∈M with E ∈ M , F ⊂N ∈N . Using, F c = N c ∪(N \ F) gives (E ∪F)c = Ec ∩F c = (E ∩N)c ∪(N ∩F c ∩Ec).
The first set on the right hand side is in M and the second is a subset of N. Thus the union is in M as desired.
To prove that µ is well defined, suppose G = E ∪F = E′ ∪F ′ for E, E′ ∈M and F, F ′ ∈N . In particular, there exists µ-null sets N, N ′ ∈M with F ⊂N and F ′ ⊂N ′.
Observe that E \ E′ ⊂G \ E′ ⊂F ′ ⊂N ′.
Thus µ(E \ E′) = 0. By symmetry, µ(E′ \ E) = 0. On the other hand, E = (E ∩E′) ∪(E \ E′).
Thus, µ(E) = µ(E ∩E′). By symmetry, µ(E′) = µ(E′ ∩E).
The proof that µ is a complete measure on M which extends µ, is left as an exercise △ 3. Outer measures and the Caratheodory Extension Theorem 3.1. Outline. The point of the construction of Lebesgue measure on the real line is to extend the naive notion of length for intervals to a suitably large family of subsets of R. We will accomplish this objective via the Caratheodory Extension Theorem (Theo-rem 3.3) Definition 3.1. Let X be a nonempty set. A function µ∗: 2X →[0, +∞] is an outer measure if: (i) µ∗(∅) = 0, (ii) (Monotonicity) µ∗(A) ≤µ∗(B) whenever A ⊂B, (iii) (Subadditivity) if {Aj}∞ j=1 ⊂2X, then µ∗ ∞ [ j=1 Aj !
≤ ∞ X j=1 µ∗(Aj).
◁ NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 9 Definition 3.2. If µ∗is an outer measure on X, then a set E ⊂X is outer measurable (or µ∗-measurable, measurable with respect to µ∗, or just measurable) if µ∗(A) = µ∗(A ∩E) + µ∗(A ∩Ec) (5) for every A ⊂X.
◁ The significance of outer measures and (outer) measurable sets stems from the following theorem: Theorem 3.3 (Caratheodory Extension Theorem). If µ∗is an outer measure on X, then the collection M of outer measurable sets is a σ-algebra, and the restriction of µ∗ to M is a complete measure.
The proof will be given later in this section; we first explain how to construct outer measures. In fact, all of the outer measures we consider will be constructed using the following proposition.
Proposition 3.4. Let E ⊂2X and µ0 : E →[0, +∞] be such that ∅, X ∈E and µ0(∅) = 0. The function µ∗: 2X →[0, ∞] defined by µ∗(A) = inf ( ∞ X n=1 µ0(En) : En ∈E and A ⊂ ∞ [ n=1 En ) (6) is an outer measure.
† Note that we have assumed X ∈E , so there is at least one covering of A by sets in E (take E1 = X and all other Ej empty), so the definition (6) makes sense. To prove the Proposition we need the following lemma on double sums, whose proof is left as an (important!) exercise.
Lemma 3.5. Let (am,n)∞ m,n=1 be a doubly indexed sequence of nonnegative real numbers.
Suppose there is a real number C so that for every finite set F ⊂N × N, X (m,n)∈F am,n ≤C.
(7) Then for each n and each m, the sums P∞ m=1 am,n, P∞ n=1 am,n are convergent, and ∞ X m=1 ∞ X n=1 am,n = ∞ X n=1 ∞ X m=1 am,n ≤C.
† Proof of Proposition 3.4. It is immediate from the definition that µ∗(∅) = 0 (cover the empty set by empty sets) and that µ∗(A) ≤µ∗(B) whenever A ⊂B (any covering of B is also a covering of A).
To prove countable subadditivity, we make our first use of the “ϵ/2n” trick.
Let (An) be a sequence in 2X and let ϵ > 0 be given.
If P∞ n=1 µ∗(An) = +∞there is nothing to prove, so we assume this sum is convergent. In particular µ∗(An) < +∞for each n. It follows from the definition of µ∗that for each 10 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 n ≥1 there exists a countable collection of sets (En,k)∞ k=1 in E such that An ⊂S∞ k=1 En,k and ∞ X k=1 µ0(En,k) < µ∗(An) + ϵ2−n.
One may now verify that the numbers an,k = µ0(En,k) satisfy the hypothesis of Lemma 3.5.
We now have that the countable collection (En,k)∞ n,k=1 covers A = S∞ n=1 An, and using Lemma 3.5 µ∗(A) ≤ ∞ X k,n=1 µ0(En,k) < ∞ X n=1 (µ∗(An) + ϵ2−n) = ϵ + ∞ X n=1 µ∗(An).
Since ϵ > 0 was arbitrary, we conclude µ∗(A) ≤P∞ n=1 µ∗(An).
△ Example 3.6. [Lebesgue outer measure] Let E ⊂2R be the collection of all open inter-vals (a, b) ⊂R, with −∞< a < b < +∞, together with ∅and R. Define m0((a, b)) = b −a, the length of the interval; µ0(∅) = 0; and µ0(R) = +∞. The corresponding outer measure is Lebesgue outer measure and it is the mapping m∗: 2R →[0, ∞] defined, for A ∈2X, by m∗(A) = inf ( ∞ X n=1 (bn −an) : A ⊂ ∞ [ n=1 (an, bn) ) (8) where we allow the degenerate intervals R = (−∞, +∞) and ∅. The value m∗(A) is the Lebesgue outer measure of A. In the next section we will construct Lebesgue measure from m∗via the Caratheodory Extension Theorem; the main issues will be to show that the outer measure of an interval is equal to its length, and that every Borel subset of R is outer measurable. The other desirable properties of Lebesgue measure (such as translation invariance) will follow from this construction.
△ Before proving Theorem 3.3 we make one observation, which will be used repeatedly.
Namely, if µ∗is an outer measure on a set X, to prove that a subset E ⊂X is outer measurable, it suffices to prove that µ∗(A) ≥µ∗(A ∩E) + µ∗(A \ E) for all A ⊂X, since the opposite inequality for all A is immediate from the subadditivity of µ∗.
We also need one lemma, which will be used to obtain completeness. A set E ⊂X is called µ∗-null if µ∗(E) = 0.
Lemma 3.7. Every µ∗-null set is µ∗-measurable.
† Proof. Let E be µ∗-null and A ⊂X. By monotonicity, A ∩E is also µ∗-null, so by monotonicity again, µ∗(A) ≥µ∗(A \ E) = µ∗(A ∩E) + µ∗(A \ E).
Thus the lemma follows from the observation immediately preceding the lemma.
△ NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 11 Proof of Theorem 3.3. We first show that M is a σ-algebra. It is immediate from Defi-nition 3.2 that M contains ∅and X, and since (5) is symmetric with respect to E and Ec, M is also closed under complementation. Next we check that M is closed under finite unions (which will prove that M is a Boolean algebra). So, let E, F ∈M and fix an arbitrary A ⊂X. Since F is measurable, µ∗(A ∩Ec) = µ∗((A ∩Ec) ∩F) + µ∗((A ∩Ec) ∩F c).
By subadditivity and the set equality A ∩(E ∪F) = (A ∩E) ∪(A ∩(F ∩Ec)), µ∗(A ∩(E ∪F)) ≤µ∗(A ∩E) + µ∗(A ∩(F ∩Ec)).
Using the last two displayed equations and the outer-measurability of E, µ∗(A ∩(E ∪F))+µ∗(A ∩((E ∪F)c)) ≤µ∗(A ∩E) + µ∗(A ∩(F ∩Ec)) + µ∗(A ∩(F c ∩Ec)) =µ∗(A ∩E) + µ∗(A ∩Ec) =µ∗(A).
Hence E ∪F is outer-measurable.
It remains to prove that µ∗(A) ≥µ∗(A ∩ ∞ [ n=1 En) + µ∗(A \ ∞ [ n=1 En).
For each N ≥1, we have already proved that SN n=1 En is outer measurable, and therefore µ∗(A) ≥µ∗(A ∩ N [ n=1 En) + µ∗(A \ N [ n=1 En).
By monotonicity, we also know that µ∗(A \ SN n=1 En) ≥µ∗(A \ S∞ n=1 En), so it suffices to prove that lim N→∞µ∗(A ∩ N [ n=1 En) ≥µ∗(A ∩ ∞ [ n=1 En).
(9) (The limit exists as an extended real number since the sequence is increasing by mono-tonicity of the outer measure.) By the outer measurability of SN n=1 En µ∗(A ∩ N+1 [ n=1 En) = µ∗(A ∩ N [ n=1 En) + µ∗(A ∩EN+1) for all N ≥1 (the disjointness of the En was used here). Iterating this identity gives µ∗(A ∩ N+1 [ n=1 En) = N+1 X k=0 µ∗(A ∩Ek) (10) and taking limits, lim N→∞µ∗(A ∩ N [ n=1 En) = ∞ X N=0 µ∗(A ∩EN).
12 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 But also, by countable subadditivity, we have lim N→∞µ∗(A ∩ N [ n=1 En) = ∞ X N=0 µ∗(A ∩EN) ≥µ∗(A ∩ ∞ [ n=1 En) which proves (9), and thus M is a σ-algebra. It also shows that µ∗(A ∩ ∞ [ n=1 En) ≥ ∞ X N=0 µ∗(A ∩EN+1).
In particular, choosing A = S∞ n=1 En proves that µ∗is countably additive. Since it is immediate that µ∗(∅) = 0, we conclude that µ∗|M is a measure.
Finally, that µ∗is a complete measure on M is an immediate consequence of Lemma 3.7.
△ 4. Construction of Lebesgue measure In this section, by an interval we mean any set I ⊂R of the from (a, b), [a, b], (a, b], [a, b), including ∅, open and closed half-lines and R itself. We write |I| = b −a for the length of I, interpreted as +∞in the line and half-line cases and 0 for ∅. Recall the definition of Lebesgue outer measure of a set A ⊂R from Example 8: m∗(A) = inf ( ∞ X n=1 |In| : A ⊂ ∞ [ n=1 In ) where the In are open intervals, or empty.
Theorem 4.1. If I ⊂R is an interval, then m∗(I) = |I|.
Proof. We first consider the case where I is a finite, closed interval [a, b]. For any ϵ > 0, the single open interval (a −ϵ, b + ϵ) covers I, so m∗(I) ≤(b −a) + 2ϵ = |I| + 2ϵ, and thus m∗(I) ≤|I|. For the reverse inequality, again choose ϵ > 0, and let (In) be a cover of I by open intervals such that P∞ n=1 |In| < m∗(I) + ϵ. Since I is compact, there is a finite subcollection (Ink)N k=1 of the In which covers I. Then N X k=1 |Ink| > b −a = |I|.
(11) To verify this statement, observe that by passing to a further subcollection, we can assume that none of the intervals Ink is contained in another one. Then re-index I1, . . . IN so that the left endpoints a1, . . . aN are listed in increasing order. Since these intervals cover I, and there are no containments, it follows that a2 < b1, a3 < b2, ...aN < bN−1.
(Draw a picture.) Therefore N X k=1 |Ik| = N X k=1 (bk −ak) = bN −a1 + N−1 X k=1 (bk −ak+1) ≥bN −a1 > b −a = |I|.
NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 13 From the inequality (11) we conclude that |I| < m∗(I) + ϵ, and since ϵ was arbitrary we have proved m∗(I) = |I|.
Now we consider the cases of bounded, but not closed, intervals (a, b), (a, b], [a, b).
If I is such an interval and I = [a, b] its closure, then since m∗is an outer measure we have by monotonicity m∗(I) ≤m∗(I) = |I|. On the other hand, for all ϵ > 0 sufficiently small we have Iϵ := [a+ϵ, b−ϵ] ⊂I, so by monotonicity again m∗(I) ≥m∗(Iϵ) = |I|−2ϵ and letting ϵ →0 we get m∗(I) ≥|I|.
Finally, the result is immediate in the case of unbounded intervals, since any un-bounded interval contains arbitrarily large bounded intervals.
△ Theorem 4.2. Every Borel set E ∈BR is m∗-measurable.
Proof. By the Caratheodory extension theorem, the collection of m∗-measurable sets is a σ-algebra, so by Propositions 1.10 and 1.7, it suffices to show that the open rays (a, +∞) are m∗-measurable. Fix a ∈R and an arbitrary set A ⊂R. We must prove m∗(A) ≥m∗(A ∩(a, +∞)) + m∗(A ∩(−∞, a]).
To simplify the notation put A1 = A ∩(a, +∞), A2 = A ∩(−∞, a]. Let (In) be a cover of A by open intervals. For each n let I′ n = In ∩(a, +∞) and I′′ n = In ∩(−∞, a]. The families (I′ n), (I′′ n) are intervals (not necessarily open) that cover A1, A2 respectively. Now ∞ X n=1 |In| = ∞ X n=1 |I′ n| + ∞ X n=1 |I′′ n| (12) = ∞ X n=1 m∗(I′ n) + ∞ X n=1 m∗(I′′ n) (by Theorem 4.1) (13) ≥m∗( ∞ [ n=1 I′ n) + m∗( ∞ [ n=1 I′′ n) (by subadditivity) (14) ≥m∗(A1) + m∗(A2) (by monotonicity).
(15) Since this inequality holds for all coverings of A by open intervals, taking the infimum we conclude m∗(A) ≥m∗(A1) + m∗(A2).
△ Definition 4.3. A set E ⊂R is called Lebesgue measurable if m∗(A) = m∗(A ∩E) + m∗(A ∩Ec) (16) for all A ⊂R. The restriction of m∗to the Lebesgue measurable sets is called Lebesgue measure, denoted m.
◁ By Theorem 3.3, m is a measure; by Theorem 4.2, every Borel set is Lebesgue measurable, and by Theorem 4.1 the Lebesgue measure of an interval is its length. It should also be evident by now that m is σ-finite. So, we have arrived at the promised extension of the length function on intervals to a measure.
(It turns out that this extension is unique, but we do not prove this here.) 14 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 Next we prove that m has the desired invariance properties. Given E ⊂R, x ∈R, and t > 0, let E+x = {y ∈R : y−x ∈E}, −E = {y ∈R : −y ∈E}, and tE = {y ∈R : y/t ∈E}.
It is evident that µ∗(E + x) = µ∗(E), µ∗(−E) = µ∗(E) and, µ∗(tE) = tµ∗(E) since, if I is an interval, then |I + x| = |I|, | −I| = |I| and |tI| = t|I|. In particular, if both E and E + x are Lebesgue measurable, then m(E + x) = m(E).
Theorem 4.4. If E ⊂R is Lebesgue measurable, x ∈R, and t > 0, then the sets E +x, −E, and tE are Lebesgue measurable. Moreover m(E + x) = m(E), m(−E) = m(E), and m(tE) = tm(E).
Proof. We give the proof for E + x; the others are similar and left as exercises. Accord-ingly, suppose E is measurable. To prove E + x is measurable, let A ⊂R be given and observe that A ∩(E + x) = ((A −x) ∩E) + x and A ∩(E + x)c = ((A −x) ∩Ec) + x.
Thus, m∗(A) = m∗(A −x) (17) = m∗((A −x) ∩E) + m∗((A −x) ∩Ec) (18) = m∗((A −x) ∩E + x) + m∗((A −x) ∩Ec + x) (19) = m∗(A ∩(E + x)) + m∗(A ∩(E + x)c), (20) where measurability of E is used in the second equality.
Hence E + x is Lebesgue measurable and m(E + x) = m(E).
△ The condition (16) does not make clear which subsets of R are Lebesgue measurable.
We next prove two fundamental approximation theorems, which say that 1) if we are willing to ignore sets of measure zero, then every Lebesgue measurable set is a Gδ or Fσ, and 2) if we are willing to ignore sets of measure ϵ, then every set of finite Lebesgue measure is a union of intervals. (Recall that a set in a topological space is called a Gδ-set if it is a countable intersection of open sets, and an Fσ-set if it is a countable union of closed sets.) Theorem 4.5. Let E ⊂R. The following are equivalent: (a) E is Lebesgue measurable.
(b) For every ϵ > 0, there is an open set U ⊃E such that m∗(U \ E) < ϵ.
(c) For every ϵ > 0, there is a closed set F ⊂E such that m∗(E \ F) < ϵ.
(d) There is a Gδ set G such that E ⊂G and m∗(G \ E) = 0.
(e) There is an Fσ set F such that E ⊃F and m∗(E \ F) = 0.
Proof. To prove (a) implies (b) let E a (Lebesgue) measurable set and ϵ > 0 be given.
Further, suppose for the moment that m(E) < ∞. There is a covering of E by open intervals In such that P∞ n=1 |In| < m(E) + ϵ. Put U = S∞ n=1 In. By subadditivity of m, m(U) ≤ ∞ X n=1 m(In) = ∞ X n=1 |In| < m(E) + ϵ.
NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 15 Since U ⊃E and m(E) < ∞(and both U and E are Lebesgue measurable), we conclude from Theorem 2.4 that m∗(U \ E) = m(U \ E) = m(U) −m(E) < ϵ.
To remove the finiteness assumption on E, we apply the ϵ/2n trick: for each n ∈Z let En = E ∩(n, n + 1]. The En are disjoint measurable sets whose union is E, and m(En) < ∞for all n. For each n, by the first part of the proof we can pick an open set Un so that m(Un \ En) < ϵ/2|n|. Let U be the union of the Un; then U is open and U \ E ⊂S∞ n=1(Un \ En). From the subadditivity of m, we get m∗(U \ E) = m(U \ E) < P n∈Z ϵ2−|n| = 3ϵ.
To prove that (b) implies (d), let E ⊂R be given and for each n ≥1 choose (using (b)) an open set Un ⊃E such that m∗(Un \ E) < 1 n. Put G = T∞ n=1 Un; then G is a Gδ containing E, and G \ E ⊂Un \ E for every n. By monotonicity of m∗we see that m∗(G \ E) < 1 n for every n and thus m∗(G \ E) = 0. (Note that in this portion of the proof we cannot (and do not!) assume E is measurable.) To prove (d) implies (a), suppose G is a Gδ set such that E ⊂G and µ∗(G\E) = 0.
Since G is a Gδ, it is a Borel set and hence Lebesgue measurable by Theorem 4.2. By Lemma 3.7, every m∗-null set is Lebesgue measurable, so G \ E, and therefore also E = G \ (G \ E), is Lebesgue measurable.
To prove that (a) implies (c), suppose E is Lebesgue measurable and let ϵ > 0 be given. Thus Ec is Lebesgue measurable and, by the already established implication (a) implies (b), there is an open set U such that Ec ⊂U and m(U \ Ec) < ϵ. Since U \ Ec = U ∩E = E \ U c, it follows that µ(E \ U c) < ϵ. Observing that U c is closed completes the proof.
Now suppose E ⊂R and (c) holds. Choose a sequence of closed sets (Fn) such that Fn ⊂E and µ∗(E \ Fn) < 1 n. The set F = ∪∞ j=1Fj is an Fσ and, by Monotonicity, for each n we have µ∗(E \F) ≤µ∗(E \Fn) < 1 n. Hence µ∗(E \F) = 0. Thus (c) implies (e).
Finally, if (e) holds, then E = F ∪(E\F) for some closed set F ⊂E with µ∗(E\F) = 0. Thus, E is the union of a closed (and hence Lebesgue) set and a set of outer measure zero (which is thus Lebesgue). Since the Lebesgue sets are closed under union, E is Lebesgue and the proof is complete.
△ Remark 4.6. The conditions in the theorem are regularity conditions which link the topology on R to Lebesgue measure. This theme leads to the notion of a regular Borel measure, discussed in further detail below.
⋄ Theorem 4.7. If E is Lebesgue measurable and m(E) < ∞, then for each ϵ > 0 there exists a set A which is a finite union of open intervals such that m(E∆A) < ϵ.
Proof. Let (In) be a covering of E by open intervals such that ∞ X n=1 |In| < m(E) + ϵ/2.
(21) 16 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 Since the sum is finite there exists an integer N so that ∞ X n=N+1 |In| < ϵ/2.
(22) Let U = S∞ n=1 In and A = SN n=1 In. Then A\E ⊂U \E, so m(A\E) ≤m(U)−m(E) < ϵ/2 by (21). Similarly E \A ⊂U \A ⊂S∞ n=N+1 In, so m(E \A) < ϵ/2 by (22). Therefore m(E∆A) < ϵ.
△ Thus, while the “typical” measurable set can be quite complicated in the set-theoretic sense (i.e. in terms of the Borel hierarchy), for most questions in analysis this complexity is irrelevant. In fact, Theorem 4.7 is the precise expression of a useful heuristic: Littlewood’s First Principle of Analysis: Every measurable set E ⊂R with m(E) < ∞is almost a finite union of intervals.
Definition 4.8. Let X be a topological space. A neighborhood U of a point x ∈X is an open set such that x ∈U.
A topological space X is locally compact if for each x ∈X there is a neighborhood Ux of x and a compact set Cx such that x ∈Ux ⊂Cx.
A topological space is Hausdorffif given x, y ∈X with x ̸= y, there exists neigh-borhoods U and V of x and y respectively such that U ∩V = ∅. (Distinct points can be separated by open sets.) A Borel measure is a measure on the Borel σ-algebra of a locally compact Hausdorff space.
A Borel measure µ is outer regular if, for all E ∈Σ, µ(E) = inf{µ(U) : U ⊃E and U is open} and is inner regular if µ(E) = sup{µ(K) : K ⊂E and K is compact}.
Finally µ is regular if it is both inner and outer regular.
◁ Theorem 4.9. If E ⊂R is Lebesgue measurable, then m(E) = inf{m(U) : U ⊃E and U is open} = sup{m(K) : K ⊂E and K is compact} That is, m is a regular Borel measure.
Proof. The first equality follows from monotonicity in the case m(E) = +∞, and from Theorem 4.5(b) (together with the additivity of m) in the case m(E) < ∞.
For the second equality, let ν(E) be the value of the supremum on the right-hand side. By monotonicity m(E) ≥ν(E). For the reverse inequality, first assume m(E) < ∞ NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 17 and let ϵ > 0. By Theorem 4.5(c), there is a closed subset F ⊂E with m(E \ F) < ϵ/2.
Since m(E) < ∞, we have by additivity m(E) < m(F) + ϵ/2, so m(F) > m(E) −ϵ/2.
However this F need not be compact. To fix this potential shortcoming, for each n ≥1 let Kn = F ∩[−n, n]. Then the Kn are an increasing sequence of compact sets whose union is F. By monotone convergence for sets (Theorem 2.4(c)), there is an n so that m(Kn) > m(F) −ϵ/2. It follows that m(Kn) > m(E) −ϵ, and thus ν(E) ≥m(E). The case m(E) = +∞is left as an exercise.
△ Remark 4.10. In the preceding theorem, for any set E ⊂R (not necessarily measurable) the infimum of m(U) = m∗(U) over open sets U ⊃E defines the Lebesgue outer measure of E. One can also define the Lebesgue inner measure of a set E ⊂R as m∗(E) = sup{m∗(K) : K ⊂E and K is compact}.
By monotonicity we have m∗(E) ≤m∗(E) for all E ⊂R. One can then prove that if E ⊂R and m∗(E) < ∞, then E is Lebesgue measurable if and only if m∗(E) = m∗(E), in which case this quantity is equal to m(E).(Some care must be taken in trying to use this definition if m∗(E) = +∞. Why?) ⋄ Example 4.11. [The Cantor set] Recall the usual construction of the “middle thirds” Cantor set. Let E0 denote the unit interval [0, 1]. Obtain E1 from E0 by deleting the middle third (open) subinterval of E0, so E1 = [0, 1 3] ∪[ 2 3, 1]. Continue inductively; at the nth step delete the middle thirds of all the intervals present at that step. So, En is a union of 2n closed intervals of length 3−n. The Cantor set is defined as the intersection C = T∞ n=0 En. It is well-known (though not obvious) that C is uncountable; we do not prove this fact here. It is clear that C is a closed set (hence Borel) but contains no intervals, since if J is an interval of length ℓand n is chosen so that 3−n < ℓ, then J ̸⊂En and thus J ̸⊂C. The Lebesgue measure of En is (2/3)n, which goes to 0 as n →∞, and thus by monotonicity m(C) = 0. So, C is an example of an uncountable, closed set of measure 0. Another way to see that C has measure zero, is to note that at the nth stage (n ≥1) we have deleted a collection of 2n−1 disjoint open intervals, each of length 3−n. Thus the Lebesgue measure of [0, 1] \ C is ∞ X n=1 2n−13−n = 1 2 2 3 1 −2 3 = 1.
Thus m(C) = 0.
△ Example 4.12. [Fat Cantor sets] The standard construction of the Cantor set can be modified in the following way: fix a number 0 < c < 1 3. Imitate the construction of the Cantor set, except at the nth stage delete, from each interval I present at that stage, an open interval centered at the midpoint of I of length cn. (In the previous construction we had c = 1 3.) Again at each stage we have a set En which is a union of 2n closed intervals.
18 NOTES ON MEASURE THEORY AND THE LEBESGUE INTEGRAL MAA5229, SPRING 2015 The total length of the intervals removed up to that stage is Pn k=1 2n−1cn = c 1−(2c)n 1−2c , so the measure of En is 1 −c1 −(2c)n 1 −2c .
Let F = T∞ n=0 En. One can prove (in much the same way as for C) that 1) F is an uncountable, closed set; 2) F contains no intervals; and 3) m(F) = 1−3c 1−2c > 0. Thus, F is a closed set of positive measure, but which contains no intervals.
△ Example 4.13. [A Lebesgue non-measurable set] Define an equivalence relation on R by declaring x ∼y if and only if x −y ∈Q. This relation partitions R into disjoint equivalence classes whose union is R. In particular, for each x ∈R its equivalence class is the set {x + q : q ∈Q}. Since Q is dense in R, each equivalence class C contains an element of the closed interval [0, 1]. By the axiom of choice, there is a set E ⊂[0, 1] that contains exactly one member xC from each class. We claim the set E is not Lebesgue measurable.
To prove the claim, let y ∈[0, 1]. Then y belongs to some equivalence class C, and hence y differs from xC by some rational number in the interval [−1, 1]. Hence [0, 1] ⊂ [ q∈[−1,1]∩Q (E + q).
On the other hand, since E ⊂[0, 1] and |q| ≤1 we see that [ q∈[−1,1]∩Q (E + q) ⊂[−1, 2].
Finally, by the construction of E the sets E + p and E + q are disjoint if p, q are distinct rationals. So if E were measurable, the the sets E +q would be also, and we would have by the countable additivity and monotonicity of m 1 ≤ X q∈[−1,1]∩Q m(E + q) ≤3 But by translation invariance, all of the m(E+q) must be equal, which is a contradiction.
△ Remark: The above construction can be modified used to show that if F is any Lebesgue set with m(F) > 0, then F contains a nonmeasurable subset.
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How many ways can we choose three items from a set of five items?
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Aditya Acharya
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·3y
Originally Answered: How many ways can three items be selected from a group of six items? ·
1,2,3,4,5,6
123,124,125,126,134,135,136,145,146,156
234,235,236,245,246,256
345,346,356,
456
Total 20 combination you can select in group of 3 from 6 items.
Please upvote if you like.😃
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William La Chenal
one-time analytical chemist.Now into science of homeopathy. · Author has 2.6K answers and 5.6M answer views
·3y
Originally Answered: Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five? ·
Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five?
One way.
A good manager chooses the right combination; the rest would be wrong.
Only an idiot would would this out with permutations and combinations.
Stoopid question.
This reminds me of an object lesson that used to be taught to accountants at senior level:
A stationery company wants to cut costs. The bean counter sees one line of special paper which sells only two or three boxes a year, and persuades the new manager to drop that line.
The c
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Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five?
One way.
A good manager chooses the right combination; the rest would be wrong.
Only an idiot would would this out with permutations and combinations.
Stoopid question.
This reminds me of an object lesson that used to be taught to accountants at senior level:
A stationery company wants to cut costs. The bean counter sees one line of special paper which sells only two or three boxes a year, and persuades the new manager to drop that line.
The company goes to the wall. They all lose their jobs, and their reputations.
That line of special paper was used by their major investor.
~
[Please look up the separate topic of study methods, such as repeated revision, and mind-mapping. Know that you need to practice, do the exercises, to learn properly and pass exams easily Also, as a student, you need to look after your health. Asking others to do your homework is cheating yourself.]
The combinatorics question is different - it has a theoretical context where all choices are equal, an none of them is chosen more than once. All choices being equal is an essential condition for the unbiased calculation. If you could choose just one several times, you would get a different answer.
The insight needed is to divide the choice into steps. First you choose one of the five. Then you choose the second. Then you choose the third.
How many ways could you choose just one out of 5? (Assuming no bias)
# # #
Think. Write that number down.
Now you have removed that one, how many are left? Prepare to make the next choice, given that you cannot choose one which is already chosen (another important assumption).
# #
How many ways are there to choose the second one?
Now, for each of the first choices, you would have that many second choices.
How many choices so far?
You can multiply the number of first choices by the number of available second choices. That gives the number of ways to choose two out of three.
Continue in a similar way.
Next, how many ways would there be to choose the third, from what remains ?
#
Multiply again; the total number of ways to choose two out of three, times the number of ways to choose the third.
There is your theoretical answer, give the assumed conditions.
~
This is the way to think so that you don’t need to remember the formula like a robot.
Mathematics is always about the process, not remembering formulas.
Although in preparation for a higher level, one should remember the names of theorems (and how they were proven, and by whom) out of respect for those that went before, and to be able to show the basis for further results.
The process described above does have a notation, and formulae to connect ways of choosing in general. Learn the process, rather than than remembering individual results parrot fashion.
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To determine how many ways we can choose three items from a set of five items, we can use the combination formula:
C(n,r)=n!r!(n−r)!C(n,r)=n!r!(n−r)!
where:
n n is the total number of items (5 in this case),
r r is the number of items to choose (3 in this case),
!! denotes factorial, which is the product of all positive integers up to that number.
Plugging in the values:
C(5,3)=5!3!(5−3)!=5!3!⋅2!C(5,3)=5!3!(5−3)!=5!3!⋅2!
Calculating the factorials:
5!=5×4×3×2×1=120 5!=5×4×3×2×1=120
3!=3×2×1=6 3!=3×2×1=6
2!=2×1=2 2!=2×1=2
Now substituting these values back into the formula:
C(5,3 C(5,3
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To determine how many ways we can choose three items from a set of five items, we can use the combination formula:
C(n,r)=n!r!(n−r)!C(n,r)=n!r!(n−r)!
where:
n n is the total number of items (5 in this case),
r r is the number of items to choose (3 in this case),
!! denotes factorial, which is the product of all positive integers up to that number.
Plugging in the values:
C(5,3)=5!3!(5−3)!=5!3!⋅2!C(5,3)=5!3!(5−3)!=5!3!⋅2!
Calculating the factorials:
5!=5×4×3×2×1=120 5!=5×4×3×2×1=120
3!=3×2×1=6 3!=3×2×1=6
2!=2×1=2 2!=2×1=2
Now substituting these values back into the formula:
C(5,3)=120 6×2=120 12=10 C(5,3)=120 6×2=120 12=10
Therefore, there are 10 ways to choose three items from a set of five items.
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Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five?
Dega Manvith Yadav
B.Tech in Information Technology&Business, Indian Institute of Information Technology, Allahabad (IIIT Allahabad) (Graduated 2023)
·4y
Originally Answered: How many ways can three items be selected from a group of six items? ·
The selection of r items from n items is a combination (not an arrangement) of r items from that n items.
This means the number of ways of selecting r items from n items = n C r n C r
And 6 C 3 6 C 3 is 20
So, there are 20 ways to select three items from a group of six items.
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Franco Ariatti
Former Vice Central Director at Enel (company) · Author has 1.7K answers and 921.1K answer views
·3y
Originally Answered: Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five? ·
(5C3)=(543)/(123)=10 ways. If the products are A,B,C,D,E, the possible choices of three products out of five are: ABC, ABD, ABE, ACD; ACE, ADE, BCD, BCE, BDE, CDE.
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Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR · Author has 5.6K answers and 10.7M answer views
·2y
Originally Answered: How many ways can three items be selected from a group of seven items? ·
Either calculate it by the formula:
C(7, 3) = 7!/[3!(7 - 3)!] = 7!/(3!4!) = 567/(23) = 35
Or read off the result at the position No 3 of the Row No 7 of Pascal’s Triangle:
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Either calculate it by the formula:
C(7, 3) = 7!/[3!(7 - 3)!] = 7!/(3!4!) = 567/(23) = 35
Or read off the result at the position No 3 of the Row No 7 of Pascal’s Triangle:
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Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional.
Ambiguous (“Squinting”) modifiers
Incorrect: Listening to loud music slowly gives me a headache.
Correct: When I listen to loud music, I slowly develop a headache.
A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly gett
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Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional.
Ambiguous (“Squinting”) modifiers
Incorrect: Listening to loud music slowly gives me a headache.
Correct: When I listen to loud music, I slowly develop a headache.
A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly getting a headache? To correct a squinting modifier, move its position in the sentence to clarify to the reader which word you intend to modify.
Misuse of lie/lay
Incorrect: He was laying on the couch.
Correct: He was lying on the couch.
If you plan to place or put an object somewhere, such as a plate on a table, you should use “lay.” If you intend to stretch out on a bed for a nap, you should use “lie.” The verb “lie” is an intransitive verb, which means it does not need an object. The transitive verb “lay” requires an object.
It may take some getting used to this “lay” or “lie” business; after all, misuse of these verbs is common. But if you remember to lay down your fork before you’re full, then you won’t have to lie down later from overeating.
Comma splices
Incorrect: He was very hungry, he ate a whole pizza.
Correct: He was very hungry. He ate a whole pizza.
He was very hungry, so he ate a whole pizza.
To splice means to connect or join. When a writer joins two independent sentences with a comma instead of separating them with a period or coordinating conjunction, that’s a comma splice.
The comma has its jobs to do, but connecting two independent sentences isn’t one of them. Besides, the period gets testy when his sister, the comma, steals his thunder. Periods have their jobs, and so do commas, but never the twain shall meet—unless it’s in the form of a semicolon. Semicolons can also take the place of a coordinating conjunction, such as “and,” “but,” or “so,” among others.
Run-on sentences
Incorrect: Lila enjoyed the bouquet of tulips John gave her on prom night however she prefers roses.
Correct: Lila enjoyed the bouquet of tulips John gave her on prom night; however, she prefers roses.
Run-on sentences, also known as fused sentences, occur when two complete sentences are squashed together without using a coordinating conjunction or proper punctuation, such as a period or a semicolon. Run-on sentences can be short or long. A long sentence isn’t necessarily a run-on sentence.
To avoid run-on sentences, see if there is more than one idea communicated by two or more independent clauses. In our examples, there are two complete sentences:
Example: Lily enjoyed the bouquet of tulips John gave her on prom night.
Example: She prefers roses.
Both sentences are complete ideas by themselves; therefore, use a semicolon or a period to indicate that they are separate independent clauses.
Using “could of” instead of “could have”
Incorrect: Sam could of received an A on his essay, but he made too many grammatical mistakes.
Correct: Sam could have received an A on his essay, but he made too many grammatical mistakes.
“Could have” is always correct; “could of” never is. Writers probably make this grammar gaffe because, when we speak, the contraction “could’ve” sounds an awful lot like “could of.”
Tautologies
Incorrect: Jack made a water pail with his own hands for Jill.
Correct: Jack made a water pail for Jill.
Tautologies express the same thing twice with different words. In our example, the word “made” implies that Jack used his own two hands to create the pail. The prepositional phrase “with his own hands” creates a redundancy. Once you know what they are, it’s fun to discover tautologies: dilapidated ruins, close proximity, added bonus, large crowd...The list goes on and on!
After reading through this list of common grammatical mistakes, you might be wondering how to remember all these rules as you write. The free Grammarly for Windows and Mac is here to help. It provides a second set of eyes on your writing in real-time, so you can avoid everyday grammar and spelling errors. Moreover, Grammarly Premium offers features that evaluate conciseness and readability as well as vocabulary enhancement suggestions and genre-specific writing style checks. These tools can help you identify when you are making these common writing mistakes so you can proactively learn and improve your writing.
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Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR · Author has 5.6K answers and 10.7M answer views
·2y
Originally Answered: How many ways are there to choose three objects out of five? ·
This kind of arrangement is called a ‘partial permutation’.
The number of partial permutations of n items taken k at a time equals the permutations of n items divided by the permutations of (n - k) items.
P(n, k) = P(n)/P(n -K) = n!/(n - k)!
For P(5, 3) this computes out to:
P(5, 3) = 5!/(5 - 3)! = 5!/2! = 120/2 = 60
Upvote ·
Calvin Campbell
B.Sc. in Computer Science&Mathematics, University of the West Indies (Graduated 1984) · Author has 3.8K answers and 4M answer views
·2y
If order matters, then this is a permutations problem, and the number of permutations of 5 items taking 3 at a time is 5P3 = 5!/(5 - 3)! = 54321/(21) = 543 = 60.
If order does not matter, then this is a combinations problem, and the number of combinations of 5 items taking 3 at a time is 5C3 = 5!/((5 - 3)!3!) = 543!/(2!3!) = 54/2 = 10.
Good luck!
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Nitin Anand
PhD in CSE from National Institute of Technology, Manipur (Graduated 2025) · Author has 1.7K answers and 403.9K answer views
·3y
Originally Answered: Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five? ·
The manager can choose 3 products out of 5 products in 5C3 ways.
We know that
nCr=n!/{ r!(n-r)!}
Putting n=5 and r=3,
5C3=5!/{3!(5–3)!}
5C3=5!/(3!2!)
5C3=10.
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Anonymous
3y
Originally Answered: Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five? ·
I love word problems. They are invariably intended to map to particular mathematical calculations, but they always make assumptions about the real world that aren’t valid. The assumption in this problem is that the decision is really entirely in the manager’s control and is a one time event. Even if the manager chooses products A, B, and C to take into production, any combination of the products could end up going into production, including fewer or more than 3.
There’s actually a second assumption as well. That is that order doesn’t matter. It actually does. The odds that there will be 3 plant
Continue Reading
I love word problems. They are invariably intended to map to particular mathematical calculations, but they always make assumptions about the real world that aren’t valid. The assumption in this problem is that the decision is really entirely in the manager’s control and is a one time event. Even if the manager chooses products A, B, and C to take into production, any combination of the products could end up going into production, including fewer or more than 3.
There’s actually a second assumption as well. That is that order doesn’t matter. It actually does. The odds that there will be 3 plants all of which are ready to begin production at the same time and each of which is equally suited to switch to production of any of the products isn’t justified.
Finally, reality should sink in here. The manager could get the news that the company has just be purchased by another company. 3 of the products are being cancelled because they would compete with the buyer’s existing products. However, they are throwing resources behind the other two because those are exactly why they made the purchase in the first place.
Upvote ·
9 1
Calvin Campbell
B.Sc. in Computer Science&Mathematics, University of the West Indies (Graduated 1984) · Author has 3.8K answers and 4M answer views
·2y
Related
How many ways can you select three things from five without repetition (without using algebra)?
Example : Selecting 3 numbers from a set of 5 numbers, {1, 2, 3, 4, 5}, if order does not matter:
1, 2, 3.
1, 2, 4.
1, 2, 5.
1, 3, 4.
1, 3, 5.
1, 4, 5.
2, 3, 4.
2, 3, 5.
2, 4, 5.
3, 4, 5.
If order is important, each of these 10 subsets can be arranged in 6 ways. Example: 1, 2, 3 can be arranged as:
1, 2, 3.
1, 3, 2.
2, 1, 3.
2, 3, 1.
3, 1, 2.
3, 2, 1.
Therefore, there are 10 combinations of 5 different items taking 3 items at a time, and 60 possible arrangements (or permutations) of 5 different items taking 3 items at a time.
Note: Solving this problem by listing the possibilities in each case is feasible. Howev
Continue Reading
Example : Selecting 3 numbers from a set of 5 numbers, {1, 2, 3, 4, 5}, if order does not matter:
1, 2, 3.
1, 2, 4.
1, 2, 5.
1, 3, 4.
1, 3, 5.
1, 4, 5.
2, 3, 4.
2, 3, 5.
2, 4, 5.
3, 4, 5.
If order is important, each of these 10 subsets can be arranged in 6 ways. Example: 1, 2, 3 can be arranged as:
1, 2, 3.
1, 3, 2.
2, 1, 3.
2, 3, 1.
3, 1, 2.
3, 2, 1.
Therefore, there are 10 combinations of 5 different items taking 3 items at a time, and 60 possible arrangements (or permutations) of 5 different items taking 3 items at a time.
Note: Solving this problem by listing the possibilities in each case is feasible. However, for larger sets of items we could get the answers by using the combinations formula , nCr = n!/(n-r)!r!), and the permutations formula, nPr = n!/(n-r)! , where n is the total number of items, and r is the number of items taken at a time.
Good luck!
Upvote ·
9 1
Herman Segal
JD & CPA in Law (college major), Columbia School of Law (Graduated 1981) · Author has 90 answers and 68.6K answer views
·3y
Originally Answered: Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five? ·
is this a math Q? Or a business Q, on how the manager should choose the 3 out of 5 products to manufacture?
Is there 1 machine that makes all the products, or insufficient staff to make the 3 products simultaneously? Is the Q also related to the order of production (as in A,B,C & also A,C,B)?
Upvote ·
9 1
9 2
Larry Scholnick
Senior Accountant at LS Capital (2013–present) · Author has 9.4K answers and 6.5M answer views
·3y
Originally Answered: Manager has to choose three products out of five possibilities to take into production. How many different ways are there to choose three products out of five? ·
One could decide on the 3 most profitable products. Or one could decide on the single most profitable product along with 2 others that are complementary (go together).
If the company already makes other products, one might choose new products that are complementary to existing products.
Another possible basis is ease of production; one might not choose potentially profitable products that are difficult to produce, or time-consuming to produce, or are quickly perishable.
Upvote ·
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Tom Apostol mathematical analysis exercise 5.28 Prove the following theorem do not use L‘Hospital rule
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5.28 Let f f and g g be two functions having finite nth derivatives in (a,b).for some interior point c in (a,b),assumes that f(c)=f′(c)=⋯=f n−1(c)=0 f(c)=f′(c)=⋯=f n−1(c)=0,and that g(c)=g′(c)=⋯=g n−1(c)=0 g(c)=g′(c)=⋯=g n−1(c)=0,but g n(x)g n(x)is never zero in (a,b),show that
lim x→c f(x)g(x)=f n(c)g n(c)lim x→c f(x)g(x)=f n(c)g n(c)
Note. f(n)f(n) and g(n)g(n) are not assumed to be continous at c. H i n t.H i n t. Let
F(x)=f(x)−(x−c)n−1 f(n−1)(c)(n−1)!,F(x)=f(x)−(x−c)n−1 f(n−1)(c)(n−1)!,
I research for n=1.
F(x)=f(x)−f(c),G(x)=g(x)−g(c)F(x)=f(x)−f(c),G(x)=g(x)−g(c),use Cauchy mean value theorem. I got
f(x)g′(x 1)=f′(x 1)g(x)f(x)g′(x 1)=f′(x 1)g(x) where x 1 x 1 is interior to the interval joining x x and c c
|f(x)g(x)−f′(c)g′(c)|=|f′(x 1)g′(x 1)−f′(c)g′(c)||f(x)g(x)−f′(c)g′(c)|=|f′(x 1)g′(x 1)−f′(c)g′(c)|
But f′(x),g′(x)f′(x),g′(x) are not assumed to be continuous at c c. I don’t know how to continue.Could you help me ?pleasa do not use L Hospital because this exercise in which chapter isn’t teach L Hospital.
Thanks in advance!!!
PS,The problem in the book where emphasis that f n(x),g n(x)f n(x),g n(x) is not assumed to be continuous at point c
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edited Aug 8, 2019 at 15:06
miracle173
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asked Aug 8, 2019 at 11:21
user682113 user682113
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Please do not post text as pictures miracle173 –miracle173 2019-08-08 11:25:53 +00:00 Commented Aug 8, 2019 at 11:25
@miracle173 I edit again,sorry for my picture.could you give me some hints for my problem ?user682113 –user682113 2019-08-08 11:37:18 +00:00 Commented Aug 8, 2019 at 11:37
now the hint disappeared miracle173 –miracle173 2019-08-08 11:42:27 +00:00 Commented Aug 8, 2019 at 11:42
Why? I am sorry for my picture. My typewriting is too slow,and my knowledge for latex is too little ,so l used picture for edit question faster.sorry.user682113 –user682113 2019-08-08 11:46:51 +00:00 Commented Aug 8, 2019 at 11:46
You must additionally assume that g(x)≠0 g(x)≠0 in a pointed neighborhood of c c.Paul Frost –Paul Frost 2019-08-08 12:46:03 +00:00 Commented Aug 8, 2019 at 12:46
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The expressions f(k)(c)f(k)(c) and g(k)(c)g(k)(c) for k≥1 k≥1 only make sense if f(k−1)f(k−1) and g(k−1)g(k−1) exist in a neighborhood of c c. And you say that g(n)(x)g(n)(x) is never 0 0 in (a,b)(a,b) which means that g g must be n n-times differentiable on (a,b)(a,b). So we may assume w.l.o.g. that bothf,g f,g are n n-times differentiable on (a,b)(a,b).
The mean value theorem implies that for x∈J=(a,b)∖{c}x∈J=(a,b)∖{c}
g(n−1)(x)=g(n−1)(x)−g(n−1)(c)=g(n)(ξ)(x−c)g(n−1)(x)=g(n−1)(x)−g(n−1)(c)=g(n)(ξ)(x−c)
for some ξ ξ between x x and c c. Since g(n)(ξ)≠0 g(n)(ξ)≠0 on J⊂(a,b)J⊂(a,b), this shows that g(n−1)(x)≠0 g(n−1)(x)≠0 for x∈J x∈J. Proceeding inductively shows that also g(n−2),…,g(1)=g′,g(0)=g g(n−2),…,g(1)=g′,g(0)=g are nonzero on J J. Hence the quotients
f(k)(x)g(k)(x)f(k)(x)g(k)(x)
are defined on J J for k=0,…,n k=0,…,n.
Let us first treat the case n=1 n=1. We have f′(c)=lim x→c f(x)−f(c)x−c=lim x→c f(x)x−c f′(c)=lim x→c f(x)−f(c)x−c=lim x→c f(x)x−c and g′(c)=lim x→c g(x)x−c g′(c)=lim x→c g(x)x−c, hence
f′(c)g′(c)=lim x→c f(x)x−c lim x→c g(x)x−c=lim x→c f(x)x−c g(x)x−c=lim x→c f(x)g(x).f′(c)g′(c)=lim x→c f(x)x−c lim x→c g(x)x−c=lim x→c f(x)x−c g(x)x−c=lim x→c f(x)g(x).
Applying this to f(n−1)f(n−1) and g(n−1)g(n−1) yields
f(n)(c)f(n)(c)=lim x→c f(n−1)(x)g(n−1)(x).f(n)(c)f(n)(c)=lim x→c f(n−1)(x)g(n−1)(x).
The extended mean value theorem (Cauchy's mean value theorem) says that for each x∈J x∈J there exists ξ ξ between x x and c c such that
f(n−2)(x)g(n−1)(ξ)=(f(n−2)(x)−f(n−2)(c))g(n−1)(ξ)=(g(n−2)(x)−g(n−2)(c))f(n−1)(ξ)=g(n−2)(x)f(n−1)(ξ).f(n−2)(x)g(n−1)(ξ)=(f(n−2)(x)−f(n−2)(c))g(n−1)(ξ)=(g(n−2)(x)−g(n−2)(c))f(n−1)(ξ)=g(n−2)(x)f(n−1)(ξ).
Therefore
f(n−2)(x)g(n−2)(x)=f(n−1)(ξ)g(n−1)(ξ).f(n−2)(x)g(n−2)(x)=f(n−1)(ξ)g(n−1)(ξ).
If x→c x→c, then also ξ→c ξ→c and we conclude
lim x→c f(n−2)(x)g(n−2)(x)=l i m ξ→c f(n−1)(ξ)g(n−1)(ξ)=f(n)(c)g(n)(c).lim x→c f(n−2)(x)g(n−2)(x)=l i m ξ→c f(n−1)(ξ)g(n−1)(ξ)=f(n)(c)g(n)(c).
Proceeding inductively we get
lim x→c f(x)g(x)=f(n)(c)g(n)(c).lim x→c f(x)g(x)=f(n)(c)g(n)(c).
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edited Aug 9, 2019 at 9:33
answered Aug 8, 2019 at 14:06
Paul FrostPaul Frost
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Why lim ξ→c f n(x)g n(x)=f n(c)g n(c)lim ξ→c f n(x)g n(x)=f n(c)g n(c),they are not continuous at c user682113 –user682113 2019-08-08 14:28:36 +00:00 Commented Aug 8, 2019 at 14:28
I edited my answer.Paul Frost –Paul Frost 2019-08-08 16:08:30 +00:00 Commented Aug 8, 2019 at 16:08
Your proof is so great!! I got it，Thank you very much!user682113 –user682113 2019-08-08 23:42:31 +00:00 Commented Aug 8, 2019 at 23:42
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http://www.kurims.kyoto-u.ac.jp/EMIS/journals/INTEGERS/papers/q50/q50.pdf
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A50 INTEGERS 16 (2016) CHARACTERIZING CONGRUENCE PRESERVING FUNCTIONS Z/nZ →Z/mZ VIA RATIONAL POLYNOMIALS Patrick C´ egielski 1 LACL, EA 4219, Universit´ e Paris-Est Cr´ eteil, France IUT S´ enart-Fontainebleau cegielski@u-pec.fr Serge Grigorieff1 IRIF, CNRS and Universit´ e Paris-Diderot, France seg@irif.univ-paris-diderot.fr Ir` ene Guessarian1 2 IRIF, CNRS and Universit´ e Paris-Diderot, France ig@irif.univ-paris-diderot.fr Received: 3/19/15, Revised: 10/26/15, Accepted: 6/15/16, Published: Abstract Using a simple basis of rational polynomial-like functions P0, . . . , Pn−1 for the free module of functions Z/nZ →Z/mZ, we characterize the subfamily of congruence preserving functions as the set of linear combinations of the products lcm(k) Pk where lcm(k) is the least common multiple of 2, . . . , k (viewed in Z/mZ). As a consequence, when n ≥m, the number of such functions is independent of n.
Introduction The notion of a congruence preserving function on rings of residue classes was introduced in Chen and studied in Bhargava .
Definition 1.1. Let m, n ≥1.
A function f : Z/nZ →Z/mZ is said to be congruence preserving if for all d dividing m for all a, b ∈{0, . . . , n −1} a ≡b (mod d) implies f(a) ≡f(b) (mod d).
(1) Remark 1.2. 1. If n ∈{1, 2} or m = 1 then every function Z/nZ →Z/mZ is trivially congruence preserving.
1Partially supported by TARMAC ANR agreement 12 BS02 007 01.
2Emeritus at UPMC Universit´ e Paris 6. Corresponding author INTEGERS: 16 (2016) 2 2. Observe that since d is assumed to divide m, equivalence modulo d is a congruence on (Z/mZ, +, ×). However, since d is not supposed to divide n, equivalence modulo d may not be a congruence on (Z/nZ, +, ×).
Example 1.3. 1. For functions Z/6Z →Z/3Z, condition (1) reduces to the condi-tions f(3) ≡f(0) (mod 3), f(4) ≡f(1) (mod 3), f(5) ≡f(2) (mod 3).
For functions Z/6Z →Z/8Z, condition (1) reduces to f(2) ≡f(0) (mod 2), f(3) ≡f(1) (mod 2), f(4) ≡f(0) (mod 4), f(5) ≡f(1) (mod 4).
In this paper, we characterize congruence preserving functions Z/nZ →Z/mZ.
We denote by Z the set of integers and by N the set of nonnegative integers (including zero).
Definition 1.4. The unary lcm function N →N maps 0 to 1 and k ≥1 to the least common multiple of 1, 2, . . . , k.
A natural way to associate with each map from N to Z a map from Z/nZ to Z/mZ is to restrict F to {0, · · · , n −1} and take its values modulo m.
Definition 1.5. With each map F : N →Z, we associate the map f : Z/nZ → Z/mZ defined by f = πm ◦F ◦ιn, where πm(x) = x (mod m), and ιn(z) is the unique element of π−1 n (z) ∩{0, . . . , n −1}.
Definition 1.5 is best pictured by the commutativity of diagram (2).
N F ! Z πm " Z/nZ ιn # f ! Z/mZ (2) Applying Definition 1.5 to binomial coefficients, we obtain a basis of the (Z/mZ)-module of functions Z/nZ →Z/mZ.
Proposition 1.6. Let Pk : Z/nZ →Z/mZ be associated with the N →N binomial function x )→ !x k " . For every function f : Z/nZ →Z/mZ there is a unique sequence (a0, . . . , an−1) of elements of Z/mZ such that f = k=n−1 # k=0 ak Pk .
(3) In other words, the family {P0, . . . , Pn−1} is a basis of the (Z/mZ)-module of func-tions Z/nZ →Z/mZ.
Our main result can be stated as INTEGERS: 16 (2016) 3 Theorem 1.7. A function f : Z/nZ →Z/mZ is congruence preserving if and only if, for each k = 0, . . . , n −1, in equation (3) the coefficient ak is a multiple of the residue of lcm(k) in Z/mZ.
The paper is organized as follows.
Proposition 1.6 is proved in Section 2 where, after recalling Chen’s notion of a polynomial function Z/nZ →Z/mZ (cf. ), we extend it to a notion of a rational polynomial function.
The proof of our main result, Theorem 1.7, is given in Section 3. We adapt the techniques of our paper , exploiting similarities between Definition 1.1 and the condition studied in for functions f : N →Z (namely, x−y divides f(x)−f(y) for all x, y ∈N). As a consequence of Theorem 1.7, the number of congruence preserving functions is independent of n for n ≥m and even for n ≥gpp(m) (the greatest prime power dividing m). Also, every congruence preserving function f : Z/nZ →Z/mZ is a rational polynomial for a polynomial of degree strictly less than the minimum of n and gpp(m).
In Section 4 we use our main theorem to count the congruence preserving func-tions Z/nZ →Z/mZ. We thus get an expression equivalent to that obtained by Bhargava in and which makes apparent the fact that, for n ≥gpp(m) (hence for n ≥m), this number depends only on m and is independent of n.
Representing Functions Z/nZ →Z/mZ by Rational Polynomials In [3, 1], congruence preserving functions Z/nZ →Z/mZ are introduced and studied together with an original notion of polynomial function Z/nZ →Z/mZ.
Definition 2.1 (Chen ). A function f : Z/nZ →Z/mZ is polynomial if it is associated (in the sense of Definition 1.5) with a function F : N →Z given by a polynomial in Z[X].
Polynomial functions Z/nZ →Z/mZ are obviously congruence preserving. Are all congruence preserving functions polynomial? Chen observed that this is not the case for some values of n, m, for instance n = 6, m = 8. He also proves that a stronger identity holds for infinitely many ordered pairs ⟨n, m⟩: every function Z/nZ →Z/mZ is polynomial if and only n is not greater than the first prime factor of m (in particular, this is the case when n = m and m is prime, cf. Kempner ).
Using counting arguments, Bhargava characterizes the ordered pairs ⟨n, m⟩such that every congruence preserving function f : Z/nZ →Z/mZ is polynomial.
Some polynomials in Q[X] (i.e., polynomials with rational coefficients) happen to map integers into integers.
INTEGERS: 16 (2016) 4 Definition 2.2. For k ∈N, let Pk ∈Q[X] be the following polynomial: Pk(x) = $x k % = &k−1 i=0 (x −i) k!
.
We will use the following examples later on: P0(x) = 1, P1(x) = x, P2(x) = x(x −1)/2, P3(x) = x(x −1)(x −2)/6, P4(x) = x(x −1)(x −2)(x −3)/24, P5(x) = x(x −1)(x −2)(x −3)(x −4)/120.
In , P´ olya used the Pk’s to give the following very elegant and elementary char-acterization of polynomials in Q[X] mapping integers to integers.
Theorem 2.3 (P´ olya). A polynomial in Q[X] is integer-valued on Z if and only if it can be written as a Z-linear combination of the polynomials Pk, k = 0, 1, 2, . . ..
It turns out that the representation of functions N →Z as Z-linear combinations of the Pk’s used in also fits in the case of functions Z/nZ →Z/mZ : every such function is a (Z/mZ)-linear combination of the Pk’s.
Definition 2.4. 1. A function f : Z/nZ →Z/mZ is rat-polynomial if is associated in the sense of Definition 1.5 with some polynomial in Q[X].
The degree of a rat-polynomial function is the smallest degree of an associated polynomial in Q[X].
We denote by P n,m k the rat-polynomial function Z/nZ →Z/mZ associated with the polynomial Pk of Definition 2.2 in the sense of Definition 1.5. When there is no ambiguity, P n,m k will be denoted simply as Pk.
Remark 2.5. In Definition 2.4, the polynomial crucially depends on the choice of representatives of elements of Z/nZ: e.g., for n = m = 6, 0 ≡6 (mod 6) but 0 = P2(0) ̸≡P2(6) = 3 (mod 6). The chosen representatives for elements of Z/nZ will always be 0, 1, . . . , n −1.
We now prove the representation result by the Pk’s.
Proof of Proposition 1.6. Let us start with uniqueness. We have f(0) = a0, and hence a0 is f(0).
We have f(1) = a0 + a1, and hence a1 = f(1) −f(0).
By induction, letting Qk = 'ℓ=k−1 ℓ=0 aℓPℓ, and noting that Pk(k) = 1, we have f(k) = Qk(k) + akPk(k) = Qk(k) + ak, and hence ak = f(k) −Qk(k). We thus are able to determine ak in Z/mZ.
For existence, argue backwards to see that this sequence suits.
Remark 2.6. The evaluation of ak Pk(x) in Z/mZ has to be done as follows: for x an element of Z/nZ, we consider it as an element of {0, . . . , n −1} ⊆N and we evaluate Pk(x) = 1 k!
&k−1 i=0 (x−i) as an element of Z, then we consider the remainder modulo m, and finally we multiply the result by ak in Z/mZ. For instance, for INTEGERS: 16 (2016) 5 n = m = 8, we have 4 P2(3) = 4 × 3 × 2 2 = 4 × 3 = 4, but we might be tempted to evaluate it as 4 P2(3) = 4 × 3 × 2 2 = 0 2 = 0, which does not correspond to our definition. However, dividing ak by a factor of the denominator is allowed.
Corollary 2.7. 1. Every function f : Z/nZ →Z/mZ is rat-polynomial with degree less than n.
The family of rat-polynomial functions {Pk | k = 0, 1, . . ., n −1} is a basis of the (Z/mZ)-module of functions Z/nZ →Z/mZ.
Example 2.8. The function f : Z/6Z →Z/6Z such that f(0) = 0, f(1) = 3, f(2) = 4, f(3) = 3, f(4) = 0, f(5) = 1, is represented by the rational polynomial Pf(x) = 3x+ 4 x(x −1) 2 which can be simplified to Pf(x) = 3x−x(x−1) on Z/6Z.
Example 2.9. The function f : Z/6Z →Z/8Z given by Chen as a non-polynomial congruence preserving function, namely the function such that f(0) = 0, f(1) = 3, f(2) = 4, f(3) = 1, f(4) = 4, f(5) = 7, is represented by the rational polynomial with coefficients a0 = 0, a1 = 3, a2 = 6, a3 = 2, a4 = 4, a5 = 4. Thus, f(x) = 3x + 6 x(x −1) 2 + 2 x(x −1)(x −2) 2 + 4 x(x −1)(x −2)(x −3) 8 + 4 x(x −1)(x −2)(x −3)(x −4) 8 = 3x + 3x(x −1) + x(x −1)(x −2) + x(x −1)(x −2)(x −3) 2 + x(x −1)(x −2)(x −3)(x −4) 2 .
Characterizing Congruence Preserving Functions Z/nZ →Z/mZ Congruence preserving functions f : Z/nZ →Z/mZ can be characterized by a sim-ple condition on the coefficients of the rat-polynomial representation of f given in Proposition 1.6.
3.1. Proof of Theorem 1.7 For proving Theorem 1.7 we will need some relations involving binomial coefficients and the unary lcm function; these relations are stated in the next three lemmata.
The proofs are elementary but technical and can be found in our paper .
Lemma 3.1. If 0 ≤n −k < p ≤n then p divides lcm(k) !n k " in N.
Lemma 3.2. If k ≤b then n divides An k,b = lcm(k) (!b+n k " − !b k ") in N.
INTEGERS: 16 (2016) 6 The following is an immediate consequence of Lemma 3.2 (set a = b + n).
Lemma 3.3. If a ≥b and k ≤b, then a −b divides lcm(k) (!a k " − !b k ") in N.
Besides these lemmata which deal with divisibility on integers, we shall use a classical result in Z/mZ. For x, y ∈Z we say x divides y in Z/mZ if and only if the residue class of x divides the residue class of y in Z/mZ.
Lemma 3.4. Let 1 ≤c1, . . . , ck ≤m and let c be their least common multiple in N. If c1, . . . , ck all divide a in Z/mZ then so does c.
Proof. It suffices to consider the case k = 2 since the passage to any k is done via a straightforward induction.
Let c = c1b1 = c2b2 with b1, b2 coprime.
Let t, u be such that a = c1t = c2u in Z/mZ. Then a ≡c1t ≡c2u (mod m). Using B´ ezout’s identity, let α, β ∈Z be such that αb1 + βb2 = 1. Then c(tα + uβ) = c1b1tα + c2b2uβ ≡aαb1 + aβb2 (mod m), and hence c(tα + uβ) ≡a (mod m), proving that c divides a in Z/mZ.
Proof of the “only if” part of Theorem 1.7. Assume f : Z/nZ →Z/mZ is con-gruence preserving and consider its decomposition f(x) = 'n−1 k=0 akPk(x) given by Proposition 1.6. We show that lcm(k) divides ak in Z/mZ for all k < n. The cases k = 0 and k = 1 are trivial since lcm(0) = lcm(1) = 1.
Claim 1. For all 2 ≤k < n, k divides ak in Z/mZ.
Proof. Recall that f(k) = 'n−1 i=0 ai !k i " = 'k i=0 ai !k i " since !k i " = 0 for i > k. We argue by induction on k ≥2.
Base case k = 2. If 2 does not divide m then 2 and m are coprime, and hence 2 is invertible and divides a2 in Z/mZ. Assume 2 divides m. As 2 divides 2−0 and f is congruence preserving, 2 also divides f(2) −f(0) = 2a1 + a2, and hence 2 divides a2.
Inductive step. Let 2 < k < n −1. The inductive hypothesis ensures that ℓ divides aℓin Z/mZ for every ℓ≤k. Let aℓ≡ℓqℓ(mod m) for 0 ≤ℓ≤k. We prove that k + 1 divides ak+1 in Z/mZ. First, observe that f(k + 1) −f(0) = (k + 1)a1 + k # i=2 $k + 1 i % ai + + ak+1 ≡ (k + 1)a1 + k # i=2 $k + 1 i % iqi + + ak+1 (mod m) f(k + 1) −f(0) = (k + 1)a1 + k # i=2 (k + 1) $ k i −1 % qi + + αm + ak+1 (4) INTEGERS: 16 (2016) 7 for some α. Let d = gcd(k + 1, m). Since d divides m and k + 1 −0 and f is congruence preserving, d also divides f(k + 1) −f(0). Using equality (4), we see that d divides the last term ak+1 of the sum. Using B´ ezout’s identity, let u, v be such that u(k + 1) + vm = d. Then u(k + 1) ≡d (mod m), and hence k + 1 divides d in Z/mZ. Since d divides ak+1, we conclude that k +1 divides ak+1 in Z/mZ.
Claim 2. (i) For all 2 ≤p ≤k < n, p divides ak in Z/mZ.
(ii) For all 2 ≤k < n, lcm(k) divides ak in Z/mZ.
Proof. Assertion (ii) is a direct application of Lemma 3.4 and assertion (i). We prove (i) by induction on p ≥2. Both the base case and the inductive step of this induction are proved by induction on k.
Base case p = 2. We have to prove that 2 divides ak for all k ≥2. If 2 does not divide m, then 2 is invertible and divides all numbers in Z/mZ. Assume now that 2 divides m. We argue by induction on k ≥2.
Base case. Apply Claim 1: 2 divides a2.
Inductive step. Let k < n −1. Assuming that 2 divides ai for all 2 ≤i ≤k, we prove that 2 divides ak+1. Two cases can occur.
Subcase 1: k + 1 is odd. Then 2 divides k and hence, by congruence preserva-tion, 2 divides f(k + 1) −f(1). As f(k + 1) −f(1) = ka1 + ('k i=2 ai !k+1 i ") + ak+1, and 2 divides k and also, by the induction hypothesis, 2 divides ai for 2 ≤i ≤k, we see that 2 divides ak+1.
Subcase 2: k + 1 is even. By congruence preservation, 2 divides f(k + 1) − f(0) = (k+1)a1+ ('k i=2 ai !k+1 i ") +ak+1. Since 2 divides k+1 and ai for 2 ≤i ≤k (induction hypothesis), we infer that 2 divides ak+1.
Inductive step. Let 2 ≤p < n −1 and assume that for all q ≤p and all ℓsuch that q ≤ℓ< n, q divides aℓin Z/mZ.
(5) By induction on k ≥p + 1, we prove that p + 1 divides ak for all k such that p + 1 ≤k < n.
Base case k = p + 1. Apply Claim 1: p + 1 divides ap+1.
Inductive step. Let k < n −1. Assuming that p + 1 divides ai in Z/mZ for all i INTEGERS: 16 (2016) 8 such that p + 1 ≤i ≤k, we prove that p + 1 divides ak+1 in Z/mZ. We have f(k + 1) −f(k −p) = k−p # i=1 ai $$k + 1 i % − $k −p i %% + ⎛ ⎝ k # i=k+1−p ai $k + 1 i %⎞ ⎠+ ak+1 (6) We first look at the terms of the first sum on the right side of (6) corresponding to 1 ≤i ≤p. Applying (5) with ℓ= i, we see that q divides ai in Z/mZ for all q ≤min(p, i) = i. Using Lemma 3.4, we conclude that lcm(i) divides ai in Z/mZ.
Observing that (k+1) = (k−p)+(p+1), we can apply Lemma 3.2 (with k−p, p+1 and i in place of b, n and k) and conclude that p+ 1 divides lcm(i) (!k+1 i " − !k−p i ") in N. Thus, p + 1 divides ai (!k+1 i " − !k−p i ") in Z/mZ.
We now turn to the terms of the first sum on the right side of (6) corresponding to p + 1 ≤i ≤k −p (if there are any). Each of these terms is divisible by p + 1 in Z/mZ, because the induction hypothesis on k ensures that p + 1 divides ai in Z/mZ whenever p + 1 ≤i ≤k.
Consider next the terms of the second sum on the right side of (6). For those terms corresponding to values of i such that p + 1 ≤i ≤k, divisibility by p + 1 in Z/mZ follows from the fact that, by the induction hypothesis on k, p+1 divides ai.
It remains to look at the terms associated with the i’s such that k + 1 −p ≤i ≤p (there are such i’s in case k + 1 −p < p + 1). For such i’s we have 0 ≤(k + 1) −i ≤ (k + 1) −p < p + 1 ≤k + 1 and Lemma 3.1 (used with k + 1, i and p + 1 in place of n, k and p) implies that p + 1 divides lcm(i) !k+1 i " . Now, for such i’s, the induction hypothesis (5) on p shows that lcm(i) divides ai in Z/mZ. A fortiori, p + 1 divides ai !k+1 i " in Z/mZ.
Let d = gcd(p + 1, m). As p + 1 divides in Z/mZ all terms of the two sums on the right side of (6) so does d. Since d divides m and k + 1 −(k −p) = p + 1 and f is congruence preserving, d also divides f(k + 1) −f(k −p). Using equality (6), we conclude that d divides in Z/mZ the last term ak+1. Using B´ ezout’s identity, let u, v be such that u(p + 1) + vm = d. Then u(p + 1) ≡d (mod m), and hence p + 1 divides d in Z/mZ. As d divides ak+1 in Z/mZ, we conclude that p+1 divides ak+1 in Z/mZ.
This ends the proof of the induction in the inductive step, and hence also the proof of Claim 2 and of the “only if” part of the Theorem.
Proof of the “if” part of Theorem 1.7. Assume f = 'k=n−1 k=0 ak Pk and that all of the ak’s are divisible by lcm(k) in Z/mZ. We can write f in the form f(n) = 'n k=0 cklcm(k) $n k % . We prove that f is congruence preserving, i.e., if 0 ≤b < a ≤ INTEGERS: 16 (2016) 9 n −1 and d divides both m and a −b then d also divides f(a) −f(b). Observe that f(a) −f(b) = b # k=0 cklcm(k) ($a k % − $b k %)+ + a # k=b+1 cklcm(k) $a k % .
By Lemma 3.3, a −b divides each term of the first sum. Consider the terms of the second sum. For b + 1 ≤k ≤a, we have 0 ≤a −k < a −b ≤a and Lemma 3.1 (used with a, k and a−b in place of n, k and p) shows that a−b divides lcm(k) $a k % .
Thus, a −b divides f(a) −f(b).
3.2. On a Family of Generators We now sharpen the degree of the rat-polynomial representing a congruence pre-serving function Z/nZ →Z/mZ. We first state some properties of the lcm function in N.
Lemma 3.5. Let m ≥1 be an integer with prime factorization m = pα1 1 · · · pαℓ ℓ.
Then lcm(k) = u &ℓ i=1 pαi,k i , where u is coprime with m and αi,k = max{βi | pβi i ≤ k}.
Definition 3.6. Let m ≥1 be an integer with prime factorization m = pα1 1 · · · pαℓ ℓ.
We let gpp(m) = max 0 pαi i | i ∈{1, . . ., ℓ} 1 be the greatest power of prime dividing m in N.
Lemma 3.7. The number gpp(m) is the least integer k such that m divides lcm(k).
Example 3.8. We have gpp(8) = 8, gpp(12) = 4 and gpp(14) = 7. The successive values of the residues in Z/mZ of lcm(k) are k 1 2 3 4 5 6 7 8 lcm(k) in Z/8Z 1 2 2 4 4 4 4 0 lcm(k) in Z/12Z 1 2 6 0 0 0 0 0 lcm(k) in Z/14Z 1 2 6 12 4 4 0 0 .
For all ℓ≥gpp(m), lcm(ℓ) is zero in Z/mZ.
Remark 3.9. 1. Either gpp(m) = m or gpp(m) ≤m/2.
In general, gpp(m) is greater than λ(m), the least k such that m divides k! (a function considered in ): for m = 8, gpp(m) = 8 whereas λ(m) = 4.
Using Lemma 3.7, we can get a better version of Theorem 1.7.
Theorem 3.10. A function f : Z/nZ →Z/mZ is congruence preserving if and only if it is associated in the sense of Definition 1.5 with a rational polynomial P = 'd−1 k=0 ak !x k " where d = min(n, gpp(m)) and such that lcm(k) divides ak in Z/mZ for all k < d.
INTEGERS: 16 (2016) 10 Proof. For k ≥gpp(m), m divides lcm(k) hence the coefficient ak is 0.
Theorem 3.11. (i) Every congruence preserving function f : Z/nZ →Z/mZ is rat-polynomial with degree less than gpp(m).
(ii) The family of rat-polynomial functions F = {lcm(k)Pk | 0 ≤k < min(n, gpp(m))} generates the set of congruence preserving functions Z/nZ →Z/mZ.
(iii) F is a basis of the set of congruence preserving functions if and only if m has no prime divisor p < min(n, m) (in case n ≥m this means that m is prime).
Proof. Assertions (i) and (ii) are restatements of Theorem 3.10. Let us prove (iii).
“Only If” part. Asssume m has a prime divisor p < min(n, m) and let p be the least one. Then lcm(p) = pa with a coprime with m, and hence lcm(p) ̸= 0 in Z/mZ. Since Pp(p) = 1 this shows that lcm(p) Pp is not the null function. However (m/p) lcm(p) = 0 in Z/mZ, and hence (m/p) lcm(p) Pp is the null function. As (m/p) ̸= 0 in Z/mZ, this proves that F cannot be a basis.
“If” part. Assume that m has no prime divisor p < min(n, m). We prove that F is (Z/mZ)-linearly independent. Suppose that the (Z/mZ)-linear combination L = 'min(n,gpp(m))−1 k=0 ak lcm(k) Pk is the null function Z/nZ →Z/mZ. By induction on k = 0, . . . , min(n, gpp(m)) −1 we prove that ak = 0.
• Basic cases k = 0, 1. From L(0) = a0 and L(1) = a0 + a1 we deduce a0 = a1 = 0.
• Induction step. Assuming k ≥2 and ai = 0 for i = 0, . . . , k −1, we prove that ak = 0. Observe that Pℓ(k) = !k ℓ " = 0 for k < ℓ< n. Since ai = 0 for i = 0, . . . , k−1, and Pk(k) = 1 we get L(k) = ak lcm(k ). As k < min(n, gpp(m)) ≤min(n, m) and m has no prime divisor p < min(n, m), the numbers lcm(k) and m are coprime.
Thus, lcm(k) is invertible in Z/mZ and equality L(k) = ak lcm(k) = 0 implies ak = 0.
Counting Congruence Preserving Functions We now compute the number of congruence preserving functions Z/nZ →Z/mZ.
As two different rational polynomials correspond to different functions by Proposi-tion 1.6 (uniqueness of the representation by a rational polynomial), the number of congruence preserving functions Z/nZ →Z/mZ is equal to the number of polyno-mials representing them.
Proposition 4.1. Let CP(n, m) be the number of congruence preserving functions Z/nZ →Z/mZ. Let m = pe1 1 pe2 2 · · · peℓ ℓ be the decomposition of m in powers of INTEGERS: 16 (2016) 11 primes. Let I = {i | pei i < gpp(m)} and J = {i | pei i ≥gpp(m)}. Then CP(n, m) = ⎧ ⎪ ⎨ ⎪ ⎩ p p1+p2 1+···+pe1 1 1 × · · · × p pℓ+p2 ℓ+···+p eℓ ℓ ℓ if n ≥gpp(m), & i∈I p pi+p2 i +···+p ei i i ×& i∈J p pi+p2 i +···+p ⌊logp n⌋ i +n(e−⌊logp n⌋) i if n < gpp(m).
Equivalently, writing E(p, α) instead of pα for better readability, we have CP(n, m) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ℓ & i=1 E(pi, ei ' k=1 pk i ) if n ≥gpp(m), & i∈I E(pi, ei ' k=1 pk i )×& i∈J E(pi, ( ⌊logp n⌋ ' k=1 pk i ) + n(e −⌊logp n⌋)) if n < gpp(m).
Corollary 4.2. For n ≥gpp(m), CP(n, m) does not depend on n.
Proof of Proposition 4.1. By Theorem 3.10, we must count the number of n-tuples of coefficients (a0, . . . , an−1), with, for k = 0, . . . , n −1, ak being a multiple of lcm(k) in Z/mZ.
Claim 1. For m = pe1 1 pe2 2 · · · peℓ ℓ, for all n, CP(n, m) = &ℓ i=1 CP(n, pei i ).
Proof of Claim 1. Let E(r, k) be the set of multiples in Z/rZ of lcm(k) and λ(r, k) be the cardinal of E(r, k). The Chinese remainder theorem shows that the map ρ : z )→ !
z (mod pei i ) " i=1,...,ℓis an isomorphism and also that ρ maps the set E(m, k) onto the Cartesian product P = &ℓ i=1 E(pei i , k). Indeed, let (ti)i=1,...,ℓ∈P. For each i = 1, . . . , ℓ, there is 0 ≤qi < pei i such that ti ≡qi lcm(k) (mod pei i ). Applying the Chinese remainder theorem, there are 0 ≤t, q < m such that t ≡ti (mod pei i ) and q ≡qi (mod pei i ). Then t ≡q lcm(k) (mod m), and hence ρ(t) = (ti)i=1,...,ℓ.
This proves that λ(m, k) = &ℓ i=1 λ(pei i , k) for each k. Thus, the number CP(n, m) of n-tuples (a0, . . . , an−1) such that lcm(k) divides ak is equal to CP(n, m) = 6 k<n λ(m, k) = 6 k<n ℓ 6 i=1 λ(pei i , k) = ℓ 6 i=1 6 k<n λ(pei i , k) = ℓ 6 i=1 CP(n, pei i ).
✷ Claim 1 reduces the problem to that of counting the congruence preserving func-tions Z/nZ →Z/pei i Z. We will use Theorem 3.10 to this end.
Claim 2. Letting ℓ= ⌊logp n⌋(and using the E(p, α) notation for pα), we have CP(n, pe) = 7 E(p, p + p2 + · · · + pe) if n ≥pe, E(p, p + p2 + · · · + pℓ+ (e −ℓ)n) if pℓ≤n < pe.
Proof of Claim 2. By Theorem 3.10, as gpp(pe) = pe, letting ν = inf(n, pe), we have CP(n, pe) = CP(ν, pe) = &ν−1 k=0 λ(pe, k). As we noted in the proof of Claim 1, for INTEGERS: 16 (2016) 12 pj ≤k < pj+1, the order λ(pe, k) of the subgroup generated by lcm(k) in Z/peZ is pe−j, and there are pj+1 −pj such k’s. For k = 0, lcm(0) = 1 yields λ(pe, 0) = pe.
• If n ≥pe then CP(n, pe) = CP(pe, pe) = pe &e−1 j=0 &pj+1−1 k=pj pe−j = pM with M = e + e−1 # j=0 (e −j)(pj+1 −pj) = p + p2 + · · · + pe • If n < pe then pℓ≤n < pe and CP(n, pe) = &n−1 k=0 λ(pe, k) = pe(&ℓ−1 j=0 &pj+1−1 k=pj pe−j)(&n−1 k=pℓpe−ℓ) = pM with M = e + ℓ−1 # j=0 (e −j)(pj+1 −pj) + n−1 # k=pℓ (e −ℓ) = (e −ℓ)pℓ+ (p + p2 + · · · + pℓ) + (n −pℓ)(e −ℓ) = (p + p2 + · · · + pℓ) + n(e −ℓ) ✷ This finishes the proof of Proposition 4.1.
Remark 4.3. In the number of congruence preserving functions Z/nZ →Z/peZ is shown to be equal to E(p, en −'n−1 k=1 min{e, ⌊logp k⌋}).
For pi ≤k < pi+1, we have ⌊logp k⌋= i, and hence min{e, ⌊logp k⌋} = ⌊logp k⌋for k ≤pe, and min{e, ⌊logp k⌋} = e for k ≥pe. Thus, we have • if n ≥pe, then 'n−1 k=1 min{e, ⌊logp k⌋} = 'pe−1 k=1 ⌊logp k⌋+ 'n−1 k=pe e = 'e−1 j=0 j(pj+1 −pj)+ e(n−pe) = −(p+· · ·+pe)+epe+e(n−pe), and hence en−'n−1 k=1 min{e, ⌊logp k⌋} = p+· · ·+pe.
This coincides with our counting in Claim 2.
• if n < pe, and l = ⌊logp n⌋, then, similarly, 'n−1 k=1⌊logp k⌋= 'ℓ−1 k=1⌊logp k⌋+ 'n−1 k=l ⌊logp k⌋= 'ℓ−1 j=0 j(pj+1 −pj) + ℓ(n −pℓ) = −(p + · · · + pℓ) + nℓ, and hence en −'n−1 k=1⌊logp k⌋= p + · · · + pℓ+ (e −ℓ)n. Again, this coincides with our counting in Claim 2.
Conclusion We proved that the rational polynomials lcm(k) Pk generate the Z/mZ submodule of congruence preserving functions Z/nZ →Z/mZ.
When n is larger than the greatest prime power dividing m, the number of functions in this submodule is independent of n. An open problem is the existence of a basis of this submodule.
Acknowledgments. We are grateful to the anonymous referee for insightful read-ing and valuable comments which helped improve the paper. We are also thankful INTEGERS: 16 (2016) 13 to the managing editor Bruce Landman whose advices improved the English, the typographic style, and the general readability.
References M. Bhargava, Congruence preservation and polynomial functions from Zn to Zm, Discrete Math. 173 (1997), 15–21.
P. C´ egielski, S. Grigorieff, I. Guessarian, Newton expansion of functions over natural integers having integral difference ratios, Int. J. Number Theory, 11 No 7 (2015), 2109–2139.
Z. Chen, On polynomial functions from Zn to Zm, Discrete Math. 137 (1995), 137–145.
A.J. Kempner, Polynomials and their residue systems, Amer. Math. Soc. Trans. 22 (1921), 240–288.
G. P´ olya, ¨ Uber ganzwertige ganze Funktionen, Rend. Circ. Mat. Palermo 40 (1915), 1–16.
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Algebraic Integers January 17, 2007 1 The ring of algebraic integers An algebraic number is called an algebraic integer if it is a root of polynomial xn + a1xn−1 + ... + an with integer coefficients.
Lemma 1.1. Let γ be any algebraic number. Then there exists a natural number a such that aγ is an algebraic integer.
Proof. We have a0γn + ... + an = 0 for some integers ai. Multiply by an−1 o , and get (a0γ)n + a1(a0γ)n−1 + ... + anan−1 0 = 0. Hence a0γ is an algebraic integer.
Lemma 1.2. Let γ1, ..., γn be non-zero complex numbers and let M be the Z-module generated by them, i.e. {a1γ1 + ... + anγn, ai ∈Z. Suppose that α has the property that αγi ∈M for each i. Then α is an algebraic integer.
Proof. We have αγi = P ci,jγj for every i and some integers ci,j. Hence γi are the solutions of the system of linear equations C⃗ γ −α⃗ γ = 0 where ⃗ γ = [γ1, ..., γn], C = Therefore the determinant det(C −αI) is 0. That determinant is a polynomial in α with integer coefficients and the highest coefficient 1.
Theorem 1.3. The set of algebraic integers is a ring.
Proof. Let α be a root of f(x), β be a root of g(x), both algebraic integers.
Let f be of degree m, g be of degree n. Consider the Z-module generated by αiβj, 0 ≤i ≤m −1, 0 ≤j ≤n −1. Then (α ± β)M ⊆M and (αβ)M ⊆M.
Hence both α ± β and αβ are algebraic integers by Lemma 1.2.
1 2 Rings of integers in finite extensions of Q Let K be a finite extension of Q. The ring of integers O = O(K) is the intersection of the ring of all algebraic integers with K.
Note that K = Q[α] for some α where the degree of the minimal polyno-mial of α = the degree of the extension = the number of different embeddings of K into ¯ Q. The embeddings σ1, ..., σn extend the maps α →αi where αi are the roots of the minimal polynomial of α.
For every β ∈K, the norm NK(β) (the trace trK(β)) is the product (the sum) of all σi(β). Both are rational numbers since these are the coefficients of the minimal polynomial for β. Note that the norm is multiplicative. As usual, the trace allows us to define a bi-linear non-degenerate form on K: (α, β) = tr(αβ). The properties are easy to verify, including tr(α2) ̸= 0 if α ̸= 0. Note that for α, β ∈O(K) we have tr(αβ) ∈Z.
For every α1, ..., αn ∈O(K) let DK(α1, ..., αn) = det(σj(ai))2 = det(trK(αiαj))) ∈Z.
The DK is the discriminant of the numbers αi.
Note that Theorem 2.1. Let I be an ideal in O(K) and let α1, ..., αn be numbers in I that are linearly independent over Q such that |DK(α1, ..., αn)| ∈N is minimal. Then I is the Z-module generated by αi’s.
Proof. Any α ∈I is (unique) linear combination a1α1 + ... + anαn. It is enough to show that ai ∈Z. Suppose that, say, a1 ̸∈Z. Then a1 = b + θ, θ the fractional part of a1 (strictly between 0 and 1).
Consider a new collection of numbers β1, ..., βn from O(K): ⃗ β = ⃗ α θ 0 0 ...
0 a2 1 0 ...
0 .
.
.
...
.
.
.
.
...
.
.
.
.
...
.
an 0 0 ...
1 .
Note that βi = αi ∈O(K) if i > 1 and β1 = θα1 + a2α2 + ... + anαn = β −bα1 ∈O(K). Clearly, βi are linearly independent since θ ̸= 0.
Then, computing the determinant in the definition of DK, we get |DK(⃗ β)| = θ2|DK(⃗ α)| < |DK(⃗ α)|, 2 since θ < 1, a contradiction.
Definition 2.2. The discriminant of K is DK(⃗ α) for some (=any) integral basis ⃗ α of O(K).
More properties of O(K).
Proposition 2.3. For every ideal I in O(K), O(K)/I is finite.
Proof. I is an n-dim free Abelian group inside another n-dim. free Abelian group O(K).
Proposition 2.4. O(K) is Noetherian.
Proposition 2.5. Every prime ideal P of O(K) is maximal.
Proof. Indeed, O(K)/P is finite domain, hence a field.
Definition 2.6. If A and B are two ideals in O(K) then we say that A is equivalent to B (denote A ∼B) if (α)A = (β)B for some non-zero α and β from O(K).
Remark 2.7. ∼is an equivalence relation.
Remark 2.8. (α)A = αA.
Remark 2.9. A ∼(1) = O(K) iffA is principal. Indeed, A ∼(1) means αA = (β) for some α, β ∈O(K). That implies β = αδ, δ ∈A ⊆O(K).
Hence A = (δ). The converse statement is obvious.
The equivalence classes of ideals are called ideal classes of K. We shall show that the number of ideal classes is finite.
Proposition 2.10 (Hurwitz). There exists N > 0 such that for every γ ∈ K, there exists t ≤M and θ ∈O(K) such that NK(tγ −θ) < 1.
3 Proof. Let ω1, ..., ωn be a basis of O(K). Then we have an isomorphism K →Qn (as vector spaces). If γ = P ciωi then we set ||γ|| = max i |ci|.
Then NK(γ) ≤ Y j X i |xi||σj(ωi)| !
≤||γ||nC (1) for some constant C.
Take any M > C so that m = M1/n is an integer.
For every i let ci = ai + bi where 0 ≤bi < 1, ai ∈Z. The integral part of γ, [γ] = P aiωi.
The fractional part is {γ} = γ −[γ]. Note that [γ] ∈O(K). Note also that ⃗ b is in the unit cube of Rn. Divide the unit cube into M cubes with side 1 m. Then for each j = 1, ..., M + 1, the coordinate vector of {jγ} is in one of these cubes. At least two of them must be in the same cube. Let it be {j1γ} and {j2γ}, j1 > j2. Let t = j1 −j2. Then tγ = θ + δ with θ ∈O(K), and MK(δ) ≤||δ||nC ≤ 1 m n C = C M < 1 by (1) as required.
Theorem 2.11. The number of ideal classes is finite.
Proof. Let A be an ideal of O. Choose β ∈A with minimal |NK(β)|.
Take any α ∈O(K).
Then by the proposition, there exists natural number t ≤M and θ ∈O(K) such that NK(t α β −θ) < 1. Multiply by NK(β): |NK(tα −θβ)| < |NK(β)|.
Hence for every α ∈A, tα ∈(β). Since t divides M!, we get that M!A ⊆(β) Therefore B = M!
β A ⊆O(K).
4 Note that B is certainly an ideal of O(K). Moreover (M!)A = (β)B, so A ∼B.
Since β ∈A, we have M! ∈B. Hence (M!) ⊆B. But there are only finitely many ideals bigger than (M!) (the factor-ring is finite), so there are finitely many choices for B.
Theorem 2.12. The classes of ideals form a group under multiplication.
Lemma 2.13. A = AB →B = (1).
Proof. Suppose that A = spanZ(α1, ..., αn). Since A = AB, we get αi = P βi,jαj, βi,j ∈B. Hence ⃗ α is the 1-eigenvector of the matrix U = .
Thus det(U −I) = 0. Expanding the determinant, we get 1 ∈B.
Lemma 2.14. (β)A = AB →B = (β).
Proof. Let δ ∈B. Then δA ⊆(β). Hence δ βA ⊆A. By Lemma 1.2, δ β ∈O.
Hence B ⊆(β) and β−1B ⊆O. Then A = (β−1B)A, hence β−1B = O(K), i.e. (β) = B.
Lemma 2.15. Am ∼(1) for some m.
Proof. Ai ∼Ai+j means (α)Ai = (β)Ai+j.
Hence (α)Ai = ((β)Aj)Ai.
Therefore by the previous lemma, (β)Aj = (α), so Aj ∼(1).
Proof of Theorem 2.12. It is easy to see that ∼is stable under multi-plication. So the set of classes is a finite semigroup. The previous lemma implies existence of inverses.
Corollary 2.16. Let hK be the ideal class number of K. Then AhK ∼(1).
Theorem 2.17 (Fundamental theorem of ideal theory.). Every ideal of O(K) can be written as a product of prime ideals; it can be written in a unique way except for the order of factors.
Lemma 2.18. AB = AC →B = C.
Proof. Ah = (α). Hence αB = αC which trivially implies B = C.
Lemma 2.19. A ⊆B implies that there exists an ideal C such that A = BC.
Proof. Bh = (β). Then Bh−1A ⊆(β). Hence C = 1 βBh−1A ⊆O(K). Then BC = 1 βBhA = A.
5 Proof of Theorem 2.17. O(K)/A is finite, hence there exists a maximal ideal P1 > A. Then by the previous lemma there exists A1: P1A1 = A. We have A ⊂A1. If A1 ̸= O(K), we can continue. Hence A = P1P2...Pm for some m (by the Noetherian property).
For uniqueness: if A = Q1...Ql then P1 must divide one of the Qi, etc.
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Plant-Associated Bacterial Degradation of Toxic Organic Compounds in Soil - PMC
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Int J Environ Res Public Health
. 2009 Aug 12;6(8):2226–2247. doi: 10.3390/ijerph6082226
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Plant-Associated Bacterial Degradation of Toxic Organic Compounds in Soil
Martina McGuinness
Martina McGuinness
1 Department of Science and Health, Institute of Technology Carlow, Kilkenny Road, Carlow, Ireland; E-Mail: Martina.McGuinness@itcarlow.ie
Find articles by Martina McGuinness
1, David Dowling
David Dowling
1 Department of Science and Health, Institute of Technology Carlow, Kilkenny Road, Carlow, Ireland; E-Mail: Martina.McGuinness@itcarlow.ie
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1,
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1 Department of Science and Health, Institute of Technology Carlow, Kilkenny Road, Carlow, Ireland; E-Mail: Martina.McGuinness@itcarlow.ie
Author to whom correspondence should be addressed; E-Mail: David.Dowling@itcarlow.ie; Tel.: +353-5991-75507; Fax: +353-5991-75501
Received 2009 Jul 28; Accepted 2009 Aug 7; Issue date 2009 Aug.
© 2009 by the authors; licensee Molecular Diversity Preservation International, Basel, Switzerland.
This article is an open-access article distributed under the terms and conditions of the Creative Commons Attribution license (
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PMCID: PMC2738884 PMID: 19742157
Abstract
A number of toxic synthetic organic compounds can contaminate environmental soil through either local (e.g., industrial) or diffuse (e.g., agricultural) contamination. Increased levels of these toxic organic compounds in the environment have been associated with human health risks including cancer. Plant-associated bacteria, such as endophytic bacteria (non-pathogenic bacteria that occur naturally in plants) and rhizospheric bacteria (bacteria that live on and near the roots of plants), have been shown to contribute to biodegradation of toxic organic compounds in contaminated soil and could have potential for improving phytoremediation. Endophytic and rhizospheric bacterial degradation of toxic organic compounds (either naturally occurring or genetically enhanced) in contaminated soil in the environment could have positive implications for human health worldwide and is the subject of this review.
Keywords: toxic organics, soil, bacteria, bioremediation
1. Introduction
Synthetic organic compounds are ubiquitous in our modern environment. They are found in our homes, workplaces, public spaces and in agriculture. These organic compounds can enter soil, air and water through either local or diffuse contamination and can often be found far from their source of origin. Local or point source pollution involves discrete locations of pollution, e.g., industrial waste via factory or sewer pipes. Diffuse or nonpoint source pollution involves pollution from multiple cumulative inputs over a large area, e.g., agricultural waste (from farms) and municipal waste. While a large number of synthetic organic compounds are harmless, some are toxic and pose serious environmental and human health risks. Effects of contamination of environmental soil with toxic synthetic organic compounds include the poisoning of animals and plants, altering of ecosystems, and human health risks. International and national legislation attempts to address local sources of pollution by targeting industrial discharge. However, diffuse sources of pollution are more difficult to monitor and usually have a greater impact on the quality of the environment.
Many toxic synthetic organic compounds are persistent and are stored in fat tissue, due to their hydrophobic properties, resulting in bioaccumulation. Therefore, organisms at higher levels in food chains (e.g., humans) tend to have greater concentrations of these bioaccumulated toxins stored in their fat tissue than those at lower levels resulting in biomagnification of the physiological effects of the toxins in higher organisms. At the highest level in the food chain, i.e., humans, these toxic organic compounds can be passed from mother to child either in utero via the placenta or post-natally via breast milk.
Synthetic organic compounds of concern as environmental contaminants include polychlorinated biphenyls (PCBs), pesticides, industrial solvents, petroleum products, dioxins and furans, explosives, and brominated flame retardants. Twelve organic compounds were listed as persistent organic pollutants (POPS) by the Stockholm Convention on Persistent Organic Pollutants, under the auspices of the United Nations Environment Programme (UNEP), an international agreement enforced in 2004 . The 12 POPs listed by the Stockholm Convention include PCBs, nine chlorinated organic pesticides [aldrin, chlordane, dichlorodiphenyltrichloroethane (DDT), dieldrin, endrin, mirex, heptachlor, hexachlorobenzene, and toxaphene], and dioxins and furans. Although the use of these POPs worldwide has been generally phased out because of their toxicity and persistence, they can still be found as contaminants in the natural environment due to their past use and continue to pose a threat to human health.
Traditional technologies routinely used for the remediation of contaminated environmental soil include excavation, transport to specialized landfills, incineration, stabilization and vitrification. Recently, however, there has been much interest in bioremediation technologies which use plants and microorganisms (including bacteria) to degrade toxic contaminants in environmental soil into less-toxic and/or non-toxic substances. The United States Environmental Protection Agency (USEPA) defines bioremediation as a treatability technology which uses biological activity to reduce the concentration and/or toxicity of a pollutant. Bioremediation technologies offer many advantages over traditional remediation technologies as they can be applied in situ without the need for removal and transport of contaminated soil, are usually less expensive and less labour-intensive relying on solar energy, have a lower carbon footprint, and have a high level of public acceptance. Phytoremediation, the use of plants to degrade toxic contaminants in the environment involving a number of processes including phytoextraction, phytotransformation, phytostabilization, phytovolatilization and rhizofiltration, has been reviewed extensively and the reader is directed to a number of recent reviews [2–4]. Phytoextraction (or phytoaccumulation) involves the uptake and concentration of pollutants into harvestable biomass for sequestration or incineration. Phytotransformation involves enzymatic modification resulting in inactivation, degradation (phytodegradation), or immobilization (phytostabilization) of pollutants. Phytovolatilization involves the removal of pollutants from soil and their release through leaves via evapotranspiration processes and rhizofiltration involves the filtering of water through a mass of roots to remove pollutants. While some success has been reported using plants alone in bioremediation [2–4], the use of plants in conjunction with plant-associated bacteria offers much potential for bioremediation. Degradation of toxic organic compounds in environmental soil by plant-associated bacteria can involve endophytic and rhizospheric bacteria. Endophytic bacteria are non-pathogenic bacteria that occur naturally in the internal tissues of plants and can promote plant growth, be beneficial to the plant host by producing a range of natural products, and contribute to enhanced biodegradation of environmental soil pollutants [5,6]. Almost all 300,000 plant species identified have at least one species of endophyte . Endophytic bacterial species isolated from plants, to date, include Acetobacter, Arthrobacter, Bacillus, Burkholderia, Enterobacter, Herbaspirillum and Pseudomonas, as reviewed by Lodewyckx et al. . In contrast to endophytes, rhizospheric bacteria are associated with the rhizosphere of plants, i.e., the area of soil surrounding plant roots, where complex microbial communities are supported by root exudates, mucilage, and decaying root cells . Rhizospheric soil typically has 10–100 times more microbes, on a per gram basis, than unvegetated soil . Rhizospheric microbial communities can benefit the plant by synthesizing compounds that protect plants by decreasing plant stress hormone levels, delivering key plant nutrients, protecting against plant pathogens, and degrading contaminants before they negatively impact the plants as reviewed by Gerhardt et al. . Phytoremediation has been reported to be approximately 10-fold less expensive than traditional remediation technologies and can include the use of buffers, vegetation filters, in situ phytoremediation plantings, and percolation controlling vegetative caps as described by Licht and Isebrands . Therefore, the use of endophytic and rhizospheric bacteria capable of degrading toxic synthetic organic compounds in combination with specific plants (chosen to suit the environment to be remediated and/or because of their relationship with the endophytic or rhizospheric bacteria to be used) could offer an efficient, economic and sustainable remediation technology for the twenty first century.
The spectrum of toxic synthetic organic compounds identified as contaminants in environmental soil and the use of plant-associated endophytic bacteria and rhizospheric bacteria to degrade these toxic organic compounds will be the subject of this review. The use of biotechnology to engineer plant-associated bacteria to produce specific enzymes capable of degrading these toxic substances and the potential of these bacteria to contribute to bioremediation will also be discussed.
2. Toxic Synthetic Organic Compounds in Environmental Soil and Associated Human Health Risks
Because of their chemical structure many synthetic organic compounds are extremely resistant to natural breakdown processes and once released into the environment may persist for years and even decades. Many of these organic compounds are toxic and are associated with serious human health risks as described below.
2.1. PCBs and Synthetic Organic Pesticides
PCBs are toxic synthetic aromatic compounds notorious for their persistence and potential toxicity and were widely used in industry in the twentieth century. They are a group of polychlorinated biphenyl compounds with 209 different congeners or related chemicals, some containing up to 10 chlorine atoms. However, commercially available PCB mixtures (e.g., Arochlor) typically contain only 20–60 congeners. Because of the chemical stability, electronic insulating properties, thermal stability and non-flammability, PCBs were widely used in flame retardants, dielectric fluids in capacitors, transformers, hydraulic fluids, surface coating, adhesives and dyes. The manufacture of PCBs was banned in the USA in 1978 because of their toxicity. Although the manufacture of PCBs has been banned they remain a problem in the environment because of their persistence. Since PCBs were first synthesized in 1864, it is estimated that approximately 1 million tonnes have been manufactured worldwide and that approximately 30% of all PCBs manufactured has been released into the natural environment resulting in the contamination of soils and sediments . PCBs have been detected in polar bears in the Arctic, an environment far removed from industry, providing evidence of the dispersal of these toxic synthetic organic compounds in the natural environment [15,16].
DDT is one of the best known toxic chlorinated organic pesticides and although its use as an agricultural insecticide worldwide was banned by the Stockholm Convention, it is still used to control malaria in some parts of the world and remains controversial. DDT was used worldwide as an insecticide from the 1940s until the 1970s, when it was banned in the USA and other countries. In 1962, Rachel Carson in her popular book ‘Silent Spring’ suggested that DTT and other pesticides were associated with cancer and that their agricultural use was a threat to wildlife, particularly birds . Along with the passing of the Endangered Species Act in 1973, the US ban on DTT is cited by scientists as a major factor in the comeback of the bald eagle in the US . However, when a global ban on DDT was proposed in 2001, several countries in Africa claimed that DDT was still needed as an inexpensive and effective means for control of the vector associated with malaria. Although DDT is generally not toxic to human beings and was banned mainly for ecological reasons, subsequent research has shown that exposure to DDT at amounts that would be needed to control malaria might cause preterm birth and early weaning, eliminating the benefit of reducing infant mortality from malaria [19,20]. Therefore, the use of DDT to help control malaria remains controversial because of its associated human health risks [19,20].
Two of the most commonly used pesticides in agriculture worldwide, the chlorinated organic compounds 2,4-dichlorophenoxyacetic acid (2,4-D) and atrazine, are not listed by the Stockholm Convention as POPs but have been listed by the USEPA as toxic and are associated with human health risks . Both 2,4-D and atrazine are broad-leaf herbicides protecting many of the world’s important crops, such as wheat, corn, and rice, which are cereal grains. Since its introduction in 1946, 2,4-D remains the most widely used herbicide worldwide. Although 2,4-D is biodegradable, it may persist in soil and water if microbes with the required capacity for biodegradation are not present in sufficient numbers. Atrazine has been banned by the European Union since 2007, but is still used in many parts of the world, and can remain in soil for greater than one year after use and leach into groundwater contaminating private and community wells .
Other synthetic chlorinated organic pesticides of concern as contaminants of environmental soil include tetrachlorophenol (TCP), pentachlorophenol (PCP), and the tin-containing pesticide, tributylin (TBT). TCP is an insecticide and a bactericide and is widely used as a preservative for latex, wood, and leather. PCP is a disinfectant, a fungicide, and an extremely effective preservative for wood. In addition, PCP and its products are toxic to plants, facilitating their use as defoliants and general herbicides. TCP and PCP can be released into the environment as a result of their manufacture, storage, transport, or use as an industrial wood preservative. Their use at sawmills has lead to extensive groundwater contamination . TCP and PCP are strong irritants and can produce skin and eye irritation upon contact. They are readily absorbed through the skin and can produce systemic effects. Acute exposure to TCP and PCP in animals is associated with convulsant activity and inhibition of oxidative phosphorylation . The tin-containing biocide TBT is used to control a wide variety of organisms. It is used in wood preservatives, as an anti-fouling pesticide in marine paints, and as an antifungal agent in industrial water systems. TBT compounds bioaccumulate as they move up the marine food chain and have been associated with toxicity to a number of marine organisms such as molluscs, otters, dolphins and whales [25,26]. Chlorobenzoates, toxic metabolic intermediates produced from biodegradation of a variety of compounds including PCBs and chlorinated aromatic pesticides, are also considered environmental contaminants.
Glyphosate is an organophosphate broad-spectrum herbicide originally sold in the 1970s under the tradename Roundup™. Today, glyphosate is the most widely used herbicide in the USA. Although glyphosate is less toxic than chlorinated organic pesticides, it is a suspected endocrine disruptor. A review of at least 58 studies on the effects of Roundup™ suggests that non-target organisms were exposed to only minimal acute and chronic risk . However, more recent research reports that glyphosate induces a variety of functional abnormalities in the specific activity of the enzymes in the liver, heart and brain, in pregnant rats and their fetuses . Glyphosate was also reported to interfere with an enzyme involved in testosterone production in mouse cell culture and to interfere with an oestrogen biosynthesis enzyme in cultures of human placental cells .
A number of synthetic organic compounds, e.g., PCBs and many pesticides, are suspected endocrine disruptors and have been associated with the feminization of males. A number of scientists attribute the striking drop in sperm counts among men worldwide to these endocrine disruptors. In a landmark study, Danish researchers reviewed 61 studies and reported in 1992 that the number and motility of sperm in men’s semen had declined by 50% since 1938 . Other studies have linked endocrine disruptors and rising rates of testicular cancer and breast cancer [33,34]. Exposure of humans to PCBs and a number of chlorinated organic pesticides has been associated with an increased risk of developing cancer and with developmental disabilities in children. A recent study reported an association between exposure to pesticides and an increased incidence of Parkinson’s disease .
2.2. Volatile Organic Compounds (VOCs)
VOCs are vapours emitted by various solids or liquids, e.g., petrol, diesel, pesticides, paint, cleaning supplies and adhesives, many of which have short- and long-term adverse health effects. Benzene, toluene, ethylbenzene and xylene (BTEX) compounds are a family of VOCs based on the benzene structure and are found in petroleum products. Refineries adjust the amounts of BTEX compounds in petrol so as to meet vapour pressure and octane standards. BTEX compounds are major contaminants of environmental soil and groundwater and are usually found near petroleum and natural gas production sites, petrol stations and other sites with underground or above-ground storage tanks containing petroleum products. Exposure of humans to BTEX can occur by either ingestion (drinking water from contaminated wells), or by inhalation (exposure to BTEX contaminated water via showering or laundering). Acute exposure to petrol and its BTEX components has been associated with skin and sensory irritation, central nervous system depression, and effects on the respiratory system in humans while long-term exposure to BTEX compounds affects the kidney, liver and blood systems . According to the USEPA, there is evidence from both human epidemiological and animal studies that benzene is a human carcinogen, and that workers exposed to high levels of benzene in occupational settings were found to have increases rates of leukaemia .
Methyl tertiary butyl ether (MTBE), also a VOC, is used as a fuel oxygenate, i.e., a chemical containing oxygen that is added to fuels, especially petrol, to make them burn more efficiently. It can be a major contaminant of groundwater as a result of the widespread spillage or leakage of MTBE-containing petrol from underground storage tanks at petrol stations. The USEPA concluded that MTBE was a potential human carcinogen at high doses .
Trichloroethylene (TCE), and tetrachloroethene [also known as perchloroethylene (PCE)], are chlorinated VOCs. TCE was widely used as a volatile anaesthetic and also as an industrial solvent during the first half of the twentieth century. As an anaesthetic, TCE was originally thought to be less toxic to the liver than chloroform, and to be less pungent and flammable than ether. However, TCE was subsequently found to be associated with serious health risks and was replaced as an anaesthetic by halothane in the 1950s. The symptoms of acute exposure to TCE are similar to those of alcohol intoxication, beginning with headache, dizziness, and confusion, progressing with increasing exposure to unconsciousness . Much of what is known about the human health effects of TCE is based on occupational exposures. Beyond the effects to the central nervous system, workplace exposure to TCE has been associated with toxic effects in the liver and kidney . Over time, occupational exposure limits on TCE have tightened, resulting in more stringent ventilation controls and personal protective equipment use by workers. TCE was also used as a dry cleaning solvent until it was replaced by PCE in the 1950s. More recently, TCE was used as a cleaning solvent to clean military weapons during the Gulf War. As a result of this exposure of military personnel to TCE, an association with the neurological disorder amyotrophic lateral sclerosis (Lou Gehrig’s disease) was reported by Kasarskis et al. , and an association with a neurologic syndrome resembling Parkinson’s disease by Gash et al. . PCE is still widely used as a solvent in the dry-cleaning industry and is a common environmental soil contaminant, associated with central nervous system dysfunction . PCE contamination of soil usually results from spillage, overfilling, sewer leakage, or illegal disposal by commercial dry cleaning facilities. Because of the mobility of PCE in groundwater, its toxicity at low levels, and its density (which causes it to sink below the water table), cleanup activities tend to be especially problematic compared to the cleanup of oil spills.
2.3. Hydrocarbons
Hydrocarbons contain hydrogen and carbon, and can be found in the environment as gases, tiny particles, or droplets. Hydrocarbons, primarily measured as total petroleum hydrocarbons, are the majority of organic compounds in most crude oils and contain hundreds of individual components. Most hydrocarbons in the environment are associated with the use of petrol, diesel, crude oil, and oil products in vehicles used for transportation. Hydrocarbons can be gases (e.g., methane and propane), liquids (e.g., hexane and benzene), waxes or low melting solids (e.g., paraffin wax and naphthalene), or polymers (e.g., polyethylene, polypropylene and polystyrene). There are three major categories of aromatic hydrocarbons of concern as contaminants of environmental soil. They are: (i) polycyclic aromatic hydrocarbons (PAHs), (ii) heterocyclic aromatic hydrocarbons, and (iii) alkyl PAHs, as described in more detail below.
2.3.1. Polycyclic Aromatic Hydrocarbons (PAHs)
All PAHs contain at least two fused aromatic rings in linear, angular or cluster arrangements [45,46] and can be produced by petroleum production sites and combustion processes. PAHs are generally more difficult to degrade than many other organic compounds and are persistent in environmental soil. The USEPA has listed 16 PAHs as priority pollutants because of their persistence and carcinogenicity based on toxicity, potential for human exposure, frequency of occurrence at hazardous waste sites, and the extent of information available . The USEPA considers seven of these 16 priority PAHs as probable human carcinogens . They are benzo(a)anthracene, chrysene, benzo(a)pyrene, benzo(b)fluoranthene, benzo(k)fluoranthene, dibenz(a,h)anthracene, and indeno(1,2,3-cd)pyrene. Naphthalene, a PAH with two fused benzene rings, is produced commercially from either coal tar or petroleum. It is used mainly in the production of other chemicals including plasticizers, dyes, and insecticides, and is a major component of creosote and mothballs. Exposure to naphthalene is associated with haemolytic anaemia (abnormal breakdown of red blood cells) , cataracts and retinal damage .
2.3.2. Heterocyclic Aromatic Hydrocarbons: Dioxins and Furans
Heterocyclic aromatic hydrocarbons include dioxins and furans, both listed by the Stockholm Convention as POPs . Dioxins are produced unintentionally by industry due to incomplete combustion, as well as during the manufacture of certain pesticides and other chemicals, metal recycling and pulp and paper bleaching. Dioxins have also been found in automobile exhaust, tobacco smoke and wood and coal smoke and in commercial mixtures of PCBs. Dioxins are a group of 75 related chemical compounds known as polychlorinated dibenzo-p-dioxins. Each of the 75 compounds differs in the number and location of chlorine atoms on a basic three-ringed structure of carbon, hydrogen and oxygen atoms. Furans are a group of 135 related heterocyclic aromatic hydrocarbons called polychlorinated dibenzofurans. Of these, 17 [including the most toxic, 2,3,7,8-tetrachloro-dibenzo-p-dioxin (2,3,7,8-TCDD)] pose a major health risk. Similar in chemical structure and biological properties, dioxins and furans are usually found together in the environment as complex mixtures. The toxicity of each compound depends on the number and position of the chlorine atoms within the molecules. 2,3,7,8-TCDD was the contaminant in the weedkiller 2,4,5-trichlorophenoxy-acetic acid (2,4,5-T) used by the U.S. military as a defoliant in the early 1970s in the Vietnam War in ‘Agent Orange’. Agent Orange was equal parts 2,4,5-T and 2,4-D (see section 2.1). Dioxins and furans do not dissolve in water and can attach to particles of soil, dust and sediment. As a result, they can persist unchanged in the environment, mainly in soil and sediment, for years. A number of studies have been carried out on populations after accidental environmental exposure to high levels of dioxins and furans and report that chloracne, a skin disorder, is the most common human health effect . The facial images of the Ukranian presidential candidate, Victor Yuschenko, widely circulated by the media in 2004 clearly showed the effect of chloracne as a result of deliberate 2,3,7,8-TCDD poisoning. Extreme exposures also lead to other effects on the skin, liver, immune system, reproduction system, and the central nervous system .
2.3.3. Alkyl PAHs
Alkyl PAHs are PAHs with alkyl group substitution on their ring structures. The alkyl groups generally have one to four saturated carbon atoms, and thus can produce many different structural isomers. Alkylated PAHs are more abundant, persist for a longer time, tend to bioaccumulate to a greater degree (alkyl substitution usually decreases water solubility), and are sometimes more toxic than the parent PAHs. Within an aromatic series, acute toxicity increases with increasing alkyl substitution on the aromatic nucleus. Crude oils contain primarily the alkyl homologues of aromatic compounds and relatively small quantities of unsubstituted PAHs. Usually, the most significant compounds when assessing environmental damage associated with oil spills are PAHs and alkylated PAHs.
2.4. Explosives
Organic explosives including trinitrotoluene (TNT), hexahydrotrinitrotriazine or Royal Demolition Explosive (RDX), and octahydro-tetranitrotetraocine or High Melting eXplosive, (HMX) can contaminate environmental soil. TNT has been associated with aplastic anaemia and hepatitis, while RDX has been shown to affect the central nervous system . While information on the health risks associated with HMX is limited, studies in laboratory rats, mice, and rabbits indicate that HMX may be harmful to the liver and central nervous system .
2.5. Brominated Flame Retardants
Brominated flame retardants are POPs and are the most widely used flame retardants because of their efficiency and low production costs . The major brominated flame retardants used worldwide are tetrabromobisphenol A, hexabromocyclododecane, and polybromodiphenyl ethers . These compounds can enter the environment locally via wastewaters of industrial facilities, through volatilization, leaching and combustion. Flame retardants have been found in air, water, soils and sediments far from where they are produced or used [55–57], again providing evidence of the wide dispersal of toxic synthetic organic compounds in the environment.
Reducing the levels of toxic synthetic organic compounds in the environment is an issue of growing concern as the effects of these chemicals on human heath become more widely understood. Therefore, the use of plant-associated endophytic and rhizospheric bacteria to degrade toxic organic compounds in contaminated environmental soil could have positive implications for human health worldwide.
3. Remediation Technologies
3.1. Traditional Technologies for the Remediation of Contaminated Soil
It is estimated that traditional global remediation costs are in the range of $US25-50 billion annually [58,59]. Unfortunately, this high cost of remediation contributes to the abandonment worldwide of a large number of polluted commercial sites or brownfields. For example, in the USA, the USEPA Office of Underground Storage Tanks reported that 34% of known contaminated sites choose to be in non-compliance in 2008 . Some of the reasons for non-compliance are typically (i) non-compliance has a lower immediate cash cost than compliance using traditional remediation technologies, and (ii) the demand for green remediation is not yet powerful enough to drive action and does not impact on sales and revenue. However, with: (i) an increasing public awareness of the need to move towards a low carbon economy, (ii) the introduction of regulations with an increased emphasis on a low carbon economy, and (iii) the development of sustainable bioremediation metrics, there is an increased interest in moving away from traditional technologies for the remediation of contaminated soil (e.g., excavation, transport to specialized landfills, incineration, stabilization and vitrification) towards bioremediation technologies by regulators, consultants and representatives from industry.
3.2. Bioremediation Technologies
In remediation of the environment, bioremediation is a treatment process that uses microorganisms (including bacteria) and plants to degrade toxic contaminants into less toxic or non-toxic substances. According to the U.S. Sustainable Remediation Forum (US SURF), sustainable remediation is broadly defined as a remedy or combination of remedies whose net benefit on human health and the environment is maximized through the judicious use of limited resources . Because bioremediation technologies can be applied in situ without the need for removal and transport of contaminated soil, are usually less expensive and less labour-intensive (relying on solar energy), have a lower carbon footprint, and have a high level of public acceptance than traditional remediation technologies, they potentially offer a sustainable solution to the problem of contaminated environmental soil. However, conditions in the contaminated environmental soil need to be optimized for effective biodegradation of the target contaminants, i.e., the levels of moisture, pH and temperature in the soil will dictate survival ranges for microorganisms and plants used for bioremediation, abundant oxygen will facilitate mineralization of soil contaminants, concentrations of nutrient and hydrocarbons in the soil will need to be balanced for efficient bioremediation, and suitable microorganisms and plants will be required to degrade and/or mineralize target contaminants. Although it appears that the advantages associated with the use of bioremediation technologies clearly outweigh the disadvantages, when compared to traditional remediation technologies, other factors to consider when using bioremediation technologies include the length of time required (months or years), geographic limitations on the use of specific plants, and the seasonal limitations associated with the use of specific plants. Choosing a technology for sustainable remediation of contaminated environmental soil requires detailed analyses of the environmental impact. Sustainable remediation metrics include economic, societal and environmental metrics for comparing and selecting remedies and monitoring success and include important elements such as water use, worker safety, community impact, and the net environmental benefit . Bioremediation technologies compare favourably with traditional remediation technologies when analysed using sustainability remediation metrics
4. Biodegradation of Toxic Organic Compounds in Environmental Soil
A number of bacterial strains have been identified in a wide variety of contaminated environments with enzymes capable of degrading toxic organic compounds. Anaerobic bacteria can convert highly chlorinated PCB congeners into less chlorinated biphenyls by reductive dechlorination [62,63]. Aerobic bacteria, e.g., Burkholderia xenovorans LB400 and Rhodococcus sp. strain RHA1 , can then cleave lesser chlorinated biphenyl rings to yield chlorinated benzoates and pentanoic acid derivatives which are often degradable by other bacteria. The dechlorinating bacteria Dehalococcoides ethenogenes, Dehalobacter restrictus, Desulfitobacterium dehalogenans, Dehalospirillum multivorans, Desulfuromonas chloroethenica, and Desulfomonile tiedjei are capable of dehalogenating PCE and other chlorinated aromatic compounds . Dehalococcoides ethenogenes strain 195, is the only bacterial strain which completely dechlorinates PCE to yield ethylene and is of interest because of its potential use in the bioremediation of TCE and PCE contaminated sites . Mannisto et al. identified bacterial strains Herbaspirullum sp K1, Sphingomonas strains K74 and MT1, Nocardioides sp K44, that could degrade TCP faster at low temperature than at room temperature. Sphingobium chlorophenolicum strain ATCC 39723 can completely mineralize PCP . Bacteria involved in the biodegradation of petroleum products in a number of different environmental soil types have also been identified [71,72]. However, for the purpose of this review article only plant-associated endophytic and rhizospheric bacteria associated with the degradation of toxic organic compounds in the environment will be discussed.
4.1. Endophytic Bacteria and Phytoremediation
A number of endophytic bacteria reported to contribute to degradation of environmental pollutants in planta are listed in Table 1.
Table 1.
Reported cases of successful bioremediation using endophytic bacteria (adapted from Table 2 in Ryan et al. ).
| Compound | Plants used | Microbes used | Reference |
:---: :---: |
| PCBs, TCP | Wheat (Triticum spp.) | Herbaspirillum sp. K1 | Mannisto et al. |
| Chlorobenzoic acids | Wild rye (Elymus dauricus) | Pseudomonas aeruginosa R75 Pseudomonas savastanoi CB35 | Siciliano et al. |
| Pesticide 2,4-D | Pea (Pisum sativum) | Pseudomonas putida VM1450 | Germaine et al. |
| VOCs and toluene Toluene MTBE, BTEX, TCE | Yellow lupine (Lupinus luteus L.) Poplar (Populus) Poplar (Populus cv. Hazendans and cv. Hoogvorst) | Burkholderia cepacia G4 Burkholderia cepacia Bu61 (pTOM-Bu61) Pseudomonas sp. | Barac et al. Taghavi et al. Germaine et al. Porteus-Moore et al. |
| HCs Naphthalene | Pea (Pisum sativum) | Pseudomonas putida VM1441 (pNAH7) | Germaine et al. |
| Explosives TNT, RDX, HMX | Poplar tissues (Populus deltoidesnigra DN34) | Methylobacterium populi BJ001 | Van Aken et al. [81,82] |
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Plant-associated endophytes with potential for bioremediation identified, to date, include endophytes of poplar trees as shown in Table 1. Van Aken et al. [81,82] describe a methylotrophic endophytic bacterium isolated from hybrid poplar trees (Populus deltoides X Populus nigra DN34) that was capable of degrading the explosives TNT, RDX and HMX, mineralizing approximately 60% of the RDX and HMX to carbon dioxide in approximately two months, suggesting that these endophytes may have potential for remediation of environmental soil containing these explosive nitroaromatic compounds. Endophytes isolated from hybrid poplar trees (P. trichocarpa X P. deltoides cv. Hazendens and Hoogvorst) growing on a BTEX-contaminated site in Belgium have been shown to be capable of degrading VOCs (toluene and naphthalene) as well as a chlorinated organic herbicide (2,4-D) [75,78,79]. Porteus Moore et al. described 121 endophytic strains isolated from these hybrid poplar trees, and identified 34 of these strains as having potential to enhance phytoremediation. Germaine et al. reported that when pea (Pisum sativum) plants were inoculated with Pseudomonas endophytes, isolated from hybrid poplars (P. trichocarpa X P. deltoides cv. Hoogvorst) and capable of degrading 2,4-D, the pea plants showed no accumulation of 2,4-D in their tissues and showed little or no signs of phytotoxicity when compared to uninoculated controls suggesting that these endophytes have potential for bioremediation of environmental soil contaminated with 2,4-D.
In a recent review, Ryan et al. listed some of the advantages associated with the use of endophytic bacteria in phytoremediation of contaminated environmental soil when compared with the use of plants alone. They include (i) quantitative gene expression of bacterial pollutant catabolic genes can be used to assess the efficiency of the remediation process, (ii) genetic engineering of a bacterial catabolic pathway is easier to manipulate than a plant catabolic pathway, and (iii) toxic pollutants taken up by the plant may be degraded in planta by endophytic degraders reducing the toxic effects of contaminants in environmental soil on flora and fauna. However, some disadvantages associated with the use of bacteria in plant-associated bioremediation of contaminated environmental soil, were also given by Ryan et al. . They include (i) this technology is limited to shallow contaminants in environmental soil, (ii) it is slower than traditional remediation technologies, (iii) the choice of plant can mean that it is only seasonally effective, (iv) it is associated with phytotoxic effects of contaminants, and (v) there is potential for the environmental contaminants or their metabolites to enter the food chain if contaminants are not completely detoxified and if the plants are consumed by local fauna. More recently, Weyens et al. reviewed the benefits of using plant-associated endophytes in bioremediation and emphasized that although successfully applied in several laboratory-scale experiments, the large-scale field application of this technology is limited by a number of issues including (i) the levels of contaminants tolerated by plants, (ii) limited bioavailability of organic contaminants, and (iii) unacceptable levels of evaportranspiration of VOCs into the atmosphere. Despite the disadvantages associated with the use of plant-associated endophytic bacteria to degrade toxic organic compounds in environmental soil, it is clear that there is potential for these bacteria to make a significant contribution to sustainable bioremediation. Doty , in a recent review, claims that a major advantage of using endophytic bacteria over rhizospheric bacteria in phytoremediation is that while a rhizospheric bacterial population is difficult to control, and competition between rhizospheric bacterial strains often reduces the number of the desired strains (unless metabolism of the pollutant is selective), the use of endophytes that naturally inhabit the internal tissues of plants reduces the problem of competition between bacterial strains.
4.2. Rhizospheric Bacteria and Phytoremediation (Rhizoremediation)
Rhizoremediation is a specific form of phytoremediation involving plants and their associated rhizospheric microorganisms (bacteria and fungi). Rhizoremediation can either occur naturally or can be facilitated by inoculating soil with microorganisms capable of degrading environmental contaminants. To date, a number of toxic organic compounds in soil have been successfully remediated using rhizospheric bacteria as shown in Table 2. For example, Kuiper et al. [85,86] reported that a grass species combined with a naphthalene-degrading Pseudomonas species protected the grass seed from the toxic effects of naphthalene, and the growing roots propelled the naphthalene-degrading bacteria into soil that would have been too deep in the absence of roots.
Table 2.
Reported cases of successful bioremediation using rhizospheric bacteria (adapted from Table 1 in Liu, ).
| Compound | Plants used | Microbes used | Reference |
:---: :---: |
| PCBs | Alfalfa (Medicago sativa) Sugar beet (Beta vulgaris L.) | Pseudomonas fluorescens | Brazil et al. |
| Rockcress (Arabidopsis) | Pseudomonas putida Flav1-1 Pseudomonas putida PML2 | Narasimhan et al. |
| Switchgrass (Panicum virogatum L.) | Indigenous degraders | Chekol et al. |
| Alfalfa (Medicago sativa) Sugar beet (Beta vulgaris L.) | Pseudomonas fluorescens | Villacieros et al. |
| Pesticides | | | |
| 2,4-D | Barley (Hordeum sativum L.) | Burkholderia cepacia | Jacobsen et al. |
| | Red Clover (Trifolium pratense) Ryegrass (Lolium perenne L.) | Indigenous degraders | Shaw et al. |
| PCP | Ryegrass (Lolium perenne L.) | Indigenous degraders | He et al. |
| VOCs | | | |
| TCE | Wheat (Triticum spp.) | Pseudomonas fluorescens | Yee et al. |
| HCs | | | |
| Petroleum products | White mustard (Sinapsis alba L.) | Indigenous degraders | Liste et al. |
| Crude oil | Wheat (Triticum spp.) | Azospirillum lipoferum spp | Muratova et al. Shaw et al. |
| PAHs | Tall fescue grass (Festuca arundinacea) | Azospirillum brasilense Cd Enterobacter cloacae CAL 2 Pseudomonas putida UW3 | Huang et al. |
| Naphthalene | Barmultra grass (Lolium multiflorum) | Pseudomonas putida PCL1444 | Kuiper et al. [85,86] |
| Phenanthracene | Barley (Hordeum sativum L.) | Degrading rhizosphere colonizing Pseudomonas | Ankohina et al. |
| Chrysene | White Clover (Trifolium repens L.) | PAH tolerant Rhizobium leguminosarum | Johnson et al. |
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4.3. Enhancement of Bacterial Degradation of Toxic Organic Compounds
Using biotechnology, bacterial strains can be engineered to produce specific enzymes capable of degrading toxic organic substances. Bacteria (rhizospheric and/or endophytic) can be engineered, via natural gene transfer or recombinant DNA technology, to produce specific enzymes, capable of degrading toxic organic pollutants found in the environment. Genetic engineering of endophytic and rhizospheric bacteria for use in plant-associated degradation of toxic compounds in soil is considered one of the most promising new technologies for remediation of contaminated environmental sites.
Studies using two genetically modified strains of the rhizospheric bacteria Pseudomonas fluorescens F113, i.e., Pseudomonas fluorescens F113 rifbph (with a single chromosomal insertion of the bph operon) and Pseudomonas fluorescens F113: 1180 (with a single chromosomal insertion of the bph operon under the control of the Sinorhizobium meliloti nod regulatory system) reported that (i) the modified rhizospheric bacteria colonized roots as effectively as the wildtype rhizospheric bacteria, (ii) bph genes were expressed in situ in soil, and (iii) the modified rhizospheric bacteria could degrade PCBs more efficiently than the wildtype rhizospheric bacteria, indicating considerable potential for the manipulation of the rhizosphere as a useful strategy for bioremediation. Pseudomonas fluorescens F113: 1180 does not contain antibiotic resistance genes from the vector making this strain more suitable for in situ applications. Since the bph element in Pseudomonas fluorescens f113: 1180 is stable, lateral transfer of the bph element to a homologous recipient would not be expected to occur at detectable frequencies in the rhizosphere .
Dzantor recently reviewed the use of biotechnology to enhance rhizospheric microbial degradation of POPs. However, because toxic organic compounds can enter the root xylem from the soil before they are degraded, and these contaminants can remain in the xylem for up to two days , plant-associated endophytes genetically enhanced so as to degrade toxic organic compounds appear to offer more potential than rhizospheric bacteria for reducing phytotoxicity. Endophytic bacteria can be isolated from host plants of interest (e.g., plants native to a geographical region) and genetically enhanced to contain degradation pathways or genes to degrade target contaminants before being reinoculated back into the host plant for bioremediation purposes.
Germaine et al. reported that a genetically enhanced endophytic strain of the poplar endophyte Pseudomonas putida VM1441, i.e., Pseudomonas putida VM1441 (pNAH7), could protect inoculated pea plants from the toxic effects of naphthalene. They also showed that inoculation of plants with this strain facilitated higher (40%) naphthalene degradation rates compared with uninoculated plants in artificially contaminated soil . Barac et al. reported that a genetically enhanced endophytic strain of the soil bacterium Burkholderia cepacia G4 could increase inoculated yellow lupine plant tolerance to toluene, and decrease phytovolatilization of toluene from the plant into the atmosphere by 50-70% in laboratory scale experiments. In this study, the plasmid, pTOM, which encodes a pathway for the degradation of toluene, was transferred via conjugation to the natural endophyte, providing the genes for toluene degradation. Later, Taghavi et al. extended this work to poplar trees and showed that this degradative plasmid, pTOM, could transfer naturally, via horizontal gene transfer, to a number of different endophytes in planta, promoting more efficient degradation of toluene in poplar plants. Horizontal gene transfer results in the natural endophyte population having the capacity to degrade environmental pollutants without the need to establish the inoculants strain long-term. Endophytes that have been engineered by horizontal gene transfer, have the distinct advantage that they may not be considered to be genetically modified microorganisms (GMMs) and could, therefore, be exempt from current international and national GM legislation thus facilitating the testing of these microorganisms in the field at an accelerated pace.
In our laboratory, bacteria expressing a specific bacterial glutathione-S-transferase (GST) isolated from Burkholderia xenovorans LB400, BphK LB400 [wildtype and mutant (Ala180Pro)], capable of dehalogenating toxic chlorinated organic pesticides were shown to protect inoculated pea plants from the effects of a chlorinated organic pesticide, chloromequat chloride . Previously, it had been shown that mutating the conserved amino acid at position 180 in BphK LB400 from Ala to Pro resulted in an approximate 2-fold increase in GST activity towards a number of chlorinated organic substrates tested including commonly used pesticides [104,105]. These data suggest that BphK LB400 [wildtype and mutant (Ala180Pro)], when inserted into endophytic or rhizospheric bacteria, could have potential for bioremediation of chlorinated organic pollutants in environmental soil.
4.4. Transgenic Plants and Phytoremediation
An exciting alternative to the use of plant-associated bacteria to degrade toxic organic compounds in soil is the use of recombinant DNA technology to generate transgenic plants expressing bacterial enzymes resulting in improved plant tolerance and metabolism of toxic organic compounds in soil. However, as this topic is beyond the scope of the current review, the reader is directed to a number of recent reviews where the development of transgenic plants capable of detoxifying herbicides , organic explosives , TCE , and PCBs , using bacterial genes encoding enzymes involved in the detoxification of the target organic contaminant, is described in detail.
5. Conclusions
Much work remains to be done in carrying out field studies based on laboratory-scale experiments before commercially viable systems are available using plant-associated endophytic and rhizospheric bacteria to degrade a wide range of toxic organic compounds of concern in environmental soil. Plant-associated endophytes may offer more potential for bioremediation than plant-associated rhizospheric bacteria since: (i) the use of endophytes that are native to the host plant reduces competition between bacterial strains and may eliminate the need for reinoculation, (ii) toxic organic contaminants can remain in the plant xylem for up to two days facilitating their degradation by endophytes, and (iii) endophytes can be isolated from host plants of interest and genetically enhanced with genes encoding degradation enzymes of interest before reinoculation for bioremediation. Emphasis should be placed, when developing bioremediation systems using plant-associated bacteria, to choose wildtype bacteria, or bacteria enhanced using natural gene transfer, to avoid the complications of national and international legislation restricting and monitoring the use of GMMs. However, with a global political shift towards sustainable and green bioremediation technologies, the use of plant-associated bacteria to degrade toxic synthetic organic compounds in environmental soil may provide an efficient, economic, and sustainable green remediation technology for our twenty first century environment.
Acknowledgments
This work was supported by an Environmental Protection Agency STRIVE Research Fellowship Award to MMcG and the HEA TSR III Agribiotics Project. The authors would like to thank the members of the EnviroCore research group at the Institute of Technology Carlow for critical reading of the manuscript.
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https://www.routledge.com/Fundamentals-of-Domination-in-Graphs/Haynes-Hedetniemi-Slater/p/book/9780824700331?srsltid=AfmBOopzYZnEfCIXoYLc8UOfKwUwGPWJtDB88n6DbtjIrxqS7ntJq2kT
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1st Edition
Fundamentals of Domination in Graphs
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"Provides the first comprehensive treatment of theoretical, algorithmic, and application aspects of domination in graphs-discussing fundamental results and major research accomplishments in an easy-to-understand style. Includes chapters on domination algorithms and NP-completeness as well as frameworks for domination."
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Teresa W. Haynes, Stephen Hedetniemi, Peter Slater
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https://www.accessdata.fda.gov/drugsatfda_docs/label/2014/017892s047lbl.pdf
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HALCION® triazolam tablets, USP CIV DESCRIPTION HALCION Tablets contain triazolam, a triazolobenzodiazepine hypnotic agent. Triazolam is a white crystalline powder, soluble in alcohol and poorly soluble in water. It has a molecular weight of 343.21. The chemical name for triazolam is 8-chloro-6-(o-chlorophenyl)-1-methyl-4H-s-triazolo-[4,3-α] [1,4] benzodiazepine. The structural formula is represented below: structural formula Each HALCION Tablet, for oral administration, contains 0.25 mg of triazolam. Inactive ingredients: 0.25mg—cellulose, corn starch, docusate sodium, FD&C Blue No. 2, lactose, magnesium stearate, silicon dioxide, sodium benzoate. CLINICAL PHARMACOLOGY Triazolam is a hypnotic with a short mean plasma half-life reported to be in the range of 1.5 to 5.5 hours. In normal subjects treated for 7 days with four times the recommended dosage, there was no evidence of altered systemic bioavailability, rate of elimination, or accumulation. Peak plasma levels are reached within 2 hours following oral administration. Following recommended doses of HALCION, triazolam peak plasma levels in the range of 1 to 6 ng/mL are seen. The plasma levels achieved are proportional to the dose given. Triazolam and its metabolites, principally as conjugated glucuronides, which are presumably inactive, are excreted primarily in the urine. Only small amounts of unmetabolized triazolam appear in the urine. The two primary metabolites accounted for 79.9% of urinary excretion. Urinary excretion appeared to be biphasic in its time course. Reference ID: 3629884 1 This label may not be the latest approved by FDA. For current labeling information, please visit HALCION Tablets 0.5 mg, in two separate studies, did not affect the prothrombin times or plasma warfarin levels in male volunteers administered sodium warfarin orally. Extremely high concentrations of triazolam do not displace bilirubin bound to human serum albumin in vitro. Triazolam 14C was administered orally to pregnant mice. Drug-related material appeared uniformly distributed in the fetus with 14C concentrations approximately the same as in the brain of the mother. In sleep laboratory studies, HALCION Tablets significantly decreased sleep latency, increased the duration of sleep, and decreased the number of nocturnal awakenings. After 2 weeks of consecutive nightly administration, the drug’s effect on total wake time is decreased, and the values recorded in the last third of the night approach baseline levels. On the first and/or second night after drug discontinuance (first or second post-drug night), total time asleep, percentage of time spent sleeping, and rapidity of falling asleep frequently were significantly less than on baseline (predrug) nights. This effect is often called "rebound" insomnia. The type and duration of hypnotic effects and the profile of unwanted effects during administration of benzodiazepine drugs may be influenced by the biologic half-life of administered drug and any active metabolites formed. When half-lives are long, the drug or metabolites may accumulate during periods of nightly administration and be associated with impairments of cognitive and motor performance during waking hours; the possibility of interaction with other psychoactive drugs or alcohol will be enhanced. In contrast, if half-lives are short, the drug and metabolites will be cleared before the next dose is ingested, and carry-over effects related to excessive sedation or CNS depression should be minimal or absent. However, during nightly use for an extended period pharmacodynamic tolerance or adaptation to some effects of benzodiazepine hypnotics may develop. If the drug has a short half-life of elimination, it is possible that a relative deficiency of the drug or its active metabolites (ie, in relationship to the receptor site) may occur at some point in the interval between each night’s use. This sequence of events may account for two clinical findings reported to occur after several weeks of nightly use of rapidly eliminated benzodiazepine hypnotics: 1) increased wakefulness during the last third of the night and 2) the appearance of increased daytime anxiety after 10 days of continuous treatment. In a study of elderly (62–83 years old) versus younger subjects (21–41 years old) who received HALCION at the same dose levels (0.125 mg and 0.25 mg), the elderly experienced both greater sedation and impairment of psychomotor performance. These effects resulted largely from higher plasma concentrations of triazolam in the elderly. INDICATIONS AND USAGE HALCION is indicated for the short-term treatment of insomnia (generally 7–10 days). Use for more than 2– 3 weeks requires complete reevaluation of the patient (see WARNINGS). Prescriptions for HALCION should be written for short-term use (7–10 days) and it should not be prescribed in quantities exceeding a 1-month supply. CONTRAINDICATIONS HALCION Tablets are contraindicated in patients with known hypersensitivity to this drug or other benzodiazepines. Benzodiazepines may cause fetal damage when administered during pregnancy. An increased risk of congenital malformations associated with the use of diazepam and chlordiazepoxide during the first trimester of pregnancy has been suggested in several studies. Transplacental distribution has resulted in neonatal CNS Reference ID: 3629884 2 This label may not be the latest approved by FDA. For current labeling information, please visit depression following the ingestion of therapeutic doses of a benzodiazepine hypnotic during the last weeks of pregnancy. HALCION is contraindicated in pregnant women. If there is a likelihood of the patient becoming pregnant while receiving HALCION, she should be warned of the potential risk to the fetus. Patients should be instructed to discontinue the drug prior to becoming pregnant. The possibility that a woman of childbearing potential may be pregnant at the time of institution of therapy should be considered. HALCION is contraindicated withmedications that significantly impair the oxidative metabolism mediated by cytochrome P450 3A (CYP 3A) including ketoconazole, itraconazole, nefazodone, and several HIV protease inhibitors, (see WARNINGS and PRECAUTIONS–Drug Interactions). WARNINGS Because sleep disturbances may be the presenting manifestation of a physical and/or psychiatric disorder, symptomatic treatment of insomnia should be initiated only after a careful evaluation of the patient. The failure of insomnia to remit after 7 to 10 days of treatment may indicate the presence of a primary psychiatric and/or medical illness that should be evaluated. Worsening of insomnia or the emergence of new thinking or behavior abnormalities may be the consequence of an unrecognized psychiatric or physical disorder. Such findings have emerged during the course of treatment with sedative-hypnotic drugs. Because some of the important adverse effects of sedative-hypnotics appear to be dose related (see Precautions and Dosage and Administration), it is important to use the smallest possible effective dose, especially in the elderly. Complex behaviors such as “sleep-driving” (i.e., driving while not fully awake after ingestion of a sedative-hypnotic, with amnesia for the event) have been reported. These events can occur in sedative-hypnotic-naïve as well as in sedative-hypnotic-experienced persons. Although behaviors such as sleep-driving may occur with sedative-hypnotics alone at therapeutic doses, the use of alcohol and other CNS depressants with sedative-hypnotics appears to increase the risk of such behaviors, as does the use of sedative-hypnotics at doses exceeding the maximum recommended dose. Due to the risk to the patient and the community, discontinuation of sedative-hypnotics should be strongly considered for patients who report a “sleep-driving” episode. Other complex behaviors (e.g., preparing and eating food, making phone calls, or having sex) have been reported in patients who are not fully awake after taking a sedative-hypnotic. As with sleep-driving, patients usually do not remember these events. Severe anaphylactic and anaphylactoid reactions Rare cases of angioedema involving the tongue, glottis or larynx have been reported in patients after taking the first or subsequent doses of sedative-hypnotics, including HALCION. Some patients have had additional symptoms such as dyspnea, throat closing, or nausea and vomiting that suggest anaphylaxis. Some patients have required medical therapy in the emergency department. If angioedema involves the tongue, glottis or larynx, airway obstruction may occur and be fatal. Patients who develop angioedema after treatment with HALCION should not be rechallenged with the drug. Reference ID: 3629884 3 This label may not be the latest approved by FDA. For current labeling information, please visit Central nervous system manifestations An increase in daytime anxiety has been reported for HALCION after as few as 10 days of continuous use. In some patients this may be a manifestation of interdose withdrawal (see CLINICAL PHARMACOLOGY). If increased daytime anxiety is observed during treatment, discontinuation of treatment may be advisable. A variety of abnormal thinking and behavior changes have been reported to occur in association with the use of benzodiazepine hypnotics including HALCION. Some of these changes may be characterized by decreased inhibition, eg, aggressiveness and extroversion that seem excessive, similar to that seen with alcohol and other CNS depressants (eg, sedative/hypnotics). Other kinds of behavioral changes have also been reported, for example, bizarre behavior, agitation, hallucinations, depersonalization. In primarily depressed patients, the worsening of depression, including suicidal thinking, has been reported in association with the use of benzodiazepines. It can rarely be determined with certainty whether a particular instance of the abnormal behaviors listed above is drug induced, spontaneous in origin, or a result of an underlying psychiatric or physical disorder. Nonetheless, the emergence of any new behavioral sign or symptom of concern requires careful and immediate evaluation. Because of its depressant CNS effects, patients receiving triazolam should be cautioned against engaging in hazardous occupations requiring complete mental alertness such as operating machinery or driving a motor vehicle. For the same reason, patients should be cautioned about the concomitant ingestion of alcohol and other CNS depressant drugs during treatment with HALCION Tablets. As with some, but not all benzodiazepines, anterograde amnesia of varying severity and paradoxical reactions have been reported following therapeutic doses of HALCION. Data from several sources suggest that anterograde amnesia may occur at a higher rate with HALCION than with other benzodiazepine hypnotics. Triazolam interaction with drugs that inhibit metabolism via cytochrome P450 3A: The initial step in triazolam metabolism is hydroxylation catalyzed by cytochrome P450 3A (CYP 3A). Drugs that inhibit this metabolic pathway may have a profound effect on the clearance of triazolam. Consequently, triazolam should be avoided in patients receiving very potent inhibitors of CYP 3A. With drugs inhibiting CYP 3A to a lesser but still significant degree, triazolam should be used only with caution and consideration of appropriate dosage reduction. For some drugs, an interaction with triazolam has been quantified with clinical data; for other drugs, interactions are predicted from in vitro data and/or experience with similar drugs in the same pharmacologic class. The following are examples of drugs known to inhibit the metabolism of triazolam and/or related benzodiazepines, presumably through inhibition of CYP 3A. Potent CYP 3A inhibitors: Potent inhibitors of CYP 3A that should not be used concomitantly with triazolam include ketoconazole, itraconazole, nefazodone and several HIV protease inhibitors including ritonavir, indinavir, nelfinavir, saquinavir and lopinavir. Although data concerning the effects of azole-type antifungal agents other than ketoconazole and itraconazole on triazolam metabolism are not available, they should be considered potent CYP 3A inhibitors, and their coadministration with triazolam is not recommended (see CONTRAINDICATIONS). Drugs demonstrated to be CYP 3A inhibitors on the basis of clinical studies involving triazolam (caution and consideration of dose reduction are recommended during coadministration with triazolam): Reference ID: 3629884 4 This label may not be the latest approved by FDA. For current labeling information, please visit Macrolide Antibiotics—Coadministration of erythromycin increased the maximum plasma concentration of triazolam by 46%, decreased clearance by 53%, and increased half-life by 35%; caution and consideration of appropriate triazolam dose reduction are recommended. Similar caution should be observed during coadministration with clarithromycin and other macrolide antibiotics. Cimetidine—Coadministration of cimetidine increased the maximum plasma concentration of triazolam by 51%, decreased clearance by 55%, and increased half-life by 68%; caution and consideration of appropriate triazolam dose reduction are recommended. Other drugs possibly affecting triazolam metabolism: Other drugs possibly affecting triazolam metabolism by inhibition of CYP 3A are discussed in the PRECAUTIONS section (see PRECAUTIONS–Drug Interactions). PRECAUTIONS General: In elderly and/or debilitated patients it is recommended that treatment with HALCION Tablets be initiated at 0.125 mg to decrease the possibility of development of oversedation, dizziness, or impaired coordination. Some side effects reported in association with the use of HALCION appear to be dose related. These include drowsiness, dizziness, light-headedness, and amnesia. The relationship between dose and what may be more serious behavioral phenomena is less certain. Specifically, some evidence, based on spontaneous marketing reports, suggests that confusion, bizarre or abnormal behavior, agitation, and hallucinations may also be dose related, but this evidence is inconclusive. In accordance with good medical practice it is recommended that therapy be initiated at the lowest effective dose (see DOSAGE AND ADMINISTRATION). Cases of "traveler’s amnesia" have been reported by individuals who have taken HALCION to induce sleep while traveling, such as during an airplane flight. In some of these cases, insufficient time was allowed for the sleep period prior to awakening and before beginning activity. Also, the concomitant use of alcohol may have been a factor in some cases. Caution should be exercised if HALCION is prescribed to patients with signs or symptoms of depression that could be intensified by hypnotic drugs. Suicidal tendencies may be present in such patients and protective measures may be required. Intentional over-dosage is more common in these patients, and the least amount of drug that is feasible should be available to the patient at any one time. The usual precautions should be observed in patients with impaired renal or hepatic function, chronic pulmonary insufficiency, and sleep apnea. In patients with compromised respiratory function, respiratory depression and apnea have been reported infrequently. Information for patients: The text of a Medication Guide for patients is included at the end of this insert. To assure safe and effective use of HALCION, the information and instructions provided in this Medication Guide should be discussed with patients. “Sleep-driving” and other complex behaviors: There have been reports of people getting out of bed after taking a sedative-hypnotic and driving their cars while not fully awake, often with no memory of the event. If a patient experiences such an episode, it should be reported to his or her doctor immediately, since “sleep-driving” can be dangerous. This behavior is more Reference ID: 3629884 5 This label may not be the latest approved by FDA. For current labeling information, please visit likely to occur when sedative-hypnotics are taken with alcohol or other central nervous system depressants (see WARNINGS). Other complex behaviors (e.g., preparing and eating food, making phone calls, or having sex) have been reported in patients who are not fully awake after taking a sedative hypnotic. As with sleep-driving, patients usually do not remember these events. Laboratory tests: Laboratory tests are not ordinarily required in otherwise healthy patients. Drug interactions: Both pharmacodynamic and pharmacokinetic interactions have been reported with benzodiazepines. In particular, triazolam produces additive CNS depressant effects when coadministered with other psychotropic medications, anticonvulsants, antihistamines, ethanol, and other drugs which themselves produce CNS depression. Drugs that inhibit triazolam metabolism via cytochrome P450 3A: The initial step in triazolam metabolism is hydroxylation catalyzed by cytochrome P450 3A (CYP 3A). Drugs which inhibit this metabolic pathway may have a profound effect on the clearance of triazolam (see CONTRAINDICATIONS and WARNINGS for additional drugs of this type). HALCION is contraindicated with ketoconzaole, itraconazole, nefazodone, and several HIV protease inhibitors. Drugs and other substances demonstrated to be CYP 3A inhibitors of possible clinical significance on the basis of clinical studies involving triazolam (caution is recommended during coadministration with triazolam): Isoniazid—Coadministration of isoniazid increased the maximum plasma concentration of triazolam by 20%, decreased clearance by 42%, and increased half-life by 31%. Oral contraceptives—Coadministration of oral contraceptives increased maximum plasma concentration by 6%, decreased clearance by 32%, and increased half-life by 16%. Grapefruit juice—Coadministration of grapefruit juice increased the maximum plasma concentration of triazolam by 25%, increased the area under the concentration curve by 48%, and increased half-life by 18%. Drugs demonstrated to be CYP 3A inhibitors on the basis of clinical studies involving benzodiazepines metabolized similarly to triazolam or on the basis of in vitro studies with triazolam or other benzodiazepines (caution is recommended during coadministration with triazolam): Available data from clinical studies of benzodiazepines other than triazolam suggest a possible drug interaction with triazolam for the following: fluvoxamine, diltiazem, and verapamil. Data from in vitro studies of triazolam suggest a possible drug interaction with triazolam for the following: sertraline and paroxetine. Data from in vitro studies of benzodiazepines other than triazolam suggest a possible drug interaction with triazolam for the following: ergotamine, cyclosporine, amiodarone, nicardipine, and nifedipine. Caution is recommended during coadministration of any of these drugs with triazolam (see WARNINGS). Drugs that affect triazolam pharmacokinetics by other mechanisms: Ranitidine—Coadministration of ranitidine increased the maximum plasma concentration of triazolam by 30%, increased the area under the concentration curve by 27%, and increased half-life by 3.3%. Caution is recommended during coadministration with triazolam. Carcinogenesis, mutagenesis, impairment of fertility: No evidence of carcinogenic potential was observed in mice during a 24-month study with HALCION in doses up to 4,000 times the human dose. Pregnancy: Reference ID: 3629884 6 This label may not be the latest approved by FDA. For current labeling information, please visit 1. Teratogenic effects: Pregnancy category X (see CONTRAINDICATIONS). 2. Non-teratogenic effects: It is to be considered that the child born of a mother who is on benzodiazepines may be at some risk for withdrawal symptoms from the drug, during the postnatal period. Also, neonatal flaccidity has been reported in an infant born of a mother who had been receiving benzodiazepines. Nursing mothers: Human studies have not been performed; however, studies in rats have indicated that HALCION and its metabolites are secreted in milk. Therefore, administration of HALCION to nursing mothers is not recommended. Pediatric use: Safety and effectiveness of HALCION in individuals below 18 years of age have not been established. Geriatric use: The elderly are especially susceptible to the dose related adverse effects of HALCION. They exhibit higher plasma triazolam concentrations due to reduced clearance of the drug as compared with younger subjects at the same dose. To minimize the possibility of development of oversedation, the smallest effective dose should be used (see CLINICAL PHARMACOLOGY, WARNINGS, PRECAUTIONS, and DOSAGE AND ADMINISTRATION). Tolerance/Withdrawal Phenomena Some loss of effectiveness or adaptation to the sleep inducing effects of these medications may develop after nightly use for more than a few weeks and there may be a degree of dependence that develops. For the benzodiazepine sleeping pills that are eliminated quickly from the body, a relative deficiency of the drug may occur at some point in the interval between each night’s use. This can lead to (1) increased wakefulness during the last third of the night, and (2) the appearance of increased signs of daytime anxiety or nervousness. These two events have been reported in particular for HALCION. There can be more severe ‘withdrawal’ effects when a benzodiazepine sleeping pill is stopped. Such effects can occur after discontinuing these drugs following use for only a week or two, but may be more common and more severe after longer periods of continuous use. One type of withdrawal phenomenon is the occurrence of what is known as ‘rebound insomnia’. That is, on the first few nights after the drug is stopped, insomnia is actually worse than before the sleeping pill was given. Other withdrawal phenomena following abrupt stopping of benzodiazepine sleeping pills range from mild unpleasant feelings to a major withdrawal syndrome which may include abdominal and muscle cramps, vomiting, sweating, tremor, and rarely, convulsions. ADVERSE REACTIONS During placebo-controlled clinical studies in which 1,003 patients received HALCION Tablets, the most troublesome side effects were extensions of the pharmacologic activity of triazolam, eg, drowsiness, dizziness, or light-headedness. The figures cited below are estimates of untoward clinical event incidence among subjects who participated in the relatively short duration (i.e., 1 to 42 days) placebo-controlled clinical trials of HALCION. The figures cannot be used to predict precisely the incidence of untoward events in the course of usual medical practice where patient characteristics and other factors often differ from those in clinical trials. These figures cannot be compared with those obtained from other clinical studies involving related drug products and placebo, as each group of drug trials is conducted under a different set of conditions. Comparison of the cited figures, however, can provide the prescriber with some basis for estimating the relative contributions of drug and nondrug factors to the untoward event incidence rate in the population Reference ID: 3629884 7 This label may not be the latest approved by FDA. For current labeling information, please visit studied. Even this use must be approached cautiously, as a drug may relieve a symptom in one patient while inducing it in others. (For example, an anticholinergic, anxiolytic drug may relieve dry mouth [a sign of anxiety] in some subjects but induce it [an untoward event] in others.) HALCION PLACEBO Number of Patients 1003 997 % Patients Reporting: Central Nervous System Drowsiness 14.0 6.4 Headache 9.7 8.4 Dizziness 7.8 3.1 Nervousness 5.2 4.5 Light-headedness 4.9 0.9 Coordination disorders/ataxia 4.6 0.8 Gastrointestinal Nausea/vomiting 4.6 3.7 In addition to the relatively common (i.e., 1% or greater) untoward events enumerated above, the following adverse events have been reported less frequently (i.e., 0.9% to0.5%): euphoria, tachycardia, tiredness, confusional states/memory impairment, cramps/pain, depression, visual disturbances. Rare (i.e., less than 0.5%) adverse reactions included constipation, taste alterations, diarrhea, dry mouth, dermatitis/allergy, dreaming/nightmares, insomnia, paresthesia, tinnitus, dysesthesia, weakness, congestion, death from hepatic failure in a patient also receiving diuretic drugs. In addition to these untoward events for which estimates of incidence are available, the following adverse events have been reported in association with the use of HALCION and other benzodiazepines: amnestic symptoms (anterograde amnesia with appropriate or inappropriate behavior), confusional states (disorientation, derealization, depersonalization, and/or clouding of consciousness), dystonia, anorexia, fatigue, sedation, slurred speech, jaundice, pruritus, dysarthria, changes in libido, menstrual irregularities, incontinence, and urinary retention. Other factors may contribute to some of these reactions, eg, concomitant intake of alcohol or other drugs, sleep deprivation, an abnormal premorbid state, etc. Other events reported include: paradoxical reactions such as stimulation, mania, an agitational state (restlessness, irritability, and excitation), increased muscle spasticity, sleep disturbances, hallucinations, delusions, aggressiveness, falling, somnambulism, syncope, inappropriate behavior and other adverse behavioral effects. Should these occur, use of the drug should be discontinued. The following events have also been reported: chest pain, burning tongue/glossitis/stomatitis. Laboratory analyses were performed on all patients participating in the clinical program for HALCION. The following incidences of abnormalities were observed in patients receiving HALCION and the corresponding placebo group. None of these changes were considered to be of physiological significance. HALCION PLACEBO Number of Patients 380 361 % of Patients Reporting: Low High Low High Hematology Hematocrit Hemoglobin Reference ID: 3629884 8 This label may not be the latest approved by FDA. For current labeling information, please visit Total WBC count 1.7 2.1 1.3 Neutrophil count 1.5 1.5 3.3 1.0 Lymphocyte count 2.3 4.0 3.1 3.8 Monocyte count 3.6 4.4 1.5 Eosinophil count 10.2 3.2 9.8 3.4 Basophil count 1.7 2.1 1.8 Urinalysis Albumi — 1.1 — Sugar — — RBC/HPF — 2.9 — 2.9 WBC/HPF — 11.7 — 7.9 Blood chemistry Creatinine 2.4 1.9 3.6 1.5 Bilirubin 1.5 1.0 SGOT 5.3 4.5 Alkaline phosphatase 2.2 2.6 Less than 1% When treatment with HALCION is protracted, periodic blood counts, urinalysis, and blood chemistry analyses are advisable. Minor changes in EEG patterns, usually low-voltage fast activity, have been observed in patients during therapy with HALCION and are of no known significance. DRUG ABUSE AND DEPENDENCE Abuse and addiction are separate and distinct from physical dependence and tolerance. Abuse is characterized by misuse of the drug for non-medical purposes, often in combination with other psychoactive substances. Physical dependence is a state of adaptation that is manifested by a specific withdrawal syndrome that can be produced by abrupt cessation, rapid dose reduction, decreasing blood level of the drug and/or administration of an antagonist. Tolerance is a state of adaptation in which exposure to a drug induces changes that result in a diminution of one or more of the drug’s effects over time. Tolerance may occur to both the desired and undesired effects of drugs and may develop at different rates for different effects. Addiction is a primary, chronic, neurobiological disease with genetic, psychosocial, and environmental factors influencing its development and manifestations. It is characterized by behaviors that include one or more of the following: impaired control over drug use, compulsive use, continued use despite harm, and craving. Drug addiction is a treatable disease, utilizing a multidisciplinary approach, but relapse is common. Controlled Substance: Triazolam is a controlled substance under the Controlled Substance Act, and HALCION Tablets have been assigned to Schedule IV. Abuse, Dependence and Withdrawal: Withdrawal symptoms, similar in character to those noted with barbiturates and alcohol (convulsions, tremor, abdominal and muscle cramps, vomiting, sweating, dysphoria, perceptual disturbances and insomnia), have occurred following abrupt discontinuance of benzodiazepines, including HALCION. The more severe symptoms are usually associated with higher dosages and longer usage, although patients at therapeutic dosages given for as few as 1–2 weeks can also have withdrawal symptoms and in some patients there may be withdrawal symptoms (daytime anxiety, agitation) between nightly doses (see CLINICAL PHARMACOLOGY). Consequently, abrupt discontinuation should be Reference ID: 3629884 9 This label may not be the latest approved by FDA. For current labeling information, please visit avoided and a gradual dosage tapering schedule is recommended in any patient taking more than the lowest dose for more than a few weeks. The recommendation for tapering is particularly important in any patient with a history of seizure. The risk of dependence is increased in patients with a history of alcoholism, drug abuse, or in patients with marked personality disorders. Such dependence-prone individuals should be under careful surveillance when receiving HALCION. As with all hypnotics, repeat prescriptions should be limited to those who are under medical supervision. OVERDOSAGE Because of the potency of triazolam, some manifestations of overdosage may occur at 2 mg, four times the maximum recommended therapeutic dose (0.5 mg). Manifestations of overdosage with HALCION Tablets include somnolence, confusion, impaired coordination, slurred speech, and ultimately, coma. Respiratory depression and apnea have been reported with overdosages of HALCION. Seizures have occasionally been reported after overdosages. Death has been reported in association with overdoses of triazolam by itself, as it has with other benzodiazepines. In addition, fatalities have been reported in patients who have overdosed with a combination of a single benzodiazepine, including triazolam, and alcohol; benzodiazepine and alcohol levels seen in some of these cases have been lower than those usually associated with reports of fatality with either substance alone. As in all cases of drug overdosage, respiration, pulse, and blood pressure should be monitored and supported by general measures when necessary. Immediate gastric lavage should be performed. An adequate airway should be maintained. Intravenous fluids may be administered. Flumazenil, a specific benzodiazepine receptor antagonist, is indicated for the complete or partial reversal of the sedative effects of benzodiazepines and may be used in situations when an overdose with a benzodiazepine is known or suspected. Prior to the administration of flumazenil, necessary measures should be instituted to secure airway, ventilation and intravenous access. Flumazenil is intended as an adjunct to, not as a substitute for, proper management of benzodiazepine overdose. Patients treated with flumazenil should be monitored for resedation, respiratory depression, and other residual benzodiazepine effects for an appropriate period after treatment. The prescriber should be aware of a risk of seizure in association with flumazenil treatment, particularly in long-term benzodiazepine users and in cyclic antidepressant overdose. The complete flumazenil package insert including CONTRAINDICATIONS, WARNINGS and PRECAUTIONS should be consulted prior to use. Experiments in animals have indicated that cardiopulmonary collapse can occur with massive intravenous doses of triazolam. This could be reversed with positive mechanical respiration and the intravenous infusion of norepinephrine bitartrate or metaraminol bitartrate. Hemodialysis and forced diuresis are probably of little value. As with the management of intentional overdosage with any drug, the physician should bear in mind that multiple agents may have been ingested by the patient. The oral LD50 in mice is greater than 1,000 mg/kg and in rats is greater than 5,000 mg/kg. DOSAGE AND ADMINISTRATION It is important to individualize the dosage of HALCION Tablets for maximum beneficial effect and to help avoid significant adverse effects. Reference ID: 3629884 10 This label may not be the latest approved by FDA. For current labeling information, please visit The recommended dose for most adults is 0.25 mg before retiring. A dose of 0.125 mg may be found to be sufficient for some patients (e.g., low body weight). A dose of 0.5 mg should be used only for exceptional patients who do not respond adequately to a trial of a lower dose since the risk of several adverse reactions increases with the size of the dose administered. A dose of 0.5 mg should not be exceeded. In geriatric and/or debilitated patients the recommended dosage range is 0.125 mg to 0.25 mg. Therapy should be initiated at 0.125 mg in these groups and the 0.25 mg dose should be used only for exceptional patients who do not respond to a trial of the lower dose. A dose of 0.25 mg should not be exceeded in these patients. As with all medications, the lowest effective dose should be used. HOW SUPPLIED HALCION Tablets are available in the following strengths and package sizes: 0.25 mg (powder blue, elliptical, scored, imprinted HALCION 0.25): Reverse numbered Unit Dose (100) NDC 0009-0017-55 10–10 Tablet Bottles NDC 0009-0017-59 Bottles of 10 NDC 0009-0017-58 Bottles of 500 NDC 0009-0017-02 Store at controlled room temperature 20° to 25°C (68° to 77°F) [see USP]. company logo LAB-0666-5.0 Revised September 2014 Reference ID: 3629884 11 This label may not be the latest approved by FDA. For current labeling information, please visit MEDICATION GUIDE HALCION Tablets/ C-IV Read this Medication Guide before you start taking HALCION and each time you get a refill. There may be new information. This Medication Guide does not take the place of talking to your doctor about your medical condition or treatment. You and your doctor should talk about the SEDATIVE-HYPNOTIC when you start taking it and at regular checkups. What is the most important information I should know about HALCION? After taking a SEDATIVE-HYPNOTIC, you may get up out of bed while not being fully awake and do an activity that you do not know you are doing. The next morning, you may not remember that you did anything during the night. You have a higher chance for doing these activities if you drink alcohol or take other medicines that make you sleepy with a SEDATIVE-HYPNOTIC. Reported activities include: driving a car ("sleep-driving") making and eating food talking on the phone having sex sleep-walking Important: 1. Take HALCION exactly as prescribed Do not take more HALCION than prescribed. Take HALCION right before you get in bed, not sooner. 2. Do not take HALCION if you: drink alcohol take other medicines that can make you sleepy. Talk to your doctor about all of your medicines. Your doctor will tell you if you can take HALCION with your other medicines cannot get a full night’s sleep are pregnant or considering becoming pregnant 3. Call your doctor right away if you find out that you have done any of the above activities after taking HALCION. What are SEDATIVE-HYPNOTICS? SEDATIVE-HYPNOTICs are sleep medicines. SEDATIVE-HYPNOTICS are used in adults for the treatment of the symptom of trouble falling asleep due to insomnia. HALCION is not indicated for use in children. Elderly patients are especially susceptible to dose related adverse effects when taking HALCION. HALCION is a federally controlled substance (C-IV) because it can be abused or lead to dependence. Keep HALCION in a safe place to prevent misuse and abuse. Selling or giving away HALCION may harm others, and is against the law. Tell your doctor if you have ever abused or been dependent on alcohol, prescription medicines or street drugs. Who should not take HALCION? Do not take HALCION if you are allergic to anything in it. See the end of this Medication Guide for a complete list of ingredients in HALCION. SEDATIVE-HYPNOTICS may not be right for you. Before starting SEDATIVE-HYPNOTICS, tell your doctor about all of your health conditions, including if you: have a history of depression, mental illness, or suicidal thoughts have a history of drug or alcohol abuse or addiction have kidney or liver disease have a lung disease or breathing problems are pregnant, planning to become pregnant, or breastfeeding Tell your doctor about all of the medicines you take including prescription and nonprescription medicines, vitamins and herbal supplements. Medicines can interact, sometimes causing side effects. Do not take SEDATIVE-HYPNOTICS with other medicines that can make you sleepy. Know the medicines you take. Keep a list of your medicines with you to show your doctor and pharmacist each time you get a new medicine. HALCION should not be taken with some drugs including ketoconazole, itraconazole, nefazodone, ritonavir, indinavir, nelfinavir, saquinavir or lopinavir. How should I take HALCION? Take HALCION exactly as prescribed. Do not take more HALCION than prescribed for you. Take HALCION right before you get into bed. Or you can take the HALCION after you have been in bed and have trouble falling asleep. Reference ID: 3629884 12 This label may not be the latest approved by FDA. For current labeling information, please visit Do not take HALCION with or right after a meal. Do not take HALCION unless you are able to get a full night’s sleep before you must be active again. Call your healthcare provider if your insomnia worsens or is not better within 7 to 10 days. This may mean that there is another condition causing your sleep problem. If you take too much HALCION or overdose, call your doctor or poison control center right away, or get emergency treatment. What are the possible side effects of SEDATIVE HYPNOTICS? Serious side effects of SEDATIVE-HYPNOTICS include: getting out of bed while not being fully awake and doing an activity that you do not know you are doing. (See “What is the most important information I should know about SEDATIVE HYPNOTICS?) abnormal thoughts and behavior. Symptoms include more outgoing or aggressive behavior than normal, confusion, agitation, hallucinations, worsening of depression, and suicidal thoughts or actions. memory loss, including “traveler’s amnesia” anxiety severe allergic reactions. Symptoms include swelling of the tongue or throat, trouble breathing, and nausea and vomiting. Get emergency medical help if you get these symptoms after taking SEDATIVE-HYPNOTICS. Call your doctor right away if you have any of the above side effects or any other side effects that worry you while using the SEDATIVE-HYPNOTIC. Common side effects of HALCION include: drowsiness headache dizziness lightheadedness “pins and needles” feelings on your skin difficulty with coordination You may still feel drowsy the next day after taking HALCION. Do not drive or do other dangerous activities (including operating machinery) after taking HALCION until you feel fully awake. You may have withdrawal symptoms for 1 to 2 days when you stop taking the SEDATIVE HYPNOTIC suddenly. Withdrawal symptoms include trouble sleeping, unpleasant feelings, stomach and muscle cramps, vomiting, sweating, shakiness, and seizures. These are not all the side effects of SEDATIVE HYPNOTICS. Ask your doctor or pharmacist for more information. Call your doctor for medical advice about side effects. You may report side effects to FDA at 1-800-FDA 1088. How should I store HALCION? Store HALCION at room temperature between 68° and 77° F (20º to 25ºC). Protect from light. Keep HALCION and all medicines out of the reach of children. Do not use HALCION after the expiration date on the bottle. General Information about SEDATIVE-HYPNOTICS Medicines are sometimes prescribed for purposes not mentioned in a Medication Guide. Do not use the SEDATIVE-HYPNOTIC for a condition for which it was not prescribed. Do not give the SEDATIVE-HYPNOTIC to other people, even if they have the same condition. It may harm them and it is against the law. This Medication Guide summarizes the most important information about HALCION. If you would like more information, talk with your doctor. You can ask your doctor or pharmacist for information about HALCION that was written for healthcare professionals. If you would like more information, contact 1-800-879 3477 What are the ingredients in HALCION? Active Ingredient: Triazolam Inactive Ingredients: 0.25 mg tablet: cellulose, corn starch, docusate sodium, FD&C Blue No. 2, lactose, magnesium stearate, silicon dioxide, sodium benzoate. Reference ID: 3629884 13 This label may not be the latest approved by FDA. For current labeling information, please visit Rx only This Medication Guide has been approved by the U.S. Food and Drug Administration. September 2014 This product’s label may have been updated. For current full prescribing information, please visit www.pfizer.com. company logo LAB-0259-13.0 Revised September 2014 Reference ID: 3629884 14 This label may not be the latest approved by FDA. For current labeling information, please visit
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Clique graph
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From Wikipedia, the free encyclopedia
Graph representing structure of another graph's cliques
This article is about a graph whose vertices correspond to maximal cliques in another graph. For other uses, see Clique graph (disambiguation).
In graph theory, a clique graph of an undirected graph G is another graph K(G) that represents the structure of cliques in G.
Clique graphs were discussed at least as early as 1968, and a characterization of clique graphs was given in 1971.
Formal definition
[edit]
A clique of a graph G is a set X of vertices of G with the property that every pair of distinct vertices in X are adjacent in G.
A maximal clique of a graph G is a clique X of vertices of G, such that there is no clique Y of vertices of G that contains all of X and at least one other vertex.
Given a graph G, its clique graph K(G) is a graph such that
every vertex of K(G) represents a maximal clique of G; and
two vertices of K(G) are adjacent when the underlying cliques in G share at least one vertex in common.
That is, the clique graph K(G) is the intersection graph of the maximal cliques of G.
Characterization
[edit]
A graph H is the clique graph K(G) of another graph if and only if there exists a collection C of cliques in H whose union covers all the edges of H, such that C forms a Helly family. This means that, if S is a subset of C with the property that every two members of S have a non-empty intersection, then S itself should also have a non-empty intersection. However, the cliques in C do not necessarily have to be maximal cliques.
When H =K(G), a family C of this type may be constructed in which each clique in C corresponds to a vertex v in G, and consists of the cliques in G that contain v. These cliques all have v in their intersection, so they form a clique in H. The family C constructed in this way has the Helly property, because any subfamily of C with pairwise nonempty intersections must correspond to a clique in G, which can be extended to a maximal clique that belongs to the intersection of the subfamily.
Conversely, when H has a Helly family C of its cliques, covering all edges of H, then it is the clique graph K(G) for a graph G whose vertices are the disjoint union of the vertices of H and the elements of C. This graph G has an edge for each pair of cliques in C with nonempty intersection, and for each pair of a vertex of H and a clique in C that contains it. However, it does not contain any edges connecting pairs of vertices in H. The maximal cliques in this graph G each consist of one vertex of H together with all the cliques in C that contain it, and their intersection graph is isomorphic to H.
However, this characterization does not lead to efficient algorithms: the problem of recognizing whether a given graph is a clique graph is NP-complete.
References
[edit]
^ Hamelink, Ronald C. (1968). "A partial characterization of clique graphs". Journal of Combinatorial Theory. 5 (2): 192–197. doi:10.1016/S0021-9800(68)80055-9.
^ Jump up to: a b c d Roberts, Fred S.; Spencer, Joel H. (1971). "A characterization of clique graphs". Journal of Combinatorial Theory. Series B. 10 (2): 102–108. doi:10.1016/0095-8956(71)90070-0.
^ Szwarcfiter, Jayme L.; Bornstein, Claudson F. (1994). "Clique graphs of chordal and path graphs". SIAM Journal on Discrete Mathematics. 7 (2): 331–336. CiteSeerX 10.1.1.52.521. doi:10.1137/S0895480191223191.
^ Alcón, Liliana; Faria, Luerbio; de Figueiredo, Celina M. H.; Gutierrez, Marisa (2009). "The complexity of clique graph recognition". Theoretical Computer Science. 410 (21–23): 2072–2083. doi:10.1016/j.tcs.2009.01.018. MR 2519298.
External links
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Information System on Graph Classes and their Inclusions: clique graph
Retrieved from "
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Remind familial hypocalciuric hypercalcemia as a cause of elevated parathyroid hormone and serum calcium: 5 patients | ECE2011 | 13th European Congress of Endocrinology | Endocrine Abstracts
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ISSN 1470-3947 (print) | ISSN 1479-6848 (online)
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Endocrine Abstracts (2011) 26 P611
ECE2011Poster PresentationsClinical case reports (73 abstracts)
Remind familial hypocalciuric hypercalcemia as a cause of elevated parathyroid hormone and serum calcium: 5 patients
I E Bonapart, A Boumpaki&A Stamataki
Remind familial hypocalciuric hypercalcemia as a cause of elevated parathyroid hormone and serum calcium: 5 patients
Author affiliations
Venizeleio-Pananeio Hospital, Heraklion, Crete, Greece.
Introduction: Familial hypocalciuric hypercalcemia (FHH) is a benign disorder, with inappropriately elevated parathyroid hormone, without the complications of hypercalcemia (1). It is important to discriminate FHH from primary hyperparathyroidism (PHPT), because of the consequences on symptomatic disease and therapeutic approach. Ca/Cr clearance ratio will be <0.01 in FHH and most often >0.02 in PHPT (2). In patients with PHPT and vitamin D deficiency, low urinary calcium will increase after vitamin D repletion, thereby differentiating it from FHH.
Case reports: Five female patients, mean age 54.4 years, without specific clinical symptoms had elevated PTH (mean value 89.2 pg/ml), elevated or at the upper limit of normal serum calcium (10.3 mg/dl) and low urine calcium (mean value 80.5 mg/24 h). Their Ca/Cr clearance ratio was <0.01 (mean value 0.006). Vitamin D 25 OH deficiency coexisted in 2 patients was supplied without substantial increase in calcium urine values. The lack of symptoms and Ca/Cr clearance ratio <0.01 confirmed the diagnosis of FHH. No further evaluations and treatment was necessary.
Conclusion: In clinical practice, frequently we see elevated levels of PTH and serum calcium, which can lead to unnecessary evaluation and unsuccessful treatment if not FHH is considered.
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Remind familial hypocalciuric hypercalcemia as a cause of elevated parathyroid hormone and serum calcium: 5 patients (<1 min ago)
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Bonapart I E
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Selection of Terms in an Arithmetic Progression
Sometimes we need to assume certain number of terms in Arithmetic Progression. The following ways are generally used for the selection of terms in an arithmetic progression.
(i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a - d, a and a + d. Here common difference is d.
(ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a - 3d, a - d, a + d and a + 3d.
(iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a - 2d, a - d, a, a + d and a + 2d. Here common difference is 2d.
(iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a - 5d, a - 3d, a - d, a + d, a + 3d and a + 5d. Here common difference is 2d.
Note: From the
above explanation we understand that in case of an odd number of terms, the
middle term is ‘a’ and the common difference is ‘d’.
Again, in case of an even number of terms the middle terms
are a - d, a + d and the common difference is 2d.
Solved examples to observe how to use the selection of terms
in an arithmetic progression
1. The sum of three numbers in Arithmetic Progression is 12 and
the sum of their square is 56. Find the numbers.
Solution:
Let us assume that the three numbers in Arithmetic
Progression be a - d, a and a + d.
According to the problem,
Solved examples to observe how to use the selection of terms
in an arithmetic progression
1. The sum of three numbers in Arithmetic Progression is 12 and
the sum of their square is 56. Find the numbers.
Solution:
Let us assume that the three numbers in Arithmetic
Progression be a - d, a and a + d.
According to the problem,
| | |
--- |
| Sum = 12 and ⇒ a - d + a + a + d = 12 ⇒ 3a = 12 ⇒ a = 4 | Sum of the squares = 56 (a - d)(^{2}) + a(^{2}) + (a + d)(^{2}) = 56 ⇒ a(^{2}) - 2ad + d(^{2}) + a(^{2}) + a(^{2}) + 2ad + d(^{2}) = 56 ⇒ 3a(^{2}) + 2d(^{2}) = 56 ⇒ 3 × (4)(^{2}) + 2d(^{2}) = 56 ⇒ 3 × 16 + 2d(^{2}) = 56 ⇒ 48 + 2d(^{2}) = 56 ⇒ 2d(^{2}) = 56 - 48 ⇒ 2d(^{2}) = 8 ⇒ d(^{2}) = 4 ⇒ d = ± 2 |
Sum = 12 and
⇒ a - d + a + a + d = 12
⇒ 3a = 12
⇒ a = 4
⇒ a - d + a + a + d = 12
⇒ 3a = 12
⇒ a = 4
Sum of the squares = 56
(a - d)(^{2}) + a(^{2}) + (a + d)(^{2}) = 56
⇒ a(^{2}) - 2ad + d(^{2}) + a(^{2}) + a(^{2}) + 2ad + d(^{2}) = 56
⇒ 3a(^{2}) + 2d(^{2}) = 56
⇒ 3 × (4)(^{2}) + 2d(^{2}) = 56
⇒ 3 × 16 + 2d(^{2}) = 56
⇒ 48 + 2d(^{2}) = 56
⇒ 2d(^{2}) = 56 - 48
⇒ 2d(^{2}) = 8
⇒ d(^{2}) = 4
⇒ d = ± 2
(a - d)(^{2}) + a(^{2}) + (a + d)(^{2}) = 56
⇒ a(^{2}) - 2ad + d(^{2}) + a(^{2}) + a(^{2}) + 2ad + d(^{2}) = 56
⇒ 3a(^{2}) + 2d(^{2}) = 56
⇒ 3 × (4)(^{2}) + 2d(^{2}) = 56
⇒ 3 × 16 + 2d(^{2}) = 56
⇒ 48 + 2d(^{2}) = 56
⇒ 2d(^{2}) = 56 - 48
⇒ 2d(^{2}) = 8
⇒ d(^{2}) = 4
⇒ d = ± 2
If d = 3, the numbers are 4 – 2, 4, 4 + 2 i.e., 2, 4, 6
If d = -3, the numbers are 4 + 2, 4, 4 - 2 i.e., 6, 4, 2
Therefore, the required numbers are 2, 4, 6 or 6, 4, 2.
2. The sum of four numbers in Arithmetic Progression is 20 and the sum of their square is 120. Find the numbers.
Solution:
Let us assume that the four numbers in Arithmetic Progression be a - 3d, a - d, a + d and a + 3d.
According to the problem,
| | | |
---
| Sum = 20 ⇒ a - 3d + a - d + a + d + a + 3d = 20 ⇒ 4a = 20 ⇒ a = 5 | and | Sum of the squares = 120 ⇒ (a - 3d)(^{2}) + (a - d)(^{2}) + (a + d)(^{2}) + (a + 3d)(^{2}) = 120 ⇒ a(^{2}) - 6ad + 9d(^{2}) + a(^{2}) - 2ad + d(^{2}) + a(^{2}) + 2ad + d(^{2}) + a(^{2}) + 6ad + 9d(^{2}) = 120 ⇒ 4a(^{2}) + 20d(^{2}) = 120 ⇒ 4 × (5)(^{2}) + 20d(^{2}) = 120 ⇒ 4 × 25 + 20d(^{2}) = 120 ⇒ 100 + 20d(^{2}) = 120 ⇒ 20d(^{2}) = 120 - 100 20d(^{2}) = 20 ⇒ d(^{2}) = 1 ⇒ d = ± 1 |
Sum = 20
⇒ a - 3d + a - d + a + d + a + 3d = 20
⇒ 4a = 20
⇒ a = 5
⇒ a - 3d + a - d + a + d + a + 3d = 20
⇒ 4a = 20
⇒ a = 5
and
Sum of the squares = 120
⇒ (a - 3d)(^{2}) + (a - d)(^{2}) + (a + d)(^{2}) + (a + 3d)(^{2}) = 120
⇒ a(^{2}) - 6ad + 9d(^{2}) + a(^{2}) - 2ad + d(^{2}) + a(^{2}) + 2ad + d(^{2}) + a(^{2}) + 6ad + 9d(^{2}) = 120
⇒ 4a(^{2}) + 20d(^{2}) = 120
⇒ 4 × (5)(^{2}) + 20d(^{2}) = 120
⇒ 4 × 25 + 20d(^{2}) = 120
⇒ 100 + 20d(^{2}) = 120
⇒ 20d(^{2}) = 120 - 100
20d(^{2}) = 20
⇒ d(^{2}) = 1
⇒ d = ± 1
⇒ (a - 3d)(^{2}) + (a - d)(^{2}) + (a + d)(^{2}) + (a + 3d)(^{2}) = 120
⇒ a(^{2}) - 6ad + 9d(^{2}) + a(^{2}) - 2ad + d(^{2}) + a(^{2}) + 2ad + d(^{2}) + a(^{2}) + 6ad + 9d(^{2}) = 120
⇒ 4a(^{2}) + 20d(^{2}) = 120
⇒ 4 × (5)(^{2}) + 20d(^{2}) = 120
⇒ 4 × 25 + 20d(^{2}) = 120
⇒ 100 + 20d(^{2}) = 120
⇒ 20d(^{2}) = 120 - 100
20d(^{2}) = 20
⇒ d(^{2}) = 1
⇒ d = ± 1
If d = 1, the numbers are 5 - 3, 5 - 1, 5 + 1, 5 + 3 i.e.,
2, 4, 6, 8
If d = -1, the numbers are 5 + 3, 5 + 1, 5 - 1, 5 - 3 i.e., 8,
6, 4, 2
Therefore, the required numbers are 2, 4, 6, 8 or 8, 6, 4, 2.
3. The sum of three numbers in Arithmetic Progression is -3 and
their product is 8. Find the numbers.
Solution:
Let us assume that the three numbers in Arithmetic
Progression be a - d, a and a + d.
According to the problem,
Therefore, the required numbers are 2, 4, 6, 8 or 8, 6, 4, 2.
3. The sum of three numbers in Arithmetic Progression is -3 and
their product is 8. Find the numbers.
Solution:
Let us assume that the three numbers in Arithmetic
Progression be a - d, a and a + d.
According to the problem,
| | |
--- |
| Sum = -3 and ⇒ a - d + a + a + d = -3 ⇒ 3a = -3 ⇒ a = -1 | Product = 8 ⇒ (a - d) (a) (a + d) = 8 ⇒ (-1)[(-1)(^{2}) - d(^{2})] = 8 ⇒ -1(1 - d(^{2})) = 8 ⇒ -1 + d(^{2}) = 8 ⇒ d(^{2}) = 8 + 1 ⇒ d(^{2}) = 9 ⇒ d = ± 3 |
Sum = -3 and
⇒ a - d + a + a + d = -3
⇒ 3a = -3
⇒ a = -1
⇒ a - d + a + a + d = -3
⇒ 3a = -3
⇒ a = -1
Product = 8
⇒ (a - d) (a) (a + d) = 8
⇒ (-1)[(-1)(^{2}) - d(^{2})] = 8
⇒ -1(1 - d(^{2})) = 8
⇒ -1 + d(^{2}) = 8
⇒ d(^{2}) = 8 + 1
⇒ d(^{2}) = 9
⇒ d = ± 3
⇒ (a - d) (a) (a + d) = 8
⇒ (-1)[(-1)(^{2}) - d(^{2})] = 8
⇒ -1(1 - d(^{2})) = 8
⇒ -1 + d(^{2}) = 8
⇒ d(^{2}) = 8 + 1
⇒ d(^{2}) = 9
⇒ d = ± 3
If d = 3, the numbers are -1 - 3, -1, -1 + 3 i.e., -4, -1, 2
If d = -3, the numbers are -1 + 3, -1, -1 - 3 i.e., 2, -1, -4
Therefore, the required numbers are -4, -1, 2 or 2, -1, -4.
Therefore, the required numbers are -4, -1, 2 or 2, -1, -4.
● Arithmetic Progression
11 and 12 Grade Math
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Unsteady Heat Transfer in Semi-infinite Solids Solidification process of the coating layer during a thermal spray operation is an unsteady heat transfer problem. As we discuss earlier, thermal spray process deposits thin layer of coating materials on surface for protection and thermal resistant purposes, as shown. The heated, molten materials will attach to the substrate and cool down rapidly. The cooling process is important to prevent the accumulation of residual thermal stresses in the coating layer. S(t) δ solid liquid Coating with density ρ, latent heat of fusion: hsf Substrate, k, α Example As described in the previous slide, the cooling process can now be modeled as heat loss through a semi-infinite solid. (Since the substrate is significantly thicker than the coating layer) The molten material is at the fusion temperature Tf and the substrate is maintained at a constant temperature Ti. Derive an expression for the total time that is required to solidify the coating layer of thickness δ.
Heat transfer from the molten material to the substrate (q=q”A) • Assume the molten layer stays at a constant temperature Tf throughout the process. The heat loss to the substrate is solely supplied by the release of the latent heat of fusion.
From energy balance: solidified mass during t) (energy input) h dm dt = q"A, where m = V = AS, where S is solidified thickness dS dt = q" sf h m Q q A t sf ∆ ∆ ∆ ∆ ( " = = ρ ρ ρ Example (cont.) Identify that the previous situation corresponds to the case discussed in chapter 9-3 in the text as a semi-infinite transient heat transfer problem with a constant surface temperature boundary condition (note: the case in the textbook corresponds to an external convection case. However, it can be modelled as constant surface temperature case by setting h=∞, therefore, Ts=T∞).
s i If the surface temperature is T and the initial temperature of the bolck is T, the analytical solution of the problem can be found: The temperature distribution and the heat transfer into the block a 2 s i 0 s s re: T(x,t)-T ,where erf( ) is the Gaussian error function. T 2 2 It is defined as erf(w)= k(T ) q "(t)= s w v i x erf T t e dv T t α π πα − = − − ∫ Example (cont.) f f 0 0 2 f 2 From the previous equation k(T ) k(T ) =q"= , and 2k(T ) ( ) , therefore, t. Cooling time 4 t i i sf sf sf i f i sf dS T T dt h dS dt t h t h T t t t k T T h δ ρ πα ρ πα δρ πα δ δ ρ πα − − = − = ∝ = − ∫ ∫ Use the following values to calculate: k=120 W/m.K, α=4×10-5 m2/s, ρ=3970 kg/m3, and hsf=3.577 ×106 J/kg, Tf=2318 K, Ti=300K, and δ=2 mm Example (cont.) δ ρ πα ( ) ) .
t T h t t i sf = − = 2k(Tf 0 00304 0 0.2 0.4 0.6 0.8 1 0 0.001 0.002 0.003 0.004 δ( ) t t • δ(t) ∝t1/2 • Therefore, the layer solidifies very fast initially and then slows down as shown in the figure • Note: we neglect contact resistance between the coating and the substrate and assume temperature of the coating material stays the same even after it solidifies.
• To solidify 2 mm thickness, it takes 0.43 seconds.
Example (cont.) What will be the substrate temperature as it varies in time? The temperature distribution is: ( , ) , 2 ( , ) 2318 (300 2318) 2318 2018 79.06 2 S i S T x t T x erf T T t x x T x t erf erf t t α α − = − = + − = − Example (cont.) • For a fixed distance away from the surface, we can examine the variation of the temperature as a function of time. Example, 1 cm deep into the substrate the temperature should behave as: T x t erf x t erf t ( .
, ) .
.
= = − F H G I K J = − F H G I K J 0 01 2318 2018 79 06 2318 2018 0 79 0 2 4 6 8 10 0 400 800 1200 1600 2000 x=1 cm x=2 cm x=3 cm Time Temperature T1( ) t T2( ) t T3( ) t t • At x=1 cm, the temperature rises almost instantaneously at a very fast rate. A short time later, the rate of temp. increase slows down significantly since the energy has to distribute to a very large mass.
• At deeper depth (x=2 & 3 cm), the temperature will not respond to the surface condition until much later.
Example (cont.) We can also examine the spatial temperature distribution at any given time, say at t=1 second.
T x t erf x t erf x ( , ) .
.
= = − F H G I K J = − 1 2318 2018 79 06 2318 2018 79 06 a f 0 0.01 0.02 0.03 0.04 0.05 0 1000 2000 3000 t=1 s.
t=5 s.
t=10 s.
distance (m) Temperature (K) T1( ) x T2( ) x T3( ) x x • Heat penetrates into the substrate as shown for different time instants.
• It takes more than 5 seconds for the energy to transfer to a depth of 5 cm into the substrate • The slopes of the temperature profiles indicate the amount of conduction heat transfer at that instant.
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Calcium Pyrophosphate Deposition Disease
Edgar A. Zamora; Roopa Naik.
Author Information and Affiliations
Authors
Edgar A. Zamora1; Roopa Naik2.
Affiliations
1 Montefiore Medical Center
2 Centra Southside Community Hospital
Last Update: June 20, 2023.
Continuing Education Activity
Calcium pyrophosphate deposition disease (CPPD) is a crystal deposition arthropathy involving the synovial and periarticular tissues. Its clinical presentation may range from being asymptomatic to acute or chronic inflammatory arthritis. Chronic CPP deposition arthritis informally referred to as pseudo-rheumatoid arthritis may present with a waxing and waning clinical course that may last for several months and resemble rheumatoid arthritis involving the wrists and metacarpophalangeal (MCP) joints. The crystals involved in calcium pyrophosphate deposition disease are composed of calcium pyrophosphate dihydrate and commonly affect larger and weight-bearing joints, including the hips, knees, or shoulders. This activity reviews the etiology, presentation, evaluation, and management of calcium pyrophosphate deposition disease, and reviews the role of the interprofessional team in evaluating, diagnosing, and managing the condition.
Objectives:
Describe the pathophysiology of calcium pyrophosphate deposition disease.
Summarize the diagnostic approach for a patient presenting with potential calcium pyrophosphate deposition disease.
Review the therapeutic options available for the management and treatment of calcium pyrophosphate deposition disease.
Explain interprofessional team strategies for improving care coordination and communication regarding the management of patients with calcium pyrophosphate deposition disease.
Access free multiple choice questions on this topic.
Introduction
Calcium pyrophosphate deposition disease (CPPD) is a crystal deposition arthropathy involving the synovial and periarticular tissues. Its clinical presentation may range from being asymptomatic to acute or chronic inflammatory arthritis.
Different terms are used to describe the varied phenotypes of calcium pyrophosphate deposition disease. Acute calcium pyrophosphate (CPP) deposition arthritis, frequently referred to as “pseudogout,” presents as an acute flare of synovitis that resembles acute urate arthropathy (gout). Chronic CPP deposition arthritis informally referred to as pseudo-rheumatoid arthritis may present with a waxing and waning clinical course that may last for several months and resemble rheumatoid arthritis involving the wrists and metacarpophalangeal (MCP) joints. The term chondrocalcinosis describes the characteristic radiological finding of intraarticular fibrocartilage calcification.
The crystals involved in calcium pyrophosphate deposition disease are composed of calcium pyrophosphate dihydrate and commonly affect larger and weight-bearing joints, including the hips, knees, or shoulders. A large number of patients present with underlying joint disease or metabolic abnormalities predisposing to CPP deposition, including osteoarthritis, trauma, surgery, or rheumatoid arthritis.
Etiology
Calcium pyrophosphate deposition disease is believed to be caused by an imbalance between the production of pyrophosphate and the levels of pyrophosphatases in diseased cartilage. As pyrophosphate deposits in the synovium and adjacent tissues, it combines with calcium to form CPP.
Several comorbidities have correlations with CPPD. In a number of studies, hyperparathyroidism presented the highest positive association with CPPD, followed by gout, osteoarthritis, rheumatoid arthritis, and hemochromatosis. Other comorbidities associated with it include osteoporosis, hypomagnesemia, chronic kidney disease, and calcium supplementation.
Deposition of calcium pyrophosphate is believed to cause activation of the immune system producing inflammation and further soft tissue injury.
Epidemiology
Most patients affected by acute calcium pyrophosphate deposition arthritis are over the age of 65. Thirty to fifty percent of patients present over the age of 85 years. A cross-sectional study involving 2,157 cases of CPPD in US veterans reported a point prevalence of 5.2 per 1000, with an average age of 68 years and 95% of male prevalence. CPPD rarely presents in patients under the age of 60.
A large cross-sectional study reported a 4 % crude prevalence of radiographic chondrocalcinosis in the general population.
History and Physical
In patients presenting with acute calcium pyrophosphate arthritis, manifestations are similar to acute urate arthropathy with joint edema, erythema, and tenderness. Up to 50% of these patients may also present with a low-grade fever. The most commonly affected joint is the knee, but other weight-bearing joints may also be affected, including the hips and shoulders.
A subset of patients present with chronic CPP arthritis, often with waxing and waning episodes of non-synchronous, inflammatory arthritis affecting multiple non-weight bearing joints such as wrists and MCP joints, resembling rheumatoid arthritis.
Clinicians should suspect crystal deposition disease in elderly patients presenting with acute degenerative arthritis in weight-bearing joints. The elderly population will commonly run a milder course. Some patients will present acute flares after traumatic injuries.
Evaluation
After an appropriate physical examination, patients suspected to have calcium pyrophosphate deposition disease should undergo arthrocentesis for synovial fluid analysis, in addition to radiography of the involved joints.
Diagnostic confirmation may be via the presence of rhomboid crystals in the synovial fluid aspirate visualized under polarized microscopy. These crystals typically present positive birefringence.
While imaging findings consistent with chondrocalcinosis support the diagnosis of CPPD, its absence does not rule it out. Early signs may be evidenced through ultrasound (US) as cartilage abnormalities. Joint cartilage calcification, known as chondrocalcinosis, may be visualized through radiographic imaging.
Previous studies have demonstrated the usefulness of magnetic resonance imaging (MRI), particularly gradient-echo sequences, for the evaluation of calcium pyrophosphate crystal deposition burden in joint cartilage.
Treatment / Management
The treatment for patients with calcium pyrophosphate deposition disease is based on decreasing inflammation and stabilizing any underlying metabolic disease that may predispose them to calcium pyrophosphate crystal deposition. In patients with acute flares, with the involvement of one or two joints, the treatment of choice is usually joint aspiration and intra-articular glucocorticoid administration, unless contraindicated.
For patients presenting with acute inflammation involving three or more joints, the treatment is typically systemic, often with nonsteroidal anti-inflammatory drugs (NSAIDs). Patients with contraindications for NSAIDs may receive treatment with colchicine or systemic glucocorticoids.
Corticosteroid injection should only be performed after septic arthritis has been ruled out or considered unlikely. In patients presenting with signs and symptoms suggestive of septic arthritis, the recommendation is to defer glucocorticoid administration until synovial fluid cultures are negative.
Other measures include the application of ice packs and joint rest with restriction of weight-bearing to decrease further inflammation. Patients with recurrent episodes of acute CPP arthritis may receive daily low-dose colchicine.
While several medications decrease serum urate levels and prevent urate crystal formation, there is no current therapy directly targeting CPP crystal deposition. Hence, the treatment of CPPD relies on treating predisposing metabolic diseases, soft tissue inflammation, and symptomatology.
Differential Diagnosis
As discussed above, calcium pyrophosphate deposition disease has different phenotypes. Therefore, the differential diagnosis should be based on the clinical presentation and may include gout, rheumatoid arthritis, ankylosing spondylitis, and erosive osteoarthritis.
The intra-articular radiological calcification seen in patients with CPPD is characteristic. In knee x-rays, chondrocalcinosis presents as a thin dense intraarticular opacification that runs parallel and separated from the bone cortex or as a calcifying opacification in the menisci.
Diagnosis of CPPD is presumed in patients with symptomatic joint disease who present the characteristic radiological findings of chondrocalcinosis. As discussed, several imaging modalities may present characteristic findings of CPPD, including x-ray, US, or MRI. When needed, a definite diagnosis is achieved by polarizing light microscopy of synovial fluid, showing the presence of CPP crystals.
Prognosis
Acute calcium pyrophosphate arthritis is generally self-limited, and the inflammation usually resolves within days to weeks of treatment.
Patients with chronic CPP inflammatory arthritis may present overlapping manifestations with rheumatoid arthritis, such as morning stiffness, localized edema, and decreased range of motion. Some patients also present tenosynovitis with carpal or cubital tunnel syndrome. Multiple joints are commonly involved, and the episodes of inflammation may present in a nonsynchronous, waxing, and waning clinical course lasting several months.
Patients with underlying joint comorbidities, such as osteoarthritis, have an increased risk for acute CPP arthritis. When deposited, CPP crystals activate the immune system, promoting inflammation and fibrocartilage injury.
Complications
The molecular structure of calcium pyrophosphate has the potential of triggering inflammatory responses. The presence of chondrocalcinosis has associations with the degradation of menisci and synovial tissue.
Rarely, patients may present with palpable nodules or masses that resemble gout tophi after several episodes of acute CPP crystal arthritis. These nodules localize in the periarticular tissue and represent CPP crystal accumulations within the synovium and adjacent soft tissue; they may lead to further degradation of the affected joint.
A subset of patients may present with spinal involvement, occasionally causing clinical manifestations such as spine stiffness and bony ankylosis resembling ankylosing spondylitis. Some individuals may have clinical manifestations similar to diffuse idiopathic skeletal hyperostosis (DISH) with calcification of the posterior longitudinal ligament leading to spinal cord compression symptoms.
Deterrence and Patient Education
Patients with chondrocalcinosis or previous episodes of acute calcium pyrophosphate arthritis will benefit from knowing that certain conditions carry associations with arthritic flares, including surgical procedures or trauma. Individuals who present at a younger age with CPPD may receive screening for underlying metabolic abnormalities such as hyperparathyroidism or hemochromatosis. Obtaining an adequate family history is also necessary.
Enhancing Healthcare Team Outcomes
Chondrocalcinosis is a relatively common imaging finding in asymptomatic patients. Studies have described a higher prevalence of these findings in late adulthood without gender predilection. Furthermore, although chondrocalcinosis correlates with osteoarthritis, studies suggest that the risk of both diseases increases independently with age, meaning that chondrocalcinosis may develop without osteoarthritis. Asymptomatic older patients may not require further workup or therapy for an isolated finding of chondrocalcinosis.
Due to a broad range of clinical manifestations, calcium pyrophosphate deposition disease should be considered in patients presenting with signs and symptoms suggestive of rheumatoid arthritis or ankylosing spondylitis.
A number of specialists use the high-resolution US in the evaluation of patients with crystal arthropathies, as it may provide useful findings to help differentiate CPPD from other conditions such as gout. Chondrocalcinosis presents earlier in the US than in plain films. This modality may also find use in patients with symptomatic arthropathies of unclear etiology, or with inconclusive results in the fluid aspirate. Additionally, it could potentially decrease the need for using more expensive imaging modalities such as MRI.
Review Questions
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Figure
CPP and Urate Crystals Image courtesy of S Bhimji MD
References
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: Disclosure: Edgar Zamora declares no relevant financial relationships with ineligible companies.
: Disclosure: Roopa Naik declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
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Zamora EA, Naik R. Calcium Pyrophosphate Deposition Disease. [Updated 2023 Jun 20]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Continuing Education Activity
Introduction
Etiology
Epidemiology
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Prognosis
Complications
Deterrence and Patient Education
Enhancing Healthcare Team Outcomes
Review Questions
References
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Table 12: [Information and Equipment for Patients with AI]. - Endotext - NCBI Bookshelf
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Table 12:
Information and Equipment for Patients with AI
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Steroid Sick Day Rules
During any acute illness or stress the total oral GC dose should be doubled (>38°C) or tripled (>39°C) for at least 72 hours; if the patient remains unwell after 72 hours, they should contact the caring physician
There should always be a supply of additional oral GC on prescription for sick days
There is no need to increase the MC dose
Sick day rule 1: need to double the routine oral GC when the patient experiences fever or illness requiring bed rest; when requiring antibiotics for an infection; or before a small outpatient procedure (e.g., dental work).
Sick day rule 2: need to inject a GC preparation i.m. or i.v. in case of severe illness, trauma, persistent vomiting, when fasting for a procedure (colonoscopy!), or during surgical intervention
Provide patient with: Sufficient supply of HC and fludrocortisone (accounting for possible sick days)
Hydrocortisone emergency injection kit prescription (vials of 100 mg HC sodium, syringes, needles; alternatively, also HC or prednisolone suppositories)
Leaflet with information on AC and hospitalization to be shown to health care staff; clearly advises regarding the need to inject 100 mg HC immediately i.v. or i.m, followed by continuous infusion of 200 mg/24 h
Emergency phone number of endocrine specialist team
Steroid Emergency Pack
Every patient should be provided with this pack to keep at home
The pack contains a vial of 100mg HC or dexamethasone 4mg, a syringe and a needle
The patient and/ or any responsible family member should be educated to administer this medication i.m. or sc during an emergency situation, i.e. severe accident, significant hemorrhage, fracture, unconsciousness, diarrhea and vomiting, and they should call the emergency medical personal immediately (adults, i.m. or sc HC 100 mg; children, i.m. HC 50mg/m 2; infants, 25mg; school-age children, 50 mg; adolescents, 100 mg).
The expiry date on the pack should be checked regularly and replaced with a new pack if expired
The patient should be advised to take the pack when travelling.
Medical-Alert bracelet or pendant and emergency steroid card
Every patient should wear or carry these in which the diagnosis and daily medication should be clearly documented
Follow up
Reinforce education and confirm understanding during each follow-up visit (at least annually in a patient without specific problems or recent crises; otherwise, more frequently)
GCs: glucocorticoids; HC: hydrocortisone; i.m.: intramuscularly; i.v.: intravenously; HC: hydrocortisone; MCs: mineralocorticoids; sc: subcutaneously
From: Diagnosis and Management of Adrenal Insufficiency
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Review Pediatric Adrenal Insufficiency: Diagnosis, Management, and New Therapies.[Int J Pediatr. 2018]Review Pediatric Adrenal Insufficiency: Diagnosis, Management, and New Therapies.Bowden SA, Henry R. Int J Pediatr. 2018; 2018:1739831. Epub 2018 Nov 1.
Review Adrenal insufficiency in pregnancy: Physiology, diagnosis, management and areas for future research.[Rev Endocr Metab Disord. 2023]Review Adrenal insufficiency in pregnancy: Physiology, diagnosis, management and areas for future research.Lee JH, Torpy DJ. Rev Endocr Metab Disord. 2023 Feb; 24(1):57-69. Epub 2022 Jul 11.
Review Fertility and pregnancy in adrenal insufficiency.[Endocr Connect. 2024]Review Fertility and pregnancy in adrenal insufficiency.Green D, Dineen R, O'Reilly MW, Sherlock M. Endocr Connect. 2024 Feb 1; 13(2). Epub 2024 Jan 12.
Review Diagnosis and Management of Adrenal Insufficiency and Adrenal Crisis in the Emergency Department.[J Emerg Med. 2022]Review Diagnosis and Management of Adrenal Insufficiency and Adrenal Crisis in the Emergency Department.Lentz S, Collier KC, Willis G, Long B. J Emerg Med. 2022 Aug; 63(2):212-220. Epub 2022 Aug 26.
Review Adrenal insufficiency: diagnosis and management.[Curr Urol Rep. 2006]Review Adrenal insufficiency: diagnosis and management.Munver R, Volfson IA. Curr Urol Rep. 2006 Jan; 7(1):80-5.
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Why doesn't a catalyst change the equilibrium position?
A catalyst doesn't change the equilibrium position because it speeds up both the forward and reverse reactions equally.
In more detail, a catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy, which is the energy required to initiate the reaction. However, it's important to understand that a catalyst does not affect the energy of the reactants or the products. It simply provides an alternative reaction pathway with a lower activation energy. This means that while a catalyst can help a reaction reach equilibrium faster, it does not change the position of the equilibrium itself.
The position of equilibrium in a chemical reaction is determined by the relative energies of the reactants and products. If the products have lower energy than the reactants, the reaction will tend to go to completion, and the equilibrium will lie to the right. If the reactants have lower energy than the products, the reaction will tend not to proceed very far, and the equilibrium will lie to the left. A catalyst does not change these relative energies, so it does not change the position of equilibrium.
Moreover, a catalyst affects both the forward and reverse reactions equally. This is because a catalyst simply provides an alternative reaction pathway with a lower activation energy for both directions of the reaction. Therefore, while the catalyst speeds up the rate at which the reaction reaches equilibrium, it does not favour one direction over the other. This means that the position of equilibrium - the ratio of products to reactants at equilibrium - remains unchanged.
In conclusion, while catalysts are incredibly useful for speeding up chemical reactions and helping them reach equilibrium more quickly, they do not have any effect on the position of equilibrium in a reaction. This is a fundamental principle of chemical equilibrium that is crucial for understanding how reactions proceed.
Answered byOli - Qualified IB Tutor | BSc Natural Sciences
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4.2 An Introduction to Odds, Odds Ratios and Exponents
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4.2 An Introduction to Odds, Odds Ratios and Exponents
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Proportions and Ratios Let’s start by considering a simple association between two dichotomous variables (a 2 x 2 crosstabulation) drawing on the LSYPE dataset. The outcome we are interested in is whether students aspire to continue in Full-time education (FTE) after the age of 16 (the current age at which students in England can choose to leave FTE). We are interested in whether this outcome varies between boys and girls. We can present this as a simple crosstabulation (Figure 4.2.1). Figure 4.2.1: Aspiration to continue in full time education (FTE) after the age of 16 by gender: Cell counts and percentages We have coded not aspiring to continue in FTE after age 16 as 0 and aspiring to do so as 1. Although it is possible to code the variable with any values, employing the values 0 and 1 has advantages. The mean of the variable will equal the proportion of cases with the value 1 and can therefore be interpreted as a probability. Thus we can see that the percentage of all students who aspire to continue in FTE after age 16 is 81.6%. This is equivalent to saying that the probability of aspiring to continue in FTE in our sample is 0.816. Odds and odds ratios However another way of thinking of this is in terms of the odds. Odds express the likelihood of an event occurring relative to the likelihood of an event not occurring. In our sample of 15,431 students, 12,591 aspire to continue in FTE while 2,840 do not aspire, so the odds of aspiring are 12591/2840 = 4.43:1 (this means the ratio is 4.43 to 1, but we conventionally do not explicitly include the :1 as this is implied by the odds). The odds tell us that if we choose a student at random from the sample they are 4.43 times more likely to aspire to continue in FTE than not to aspire to continue in FTE. We don’t actually have to calculate the odds directly from the numbers of students if we know the proportion for whom the event occurs, since the odds of the event occurring can be gained directly from this proportion by the formula (Where p is the probability of the event occurring.): Thus the odds in our example are: Odds= [p/(1-p)] = .816 / (1-.816 )= .816 /.184 = 4.43. The above are the unconditional odds, i.e. the odds in the sample as a whole.However odds become really useful when we are interested in how some other variable might affect our outcome. We consider here what the odds of aspiring to remain in FTE are separately for boys and girls, i.e. conditional on gender. We have seen the odds of the event can be gained directly from the proportion by the formula odds=p/(1-p). So for boys the odds of aspiring to continue in FTE = .766/(1-.766)= 3.27 While for girls the odds of aspiring to continue in FTE = .868/(1-.868)= 6.56. These are the conditional odds, i.e. the odds depending on the condition of gender, either boy or girl. We can see the odds of girls aspiring to continue in FTE are higher than for boys. We can in fact directly compare the odds for boys and the odds for girls by dividing one by the other to give the Odds Ratio (OR). If the odds were the same for boys and for girls then we would have an odds ratio of 1. If however the odds differ then the OR will depart from 1. In our example the odds for girls are 6.53 and the odds for boys are 3.27 so the OR= 6.56 / 3.27 = 2.002, or roughly 2:1. This says that girls are twice as likely as boys to aspire to continue in FTE. Note that the way odd-ratios are expressed depends on the baseline or comparison category. For gender we have coded boys=0 and girls =1, so the boys are our natural base group. However if we had taken girls as the base category, then the odds ratio would be 3.27 / 6.56= 0.50:1. This implies that boys are half as likely to aspire to continue in FTE as girls. You will note that saying “Girls are twice as likely to aspire as boys” is actually identical to saying “boys are half as likely to aspire as girls”. Both figures say the same thing but just differ in terms of the base. Odds Ratios from 0 to just below 1 indicate the event is less likely to happen in the comparison than in the base group, odds ratios of 1 indicate the event is exactly as likely to occur in the two groups, while odds ratios from just above 1 to infinity indicate the event is more likely to happen in the comparator than in the base group. Extension D provides a table that shows the equivalence between ORs in the range 0 to 1 with those in the range 1 to infinity. Seeing the relationship as a model An interesting fact can be observed if we look at the odds for boys and the odds for girls in relation to the odds ratio (OR). For boys (our base group) the odds= 3.27 1 = 3.27 For girls the odds = 3.27 2.002 = 6.56. So another way of looking at this is that the odds for each gender can be expressed as a constant multiplied by a gender specific multiplicative factor (namely the OR). p/(1-p) = constant OR. However there are problems in using ORs directly in any modelling because they are asymmetric. As we saw in our example above, an OR of 2.0 indicates the same relative ratio as an OR of 0.50, an OR of 3.0 indicates the same relative ratio as an OR of 0.33, an OR of 4.0 indicates the same relative ratio as an OR of 0.25 and so on. This asymmetry is unappealing because ideally the odds for males would be the opposite of the odds for females. Odds, Log odds and exponents This asymmetry problem disappears if we take the ‘log’ of the odds ratio (OR). ‘Log’ doesn’t refer to some sort of statistical deforestation… rather a mathematical transformation of the odds which will help in creating a regression model. Taking the log of an OR of 2 gives the value Log(2)= +0.302 and taking the log of an OR of 0.50 gives the value Log (0.5)= -0.302. See what’s happened? The Log of the OR, sometimes called the logit (pronounced ‘LOH-jit’, word fans!) makes the relationships symmetric around zero (the OR’s become plus and minus .302). Logits and ORs contain the same information, but this difference in mathematical properties makes logits better building blocks for logistic regression. But what is a log function? How does it transform the ORs? Well, the natural log function looks like this (Figure 4.2.2): Figure 4.2.2: The natural log function So if we take the log of each side of the equation we can then express the log odds as: Log [p/(1-p)] = constant + log (OR) If the constant is labelled a, the log of the OR is labelled b, and the variable gender (x) takes the value 0 for boys and 1 for girls, then: Log [p/(1-p)] = a + bx Note that taking the log of the odds has converted this from a multiplicative to an additive relationship with the same form as the linear regression equations we have discussed in the previous two modules(it is not essential, but if you want to understand how logarithms do this it is explained in Extension E). So the log of the odds can be expressed as an additive function of a + bx. This equation can be generalised to include any number of explanatory variables: Log [p/(1-p)] = a + b 1 x 1+ b 2 x 2 + b 3 x 3 + ... + b n x n. Output from a logistic regression of gender on educational aspiration If we use SPSS to complete a logistic regression (more on this later) using the student level data from which the summary Figure 4.2.1 was constructed, we get the logistic regression output shown below (Figure 4.2.3). Figure 4.2.3: Output from a logistic regression of gender on aspiration to continue in FTE post 16 Let’s explain what this output means. The B weights give the linear combination of the explanatory variables that best predict the log odds. So we can determine that the log odds for: Male: Log [p/(1-p)] = 1.186 + (0.694 0) = 1.186 Female: Log [p/(1-p)] = 1.186 + (0.694 1) = 1.880 The inverse of the log function is the exponential function, sometimes conveniently also called the anti-logarithm (nice and logical!). So if we want to convert the log odds back to odds we take the exponent of the log odds. So the odds for our example are: Male: Exp (1.186) = 3.27 Female: Exp (1.880) = 6.55 The odds ratio is given in the SPSS output for the gender variable [indicated as Exp(B)] showing that girls are twice as likely as boys to aspire to continue in FTE. By simple algebra we can rearrange the formula odds= [p/(1-p] to solve for probabilities: Males: p= 3.27 / (1+3.27) = .766 Females: p= 6.55 / (1+6.55) = .868. These probabilities, odds and odds ratios - derived from the logistic regression model - are identical to those calculated directly from Figure 4.2.1. This is because we have just one explanatory variable (gender) and it has only two levels (girls and boys). This is called a saturated model for which the expected counts and the observed counts are identical. The logistic regression model will come into its own when we have an explanatory variable with more than two values, or where we have multiple explanatory variables. However what we hope this section has done is show you how probabilities, odds, and odds ratios are all related, how we can model the proportions in a binary outcome through a linear prediction of the log odds (or logits), and how these can be converted back into odds ratios for easier interpretation. Take the quiz to check you are comfortable with what you have learnt so far. If you are not perturbed by maths and formulae why not check out Extension Efor more about logs and exponents. Take the Logistic Regression Quiz A ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- ###### Check that you have grasped the first few pages of this tricky module.##### Navigation Home Modules Site Guide Module 4 Contents Resources
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News & Stories
Return of the Station Wagon
Apr 14, 2017In the News
By Jayme Moore
Station wagons.
Many of us grew up in them. The green 2-door Plymouth Suburban. The red and white Ford Parklane with the Thunderbird V-8. That slow-as-a-slug 6-cylinder automatic Chevy Kingswood. Like the dog, they were lovable members of the family.
If you are less than about 55 years of age, that paragraph likely means nothing. You possibly grew up in minivans or SUVs. Too bad. You missed a sweet spot in the automotive timeline. Then again, hope is not lost. Station wagons are still around…in style and utility if not in name.
They were called station wagons because they were originally devised to take people to and from stations. In the beginning, the vehicles often had hand-built wooden bodies. Then steel took over, though homage was long played to the segment’s heritage with “woodie” station wagon options.
Come 1976, station wagons were 10 percent of the vehicle market, more than 970,000 sold. They were also on their way out. The 1976 Buick in the below photo weighed just over 5100 pounds and needed a thirsty 455-cubic-inch V-8 to get the job done. Worse yet, the car’s handling has been described as that of a bathtub half-full of water.
Definitely not the vehicle for the future, given the fuel economy standards put into effect after the 1973-1974 Arab Oil Crisis, ready to hit automakers right between the headlights in 1978.
Thanks to Chrysler, minivans became the people hauler of choice after 1984 and by the mid-1990s there were civilized SUVs to tempt families. Then came the CUV. Wikipedia’s definition says it, “…is a vehicle built on a car platform and combining, in highly variable degrees, features of a sport utility vehicle (SUV) with features from a passenger vehicle, especially those of a station wagon or hatchback.”
In other words, an automaker can make up its definition of a CUV as it goes along.
It’s easy to understand why automaker’s marketeers dig deep into their cauldron of adjectives to describe a vehicle as a CUV or SUV rather than a station wagon. These days, station wagons make up just 1.1 percent of vehicles sold. CUVs, on the other hand, were 33.9 percent of vehicle sales in the first three months of 2017.
Karl Brauer, analyst at Kelly Blue Book adds, “No car maker wants to admit to making a station wagon because of the stigma still attached to that word, so most of them use phrases like ‘crossover’ or ‘sportwagon’ to avoid visions of Chevy Chase driving across America in the Griswold family’s Truckster.
“The good news is, no matter what you call them, these vehicles all offer greater utility and flexibility than a similarly sized sedan. If we all have to nod our heads and agree not to use the term ‘station wagon’ so be it.”
Arguably the most glaring example of redefining the wagon might be Mercedes-Benz’ GLA250 and AMG GLA45. The automaker labels them SUVs when it in reality, they are attractive, sweet-size station wagons with lower cladding and a bit more ground clearance.
Volvo has long been what you might call the “keeper of the keys” when it comes to station wagons. Most all of us who owned one have warm memories of them, whether a 122, the 145 or my intercooled Turbo Wagon that could chase down pony cars.
The Swedish automaker’s U.S. product manager, Anders Robertson, explains how they managed to have what they freely call a station wagon.
Their XC90 is an SUV. Designed on the same platform is the V90 Cross Country, “It’s not as extreme as an SUV, but based on a wagon with more ground clearance and all-wheel drive standard. It has basically the off-road capabilities of the SUV type of vehicles, providing the agility of an SUV with the road holding a sedan.” Volvo calls it a Crossover Wagon.
Remove the added ground clearance and the cladding along the lower sill and wheel arches and you have the V90 station wagon.
Volvo does this dance with its XC60, V60 Cross Country and V60 station wagon.
In a comment that underlines another reason other than stigma why many automakers avoid station wagons, Robertson admits they will only sell a few hundred V90 wagons. Once the XC90 and V90 Cross Country were certified for U.S. sale, so was the V90. If you want one, you custom order it from the dealer or online and wait about 3 months for the car.
So why does Volvo go through the effort? “We have a fan base,” Robertson points out. “A heritage. People who want the wagon know we have it.”
Which does nothing to explain the newest station wagon headed for the U.S. market. Buick’s Regal TourX. It is based on the station wagon version of the Insignia made by Opel, which General Motors recently sold to France’s PSA.
Images of the Insignia wagon show no lower cladding or increased ride height, but there they are on the TourX. Sorry, Buick, it’s still a station wagon…and a handsome one. Power comes from a 2.0-liter turbocharged four-cylinder engine with an estimated 250 hp and 295 lb-ft of torque. The transmission is an 8-speed automatic and all-wheel drive is standard.
We’ve assembled a gallery of images of some vehicles we consider to be station wagons. And we’d be curious to hear your opinions.
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Coenzymes: Essential catalysts in biochemical reactions
Coenzymes are vitamin-derived molecules that play a crucial role in enabling enzymes to catalyze essential biochemical reactions, reflecting their foundational importance in both early and modern metabolism.
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Coenzymes are small, organic, non-protein molecules that support enzymes in catalyzing biochemical reactions. These essential molecules are typically derived from vitamins or other critical nutrients, underscoring their nutritional importance. Coenzymes are believed to have originated in early prebiotic metabolism, existing before enzymes, polymers, and molecular replication mechanisms emerged. This ancient origin implies that coenzymes were among the earliest biomolecules involved in facilitating chemical transformations in primitive life forms 1.
Coenzymes are not proteins but rather a specialized group of cofactors, non-protein molecules that aid enzymes. Their functional role is believed to have emerged when small organic molecules with coenzyme-like properties functioned as “holoribozymes,” binding to RNA fragments and facilitating the synthesis of macromolecular templates, such as RNA. The RNA world hypothesis offers valuable insights into how early metabolic pathways may have operated prior to the evolution of modern protein enzymes 1,2.
Coenzymes enable enzymes to perform their vital roles in a wide range of biochemical processes essential for life. They serve as molecular tools that enzymes use to perform chemical transformations that would otherwise be impossible or inefficient. They generally bind to specific enzymes and play an active role in facilitating catalytic transformations 3. Their small size allows for easy diffusion within cells, and their chemical versatility enables participation in diverse reactions.
Relationship with enzymes
Some enzymes are naturally active without coenzymes, whereas others require a coenzyme for activation. This distinction reflects the evolutionary adaptation of enzymes to diverse metabolic needs and environmental conditions. Enzymes that are inactive in the absence of their coenzymes are termed apoenzymes, while the active complex formed upon binding to a coenzyme is called a holoenzyme. The reversible nature of this binding enables cells to dynamically regulate enzyme activity in response to metabolic demands 4.
Coenzyme specificity
Coenzyme specificity refers to an enzyme’s capability to preferentially bind and utilize a specific coenzyme during catalysis. This specificity can be modified via targeted mutations at the coenzyme-binding site.
For example, the enzyme malate dehydrogenase (MDH) typically shows specificity for either nicotinamide adenine dinucleotide (NAD+) or nicotinamide adenine dinucleotide phosphate (NADP+), depending on the organism and the direction of the reaction. The Bacillus subtilis malate dehydrogenase (BsMDH) is also primarily NAD+-dependent 5, 6.
A notable mutant, BsMDH-T7, shows a significantly higher affinity for nicotinamide adenine dinucleotide phosphate (NADPH) than nicotinamide adenine dinucleotide (NADH), indicating a shift in coenzyme preference. This shift not only enhances the enzyme’s catalytic efficiency but also broadens its potential for industrial applications, such as in the efficient production of malate. Therefore, the coenzyme specificity of MDH is crucial for optimizing metabolic pathways and improving biotransformation efficiency 6.
Coenzyme reusability
Coenzymes are frequently regenerated during enzymatic reactions, enabling their reuse without being expended. This reusability is crucial for efficient biochemical processes, as it ensures their participation in successive reaction cascades and facilitates continuous catalytic activity 7.
A recent study developed an innovative artificial cascade-catalytic system to convert methanol into ethylene glycol. This system, which incorporates methanol dehydrogenase, glycolaldehyde synthase, and lactaldehyde–pyruvate oxidoreductase, efficiently reuses coenzymes to simplify the metabolic pathway 8.
The system achieved a high ethylene glycol concentration within 48 hours by optimizing the ribosome-binding site (RBS), isopropyl β-D-1-thiogalactopyranoside induction, and methanol concentration, achieving a conversion rate of 58.92%. This breakthrough highlights the critical role of coenzyme reusability in driving biotransformation processes and paves the way for the industrial-scale production of ethylene glycol 8.
Types of coenzymes
Coenzymes are derived from vitamin and non-vitamin sources alike, each playing a significant role in a variety of biological processes.
Vitamin-derived coenzymes
Vitamins play a vital role in the body’s production of coenzymes, with water-soluble B-complex vitamins being particularly essential for their synthesis. B vitamin-derived coenzymes are involved in numerous enzymatic processes that support cellular functions, including brain and nervous system activities.
Key B vitamin-derived coenzymes include:
Thiamine pyrophosphate (TPP), derived from thiamine (B1).
Flavoproteins (FAD; FMN), from riboflavin (B2).
NAD and NADP, from niacin (B3).
Coenzyme A (CoA), from pantothenic acid (B5).
Pyridoxal 5’-phosphate (PLP), from vitamin B6.
Methyltetrahydrofolate, from folic acid (B9).
Methylcobalamin and 5’-deoxyadenosylcobalamin, from cobalamin (B12).
Biotin(B7).
These coenzymes facilitate important biochemical reactions, ensuring precise metabolic and physiological function 9–11.
Although B-complexes are important sources of coenzymes, vitamin C (ascorbic acid) also serves as a cofactor for enzymes such as prolyl hydroxylase, which is essential for collagen synthesis 12.
Thecoenzyme A (CoA) assay kit (ab102504) is designed for easy and accurate measurement of CoA levels in a wide range of biological samples, with a quantitative range of 2.5-250 µM and a rapid 1-hour assay time. Suitable for plasma, serum, tissue extracts, and more, this kit ensures reliable and consistent results.
TheNADP coenzyme (ab146316), with a purity of ≥93%, is designed to support redox reactions by facilitating electron transport in various cytochrome P450 and oxidase/reductase systems. It can be stored under desiccating conditions for up to 12 months, making it a versatile coenzyme suitable for a broad range of research applications.
Non-vitamin coenzymes
Non-vitamin coenzymes, derived from sources such as nucleotides, microbial metabolites, and plant-derived compounds, are necessary for various physiological processes, including matter and energy transport, blood clotting, and metabolism. These include:
Energy transduction(eg, ATP/ADP in phosphorylation reactions)13.
Electron transfer(eg, quinones in respiratory chains)13.
Group transfer(eg, S-adenosylmethionine [SAM] for methylation)14.
Blood clotting(eg, calcium ions in coagulation cascades)15.
Key examples include:
Adenosine phosphates(AMP, ADP, and ATP): Central to cellular energy currency.
Coenzyme Q 10/ubiquinone (CoQ 10): Mitochondrial electron carrier 16.
Lipoic acid: Critical in oxidative decarboxylation 17.
Glutathione: Redox regulation and detoxification 18.
Coenzymes vs. cofactors
Cofactors are non-protein chemical compounds required for the biological activity of an enzyme, broadly categorized into organic coenzymes and inorganic cofactors.
The key distinction lies in their chemical nature and origin, coenzymes(eg, NAD⁺, FAD, and CoA) are organic molecules, often derived from vitamins, that transiently bind to enzymes and participate directly in catalysis by transferring functional groups or electrons. In contrast,inorganic cofactors(eg, Mg²⁺, Zn²⁺, and Fe²⁺/Fe³⁺) are metal ions that stabilize enzyme structures and mediate redox reactions without being consumed.
For example, lactate dehydrogenase relies on the coenzyme NAD⁺ to shuttle hydride ions, while carbonic anhydrase depends on a zinc ion (inorganic cofactor) to catalyze CO₂ hydration. Some enzymes require both, pyruvate dehydrogenase uses thiamine pyrophosphate as a coenzyme and Mg²⁺ as a cofactor 19.
Although coenzymes often undergo structural changes during catalysis, they are regenerated and reused, in contrast to cofactors, which typically assist enzymes in a more general manner. Coenzymes, therefore, serve as specialized molecular tools for specific biochemical transformations 1.
Mechanism of action
By binding to enzymes and acting as carriers of electrons and chemical groups, coenzymes are molecular shuttles operating precisely through orchestrated binding and release mechanisms that maximize metabolic efficiency. Their ability to recycle (eg, NAD⁺⇌NADH) allows a small pool of molecules to drive countless reactions. This recyclability makes coenzymes extraordinarily efficient; a single NAD⁺ molecule can participate in thousands of oxidation–reduction cycles per minute in active metabolic tissues.
Enzymes often rely on non-protein molecules to help facilitate their biochemical functions, either through temporary or permanent binding to the enzyme active site. The choice between temporary and permanent binding reflects evolutionary optimization, transient interactions allow metabolic flexibility, whereas permanent attachments ensure constant catalytic readiness for essential reactions.
Temporary or permanent binding to the enzyme active site
Temporary binding, such as that between NAD+ and dehydrogenases, occurs through weaker interactions such as ionic or hydrogen bonds, facilitating the release of the coenzyme post-reaction. This reversibility is essential for metabolic regulation and pathway branching. In contrast, permanent binding involves stronger covalent bonds. An example of permanent binding is the prosthetic group FAD in succinate dehydrogenase, which remains covalently attached throughout catalysis. This enduring association is particularly crucial for enzymes embedded in membranes, such as those within the electron transport chain.
The binding capacity of coenzymes is crucial for an enzyme’s optimal structure and function. Through induced fit mechanisms, coenzyme binding often produces conformational changes in enzymes, aligning catalytic residues with substrates in a precise manner, thereby enhancing reaction specificity.
Facilitation of biochemical reactions by acting as carriers of electrons, hydrogen, and chemical groups
Coenzymes act as carriers of various chemical groups, eg, NAD+, which functions primarily as an electron carrier in oxidation-reduction reactions. Here, NAD+ accepts a hydrogen ion (H+) and two electrons from one substrate, becoming NADH. NADH then donates these electrons to a second substrate, returning to its original form, NAD+. This process allows NAD+ to transfer electrons from one substrate (which gets oxidized) to another (which becomes reduced), facilitating essential metabolic processes 20.
In addition to NAD+, many other coenzymes function as carriers for various chemical groups such as carbon dioxide (CO 2; via biotin), acyl groups (via CoA), and methyl groups (via SAM)21. These coenzymes, in conjunction with their respective enzymes, catalyze the transfer of specific chemical groups between substrates, thereby driving a range of essential biochemical pathways.
Importance in biochemical efficiency
In cellular respiration, coenzymes such as NAD+ and FAD act as electron carriers, forming NADH during glycolysis and both NADH and flavin adenine dinucleotide dihydrogen (FADH 2) during the Krebs cycle22. These molecules transport electrons to the electron transport chain, enabling efficient ATP production.
Similarly, NADP+ serves as an essential electron carrier in light-dependent reactions during photosynthesis. It accepts electrons and hydrogen ions generated by the splitting of water, forming NADPH. This NADPH then provides the reducing power required in the Calvin cycle to convert CO 2 into glucose and other carbohydrates, supporting energy storage and biosynthesis. The coordinated action of these coenzyme systems allows photosynthetic organisms to convert light energy into chemical energy with remarkable efficiency.
Role of coenzymes in metabolism
Coenzymes contribute significantly to energy production and the synthesis of specialized metabolites.
Cellular respiration
Various pathways such as glycolysis, the Krebs cycle, and the electron transport chain have the involvement of coenzymes. A detailed discussion of their involvement follows.
Glycolysis
Glycolysisis a fundamental metabolic pathway that generates energy in the form of ATP, even in the absence of oxygen. Throughout its 10-step process, various coenzymes are involved in facilitating the biochemical reactions that convert glucose into pyruvate, which can either enter the mitochondria for further ATP production or be reduced to lactate under anaerobic conditions. In glycolysis, coenzymes serve as carriers for electrons, protons, and chemical groups, facilitating the proper function of the enzymes involved in this pathway 8, 23.
One of the most vital coenzymes in glycolysis is NAD+, which participates in oxidation–reduction reactions. During the conversion of glyceraldehyde 3-phosphate (G3P) to 1,3-bisphosphoglycerate (1,3-BPG) in glycolysis, NAD+ accepts two electrons and H+ and is converted to NADH 8, 23.
This reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase (GAPDH). In the presence of oxygen, the resulting NADH can later be used to generate ATP in oxidative phosphorylation. This step is essential for maintaining the redox balance within the cell and enabling efficient energy production.
Another vital role of coenzymes in glycolysis is the participation of ATP in substrate-level phosphorylation. Coenzymes facilitate the transfer of phosphate groups to substrates via two key reactions, leading to ATP production. In the first reaction, 1,3-bisphosphoglycerate (1,3-BPG) donates a phosphate group to ADP, forming ATP and 3-phosphoglycerate (3-PG). In the second reaction, phosphoenolpyruvate (PEP) transfers a phosphate group to ADP, generating ATP and pyruvate. These reactions do not require oxygen and are essential for the anaerobic production of ATP 24, 25.
Krebs cycle
The Krebs cycle, also known as the citric acid cycle or tricarboxylic acid (TCA) cycle, is central to cellular energy production. This cycle involves eight enzyme-catalyzed reactions in the mitochondrial matrix, where acetyl-CoA is converted into CO 2 and high-energy electron carriers such as NADH and FADH 2. Coenzymes play a crucial role in facilitating these reactions by transporting electrons, protons, and chemical groups, thereby assisting in the catalysis of several steps in the cycle.
CoA is an essential coenzyme in the Krebs cycle, particularly in the conversion of acetyl-CoA, which is derived from carbohydrates, fats, and proteins, into its active form. Acetyl-CoA enters the cycle by combining with oxaloacetate to form citrate in the first step, catalyzed by the enzyme citrate synthase. This highly exergonic reaction is vital for the initiation of the Krebs cycle 26.
As with glycolysis, NAD+ plays an important role in the oxidation reactions of the Krebs cycle. During the conversion of isocitrate to α-ketoglutarate, NAD+ is reduced to NADH by the enzyme isocitrate dehydrogenase. This reaction also releases CO 2 and H+. The NADH generated in this cycle is subsequently used in the electron transport chain to produce ATP, thus contributing significantly to energy production 26, 27.
FAD is another essential coenzyme in the Krebs cycle, specifically involved in the conversion of succinate to fumarate by the enzyme succinate dehydrogenase. In this reaction, FAD accepts electrons and is reduced to FADH 2, which then enters the electron transport chain to help generate ATP. This step is unique because it is the only reaction in the Krebs cycle that occurs in the inner mitochondrial membrane, directly linking the TCA cycle to the respiratory chain 28.
CoQ 10 is part of the mitochondrial electron transport chain and is indirectly involved in the Krebs cycle. It accepts electrons from both FADH 2 (produced during the succinate to fumarate conversion) and NADH (generated in several steps of the cycle). These electrons ultimately drive ATP production through oxidative phosphorylation.
The NADH and FADH 2 molecules produced in the Krebs cycle carry high-energy electrons to the electron transport chain, where they are oxidized. This oxidation process is coupled with ATP production via chemiosmosis, a process that requires oxygen. As a result, NAD+ and FAD must be regenerated to sustain the cycle 26, 28.
Electron transport chain
The electron transport chain in mitochondria relies on coenzymes to transfer electrons and generate ATP. NADH and FADH 2 donate electrons to the chain, with NADH initiating the process at complex I and FADH 2 at complex II. As electrons pass through these complexes, NADH is oxidized to NAD+, and FADH 2 is oxidized to FAD. These electron transfers release energy, which is used to pump H+ across the mitochondrial membrane, creating an electrochemical gradient 28.
CoQ 10 carries electrons between complexes I/II and III, with cytochrome c shuttling electrons from complex III to complex IV. Oxygen serves as the final electron acceptor at complex IV, where it combines with protons to form water 21.
This electron transfer process drives proton pumping, which powers ATP synthase, ultimately generating ATP. Coenzymes such as NADH, FADH 2, coenzyme Q, and cytochrome c are vital for maintaining the flow of electrons and the establishment of the proton gradient, both of which are essential for efficient ATP production.
Other metabolic pathways
Coenzymes derived from B-complexes (B1–B9) are especially essential for the metabolism of lipids, proteins, and carbohydrates. They play key roles in both primary metabolism—encompassing protein, carbohydrate, lipid, and nucleic acid processes—and specialized metabolism, which includes the biosynthesis of terpenoids, alkaloids, and phenolics. These coenzymes may directly catalyze the formation of specialized metabolites or act indirectly by supporting the production of precursors or other coenzymes 29.
For example, NADPH, derived from vitamin B3, is essential for redox reactions in enzymes such as cytochrome P450s, which are involved in the biosynthesis of specialized metabolites. Vitamin B2 contributes coenzymes FAD and FMN, which facilitate electron transfer in enzymes such as alcohol dehydrogenases and cytochrome P450s. Vitamin B5-derived CoA carries acyl groups in specialized metabolic reactions, and vitamin B6-derived pyridoxal 5’-phosphate (PLP) is involved in amino acid biosynthesis, an indirect pathway contributing to the production of specialized metabolites 29, 30.
Vitamin B9 supports methyltransferase reactions, and vitamin B1 (thiamine diphosphate, TDP) is crucial for energy-generating reactions in photosynthesis and respiration. Additionally, TDP plays a role in the biosynthesis of isoprenoids, which are essential for plant pigments, phytohormones, and terpenoid defense compounds 14, 31, 32.
These coenzymes often interact with one another, creating interdependencies that influence both primary and specialized metabolic pathways. Moreover, coenzymes and their precursors can serve as substrates for specialized metabolites, further underscoring their importance in metabolic networks.
Health and medical significance of coenzymes
Because coenzymes are essential for facilitating the biochemical processes that sustain life and health, their defective functioning can disrupt critical metabolic pathways and bodily functions, leading to a variety of health complications.
Common deficiencies and impacts
Coenzyme deficiencies can present with symptoms such as fatigue and weakness, and may lead to more severe disorders, including pellagra, caused by a deficiency in niacin (vitamin B3), or beriberi, resulting from a lack of thiamine (vitamin B1), which is essential for the formation of thiamine pyrophosphate.
Pellagra
Pellagra is characterized by the classic triad of dermatitis, dementia, and diarrhea. Its symptoms include gastrointestinal disturbances such as nausea and diarrhea, photosensitive skin lesions, and neurological changes, including anxiety, confusion, and muscle weakness. If left untreated, pellagra can be fatal.
Historically, pellagra was prevalent in populations with limited access to niacin-rich foods, particularly in the Southern United States during the early 1900s, where corn was a primary dietary staple 33. However, corn contains niacin in a bound, nutritionally unavailable form (niacytin), which requires alkali treatment to become bioavailable 34. This limitation in niacin bioavailability contributed to the deficiency.
In addition to dietary insufficiency, niacin deficiency can result from genetic disorders such as Hartnup disease 35, malabsorptive conditions, and the use of certain medications, such as those in isoniazid therapy 36. Although pellagra has been largely eliminated in industrialized countries thanks to improved nutrition, certain populations remain at risk, including individuals with chronic alcoholism, malabsorptive diseases, and eating disorders such as anorexia nervosa.
Beriberi
Beriberi presents in two primary forms: dry and wet beriberi. Dry beriberi predominantly affects the nervous system, causing symmetrical peripheral neuropathy, muscle weakness, and sensory deficits, especially in the extremities. Wet beriberi, on the other hand, primarily impacts the cardiovascular system, leading to heart failure, edema, and fluid retention. Patients with wet beriberi often experience tachycardia and dilated cardiomyopathy 37.
A severe complication of thiamine deficiency is Wernicke’s encephalopathy, which is characterized by a triad of ocular abnormalities (nystagmus and ophthalmoplegia), gait disturbances, and confusion. In advanced stages, untreated thiamine deficiency can lead to memory loss, psychosis, and ultimately death 38.
Dietary sources of coenzymes
Good dietary sources of thiamine include peas, whole grains, nuts, bananas, and fortified cereals. Riboflavin can be found in dairy products, eggs, and mushrooms. Niacin is abundant in meat, fish, eggs, and wheat flour. Chicken, beef, liver, eggs, mushrooms, and avocados are rich in pantothenic acid. Vitamin B6 is present in pork, poultry, fish, peanuts, oats, and bananas 39.
CoQ 10 is most concentrated in meat, fish, nuts, and certain oils, with smaller amounts found in dairy products, vegetables, fruits, and cereals. A well-balanced diet incorporating these sources of both vitamin and non-vitamin coenzymes is essential for maintaining optimal cellular function and preventing a range of diseases 40.
Applications in medicine and biotechnology
Research has underscored the presence of CoQ 10 in cellular membranes and blood plasma, highlighting its potent antioxidant properties 41.These functions are essential for the clinical applications of CoQ 10. Recent studies have also demonstrated the influence of CoQ 10 on gene expression, particularly in relation to cell signaling, metabolism, and transport.
CoQ 10 deficiencies can arise from various factors, including genetic mutations, mitochondrial diseases, age-related oxidative stress, cancer, and as a side effect of statin medications 42. Low levels of CoQ 10 have been associated with numerous conditions, including neurodegenerative disorders, diabetes, cancer, fibromyalgia, and cardiovascular diseases 42, 43.
CoQ 10 supplementation is generally safe and does not typically cause serious adverse effects. New formulations have improved absorption and tissue distribution of CoQ10, enhancing its effectiveness for mitochondrial energy deficiencies and antioxidant support, often providing significant symptomatic relief 44.
The protein engineering of oxidoreductases making use of nicotinamide-based coenzymes has significantly advanced synthetic biology applications. Coenzymes such as NAD and NADP are critical for most oxidoreductases, playing key roles in both catabolic and anabolic processes. The inherent preference of these enzymes for one coenzyme over another, depending on their metabolic function, can result in imbalances that hinder the efficiency of synthetic biology pathways 45.
To address these challenges, engineering the coenzyme preference of NAD(P)-dependent oxidoreductases has become a focus, particularly for large-scale industrial applications in synthetic biology. Cutting-edge engineering techniques, including rational design, semi-rational design, random mutagenesis, and directed evolution, combined with high-throughput screening, are being used to optimize the coenzyme preference of these enzymes. Furthermore, next-generation biomimetic nicotinamide-based coenzymes have been developed as alternatives to natural coenzymes, offering improved stability and cost-effectiveness.
Future perspectives on coenzymes
Coenzymes will continue to play a critical role in advancing health and longevity by supporting essential metabolic and antioxidant processes.
Maintaining CoQ 10 homeostasis through dietary supplementation and other pro-longevity interventions can help mitigate age-related mitochondrial dysfunction and oxidative damage, contributing to healthier aging. Emerging research indicates that NAD⁺ precursors, such as NMN and NR, activate sirtuins and enhance genomic stability, providing promising anti-aging strategies 46.Furthermore, CoQ 10 supplementation has shown considerable potential in improving glycemic control and reducing risks associated with hyperglycemia-related conditions 47.
Innovative therapies targeting coenzyme activity focus on modulating key metabolic pathways, such as those involving acetyl-CoA, to disrupt cancer progression and improve treatment outcomes. These strategies aim to reduce tumor growth and address metabolic reprogramming in cancer cells by inhibiting dysregulated enzymes such as ATP citrate lyase and acetyl-CoA synthetase 48.
Preclinical evidence supports the efficacy of pharmacological agents, including statins and enzyme inhibitors, in modulating acetyl-CoA metabolism. Additional therapeutic strategies focus on the selective depletion of coenzymes, such as NADPH and glutathione, thereby disrupting redox homeostasis in cancer cells and exacerbating oxidative stress. Notably, dual-targeting approaches that combine NADPH depletion with PARP inhibition have shown synergistic anti-tumor effects 49.
Furthermore, advanced nanocarriers, such as sorafenib-loaded frameworks with polyphenol coatings, have shown potential in targeting hypoxic tumor microenvironments. These systems induce apoptosis by depleting coenzymes and generating reactive oxygen species 50.
Conclusion
From their origins as prebiotic catalysts to their contemporary functions in health and disease, coenzymes exhibit significant evolutionary conservation and functional versatility, highlighting their role as essential molecular partners in the biochemical processes that sustain life. The dynamic interaction between enzymes and coenzymes, illustrated by the apoenzyme-holoenzyme transition, enables precise control over metabolic flux, with coenzyme specificity and reusability ensuring both efficiency and regulatory complexity.
Vitamin-derived coenzymes, such as NAD⁺ and FADH₂, are integral to maintaining redox balance, and non-vitamin cofactors, including ATP and CoQ 10, regulate energy metabolism and electron transport. Clinically, deficiencies in coenzymes result in severe nutritional disorders like pellagra and beriberi, underscoring their physiological importance. Moreover, therapeutic modulation of coenzyme pathways holds promise for interventions in aging and cancer treatment.
Emerging biotechnological applications are increasingly using engineered coenzyme specificity in industrial biocatalysis, with techniques such as directed evolution enhancing metabolic engineering, exemplified by the development of enzyme mutants such as BsMDH-T7. The advent of advanced delivery systems, such as NAD⁺ precursors and CoQ 10 nanocarriers, further underscores the translational potential of coenzymes in mitochondrial medicine.
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Algebra 1
5.1.1 Understanding Exponents
Algebra 15.1.1 Understanding Exponents
Contents Contents
Highlights
Table of contents
Getting Started
Supporting All Learners
1 Linear Equations
2 Linear Inequalities and Systems
3 Two-Variable Statistics
4 Functions
5 Introduction to Exponential Functions
Unit 5 Overview and Readiness
5.1 Properties of Exponents
5.1.0 Lesson Overview
5.1.1 Understanding Exponents
5.1.2 Using Product and Quotient Properties for Exponents
5.1.3 Using Zero Exponent Property and Negative Exponents
5.1.4 Using Power Properties for Exponents
5.1.5 Simplifying Exponents
5.1.6 Practice
5.1.7 Lesson Summary
5.2 Rational Exponents
5.3 Patterns of Growth
5.4 Representing Exponential Growth
5.5 Representing Exponential Decay
5.6 Negative Exponents and Scientific Notation
5.7 Analyzing Graphs
5.8 Exponential Situations as Functions
5.9 Interpreting Exponential Functions
5.10 Looking at Rates of Change
5.11 Modeling Exponential Behavior
5.12 Reasoning about Exponential Graphs, Part 1
5.13 Reasoning about Exponential Graphs, Part 2
5.14 Which One Changes Faster?
5.15 Changes Over Equal Intervals
Project 5: Introduction to Exponential Functions
Unit 5 Wrap Up
6 Working with Polynomials
7 Introduction to Quadratic Functions
8 Quadratic Equations
9 More Quadratic Equations
Research in Practice
Appendix
Index
Search for key terms or text.
Close
5.1.1 • Understanding Exponents
Warm Up
An exponent indicates repeated multiplication of the same quantity.
Expand each exponential expression to show repeated multiplication.
4 3 4 3 4 3
Compare your answer:
4⋅4⋅4 4·4·4 4·4·4
x 5 x 5 x 5
Compare your answer:
x⋅x⋅x⋅x⋅x x·x·x·x·x x·x·x·x·x
(−8)6(−8)6(−8)6
Compare your answer:
(−8)⋅(−8)⋅(−8)⋅(−8)⋅(−8)⋅(−8)(−8)·(−8)·(−8)·(−8)·(−8)·(−8)(−8)·(−8)·(−8)·(−8)·(−8)·(−8)
(−z)3(−z)3(−z)3
Compare your answer:
(−z)⋅(−z)⋅(−z)(−z)·(−z)·(−z)(−z)·(−z)·(−z)
PreviousNext
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Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis, Jay Abramson, Sharon North, Amy Baldwin, Alyssa Howell
Publisher/website: OpenStax
Book title: Algebra 1
Publication date: Jun 4, 2025
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分母（除数）为什么不能为 0？ - 知乎
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分母（除数）为什么不能为 0？
数学总是很严谨的，正如偶数次根号下不能为负数（实数范围内）是因为找不到一个实数的平方等于负数。 小学的时候，数学老师说除数（分母）不能为零的原因是除法…显示全部
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果壳搬运，侵删。。。
如果你问苹果手机上的Siri，“零除以零等于多少”，它会显示：
但是，英文版的Siri还会用语音说这一段话：
“假如你有0块饼干，要分给0个朋友，每个人能分到几块？你看，这个问题没有任何意义吧？甜饼怪会难过，因为没有饼干吃，而你也会难过，因为你一个朋友都没有。”
（中文版也会，但言辞就没那么伤人了……）
抛开这个伤人的回答不论（有朋友谁特么会跟你聊天啊喂！），除以零确实是个困扰很多人的问题。十除以二等于五，六除以三等于二，一除以零是多少？小学数学就会告诉你，答案是不能除。但是为什么？零也是个数字，它到底哪里特殊了？
小学篇
小学算术里，这个问题很简单。那时我们把除法定义成“把一个东西分成几份”，分成一二三四五六七份都很容易想象，但是你要怎么把10个饼干分给0个人呢？想象不出来嘛！所以不能除。
敏锐的同学可能会想到，要是0个饼干分给0个人的话，本来无一物，好像就没关系了。但既然无物也无人，每个人分得多少都是可能的呀，根本无法给出一个单一确定的数值。
这结论没错，但这都是凭直觉而得到的东西。你想象不出来，不一定意味着它没有。远古时代的数学是建立在直觉上的，买菜是够用了，但要进一步发展，就必须要有定义和证明——所以，我们上了中学。
初中篇
现在我们开始接触最最基本的代数学——也就是解方程。我们发现，除法和乘法互为逆运算，所以问
1 / 0 = ?
就等于是解方程
0 x = 1
好了，按照定义，0乘以任何数都是0，不可能等于1，所以满足x的数字不存在，所以不能除。
同样，如果问
0 / 0 = ?
就等于是解方程
0 x = 0
同理，任何数字都可以满足x，所以也不能除——无法确定一个单一的答案。
高中篇
等到接触了基本的形式逻辑，我们又会发现另一种证明方式：反证法。
一堆真的表述，不能推出一个假的表述，所以如果我们用“能够正常地除以零”加上别的一堆真表述，最后推出假的来，那只能说明“除以零”这件事情不成立了。
所以，已知
0 1 = 0
0 2 = 0
推出 0 1 = 0 2
两边同时除以零，得到 ( 0 / 0 ) 1 = ( 0 / 0 ) 2
化简得到 1 = 2。这显然是错的啦。
那么，问题解决了吧！其实还没有。想想另一个问题：-1的平方根是多少？
你可能会说，-1不能开平方根，因为所有数的平方都是非负的。但是这说的是实数，我要是增加一个定义呢？定义i^2=-1，这就创造出了虚数，于是-1也能开平方根了。
那么，为何不能定义一个“新”的数，让 1 / 0 也等于它，并为这个数设立一套运算法则呢？这就得去大学里回答了。
大一篇
刚学微积分课程就会立刻接触到∞这个符号。咦，这不就是“无限”嘛。我们都学了极限的概念了，那么我令b趋向于0，然后把a/b的极限定义为无穷，不行吗？
这就立刻遇到一个问题，它的左极限和右极限不一样啊。b是从负的那头靠近0，还是正的那头？这一个是越来越负，一个是越来越正，碰不到一起去。这样的极限是没法定义的。
因此，微积分课程里会反复说，虽然用到了∞这个符号，但是这只是代表一个趋势，绝对不是一个真正的数，不可参与运算。
大二篇
那么吸取教训，我不用现成符号了，我直接定义 1 / 0 = w，w是个“无限大”的数，不碰什么极限，你总没话说了吧！
然而，定义不是说来就来的，你虽然可以随便定义东西，但定义完了如果和现有的其他系统矛盾，那就不能用，或者很不好用。
而我们面对w立刻就遇到了问题。首先，w要怎么放入基本的加减乘除体系里？1 + w等于多少？w - w等于多少？如果你造了一个数，却连加减乘除都不能做，那就不是很有用对吧。
比如直觉上，1 + w 应该等于 w，它都无限了嘛！ 而 w - w 则等于0，自己减自己嘛！
但这样立刻会和加法里极其重要的“结合律”产生矛盾： 1 + ( w - w ) = 1 + 0 = 1，可是( 1 + w ) - w = w - w = 0。结合律是加法里非常基本的东西，为了一个w，连结合律都不要了，这成本有点大——不光是结合律本身，多少数学定理证明过程中不自觉都用了它，扔了它就都得重来，建立新体系。新体系不是不能建，但是费心费力又（暂时）无卵用，所以大家还是在老实用旧的——而旧的里面，为了保住结合律，就不能这么玩。
欢迎读者们发挥自己的想象力，尝试为 w 给出运算方式。但是你会发现，无论怎么规定w和别的数字之间的关系，只要你还坚持 1 / 0 = w，你就没法让它和你从小学习的基本数学不矛盾。还是那句话，你可以另立门户，在w的基础上建立起你的新数学，但它和大部分传统数学是不相容的，而且肯定会非常不好用，所以我们用了一个不能除以零的体系是非常合理的。
大三篇
你可能会提出反对：有那么多的定义方式，我都试过？要是没试过，我怎么知道不会某一天冒出来一个能够自洽的办法？
“新发现推翻旧结论”这种事情，在生物里可以有，化学里可以有，物理里可以有，唯独数学里没有。因为数学建立在逻辑上，个案有例外，逻辑没有例外。当然我们的数学还没有完成最终公理化，还要面对哥德尔的幽灵，但至少在这个例子里，如果w是一个真正的数，那它就违反了一些非常重要的公理，而这些公理的地位可是非常之深。
比如有一组基本的公理叫“皮亚诺公理”，其中有一条说，每一个确定的自然数都有一个确定的后继，后继也是自然数；另一条说，自然数b=c，当且仅当b的后继=c的后继。
那w是谁的后继呢——或者说，谁加上1能得到w呢？显然所有其他的数字都已经有了自己的后继，w在其中没有位置，没有任何其他的数加上1能成为w。那么就只能是1+w=w了，可那就直接和第二句话矛盾。而没有皮亚诺公理，整个自然数的体系都不能成立。
这里假定w是自然数。其他情况会略微复杂一些，但无论如何，类似的事情发生在w的各种定义里。如果你想把w当成一个数，那就没法和我们现有的实数兼容。所以我们在几乎所有场合下都只能宣布，不能除以0。
大四以上篇
既然我们之前说了个“几乎”，那就是有例外的——在个别奇葩场合下，可以。
比如有一个东西叫做“复无穷”，它是扩充复平面上的一个点，真的是有定义的一个点。在这个特殊的规则下你可以写下 1 / 0 = ∞ 这样一个表达式。这么做的原因就说来话长了，但它不是平常意义上的运算——比如你不能把0拿回来，不能写 1 = 0 ∞。
另外，“无穷”二字在一些别的场合下是可以当成一个“东西”去对待的。比如当你衡量一个集合的大小的时候，它可以是无穷大的。但这就有很多种不同的无穷大了——自然数是无穷多的，有理数是无穷多的，实数也是无穷多的，可是奇数和偶数和正整数和负整数和自然数和有理数都一样多，而实数却比它们都多！同样是无穷，有的无穷比别的无穷更无穷。但这就是另一个话题了，打住。
总结篇
所以，当我们说不能除以零的时候，理由……竟然出乎意料地充足。有许多直觉在数学里被推翻了，但是这一条没有。我们有种种数学上的方式去证明它无法成立的原因，虽然也许听起来不如Siri的回答那么心暖（或者心寒），但这些理性的愉悦也是一种美丽，对吧？
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除数，是为了处理分配情况的。
举例，6个苹果，分给每人两个，可以分给3个人，分完之后苹果没有剩余。
如果6个苹果，分给每人0个，可以分给无限个人，但是却分不完，最后还是有6个苹果剩余。因此0不能作为除数或分母。
发布于 2013-02-23 12:09
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2019 科学季
如果在一个集合 S S 中定义一种运算 f f ，并且该运算满足：
f(f(a,b),c)=f(a,f(b,c))f(f(a, b), c)=f(a, f(b, c))
∀a,b∈S|f(a,b)∈S\forall a, b\in S|f(a, b)\in S
则称 ⟮S,f⟯\lgroup S, f\rgroup 为半群，若 ∃e∈S|f(a,e)=f(e,a)=a\exists e\in S|f(a, e)=f(e, a)=a 则称⟮S,f⟯\lgroup S, f\rgroup 为含幺半群（简称幺半群）， e e 被称作幺元。若 ∃o∈S|f(o,a)=f(a,o)=o\exists o\in S|f(o, a)=f(a, o)=o ，则称 o o 为半群⟮S,f⟯\lgroup S, f\rgroup 的零元。
设含幺半群⟮S,f⟯\lgroup S, f\rgroup 的幺元为 e e ，若 ∀a∈S|(∃x∈S|f(a,x)=f(x,a)=e)\forall a\in S|(\exists x\in S|f(a, x)=f(x, a)=e) ，则称 x x 是 a a 的逆元，暂且将其记作 a−1 a^{-1} ，这时含幺半群就变成了群。
实际上这个问题就是问，为什么零元不存在逆元。
零元不存在逆元，可以符号化为（假设幺元不等于零元）：
\forall a\in S|f(o, a)=f(a, o)=o\rightarrow\forall x\in S|x\neq o^{-1}
这个命题不太好证，我们用假言易位律对上述命题做等价变化，得
\exists x\in S|x=o^{-1}\rightarrow\exists a\in S|f(o, a)=f(a, o)\neq o
然后我们证明这个命题，就可以说明零元不存在逆元（这就是反证法）
\because\exists x\in S|x=o^{-1}
\therefore f(o, x)=f(x, o)=e\neq o
\therefore\exists a\in S|f(o, a)=f(a, o)\neq o
也就是说，零元必须和任何元经过运算 f 都得到零元，但与其逆经过运算 f 又必须得到幺元，只有幺元和零元相等才有可能出现这种情况，而在数域中， 0 就是 0 ， 1 就是 1 ，所以 0 自然不存在乘法逆元。
定义，若群\lgroup S, f\rgroup 是阿贝尔群，则与一个数的逆之间进行运算 f 称作和这个数进行 f 的逆运算。（也就是说，小学学习的分数除法的运算方法，实际上是用来定义除法的。之所以除以一个数就是乘上这个数的倒数，乃是因为乘上一个数的倒数被定义为除以这个数，这不是循环定义，因为分数不是自然数，只能通过将自然数的乘法半群扩充成群来定义）
既然 0 没有乘法逆元，自然 0 就不存在倒数，也就不存在与 0 的倒数相乘的情形，也就不存在除以 0 的情形， 0 也就做不了除数。
不过，在微积分中，“零除数”另有用途，但只适用于形式化极限计算，而不能进行严格的数学计算和推理证明。虽然我们可以认为 \frac{1}{0}=\infty ，但这并不意味着 \infty 是“实数乘法群”中 0 的逆元，计算机的浮点数运算都会做这样的处理：正数除以零得正无穷，负数除以零得负无穷，零除以零得到非数。
形式化计算有助于我们判定极限是可以直接通过函数的连续性来求，还是必须要经过一定的处理来求，同时也有助于计算极限，通常认为 \frac{a}{\infty}=0(a\neq\infty) 。
无穷大和无穷小的运算法则是（设 a\neq0\wedge a\neq\infty ）
0\pm0=0
0\pm a=\pm a, a\pm0=a
0\pm\infty=\pm\infty, \infty\pm0=\infty
a\pm\infty=\pm\infty, \infty\pm a=\infty
\infty\pm\infty 是未定式
0\times0=0, 0\times a=0
a\times\infty=\infty, \infty\times\infty=\infty
0\times\infty 是未定式（也就是说0和无穷大相乘不一定是1）
\frac{0}{a}=0, \frac{0}{\infty}=0, \frac{a}{\infty}=0
\frac{a}{0}=\infty, \frac{\infty}{a}=\infty, \frac{\infty}{0}=\infty
\frac{0}{0} 和 \frac{\infty}{\infty} 均是未定式
无穷大不是一个数，它的运算不符合数的运算的特性，自然我们也不可能将之放入群中，更不可能放入域中，数域中的零还是不能做除数。零分母和无穷大，只能用于形式计算。
关于比的后项
如果定义两个数相除又叫两个数的比，那么零自然不能作为比的后项，但是比和除法的功能毕竟不一样。除法是为了已知积和一个乘数求另一个乘数（有时也叫因数），而比是为了反映两个量之间的倍分关系，只不过前项除以后项得到的商可以被称作比值。实际上，比应该这样定义：
设 a\geqslant0, b\geqslant0 ，若 a=mx, b=nx 则 a:b=m:n 。这样实际上比的后项可以是零，它表示处在后项上的量等于零。比如电压比 1:0 表示一个元器件两端有电压，另一个元器件两端没有电压。而 0:0 表示两个量均为零。 0 可以做比的后项，只是没有比值（当前项不为零）或者比值不确定（当前项为零）而已。
这样实际上方便我们表示直线的点向式方程，我对直线下的定义是：若图形各点位矢满足 \vec{r}=\vec{r}_0+\vec{l}t(\vec{l}\neq\vec{0}) ，那么该图形就是直线。也就是说，点对应的位矢的各个分量与定点对应的位矢的各个分量之差与方向向量的各个分量成正比。而方向向量虽然不是零向量，亦即方向向量各分量不全为零，但不全为零不是全不为零，所以还是会有几个分量等于零的。因而我们需要零分母帮我们写出这样的直线的点向式方程，此时的零分母实际上就是等于零得比的后项，此时比的前项也必须为零。
0:0 的比值不确定，可以认为这个比值是任意的，但如果有两个比值同时等于 0:0 ，那么这两个比值必须相等，也就是说，如果 a:b=c:d=0:0 （ b\neq0, d\neq 0 ），那么必须有 \frac{a}{b}=\frac{c}{d}
这样类似 \frac{x-x_0}{a}=\frac{y-y_0}{0}=\frac{z-z_0}{c} 的直线方程就也可以表示一条直线了，其中 y\equiv y_0
P. S.：在群论中，零元很好理解，幺元这个词有点意思。幺元也称单位元，单位元这个词也好理解，因为 1 表示的就是单位量。而“幺”这个字来源于我们平时对数 1 的称呼，有的时候我们看到1不读“一”而是读作“幺”，比如手机号码187读作“幺八七某某某某某某某某”，而且如果说单位元是“一元”听着也别扭，所以在群论中，单位元也就有了幺元这个称呼。
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Dividing Fractions | A Complete Step-By-Step Guide (Learn Everything You Need to Know)
Math with Mr. J
1700000 subscribers
2033 likes
Description
164992 views
Posted: 15 Jun 2023
Welcome to Dividing Fractions: A Complete Step-by-Step Guide with Mr. J! Need help with how to divide fractions? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with dividing fractions. Mr. J will go through examples of dividing fractions by fractions, dividing whole numbers and fractions, dividing mixed numbers, dividing fractions using cancellation, and dividing improper fractions.
✅ Chapters and Timestamps
00:00 - Dividing a Fraction by a Fraction
4:39 - Dividing a Fraction by a Whole Number
7:37 - Dividing a Whole Number by a Fraction
11:16 - Diving a Mixed Number by a Mixed Number
16:22 - Dividing a Mixed Number by a Whole Number
20:25 - Dividing a Whole Number by a Mixed Number
24:38 - Dividing a Mixed Number by a Fraction
28:13 - Dividing a Fraction by a Mixed Number
31:01 - Dividing Fractions Using Cancellation
37:40 - Dividing Mixed Numbers Using Cancellation
43:18 - Dividing an Improper Fraction by an Improper Fraction
46:29 - Dividing an Improper Fraction by a Whole Number
48:24 - Dividing a Whole Number by an Improper Fraction
51:30 - Why Do We Use Keep, Switch, Flip?
About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free. #MathWithMrJ
Click Here to Subscribe to the Greatest Math Channel On Earth:
Follow Mr. J on Twitter: @MrJMath5
Email: math5.mrj@gmail.com
Music:
Hopefully this video is what you're looking for when it comes to dividing fractions.
Have a great rest of your day and thanks again for watching! ✌️✌️✌️
161 comments
Transcript:
Dividing a Fraction by a Fraction welcome to math with Mr Jay [Music] in this video I'm going to cover everything you need to know about dividing fractions we will go through how to divide fractions by fractions fractions and whole numbers mixed numbers improper fractions how to use cross cancellation and we will go through a bunch of examples everything is time stamped and put into chapters so feel free to jump around check the description for the list of chapters and time stamps let's start by taking a look at dividing fractions by fractions now when working with fractions and division we need to remember three steps keep switch flip we can also think multiply by the reciprocal when we flip a fraction that's called the reciprocal so this means we're going to use multiplication to solve these this actually makes the process simpler now how and why this works that's something I'll cover later in this video Let's jump into number one where we have three fourths divided by one-seventh so we are seeing how many one-sevenths we can make out of three-fourths or how many one-sevenths are in three-fourths so to speak so let's start by rewriting this problem using keep switch flip so we always keep the first fraction or number so we keep three-fourths as is we switch to the opposite or inverse of division which is multiplication and since we switch to the opposite or inverse of division we're going to flip the second fraction so the 7 is going to be our numerator and the one is going to be our denominator so we end up with seven over one once we get to this point it's a multiplying fractions problem so we can multiply straight across three times seven is twenty one and then four times one is four and we end up with 21 4. now that's an improper fraction so we do not want to leave it we need to divide our numerator of 21 by 4. how many whole groups of four are in 21 well 5 that gets us to 20. so five whole groups of four that's our whole number of the mixed number here now that doesn't hit 21 exactly we have a remainder of 1 so that's our numerator in our fractional part of the mixed number and we keep our denominator of 4 the same always look to see if you can simplify the fractional part of a mixed number one fourth is in simplest form so we are done five and one fourth is our answer for number one so three fourths divided by one-seventh equals five and one fourth let's move on to number two where we have two-thirds divided by two-fifths keep switch flip so keep two-thirds here switch to multiplication and then flip that second fraction so five over two and now we can multiply straight across two times five is ten three times two is six so we get ten sixths so that's an improper fraction we need to convert to a mixed number so 10 divided by six how many whole groups of six out of ten well one and we get a remainder of four keep our denominator of six the same we can simplify 4 6 we have a common factor of two so let's divide each of these by two here and we get one and two thirds so our answer for number two is one and two thirds two-thirds divided by two-fifths equals one and two-thirds there's how to divide a fraction by a fraction let's move on to dividing a fraction by a whole number now we will take a look at dividing a Dividing a Fraction by a Whole Number fraction by a whole number let's jump into number one where we have one-half divided by three so we are taking one half and dividing it into three equal groups or parts let's rewrite the problem using keep switch flip in order to solve this so always keep the first fraction or number so let's rewrite the problem so one half we are going to keep then we switch to the opposite or inverse of division which is multiplication and since we switched to multiplication we need to flip the second number or fraction in the case of number one we have a whole three here so how do we flip this well let's put that whole number over one that way we have a numerator and a denominator that does not change the value of the problem that still has a value of three so now we can flip so the one is going to go up to the top and be our numerator and the three is going to be the denominator so we flipped that second number once we get to this point we have a multiplying fractions problem so we can multiply straight across 1 times 1 is 1. and 2 times 3 is 6. so 1 6 is our final answer always check to see if you can simplify 1 6 is in simplest form so we are done one-half divided by three equals 1 6. let's move on to number two where we have four sevenths divided by six keep switch flip so let's rewrite the problem keep four sevenths switch to multiplication and then we need to flip so we can write that six in fractional form and now we can flip so the one is now our numerator and the six is now our denominator and we can multiply straight across four times one is four seven times six is forty-two so we get 4 over 42 which can be simplified 4 and 42 have a greatest common factor of two so let's divide each of these by two and we get 2 over 21 for our final simplified answer so there's how to divide a fraction by a whole number let's move on to dividing a whole number by a fraction now we will Dividing a Whole Number by a Fraction take a look at dividing a whole number by a fraction let's jump into number one where we have 2 divided by 1 4. now we can think of this as how many one-fourths are in two or how many one-fourths can we make out of two we are dividing two by one fourth let's rewrite our problem using keep switch flip also known as multiplying by the reciprocal so we start with keep we always keep the first number or fraction in the case of number one we have a whole two so we need to keep that as two but we're going to rewrite it as a fraction whenever you rewrite a whole number and put it in fractional form all you need to do is put it over one now the purpose of this we want to have a numerator and a denominator so we kept the value of 2 there we just put it in fractional form then we switch from division to multiplication and since we switched to the inverse or opposite of division we're going to need to flip the second number or fraction in the case of number one here we have a fraction so the denominator is now going to be the numerator and the numerator is now going to be the denominator so we have 4 over 1. once we get to this point we have a multiplying fractions problem which we just multiply straight across when we multiply fractions so 2 times 4 is 8 and 1 times 1 is 1. so we get to the answer of 8 over 1 which is an improper fraction so we do not want to leave it like that this is going to be eight our final answer is eight two divided by one fourth equals eight let's move on to number two where we have seven divided by two fifths so we are finding out how many two-fifths are in seven so to speak so keep switch flip we will keep our seven but put it in fractional form by putting it over one switch to multiplication and then flip our second number or fraction now we can multiply straight across seven times five is 35 and 1 times 2 equals two so our answer is 35 over 2. now we don't want to leave it improper we want to convert this to a mixed number so we do 35 divided by 2. how many whole groups of two are in 35 well 17. so that's our whole number of our mixed number now we didn't hit 35 exactly we had a remainder of 1 so that's our numerator of the fractional part and then we keep our denominator of 2 the same so 17 and a half always look to see if you can simplify the fractional part one-half is in simplest form so we are done our final answer 17 and a half so there's how to divide a whole number by a fraction let's move on to dividing a mixed number by a mixed number now we will take a look at dividing a Diving a Mixed Number by a Mixed Number mixed number by a mixed number let's jump into our example where we have 4 and 3 4 divided by two and one-third now the first thing that we want to do is to convert these mixed numbers to improper fractions that way we just have a numerator and a denominator let's start with four and three fourths and we start at the bottom and work our way to the top so we multiply then add so we do 4 times 4 so the denominator times the whole number four times four is sixteen then we add the numerator sixteen plus three is nineteen so nineteen is the numerator of our improper fraction we keep our denominator of 4 the same then we can bring our division sign down and convert our second mixed number so again we multiply then add so three times two is six plus one is seven that's our numerator and then keep the denominator of 3 the same so 7 3 19 4 is equivalent to four and three-fourths and then seven thirds is equivalent to two and one-third so we're not changing the value of the problem at all when we convert to improper fractions mixed numbers and improper fractions are equivalent they're just in different forms now again we're converting to improper fractions so we just have a numerator and a denominator and then we can go through our steps for dividing these fractions which we're going to use keep switch flip also known as Keep Change Flip so no matter how you think of it they mean the same thing I know there's different words out there for these steps but in the end they all mean the same thing so let's keep our first fraction here our first improper fraction so nineteen fourths then we switch or change our division to multiplication and then we flip our second fraction our second improper fraction here so the denominator becomes the numerator and the numerator becomes the denominator this is called the reciprocal so we have three sevenths once we get to this point we can multiply straight across so numerator times numerator and then denominator times denominator let's do the numerators first so 19 times 3 that gives us 57 and then we have 4 times 7 which is twenty eight now that's our answer as an improper fraction we're going to convert it to a mixed number so we need to do our numerator 57 divided by our denominator 28. now we can do this using mental math or we can write everything out some problems are going to be a little simpler and we're going to be able to use Mental Math and in some situations we'll have to write out the problem I'll go through both here so you can see what that looks like let's use Mental Math here first so how many whole groups of 28 out of 57 well if you're able to think about that it's going to be two whole groups of 28 because that gets us to 56 so two whole groups of 28 now we do not hit 57 exactly we have something left over a remainder so the difference between 57 and 56 is one so we have a remainder of 1 and we keep our denominator of 28 the same let me write that out as well so we have 57 divided by 28. so we need to think how many whole groups of 28 out of 57 well we said that's two that gets us to 56 so 2 2 times 28 is 56 so we get a remainder of 1. this 2 is going to be the whole number part of our mixed number 2 whole 28 out of 57 and then this one the remainder is going to be the numerator and then we keep our denominator of 28 the same once we get to our mixed number we can check to see if we can simplify the fractional part 1 28 is in simplest form the only common factor between 1 and 28 is one so we are done 2 and 1 28th so there's how to divide a mixed number by a mixed number let's move on to dividing a mixed number by a whole number now let's take a look at dividing a mixed number by a whole number let's Dividing a Mixed Number by a Whole Number jump into our example where we have 5 and 4 9 divided by three now the first thing that we need to do is rewrite everything in fractional form so we're going to convert everything to improper fractions that way we just have numerators and denominators so we can convert 5 and 4 9 to an improper fraction first we start at the bottom and multiply and then add so take the denominator and multiply it by the whole number so 9 times 5 that gives us 45 and then we add the numerator so add 4 45 plus 4 gives us 49. and we keep the denominator of 9 the same divided by three so how do we write 3 in fractional form well we can write any whole number in fractional form by putting it over one so now we have 49 9 divided by three over one now let's keep switch flip so keep 49 9 switch to multiplication and then flip 3 over 1. so 1 is now the numerator and 3 is now the denominator so 49 9 times 1 3 Let's multiply straight across starting with the numerators so 49 times 1 gives us 49. now for the denominators 9 times 3 gives us 27 so we end up with 49 27 which is our answer but it's an improper fraction so let's convert it to a mixed number we do that by dividing the numerator 49 by the denominator 27 so 49 divided by 27 how many whole groups of 27 in 49 well one now we don't hit 49 exactly we have a remainder a remainder of 22. that is the numerator of the fractional part of the mixed number and then we keep the denominator of 27 the same so we end up with 1 and 22 27 we can always look to see if we can simplify the fractional part of a mixed number the only common factor between 22 and 27 is one so that is in simplest form we are done 1 and 22 27 just for a quick recap I'm going to write out how we went from that improper fraction of 49 27 to the mixed number of 1 and 22 27 so we divided the numerator 49 by the denominator 27 so 49 divided by 27. how many whole groups of 27 in 49 well one that is the whole number of our mixed number 1 times 27 is 27. subtract so 49 minus 27 9 minus 7 is 2 4 minus 2 is 2 as well so we get a remainder of 22. that is the numerator of the fractional part of the mixed number and then we keep the denominator of 27 the same so 1 and 22 27. so there's how to divide a mixed number by a whole number let's move on to dividing a whole number by a mixed number Dividing a Whole Number by a Mixed Number now let's take a look at dividing a whole number by a mixed number let's jump into our example where we have 10 divided by 4 and a half now the first thing that we need to do is rewrite this problem with everything in fractional form so everything Rewritten as an improper fraction that way we just have numerators and denominators and we can go through our steps keep switch flip so let's start by rewriting 10 in fractional form as an improper fraction and we do that by putting it over one remember we can rewrite any whole number in fractional form by putting it over one divided by and then as far as 4 and a half we need to convert that to an improper fraction by working our way up from the bottom so start at the bottom and work our way up we multiply and then add so multiply the denominator by the whole number so 2 times 4 is 8 and then add the numerator 8 plus 1 is 9. that's the numerator of our improper fraction and then keep the denominator of 2 the same so we have 10 over 1 divided by nine over two now we just have numerators and denominators and we are able to keep switch flip so let's work through our steps we keep 10 over 1 switch to multiplication and then flip nine over two so two is now the numerator and 9 is now the denominator now we can multiply straight across let's start with the numerators 10 times 2 is 20. now for the denominators one times nine this is nine so we end up with 29 which is our answer but it's an improper fraction so let's convert it to a mixed number by dividing the numerator 20 by the denominator 9. so we have 20 divided by 9. how many whole groups of nine in 20 well two whole groups of nine in 20. that's the whole number of our mixed number now we don't hit 20 exactly two whole groups of nine gets us to 18. so we have a remainder of two that is the numerator of the fractional part of the mixed number and then we keep the denominator of nine the same so we end up with two and two ninths we can always look to see if we can simplify the fractional part of a mixed number the only common factor between two and nine is one so this is in simplest form and we are done two and two ninths now I do want to go through a quick recap as far as how we went from that improper fraction of twenty-ninths to the mixed number of 2 and 2 9. so again we divided the numerator 20 so I'm going to write this out off to the side here so numerator 20 divided by the denominator 9. so how many whole groups of 9 and 20 20 divided by 9. well 2 and that 2 is the whole number of the mixed number now we multiply two times nine is 18. subtract 20 minus 18 is 2. so we get a remainder of 2. that remainder is the numerator of the fractional part of the mixed number and then we keep the denominator of 9 the same so there's how to divide a whole number by a mixed number let's move on to dividing a mixed number by a fraction now let's take a Dividing a Mixed Number by a Fraction look at dividing a mixed number by a fraction let's jump into our example where we have 2 and 4 7 divided by one-fifth now the first thing that we need to do is convert the mixed number to an improper fraction that way we just have a numerator and a denominator and we can go through our dividing fractions steps keep switch flip so let's convert 2 and 4 7 to an improper fraction so we start at the bottom we multiply and then add so the denominator times the whole number 7 times 2 is 14 and then add the numerator 14 plus 4 is 18. that is our numerator of the improper fraction and then we keep the denominator of 7 the same divided by one-fifth so I rewrote the problem with the mixed number as an improper fraction now let's go through our steps keep switch flip so keep 18 7. switch to multiplication and then flip one-fifth so 5 is now the numerator and one is now the denominator now we can multiply straight across let's start with the numerators 18 times 5 that gives us 90. now for the denominators 7 times 1 that gives us seven so we end up with ninety-sevenths which is our answer but it's an improper fraction so let's convert it to a mixed number we do that by dividing the numerator 90 by the denominator seven so we have 90 divided by seven so how many whole groups of seven in ninety well let's work through this off to the side here so 90 divided by seven nine divided by seven we have one one whole group of seven in nine multiply one times seven is seven subtract nine minus seven is two bring down and now we repeat so we divide twenty divided by seven how many whole groups of seven in twenty well two two times seven is fourteen subtract 20 minus fourteen gives us a remainder of 6. so 12 whole groups of seven in ninety that's the whole number of our mixed number and then we have a remainder of six we don't hit ninety exactly that remainder is the numerator of the mixed number and then we keep the denominator of 7 the same so we end up with 12 and 6 7. we can always look to see if we can simplify the fractional part of a mixed number 6 7 is in simplest form the only common factor between six and seven is one so we are done 12 and 6 7. so there's how to divide a mixed number by a fraction let's move on to dividing a fraction by a mixed number now let's take a look at dividing a Dividing a Fraction by a Mixed Number fraction by a mixed number let's jump into our example where we have three fourths divided by one and five sixths now the first thing that we need to do is rewrite this problem with the mixed number written as an improper fraction so we need to convert that mixed number to an improper fraction that way we just have numerators and denominators and we can go through our steps keep switch flip so let's rewrite this problem we have three fourths divided by and then as far as 1 and 5 6 let's convert this to an improper fraction we start at the bottom and work our way up we multiply and then add so take the denominator and multiply it by the whole number so 6 times 1 is 6 and then add the numerator 6 plus 5 gives us 11. that's the numerator of the improper fraction and then we keep the denominator of six the same so now we have 3 4 divided by 11 6. now let's go through our steps keep switch flip so keep three-fourths switch to multiplication and then flip 11 6 so 6 is now the numerator and 11 is now the denominator so we have three fourths times six elevenths now let's multiply straight across starting with the numerators so three times six is 18. now for the denominators 4 times 11 is 44. so we get 18 44 which is our answer but we can simplify this fraction we have a common factor other than 1 between 18 and 44. we have a greatest common factor of two so let's divide 18 by 2 and 44 by 2. 18 divided by 2 is 9. 44 divided by 2 is 22 so we end up with 9 20 seconds or 9 over 22. the only common factor between 9 and 22 is 1 so this is in simplest form and we are done so there's how to divide a fraction by a mixed number let's move on to using cross cancellation when we have a division Dividing Fractions Using Cancellation problem involving fractions now we will take a look at using cross-cancellation when we have a division problem involving fractions cancellation gives us smaller numbers in value to work with and easier numbers to work with this is a way to simplify the problem before multiplying and remember when we have a division problem involving fractions we end up multiplying let's jump into number one where we have 4 9 divided by 8 15. we can't use cancellation quite yet we can't use it right away we use cancellation when multiplying fractions so we need to wait until we go through the steps we use to divide fractions keep switch Flip or Keep Change Flip so let's keep the first fraction of 4 9 switch to multiplication and then flip the second fraction of 8 15. so 15 is now the numerator and 8 is now the denominator now we can multiply straight across but before doing so we can look to use cancellation when we use cancellation we look diagonally for common factors other than one that we can divide by think of this like simplifying fractions but we are looking diagonally so for number one we look at four and eight and then 9 and 15. let's look at four and eight first are there any common factors other than one that we can divide four and eight by yes let's use the greatest common factor of four so we can divide four and eight by four four divided by four is one eight divided by four is two now we can take a look at nine and 15. are there any common factors other than one that we can divide 9 and 15 by yes 3 is a common factor between 9 and 15 and it's the greatest common factor so let's divide 9 and 15 by 3. 9 divided by three is three fifteen divided by three is five now we can multiply straight across and we have smaller numbers in value to work with easier numbers to work with let's start with the numerators one times five is five now for the denominators three times two is six so we end up with five sixths the greatest common factor between five and six is one so we are in simplest form and this is our final simplified answer again five sixths now let's work through that problem without using cancellation and see what happens so keep switch flip so keep four ninths switch to multiplication and then flip 8 15 so now we have fifteen eighths let's multiply straight across 4 times 15 is 60. nine times eight is seventy-two so we get 60 70 seconds now that's different than the answer above that's because we haven't simplified yet are there any common factors between 60 and 72 other than one yes the greatest common factor between 60 and 72 is 12. so let's divide both of them by 12. 60 divided by 12 is 5. 72 divided by 12 is 6 so you can see that we get 5 6 that way as well we get the same answer either way now do we have to use cancellation no but it's a good strategy to be familiar with and use when possible it can be helpful let's move on to number two where we have 12 divided by 6 7. let's keep switch flip so keep 12 and I'm going to write it in fractional form that way we have a numerator and a denominator and we can multiply straight across remember we can rewrite any whole number in fractional form by putting it over one then we can switch to multiplication and flip so seven sixths now we can multiply straight across but before doing so we can look to use cancellation so let's look diagonally for common factors other than one let's look at 12 and 6 first are there any common factors other than one that we can divide by yes 6 is the greatest common factor between 12 and 6 so let's divide by six twelve divided by 6 is 2. six divided by six is one now let's take a look at one and seven are there any common factors between one and seven other than one that we can divide by no so we can't use cancellation going that way so now we're ready to multiply straight across let's start with the numerators two times seven is fourteen now for the denominators one times one is one so we end up with fourteen over one which is 14. and that's our final answer so there you have it there's how to use cancellation when we have a division problem involving fractions now can we use cancellation with every single problem no we have to have common factors other than one that we can divide by diagonally and do we have to use cancellation in order to get to the correct answer no but it is a good strategy to be familiar with and it can be helpful so there's how to use cross cancellation when we have a division problem involving fractions let's move on to using cross cancellation when dividing mixed numbers now we will take a look at using cross Dividing Mixed Numbers Using Cancellation cancellation when dividing mixed numbers let's jump into our example where we have 8 and 4 7 divided by two and two-fifths now we use cross cancellation when we get to the multiplying step so keep that in mind now the first thing that we want to do is convert these mixed numbers to improper fractions that way we just have a numerator and a denominator and we can move forward with our steps so we'll start with 8 and 4 7. we'll start at the bottom and work our way to the top we multiply then add so we do our denominator times the whole number 7 times 8 is 56 plus our numerator of 4. so 56 plus 4 that gives us 60. we keep our denominator of 7 the same let's bring our division sign down then do the second mixed number here so start at the bottom multiply then add 5 times 2 is 10 plus 2 is 12. so we get 12 fifths now that we have our mixed numbers converted to improper fractions we can go through our dividing fractions steps keep switch flip also known as Keep Change Flip there's different words out there for the steps but they all mean the same thing one thing I do want to mention before moving forward these improper fractions are equivalent to the mixed numbers so 67 is equivalent to eight and four sevenths twelve fifths is equivalent to two and two-fifths so when we convert to improper fractions we're not changing the value of the problem at all all right let's go through our steps keep switch flip so we always keep the first fraction so 60 over seven then we switch or change to multiplication and then flip our second fraction so the denominator becomes the numerator and the numerator becomes the denominator this is called the reciprocal so we have five twelfths now we can multiply straight across but in this problem we have an opportunity for cross-cancellation cross cancellation is a way to simplify fractions before multiplying it gives us smaller and easier numbers to work with therefore a simpler problem to solve we cross cancel by looking for common factors between the numerators and denominators so the top and bottom think of it like simplifying fractions but we can look diagonally as well for example we have common factors other than one between 60 and 12 and the greatest common factor of 12 so let's divide 60 and 12 by 12. we do that by Crossing them out and dividing by 12. so 60 divided by 12 is 5 12 divided by 12 is 1. we can also look diagonally the other way so we have 7 and 5. the only common factor between 7 and 5 is 1 so we can't use cross cancellation with the 7 and the five once we cross cancel we can multiply straight across so we have 5 times 5 which is 25 and then we have 7 times 1 which is seven so we get to 25 7. that's an improper fraction let's convert it to a mixed number so we do that by doing the numerator 25 divided by 7. so 25 divided by 7 let's think how many whole groups of seven can we pull out of 25 well three whole groups of seven that's our whole number now we do not hit 25 exactly we have something left over a remainder so the difference between 25 and 21 is 4 that's our remainder so that's our numerator and then we keep our denominator of 7 the same always look to see if you can simplify the fractional part of a mixed number 4 7 is in simplest form so this is our final answer 3 and 4 7. now do we have to use cross cancellation no we can multiply straight across and do 60 times 5 and 7 times 12 then go from there but cross cancellation again gave us smaller and easier numbers to work with it's a useful tool to use when looking at multiplication problems that involve fractions and another thing I want to mention cross cancellation can't be used with every problem we have to have common factors other than one between our numerators and denominators that we can divide by if the only common factor is one then cross cancellation cannot be used so there's how to use cross cancellation when dividing mixed numbers let's move on to dividing improper fractions by improper fractions Dividing an Improper Fraction by an Improper Fraction now we can take a look at dividing an improper fraction by an improper fraction now when we have a division problem involving improper fractions we go through the same steps we use when we have division problems involving proper fractions keep switch flip so don't let a problem hold you up if it involves improper fractions again just go through the same steps let's jump into our example where we have eight thirds divided by 11 fifths so keep switch flip so keep eight thirds switch to multiplication and then flip 11 fifths so five is now the numerator and 11 is now the denominator now we can multiply straight across let's start with the numerators eight times five is forty now for the denominators 3 times 11 is 33 so we end up with 40 over 33 or 40 30 thirds now that is our answer but it's an improper fraction so let's convert it to a mixed number by dividing the numerator 40 by the denominator 33. so how many whole groups of 33 in 40. well one whole group of 33 in 40. now we don't hit 40 exactly we have a remainder and that remainder is 7. that's the numerator of the fractional part of the mixed number and then we keep the denominator of 33 the same always look to see if you can simplify the fractional part of a mixed number 7 3 is in simplest form the only common factor between 7 and 33 is one so this is our final answer one and seven thirty thirds now just as a quick recap as far as how we went from that improper fraction of 40 over 33 to the mixed number of 1 and 7 33. I'm going to write it out so again we divided the numerator 40 by the denominator 33 so 40 divided by 33. how many whole groups of 33 in 40. well 1 1 times 33 is 33. subtract and we get that remainder of 7. 1 is the whole number part of the mixed number and then seven the remainder is the numerator of the fractional part and then we keep the denominator of 33 the same so there's how to divide an improper fraction by an improper fraction let's move on to dividing an improper fraction by a whole number Dividing an Improper Fraction by a Whole Number now we will take a look at dividing an improper fraction by a whole number let's jump into our example where we have 10 4 divided by nine so keep switch flip let's keep ten fourths switch to multiplication and then we need to flip that nine well we flip that 9 by writing it in fractional form and remember we can write any whole number in fractional form by putting it over one and now we can flip it so one is the numerator and 9 is the denominator so keep switch flip and then we can multiply straight across now we will start with the numerators 10 times 1 is 10. now for the denominators 4 times 9 is 36 so we get 10 36 or 10 over 36 which is our final answer but we can simplify we have a common factor other than 1 between 10 and 36. there's a greatest common factor of two so let's divide ten by two and 36 by 2 in order to simplify 10 divided by 2 is 5 36 divided by 2 is 18. the only common factor between 5 and 18 is one so this is in simplest form final simplified answer 5 18. so there's how to divide an improper fraction by a whole number let's move on to dividing a Dividing a Whole Number by an Improper Fraction whole number by an improper fraction now let's take a look at dividing a whole number by an improper fraction let's jump into our example where we have eight divided by 15 8. so keep switch flip let's keep eight and I'm going to write it in fractional form that way we have a numerator and a denominator and remember we can write any whole number in fractional form by putting it over one so keep eight and again we wrote it in fractional form switch to multiplication and then flip 15 8 so 8 is now the numerator and 15 is now the denominator now we can multiply straight across let's start with the numerators eight times eight is sixty-four now for the denominators one times fifteen is 15 so we get 64 15 and that is our answer but it's an improper fraction so let's convert it to a mixed number by dividing the numerator 64 by the denominator 15. so 64 divided by 15. how many whole groups of 15 in 64 well four four whole groups that gets us to 60. we don't hit 64 exactly we have a remainder of 4. that's the numerator of the fractional part of the mixed number and then we keep the denominator of 15 the same we can always look to see if we can simplify the fractional part of a mixed number 4 15 is in simplest form the only common factor between 4 and 15 is one so Final Answer 4 and 4 15. now just as a quick recap as far as how we went from that improper fraction 64 15 to the mixed number of four and four fifteenths I'm going to write that out off to the side so we did the numerator 64 divided by the denominator 15. so we have 64 divided by 15. how many whole groups of 15 in 64 well 4 that's the whole number of our mixed number now we multiply 4 times 15 is 60 subtract 64 minus 60 gives us 4 so we have 4 remainder 4 that remainder is the numerator of the fractional part of the mixed number and then we keep the denominator of 15 the same so there's how to divide a whole number by an improper fraction let's move on to taking a look at why we use keep switch flip when we have division problems involving fractions now we will take a look at why we use Why Do We Use Keep, Switch, Flip? keep switch flip when we have division problems involving fractions you may also hear Keep Change Flip or multiply by the reciprocal these all mean the same thing now before we get into our dividing fractions example we need to review and fully understand a few things about fractions we'll start with fractions representing division a division problem can be set up as a fraction so for number one we have eight divided by 4 equals two now we can set that up as a fraction we can set it up as eight divided by 4 equals 2. so again we need to know that fractions represent division the second thing that we need to understand is that anything over one so a denominator of one equals itself for example five over one equals five so again anything over one equals itself under number three here we have fractions that have the same numerator and denominator for example seven over seven sevenths anytime the numerator and denominator are the same that fraction represents one whole so seven sevenths or seven over seven this equals one again same numerator same denominator equals one the fourth thing that we need to understand equivalent fractions now remember in order to keep a fraction equivalent whatever we do to the bottom we must do to the top or vice versa and we can use multiplication or division this will give us an equivalent fraction for example three-fourths equals how many eighths well we know that four times two equals eight four times two equals eight now whatever we do to the bottom we must do to the top in order to keep this equivalent so three times two gives us 6. so 3 4 is equivalent to six eighths both of these fractions have the same value so we need to understand equivalent fractions lastly number five here we have reciprocals which is flipping a fraction so the reciprocal of 4 7 is 7 4. so this is the reciprocal of four sevenths now I'm multiplying four sevenths by its reciprocal seven fourths to show that we get one whenever we multiply a fraction by its reciprocal so 4 7 times 7 4 well multiply straight across four times seven is twenty-eight now let's multiply the denominators the bottom numbers 7 times 4 is 28 so we end up with 28 28 and we know that if we have the same number on top the numerator as we do the bottom the denominator this is going to equal one whole or one so reciprocals are the last thing that we need to understand and again a reciprocal we flip the fraction the numerator becomes the denominator and the denominator becomes the numerator and when we multiply a fraction by its reciprocal we get one now we are ready to see why we keep switch Flip or maybe you refer to it as Keep Change Flip or multiply by the reciprocal remember all of those things mean the same thing let's use three-fifths divided by one-fourth for our example here we know that we can set up a division problem as a fraction fractions represent division so let's set up three-fifths divided by 1 4 as a fraction we get three-fifths yeah over 1 4. this gives us a fraction over a fraction so two fractions making up one fraction this is called a complex fraction a complex fraction is a fraction where the numerator denominator or both are a fraction now once we have our complex fraction we want to get our denominator to equal one we know that anything over one equals itself and that is going to make this much more manageable well how do we get our denominator to equal one we need to multiply it by its reciprocal flip the fraction so let's multiply 1 4 by 4 over 1. and again we are doing this so we get that denominator to equal one because we know that anything over one equals itself now since we are multiplying our denominator by four over one remember in order to keep a fraction equivalent whatever we do to the bottom we must do to the top or vice versa this keeps everything equivalent the value doesn't change so we need to multiply our numerator by four over one as well foreign so this is going to give us well 1 times 4 is 4 4 times 1 is 4. remember when we multiply fractions we multiply straight across so again 1 times 4 is 4 and then four times one is four let's do our numerator so 3 times 4 right here is going to give us 12. and then 5 times 1 is going to give us 5. so we have 12 fifths over four fourths now we know that when we have the same numerator and denominator the fraction equals one so let's rewrite this as twelve fifths over one this is our new complex fraction it's equivalent to what we started with but now we have a denominator of one and we know that anything over one just equals itself so our answer is twelve fifths now 12 fifths is an improper fraction so we want to change this to a mixed number we do that by dividing our numerator by the denominator so 12 divided by 5. how many whole groups of five can we pull out of 12. well 2. now that doesn't hit 12 exactly we have a remainder of 2 and we keep our denominator of 5 the same now two and two-fifths our fractional part is in simplest form so we are done two and two-fifths is our answer for three-fifths divided by one-fourth so you can see that that was a fairly long process that's the long way of dividing fractions it does work but we can use keep switch flip to be more efficient let's try keep switch flip as well and see how everything Compares so I'm going to rewrite the problem at the bottom of the screen here so we have three-fifths divided by 1 4. so let's keep switch flip so keep our first fraction three-fifths switch to multiplication and then flip our second fraction we get four over one and then we multiply straight across we get 12 fifths the same answer we got when we used the long way and we convert this to a mixed number which is two two and two-fifths so you can see look up here and you'll notice that we have three-fifths times four over one the same thing we have down here now keep switch flip again is a much quicker process here we're always going to end up with the same thing here and here so why not cut out all of the extra parts of that process and just jump straight to keep switch flip so there you have it there's a complete guide to dividing fractions I hope that helped thanks so much for watching until next time peace
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12.1: The lac Operon
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: Apr 9, 2022
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12: Regulation of Gene Expression
12.2: The Use of Mutants to Study the lac Operon
Page ID
: 4150
Todd Nickle and Isabelle Barrette-Ng
Mount Royal University & University of Calgary
( \newcommand{\kernel}{\mathrm{null}\,})
Early insights into mechanisms of transcriptional regulation came from studies of E. coli by researchers Francois Jacob & Jacques Monod. In E. coli, and many other bacteria, genes encoding several different proteins may be located on a single transcription unit called an operon. The genes in an operon share the same transcriptional regulation, but are translated individually. Eukaryotes generally do not group genes together as operons (exception is C. elegans and a few other species).
Basic lac Operon structure
E. coli encounters many different sugars in its environment. These sugars, such as lactose and glucose, require different enzymes for their metabolism. Three of the enzymes for lactose metabolism are grouped in the lac operon: lacZ, lacY, and lacA (Figure 12.1.1). LacZ encodes an enzyme called β-galactosidase, which digests lactose into its two constituent sugars: glucose and galactose. lacY is a permease that helps to transfer lactose into the cell. Finally, lacA is a trans-acetylase; the relevance of which in lactose metabolism is not entirely clear. Transcription of the lac operon normally occurs only when lactose is available for it to digest. Presumably, this avoids wasting energy in the synthesis of enzymes for which no substrate is present. A single mRNA transcript includes all three enzyme-coding sequences and is called polycistronic. A cistron is equivalent to a gene.
cis- and transRegulators
In addition to the three protein-coding genes, the lac operon contains short DNA sequences that do not encode proteins, but are instead binding sites for proteins involved in transcriptional regulation of the operon. In the lac operon, these sequences are called P (promoter), O (operator), and CBS (CAP-binding site). Collectively, sequence elements such as these are called cis-elements because they must be located on the same piece of DNA as the genes they regulate. On the other hand, the proteins that bind to these cis-elements are called trans-regulators because (as diffusible molecules) they do not necessarily need to be encoded on the same piece of DNA as the genes they regulate.
lacI is an allosterically regulated repressor
One of the major trans-regulators of the lac operon is encoded by lacI. Four identical molecules of lacI proteins assemble together to form a homotetramer called a repressor (Figure 12.1.2). This repressor binds to two operator sequences adjacent to the promoter of the lac operon. Binding of the repressor prevents RNA polymerase from binding to the promoter (Figure 12.1.3). Therefore, the operon will not be transcribed when the operator is occupied by a repressor.
Besides its ability to bind to specific DNA sequences at the operator, another important property of the lacI protein is its ability to bind to lactose. When lactose is bound to lacI, the shape of the protein changes in a way that prevents it from binding to the operator. Therefore, in the presence of lactose, RNA polymerase is able to bind to the promoter and transcribe the lac operon, leading to a moderate level of expression of the lacZ, lacY, and lacA genes. Proteins such as lacI that change their shape and functional properties after binding to a ligand are said to be regulated through an allosteric mechanism. The role of lacI in regulating the lac operon is summarized in Figure 12.1.4.
CAP is an allosteric activator of the lac operon
A second aspect of lac operon regulation is conferred by a trans-factor called cAMP binding protein (CAP, Figure 12.1.4). CAP is another example of an allosterically regulated trans-factor. Only when the CAP protein is bound to cAMP can another part of the protein bind to a specific cis-element within the lac promoter called the CAP binding sequence (CBS). CBS is located very close to the promoter (P). When CAP is bound to at CBS, RNA polymerase is better able to bind to the promoter and initiate transcription. Thus, the presence of cAMP ultimately leads to a further increase in lac operon transcription.
The physiological significance of regulation by cAMP becomes more obvious in the context of the following information. The concentration of cAMP is inversely proportional to the abundance of glucose: when glucose concentrations are low, an enzyme called adenylate cyclase is able to produce cAMP from ATP. Evidently, E. coli prefers glucose over lactose, and so expresses the lac operon at high levels only when glucose is absent and lactose is present. This provides another layer of logical control of lac operon expression: only in the presence of lactose, and in the absence of glucose is the operon expressed at its highest levels.
12: Regulation of Gene Expression
12.2: The Use of Mutants to Study the lac Operon
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https://www.fs.usda.gov/wildflowers/ethnobotany/Mind_and_Spirit/datura.shtml
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Solanaceae: Datura
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Solanaceae: Datura
The Powerful Solanaceae: Datura
Jimsonweed, Thornapple, Moonflower, and Sacred Daturas (Datura spp.)
“At the time of the White Dawn;
At the time of the White Dawn,
I arose and went away.
At Blue Nightfall I went away.
I ate the thornapple leaves
And the leaves made me dizzy.
I drank thornapple flowers
And the drink made me stagger…”
~Original Author Unkown
So begins the Pima “Datura Song” first translated in 1901 from a song-poem ritually performed by Tohono O’odham shamans of south-central Arizona and northern Sonora, Mexico. This song was sung to bring success with a hunt, but also in cases of sickness where the characteristic symptoms were vomiting and dizziness (Frank Russell, The Pima Indians, 1908). It illustrates the great significance and nature of datura plants in the oral and sacred traditions of many native cultures.
Datura species have been revered as sacred visionary plants among almost all cultures around the world that have encountered it. Archeological evidence shows that Datura has been in use for at least 3,000 years in the southwestern United States and even longer in other parts of the world.
Sacred Datura (Datura wrightii) is also called moonflower, a reference to the fact that the flowers open late in the day. Hawkmoths are a primary pollinator. Photo: Teresa Prendusi.
It is easy to see why the fruit of the Datura plant is often called thornapple. Photo: Teresa Prendusi.
The Chumash Indians of southern California used Datura as a sacred and medicinal plant. Archeologists have interpreted the spiny silhouette around the circles in this pictograph to be Datura fruits. Photo: Painted Cave State Historic Park.
The genus Datura, with 9 to 12 known species, occurs widely throughout the temperate and warmer parts of both the Old and New World. The plants produce large white to purple tinged trumpet flowers and spiny round seedpods (thornapple). Daturas have been used as poisons, medicines, and ritual intoxicant agents since time immemorial. Each species has its own characteristic strong narcotic tropane alkaloids (typically scopolamine, atropine, daturine, and stramonine), which also make them among the most dangerous and poisonous plants known. Meteloidine is the principal hallucinogen, but it is only abundantly found in one of two species. Ingestion of even small quantities of any plant parts can cause delirium, frenzy, powerful hallucinations, coma, and death.
Jimson weed (Datura stramonium). Photo: CalPhotos.
Early American Shaker ad for Extract of Thorn Apple that includes Datura stramonium.
Early American Shaker ad for Pain King that includes Datura.
In the United States, the common name for the thornapple is Jimson weed, a contraction of Jamestown Weed. Robert Beverly in his book History and Present State of Virginia (1705), tells of an incident that occurred in colonial Jamestown to a group of unwitting soldiers who ate thorn apple leaves in a salad.
“Some of them eat plentifully of it, the Effect of which was a very pleasant Comedy; for they turn’d natural Fools upon it for several Days… A thousand such simple Tricks they play’d, and after Eleven Days, return’d to themselves again, not remembering anything that had pass’d.”
Jimsonweed is a major cause of accidental poisonings and death by those looking to get a cheap high.
Did You Know?
Angel trumpets (Brugmansia spp.) are closely related to Daturas and are found in the Andean tropics of western South America. Highly toxic like their North American counterparts, these beautiful woody trees and shrubs are nevertheless popular ornamentals throughout the world.
Some scholars believe that Delphic oracles in ancient Greece used Datura to induce their legendary visions.
The Aztecs used Datura as a painkiller in initiation rituals and as a narcotic for ritual sacrifices.
Datura has been used as an ingredient in the ritual practice of zombification in the Caribbean and is locally known as “Zombie-cucumber.”
Until recently, Datura cigarettes were prescribed to asthma sufferers.
Datura plants have been studied for their capacity to absorb heavy metals in bioremediation projects.
Some Datura plants produce addictive narcotic nectar to entice hawk moths to their flowers.
Browsing animals usually avoid eating the plants due to its foul odor and taste.
The Chumash Indians of Southern California integrated Datura in their creation mythology. The old grandmother “Momoy” turned herself into the plant Datura after the great flood, which marked the transition between mythical times and the modern world.
Georgia O’Keeffe often depicted Datura flowers in her world-famous paintings. She said, “when I think of the delicate fragrance of the flowers, I almost feel the coolness and sweetness of the evening.”
Brugmansia vesicolor is a long-lived Andean woody shrub or tree with large, pendulant flowers up to 10 inches long. It contains the same highly toxic tropane alkaloids found in Datura species.
The Chumash saw sacred Datura as the powerful ancestor-woman, Momoy. Photo: Santa Barbara Museum of Natural History.
Beautiful Datura flowers were a favorite subject of the famous painter Georgia O’Keeffe.
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"pulchritude": Physical beauty; attractiveness of appearance - OneLook
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Usually means: Physical beauty; attractiveness of appearance.
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Definitions from Wiktionary ( pulchritude ) ▸ noun:Physical beauty. Similar: venusty, beautiness, philauty, kalon, elegance, urbanity, exaltation, luster, form, splendor, more... Opposite: ugliness, unattractiveness, plainness Types: beauty, attractiveness, loveliness, appeal, charm, more... Phrases: International Pageant of Pulchritude, more... Adjectives: feminine, female, personal, male, physical, moral, masculine, much, awful, less, spiritual Colors: lavender, rose, coral, turquoise, emerald, more... Found in concept groups: Beauty Test your vocab: BeautyView in Idea Map From "Reet Petite and Gone" by Louis Jordan: > That chick has changed my attitude > > Whe ever I gaze on her pulchritude ▸ Word origin ▸ Words similar to pulchritude ▸ Usage examples for pulchritude ▸ Idioms related to pulchritude ▸ Wikipedia articles (New!) ▸ Popular adjectives describing pulchritude ▸ Words that often appear near pulchritude ▸ Rhymes of pulchritude ▸ Invented words related to pulchritude
Similar:
venusty, beautiness, philauty, kalon, elegance, urbanity, exaltation, luster, form, splendor, more...
Opposite:
ugliness, unattractiveness, plainness
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beauty, attractiveness, loveliness, appeal, charm, more...
Phrases:
International Pageant of Pulchritude, more...
Adjectives:
feminine, female, personal, male, physical, moral, masculine, much, awful, less, spiritual
Colors:
lavender, rose, coral, turquoise, emerald, more...
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cage intellectual general army fighting public field force freedom
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cage fighting: The practise of participating in hand-to-hand combat inside of a usually steel cage.
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general public: Those members of the public who have no special role in a particular area, such as an airport, hospital or railway station; there will typically be restrictions on their access.
public intellectual: (idiomatic) A well-known, intelligent, learned person whose written works and other social and cultural contributions are recognized not only by academic audiences and readers, but also by many members of society in general.
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https://www.geogebra.org/m/y7yafkzx
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Angle Bisector, Perpendicular Bisector and Median – GeoGebra
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https://proofwiki.org/wiki/Definition:Relative_Complement
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Definition:Relative Complement
From ProofWiki
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Contents
1 Definition
2 Illustration by Venn Diagram
3 Notation
4 Also known as
5 Examples
5.1 Examples
6 Also see
7 Linguistic Note
8 Technical Note
9 Sources
Definition
Let $S$ be a set, and let $T \subseteq S$, that is: let $T$ be a subset of $S$.
Then the set difference $S \setminus T$ can be written $\relcomp S T$, and is called the relative complement of $T$ in $S$, or the complement of $T$ relative to $S$.
Thus:
: $\relcomp S T = \set {x \in S : x \notin T}$
Illustration by Venn Diagram
The relative complement $\relcomp S T$ of the set $T$ with respect to $S$ is illustrated in the following Venn diagram by the red area:
Notation
Different notations for the relative complement $\relcomp S T$ mainly consist of variants of the $\complement$:
: $\map {\CC_S} T$
: $\map {\mathbf C_S} T$
: $\map {c_S} T$
: $\map {C_S} T$
: $\map {\mathrm C_S} T$
or sometimes:
: $\map {T^c} S$
: $\map {T^\complement} S$
and sometimes the brackets are omitted:
: $C_S T$
If the superset $S$ is implicit, then it can be omitted: $\map \complement T$ etc.
See the notation for set complement.
Some sources do not bother to introduce a specific notation for the relative complement, and instead just use one of the various notations for set difference:
: $S \setminus T$
: $S / T$
: $S - T$
Also known as
Some authors refer to the relative complement as just the (set) complement, and use the term relative complement for the set difference $S \setminus T$ when the stipulation $T \subseteq S$ is not required.
Others emphasize the connection with set difference by referring to the relative complement as a proper difference.
Thus, in this view, the relative complement is a specific case of a set difference.
Examples
Examples
Let $E$ denote the set of all even integers.
Then the relative complement of $E$ in $\Z$:
: $\relcomp \Z E$
is the set of all odd integers.
Also see
Definition:Set Difference
Definition:Set Complement
Results about relative complement can be found here.
Linguistic Note
The word complement comes from the idea of complete-ment, it being the thing needed to complete something else.
It is a common mistake to confuse the words complement and compliment.
Usually the latter is mistakenly used when the former is meant.
Technical Note
The $\LaTeX$ code for (\relcomp {S} {T}) is \relcomp {S} {T} .
This is a custom construct which has been set up specifically for the convenience of the users of $\mathsf{Pr} \infty \mathsf{fWiki}$.
Note that there are two arguments to this operator: the subscript, and the part between the brackets.
If either part is a single symbol, then the braces can be omitted, for example:
: \relcomp S T
Sources
1951: J.C. Burkill: The Lebesgue Integral ... (previous) ... (next): Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets
1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
1961: John G. Hocking and Gail S. Young: Topology ... (previous) ... (next): A Note on Set-Theoretic Concepts
1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
1965: A.M. Arthurs: Probability Theory ... (previous) ... (next): Chapter $1$: Set Theory: $1.3$: Set operations
1965: Claude Berge and A. Ghouila-Houri: Programming, Games and Transportation Networks ... (previous) ... (next): $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 1.6$. Difference and complement
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets
1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 2$: Sets and functions: Sets
1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 2$: Sets and functions: Sets
1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement
1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): Notation and Terminology
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 6$: Subsets
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The Definition of a Series
The Definition of a Series
Return to the Series, Convergence, and Series Tests starting page
Let
be a sequence. We call the sum
an infinite series (or just a series) and denote it as
.
We define a second sequence, s[n], called the partial sums, by
,
,
,
or, in general,
.
We then define convergence as follows:
Definition of Convergence, or the Limit of a Series
Given a series
let s[n] denote its n th partial sum:
.
If the sequence s[n] has a limit, that is, if there is some s such that for all > 0 there exists some N> 0 such that |s[n] - s| <, then the series is called convergent, and we say the series converges. We write
or
.
The number s is called the sum of the series. If the series does not converge, the series is called divergent, and we say the series diverges.
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MIT OpenCourseWare 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: 1 Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 141 Reading: • Proakis & Manolakis, Chapter 3 (The z-transform) • Oppenheim, Schafer & Buck, Chapter 3 (The z-transform) The Discrete-Time Transfer Function Consider the discrete-time LTI system, characterized by its pulse response {hn}: We saw in Lec. 13 that the output to an input sequence {fn} is given by the convolution sum: ∞ ∞ yn = fn ⊗hn = fkhn−k = hkfn−k, k=−∞ k=−∞ where {hn} is the pulse response. Using the convolution property of the z-transform we have at the output Y (z) = F (z)H(z) where F (z) = Z {fn}, and H(z) = Z {hn}. Then Y (z) H(z) = F (z) is the discrete-time transfer function, and serves the same role in the design and analysis of discrete-time systems as the Laplace based transfer function H(s) does in continuous systems. 1copyright ⃝ c D.Rowell 2008 14–1 2 In general, for LTI systems the transfer function will be a rational function of z, and may be written in terms of z or z−1, for example N(s) b0 + b1z−1 + b2z−2 + . . . + bM z−M H(z) = = D(s) a0 + a1z−1 + a2z−2 + . . . + aN z−N where the bi, i = 0, . . . , m, ai, i = 0, . . . , n are constant coefficients. The Transfer Function and the Difference Equation As defined above, let Y (z) b0 + b1z−1 + b2z−2 + . . . + bM z−M H(z) = = F(z) a0 + a1z−1 + a2z−2 + . . . + aN z−N and rewrite as a0 + a1z −1 + a2z −2 + . . . + aN z −N Y (z) = b0 + b1z −1 + b2z −2 + . . . + bM z −M F(z) If we apply the z-transform time-shift property Z {fn−k} = z −kF(z) term-by-term on both sides of the equation, (effectively taking the inverse z-transform) a0yn + a1yn−1 + a2yn−2 + . . . + aN yn−N = b0fn + b1fn−1 + b2fn−2 + . . . + bM fn−M and solve for yn 1 1 yn = − (a1yn−1 + a2yn−2 + . . . + aN yn−N ) + (b0fn + b1fn−1 + b2fn−2 + . . . + bM fn−M ) a0 a0 N M −ai bi = yn−i + fn−i a0 a0 i=1 i=0 which is in the form of a recursive linear difference equation as discussed in Lecture 13. The transfer function H(z) directly defines the computational dif ference equation used to implement a LTI system. Example 1 Find the difference equation to implement a causal LTI system with a transfer function (1 − 2z−1)(1 − 4z−1) H(z) = z(1 − 1 2 z−1) Solution: z−1 − 6z−2 + 8z−3 H(z) = 1 − 1 z−1 2 14–2 from which 1 yn − yn−1 = fn−1 − 6fn−2 + 8fn−3, 2 or 1 yn = yn−1 + (fn−1 − 6fn−2 + 8fn−3). 2 The reverse holds as well: if we are given the difference equation, we can define the system transfer function. Example 2 Find the transfer function (expressed in powers of z) for the difference equation yn = 0.25yn−2 + 3fn − 3fn−1 and plot the system poles and zeros on the z-plane. Solution: Taking the z-transform of both sides Y (z) = 0.25z −2Y (z) + 3F (z) − 3z −1F (z) and reorganizing Y (z) 3(1 − z−1) 3z(z − 1) H(z) = = = F (z) 1 − 0.25z−2 z2 − 0.25 which has zeros at z = 0, 1 and poles at z = −0.5, 0.5: ! " # ! " # 14–3 3 Introduction to z-plane Stability Criteria The stability of continuous time systems is governed by pole locations - for a system to be BIBO stable all poles must lie in the l.h. s-plane. Here we do a preliminary investigation of stability of discrete-time systems, based on z-plane pole locations of H(z). Consider the pulse response hn of the causal system with z 1 H(z) = = z − a 1 − az−1 with a single real pole at z = a and with a difference equation yn = ayn−1 + fn. % % % ! ! % % $ Clearly the pulse response is 1 n = 0 hn = an n ≥ 1 The nature of the pulse response will depend on the pole location: 0 < a < 1: In this case hn = an will be a decreasing function of n and limn→∞ hn = 0 and the system is stable. a = 1: The difference equation is yn = yn−1 + fn (the system is a summer and the impulse response is hn = 1, (non-decaying). The system is marginally stable. a > 1: In this case h = an will be a increasing function of n and limn→∞ h = ∞ and the n n system is unstable. −1 < a < 0: In this case hn = an will be a oscillating but decreasing function of n and limn→∞ hn = 0 and the system is stable. a = −1: The difference equation is yn = −yn−1 + fn and the impulse response is hn = (−1)n , that is a pure oscillator. The system is marginally stable. a < −1: In this case hn = an will be a oscillating but increasing function of n and limn→∞ |hn| = ∞ and the system is unstable. 14–4 4 This simple demonstration shows that this system is stable only for the pole position −1 < a < 1. In general for a system M (z − zk) H(z) = K k=1 N (z − pk) k=1 having complex conjugate poles (pk) and zeros (zk) : A discrete-time system will be stable only if all of the poles of its transfer function H(z) lie within the unit circle on the z-plane. The Frequency Response of Discrete-Time Systems Consider the response of the system H(z) to an infinite complex exponential sequence fn = A ej ωn = A cos(ωn) + jA sin(ωn), where ω is the normalized frequency (rad/sample). The response will be given by the con volution ∞ ∞ yn = hkfn−k = hk A ej ω(n−k) k=−∞ k=−∞ ∞ = A hk e −j ωk ej ωn k=−∞ = AH(ej ω)ej ωn where the frequency response function H(ej ω) is H(ej ω) = H(z)| j ω z=e that is The frequency response function of a LTI discrete-time system is H(z) evaluated on the unit circle - provided the ROC includes the unit circle. For a stable causal system this means there are no poles lying on the unit circle. ' & & & ! ( 14–5 ) Alternatively, the frequency response may be based on a physical frequency Ω associated with an implied sampling interval ΔT, and H(ej ΩΔT ) = H(z)| j ΩΔT z=e which is again evaluated on the unit circle, but at angle ΩΔT. ' ) ' & & & ! ( From the definition of the DTFT based on a sampling interval ΔT ∞ H ∗ (jΩ) = h −mjnΩΔT = H(z)| n e z= e−mjnΩΔT n=0 we can define the mapping between the imaginary axis in the s-plane and the unit-circle in the z-plane s = j Ωo ← → z = ej ΩoΔT & & & ! ) ' & & ' & & ( + ' ' + ' The periodicity in H( ej ΩΔT ) can be clearly seen, with the “primary” strip in the s-plane (defined by − π/ΔT < Ω < π/ΔT) mapping to the complete unit-circle. Within the primary strip, the l.h. s-plane maps to the interior of the unit circle in the z-plane, while the r.h. s-plane maps to the exterior of the unit-circle. 14–6 5 Aside: We use the argument to differentiate between the various classes of transfer functions: H(s) H(jΩ) H(z) H( ej ω) ⇕ ⇕ ⇕ ⇕ Continuous Continuous Discrete Discrete Transfer Frequency Transfer Frequency Function Response Function Response The Inverse z-Transform The formal definition of the inverse z-transform is as a contour integral in the z-plane, 1 ∞ F(z)z n−1 dz 2πj −∞ where the path is a ccw contour enclosing all of the poles of F(z). Cauchy’s residue theorem states 1 ∞ F(z) dz = Res [F(z), pk] 2πj −∞ k where F(z) has N distinct poles pk, k = 1, . . . , N and ccw path lies in the ROC. For a simple pole at z = zo Res [F(z), zo] = lim (z − zo)F(z), z→zo and for a pole of multiplicity m at z = zo 1 dm−1 Res [F(z), zo] = lim (z − zo)mF(z) z→zo (m − 1)! dzm−1 The inverse z-transform of F(z) is therefore fn = Z−1 {F(z)} = Res F(z)z n−1 , pk . k Example 3 A first-order low-pass filter is implemented with the difference equation yn = 0.8yn−1 + 0.2fn. Find the response of this filter to the unit-step sequence {un}. 14–7 Solution: The filter has a transfer function Y (z) 0.2 0.2z H(z) = = = F (z) 1 − 0.8z−1 z − 0.8 The input {fn} = {un} has a z-transform z F (z) = z − 1 so that the z-transform of the output is 0.2z2 Y (z) = H(z)U(z) = (z − 1)(z − 0.8) and from the Cauchy residue theorem yn = Res Y (z)z n−1 , 1 + Res Y (z)z n−1 , 0.8 = lim(z − 1)Y (z)z n−1 + lim (z − 0.8)Y (z)z n−1 z→1 z→0.8 0.2zn+1 0.2zn+1 = lim + lim z→1 z − 0.8 z→0.8 z − 1 = 1 − 0.8n+1 which is shown below ! , - . / ! , - . / ! ! " , ! " - ! " . ! " / Example 4 Find the impulse response of the system with transfer function 1 z2 z2 H(z) = = = 1 + z−2 z2 + 1 (z + j 1)(z − j 1) 14–8 Solution: The system has a pair of imaginary poles at z = ±j 1. From the Cauchy residue theorem hn = Z−1 {H(z)} = Res H(z)z n−1 , j 1 + Res H(z)z n−1 , −j 1 n+1 n+1 z z = lim + lim z→j1 z + j 1 z→−j 1 z − j 1 1 1 = (j 1)n+1 − (−j 1)n+1 j 2 j 2 = j n 1 + (−1)n+1 2 0 n odd hn = (−1)n/2 n even = cos(nπ/2) where we note that the system is a pure oscillator (poles on the unit circle) with a frequency of half the Nyquist frequency. , - . / ! , - . / ! " # ! ! " # Example 5 Find the impulse response of the system with transfer function 1 z2 z2 H(z) = = = 1 + 2z + z−2 z2 + 2z + 1 (z + 1)2 Solution: The system has a pair of coincident poles at z = −1. The residue at z = −1 must be computed using 1 dm−1 Res [F (z), zo] = lim (z − zo)mF (z). z→zo (m − 1)! dzm−1 With m = 2, at z = −1, Res H(z)z n−1 , −1 = lim 1 d (z − 1)2H(z)z n−1 z→−1 (1)! dz d n+1 = lim z z→−1 dz = (n + 1)(−1)n 14–9 The impulse response is hn = Z −1 { H(z)} = Res H(z)z n−1 , − 1 = (n + 1)(− 1)n . , - . / ! , - . / , ! ! ! ! , ! Other methods of determining the inverse z-transform include: Partial Fraction Expansion: This is a table look-up method, similar to the method used for the inverse Laplace transform. Let F (z) be written as a rational function of z−1: M biz−k F (z) = k=0 N −k aiz k=0 M −1) (1 − ciz k=1 = N (1 − diz−1) k=1 If there are no repeated poles, F (z) may be expressed as a set of partial fractions. N Ak F (z) = 1 − dkz−1 k=1 where the Ak are given by the residues at the poles Ak = lim (1 − dkz −1)F (z). z→dk Since Z 1 (dk)n un ← → 1 − dkz−1 N fn = Ak (dk)n un. k=1 14–10 Example 6 Find the response of the low-pass filter in Ex. 3 to an input fn = (−0.5)n Solution: From Ex. 3, and from the z-transform of {fn}, 1 0.2 F (z) = , H(z) = 1 − 0.5z−1 1 − 0.8z−1 so that 0.2 Y (z) = (1 + 0.5z−1)(1 − 0.8z−1) A1 A2 = + 1 + 0.5z−1 1 − 0.8z−1 Using residues 0.2 0.1 A1 = lim = z→−0.5 1 − 0.8z−1 1.3 0.2 0.16 A2 = lim = z→0.8 1 + 0.5z−1 1.3 and yn = 0.1 Z−1 1 + 0.16 Z−1 1 1.3 1 + 0.5z−1 1.3 1 − 0.8z−1 0.1 0.16 = (−0.5)n + (0.8)n 1.3 1.3 Note: (1) If F (z) contains repeated poles, the partial fraction method must be ex tended as in the inverse Laplace transform. (2) For complex conjugate poles – combine into second-order terms. Power Series Expansion: Since ∞ F (z) = fnz −n n=−∞ if F (z) can be expressed as a power series in z−1, the coefficients must be fn. 14–11 Example 7 Find Z−1 {log(1 + az−1)}. Solution: F (z) is recognized as having a power series expansion: ∞ a F (z) = log(1 + az −1) = (−1)n+1 n z −n for |a| < |z| n n=1 Because the ROC defines a causal sequence, the samples fn are ⎧ ⎨0 for n ≤ 0 fn = (−1)n+1 n a ⎩ for n ≥ 1. n Polynomial Long Division: For a causal system, with a transfer function written as a rational function, the first few terms in the sequence may sometimes be computed directly using polynomial division. If F (z) is written as N(z−1) −1 −2 −2 F (z) = D(z−1) = f0 + f1z + f2z + f2z + · · · the quotient is a power series in z−1 and the coefficients are the sample values. Example 8 Determine the first few terms of fn for 1 + 2z−1 F (z) = 1 −2z−1 + z−2 using polynomial long division. Solution: 1 + 4z−1 + 7z−2 + · · · 1 −2z−1 + z−2 1 + 2z−1 1 −2z−1 + z−2 4z−1 −z−2 4z−1 −8z−2 + 4z−3 7z−2 −4z−3 so that 1 + 2z−1 F (z) = 1 −2z−1 + z−2 = 1 + 4z −1 + 7z −2 + · · · and in this case the general term is fn = 3n + 1 for n ≥ 0. 14–12 In general, the computation can become tedious, and it may be difficult to recognize the general term from the first few terms in the sequence. 14–13
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3 Thrilling Activities to Add and Subtract Decimals
By Tammy
Math, Planning
When I mention “add and subtract decimals,” your mind likely immediately jumps to money. That’s great! … For us, anyway. Your kiddos, however, don’t yet have that real-world connection to drive home the importance of decimals.
Decimals are one of those mathematical concepts that tend to evoke a feeling of apprehension or even outright fear in upper elementary students, just like fractions. By the time they get to your classroom, they’ll have heard all about the horrors of decimals on the playground. It’s like long division, but worse—there are no easy remainders here!
Despite the moaning and groaning that’s sure to fill your math block, students will still have to learn to add and subtract decimals up to the hundredths place in order to comply with Common Core. Since it’s not a question of if, it becomes a question of how–how can we help students learn to add and subtract decimals with the least amount of pain possible?
Thankfully, I have answers. Three of them, in fact! Three engaging math activities that are hands-on with real-world connections; perfect for replacing your students’ anxiety with enthusiasm! Setting your kiddos up for success means building a strong foundation for lifelong learners rather than allowing them to live in fear of the daunting decimal.
Let’s dive in to a few ways you can teach your math classroom to add and subtract decimals with minimal struggle!
Activities for Learning to Add and Subtract Decimals
1. Decimal Shopping Spree
In this activity, we’ll simulate a shopping trip in order to demonstrate the real-world applications of learning to add and subtract decimals to the hundredths place—or, in other words, learning to use money!
To carry out this shopping spree, you’ll need play money with decimal denominations, such as $0.10, $0.50, $1.00, and $5.00. You can find these on Amazon or simply make your own! There are plenty of lovely clipart sets floating around out there!
You’ll also need paper and pencil for price tags as well as calculations. Finally, you’ll need a variety of classroom items, handmade items, or even just printed pictures of items to serve as “goods” that students can purchase.
To start off, each student should receive an equal amount of money and then be divided into small groups. Each group will be assigned a “shop” with a unique theme or type of product that they sell. Students can create their own products or services for the shop by drawing pictures (or using images provided by you).
Each “shop” should also set prices for their product, making sure to include decimals (e.g., $2.75 rather than just $3.00). Make sure they keep in mind everyone’s starting budget so nothing is impossible to buy! Then, students will rotate between the role of shopkeeper and customer. As customers, they can browse shops and use their play money to purchase items.
In order to purchase items from a shop, students must calculate the total cost of their products and hand over the correct amount of money. The shopkeeper must then provide the correct amount of change in return.
Once you feel enough time has elapsed and everyone has had an opportunity to play both customer and shopkeeper, students should return to their groups to calculate their total spending. They will practice adding decimals to find their total amounts spent, and subtracting decimals to determine how much money is left. Then, groups should work together to calculate their total sales and the change they made in order to find their net profit.
Afterward, host a classroom discussion about the experience. Each “shop” should share out their total sales, the change they made, and their net profit, demonstrating their decimal calculations. Then, brainstorm reasons why it’s important to accurately add and subtract decimals, especially when it comes to money!
You can also bring up rounding. Ask probing questions, such as “When is it okay to round when you’re using money? Would rounding be helpful or harmful to the customer? What about the business?” This highlights the importance of using specific decimals rather than simply rounding to whole numbers for easier calculations.
2. Decimal Planet Exploration
Since money is one of the best ways to explore that real-world connection with decimals out to the hundredths place, let’s continue that trend! For this activity, you’ll need more of the same play money as described above, but in larger denominations. Why? Because your kiddos are about to blast off!
Students are going to take on the role of astronauts on a mission to explore a distant “decimal planet.” Divide your classroom into small teams of 4-5 astronauts, each representing a crew with different roles, such as navigator, engineer, scientist, and so on. Each crew will be given a budget with decimal values (so something like $50.75 rather than simply $50.00) to purchase equipment and supplies for their mission, provided as play money.
Then, you’ll set up different “supply stations” around the classroom, each with items priced in decimals, such as oxygen tanks for $5.50 or space food for $2.25. Teams must decide what they need for a successful mission and calculate the total cost of their items in order to stay within their budget. Of course, students should be sure not to spend all their money on preparation… Who knows what might happen during their adventure?
Once students blast off, hand out a series of “mission scenarios” that teams will encounter on “Decimal Planet.” Each scenario should require teams to solve problems using addition and subtraction of decimals. Scenarios can include situations like fuel consumption calculations, resource allocations, unexpected expenses, or scientific discoveries with measurements in decimals.
Ultimately, as each crew progresses through their scenarios, they should keep a “journey log” recording their decisions, calculations, and remaining budget. The team should work together by checking each other’s calculations for accuracy and discussing different strategies for managing their resources. No one should be the only one doing all the math! If that’s something you’re concerned about, maybe check out a few different ways to group your students.
At the end of the activity, each team should present a summary of their mission: the equipment they chose, the challenges they faced regarding decimal calculations, and how they were able to add and subtract decimals to solve their problems. They should also present their final budget, showing how much money they spent as well as how much is left over.
After the activity presentations, facilitate a class discussion (using student discourse!) on the importance of decimal knowledge in real-world scenarios such as budgeting and resource management. Sure, humans aren’t currently blasting off to “Decimal Planet,” but using a budget to prepare for upcoming expenses is something every business faces on the daily! Students should also reflect on the teamwork aspect of the activity and explore how cooperative learning helped them with problem-solving.
3. Decimals in Motion
Now, we’ve been sitting long enough… Let’s get those kiddos moving! Combining decimals and dancing may seem a bit unintuitive at first glance, but your students will be engaged with every step.
For this activity, you’ll need to develop a list of dance moves with assigned decimal values. For example, a spin may be 0.25, a jump could be 0.10, and a step could be 0.05. There should be enough variety to prevent confusion, but the moves shouldn’t be overly-complicated, either. Your kiddos aren’t professional dancers!
Once again, divide your students into small groups. Each group will create their own unique dance sequence after some discussion, being sure to allow every member to contribute to the choreography. For each group, come up with a unique total number that the decimal value of the moves should add up to, such as exactly 1.00 or 1.50. Be sure to encourage creativity and inclusion so that every group member is actively participating… But your kiddos already do that, right? 😉
After allowing enough time for dance practice and appropriate decimal calculations, each group should perform their dance in front of the class. Afterwards, other students should try to guess the total decimal value of the dance based on the dance moves they used. Then, the performing group reveals the actual sum and then performs their dance again, with the audience adding up the dance moves to confirm that the group reached their given sum.
Once every dance squad has performed and calculated their sums, engage the class in a discussion about the experience. Focus on how they approached the different methods to add and subtract decimals in a practical, physical context. What challenges did they face? How did they overcome them?
Ultimately, providing your math classroom with real-life scenarios and unique hands-on experiences is sure to teach them to add and subtract decimals in a manner they won’t soon forget. Engagement with math is a necessity, especially when it comes to daunting topics such as learning to add and subtract decimals.
Still, you can’t simply dance the decimals away—they’re here to stay! When your kiddos are ready for pencil and paper action, consider my math workshop unit geared to explore how to add and subtract decimals. Or, since it’s that time of year… Maybe a holiday-themed activity is more apt: add and subtract decimals to build this Christmas tree!
Regardless of how you choose to present ways to add and subtract decimals to your math classroom, I have no doubt that you’ll do what’s best by your kiddos. After all, you know your classroom and its needs best! Engage your upper elementary students and decimate those daunting decimals!
What decimal activities do you use in your classroom? Let me know in the comments!
add and subtract decimals, decimals
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Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 283 ISSN 1989 – 9572 DOI: 10.47750/jett.2022.13.05.026 An Error Analysis of Students’ Misconceptions and Skill Deficits in Pre-Calculus Subjects Cristo Mark D. Ancheta1 Journal for Educators, Teachers and Trainers, Vol. 13 (5) Date of reception: 15 Aug 2022 Date of revision: 08 oct 2022 Date of acceptance: 10 oct 2022 Cristo Mark D. Ancheta (2022). An Error Analysis of Students’ Misconceptions and Skill Deficits in Pre-Calculus SubjectsJournal for Educators, Teachers and Trainers,Vol. 13(5). 283-295. 1Faculty member, Nueva Ecija University of Science and Technology, Philippines Journal for Educators, Teachers and Trainers, Vol. 13 (5) ISSN 1989 – 9572 Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 284 An Error Analysis of Students’ Misconceptions and Skill Deficits in Pre-Calculus Subjects Cristo Mark D. Ancheta Faculty member, Nueva Ecija University of Science and Technology, Philippines Email: c_mark14@yahoo.com ABSTRACT The study determined the students’ misconceptions and skill deficits in Pre-Calculus subjects utilizing error analysis. It determined the students’ misconceptions, skill deficits, and their reasons, and propose reinforcement activities addressing the problem in Pre-Calculus subjects. The quantitative and qualitative approaches were conducted to come up with a high reliability score result. The respondents were the 50 second-year Engineering students of Nueva Ecija University of Science and Technology who have difficulty in pre-requisite subjects of Calculus. Based on the findings, students had misconceptions about the basic concepts and laws in Algebra necessary for performing fundamental algebraic operations. They also lacked skills in performing algebraic operations. Students also had misconceptions about Trigonometry, particularly on trigonometric functions and their inverses which were pre-requisite concepts and skills in higher mathematics subjects. Moreover, students had misconceptions about Analytic Geometry, particularly in the slope of a line, equations of a line, and graphing functions and conics. The reasons for their misconceptions and skill deficits were due to their lack of knowledge retention, teachers’ methodology, teachers’ lack of knowledge, and lots of school activities. Keywords: Analytic Geometry, Calculation Error, Deficit, Error Analysis, Pre-Calculus INTRODUCTION In the mathematics curriculum, subjects are sequenced in order of requisites to prepare students to be more receptive to new concepts, approaches, and applications in mathematics offered in the upcoming years. Calculus is one of the subjects is taught in engineering, education, and other technical courses requiring pre-requisite mathematics subjects. The success or failure of any Calculus student is dependent on the extent of competency gained in Pre-Calculus subjects like Algebra, Trigonometry, and Analytic Geometry. Students’ prior knowledge of Pre-Calculus is essential in the learning of concepts, definitions, theorems, and applications of Calculus (Beng, YH. and Yunus, A. A., 2013). Students who have a deficiency in relevant prior knowledge will have difficulty in acquiring new knowledge and new understandings. Inadequate content knowledge of students in these subjects may cause their pitfall in Calculus. Students build more advanced knowledge from prior understandings(Smith III, J., Sessa, A. and Roschelle, J., 2009). According to Mwavita (2005), the background characteristics of students that have been examined by researchers is the role of prior mathematics preparation in relation to calculus performance. Students who passed Pre-Calculus subjects with mediocrity are likely to grapple with Calculus as this subject requires prior knowledge and skill on these subjects. In the study of Beng and Yumus, (2013) the students who passed calculus were the students with high marks in pre-calculus subjects and the students with low marks in pre-calculus failed in calculus. Relative to this, a mathematics teacher must know how mathematical instruction in pre-Calculus subjects is received, processed, and retained in the students’ minds. By doing so, appropriate teaching and learning strategies can be planned and executed to attain more meaningful student learning outcomes. Viable approaches can be employed to probe students’ conceptions of mathematical knowledge and skills deficit. According to Kingsdorf and Krawec (2014) the error analysis approach has proven to be an effective tool for analyzing students’ errors in mathematics. Error analysis is a type of diagnostic assessment that can help a teacher determine what types of errors a student is making and why. More specifically, it is the process of identifying and reviewing students’ errors to determine whether an error pattern exists—that is, whether a student is making the same type of error consistently. Error analysis involves the analysis of error patterns to identify about students have with facts, concepts, strategies, and procedures. Identifying the type of error allows teachers to address the needs of the learners more efficiently (Kundu, U. and Sengupta, D, 2014). If a pattern does exist, the teacher can identify students’ misconceptions or skill deficits. A student has skill deficits if he does not know how to perform the desired skills. Students who are found weak in basic conceptual knowledge in mathematics commit errors in solving Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 285 problems (Kaur, N. A, 2011) and (Luneta, 2015). According to Elbrink(2008) and Marshall(1983), the general error categories in mathematics include processing language information, interpreting spatial information, selecting appropriate procedures, making concept associations, and using irrelevant rules or information, calculation errors, procedural errors, and symbolic errors. From the foregoing discussion, it is evident that students commit errors in mathematics due to their misconceptions and skill deficits, and thus important to analyze the nature and sources of these errors. If this is done, measures and interventions can be adopted to overcome them. Hence, a study is needed to analyze students’ misconceptions and skill deficits about subjects. This study analyzed the students’ misconceptions and skill deficits in Pre-Calculus subjects using an error analysis approach. Specifically, it answered the following research objectives, identify students’ misconceptions and skill deficits in pre-Calculus subjects, determine the reasons causing students’ misconceptions and skill deficits in Pre-Calculus subjects, and propose reinforcement activities to help students with mathematical misconceptions and skill deficits. The students’ misconceptions and skill deficits in pre-calculus subjects are two hindrances to the success of students in the calculus subject (Jones, 2016). To address such problems, identification, and analysis of students’ specific needs in every pre-calculus subject must ensure. Also, the factors influencing the students’ misconceptions and skill deficits need to be analyzed. Figure 1. Research Paradigm Figure 1 shows the paradigm of the study. The conceptual framework of the study used the diagram of the Error Analysis Approach. The diagram starts with the error analysis of the students’ performance in the pre-calculus subjects which include the subjects of Algebra, Trigonometry, and Analytic Geometry. The students’ errors were described and analyzed based on the result of the Pre-Calculus Test. From the analysis, the misconceptions and skill deficits of the students were identified. The identified misconceptions and skill deficits served as the basis for proposing reinforcement activities in addressing their misconceptions and skill deficits. Research on error analysis is not new. Researchers around the world have been conducting studies on this topic for decades. Error analysis is an effective method for identifying patterns of mathematical errors for any student, with or without disabilities, who are struggling in mathematics4. Error analysis can identify the patterns of errors or mistakes that students make in their work, understand why students make the errors, and provide targeted instruction to correct the errors (Cohen and Spenciner, 2016). This study adopted quantitative and qualitative approaches. Quantitative research specifically the descriptive method was used in the description of misconceptions and skill deficits of the students. Quantitative research is the numerical representation and manipulation of observations for the purpose of describing and explaining the phenomena that those observations reflect (Beng, Y. H and Yumus A.S, 2013). Moreover, the study also made use of a qualitative research design, a case study type. When: (a) the purpose of the study is to answer "how" and "why" questions; (b) you cannot manipulate the behavior of those involved in the study; (c) you want to cover contextual conditions because you believe they are relevant to the phenomenon under study; or (d) the boundaries between the phenomenon and context are unclear. This method is required when analyzing the underlying causes of students' misconceptions and skill deficits.The above premises justified the use of quantitative and qualitative approaches since they determined the students’ misconceptions and skill deficits in Pre-Calculus subjects and the underlying reasons behind these were analyzed. The participants of the study were 50 second-year Engineering students; 30 Civil, 10 Electrical, and 10 Mechanical. They were purposely chosen from a population of 450 students who have difficulties in Algebra, Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 286 Trigonometry, and Analytic Geometry. Three sets of instruments were used in gathering the data needed. One is the Pre-Calculus Test (PCT), which consisted of three sub-tests on Algebra, Trigonometry, and Analytic Geometry. The contents of the test were drawn from the learning competencies in the syllabi of instruction in these subjects. Mapping of competencies and skills developed among students in Pre-Calculus and Calculus subjects served as the first step in the development of the test. This mapping was the guide in doing the questions in the Pre-Calculus Test. The items were constructed to measure misconceptions and skill deficits. The second is the interview guide for students. It was used to determine the underlying reasons for their misconceptions and skill deficits. This was administered after the results of the Pre-Calculus test were initially analyzed. The answers and solutions provided by the students were also clarified during the interview. The third is the interview guide for teachers. It was utilized to clarify the identified misconceptions and skill deficits identified from the results of the Pre-Calculus test. The results of the interview provided consistency with the answers drawn from the other instruments. METHODS The figure below is the block diagram used as the error analysis model of the study. Figure 2. The Error Analysis Model The first step is the identification of participants. Coordination was made among teachers of Calculus to identify students who have difficulty in Pre-Calculus subjects. From a total of 450 engineering students, a total of 50 students comprising 10 electrical, 10 mechanical, and 30 civil engineering students were purposely chosen to compose the samples of the study. The purpose of the study was explained to the students. They were requested to voluntarily participate in the study. The Pre-Calculus test was administered to the participants in coordination with their subject teachers. After the retrieval of the tests, answers were checked to determine the items where the students had misconceptions and skill deficits (Brown, J., and Skow, K., 2014). The answers of the students in each item were analyzed by identifying the errors committed in Pre-Calculus subjects. Interviews with the participants were conducted after the test was checked to clarify their solutions and obtain relevant information from their answers. Moreover, interviews with the mathematics teachers were also conducted to gather additional information regarding the errors committed pertaining to their misconceptions and skill deficits in Pre-Calculus subjects. Interviews with students and teachers were conducted and analyzed to determine the reasons causing the misconceptions and skill deficits. Their answers were categorized according to the themes that emerged from their answers. The classification of answers by themes was done by analyzing their responses individually. Reinforcement activities in each of the Pre-requisite subjects were made based on the identified students’ misconceptions and skill deficits (Dawkins, P., 2016). Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 287 RESULTS AND DISCUSSIONS Students’ Misconceptions and Skill Deficits in Pre-Calculus Subjects The students’ misconceptions and skill deficits in Pre-Calculus subjects were presented and discussed according to the subject: Algebra, Trigonometry, and Analytic Geometry. Students’ Misconceptions in Algebra The students’ misconceptions in Algebra were categorized into algebraic expressions and functions, special products and factoring rational expressions, exponents and radicals, linear equations, quadratic and word problems, and exponential and logarithmic functions. Table 1 Number and Percentage of Students with Misconceptions in Algebra MAIN TOPIC EE ME CE TOTAL % Qualitative Description of Students with Misconceptio ns n =10 n= 10 n=30 n=50 Algebraic Expressions and Functions 7.7 7.0 16.3 31 62 Many Special Products and Factoring 7.3 8.0 19.3 35 69 Many Rational Expressions 6.8 5.0 19.3 31 62 Many Exponents and Radicals 7.0 6.0 20.3 33 67 Many Linear Equations, Quadratic and Word Problems 6.5 6.5 18.5 32 63 Many Logarithms 6.0 6.5 18.0 31 61 Many Average 6.9 6.5 18.6 32 64 Many Table 1 shows the number and percentage of students with misconceptions on selected topics in Algebra. On Algebraic Expressions and Functions, students had misconceptions in identifying similar terms, identifying algebraic expressions which were not polynomials, and defining relation and function. Many of the students (n = 31, 62%) in general have misconceptions about algebraic expressions and functions. Many of the students (n = 35, 69%) have misconceptions about special products and factoring. Many of the students from electrical engineering, mechanical engineering, and civil engineering have misconceptions about rational expression (n = 31, 62%), exponents and radicals (n = 33, 67%), linear equations, quadratic, and word problems (n = 32, 63%), and logarithms (n = 31, 61%). The results would indicate that many of the respondents have misconceptions about each of the topics in Algebra (n = 32, 64%). Some of them did not realize the difference between polynomial expression and algebraic expression, they did not distinguish that function is a type of relation (Kaur, 2013). The students easily forgot the patterns in factoring as they did not identify polynomial expressions which are not factorable and many other reasons why there is a big percentage of students who have misconceptions in algebra (Prudchenko, K., 2012). Students’ Skill Deficits in Algebra The students’ skill deficits in Algebra are categorized into algebraic expressions and operations, special products and factoring, rational expressions, exponents, and radicals, solving of linear, quadratic, rational, exponential, and logarithmic equations, and solving of word problems (Seed, 2014). Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 288 Table 2 Number and Percentage of Students with Incorrect Answers in the Skills of Algebra MAIN TOPIC EE ME CE TOTAL % Qualitative Description of Students with Skill Deficits n =10 n= 10 n=30 n=50 Algebraic Expressions and Functions 9.0 5.5 16.5 31 62 Many Special Products and Factoring 8.8 9.6 28.4 47 94 Almost All Rational Expressions 4.0 7.7 15.7 27 55 Some Exponents and Radicals 7.2 7.5 19.7 34 69 Many Solving of Exponential and Logarithmic Equations 9.0 9.0 21.5 40 79 Many Solving worded Problems 9.7 7.3 25.7 43 85 Almost All Average 7.9 7.8 21.2 37 74 Many Table 2 shows the number of students who got incorrect answers to skill items on selected topics in Algebra. Some of the respondents have a deficit in rational expression (n = 27, 55%). Many (n = 31, 62%) have a deficit in Algebraic Expressions and Operations, exponents, and radicals (n = 34, 69%), and solving of exponential and logarithmic equations (n = 40, 79%). The reason for their mistakes is, sign problems, they did not simplify the results, putting a group symbol, they did not recall the properties of the exponential function, and they did not know what to do in solving the problem. Almost all respondents have a deficit in special products and factoring (n = 47, 94%), and solving worded problems (n = 43, 85%). Most of them do not know what to do in factoring and solving word problems. The student had an incomplete solution and only wrote the first equation but failed to write the second (Jones, 2016) and (Vella, 2014). Students formed an incorrect equation and translated the words in math symbols incorrectly. Lastly, most of the students did not know how to translate the words in math symbols correctly due to the wrong analysis of the problem19. Students’ Misconceptions in Trigonometry The students’ misconceptions in Trigonometry are categorized into angle and angle measure, trigonometric functions, trigonometric identities, inverse trigonometric functions, graphs of trigonometric and inverse trigonometric functions, and solutions of triangles. Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 289 Table 3 Number and Percentage of Students with Misconceptions in Trigonometry MAIN TOPIC EE ME CE TOTAL % Qualitative Description of Students with Misconceptions n =10 n= 10 n=30 n=50 Angle Measure 8.0 6.0 24.0 38 76 Many Trigonometric Functions 6.0 7.4 18.7 32 64 Many Trigonometric Identities 3.0 7.0 20.5 31 61 Many Inverse Trigonometric Functions 8.0 7.0 23.6 39 77 Many Graph of Trigonometric and Inverse Trigonometric Functions 7.2 7.3 19.2 34 67 Many Solutions of Triangles 8.0 7.7 21.0 37 73 Many Average 6.7 7.1 21.2 35 70 Many Table 3 shows the number and percentage of students with misconceptions on selected topics in Trigonometry. Many students have misconceptions about angle measure (n = 38, 76%), trigonometric functions (n = 32, 64%), trigonometric identities (n = 31, 61%), inverse trigonometric functions (n = 39, 77%), graph of trigonometric and inverse trigonometric functions (n = 34, 67%), and solutions of triangle (n = 37, 73%). The reasons are, that they assumed that the representation of radian is π which is wrong since decimal can also be used to represent a radian measure. Students were not comfortable using the radian measure since they did not understand its meaning. In measuring angle measures, they frequently used degree measure instead of radian measure. Students were always uncertain when it comes to trigonometric functions. They have difficulty recalling the concept of even function. Students always forgot the values of trigonometric functions since they did not understand how to obtain the values (Seed, 2014). They only memorized it for the sake of compliance. Students said that they did not consider the reference angle and they only assumed that the horizontal side was the adjacent side while the vertical side of the triangle was the opposite side. They easily forget the trigonometric functions (Dawkins, 2016). The students admitted that they were confused about the signs of trigonometric functions in quadrants. For them, the co-function of sine is cosecant. The students always use this in their solutions to trigonometric functions (Seed, 2014). They also added that the co-function of cosine is secant. These students did not realize that what they did was reciprocal functions and not co-functions. The co-function of sine is cosine since if their corresponding angles are complementary then their values are equal. They said also that the reciprocal of cosecant was the secant function. students said that these functions are always partners. Students did not realize what they thought was the co-function. The reciprocal of cosecant is the sine function since cosecant is equal to 1 over the sine function. The students always forget not only the range but also the domain of trigonometric functions. The students said that the inverse of the cosine function is the sine function. They were not aware of the relations between the trigonometric functions and their inverse. The students forgot the trigonometric function values, so they tend to only guess their answers. The students do not understand the concept of symmetry. They always forget the conditions when using the different laws of trigonometric functions. They added that aside from the conditions, students also forgot the formulas of the laws of trigonometry. Students’ Skill Deficits in Trigonometry The students’ skill deficits in Trigonometry are categorized into trigonometric functions, inverse trigonometric functions, proving identities, graphs of trigonometric and inverse trigonometric functions, and application of trigonometric functions (Dawkins, 2016). Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 290 Table 4 Number and Percentage of Students with Incorrect Answers in Skills of Trigonometry MAIN TOPIC EE ME CE TOTAL % Qualitative Description of Students with Skill Deficits n =10 n= 10 n=30 n=50 Trigonometric Functions 5.7 9.3 22.3 37 75 Many Inverse Trigonometric Functions 6.7 9.7 23.7 40 80 Many Trigonometric Identities 4.7 7.7 16.7 29 58 Some Graph of Trigonometric and Inverse Trigonometric Functions 8.6 10.0 28.4 47 94 Almost All Application of Trigonometric Functions 9.0 9.7 28.0 47 93 Almost All Average 6.9 9.3 23.8 40 80 Many Table 4 shows the number of students who got incorrect answers to items on selected topics in Trigonometry. Some of them have skills deficits in trigonometric identities (n = 29, 58%). Many of the have skills deficits in trigonometric functions (n = 37, 75%), and inverse trigonometric functions (n = 40, 80%). Almost all of them have skills deficits in graphing trigonometric and inverse trigonometric functions (n = 47, 94%), and in the application of trigonometric functions (n = 47, 93%). The reasons are, that students forgot the trigonometric values as they assumed the value of cos π as zero instead of -1. The students cannot recall the trigonometric function values for sine and cosine function in degree measure while taking the examination. They forgot the steps in finding the value of the trigonometric function. The students obtained an error in evaluating trigonometric values for sine and cosine using the operation of addition (Ozkan and Ozkan, 2012). The students need to apply first the inverse sine function since the condition was to find an angle x for the sine function that makes it equal to one half. The students also need to recall the values of trigonometric functions to obtain the correct answer. The students also forgot the steps in solving inverse trigonometric functions. In proving trigonometric equations, students need to recall the different fundamental identities which in this case, the use of Pythagorean and ratio identities. The students also need to have a critical observation of the problem so that the problem will be solved. In proving trigonometric equations, students need to recall the different fundamental identities which in this case, the use of double angle identity for the sine function20. The students also need to have a critical observation of the problem so that the problem will be solved. In proving trigonometric equations, students need to recall the different fundamental identities which in this case, the use of Pythagorean identity (Idris, 2011). The students also need to have a critical observation of the problem so that the problem will be solved. The students were not able to graph the function since they forgot the steps in graphing trigonometric functions. They were able to solve the hypotenuse of the triangle but did not obtain the final answer since they were taught that this was the final answer. The students need to obtain the given information and determine what is being asked in the problem16. The use of an oblique triangle involving cosine law is needed to obtain the correct answer. The student can draw the figure correctly, but he used the right triangle instead of the oblique triangle (Idris, 2011). Students’ Misconceptions in Analytic Geometry The students’ misconceptions in Analytic Geometry are categorized into a rectangular coordinate system, the slope of a line, equations of a line, conic sections, and plane and solid figures. Table 6 shows the number and percentage of students with misconceptions on selected topics in Analytic Geometry. Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 291 Table 5 Number and Percentage of Students with Misconceptions in Analytic Geometry MAIN TOPIC EE ME CE TOTAL % Qualitative Description of Students with Misconceptions n =10 n= 10 n=30 n=50 Rectangular Coordinate System 6.8 5.8 16.8 29 59 Some Slope of a Line 6.2 6.3 19.5 32 64 Many Equations of a Line 5.3 6.7 18.9 31 62 Many Conic Sections 6.5 6.5 21.1 34 68 Many Plane and Solid Figures 5.0 5.5 17.5 28 56 Some Average 5.9 6.2 18.7 31 62 Many Some of the respondents have misconceptions about the rectangular coordinate system (n = 29, 59%), and the plane and solid figures (n = 28, 56%). Many of the respondents have misconceptions about the slope of a line (n = 32, 64%), the equations of a line (n = 31, 62%), and the conic sections (n = 34, 68%). The reason is, that the distance of a point P from the y-axis is the y-coordinate which is wrong since it is the x-coordinate. They did not analyze the problem carefully. Since the y-axis is on the problem, they already assumed that it is the y-coordinate in complement of the y-axis. The students have misconceptions about defining the x-coordinate (Jones, 2016). They said that students frequently interchanged the definition of x and y-coordinates. The students were not aware of the abscissa and ordinate. They could hardly recognize which is the x or the y-coordinate. Often, the two are interchanged. They could not define the origin. If the definition is given to them, they would surely memorize it without understanding16. The students did not understand the concept of negative slope in a graph of a line. They were merely making a wild guess in identifying the slope of a line. The students were not aware of the sign of the slope of the line whether it is parallel or perpendicular to the axes. They were not aware of the slope of a line whether it is parallel or perpendicular to the axes. The students frequently made a mistake in finding the slope of a line since they used incorrect formula for slope (Ozkan and Ozkan, 2012). It simply shows that students did not have a strong conceptual understanding of the slope of a line. The students were confused about the different equations of a line. They did not recognize the specific formula to be used in finding the equation of a line. The students were confused about zero slopes since they did not understand the concept of slope in an equation (Idris, 2011). The students have misconceptions in identifying the perimeter of rectangles since they were always confused about formulas. They used to forget the formula by not having twice its length and width. Instead, they only take its product as in the formula of area (Beng, Y. H and Yumus A.S, 2013). They were confused about the formula of perimeter and area. Students’ Skill Deficits in Analytic Geometry Students’ skill deficits in Analytic Geometry are categorized into the slope of a line, graphing of functions and relations, equation of a line, and curve sketching. Table 6 Number and Percentage of Students’ Skill Deficits in Analytic Geometry MAIN TOPIC EE ME CE TOTAL % Qualitative Description of Students with Skills Deficits n =10 n= 10 n=30 n=50 Finding Slope of a Line 7.0 7.3 21.0 35 71 Many Graphing of Functions and Relations 7.6 9.0 27.6 44 88 Almost All Solving Equations of a Line 8.0 9.0 26.0 43 86 Almost All Drawing of Curves 4.0 3.0 18.0 25 50 Some Average 6.7 7.1 23.2 37 74 Many Table 6 shows the number of students who got incorrect answers to the items on selected topics in Analytic Geometry. Some of the respondents have skills deficits in drawing curves (n = 25, 50%). Many of the respondents have skills deficits in finding the slope of a line (n = 35, 71%). Almost all the respondents have skills deficits in graphing functions and relations (n = 44, 88%), and solving equations of a line (n = 43, 86%). The reasons are, that the student assumed that the x and y-axis was the slope and used the derivative in finding Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 292 the slope but failed to obtain the answer since the derivative of a constant was incorrect. The students forgot the steps in finding the slope of a line. The students don’t know how to manipulate the equation in different forms to obtain the slope (Kaur, 2013). The equation is required to change into slope-intercept form. They solved the x and y-intercepts and made these the x and y coordinates of a point. They don’t know how to graph quadratic equations (Prudchenko, 2012). The students forgot the steps in graphing rational functions. The students solved for x and y intercepts and plotted this to the Cartesian plane and mistakenly connected the points as a line instead of the parabola. They forgot the steps in finding the equation of a line and were able to find the slope but did not find the equation (Vella, 2014). Students’ Reasons that Caused Misconceptions and Skill Deficits in Algebra The reasons for students’ misconceptions and skill deficits in Algebra were categorized based on the description of students’ reasons. Table 7 Students’ Reasons for Misconceptions and Skill Deficits in Algebra MAIN TOPIC Description of Students' Reason Algebraic Expressions and Functions Students’ Poor Knowledge Retention Students’ Attitude Toward the Lesson Special Products and Factoring Students’ Readiness Students’ Absences Teachers’ methodology Students’ Poor Knowledge Retention Rational Expressions Students’ Poor Knowledge Retention Time of Schedule of Classes Teachers’ methodology Exponents and Radicals Students’ Poor Knowledge Retention Students’ Tardiness Linear Equations, Quadratic and Word Problems Students’ Poor Knowledge Retention Teachers’ methodology Curriculum/Time of Classes Students’ Language Deficit Students’ Poor Analysis Logarithms Students’ Poor Knowledge Retention Teachers’ methodology Table 7 shows the students’ reasons for misconceptions and skill deficits in Algebra. Generally, the students’ reasons for causing misconceptions and skill deficits were poor knowledge retention and attitudes toward the lesson (Tall, 2002). In Algebra, the students’ reason was due to poor knowledge retention. It was evident in the statement “I forgot the topic” which means that they had already forgotten the concepts and skills in identifying polynomial expressions. One teacher suggested that the students should always review their notes and stay focused. The other statements were “hindi naintindihan”, “nakakalito” and “a bit confusing”. The student’s poor knowledge retention was observed in translating words to math symbols. They have also problems understanding the English language used in problem-solving. Students’ Reasons Causing Misconceptions and Skill Deficits in Trigonometry The reasons for students’ misconceptions and skill deficits in Trigonometry were categorized based on the description of students’ reasons. Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 293 Table 8 Students’ Reasons for Misconceptions and Skill Deficits in Trigonometry MAIN TOPIC Description of Students' Reason Angle Measures Students’ Poor Knowledge Retention Trigonometric Functions Students attitude toward listening School Activities/Programs Students’ Poor Knowledge Retention Trigonometric Identities Students’ Poor Knowledge Retention Time of Schedule of Classes Teachers’ methodology Inverse Trigonometric Functions Students’ Poor Knowledge Retention Students attitude toward listening Teachers’ Methodology Graph of Trigonometric and Inverse Trigonometric Functions Students’ Poor Knowledge Retention Teachers’ methodology Students attitude toward listening Solutions of Triangles Students’ Poor Knowledge Retention Students attitude toward listening Students’ Poor Analysis On Trigonometry, the reason for misconceptions and skill deficits was due to students’ poor knowledge retention. In trigonometry, the statements “I forgot the topic”, “nakalimutan”, “nakalimutan na”, “hindi ko po kabisado yung conversion formula” and “medyo nakalimutan na rin” mean that the students’ retention of the concepts and skills was poor. This was evident in the statements of the students “I forgot the topic”, “nakalimutan”, “hindi ko sure kung napagbaliktad ko na yung co-function at reciprocal”, “hindi ko po kabisado, lagi kong nakakalimutan”, “nakalimutan na sa dami ng formula”, “it always rumbles in my mind”, “I forgot”, and “nakalimutan ko na dahil karaming rules na dapat tandaan at nakakalito”. Another reason for students’ misconceptions and skill deficits was due to teachers’ methodology. From the statements of the student, “nalilito po ako hindi po kasi tinuro ng prof” and “because I don’t know to my teacher, she always said the topics are easy, but she can’t teach these to us”, it can be deduced that the methods of teaching were not fitted to them. Students’ Reasons Causing Misconceptions and Skill Deficits in Analytic Geometry The reasons for students’ misconceptions and skill deficits in Analytic Geometry were categorized based on the description of students’ reasons. Table 9 Students’ Reasons for Misconceptions and Skill Deficits in Analytic Geometry MAIN TOPIC Description of Students' Reason Rectangular Coordinate System Students’ Poor Knowledge Retention Slope of a Line Students’ Poor Knowledge Retention School Activities/Programs Equations of a Line Students’ Poor Knowledge Retention Teachers’ methodology Conic Sections Students’ Poor Knowledge Retention Students attitude toward listening Students’ Lack Concentration Students’ Absences Students' Poor Graphic Skills Students Lack of Vocabulary Teachers’ Methodology Plane and Solid Figures Students’ Poor Knowledge Retention Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 294 In Analytic Geometry, the reason for misconceptions and skill deficits was due to students’ poor knowledge retention. The students’ poor knowledge retention was observed in defining the second coordinate, identifying the first coordinate, defining the origin, and identifying the term for the first coordinate. It was evident in their statements such as “I forgot” and “nalilito”. The students’ poor knowledge retention was observed in identifying the slope formula, identifying the slope of perpendicular lines, identifying the slope of the horizontal line, identifying the slope of vertical lines, and identifying the negative slope. The statements of the students such as “nakalimutan po”, “I forgot the lesson”, “nalilimutan yung mga iba”, “naghalo-halo yung mga formulas, hindi ko na alam kung ano gagamitin” means that the students did not recall the concepts and skill needed in the question. The many school activities can be seen also in identifying negative slopes. The statement “we didn’t meet our teacher because of the school activities” means that they did not tackle the lesson due to the school events. Proposed Reinforcement Activities The proposed reinforcement activities in Pre-Calculus were divided into subject categories, namely: Algebra Trigonometry and Analytic Geometry. Each subject was composed of different activities to reinforce the concepts and skills found to have misconceptions and skill deficits. The approach of the activities was made funny and enjoyable for the benefit of the students with difficulties in understanding concepts and skills in Pre-Calculus subjects22. Each activity has this format:(1) introduction, which will give students the idea about the activity they will deal with, (2) learning outcome/s, that will need to meet at the end of the activity, and (3) list of materials needed in the activity, (4) a step-by-step procedure, that serves as a guide for the teachers, (5) assessment, that will assess the retention of the concepts and skills gained by the students in the activity, and (6) references for additional search about the lesson. In Algebra, there were 12 activities which include the following: Pick Me Up! Factoring Battle, Common Multiple War, Pick Me Up 2!, Pick a Statement, Ex-Log War, Polynomial Division Flip, Polynomial Tag, Speed Dating, Row Game, Rational and Quadratic Equation Hunt, and Word Problem Battle. In Trigonometry, the six (6) activities include the following: Measure Me! Follow me! The Biorhythm Cycle, Trigonometry Square 1 and 2, and Graphing Inverse Trigonometric Functions. Analytic Geometry also has six (6) activities namely: Solve a graph Puzzle, Slope of a Line, Story of My Equation, Camera Activity, Human Conics, and Measure Me2! CONCLUSIONS AND RECOMMENDATIONS Based on the findings, the following conclusions are drawn: Students have misconceptions about the basic concepts and laws in Algebra that are necessary for performing fundamental algebraic operations. They also lack the required skills in performing algebraic operations. Students have misconceptions about the basic concepts of Trigonometry. These basic concepts are necessary in the study of Analytic Geometry. The students’ misconceptions and skill deficits in Analytic Geometry are attributed to the students’ misconceptions and skill deficits in Algebra and Trigonometry. The underlying reasons for students’ misconceptions are due to students’ lack of conceptual understanding and retention in their schema. The students easily forget what they previously learned because of rote learning. They tend to memorize the concepts needed for skillful mathematical operation instead of grasping the meaning of these concepts. Some skill deficits are caused by carelessness and incorrect application of laws and principles. Mathematics teachers are encouraged to provide immediate feedback to students and their peers on students’ misconceptions and skill deficits so that these can be addressed earlier. In doing so, the escalation of these problems in higher mathematics subjects can be minimized. Additional reinforcement activities in Algebra, Trigonometry, and Analytic Geometry are proposed to be given in addition to the learning tasks and examples provided during classes. Mathematics teachers are encouraged to reflect on and or revisit their teaching practices as these may provide them sound bases in utilizing other teaching methods that may redound to improve students’ mathematics performance. A remediation program can be designed to help students with misconceptions and skill deficits in Pre-calculus subjects. The underlying reasons causing students’ misconceptions and deficit skills in Pre-Calculus subjects be shared with the other mathematics teachers and students so that they may know what to focus on and emphasize during teaching and learning. The students may also be given training/seminar on time management so they can balance their studies and other extra-curricular activities participated in. The proposed reinforcement activities are recommended for implementation in mathematics classes to improve identified deficiencies in content. Similar studies are recommended to be conducted on the students’ misconceptions and skill deficits in other subject areas of mathematics. Journal for Educators, Teachers and Trainers JETT, Vol. 13 (5); ISSN: 1989-9572 295 REFERENCES 1. Beng, YH. and Yunus, A(2013). A holistic approach to activate and enhance prior knowledge of tertiary learners in the upcoming lectures of calculus, Department of Physical and Mathematical Science, Universiti Tunku Abdul Rahman, Jalan Universiti, Bandar Barat 31900 Kampar, Perak, Malaysia 156 2. Smith III, J., Sessa, A. and Roschelle, J. (2009). Misconceptions Reconceived: a constructivist analysis of knowledge in transition. Published online: 17, page 115-163 3. Mwavita, M (2005). Factors influencing calculus course success among freshmen engineering students, 2005. 4. Kingsdorf, S. and Krawek J.(2014). Error analysis of mathematical word problem solving across students with and without learning disabilities, Learning Disabilities Research and Practice, 29(2),pp 66-74 5. Kundu, U. and Sengupta, D (2014). Error analysis in mathematics in Relation to secondary school students. Indian Journal of Educational Research, Volume-III, 2014, pp. 105-125 6. Kaur, N. (2011). A diagnostic study of errors in mathematics among ninth grade students at different levels of intelligence. Researcher’s Tandem, 2(8), 22-31. 7. Luneta, K (2015). Understanding students’ misconceptions: an analyss of final grade 12 examination questions in geometry. Pythagoras; Vol 36, No 1, 11 pages. doi: 10.4102/pythagoras.v36i1.261 8. Elbrink, M. (2008). Analyzing and addressing common mathematical errors in secondary education. B.S. Undergraduate Mathematics Exchange, 50(1), 2-4. Retrieved from 9. Marshall, S.P. (1983). Sex differences in mathematical errors: An analysis of distracter choices. Journal for Research in Mathematics Education, 14, 325-336. 10. Jones, J. (2016). Skills needed for success in calculus, Illinois Mathematical Association of Richland Community College(IMACC), 2014, Retrieved from July 2016 11. Cohen, L. and Spenciner, L. (2016). Error analysis of mathematics, Pearson Allyn Bacon Prentice Hall, 2010, Retrieved from July 2016 12. Brown, J., Skow, K.(2014). The IRIS Center. Mathematics: Identifying and addressing students errors, Claremont Graduate University Vanderbilt Peabody College, 2014, Retrieved from July 2016 13. Dawkins, P. (2016).Paul’s online math notes: algebra/trigonometry review, 2016, Retrived from 14. Kaur, G, A (2013). Review of selected literatures on causative agents and identification strategies of students’ misconceptions, University School of Education, Desh Bhagat University, Mandi Gobindgarh, Punjab, India, 2013, Vol. 2, No. 11, Nov. 2013. 15. Prudchenko, K. (2012). What math courses should students take prior to calculus?, Demand Meia, 2012, Retrived from 16. Seed, K. (2014). Reasons for errors of students in solving mathematical problems, Educational Research Article, 2014. Retrieved from 17. Vella, AD.(2014).What is the prerequisite to learn calculus?, Austine Community College, 2014, Retrieved from 18. Dejarnette, A. (2014). Students’ conceptions of trigonometric functions and positioning practices during pair work with ETOYS, Dissertation, University of Illinois at Urbana-Champaign 19. Tall, D. (2002). Students’ difficulties in calculus, Mathematics Education Research Center University of Warwick, Plenary presentation in working group 3, ICME, Quebec 20. Idris, S. (2011)Error patterns in addition and subtraction for fractions among form two students, Journal of mathematics Education, 2011, 4(2), pp 35-54 21. Ozkan, EM. and Ozkan (2012). A, Misconception in Exponential numbers in IST and IIND level primary school mathematics, procedia-social and behavioral sciences, 2012, 46(2012)65-69, www.sciencedirect.com
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Using the reaction quotient (video) | Khan Academy
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AP®︎/College Chemistry
Course: AP®︎/College Chemistry>Unit 7
Lesson 6: Using the reaction quotient
Using the reaction quotient
Worked example: Using the reaction quotient to find equilibrium partial pressures
Worked example: Using the reaction quotient to predict a pressure change
Using the reaction quotient
Science>
AP®︎/College Chemistry>
Equilibrium>
Using the reaction quotient
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Using the reaction quotient
AP.Chem: TRA‑7 (EU), TRA‑7.F (LO), TRA‑7.F.1 (EK), TRA‑8 (EU), TRA‑8.B (LO)
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About About this video Transcript
By comparing the reaction quotient to the equilibrium constant, we can predict the direction a reaction will proceed to reach equilibrium. If Q<K, the reaction will proceed towards the products. If Q>K, the reaction will proceed towards the reactants. If Q = K, the reaction is already at equilibrium and will not change.Created by Jay.
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Forever Learner 3 years ago Posted 3 years ago. Direct link to Forever Learner's post “I don't see how you could...” more I don't see how you could get a different Qc value than your Kc value if you calculate them the exact same way. I guess sometimes the actual experimental Kc value is different than the Kc calculated? I just want to know why it would be different.....A big bucket of thanks to anybody who helps me with this! ;)
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Richard 3 years ago Posted 3 years ago. Direct link to Richard's post “So the reaction quotient,...” more So the reaction quotient, Qc, the equilibrium constant, Kc, are calculated using the same equation for any particular reaction, but the key is that we input different concentrations at different time periods for the two values.
The equilibrium constant is a single value (at a particular temperature) in which the forward and reverse reaction rates are equal. Realistically this is observed at constant concentrations for all chemical species involved in the reaction. So it's a single set of concentrations which yields a single unchanging value.
The reaction quotient meanwhile is taken at any instant of time and can have infinitely many possible values. The purpose of the reaction quotient is just to compare to the equilibrium constant to judge which reaction rate will be favored; the forward or the reverse.
Hope that helps.
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Jaden Leggett a year ago Posted a year ago. Direct link to Jaden Leggett's post “In the last example how d...” more In the last example how do you know what [CO] and [Cl2] equals? is this always a value that will be given and you just plug it in or ?
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Richard a year ago Posted a year ago. Direct link to Richard's post “In this example, Jay simp...” more In this example, Jay simply selected three random nonequilibrium concentrations for the reactants and products to demonstrate that the reaction quotient was not equal to the equilibrium constant and that the reaction is not at equilibrium. Some problems will be more simple where the concentrations are provided and you simply plug into Q like you mentioned, but others require that find the concentrations somehow first.
Hope that helps.
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John Smith 3 years ago Posted 3 years ago. Direct link to John Smith's post “Why does each dot (blue/r...” more Why does each dot (blue/red) represent 0.1M? Is this arbitrary for explanation purposes?
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Richard 3 years ago Posted 3 years ago. Direct link to Richard's post “Yeah you're right, it's a...” more Yeah you're right, it's arbitrary. Jay could have picked any value for each of the dots and the process would be identical.
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JegoatEM 6 months ago Posted 6 months ago. Direct link to JegoatEM's post “Wouldnt it be Qp and Kp s...” more Wouldnt it be Qp and Kp since we are dealing with gases
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Richard 6 months ago Posted 6 months ago. Direct link to Richard's post “We use Kp/Qp when were ar...” more We use Kp/Qp when were are expressing the concentrations of a reaction’s chemical species as partial pressures. And Kc/Qc are used when the concentrations are expressed as molarities. Gas concentrations can be expressed as either partial pressures or molarities. In this case, since the gas concentrations are expressed as molarities, the use of Kc/Qc is correct.
Hope that helps.
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Winter 2 years ago Posted 2 years ago. Direct link to Winter's post “Is it ever possible for Q...” more Is it ever possible for Qc > Kc in the example with the red and blue dots from the video?
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Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “Well yeah, that simply me...” more Well yeah, that simply means that we have more products than is allowed at equilibrium. Any reaction is capable of having too much of either reactants or products for equilibrium.
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Jordan 8 months ago Posted 8 months ago. Direct link to Jordan's post “3:56 where did he get the...” more 3:56 where did he get the concentrations from or was it a part of the problem
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Richard 8 months ago Posted 8 months ago. Direct link to Richard's post “Jay explained this at 0:5...” more Jay explained this at 0:56 .
2 comments Comment on Richard's post “Jay explained this at 0:5...”
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aisha mohamed a year ago Posted a year ago. Direct link to aisha mohamed's post “why did he change from 0....” more why did he change from 0.1 M to 1 M at the second and the third reaction
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Richard a year ago Posted a year ago. Direct link to Richard's post “He simplified the math si...” more He simplified the math since you get the same ratios by counting the colored dots in the particulate diagrams as you ould if you actually used molarities. For exmaple, 0.5/0.3 = 5/3
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Video transcript
[Instructor] The reaction quotient is symbolized by the capital letter Q. And it tells us whether a reaction is at equilibrium or not. If the reaction is not at equilibrium, it also allows us to predict which direction the net reaction will go to reach equilibrium. For example, let's look at the hypothetical reaction where A gas turns into gas B. Gas A will be represented by red circles or red spheres, and gas B will be represented by blue spheres. The equilibrium constant for this hypothetical reaction is equal to 3 at 25 degrees Celsius. Let's start by writing out the expression for the reaction quotient. So we would write out here QC is equal to, and this has the same form as the equilibrium constant expression. So we would put concentration of B to the first power, divided by the concentration of A, also to the first power. Let's look at our first particulate diagram here, and let's think about each particle representing 0.1 moles of a substance. And the volume of the container is one liter. So let's first find the concentration of B so we can plug it into our expression for Q. B are the blue spheres. So we count up one, two, three blue spheres. Each sphere or each particle represents 0.1 moles. So three times 0.1 is 0.3. And then we divide that by the volume of one liter to get a concentration of 0.3 moles. So we can go ahead and plug in 0.3 mole for the concentration of B. Next, we divide that by the concentration of A, and since there are five red particles, and each particle represents 0.1 moles, five times 0.1 is 0.5 moles of A divided by a volume of one liter, it gives a concentration of 0.5 mole for A. Notice that we could have just counted the number of particles, three blues and five reds, and just done three divided by five and get the same value for the reaction quotient Q. So QC is equal to three fists or 0.6. And KC remember was equal to three. So Q is not equal to K. In this case, QC is less than KC. Since Q is not equal to K, at this moment and time, the reaction is not at equilibrium. In order for this reaction to reach equilibrium, Q needs to be equal to K. And since Q is a lot smaller than K, if you think about it, we need to increase the numerator and decrease the denominator. So that means, we have too many reactants and not enough products. And so the net reaction, if go back up here to the equation here, the net reaction is going to move to the right to make more products. So the net reaction moves to the right to make more blue particles and therefore the number of red particles would decrease. We can see that comparing these first two particulate diagrams here. So let's compare the first particular diagram to the second particular diagram. And the first particulate diagram, we had three blues and five reds. And the second particulate diagram, we have five blues and only three reds. So that shows the reaction has moved to the right to increase the amount of products and to decrease the amount of reactants. Let's calculate QC at this moment and time for our second particular diagram to see if we've reached equilibrium yet. Well, we have five blue particles and only three reds. So QC would be equal to five over three or five thirds. Remember that the equilibrium constant KC, is equal to three. Therefore, QC is still not equal to KC. And therefore the reaction is not at equilibrium. And Q is actually still less than K. Therefore the net reaction is going to move to the right again to increase the amount of products and to decrease the amount of reactants. Let's compare our second particulate diagram to our third particulate diagram. And the second particulate diagram, we had five blues and three reds. And then the third one here, we have one, two, three, four, five, six blues, and two reds. So we've increased in the amount of blue and we've decreased in the amount of red. Let's calculate QC for the moment of time represented by our third particulate diagram. Well, there are six blues and only two reds. So QC is equal to six divided by two, which is equal to three. So QC is equal to three and remember K is also equal to three. So QC is equal to KC and therefore this reaction is now at equilibrium. And at equilibrium, the reactions turn into the products at the same rate the products turn back into the reactants. And therefore the concentration of both reactants and products remained constant at equilibrium. So when Q is less than K, the reaction is not at equilibrium. There are too many reactants and not enough products. Therefore the net reaction goes to the right to increase the amount of products. The net reaction continues to go to the right until Q is equal to K and the reaction has reached equilibrium. At that point, the concentrations of reactants and products stop changing and they remain constant. It's also possible for Q to be greater than K. And if that's true, the reaction is not at equilibrium, but in this case you have too many products and not enough reactants. Therefore the net reaction goes to the left to increase the amount of reactants and to decrease the amount of products. The net reaction will continue to go to the left until Q is equal to K and the reaction reaches equilibrium. Let's look at another reaction, which is the decomposition of phosgene to form carbon monoxide and chlorine gas. KC for this reaction is equal to 2.2 times 10 to the negative 10th at 100 degrees Celsius. And let's say we're given concentrations of phosgene carbon monoxide and chlorine at a moment in time, and asked if the reaction is at equilibrium or not. And if the reaction's not at equilibrium, we need to predict which direction the net reaction will go to reach equilibrium. Our approach is gonna be to calculate QC at that moment in time, and then compare QC to KC. So first we need to write our QC expression, and this is equal to the concentration of carbon monoxide raised to the first power times the concentration of chlorine raised to the first power and that's divided by the concentration of phosgene. So the concentration of COCL2. So at this moment of time, the concentration of carbon monoxide is 3.4 times 10 to the negative six mole. The concentration of chlorine is 6.0 times, times 10 to the negative six mole, and the concentration of phosgene is equal to 2.0 times 10 to the negative third mole. When we plug all those into our Q expression and solve, we get that QC is equal to 1.0 times 10 to the negative eighth. So in this case, QC is greater than KC because QC is equal to 1.0 times 10 to the negative eight and KC is equal to 2.2 times 10 to the negative 10th. And when QC is greater than KC, we have too many products and not enough reactants. Therefore the net reaction is going to go to the left and there's going to be an increase in the amount of phosgene. The reaction will continue to go to the left until Q is equal to K, and the reaction reaches equilibrium.
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Antipsychotic Medications
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Author: Krutika Chokhawala
Editor: Lee Stevens
Updated: 2/26/2023 5:58:45 PM
Indications
First-generation antipsychotics are dopamine receptor antagonists and are known as typical antipsychotics. They include phenothiazines (trifluoperazine, perphenazine, prochlorperazine, acetophenazine, triflupromazine, mesoridazine), butyrophenones (haloperidol), thioxanthenes (thiothixene, chlorprothixene), dibenzoxazepines (loxapine), dihydroindoles (molindone), and diphenylbutylpiperidines (pimozide).
Second-generation antipsychotics are serotonin-dopamine antagonists and are also known as atypical antipsychotics. The Food and Drug Administration (FDA) has approved 12 atypical antipsychotics as of the year 2016. They are risperidone, olanzapine, quetiapine, ziprasidone, aripiprazole, paliperidone, asenapine, lurasidone, iloperidone, cariprazine, brexpiprazole, and clozapine.
Indications
Schizophrenia and Schizoaffective disorders: First and second-generation antipsychotics (except clozapine) are indicated for the treatment of an acute episode of psychoses and maintenance therapy of schizophrenia and schizoaffective disorders. First-generation antipsychotics are better for treating positive symptoms of schizophrenia, eg, hallucinations, delusions, among others. They also decrease the risk of a repeat episode of psychosis. Second-generation antipsychotics treat both positive symptoms and negative symptoms of schizophrenia, eg, withdrawal and ambivalence, among others, and are known to reduce relapse rates.
Acute Mania: First-generation antipsychotics are effective in the treatment of acute mania with psychotic symptoms. All second-generation antipsychotics except clozapine can also be used as a treatment of symptoms of acute mania. Antipsychotics are used with mood stabilizers like lithium, valproic acid, or carbamazepine initially, and then after symptoms stabilize, they can be gradually decreased and withdrawn.
Major Depressive Disorder with Psychotic features: First or second-generation antipsychotics, along with an antidepressant, are the treatment of choice for depression with psychotic features. Olanzapine and fluoxetine, as a combination therapy, have FDA approval for treatment-resistant depression.
Delusional Disorder: First-generation antipsychotics are indicated in the treatment of delusional disorder and paranoia associated with personality disorders.
Severe Agitation: Severely agitated, irritable, hostile, and hyperactive patients can be treated with a short-term course of first-generation antipsychotics, irrespective of the etiology of the behavioral disturbance. Second-generation antipsychotics can also be used for treating acute agitation. Antipsychotics can also be used in children with severe autism exhibiting behavioral disturbances, though repeatedly giving antipsychotics is not preferred. Risperidone and olanzapine are useful for controlling aggression in children.
Tourette Disorder: Haloperidol and pimozide are the antipsychotics most commonly used for this syndrome. Tourette disorder is an off-label indication for second-generation antipsychotics.
Borderline Personality Disorder: This type of personality disorder can have symptoms of psychosis and paranoia. Both first and second-generation antipsychotics are used for the treatment of these symptoms.
Dementia and Delirium: A low dose of high-potency first-generation antipsychotics like haloperidol is recommended for treating agitation in delirium and dementia. It is essential to use caution in elderly patients, as the antimuscarinic effects can cause significant adverse effects in this population. Second-generation antipsychotics can also be used for treating behavioral disturbances in dementia. Off-label use of second-generation antipsychotics is acquired immunodeficiency syndrome-related dementia.
Substance-induced psychotic disorder: In cases of severe psychosis secondary to substance use, antipsychotics can be used to control agitation symptoms. Caution is necessary when using first-generation antipsychotics in alcohol withdrawal and phencyclidine intoxication.
Childhood Schizophrenia: Recent studies have shown the benefit of clozapine in treating early-onset schizophrenia.
Other Indications
Huntington disease, Parkinson disease, Lesch-Nyhan syndrome, and pervasive developmental disorder are some other conditions where antipsychotics are an option, though it is not the primary drug of choice.
Clozapine
This drug is the agent of choice when the patient has failed multiple trials of standard antipsychotic therapies. Clozapine is also useful in the treatment of tardive dyskinesia. Indications for the use of clozapine include treatment-resistant mania, severe psychotic features, obsessive-compulsive disorder, pervasive developmental disorders, childhood autism, Parkinson disease, Huntington disease, and suicidal patients with schizophrenia or schizoaffective disorder.
Mechanism of Action
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Mechanism of Action
The first-generation antipsychotics work by inhibiting dopaminergic neurotransmission; their effectiveness is best when they block about 72% of the D2 dopamine receptors in the brain. They also have noradrenergic, cholinergic, and histaminergic blocking action. Second-generation antipsychotics work by blocking D2 dopamine receptors as well as serotonin receptor antagonist action. 5-HT2A subtype of serotonin receptor is most commonly involved.
Administration
First-Generation Antipsychotics
All dopamine receptor antagonists are available and can be administered in oral form. Except for thioridazine, pimozide, and molindone, all other first-generation antipsychotics can also be given parenterally. Haloperidol and fluphenazine can be delivered in long-acting depot parenteral form.
Second-Generation Antipsychotics
These can be administered orally or parenterally. Risperidone, olanzapine, aripiprazole, and paliperidone are extended-release or long-acting injectable forms. Clozapine, asenapine, and olanzapine are available in the sublingual formulation.
Adverse Effects
First-generation antipsychotics (FGAs) are associated with significant extrapyramidal side effects. Anticholinergic adverse effects like dry mouth, constipation, and urinary retention are common with low-potency dopamine receptor antagonists like chlorpromazine and thioridazine. The action of H1 histamine blocking by first-generation antipsychotics causes sedation. Chlorpromazine is the most sedating, while fluphenazine, haloperidol, and pimozide are less sedating. First-generation antipsychotics can also lower the seizure threshold, and chlorpromazine and thioridazine are more epileptogenic than others. Haloperidol can cause abnormal heart rhythm, ventricular arrhythmia, torsades de pointes, and even sudden death if injected intravenously. Other FGAs can cause prolongation of QTc interval, prolonged atrial and ventricular contraction, and other cardiac conduction abnormalities. Thioridazine has an FDA-backed warning for sudden cardiac death. Low-potency FGAs, like chlorpromazine or thioridazine, commonly cause orthostatic hypotension. This adverse effect caused by alpha-adrenergic block usually occurs when starting treatment, and patients often develop a tolerance. It is important to avoid treating hypotension with epinephrine. Leukopenia, thrombocytopenia, and blood dyscrasia are rare side effects of treatment with FGAs. Increased serum prolactin concentrations along with galactorrhea, breast enlargement, amenorrhea, impotence in men, and anorgasmia in women are known adverse effects due to the action of the dopamine receptor block in the tuberoinfundibular tract. Allergic dermatitis and photosensitivity can occur with chlorpromazine. Chlorpromazine is also associated with blue-gray skin discoloration and benign pigmentation of the lens and cornea. Thioridazine can cause retinal pigmentation, which can continue even after discontinuing the drug.
Neuroleptic malignant syndrome is a rare but fatal adverse effect that can occur at any time during treatment with FGAs. The onset of symptoms is over 24 to 72 hours with increased temperature, severe muscular rigidity, confusion, agitation, elevation in white blood cell count, elevated creatinine phosphokinase concentrations, elevated liver enzymes, myoglobinuria, and acute renal failure. The antipsychotic should be immediately discontinued, and dantrolene 0.8 to 2.5 mg/kg every 6 hours up to 10 mg per day is the drug of choice. Adequate hydration and cooling should be closely monitored, as well as vital signs and serum electrolytes. Though the risk of neuroleptic malignant syndrome is higher with first-generation antipsychotics, second-generation antipsychotics also cause this adverse effect.
Second-generation antipsychotics have a decreased risk of extrapyramidal side effects as compared to first-generation antipsychotics. Second-generation antipsychotics are associated with significant weight gain and the development of metabolic syndrome. The FDA recommends monitoring personal and family history of diabetes mellitus, dyslipidemia, weight and height, waist circumference, blood pressure, fasting plasma glucose, and fasting lipid profile for all patients. Risperidone is associated with dizziness, anxiety, sedation, and extrapyramidal side effects. Paliperidone can cause temperature sensitivity to hot or cold temperatures and QTc prolongation. Olanzapine is most frequently associated with weight gain, increased appetite, and somnolence. Quetiapine is the least likely to cause extrapyramidal side effects. The most common side effects of quetiapine are somnolence, orthostatic hypotension, and dizziness. Ziprasidone has almost no weight gain but can cause prolongation of QTc. Aripiprazole is the most common side effect of agitation, headache, and akathisia-like restlessness.
Asenapine can cause an increase in serum prolactin concentrations, weight gain, and prolongation of QTc. Clozapine can cause hypersalivation, tachycardia, hypotension, and anticholinergic side effects. Clozapine is unusual in that it suppresses dyskinesia. Clozapine can cause clinically important agranulocytosis and leukopenia and, therefore, requires monitoring of white blood cells and absolute neutrophil count. The FDA guidelines indicate monitoring absolute neutrophil count weekly for the first 6 months and, if normal, can be monitored every 2 weeks after that. Treatment with clozapine should be discontinued if absolute neutrophil count drops below 1000 cells per cubic millimeter or below 500 cells per cubic millimeter in those with benign ethnic neutropenia. Clozapine can also cause the rare side effects of cardiomyopathy and myocarditis. FDA boxed warning: Observational studies suggest that similar to second-generation antipsychotics, treatment with first-generation antipsychotics may increase mortality risk in elderly patients with dementia-related psychosis. Larger doses of antipsychotics increase the risk of adverse drug effects.
Contraindications
First-generation antipsychotics are contraindicated in the following situations:
History of severe allergy
Use of central nervous system depressants like barbiturates, benzodiazepines, opioids
With anticholinergic medication like scopolamine or the use of phencyclidine
Severe cardiac abnormalities
History of seizure disorder
Narrow-angle glaucoma or prostatic hypertrophy
History of or ongoing tardive dyskinesia
Second-generation antipsychotics carry the FDA-boxed warning of increased incidence of stroke in elderly patients with dementia. The recommendation is to avoid the use of second-generation antipsychotics along with other drugs that prolong the QTc interval. Antipsychotics should be avoided during pregnancy, especially in the first trimester, and should be used only if the benefits outweigh the risks of treatment. Antipsychotics are secreted in breast milk, and it is advisable to avoid breastfeeding.
Monitoring
Some antipsychotics can be monitored for a therapeutic range in the plasma. Plasma concentrations should be monitored at a trough, at least 12 hours after the prior dose, and best at 20 to 24 hours after the last dose. Most antipsychotics do not have a well-defined dose-response curve. Haloperidol has a therapeutic range of 2 to 15 ng/ml, chlorpromazine range of 30 to 100 ng/ml, and perphenazine range of 0.8 to 2.4 ng/ml. Clozapine shows a better treatment response at a plasma concentration of 350 micrograms per milliliter or higher.
Enhancing Healthcare Team Outcomes
Antipsychotics are widely used medications for a variety of mental health disorders. While effective, these drugs do have many potential side effects. Healthcare workers, working as an interprofessional team, need to be aware of the adverse effects because they can seriously affect the quality of life. To avoid the metabolic effects of these drugs, the patient needs to receive information regarding lifestyle changes. Regular exercise, discontinuation of smoking, and eating a healthy diet are essential. Also, the patient's body weight, blood cholesterol, blood sugar, and blood pressure should be recorded at each clinic visit. If the patient is on clozapine, then regular monitoring of the white cell count is required. Only through strict monitoring can the high morbidity of these drugs be limited.
When the clinician initiates therapy, they should explain the treatment to the patient. Nurses can reiterate this education, answer questions, and assess patient progress and adherence to subsequent visits. Lack of drug adherence can be a significant cause of treatment failure. The pharmacist must check for all potential drug-drug interactions and verify appropriate dosing. A board-certified psychiatric pharmacist can also consult on the case with the clinician, offering possible alternatives if treatment does not progress as hoped, eg, recommending a long-acting dosage form. With these interprofessional interactions, antipsychotic therapy can achieve optimal outcomes for the patient.
References
Drummond N, McCleary L, Freiheit E, Molnar F, Dalziel W, Cohen C, Turner D, Miyagishima R, Silvius J. Antidepressant and antipsychotic prescribing in primary care for people with dementia. Canadian family physician Medecin de famille canadien. 2018 Nov:64(11):e488-e497
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Jennings AA, Guerin N, Foley T. Development of a tool for monitoring the prescribing of antipsychotic medications to people with dementia in general practice: a modified eDelphi consensus study. Clinical interventions in aging. 2018:13():2107-2117. doi: 10.2147/CIA.S178216. Epub 2018 Oct 23
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Faden J, Citrome L. Resistance is not futile: treatment-refractory schizophrenia - overview, evaluation and treatment. Expert opinion on pharmacotherapy. 2019 Jan:20(1):11-24. doi: 10.1080/14656566.2018.1543409. Epub 2018 Nov 8
[PubMed PMID: 30407873]
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Haddad PM,Correll CU, The acute efficacy of antipsychotics in schizophrenia: a review of recent meta-analyses. Therapeutic advances in psychopharmacology. 2018 Nov
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Yazici E, S Cilli A, Yazici AB, Baysan H, Ince M, Bosgelmez S, Bilgic S, Aslan B, Erol A. Antipsychotic Use Pattern in Schizophrenia Outpatients: Correlates of Polypharmacy. Clinical practice and epidemiology in mental health : CP & EMH. 2017:13():92-103. doi: 10.2174/1745017901713010092. Epub 2017 Aug 11
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Ross E, Barnett R, Tudhope R, Vasudev K. Can We Improve Physical Health Monitoring for Patients Taking Antipsychotics on a Mental Health Inpatient Unit? Journal of clinical psychopharmacology. 2018 Oct:38(5):447-453. doi: 10.1097/JCP.0000000000000931. Epub
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Crouse EL, Alastanos JN, Bozymski KM, Toscano RA. Dysphagia with second-generation antipsychotics: A case report and review of the literature. The mental health clinician. 2017 Mar:7(2):56-64. doi: 10.9740/mhc.2017.03.056. Epub 2018 Mar 23
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Lozupone M,La Montagna M,D'Urso F,Piccininni C,Sardone R,Dibello V,Giannelli G,Solfrizzi V,Greco A,Daniele A,Quaranta N,Seripa D,Bellomo A,Logroscino G,Panza F, Pharmacotherapy for the treatment of depression in patients with alzheimer's disease: a treatment-resistant depressive disorder. Expert opinion on pharmacotherapy. 2018 Jun
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Brodaty H, Aerts L, Harrison F, Jessop T, Cations M, Chenoweth L, Shell A, Popovic GC, Heffernan M, Hilmer S, Sachdev PS, Draper B. Antipsychotic Deprescription for Older Adults in Long-term Care: The HALT Study. Journal of the American Medical Directors Association. 2018 Jul:19(7):592-600.e7. doi: 10.1016/j.jamda.2018.05.002. Epub
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Pathak S, Duff E. Antipsychotic use in older adults: Canadian best practices. The Nurse practitioner. 2018 Jun 11:43(6):50-55. doi: 10.1097/01.NPR.0000532766.14251.c7. Epub
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Volume 31, Number 1—January 2025
Volume 31, Number 1—January 2025
Research Letter
Spread of Antifungal-Resistant Trichophyton indotineae, United Kingdom, 2017–2024
Cite This Article
Abstract
We describe 157 cases of Trichophyton indotineae infection in the United Kingdom, mostly in patients linked to southern Asia. T. indotineae is spreading in the United Kingdom and accounts for 38% of dermatophyte isolates referred to the UK National Mycology Reference Laboratory. Clinicians should suspect T. indotineae in tinea corporis cases.
Outbreaks of superficial skin infections caused by the emergent dermatophyte Trichophyton indotineae (Trichophyton mentagrophytes genotype VIII) were reported in southern Asia starting in 2014 (1–4). Typically, T. indotineae infections initially involve the groin (tinea cruris) and respond poorly to treatment, resulting in widespread lesions affecting multiple body sites. Many isolates exhibit in vitro resistance to terbinafine, and most infections are clinically resistant to that drug (1–5). Infections spread easily from person to person (1–8), and some reports suggest sexual transmission (9).
T. indotineae is endemic across Asia, but cases have been reported worldwide (4), including in Europe (5–7), Canada (8), and the United States (9). Mounting evidence suggests infection acquisition and transmission outside original areas of endemicity (5,7,9,10). Occasional cases of T. indotineae infection have been reported from the United Kingdom (10). We describe all cases of T. indotineae identified at the UK National Mycology Reference Laboratory (MRL) during a 7-year period.
We reviewed laboratory records from August 2017–July 2024 for dermatophytes identified as T. indotineae. When available, we extracted clinical and epidemiologic data from requisition forms. Dermatophyte identification was determined by whole-genome sequencing (WGS) or internal transcribed spacer sequencing, combined with phenotypic identification (Appendix Table). Isolates received after 2021 were identified using phenotypic features alone. A key defining microscopic feature was abundant fusiform to clavate, thin smooth-walled macroconidia with an acute apical tip, as well as other macroscopic and microscopic characteristics (Appendix Figure 1). We performed susceptibility testing by broth microdilution according to Clinical and Laboratory Standards Institute standards (Appendix). In the absence of an established clinical breakpoint for terbinafine, we used an MIC of >0.5 mg/L to identify non–wild-type isolates.
The first WGS-confirmed case we noted was from October 2018. In nearly half (42.7%, 67/157) of identified cases, the groin, buttocks, and thighs were directly involved, and neighboring body sites (abdomen and back) were implicated in another 18 cases (Table). Most (84.7%) patients had links to endemic areas, including South Asian ethnic background (n = 97), recent travel to the Indian subcontinent or Middle East (n = 41), or both (n = 36). Household spread was noted in 5 cases (Appendix Table).
Figure
Figure. Numbers and percentages of isolates per year in study of spread of antifungal-resistant Trichophyton indotineae, United Kingdom, 2017–2024. Numbers of isolates of T. indotineaeand all other...
Before 2023, most (27/36) cases were identified in London, which was the most affected city according to total case numbers. Since 2023, increasing numbers of cases were found in an additional 27 cities in the United Kingdom and Ireland, and isolate numbers outside London exceed those in London (Appendix Figure 3). From 2018 to 2019, the prevalence of T. indotineae in the United Kingdom increased from 2% to 7% of all dermatophyte isolates referred to the MRL. This prevalence remained largely stable during 2019–2023 (range 5%–12%). Of note, T. indotineae comprised 38% of all dermatophyte isolates received by the MRL in 2024 up to July (Figure).
Antifungal susceptibility data for terbinafine were available for 124/157 isolates, and in vitro resistance (MIC >0.5 mg/L) was documented in 92/124 (74.2%) cases, in keeping with previous reports (1,2,4,5). Of the 108 isolates in our study, 14% displayed MICs >0.5 mg/L to itraconazole; however, a breakpoint for itraconazole with T. indotineae is lacking. Fifty (31.8%) of 157 cases had documented treatment failure, 34 (21.7%) cases had terbinafine failure, and 7 (4.5%) cases had poor response to itraconazole.
In this study, London had the highest caseloads before 2023, likely because of absolute population numbers, comprehensive travel links to the Asian subcontinent through major London airports, and enhanced access to private dermatology clinics. The largely stable prevalence from 2019 through 2023 is probably because of COVID-19 prevention measures, which reduced population mixing and subsequent spread of T. indotineae. Our findings suggest that infections were acquired either directly in southern Asia and imported into the United Kingdom or from contacts with recent travel to such areas.
The first limitation of this study is underestimation of T. indotineae prevalence because of limited awareness among medical practitioners and microbiology laboratorians, likely misidentifications in routine laboratories, lack of commercial methods for rapid and accurate identification, and difficulties in obtaining skin scrapings from patients impeding laboratory identification of causative agent. Second, probable regional differences exist in awareness and identification capacity driven by regional prevalence and likelihood of prior encounter. Third, we do not have clinical information on dose or duration of terbinafine therapy for most patients with reported treatment failures; thus, we are unable to link treatment failure to elevated MIC values. Finally, only a proportion of T. indotineae isolates had genetic confirmation of identity. Despite our confidence in our methods, the identification of some cases by phenotypic methods alone could lead to some misidentification of species within the T. mentagrophytes species complex.
In conclusion, we show that T. indotineae was introduced into the United Kingdom from endemic areas and is spreading substantially. On the basis of current trends, we predict T. indotineae will rapidly become the predominant cause of tinea corporis in the United Kingdom. Clinicians and microbiology laboratorians should recognize this fungus as a predominant cause of tinea corporis.
Dr. Abdolrasouli is a clinical scientist in medical mycology at King’s College Hospital, London, United Kingdom. His primary research interests include emerging pathogens, antifungal resistance, and laboratory diagnosis of fungal infections.
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Acknowledgment
We thank Elizabeth Johnson for her interest in this work. We are also grateful to Johanna Rhodes for analyzing the whole-genome sequencing data, Daniel Kibbey for help with LIMS database searches, Sue McLachlan for assistance with isolate identification, and Sue McLachlan, Cheryl Yung, and Patricia Coll-Gutierrez for performing antifungal drug susceptibility testing of Trichophyton indotineae isolates.
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References
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DOI: 10.3201/eid3101.240923
Original Publication Date: December 17, 2024
Table of Contents – Volume 31, Number 1—January 2025
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Please use the form below to submit correspondence to the authors or contact them at the following address:
Correspondence: Andrew M. Borman, Mycology Reference Laboratory, UK Health Security Agency, Science Quarter, Southmead Hospital, Bristol BS10 5NB, UK
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| EID | Abdolrasouli A, Barton RC, Borman AM. Spread of Antifungal-Resistant Trichophyton indotineae, United Kingdom, 2017–2024. Emerg Infect Dis. 2025;31(1):192-194. |
| AMA | Abdolrasouli A, Barton RC, Borman AM. Spread of Antifungal-Resistant Trichophyton indotineae, United Kingdom, 2017–2024. Emerging Infectious Diseases. 2025;31(1):192-194. doi:10.3201/eid3101.240923. |
| APA | Abdolrasouli, A., Barton, R. C., & Borman, A. M. (2025). Spread of Antifungal-Resistant Trichophyton indotineae, United Kingdom, 2017–2024. Emerging Infectious Diseases, 31(1), 192-194. |
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Relativistic Velocity Addition In Special Relativity (Why does relative velocity never exceed c?)
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Let's explore why relative velocities don't just add up in special relativity as we learnt in Newtonian mechanics. We will use the ideas of time dilation, length contraction, and relativity of simultaneity to intuitively derive the relativistic velocity addition formula.
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this video is sponsored by Squarespace the all-in-one website building tool a girl throws a baseball at a blistering speed of 90% the speed of light but she's in a train that's already moving at 87% the speed of light so according to Newton wouldn't the total speed of baseball just add up and become more than the speed of light Breaking Einstein's rules this is the question that I asked my physics teacher back in high school and his explanation was no that doesn't happen because speeds don't add up velocities don't add up just like how Newton said instead to calculate relative velocity used this formula and if you plug in these values in this formula You'll Always Find the speed to be less than the speed of light but you see my problem with that right that's just a dumb formula I mean I can use it to solve problems but every time I look at a question like this my brain keeps saying that hey that ball should be traveling faster than light why isn't it doing that and so I thought of gaining some intuition to do that that I thought like let's look at the derivation of it and then I looked at it oh my God it just involved so much algebra and manipulation and I thought okay I'm never going to be able to get an intuition for this until now after making Einstein and Newton talk for months I think I finally get it so by the end of this video we're going to see why Newton's addition is wrong how Einstein fixed it and why relative velocities will never exceed C all intuitively so if you're ready for this let's dive in we start by asking new buddy why did you even say that the velocity adds up in the first place well Newton says if we look at the train and if we wait for 1 second that in that 1 second that ball would travel .9 C forward because that's its speed but in that 1 second the train also goes forward by 87c so look at the total distance that the ball has traveled it is 87c plus 97c giving you 1.77 C in 1 second and therefore that is its speed faster than light Einstein says well first of all let's just make c equal to one I mean it's just so annoying to keep using the word keep saying c c over and over again that's why physicists love units in which the velocity of light is just one simple right but more importantly Newton you made three boo boos first you overlooked length contraction moving things shrink in the direction of motion at 87% light speed the train would be contracted to half its length and therefore you can immediately see the ball does not travel .9 when we look at it it will travel .9 divided by 2 so the speed is already less than what Newton expected now at this point we should ask well Einstein remember this is an intuitive derivation how did you realize that at 87% light speed the length contraction Factor will be two and I says patience we'll get there but we're already beginning to see see how Newton's addition is wrong isn't it onto your second mistake Newton time dilation moving clocks take slower than clocks that are at rest at 87% light speed the clocks will take at half the rate which means when 1 second is up in our clock only5 second will be her up in her clock that means the ball wouldn't even reach the entire point9 the end of the train it would be somewhere halfway in between which means time dilation further slows down the distance that we see by a factor of two and so we'll get another two in the denominator and the speed becomes even lower than what you expected but but but Einstein we've already considered both the length contraction and time dilation and look the speed of the ball is still more than one what gives and Einstein says this takes us to the third and the final mistake that Newton made it's the it's something that we've talked about before it's it's the one that solves all the paradoxes it's the Ultimate Reality the relativity of simultaneity okay I'm having way too much fun with this but the point is if you look inside the train then the ball reaching the halfway mark is simultaneous with her clock reading5 but things that are simultaneous in one frame need not be simultaneous in a moving frame in fact from our frame we'll find that her clock will first read5 then the baseball would reach the halfway mark which means if we consider simultaneity the ball would be moving even slower than before but wait a second at this point we can ask Einstein well Einstein how did you know that it is this event that happens first and then this would happen like how did you know that well einen says his go-to method is to imagine synchronizing a couple of events for example let's say we have two firecrackers and we have have a light source right in between then we send photons outwards and they will because the light source was right in between the photons will go and hit the firecrackers at exactly the same time and let's say light can ignite the firecrackers I mean this is physics right so the firecrackers would explode at exactly the same time but from the mooving frame what would happen well again light is going to move out from that point at exactly the same speed in both the directions but this time time the train is also moving forward therefore what do you see the firecracker over here bursts first and from this we can get an intuition for the fact that the events at the back of the train happen before the events in the front of the train this is our go-to method and this would be true in general you don't really need clocks and synchonization technique that's just a method but this will be true in general and so this means if we go back over here events in the back are happening before the events in the front and that's why her clock will read 0.5 before the ball even reaches the halfway mark But Einstein how do we incorporate this over here ah good question again if we go and look at things from her frame then from her frame the ball traveled .9 units in 1 second again these two events are simultaneous but from our frame they're not simultaneous because events at the back happen first by the time the ball reaches over here this clock would have already gone past 1 second in fact if you do the math which we will do later on we will see that this would be about 1.78 so we will see that ball travel 0.9 units in not 1 second but 1.78 second which means the relativity of simultaneous slows down that ball further by 1.78 and So that's its contribution so if you factor that in boom you will find that the speed of the baseball is is less than the speed of light but wait wait I know what you're thinking you might be thinking you must be thinking what if the train was going even faster then wouldn't it give a little bit more boost to the baseball and make it go faster than light einin says challenge accepted well if we make the Train's velocity now 98 the what about the velocity of the ball would it now become faster well Einstein reminds us now all these factors will be become more extreme for example the train will now shrink to 1/5 its length the time dilation Factor would be about five you correctly plug those in and you will still see that there's no way to go faster than the speed of light okay wait a second what if we don't do that what if instead we keep the speed of the Train the same so that the length contraction time dilation is exactly the same but if we just threw the base ball slightly faster now it's got to break the speed of light right no because if now you you make it go faster events are more separated apart and as a result you will now find that the relativity of simultaneity factor will be higher so whatever you do if you increase this number this number will go up and again you will find that the total velocity will still be less than C this formula is Ironclad folks there's no way to go around it and it should not be very surprising because remember the whole relativity is built on the postulate that the speed of light is a constant and it's the same in all reference frames so it's not surprising that nothing can travel you know even if you add velocities you cannot travel faster than the speed of flight the point however is was to derive an expression for adding velocities that respected Einstein's postulates and we have found it we have found it because what we can generalize this now okay what if we what if we call this velocity U the speed of the train U and the velocity of the ball with respect to train wi then look what will be our relative velocity what how to add velocities it will be U + V divided by the length contraction Factor the time dilation factor and the relativity of simultaneity factor this is how you add velocities correctly Newton okay all right all right but at this point we could ask Einstein well Einstein so far so good but this is how far intuition can get us right I mean mean if you want to get an exact expression for this now we just have to do the algebra and I says no even for finding the exact expression there's an elegant way to do it and it can be as intuitive as say building a website using Squarespace the sponsor of this video just like with velocity Edition or any physics topic for that matter let's start from scratch how do we build a website a landing page well we need a domain so let's go to squarespace.com click on a button which says get domain buy the domain you want and boom just like that your domain becomes active and it comes with who is privacy for free which is great if you want to protect our privacy which is something we should try and do but wait neither Newton nor I have any decent coding experience I don't even know what our landing page should look like well Squarespace has incredible templates that we can start with so we can just click on say port folio then find the one that looks pretty for us customize it by clicking buttons and in a few hours in our rest frame of course the page is live but there's so much more there's incredible analytics to tell us about the performance they also have great support that will unblock us if we get stuck somewhere so I think Squarespace truly is a One-Stop solution for building any kind of website so if you want to check your website building intuition for free then go to www.squarespace.com floatheadphysics you can also use the promo code float head physics at the checkout to save 10% on your first purchase the link is also in the description all right back to the video okay Einstein show me the elegant way of looking at these things let's start with time dilation time dilation factor is gamma and we've derived this in a previous video all we needed is a photon clo and a Pythagoras over there we saw u sare v sare c square but here C is one remember that but there is more to it I mean if you look at a table of values of them you will see the same thing as we saw before at 87% light speed the time dation factor is about two 98% light speed the time relation factor is about five it comes from this formula but a more elegant way to look at this is if you square them and add them like this you will always get at one now that might seem like an algebraic manipulation of this but there's something deeper going on over here it's something it's got it's telling us something about space and time and its connection with the speed of light and this something we'll get back to in a future video but this is one of the ways in which you can look at this factor and since length contraction factor is just the same we already have two out of the three both these factors length contraction and time dilation is just gamma which leaves us with the last one the relativity of simultaneity factor how do you figure this out Einstein and Let Me Tell You Folks this is the most elegant way of looking at things I've saved the last for the best only for you because you've reached till this part of the video so you deserve it okay so let's go to relativity of simultaneity and the way to look at it is if you have two clocks or two events that are let's say one unit apart and they are also mov moving very close to the speed of light then we already know that the events over at the back would be ahead by the events ahead of the events in the front but can you make a wild guess in this particular case how much will this clock be ahead of compared to this clock just just make a wild guess it turns out to be ex pretty much by one unit this means for example if this clock read3 this would read about 1.3 I find it so fascinating because look you have two events that are traveling at one spaced apart by one are also apart in time by one isn't that elegant I mean isn't that telling us about how space and time are conspiring together whoa but we can now use this to build up a general formula for example we could ask what happens if the their speeds was not one but they were going at half the speed of light well then this just happens to be half what if they were going at 3 the speed of light well then this just becomes. 3 in other words whatever is the velocity over here the discrepancy is just that you isn't that amazing but wait this is for when the events are one unit apart in space what if the events were two units apart in space well the discrepancy would be twice as big what if the events were half as much apart then the discrepancy would be only half as much so in other words what if the events were X units apart then the discrepancy would be U times x and by the way I when I say x distance I mean the proper length the length seen from the rest frame okay so if if you now go inside the train look the two events over here are V units apart in space and therefore from our reference frame they will also be UV units apart in time therefore if this clock reads one at that same moment from our frame this clock would read 1 plus UV and therefore from our frame the ball travel distance V in not 1 second but in 1 plus UV second and that's why we have to divide it out by 1 plus UV and you can now see where that 1 plus UV in the original formula comes from so if you incorporate that the relativity of simultaneity is just one plus u and there you have it folks in its full Glory the relativistic velocity addition formula derived intuitively isn't that elegant but I know what you're thinking you're thinking well this is not the formula in our textbooks and I says yeah yeah of course of course I mean if you substitute for the value of gamma and just simplify it a little bit you will get the familiar formula that you see in your textbooks this refined version is efficient for problem solving and that's why we find it in our textbooks but for a deeper and more intuitive perspective Ive I find this raw version to be much better and I think this mirrors life itself we often seek shortcuts and efficiencies to boost our productivity but do take a moment every now and then to appreciate the raw unrefined experiences for it is these that often offer the most profound insights until we meet again
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Proofs (mathematics)
Vector Arithmetic
Distance (general)
The Triangle
Prove Mathematically
Vectors (mathematics)
5
How do you prove by vector method that the middle point of hypotenuse of a rightangled triangle equidistant from its vertices?
Doug Dillon
Ph.D. Mathematics
·
Author has 12.4K answers and 11.4M answer views
·
10mo
Related questions
Can the middle point of the hypotenuse in a right angle triangle be proved to be equidistant from its vertices using only the vectors method, without using coordinate geometry?
How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
Show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices?
How do you prove the midpoint of hypotenuse is equidistant from the vertices?
How do you prove that the midpoints of the hypotenuse of an RT triangle is equidistant from all vertices?
Richard Goldstone
PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995)
·
Author has 1.8K answers and 3.9M answer views
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10mo
The question asks explicitly for vector methods. These methods make it easy to write an expression for the median from any vertex of a triangle, and the resulting expression can be specialized to the right triangle case. As a bonus, we get a converse to the given question and a well-know theorem about angles inscribed in a semicircle.
In the following diagram (not necessarily a right triangle yet),
Let
−−→CA=v,−−→CB=w,−−→CM=m,−−→MA=h.
Since M is the midpoint of \overlin
The question asks explicitly for vector methods. These methods make it easy to write an expression for the median from any vertex of a triangle, and the resulting expression can be specialized to the right triangle case. As a bonus, we get a converse to the given question and a well-know theorem about angles inscribed in a semicircle.
In the following diagram (not necessarily a right triangle yet),
Let
−−→CA=v,−−→CB=w,−−→CM=m,−−→MA=h.
Since M is the midpoint of ¯¯¯¯¯¯¯¯AB, we have −−→MB=−h.
With these assignments,
m+h=v,m+(−h)=w,
so
This much is true for any triangle, but if is a right triangle, then and then we have which proves that is equidistant from and
Note that since the expression is valid for any triangle, we obtain a converse to the given question:
Proposition. If the median in is the same length as and then is a right triangle with right angle at
Proof. If the expression forces
Corollary. An angle inscribed in a semicircle is a right angle.
Proof. is inscribed in a semicircle if and only if the midpoint of is the center of the semicircle and are on the circumference. Since and are radii of the semicircle, they are all the same length, and then the proposition says we have a right triangle with the right angle at
Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
·
Author has 6.2K answers and 7M answer views
·
10mo
Compute magnitudes of
,
and
and check!!!
Compute magnitudes of
,
and
and check!!!
Dave Benson
trying to make maths easy.
·
Author has 6.1K answers and 2.1M answer views
·
10mo
How do you prove by vector method that the middle point of hypotenuse of a rightangled triangle equidistant from its vertices?
Let triangle be ABC and angle A be 90º and mid point of BC be M & Vectors in bold type
AB·AC = 0 (AB perpendicular to AC)
(AM+MB)(AM+MC) = 0
(AM+CM)(AM-CM) =. 0 so CM = MB
AM²-CM² = 0
AM² = CM²
AM = CM = MB Answer and proof.
How do you prove by vector method that the middle point of hypotenuse of a rightangled triangle equidistant from its vertices?
Let triangle be ABC and angle A be 90º and mid point of BC be M & Vectors in bold type
AB·AC = 0 (AB perpendicular to AC)
(AM+MB)(AM+MC) = 0
(AM+CM)(AM-CM) =. 0 so CM = MB
AM²-CM² = 0
AM² = CM²
AM = CM = MB Answer and proof.
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Related questions
Will the three medians of a triangle always intersect at a point equidistant from the three vertices of a triangle?
Can you provide a proof that the mid-point of a hypotenuse is equidistant from the other two vertices using vector form?
How is the middle point (M) of the hypotenuse equidistant to the three summits of this triangle?
How many equilateral triangles can be formed with vertices lying on 24 equidistant points on the circumference of a circle?
What is the formula for finding the area of a right angle triangle using vectors? Is there an alternative method for calculating this without using (x2-x1)?
Dean Rubine
I am obsessed with triangles, especially their area.
·
Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 10.6K answers and 23.6M answer views
·
10mo
Originally Answered: Can the middle point of the hypotenuse in a right angle triangle be proved to be equidistant from its vertices using only the vectors method, without using coordinate geometry?
·
Bot question. Every right triangle is half of a rectangle that’s divided along one of its diagonals. Clearly the center of the rectangle is the meet of its diagonals, so by symmetry the midpoint of its diagonals, equidistant from the four corners. That center is the midpoint of the common hypotenuse of the two right triangles, equidistant from the vertices, making it the circumcenter of the right triangles.
Bot question. Every right triangle is half of a rectangle that’s divided along one of its diagonals. Clearly the center of the rectangle is the meet of its diagonals, so by symmetry the midpoint of its diagonals, equidistant from the four corners. That center is the midpoint of the common hypotenuse of the two right triangles, equidistant from the vertices, making it the circumcenter of the right triangles.
Varun Kumar
Btec 2nd year in Mechanical Engineering, LNJPIT,chapra,bihar (Graduated 2023)
·
4y
Originally Answered: How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
·
Let P be the mid point of the hypo. of the right triangle ABC, right angled at B.
Draw a line parallel to BC from P meeting AB at D.
Join PB.
in triangles,PAD and PBD,
angle PDA= angle PDB (90 each due to conv of mid point theorem)
PD=PD(common)
AD=DB( as D is mid point of AB)
so triangles PAD and PBD are congruent by SAS rule.
PA=PB(C.P.C.T.)
but
PA=PC(given as P is mid point )
So,
PA=PC=PB
Let P be the mid point of the hypo. of the right triangle ABC, right angled at B.
Draw a line parallel to BC from P meeting AB at D.
Join PB.
in triangles,PAD and PBD,
angle PDA= angle PDB (90 each due to conv of mid point theorem)
PD=PD(common)
AD=DB( as D is mid point of AB)
so triangles PAD and PBD are congruent by SAS rule.
PA=PB(C.P.C.T.)
but
PA=PC(given as P is mid point )
So,
PA=PC=PB
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Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
·
Author has 6.2K answers and 7M answer views
·
4y
Originally Answered: How do you prove the midpoint of hypotenuse is equidistant from the vertices?
·
Construct a right triangle on XY plane. Let vertex A(0,b) on Y axis, B(a,0) on X axis and right angle C(0,0) on origin.
M is midpoint of AB, so M = (a/2,b/2)
AM = √[(0-a/2)^2 + (b-b/2)^2] = √(a^2/4 + b^2/4)
BM = √[(a-a/2)^2 + (0-b/2)^2] = √(a^2/4 + b^2/4)
CM = √[(0-a/2)^2 + (0-b/2)^2] = √(a^2/4 + b^2/4)
So, AM = BM = CM , hence proved.
Construct a right triangle on XY plane. Let vertex A(0,b) on Y axis, B(a,0) on X axis and right angle C(0,0) on origin.
M is midpoint of AB, so M = (a/2,b/2)
AM = √[(0-a/2)^2 + (b-b/2)^2] = √(a^2/4 + b^2/4)
BM = √[(a-a/2)^2 + (0-b/2)^2] = √(a^2/4 + b^2/4)
CM = √[(0-a/2)^2 + (0-b/2)^2] = √(a^2/4 + b^2/4)
So, AM = BM = CM , hence proved.
Henry Burek
M.Phil, B.Sc. in Ophthalmic Optics, University of Bradford (MDIS) (Graduated 1977)
·
Author has 2.2K answers and 3.3M answer views
·
10mo
How do you prove by vector method that the middle point of hypotenuse of a rightangled triangle equidistant from its vertices?
Why “vector method” when geometry gives a simple method?
The circumcircle of a right triangle has the hypotenuse as a diameter. The middle point of the hypotenuse is thus the centre of the circumcircle. Since the triangle vertices lie on the circumference they are equidistant from the circumcircle centre and therefore equidistant from the hypotenuse midpoint.
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Gaurav Kumar
Former Mathematics Learner
·
Author has 536 answers and 1.7M answer views
·
5y
Originally Answered: How do you prove the midpoint of hypotenuse is equidistant from the vertices?
·
Gopal Menon
B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010)
·
Author has 10.2K answers and 15.2M answer views
·
4y
Originally Answered: How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
·
How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
By geometry.
Let [math]ABC[/math] be a right angled triangle with [math]AC[/math] as its hypotenuse.
Let [math]D[/math] be the midpoint of [math]AC.[/math]
Draw [math]AE\parallel BC[/math] and [math]CE\parallel BA[/math] so that [math]AECB[/math] is a rectangle.
Join [math]BE.[/math]
Since the diagonals of a rectangle bisect each other and [math]D[/math] is the midpoint of [math]AC,[/math] it is clear that [math]D[/math] is the point of intersection of the two diagonals.
Further, the lengths of both the diagonals of a rectangle are equal.
[math]\Rightarrow AD=BD[/math]
How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
By geometry.
Let [math]ABC[/math] be a right angled triangle with [math]AC[/math] as its hypotenuse.
Let [math]D[/math] be the midpoint of [math]AC.[/math]
Draw [math]AE\parallel BC[/math] and [math]CE\parallel BA[/math] so that [math]AECB[/math] is a rectangle.
Join [math]BE.[/math]
Since the diagonals of a rectangle bisect each other and [math]D[/math] is the midpoint of [math]AC,[/math] it is clear that [math]D[/math] is the point of intersection of the two diagonals.
Further, the lengths of both the diagonals of a rectangle are equal.
[math]\Rightarrow AD=BD=CD=ED.[/math]
[math]\Rightarrow[/math] The midpoint of the hypotenuse of a right angled triangle is equidistant from its vertices.
By coordinate geometry.
Let [math]ABC[/math] be a right angled triangle with [math]AC[/math] as its hypotenuse.
Let [math]D[/math] be the midpoint of [math]AC.[/math]
Without loss of generality, place this right angled triangle on the Cartesian plane with [math]B[/math] at the origin, [math]A[/math] on the positive Y axis and [math]C[/math] on the positive X axis.
Let the coordinates of of [math]A[/math] and [math]C[/math] be math[/math] and math[/math] respectively.
[math]\Rightarrow[/math] The length of [math]AC[/math] is [math]\sqrt{a^2+c^2}.[/math]
[math]\Rightarrow[/math] The lengths of both [math]AD[/math] and [math]CD[/math] are [math]\frac{1}{2}\sqrt{a^2+c^2}.[/math]
Since [math]D[/math] is the midpoint of [math]AC,[/math] the coordinates of [math]D[/math] are [math]\left(\frac{c}{2},\frac{a}{2}\right).[/math]
[math]\Rightarrow[/math] The length of [math]BD[/math] is [math]\sqrt{\left(\frac{a}{2}\right)^2+\left(\frac{c}{2}\right)^2}=\frac{1}{2}\sqrt{a^2+c^2}.[/math]
[math]\Rightarrow AD=BD=CD.[/math]
[math]\Rightarrow[/math] The midpoint of the hypotenuse of a right angled triangle is equidistant from its vertices.
Bhalchandra Khare
Former Self Employed Chemical Engineer (1975–2016)
·
Author has 4.3K answers and 1.4M answer views
·
4y
Originally Answered: How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
·
In triangle ABC D is the midpoint of hypotenuse BC.
A(2,2) B(2,10) and C(14,2)
D is mid point so its coordinates are (8,6).
AD^2=(8-2)^2+(6-2)^2=36+16=52.
Thus AD=BD=CD.
In triangle ABC D is the midpoint of hypotenuse BC.
A(2,2) B(2,10) and C(14,2)
D is mid point so its coordinates are (8,6).
AD^2=(8-2)^2+(6-2)^2=36+16=52.
BD^2=(8-2)^2+(6-10)^2=36×16=52
CD^2=(8-14)^2+(6-2)^2=36+16=52
Thus AD=BD=CD.
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
3y
How can you prove analytically that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices?
It is much easier to “prove” in reverse.
The consequence of 3 points being equidistant from some fixed point is that they must lie on a circle and we are also given that two of the points must be the diameter of this circle.
So drawing this scenario we have…
… where C is ANY point on the circle (but distinct from A and B).
There is no doubt that ABC is in fact a triangle and OA = OB = OC because they are all radii of the same circle.
The only thing left is to show that ∆ABC is a right angled triangle.
Using the famous theorem from basic circle geometry…
the angle AOB = 180 degrees and this is referre
It is much easier to “prove” in reverse.
The consequence of 3 points being equidistant from some fixed point is that they must lie on a circle and we are also given that two of the points must be the diameter of this circle.
So drawing this scenario we have…
… where C is ANY point on the circle (but distinct from A and B).
There is no doubt that ABC is in fact a triangle and OA = OB = OC because they are all radii of the same circle.
The only thing left is to show that ∆ABC is a right angled triangle.
Using the famous theorem from basic circle geometry…
the angle AOB = 180 degrees and this is referred to as “the angle at the centre”.
The angle ACB is referred to as “the angle at the circumference” and so is equal to half the angle at the centre.
Angle ACB = 90 degrees so ∆ABC is a right angled triangle.
Alexandru Carausu
Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010)
·
Author has 3K answers and 874.6K answer views
·
10mo
This is a simple question with a simple answer.
If Δ ABC is rigth-angled at A , then the opposite side BC is its hypotenuse. The point M ∈ BC is the midpoint of this side if and only if d ( B , M ) = d ( M , C ) . This description can be easily “translated” in terms of bound (fixed) and free vectors, the basic elements of the VECTOR ALGEBRA.
Definition. Two bound vector AB , CD are directed line segments, and they are equivalent if they share the same direction and the same magnitude (or length).
Note 1. The specific notation for this relation is AB ~ CD , and it is an actual equivalence relation
This is a simple question with a simple answer.
If Δ ABC is rigth-angled at A , then the opposite side BC is its hypotenuse. The point M ∈ BC is the midpoint of this side if and only if d ( B , M ) = d ( M , C ) . This description can be easily “translated” in terms of bound (fixed) and free vectors, the basic elements of the VECTOR ALGEBRA.
Definition. Two bound vector AB , CD are directed line segments, and they are equivalent if they share the same direction and the same magnitude (or length).
Note 1. The specific notation for this relation is AB ~ CD , and it is an actual equivalence relation (or simply equivalence) : it is reflexive, symmetric and transitive. It is often used to write (instead of the above notation) AB = CD ; this means that the two line segments represent the same free vector v , and it is used to write AB = v instead of AB ∈ v ; the same hold for the other fixed vector CD . A free vector has infinitely many bound / fixed vectors that represent it, and it is their equivalence class.
Note 2. The magnitude of any vector AB is simply the distance from its initial point (or tail) A to its endpoint (or tip) B . As regards the direction of v = AB (more exactly v ∍ AB ) , includes not only one but two features : ( i ) the direction, as the equivalence class of all lines that are parallel to ( AB ) , the suport line of the line segment AB ; ( ii ) the orientation or sense of AB : from A to B . Hence, the sense of BA is opposite to that of AB , and it is correct to write BA = ‒ AB . It follows that AB + BA = { 0 } : the zero free vector which is represented by any point in the space.
Coming back the the above question, I can give not one but five characterization of the midpoint M of the hypotenuse BC by means of vectors (and operations with them) . For possibly less acquainted readers of my answer, let me mention that the set of vectors on a line / in a plane / in the 3D space is not a simple set : it is a vector (linear) space, with its two linear operations : the sum u + v and the multiplication of vectors by salars , λ v with λ ∈ F : the numerical field over which the vector space is defined ; in most cases, F = R (the real field), or F = C (the complex field) .
Here are given five (equivalent) characterizations, in terms of vectors and operations with them :
[ BM = MC <==> CM = MB ] ==> d ( M , B ) = d ( M , C ) .
[ AM = AB + BM <==> AM = AC + CM ] ==> d ( M , B ) = d ( M , C ) .
[ MB + MC = MB + BM = { M } = 0 or BM + MB = { B } = 0 ] ==> d ( M , B ) = d ( M , C ) .
[ BM = MC = (1/2) BC ] ==> d ( M , B ) = d ( M , C ) .
[ CM = MB = (1/2) CB ] ==> d ( M , B ) = d ( M , C ) .
Note 3. I could continue with even more proofs / conditions for M = midpoint of BC ; but I suppose that the above ones suffice. It is important to see how the equivalence of bound (fixed) vectors has been above used. In the second and third points, I have applied the rule of the triangle for the sum of teo vectors : for instance, he right side in AM = AB + BM means that the origin of the right term (vector) must coincide with the endpoint of the left term. This rule is much better than the rule of the parallelogram, from several viewpoints. Note 3. The notations above used may differ from those met in other texts : many (if not the most) manuals / textbooks of ANALYTIC GEOMETRY start with a chapter of Vector Algebra. These notations were used in a textbook of Vector Algebra, Analytic & Differential Geometry, printed by a publisihing of my city Iasi (in 2003). But I have not quoted it since the self-quotations might be assimilated with plagiarism by the QUORA team. I gave classes in this domain to the students in Civil Engineering at the Technical University of Iasi (with teaching in English) during 16 years, 1991–2007. Hence I have simply “extracted” them from my memory.
Final Note 4. I have offered a (probably) too long / too detailed answer. But I use / like to offer such answers to interesting and well formulated questions, like the above one. I prefer to answer one such question instead of wasting my time with (too many) improperly formulated / too simple of even incorrect questions that arrive in my inbox almost everyday.
Related questions
Can the middle point of the hypotenuse in a right angle triangle be proved to be equidistant from its vertices using only the vectors method, without using coordinate geometry?
How do I show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices (the answer must be proved by Cartesian system of rectangular coordinate)?
Show that the middle point of the hypotenuse of a right angled triangle is equidistant from its vertices?
How do you prove the midpoint of hypotenuse is equidistant from the vertices?
How do you prove that the midpoints of the hypotenuse of an RT triangle is equidistant from all vertices?
Will the three medians of a triangle always intersect at a point equidistant from the three vertices of a triangle?
Can you provide a proof that the mid-point of a hypotenuse is equidistant from the other two vertices using vector form?
How is the middle point (M) of the hypotenuse equidistant to the three summits of this triangle?
How many equilateral triangles can be formed with vertices lying on 24 equidistant points on the circumference of a circle?
What is the formula for finding the area of a right angle triangle using vectors? Is there an alternative method for calculating this without using (x2-x1)?
Why don't we use the formula of triangle law of vector addition rather than triangle law of vector subtraction for determining the length of the third side of a triangle (in geometry) given first two sides and the angle between them?
The medians of the triangle ABC intersect at the point O. What is the sum of the vector OA, OB, and OC?
How would I get the length of the two other sides of a right triangle if I only had two coordinates, the start(A) and end(B) of the hypotenuse?
How do I add two parallel vector using triangle method?
What are the coordinates of the triangle of the midpoint of the hypotenuse triangle whose vertices are A (1,1) B (5,2) and C (4,6) and show that is equidistant from each of the vertices find the area?
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oct() | Python’s Built-in Functions – Real Python
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Table of Contents
oct() Signature
Arguments
Return Value
Related Resources
oct() Examples
oct() Common Use Cases
oct() Real-World Example
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Reference
Python’s Built-in Functions/
oct()
The built-in oct() function converts a given integer into its octal string representation, prefixed with 0o. The function’s output is a string:
Python
```
oct(42)
'0o52'
```
oct() Signature
Python Syntax
oct(x)
Arguments
| Argument | Description |
--- |
| x | An integer value or an object that defines an .__index__() method returning an integer |
Return Value
Returns a string holding the octal representation of the input integer. If the integer is negative, then the string will include the negative sign.
oct() Examples
With a positive integer as an argument:
Python
```
oct(8)
'0o10'
```
With a negative integer as an argument:
Python
```
oct(-56)
'-0o70'
```
With an object of a class that has a .__index__() method:
Python
```
class Number:
... def init(self, value):
... self.value = value
... def index(self):
... return self.value
...
number = Number(10)
oct(number)
'0o12'
```
oct() Common Use Cases
The most common use cases for the oct() function include:
Converting decimal numbers to their octal representation for use in systems or applications that require octal numbers.
Formatting output to display numbers in octal format.
Preparing data for systems that operate primarily on octal numbers, such as some file permission systems in Unix.
oct() Real-World Example
Say you’re working on a Unix-like system and need to display file permissions in octal format for auditing purposes. You can use oct() to convert permission values to octal:
Python
```
permissions = 493
f"File permissions in octal: {oct(permissions)}"
'File permissions in octal: 0o755'
```
In this example, the oct() function helps you convert numeric file permissions into a human-readable octal format, which is commonly used in Unix-based systems.
Related Resources
Tutorial
Python's Built-in Functions: A Complete Exploration ---------------------------------------------------
In this tutorial, you'll learn the basics of working with Python's numerous built-in functions. You'll explore how to use these predefined functions to perform common tasks and operations, such as mathematical calculations, data type conversions, and string manipulations.
basicspython
For additional information on related topics, take a look at the following resources:
Python String Formatting: Available Tools and Their Features (Tutorial)
Python's Magic Methods: Leverage Their Power in Your Classes (Tutorial)
Python's Built-in Functions: A Complete Exploration (Quiz)
Python String Formatting Tips & Best Practices (Course)
Python String Formatting: Available Tools and Their Features (Quiz)
Python's Magic Methods in Classes (Course)
Python's Magic Methods: Leverage Their Power in Your Classes (Quiz)
By Leodanis Pozo Ramos • Updated Aug. 11, 2025 • Reviewed by Dan Bader
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EdTech Books
Cover
Introducción
Tabla de contenidos
Repaso gramatical
El presente
Ser y estar
El verbo gustar
El pretérito
Capítulo 1
Vocabulario: La ropa
Vocabulario: Los accesorios
Vocabulario: De compras
Lectura - Capítulo 1
Actividad auditiva - Capítulo 1
Más práctica: Capítulo 1
Los pronombres de objeto directo
El posicionamiento de los pronombres
Comparativos y superlativos
Pronunciación 1: La acentuación
Cultura: Capítulo 1
Canción: Capítulo 1
Sinopsis: Capítulo 1
Capítulo 2
Vocabulario: La comida
Vocabulario: Los restaurantes
Lectura - Capítulo 2
Actividad auditiva - Capítulo 2
El imperfecto
Ejemplos de oraciones que usan el imperfecto:
Actividades de conversación
Actividad 1. Una cena en familia.
Actividad 2. Tu restaurante favorito.
Actividad 3. Preguntas personales
Actividad 4. El comienzo de una novela…
Para seguir practicando
El imperfecto versus el pretérito
Los pronombres de complemento indirecto
Los pronombres usados juntos
Otros usos de se
Pronunciación 2: ll & y
Cultura: Capítulo 2
Canción: Capítulo 2
Sinopsis: Capítulo 2
Capítulo 3
Vocabulario: La casa
Vocabulario: Los quehaceres domésticos
Lectura - Capítulo 3
Actividad auditiva - Capítulo 3
Los posesivos enfáticos
Por y para
Las formas del subjuntivo
El subjuntivo en cláusulas nominales
El indicativo y el subjuntivo en cláusulas nominales
Pronunciación 3: Los diptongos, triptongos e hiatos
Cultura: Capítulo 3
Canción: Capítulo 3
Sinopsis: Capítulo 3
Capítulo 4
Vocabulario: Películas
Vocabulario: La televisión
Vocabulario: Música, arte, literatura
Lectura - Capítulo 4
Actividad auditiva - Capítulo 4
Expresiones impersonales
Pero, sino y sino que
Cláusulas adjetivas
Pronunciación 4: La ñ
Cultura: Capítulo 4
Canción: Capítulo 4
Sinopsis: Capítulo 4
Capítulo 5
Vocabulario: Las partes del cuerpo
Vocabulario: La visita al médico
Lectura - Capítulo 5
Actividad auditiva: Capítulo 5
El futuro
El condicional
Pronunciación 5: La d y la t
Cultura: Capítulo 5
Canción: Capítulo 5
Sinopsis: Capítulo 5
Capítulo 6
Vocabulario: Los viajes
Vocabulario: Los hoteles
Vocabulario: La geografía
Lectura - Capítulo 6
Actividad auditiva - Capítulo 6
El presente perfecto
El pluscuamperfecto
Los pronombres relativos
Pronunciación 6: La diéresis
Cultura: Capítulo 6
Canción: Capítulo 6
Sinopsis: Capítulo 6
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Capítulo 2El imperfecto
El imperfecto
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"... y llovía." de Fran+Silva© tiene licencia CC BY 2.0.
Ya aprendieron a narrar hechos y eventos en el pasado usando el pretérito. El español tiene también otro tiempo verbal, conocido como el imperfecto y que también se utiliza para hablar del pasado. Como discutiremos más adelante, el imperfecto se usa generalmente para discutir acciones que eran habituales o que estaban transcurriendo en el pasado, para describir emociones, estados o para dar descripciones en el pasado.
La conjugación del imperfecto es mucho más simple que la del pretérito. No hay verbos con cambio ortográfico ni cambio de raíz y sólo hay tres verbos irregulares. Además, los verbos terminados en -er e -ir se conjugan con la misma terminación.
Verbos -ar:Verbos -er e -ir:
Hablar
yo hablaba
tú hablabas
él, ella, usted hablaba
nosotros hablábamos
vosotros hablabais
ellos, ellas, ustedes hablabanVivir
yo vivía
tú vivías
él, ella, usted vivía
nosotros vivíamos
vosotros vivíais
ellos, ellas, ustedes vivían
Los tres verbos irregulares en imperfecto son:
SerIrVer
yo era
tú eras
él, ella, usted era
nosotros éramos
vosotros erais
ellos, ellas, ustedes eran yo iba
tú ibas
él, ella, usted iba
nosotros íbamos
vosotros ibais
ellos, ellas, ustedes iban yo veía
tú veías
él, ella, usted veía
nosotros veíamos
vosotros veíais
ellos, ellas, ustedes veían
Ejemplos de oraciones que usan el imperfecto: Back to Top
Acciones habituales o repetidas en el pasado
Cuando yoeraniño, mi familiacomía en McDonald’s con frecuencia. / When I was a child, my family ate at McDonald’s frequently.
Mi padre pedía una hamburguesa, pero mi hermano prefería los sándwiches de pollo. /My father would order a hamburger, but my brother preferred chicken sandwiches.
Descripción de una escena, descripciones físicas o de estados emocionales en el pasado
Eran las 6:00, y el restaurante ya estaba lleno de clientes. /It was 6:00 and the restaurant was already full of customers.
Nuestro camarero se llamaba Pedro. /Our waiter’s name was Pedro.
Nosotros teníamos hambre. /We were hungry.
Acciones en progreso en el pasado (acciones que estaban ocurriendo)
Mientras nosotros charlábamos, el cocinero preparaba la comida. /While we were chatting, the cook was preparing the food.
Para hablar de lo que iba a ocurrir (what was going to happen) usando el imperfecto de IR+ A+ infinitivo
Nosotros íbamos a comer en una hora. /We were going to eat in an hour.
Ibas a comprar el pan. / You were going to buy the bread.
Vídeos
AIL Spanish - El pretérito imperfecto en español
Actividades de conversación Back to Top
Actividad 1.Una cena en familia. Back to Top
La familia Pérez estaba cenando en casa. Lee la descripción de la escena abajo y completa las oraciones con el imperfecto de los verbos indicados.
Todos los miembros de la familia Pérez 1. … (tener) mucha hambre. La mesa 2. … (estar) puesta, y Tomás, el padre, 3. … (cortar) el pollo. Los hijos, Carolina e Israel, 4.… (esperar) en sus sillas mientras la madre, Roberta 5. … (traer) las verduras. Ellos normalmente no 6. …. (comer) juntos porque todos 7. … (trabajar) horas diferentes. Pero ese día 8. … (ser) un día especial: el cumpleaños de Carolina. Carolina 9. … (querer) un perro para su cumpleaños, pero en el momento de la cena todavía no 10. … (saber) si 11. … (ir) a recibir uno.
Actividad 2.Tu restaurante favorito. Back to Top
Con un(a) compañero(a), conversa sobre tu restaurante favorito cuando eras niño(a). Usa las siguientes preguntas como guía, y contesta en oraciones completas.
¿Cómo se llamaba tu restaurante favorito?
¿Dónde estaba el restaurante?
¿Qué comida servía?
¿Ibas al restaurante frecuentemente? ¿Con qué frecuencia ibas?
¿Cómo eran los camareros?
¿Cuántos clientes normalmente había en el restaurante?
¿Generalmente ibas con tu familia o con tus amigos?
Actividad 3.Preguntas personales Back to Top
Con un(a) compañero(a), contesta las siguientes preguntas sobre tu vida con oraciones completas. Después vamos a compartir lo que más nos ha llamado la atención sobre las respuestas de nuestro compañero(a)
¿Cómo eras cuando eras pequeño?
¿Qué te gustaba comer?
¿A qué jugabas?
¿Qué deporte practicabas?
¿Cuál era tu clase favorita?
¿Dónde ibas de vacaciones cuando eras pequeño?
¿Tenías muchos amigos?
¿Peleabas con tus hermanos?
¿Qué querías ser de mayor?
¿Qué veías en televisión?
Actividad 4.El comienzo de una novela… Back to Top
Tu grupo está escribiendo una novela. El primer párrafo es unadescripción(en el pasado) del restaurante donde todo comienza. En grupos pequeños, escriban una descripción de la escena abajo usando el imperfecto de los verbos. Esta descripción será el primer párrafo de su libro, así que ¡traten de captar la atención de sus lectores! (Abajo tienen la primera oración del libro y unas preguntas sobre la escena para ayudarlos a comenzar.)
Comedor principal del restaurante El Bajío en Azcapotzalco, Ciudad de México de AlejandroLinaresGarcia tiene licencia CC BY-SA.
Para comenzar:Los detectives Peralta y Sánchez estaban en su restaurante favorito que se llamaba…
Preguntas sobre la escena:
¿Dónde estaba el restaurante?
¿Cómo era el restaurante?
¿Cuántas personas había?
¿Qué otras cosas había?
¿Qué hacían los detectives?
¿Cómo se sentían los detectives?
Etc.
¡Recuerda, con el imperfecto, están describiendo la escena, lo que estaba ocurriendo y lo que las personas estaban haciendo o cómo se estaban sintiendo… Eso es, todo antes de que lleguen al hecho (event) o a la acción en sí misma!
Para seguir practicando Back to Top
Si quieren seguir practicando el imperfecto pueden usar los siguientes enlaces:
Lingolia - Pretérito imperfecto
DELE Ahora - Pretérito imperfecto
Aprender español - Imperfecto indicativo
The information in this chapter, when not created by us, is adapted from:
Farmer,Zamostny,Hill, and Henderson Hollenbeck,Bienvenidos(CC BY).
This work is released under a CC BY-NC-SA license, which means that you are free to do with it as you please as long as you (1) properly attribute it, (2) do not use it for commercial gain, and (3) share any subsequent works under the same or a similar license.
This content was previously published as follows:
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Solve Equations with Fractions - Steps, Examples & Questions
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In order to access this I need to be confident with:
Adding fractions Subtracting fractions Dividing fractions Multiplying fractions Rational numbers Inverse operations Reciprocal math
Introduction
What are equations with fractions?
Common Core State Standards
How to solve equations with fractions
Solve equations with fractions examples↓
Example 1: equations with one operation Example 2: equations with one operation Example 3: equations with two operations Example 4: equations with two operations Example 5: equations with three operations Example 6: equations with three operations Example 7: equations with the unknown as the denominator Example 8: equations with the unknown as the denominator
Teaching tips for solving equations with fractions
Easy mistakes to make
Related math equations lessons
Practice solve equations with fractions questions
Solve equations with fractions FAQs
Next lessons
Still stuck?
Math resources Algebra Math equationsSolve equations with fractions
Solve equations with fractions
Here you will learn about how to solve equations with fractions, including solving equations with one or more operations. You will also learn about solving equations with fractions where the unknown is the denominator of a fraction.
Students will first learn how to solve equations with fractions in 7th grade as part of their work with expressions and equations and expand that knowledge in 8th grade.
What are equations with fractions?
Equations with fractions involve solving equations where the unknown variable is part of the numerator and/or denominator of a fraction.
The numerator (top number) in a fraction is divided by the denominator (bottom number).
To solve equations with fractions, you will use the “balancing method” to apply the inverse operation to both sides of the equation in order to work out the value of the unknown variable.
The inverse operation of addition is subtraction.
The inverse operation of subtraction is addition.
The inverse operation of multiplication is division.
The inverse operation of division is multiplication.
For example,
\begin{aligned} \cfrac{2x+3}{5} \, &\= 7\\ \colorbox{#cec8ef}{$\times \, 5$} \; & \;\; \colorbox{#cec8ef}{$\times \, 5$} \\\\ 2x+3&\=35 \\ \colorbox{#cec8ef}{$-\,3$} \; & \;\; \colorbox{#cec8ef}{$- \, 3$} \\\\ 2x & = 32 \\ \colorbox{#cec8ef}{$\div \, 2$} & \; \; \; \colorbox{#cec8ef}{$\div \, 2$}\\\\ x & = 16 \end{aligned}
What are equations with fractions?
Common Core State Standards
How does this relate to 7th grade and 8th grade math?
Grade 7: Expressions and Equations (7.EE.A.1)
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
Grade 8: Expressions and Equations (8.EE.C.7)
Solve linear equations in one variable.
Grade 8: Expressions and Equations (8.EE.C.7b)
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
How to solve equations with fractions
In order to solve equations with fractions:
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
When you have the variable on one side, you have the final answer.
Check the answer by substituting the answer back into the original equation.
[FREE] Solve Equations with Fractions Worksheet (Grade 7 and 8)
Use this worksheet to check your 7th grade and 8th grade students’ understanding of solving equations with fractions. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
x
[FREE] Solve Equations with Fractions Worksheet (Grade 7 and 8)
Use this worksheet to check your 7th grade and 8th grade students’ understanding of solving equations with fractions. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
Solve equations with fractions examples
Example 1: equations with one operation
Solve for x \text{: } \cfrac{x}{5}=4 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, the x is divided by 5.
\cfrac{x}{5}
2Apply the inverse operations, one at a time, to both sides of the equation.
The inverse of “dividing by 5 ” is “multiplying by 5 ”.
You will multiply both sides of the equation by 5.
3When you have the variable on one side, you have the final answer.
The final answer is x=20.
4Check the answer by substituting the answer back into the original equation.
You can check the answer by substituting the answer back into the original equation.
\cfrac{20}{5}=20\div5=4
Example 2: equations with one operation
Solve for x \text{: } \cfrac{x}{3}=8 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, the x is divided by 3.
\cfrac{x}{3}
Apply the inverse operations, one at a time, to both sides of the equation.
The inverse of “dividing by 3 ” is “multiplying by 3 ”.
You will multiply both sides of the equation by 3.
Write the final answer, checking that it is correct.
The final answer is x=24.
You can check the answer by substituting the answer back into the original equation.
\cfrac{24}{3}=24\div3=8
Example 3: equations with two operations
Solve for x \text{: } \cfrac{x \, + \, 1}{2}=7 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, 1 is added to x and then divided by 2 (the denominator of the fraction).
\cfrac{x \, + \, 1}{2}
Apply the inverse operations, one at a time, to both sides of the equation.
First, clear the fraction by multiplying both sides of the equation by 2.
Then, subtract 1 from both sides.
Write the final answer, checking that it is correct.
The final answer is x=13.
You can check the answer by substituting the answer back into the original equation.
\cfrac{13 \, +1 \, }{2}=\cfrac{14}{2}=14\div2=7
Example 4: equations with two operations
Solve for x \text{: } \cfrac{x}{4}-2=3 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, x is divided by 4 and then 2 is subtracted.
\cfrac{x}{4}-2
Apply the inverse operations, one at a time, to both sides of the equation.
First, add 2 to both sides of the equation.
Then, multiply both sides of the equation by 4.
Write the final answer, checking that it is correct.
The final answer is x=20.
You can check the answer by substituting the answer back into the original equation.
\cfrac{20}{4}-2=20\div4-2=5-2=3
Example 5: equations with three operations
Solve for x \text{: } \cfrac{3x}{5}+1=7 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, x is multiplied by 3, then divided by 5 , and then 1 is added.
\cfrac{3x}{5}+1
Apply the inverse operations, one at a time, to both sides of the equation.
First, subtract 1 from both sides of the equation.
Then, multiply both sides of the equation by 5.
Finally, divide both sides by 3.
Write the final answer, checking that it is correct.
The final answer is x=10.
You can check the answer by substituting the answer back into the original equation.
\cfrac{3 \, \times \, 10}{5}+1=\cfrac{30}{5}+1=6+1=7
Example 6: equations with three operations
Solve for x \text{: } \cfrac{2x-1}{7}=3 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, x is multiplied by 2, then 1 is subtracted, and the last operation is divided by 7 (the denominator).
\cfrac{2x-1}{7}
Apply the inverse operations, one at a time, to both sides of the equation.
First, multiply both sides of the equation by 7.
Next, add 1 to both sides.
Finally, divide both sides by 3.
Write the final answer, checking that it is correct.
The final answer is x=11.
You can check the answer by substituting the answer back into the original equation.
\cfrac{2 \, \times \, 11-1}{7}=\cfrac{22-1}{7}=\cfrac{21}{7}=3
Example 7: equations with the unknown as the denominator
Solve for x \text{: } \cfrac{24}{x}=6 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, x is the denominator. 24 is divided by x.
\cfrac{24}{x}
Apply the inverse operations, one at a time, to both sides of the equation.
You need to multiply both sides of the equation by x.
Then, you can divide both sides by 6.
Write the final answer, checking that it is correct.
The final answer is x=4.
You can check the answer by substituting the answer back into the original equation.
\cfrac{24}{4}=24\div4=6
Example 8: equations with the unknown as the denominator
Solve for x \text{: } \cfrac{18}{x}-6=3 .
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, x is the denominator. 18 is divided by x , and then 6 is subtracted.
\cfrac{18}{x}-6
Apply the inverse operations, one at a time, to both sides of the equation.
First, add 6 to both sides of the equation.
Then, multiply both sides of the equation by x.
Finally, divide both sides by 9.
Write the final answer, checking that it is correct.
The final answer is x=2.
You can check the answer by substituting the answer back into the original equation.
\cfrac{18}{2}-6=9-6=3
Teaching tips for solving equations with fractions
When students first start working through practice problems and word problems, provide step-by-step instructions to assist them with solving linear equations.
Introduce solving equations with fractions with one-step problems, then two-step problems, before introducing multi-step problems.
Students will need lots of practice with solving linear equations. These standards provide the foundation for work with future linear equations in Algebra I and II.
Provide opportunities for students to explain their thinking through writing. Ensure that they are using key vocabulary, such as, absolute value, coefficient, equation, common factors, inequalities, simplify, etc.
Easy mistakes to make
Forgetting that the solution to an equation can be any type of number
The unknowns do not have to be integers (whole numbers and their negative opposites). The solutions can be fractions or decimals. They can also be positive or negative numbers.
Forgetting that the unknown of an equation can be on either side of the equation
The unknown, represented by a letter, is often on the left hand side of the equations; however, it doesn’t have to be. It could also be on the right hand side of an equation.
When multiplying both sides of an equation, not multiplying each and every term
When multiplying each side of the equation by a number, it is a common mistake to forget to multiply every term.
For example,
Solve:
\cfrac{x}{2}+3=9
Here, the + 3 was not multiplied by 2, resulting in the incorrect answer.
This person has correctly multiplied each term by the denominator.
Lowest common denominator (LCD)
It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting, you need to work out the lowest/least common denominator (sometimes called the least common multiple or LCM). When you solve equations involving fractions, multiply both sides of the equation by the denominator of the fraction.
Related math equations lessons
Math equations
Rearranging equations
How to find the equation of a line
Substitution
Linear equations
Writing linear equations
Solving equations
Identity math
One step equations
Practice solve equations with fractions questions
1. Solve: \cfrac{x}{6}=3
x=9
x=36
x=12
x=18
You will multiply both sides of the equation by 6, because the inverse of “dividing by 6 ” is “multiplying by 6 ”.
The final answer is x = 18.
You can check the answer by substituting the answer back into the original equation.
\cfrac{18}{6}=18 \div 6=3
2. Solve: \cfrac{x \, + \, 4}{2}=7
x=18
x=10
x=26
x=30
First, clear the fraction by multiplying both sides of the equation by 2.
Then subtract 4 from both sides.
The final answer is x = 10.
You can check the answer by substituting the answer back into the original equation.
\cfrac{10 \, + \, 4}{2}=\cfrac{14}{2}=14 \div 2=7
3. Solve: \cfrac{x}{8}-5=1
x=40
x=64
x=48
x=56
First, add 5 to both sides of the equation.
Then multiply both sides of the equation by 8.
The final answer is x = 48.
You can check the answer by substituting the answer back into the original equation.
\cfrac{48}{8}-5=48 \div 8-5=1
4. Solve: \cfrac{3x \, + \, 2}{4}=2
x=4
x=16
x=12
x=2
First, multiply both sides of the equation by 4.
Next, subtract 2 from both sides.
Finally, divide both sides by 3.
The final answer is x = 2.
You can check the answer by substituting the answer back into the original equation.
\cfrac{3 \, \times \, 2+2}{4}=\cfrac{6 \, + \, 2}{4}=\cfrac{8}{4}=8 \div 4=2
5. Solve: \cfrac{4x}{7}-2=6
x=11
x=14
x=7
x=10
First, add 2 to both sides of the equation.
Then multiply both sides of the equation by 7.
Finally, divide both sides by 4.
The final answer is x = 14.
You can check the answer by substituting the answer back into the original equation.
\cfrac{4 \, \times \, 14}{7}-2=\cfrac{56}{7}-2=56 \div 7-2=6
6. Solve: \cfrac{42}{x}=7
x=5
x=294
x=7
x=6
You need to multiply both sides of the equation by x.
Then you divide both sides by 7.
The final answer is x = 6.
You can check the answer by substituting the answer back into the original equation.
\cfrac{42}{6}=42 \div 6=7
Solve equations with fractions FAQs
Do I still follow the order of operations when solving equations with fractions?Yes, you still follow the order of operations when solving equations with fractions. You will start with any operations in the numerator and follow PEMDAS (parenthesis, exponents, multiply/divide, add/subtract), followed by any operations in the denominator. Then you will solve the rest of the equation as usual.
The next lessons are
Inequalities
Types of graphs
Math formulas
Coordinate plane
Number patterns
Algebraic expressions
Fractions
Still stuck?
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Introduction
What are equations with fractions?
Common Core State Standards
How to solve equations with fractions
Solve equations with fractions examples↓
Example 1: equations with one operation Example 2: equations with one operation Example 3: equations with two operations Example 4: equations with two operations Example 5: equations with three operations Example 6: equations with three operations Example 7: equations with the unknown as the denominator Example 8: equations with the unknown as the denominator
Teaching tips for solving equations with fractions
Easy mistakes to make
Related math equations lessons
Practice solve equations with fractions questions
Solve equations with fractions FAQs
Next lessons
Still stuck?
x
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middle dot meaning? | Free Math Help Forum
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middle dot meaning?
Thread startercruz33
Start dateJul 19, 2013
C
cruz33
New member
Joined Jul 9, 2013 Messages 37
Jul 19, 2013
#1
What does this mathematical expression mean?
7 . 3
But, with the dot in the middle, not at the bottom. I can’tfind a way to type a dot in the middle.
Thanks!
D
DrPhil
Senior Member
Joined Nov 29, 2012 Messages 1,383
Jul 19, 2013
#2
cruz33 said:
What does this mathematical expression mean?
7 · 3
But, with the dot in the middle, not at the bottom. I can’tfind a way to type a dot in the middle.
Alt-0183 --> ·
Thanks!
Click to expand...
-In some countries the center dot is the normal position for the decimal point 7·3 = 7.3
In the US, the center dot is multiplication: 7 · 3 = 21
L
lookagain
Elite Member
Joined Aug 22, 2010 Messages 3,266
Jul 20, 2013
#3
cruz33 said:
What does this mathematical expression mean?
7 . 3
But, with the dot in the middle, not at the bottom. I can’t find a way to type a dot in the middle.
Thanks!
Click to expand...
cruz33, try typing thie following in Latex: (\displaystyle \ \ ) [ tex]7 \cdot 3[ /tex] (\displaystyle \ \ ) But don't have any spaces between the open bracket and the "t" of "tex]." Also, don't have any spaces between the open bracket and the "/" of "/tex]."
C
cruz33
New member
Joined Jul 9, 2013 Messages 37
Jul 20, 2013
#4
that's what I thouht...
DrPhil said:
-In some countries the center dot is the normal position for the decimal point 7·3 = 7.3
In the US, the center dot is multiplication: 7 · 3 = 21
Click to expand...
That's what I thought it meant, but WHY a dot is used instead of an x, like 7x3, like I have always seen it, why use a dot instead? Sorry, I am a very inquisitive person. Is there a reason why a dot is used instead of the "normal" way of writing a multiplication problem?
J
JeffM
Elite Member
Joined Sep 14, 2012 Messages 7,875
Jul 20, 2013
#5
cruz33 said:
That's what I thought it meant, but WHY a dot is used instead of an x, like 7x3, like I have always seen it, why use a dot instead? Sorry, I am a very inquisitive person. Is there a reason why a dot is used instead of the "normal" way of writing a multiplication problem?
Click to expand...
Nothing wrong with being inquisitive.
In algebra and higher mathematics, letters are used to stand for a number that is not yet known or specified. So x might be a number instead of an operator. So, to avoid confusion, a new symbol was needed for multiplication. Some people use the dot to mean multiplication; more nowadays use the asterisk.
C
cruz33
New member
Joined Jul 9, 2013 Messages 37
Jul 20, 2013
#6
got it!
JeffM said:
Nothing wrong with being inquisitive.
In algebra and higher mathematics, letters are used to stand for a number that is not yet known or specified. So x might be a number instead of an operator. So, to avoid confusion, a new symbol was needed for multiplication. Some people use the dot to mean multiplication; more nowadays use the asterisk.
Click to expand...
Interesting! Thanks for the explanation!
D
Deleted member 4993
Guest
Jul 20, 2013
#7
The middle dot - or· -is still used in vector multiplication as a sign for scalar product.
C
cruz33
New member
Joined Jul 9, 2013 Messages 37
Jul 20, 2013
#8
thank you!
Subhotosh Khan said:
The middle dot - or· -is still used in vector multiplication as a sign for scalar product.
Click to expand...
Thanks for the additional info! I had to goggle "vector multiplication" though, so I learned another mathematical concept!
D
Deleted member 4993
Guest
Jul 20, 2013
#9
cruz33 said:
Thanks for the additional info! I had to goggle "vector multiplication" though, so I learned another mathematical concept!
Click to expand...
Remember however that you just learned about the "word" vector multiplication.
Vector multiplication is pretty complicated and deep stuff - the concept is pretty difficult.
Just food for thought - although we have vector multiplication, we do not have vector division defined.
C
cruz33
New member
Joined Jul 9, 2013 Messages 37
Jul 20, 2013
#10
got it!
Subhotosh Khan said:
Remember however that you just learned about the "word" vector multiplication.
Vector multiplication is pretty complicated and deep stuff - the concept is pretty difficult.
Just food for thought - although we have vector multiplication, we do not have vector division defined.
Click to expand...
Just learning the meaning of the word is enough for me, don't want to go deep, I'm already overwhelmed with what I'm learning now in my current math class!
mmm4444bot
Super Moderator
Joined Oct 6, 2005 Messages 10,962
Jul 20, 2013
#11
JeffM said:
more [people] nowadays use the asterisk.
Click to expand...
You're speaking about texting, yes? :cool:
I never write an asterisk, when using paper and pencil. I also use centered dots when formatting math (eg: equation editors, diagrams, LaTex, et cetera).
J
JeffM
Elite Member
Joined Sep 14, 2012 Messages 7,875
Jul 20, 2013
#12
mmm4444bot said:
You're speaking about texting, yes? :cool:
I never write an asterisk, when using paper and pencil. I also use centered dots when formatting math (eg: equation editors, diagrams, LaTex, et cetera).
Click to expand...
Well actually I was thinking of calculators, spreadsheets, etc. I tend to use a dot myself when using paper and pencil, but that is because (a) I am old enough that dots were what I was taught in algebra as a child, (b) dots are quicker to write than asterisks, and (c) it is useful to have a symbol for an unspecified operation, and the asterisk was used for that purpose back when I studied abstract algebra.
mmm4444bot
Super Moderator
Joined Oct 6, 2005 Messages 10,962
Jul 20, 2013
#13
JeffM said:
actually I was thinking of calculators, spreadsheets, etc.
Click to expand...
Okay -- agreed -- more people type asterisks for that purpose than do centered dots!
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https://www.youtube.com/watch?v=zxvUdkgR6EM
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Area of a Triangle - Box Method
MooMooMath and Science
583000 subscribers
37 likes
Description
5582 views
Posted: 28 Jun 2013
Area Triangle using the box method.
The box method calculates the area of a triangle by subtracting the area of the triangle shapes from its bounding box.
For even more methods of finding the area of triangle see:
Box method for Area of a Triangle Demo by J Scoles.MOV
5 ways to find the area of a triangle
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Please watch: "Study Skills Teacher's Secret Guide to your Best Grades"
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Transcript:
hi Welcome to mo math today i'm going to teach you how to find the area Of A triangle using the box method now You're going to need to have your triangle um already drawn on the coordinate Plane And you need to know two formulas before you start you need to know the area Of A rectangle Which is length Times width And you need to then know the area Of A triangle Which is 1/2 base times height If you know all those items then You're good to go Ok here's our example I'm going to go ahead and zoom in cuz I'm going to be doing some drawing on the graph to kind of show you How all this works ok So I have a triangle ABC and I have identified the coordinates across here Ok Now the coordinates really aren't that important in this method what I'm going to do is I'm actually going to draw Myself a rectangle That is going to go all the way around or circumscribe the rectangle so see what I'm doing I'm drawing from basically vertex to vertex so from a i drew all the way over Till I got to c c iw drew all the way up Until I got to B and then it drew across back to a so I've got a rectangle right here Now what I'm going to do is I'm going to find the area of this rectangle So I need to know the length and the width of this So I'm just going to count My blocks 1 2 3 4 5 si so That is si Units Long and it's 1 2 3 4 and it's four Units high ok So Let's go back for a second to our sheet We needed to know the area of this whole rectangle and then What We're going to do is once we do that We're going to take it out the area of these smaller re or the small triangles and What We're going to be left with is the area of triangle ABC so the concept is find the area of the rectangle and then remove these three right triangles And you have the area of What's left ok So what We're going to do first is find the area of the rectangle and We just talked about a second Ago What's the area Formula for a rectangle It's a length times width So I'm just going to go 6 x 4 6 x 4 so my area is 24 and That's the area of the whole rectangle now Let's take each small triangle So I'm going to take this one Ok this top one and this triangle is 1 2 3 4 Units Long and two Units high so it's 1/2 4 x 2 Ok Which is going to be half of 4 2 is 4 OK then I'm going to do the same thing for the next triangle Over Here on the left side It is 1 2 3 4 Units high two Units Wide 4 by 2 So I'm going to take 1/2 of 4 x 2 And Again That's going to be 4 and then I'm going to find the area of this bottom triangle well that we already know this is 6 and this portion is two so 1/2 of 6 x 2 half of 6 2 is 6 So I'm going to take the area of the rectangle Which is 24 and I'm going to subtract the two triangles have an area of 4 and then the third triangle which has An area 6 and What's going to give me What's going to be left is the area of the triangle so Let's just jop that down So I have the area length times width 6 x 4 is 24 and That is the area of the rectangle and then I've got each of the three triangles So I'm going to say triangle number one is 1/2 4 x 2 so the area of that one is 4 I have another triangle who has a height and a base of 4 and two so it's area is four and then the third triangle has the base of six and a height of two so the area of that triangle is 6 I'm going to take 24 and subtract 4 subtract 4 and subtract 6 So I'm subtracting 14 so 24 - 14 leaves me with An area of 10 and That is how you find the area Of A triangle when it is in the middle Of A graph And you can't draw An easily in An altitude And you can't easily find its base so That's how you do it and remember Units and area are always unit squared and That is the box method when you have a triangle grafted on the cartesian coordinate Plane hope this video was helpful
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https://www.justia.com/law-schools/justia-case-law-outlines/property-cases-outline/
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Property Cases Outline
While statutes and regulations have shaped many areas of property law, a student of this topic also must understand how courts have interpreted common-law, statutory, and constitutional principles. Courts have addressed issues such as theories of possession and ownership, the process for transferring property interests, restrictions on the use of property, and intersections with constitutional rights. Below is an outline of key cases in property law with links to the full text of virtually every case, provided free by Justia.
Topics in Property Law
1
Possession and Ownership
2
Adverse Possession
3
Conveyances and Types of Property Interests
4
Landlord-Tenant Laws
5
Easements, Covenants, and Nuisance
6
Zoning Rules
7
Government Takings
8
Equal Protection, Due Process, and Property
9
The Right to Exclude and Constitutional Freedoms
10
Intellectual Property
Possession and Ownership
Ownership of property confers a robust set of rights, but clear title may not be easy to prove. As a pragmatic alternative, courts often link ownership to possession. For example, the occupation theory provides that physical occupation inherently confers a right to keep property as long as it is in the control of the occupier.
Pierson v. Post 一 Property in wild animals is acquired by occupancy only. This requires depriving them of their natural liberty and subjecting them to the control of their pursuer.
Commonwealth v. Agway, Inc. 一 A state does not have a property interest in fish in a state of freedom, the invasion of which would support an action in trespass for damages.
Johnson & Graham’s Lessee v. McIntosh 一 Native American inhabitants are to be considered merely occupants, to be protected while in peace in the possession of their lands, but deemed incapable of transferring absolute title to others.
City of Sherrill v. Oneida Indian Nation of New York 一 When a tribe had relinquished governmental reins over an area long before, it could not regain them through open-market purchases from current titleholders.
U.S. v. Percheman 一 Private property generally should not be confiscated and private rights annulled on a change in the sovereignty of a territory.
Tapscott v. Cobbs 一 The law protects a peaceable possession against all except the party who has the actual right to the possession.
Winchester v. City of Stevens Point 一 A plaintiff must prove their title to recover damages for a permanent injury to property.
Geragosian v. Union Realty Co. 一 An owner of land is entitled to an injunction for the removal of trespassing structures. Only when there is some estoppel or laches on the part of the plaintiff, or a refusal on their part to consent to acts necessary to the removal or abatement, will an injunction be refused.
Peters v. Archambault 一 A landowner is ordinarily entitled to mandatory equitable relief to compel removal of a structure significantly encroaching on their land, even if the encroachment was unintentional, and the cost of removal is substantial in comparison to any injury suffered by the owner of the lot on which the encroachment has taken place.
Edwards v. Sims 一 The owner of realty, unless there has been a division of the estate, is entitled to the free and unfettered control of their land above, upon, and beneath the surface.
Adverse Possession
A possessor of property may be able to claim ownership if they can prove that their possession was adverse to the true owner. Their possession must be actual and exclusive, open and notorious, continuous during the statutory period, and hostile and under claim of right, without permission from the owner.
Lessee of Ewing v. Burnet 一 An entry by one party on the land of another party is or is not an ouster of the legal possession arising from the title according to the intention with which it is done. If it is made under claim or color of right, it is an ouster; otherwise, it is a mere trespass.
Peters v. Juneau-Douglas Girl Scout Council 一 In determining the exclusive character of possession, the character and locality of the property and the uses and purposes for which it is naturally adopted are considered. Casual and occasional trespasses on the land by a stranger generally do not interrupt the continuity of adverse possession. Finally, hostility of possession means that the possessor must use and enjoy the property continuously for the required period as the average owner would use it, without the consent of the true owner.
McCarty v. Sheets 一 While maintenance activities in a residential area are a factor in a property dispute, standing alone, they are not sufficient to support a divesture of property based upon adverse possession.
Mendonca v. Cities Service Oil Corp. 一 Continuity of possession was broken when the true owner of the contested property removed fences on the property and used it for three or four weeks to store building materials and equipment.
Ennis v. Stanley 一 When the possession is up to a fixed boundary under a mistake as to the true line, and the intention of the parties is to hold only to the true line, such possession is not hostile and will not ripen into title.
Mannillo v. Gorski 一 Any entry and possession for the required time that is exclusive, continuous, uninterrupted, visible, and notorious, even though it is under mistaken claim of title, is sufficient to support a claim of title by adverse possession.
Belotti v. Bickhardt 一 All that is necessary to make an adverse possession effectual for the statutory period by successive persons is that such possession be continued by an unbroken chain of privity between the adverse possessors.
Fallon v. Davidson 一 Until an actual ouster of co-tenants had been established by conduct apart from mere use and occupation of the land, the statute giving rise to a claim of adverse possession did not begin to run.
Conveyances and Types of Property Interests
The process of conveying an interest in real estate usually involves a purchase and sale agreement. However, forming a contract does not transfer title immediately. Ownership of the property is transferred when the deed is delivered to the buyer, which usually occurs at the real estate closing.
Metzger v. Miller 一 A contract concerning real property need not be in any particular form. A letter is a sufficient memorandum of an agreement related to such property to avoid the statute of frauds.
Hayes v. Hayes 一 To take a parol grant or gift of land out of the statute of frauds, there must be not only acceptance but also such a performance in the way of making valuable improvements as would make it substantial injustice or fraud to hold the grant or gift void under the statute.
Micklethwait v. Fulton 一 An escrow deed that was wrongfully delivered and recorded before the performance of the conditions of the escrow is a valid deed of conveyance as to third parties who innocently renew or extend credit to the grantee in reliance on their record ownership of the property.
Hood v. Webster 一 When a deed is held in escrow, but another party is granted the property by another deed and records that deed before the conditions of escrow are met, the party who was granted the property by the deed in escrow may annul the later deed.
Storke v. Penn Mutual Life Insurance Co. 一 A breach of a condition subsequent does not revest title in the original grantor or their heirs. Re-entry is necessary to revest title, and a court will not aid a forfeiture when no right of re-entry is provided in the covenant.
Browning v. Sacrison 一 When a devise is made to a life tenant with a remainder conditioned on an ambiguous form of survivorship, the intent “commonly prevalent among conveyors similarly situated” is deemed to require that the remainderman survive the life tenant rather than the testator.
Brown v. Independent Baptist Church of Woburn 一 The rule against perpetuities does not apply to possibilities of reverter.
Ryan v. Beshk 一 The intention to be sought in the construction of a will is what is expressed by the language of the will, rather than what may be presumed by inference to have been in the mind of the testator. Also, a contingent remainder is a remainder limited to take effect in possession on an event that may not happen until after the termination of the particular estate.
Holbrook v. Holbrook 一 When a state has abolished joint tenancy, language purporting to create a joint tenancy creates concurrent estates for life with contingent remainders in the life tenants, the remainder to vest in the survivor.
Landlord-Tenant Laws
A tenant receives a possessory interest in property, while the landlord retains an interest expressed by rent. A landlord can regain possession if the tenant fails to pay rent or violates other terms of the lease. Some leases grant possession for a fixed time, while others may continue indefinitely unless they are terminated.
Lemle v. Breeden 一 In the lease of a dwelling, there is an implied warranty of habitability and fitness for the use intended. Since a lease is essentially a contractual relationship, remedies available to a tenant for a breach are the contract remedies of damages, reformation, and rescission.
Hilder v. St. Peter 一 In the rental of any residential dwelling unit, an implied warranty exists in the lease, whether oral or written, that the landlord will deliver over and maintain premises that are safe, clean, and fit for human habitation. This warranty of habitability is implied in tenancies for a specific period or at will.
Reste Realty Corp. v. Cooper 一 Any act or omission of the landlord or of anyone who acts under authority or legal right from the landlord, or of someone having superior title to that of the landlord, which renders the premises substantially unsuitable for the purpose for which they are leased, or which seriously interferes with the beneficial enjoyment of the premises, is a breach of the covenant of quiet enjoyment and constitutes a constructive eviction of the tenant.
Javins v. First National Realty Corp. 一 The traditional no-repair rule cannot coexist with the obligations imposed on a landlord by a modern housing code, and this rule must be abandoned in favor of an implied warranty of habitability.
Helmsley v. Borough of Fort Lee 一 A rent control ordinance must permit an efficient landlord to realize a just and reasonable return.
Chicago Board of Realtors, Inc. v. City of Chicago 一 Prior regulation of the landlord-tenant relationship dictates a lower level of scrutiny of an ordinance in this area, which must be supported by legitimate and significant purposes.
Pennell v. City of San Jose 一 Preventing unreasonable rent increases caused by a city housing shortage is a legitimate exercise of the police powers of the city government.
Braschi v. Stahl Assocs. Co. 一 When an ordinance prohibits a landlord from dispossessing a family member of a deceased tenant who was living with the tenant, “family” should be defined to include a same-sex partner of the tenant.
Easements, Covenants, and Nuisance
Sometimes a property owner holds rights regarding property owned by someone else. An easement usually makes a grant of a right, while a covenant provides a promise to do or not do something. Meanwhile, a nuisance occurs when someone else uses their property in a way that significantly undermines a property owner's use and enjoyment of their property.
Waldrop v. Town of Brevard 一 A covenant or agreement may operate as a grant of an easement if it is necessary to give it that effect to carry out the manifest intention of the parties.
Cox v. Glenbrook Co. 一 The extent of an easement created by conveyance is fixed by the conveyance, if it is clear and unambiguous. When the width is not specified, the conveying instrument must be construed in light of the facts and circumstances existing at its date and affecting the property, the intention of the parties being the object of inquiry. Also, the owner of an easement may prepare, maintain, improve, or repair the way in a manner and to an extent reasonably calculated to promote the purposes for which it was created. However, they may not cause an undue burden on the servient estate or an unwarranted interference with the independent rights of others who have a similar right of use.
Charping v. J.P. Scurry & Co., Inc. 一 The burden of showing that restrictions in a deed are covenants running with the land is on the party claiming the benefit of the restrictions. There must be an indication that the parties who imposed the restrictions intended for the covenant to run with the land.
VanSant v. Rose 一 The right to enjoin the breach of restrictive covenants does not depend on whether the covenantee will be damaged by the breach, nor on whether they own other land in the vicinity.
Streams Sports Club Ltd. v. Richmond 一 A third-party beneficiary of a contract between the original developer of a condominium complex and the condominium owners could enforce its rights under the contract.
Neponsit Property Owners’ Ass’n v. Emigrant Industrial Savings Bank 一 A corporation formed as a medium for the enjoyment of common rights of property owners may enforce the covenant upon which such common rights depend, even though the corporation has no interest of its own in the enforcement of the covenant.
Werner v. Graham 一 Parties who desire to create mutual rights in real property must say so in the written instruments exchanged between them that constitute the final expression of their understanding.
Riley v. Bear Creek Planning Committee 一 From the recordation of the first deed that imposes restrictions on the land conveyed and the land retained by the common grantor, the restrictions are binding on all subsequent grantees of affected parcels who take with notice of the restrictions, even if similar clauses are omitted from their deeds. Acknowledgment and recording of a declaration of restrictions by a grantor after a conveyance to another party will not affect property in which the grantor no longer has an interest.
Sanborn v. McLean 一 Reciprocal negative easements arise out of a benefit accorded land retained, by restrictions upon neighboring land sold by a common owner. Such a scheme of restrictions must start with a common owner; it cannot arise and fasten upon one lot by reason of other lot owners conforming to a general plan.
Snow v. Van Dam 一 The existence of a scheme for restricting the lots in a tract undergoing development to obtain substantial uniformity in building and use may be used to show an intention that the restrictions imposed upon the several lots shall be appurtenant to every other lot in the tract included in the scheme.
Russell Realty Co. v. Hall 一 Purchasers’ rights under restrictive covenants are circumscribed and confined by the territorial limits of the plat with reference to which the purchasers bought.
Rhue v. Cheyenne Homes, Inc. 一 A covenant requiring the approval of the architectural committee before the erection of a house in a subdivision is enforceable.
Davis v. Huey 一 Covenants requiring submission of plans and prior consent before construction are valid insofar as they furnish adequate notice to the property owner of the specific restriction sought to be enforced.
Ginsberg v. Yeshiva of Far Rockaway 一 Unless the character of the neighborhood has so changed as to defeat the object and purposes for which the restrictions were imposed, a restrictive covenant is enforceable against a party that purchased the restricted lot with knowledge of the covenant.
Camelback Del Este Homeowners Ass’n v. Warner 一 The test for determining whether restrictive covenants should be enforced is whether or not the conditions have changed so much that it is impossible to secure in a substantial degree the benefits intended to be secured by the covenants. The court must consider the overall relation of the various lots and not merely decide whether it may be in the best interest of particular lot owners in a subdivision to be absolved from the operation of restrictive covenants.
Morgan v. High Penn Oil Co. 一 A private nuisance exists in a legal sense when one makes an improper use of his own property and in that way injures the land or some incorporeal right of one's neighbor. A person who intentionally creates or maintains a private nuisance is liable for the resulting injury to others, regardless of the degree of care or skill exercised by him to avoid such injury.
Prah v. Maretti 一 Private nuisance law has the flexibility to protect both a landowner’s right of access to sunlight and another landowner’s right to develop land.
Boomer v. Atlantic Cement Co. 一 When there is a large disparity in economic consequences between allowing a nuisance and granting an injunction, a court may grant the injunction conditioned on the payment of permanent damages.
Spur Industries, Inc. v. Del E. Webb Development Co. 一 When the operation of a business is lawful in the first instance but becomes a nuisance by reason of a nearby residential area, the operation of the business may be enjoined in an action brought by the developer of the residential area. However, the developer of a completely new town or urban area in a previously agricultural area may be required to indemnify the operator of the business who must move or cease operation because of the presence of the residential area created by the developer.
Zoning Rules
Each state has enacted zoning enabling legislation for municipalities, although constitutional limits still apply. Zoning allows a local government to control land use. Zoning must not be arbitrary, unreasonable, or capricious, and it should not be retroactive.
Village of Euclid v. Ambler Realty Co. 一 A zoning ordinance is not unconstitutional unless it is clearly arbitrary and unreasonable, having no substantial relation to public health, safety, morals, or general welfare.
Nectow v. City of Cambridge 一 The inclusion of private land in a residential district under a zoning ordinance, with resulting inhibition of its use for business and industrial buildings to the serious damage of the owner, violates the Fourteenth Amendment if the health, safety, convenience, or general welfare of the part of the city affected will not be promoted as a result.
Village of Belle Terre v. Boraas 一 The police power is not confined to elimination of filth, stench, and unhealthy places. It is ample to lay out zones where family values, youth values, and the blessings of quiet seclusion and clean air make the area a sanctuary for people.
Pierro v. Baxendale 一 Once selections of particular uses that are congruous in residential zones are made and duly embodied in a zoning ordinance, these selections will not be nullified except upon an affirmative showing that the action taken by the municipal officials was unreasonable, arbitrary, or capricious.
Town of Preble v. Song Mountain, Inc. 一 Procedures set forth in zoning statutes must be strictly followed, but only a departure in substance from the formula prescribed by law vitiates the proceedings.
State ex rel. Stoyanoff v. Berkeley 一 Courts will not substitute their judgments for a city’s legislative body if the result is not oppressive, arbitrary, or unreasonable and does not infringe upon a valid pre-existing non-conforming use.
Anderson v. City of Issaquah 一 A design review ordinance must contain workable guidelines. Too broad a discretion permits determinations based on whim, caprice, or subjective considerations.
Southern Burlington County NAACP v. Mt. Laurel (Mt. Laurel I) 一 A municipality must presumptively make realistically possible an appropriate variety and choice of housing. It must affirmatively afford an opportunity for low and moderate income housing to the extent of its fair share of the present and prospective regional need for this housing.
Southern Burlington County NAACP v. Mt. Laurel (Mt. Laurel II) 一 A builder’s remedy generally will be granted when a developer has acted in good faith, attempted to obtain relief without litigation, and thereafter vindicated the constitutional obligation in Mount Laurel-type litigation. The proposed project must include an appropriate portion of low and moderate income housing, and it must be located and designed in accordance with sound zoning and planning concepts, including its environmental impact.
Hills Development Co. v. Bernards 一 When the governor and the legislature substantially occupied the field of enforcing the Mount Laurel doctrine, this response was entitled to deference from courts.
Government Takings
The government cannot take private property for public use without providing just compensation. A physical taking occurs when the government physically invades the property and destroys its present use. A regulatory taking frustrates investment-backed expectations, preventing an economically beneficial or productive use of the property.
Pennsylvania Coal Co. v. Mahon 一 If regulation goes too far, it will be recognized as a taking for which compensation must be paid.
Miller v. Schoene 一 When forced to make the choice, the state does not exceed its constitutional powers by deciding upon the destruction of one class of property in order to save another which, in the judgment of the legislature, is of greater value to the public.
Penn Central Transportation Co. v. New York City 一 There is no set formula for determining when justice and fairness require that economic losses caused by public action be compensated by the government. Factors to consider include the economic impact of the regulation on the property owner, the extent to which the regulation interferes with distinct investment-backed expectations, and the character of the government action.
Loretto v. Teleprompter Manhattan CATV Corp. 一 When the character of a governmental action is a permanent physical occupation of property, there is a taking to the extent of the occupation regardless of whether the action achieves an important public benefit or has only a minimal economic impact on the owner.
Keystone Bituminous Coal Ass’n v. DeBenedictis 一 Since mine operators retained the right to mine virtually all the coal in their mineral estates, a burden placed on the support estate did not constitute a taking. (The support estate is not a separate segment of property for takings law purposes.)
Hodel v. Irving 一 While the government has broad authority to adjust the rules governing the descent and devise of property, the complete abolition of both the descent and devise of a particular class of property may be a taking.
First English Evangelical Lutheran Church v. Los Angeles County 一 When the government has taken property by a land use regulation, the property owner may recover damages for the time before it is finally determined that the regulation constitutes a taking.
Nollan v. California Coastal Commission 一 Unless the permit condition serves the same governmental purpose as the development ban, the building restriction is not a valid regulation of land use.
Dolan v. City of Tigard 一 It must be determined whether an essential nexus exists between a legitimate state interest and the permit condition. If one does, it must be decided whether the degree of the exactions demanded by the permit conditions bears the required relationship to the projected impact of the proposed development. In deciding the second question, the necessary connection is “rough proportionality.”
Koontz v. St. Johns River Water Management District 一 The government’s demand for property from a land-use permit applicant must satisfy the Nollan and Dolan requirements even when the demand is for money.
Lucas v. South Carolina Coastal Council 一 When the state seeks to sustain regulation that deprives land of all economically beneficial use, it may resist compensation only if the logically antecedent inquiry into the nature of the owner’s estate shows that the proscribed use interests were not part of their title at the outset. A total taking inquiry entails an analysis of the following factors: the degree of harm to public lands and resources, or adjacent private property, posed by the claimant’s proposed activities; the social value of the claimant’s activities and their suitability to the locality in question; and the relative ease with which the alleged harm can be avoided through measures taken by the claimant and the government (or adjacent private landowners) alike.
Palazzolo v. Rhode Island 一 A Penn Central claim is not barred by the mere fact that the property owner acquired title after the effective date of the state-imposed restriction.
Tahoe-Sierra Preservation Council, Inc. v. Tahoe Regional Planning Agency 一 A permanent deprivation of all use is a taking of the parcel as a whole, but a temporary restriction causing a diminution in value is not, for the property will recover value when the prohibition is lifted.
Arkansas Game & Fish Commission v. U.S. 一 Government-induced flooding temporary in duration gains no automatic exemption from Takings Clause inspection.
Murr v. Wisconsin 一 In determining the denominator of the takings inquiry, a court must consider the treatment of the land under state and local law, the physical characteristics of the land, and the prospective value of the regulated land. The endeavor should determine whether reasonable expectations about property ownership would lead a landowner to anticipate that their holdings would be treated as one parcel or as separate tracts.
Berman v. Parker 一 It is within the power of the legislature to determine that the community should be beautiful as well as healthy, spacious as well as clean, well balanced as well as carefully patrolled. If those who govern the District of Columbia decide that the nation's capital should be beautiful as well as sanitary, there is nothing in the Fifth Amendment that stands in the way.
Hawaii Housing Authority v. Midkiff 一 Only the purpose of a taking, rather than its mechanics, must pass scrutiny under the Public Use Clause of the Fifth Amendment.
Kelo v. City of New London 一 A city’s decision to take property for the purpose of economic development satisfied the public use requirement of the Fifth Amendment.
Kaur v. New York State Urban Development Corp. 一 The exercise of eminent domain power by the New York State Urban Development Corporation to benefit a private elite education institution violated the Takings Clause.
Knick v. Township of Scott 一 Property owners may bring Fifth Amendment claims for compensation as soon as their property has been taken, regardless of any other post-taking remedies that may be available to the property owner.
Horne v. Dept. of Agriculture 一 The Fifth Amendment requires that the government pay just compensation when it takes personal property, just as when it takes real property.
Equal Protection, Due Process, and Property
The Equal Protection Clause of the Fourteenth Amendment prohibits state action to enforce discriminatory property covenants. Meanwhile, the Due Process Clause may apply to situations that involve property interests outside the scope of the Takings Clause.
Shelley v. Kraemer 一 Private agreements to exclude persons of a designated race or color from the use or occupancy of real estate for residential purposes do not violate the Fourteenth Amendment, but it is violative of the Equal Protection Clause for state courts to enforce them.
Barrows v. Jackson 一 The Fourteenth Amendment bars the enforcement of a covenant forbidding the use and occupancy of real estate by non-Caucasians by an action at law in a state court to recover damages from a co-covenantor for a breach of the covenant.
Flemming v. Nestor 一 Particularly when dealing with a withholding of a non-contractual benefit under a social welfare program, the Due Process Clause interposes a bar only if the statute manifests a patently arbitrary classification, utterly lacking rational justification.
Cleveland Board of Education v. Loudermill 一 While the legislature may elect not to confer a property interest in public employment, it may not constitutionally authorize the deprivation of such an interest, once conferred, without appropriate procedural safeguards.
The Right to Exclude and Constitutional Freedoms
A property owner typically holds a right to prevent certain people from entering their property and prevent certain activities there. However, the First Amendment and state constitutional provisions related to the freedom of speech and association may clash with this right.
Jacque v. Steenberg Homes, Inc. 一 Both the private landowner and society have much more than a nominal interest in excluding others from private land. Thus, nominal damages may support a punitive damages award in an action for intentional trespass to land.
State v. Shack 一 Rights in real property are not absolute, and private or public necessity may justify entry upon the lands of another party.
Pruneyard Shopping Center v. Robins 一 State constitutional provisions, as construed to permit people reasonably to exercise free speech and petition rights on the property of a privately owned shopping center to which the public is invited, do not violate the property rights or the free speech rights of the shopping center owner under the U.S. Constitution.
Marsh v. Alabama 一 A state cannot impose criminal penalties for distributing religious literature on the sidewalk of a company town contrary to regulations of the town management, when the town and its shopping district are accessible to the general public.
Lloyd Corp., Ltd. v. Tanner 一 A shopping center does not lose its private character merely because the public is generally invited to use it for the purpose of doing business with the tenants of the center.
State v. Schmid 一 Restrictions on private property to accommodate the rights of speech and assembly require considering the nature, purposes, and primary use of the property, as well as the extent and nature of the public’s invitation to use the property and the purpose of the expressional activity undertaken on the property in relation to the private and public use of the property.
Cedar Point Nursery v. Hassid 一 The right to exclude is a fundamental element of the property right. A regulation granting labor organizations a right to take access to an agricultural employer’s property to solicit support for unionization was a per se physical taking.
Intellectual Property
People and entities can hold property in intangible ideas rather than land. Common types of intellectual property include patents, copyrights, and trademarks, each of which is regulated by the government.
International News Service v. Associated Press 一 When a business gathers news with effort and expense for the purpose of lucrative publication, the business has a quasi-property in the results of its enterprise as against a rival in the same business, and the appropriation of those results at the expense and to the damage of one and for the profit of the other is unfair competition.
Feist Publications, Inc. v. Rural Telephone Service Co., Inc. 一 The originality requirement of copyright necessitates independent creation plus a modicum of creativity. Although a compilation of facts may possess the requisite originality because the author typically chooses which facts to include, in which order to place them, and how to arrange the data so that readers may use them effectively, copyright protection extends only to those components of the work that are original to the author.
This outline has been compiled by the Justia team for solely educational purposes and should not be treated as an independent source of legal authority or a summary of the current state of the law. Students should use this outline as a supplement rather than a substitute for course-specific outlines.
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Propulsion Principles and Engine Classification
1.5.2 Propulsive Efficiency
Using Eqs. (1.6), (1.9) the propulsive efficiency for a rocket becomes
(1.35)
The variation of propulsive efficiency with velocity ratio V0/Ve is shown for the rocket in Fig. 1.14.
For comparison, the propulsive efficiency of the propeller and turbojet is also shown on the same figure. The propulsive efficiency of a rocket is seen to be greater than that of a turbojet or propeller at the same value of V0/Ve. However, the practical operating range of V0/Ve for the rocket is lower than that of the turbojet which in turn is lower than that of the propeller. As a practical matter then, the propulsive efficiency of the rocket is lower than that of the turbojet which is lower than that of the propeller. The high exhaust speed of the rocket engine results in more power being wasted in accelerating the propellants to produce thrust.
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Book2017, Theory of Aerospace Propulsion (Second Edition)Pasquale M. Sforza
Chapter
Propulsion Principles and Engine Classification
2017, Theory of Aerospace Propulsion (Second Edition)Pasquale M. Sforza
1.3.4 Propulsive Efficiency
Remember that the flight speed V0 is constant and therefore the drag on the vehicle is equal to the thrust produced, D = F. The power required to keep the vehicle moving at constant speed V0 is DV0 = FV0. The power supplied to the fluid P = FVavg is not necessarily converted completely into useful thrust power FV0. Using Eq. (1.5) we define the propulsive efficiency ηp as the ratio of useful thrust power to total power imparted to the airstream:
(1.9)
This equation shows that the propulsive efficiency increases as the flight speed V0 approaches the jet exit velocity Ve, as shown in Fig. 1.4.
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Review article
Engineering perspective on bird flight: Scaling, geometry, kinematics and aerodynamics
2023, Progress in Aerospace SciencesTianshu Liu, ... Guowei He
2.4 Propulsive efficiency
Based on the scaling laws , and , the propulsive efficiency in bird cruise flight is estimated, i.e.,
(2.7)
For the air density , an estimate is . Since the mean total drag coefficient of birds is , the propulsive efficiency of bird flight is , which is reasonably high similar to that of propellers. The effect of the air density on the propulsive efficiency can be clearly seen in Eq. (2.7). An unsteady lifting-line theory gave the propulsive efficiency of 69–76% and a theoretical model based on prescribed wake vortex structures constrained by minimum induced power requirements gave the propulsive efficiency as high as 86% [92,93].
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Journal2023, Progress in Aerospace SciencesTianshu Liu, ... Guowei He
Chapter
Performance, Performance Testing, and Performance Optimization
2015, Gas Turbines (Second Edition)Claire Soares
Propulsive Efficiency
Performance of the jet engine is not only concerned with the thrust produced, but also with the efficient conversion of the heat energy of the fuel into kinetic energy, as represented by the jet velocity, and the best use of this velocity to propel the aircraft forward, i.e., the efficiency of the propulsive system.
The efficiency of conversion of fuel energy to kinetic energy is termed thermal or internal efficiency and, like all heat engines, is controlled by the cycle pressure ratio and combustion temperature. Unfortunately, this temperature is limited by the thermal and mechanical stresses that can be tolerated by the turbine. The development of new materials and techniques to minimize these limitations is continually being pursued.
The efficiency of conversion of kinetic energy to propulsive work is termed the propulsive or external efficiency and this is affected by the amount of kinetic energy wasted by the propelling mechanism. Waste energy dissipated in the jet wake, which represents a loss, can be expressed as [W(vJ – V)2]/2g where (vJ – V) is the waste velocity. It is therefore apparent that at the aircraft lower speed range the pure jet stream wastes considerably more energy than a propeller system and consequently is less efficient over this range. However, this factor changes as aircraft speed increases, because although the jet stream continues to issue at a high velocity from the engine its velocity relative to the surrounding atmosphere is reduced and, in consequence, the waste energy loss is reduced.
Briefly, propulsive efficiency may be expressed as:
or simply
Work done is the net thrust multiplied by the aircraft speed. Therefore, progressing from the net thrust equation given earlier, the following equation is arrived at:
In the instance of an engine operating with a non-choked nozzle, the equation becomes:
Simplified to: 2V/(V + vj)
This latter equation can also be used for the choked nozzle condition by using VJ to represent the jet velocity when fully expanded to atmospheric pressure, thereby dispensing with the nozzle pressure term (P – P0)A.
Assuming an aircraft speed (V) of 375 m.p.h. and a jet velocity (vJ) of 1230 m.p.h., the efficiency of a turbo-jet is:
On the other hand, at an aircraft speed of 600 m.p.h. the efficiency is:
Propeller efficiency at these values of V is approximately 82 and 55%, respectively, and from reference to Figure 10–9 it can be seen that for aircraft designed to operate at sea level speeds below approximately 400 m.p.h. it is more effective to absorb the power developed in the jet engine by gearing it to a propeller instead of using it directly in the form of a pure jet stream. The disadvantage of the propeller at the higher aircraft speeds is its rapid fall off in efficiency, due to shock waves created around the propeller as the blade tip speed approaches Mach 1.0. Advanced propeller technology, however, has produced a multi-bladed, swept back design capable of turning with tip speeds in excess of Mach 1.0 without loss of propeller efficiency. By using this design of propeller in a contra-rotating configuration, thereby reducing swirl losses, a “prop-fan” engine, with very good propulsive efficiency capable of operating efficiently at aircraft speeds in excess of 500 m.p.h. at sea level, can be produced.
To obtain good propulsive efficiencies without the use of a complex propeller system, the bypass principle is used in various forms. With this principle, some part of the total output is provided by a jet stream other than that which passes through the engine cycle and this is energized by a fan or a varying number of L.P. compressor stages. This bypass air is used to lower the mean jet temperature and velocity either by exhausting through a separate propelling nozzle, or by mixing with the turbine stream to exhaust through a common nozzle.
The propulsive efficiency equation for a high bypass ratio engine exhausting through separate nozzles is given below, where W1 and relate to the bypass function and W2 and to the engine main function.
Propulsive efficiency =
By calculation, substituting the following values, which will be typical of a high bypass ratio engine of triple-spool configuration, it will be observed that a propulsive efficiency of approximately 85% results.
: V = 583 m.p.h.
: W1 = 492 lb. per sec.
: W2 = 100 lb. per sec.
: = 781 m.p.h.
: = 812 m.p.h.
Propulsive efficiency can be further improved by using the rear mounted contra-rotating fan configuration of the bypass principle. This gives very high bypass ratios in the order of 15:1, and reduced “drag” results due to the engine core being “washed” by the low velocity aircraft slipstream and not the relatively high velocity fan efflux.
The improved propulsive efficiency of the bypass system bridges the efficiency gap between the turbo-propeller engine and the pure turbo-jet engine. A graph illustrating the various propulsive efficiencies with aircraft speed is shown in Figures 10–9.
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Chapter
Idealized Flow Machines
2012, Theory of Aerospace PropulsionPasquale M. Sforza
1.2.4 Propulsive efficiency
The total power expended is not necessarily converted completely into thrust power FV0, the rate at which force applied to the fluid does work. Remember that flight speed V0 is constant and therefore the drag on the vehicle is equal to the thrust produced, D = F. Then the rate at which work must be done to keep the vehicle at constant speed V0 is DV0 = FV0. However, it has been shown that the power expended is P = FVavg so that propulsive efficiency ηp may be defined as the ratio of useful thrust power to total power delivered to the airstream:
(1.6)
This equation shows that at a given flight speed the efficiency drops off with increasing exhaust velocity Ve, as shown in Figure 1.3.
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Book2012, Theory of Aerospace PropulsionPasquale M. Sforza
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Idealized Flow Machines
2012, Theory of Aerospace PropulsionPasquale M. Sforza
1.3.3 Overall efficiency
The overall efficiency in the case of thrust produced by means of heat addition may be defined as the ratio of the required thrust power to the rate of heat addition
(1.10)
However, as already shown in Equation (1.6), the propulsive efficiency is
We may then define the thermal efficiency as
(1.11)
Then the overall efficiency is the product of propulsive and thermal efficiencies
(1.12)
Thermal efficiency accounts for the fact that not all the heat added is converted to useable heat power, as some is rejected as increased internal energy in the exhaust gases. Thus, thermal efficiency illustrates the extent to which the flow machine puts the heat added to good use in increasing the kinetic energy of the exhaust stream, that is, increasing Ve, whereas propulsive efficiency illustrates the extent to which that increased kinetic energy provides thrust power to maintain flight at the speed V0. In a jet engine, these two efficiencies generally drive in different directions, with the higher exhaust velocities sustainable at high thermal efficiency leading to lower propulsive efficiencies at a given flight speed.
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Book2012, Theory of Aerospace PropulsionPasquale M. Sforza
Review article
Aerodynamics of cross-flow fans and their application to aircraft propulsion and flow control
2009, Progress in Aerospace SciencesThong Q. Dang, Peter R. Bushnell
It is seen from the above relations that while the maximum propulsive efficiency of the conventional system approaches 1 when Uj/U∞→1, the corresponding value of the wake-filling propulsion system can reach 2 when Uw/U∞→0. The fact that ηp can be greater than 1 is well known to the naval industry, and the term propulsive coefficient is sometimes used for ηp.
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Journal2009, Progress in Aerospace SciencesThong Q. Dang, Peter R. Bushnell
Review article
Special Issue on Green Aviation
2023, Progress in Aerospace SciencesS. Delbecq, ... F. Simatos
4.1.1 Engines
Improving engine efficiency, which can be measured by the decrease in Specific Fuel Consumption (SFC), relies on gains in both thermal efficiency and propulsive efficiency, the product of which defines thermopropulsive efficiency. A gain in thermal efficiency depends on the efficiency improvement of the thermodynamic cycle via the components of the turbomachinery (compressor, combustion chamber and turbine). A gain in propulsive efficiency consists in decreasing the residual kinetic energy in the exhaust jet. The highest propulsive efficiency is reached when the difference between gas jet speed and flight speed is as small as possible. The objective is therefore to generate a high air flow rate with the minimum speed difference with respect to the aircraft speed, and this is the role the fan plays in turbofan engines. The efficiency of current engines, without significant disruptive technology, can still be improved, primarily by increasing the bypass ratio and the temperature at the combustion chamber outlet.
However, technological limits are being reached. For example, the materials used to make turbine blades are already highly complex (titanium alloys) but do not alone ensure blade resistance to extreme temperatures at the combustion chamber outlet (of around 1800 K to 2000 K), which exceed the metal alloy melting temperature . It is therefore necessary to cool the first stages of the high-pressure turbine at the combustion chamber outlet with “cold” air from the compressor, a bleeding which reduces the efficiency of the engine . It is thus impossible to significantly increase the temperature at the combustion chamber outlet yet, without affecting the performance of the thermodynamic cycle. In the same way, the increase in bypass ratio faces problems such as increasing the size and weight of the engine with respect to its integration on the aircraft.
Very high bypass ratio (VHBR) and ultra high bypass ratio (UHBR) engine designs are based on the increase in bypass ratio up to around 20 (to be compared to a current bypass ratio of around 10 for the latest generation LEAP or Trent engines). An increase in bypass ratio not only increases propulsive efficiency but also reduces the noise emitted by the engine, by decreasing the exhaust gas jet speed. The expected gains are potentially around 5 to 10% on fuel consumption, while reducing the noise emitted by around 7 dB, compared to the current generation .
The increase in the bypass ratio enables turboprop engines to be particularly efficient. The (ducted) fan is replaced by an (open) propeller, which increases blade span, and thus ensures a very high bypass ratio. Therefore these engines can be used to reduce fuel consumption by around 40%. However, the propeller rotation speed, combined with the aircraft’s forward speed, may induce the appearance of intense compressible effects (especially shock waves related to the presence of local supersonic regions), which significantly deteriorate performances as soon as the flight speed or the size of the aircraft are increased. These engines are therefore limited to flight Mach numbers of around 0.65, lower than those reached with a turbofan (of around 0.85), but also mainly used to power 100-seats short-haul aircraft due to thrust limitations. A solution to offset this disadvantage is to ensure thrust using two contra-rotating propellers, which enables them to reduce their rotation speed for a given level of thrust. These open rotor engine architectures are generally called Contra-Rotating Open Rotors (CROR) , see Fig. 10. This type of engines have a higher flight Mach number limit of around 0.75 and enables substantial fuel savings of around 10% to 20% compared to the best modern turbofans .
Although a lot more exploratory, improving the turbofan thermodynamic cycle remains possible, especially by replacing the isobaric combustion (Brayton cycle) by an isochoric combustion (Humphrey cycle) . These engines, called pulsating detonation engines (PDE) or rotating detonation engines (RDE) exhibit better performances than those of isobaric engines, whether in terms of thermopropulsive efficiency or specific power. However, their operation in unsteady regime requires more research efforts, in order to adapt the components downstream of the combustion chamber (the turbine in particular) [73,74]. Despite degraded functioning of a current turbine downstream of this type of combustion chamber, recent publications show thermal efficiency gains over the entire cycle of around 5% .
To conclude, future engine improvements mainly rely on larger dimensions (UHBR, CROR), which will further increase the coupling between the aircraft’s aerodynamics, weight, and engine. It is therefore necessary, now more than ever, to more effectively integrate the propulsion system within the aircraft airframe which could represents another potential source of fuel savings. Bury et al. show the complexity of the flow in the case of a configuration where the engine is integrated under the wing, especially the presence of localized separations responsible for drag increase. Optimized aerodynamic design of the entire wing/engine integrated system is a crucial aspect for reducing fuel consumption. Integrating propulsion systems directly on the aircraft airframe and no longer under the wing is one of the disruptive technologies considered in the next ten years (see Section 4.2).
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Journal2023, Progress in Aerospace SciencesS. Delbecq, ... F. Simatos
Chapter
Propulsion Principles and Engine Classification
2017, Theory of Aerospace Propulsion (Second Edition)Pasquale M. Sforza
1.4.3 Overall Efficiency
The overall efficiency in the case of thrust produced by means of heat addition may be defined as the ratio of the thrust power required to maintain the motion to the rate of heat addition to the fluid and may be written as
(1.15)
We may then define the thermal efficiency as
(1.16)
Then, identifying V0/Vavg as the propulsive efficiency ηp defined in Eq. (1.8), the overall efficiency is expressed as the product of the propulsive and thermal efficiencies
(1.17)
The thermal efficiency accounts for the fact that not all the heat added is converted to useable heat power, since some is rejected as increased internal energy in the exhaust gases. The propulsive efficiency accounts for the fact that the increased kinetic energy of the exhaust jet represents a loss in mechanical power to maintain flight at the speed V0. In a jet engine these two efficiencies generally drive in different directions, with the higher exhaust velocities sustainable at high thermal efficiency leading to lower propulsive efficiencies at a given flight speed. Whereas a propeller-driven aircraft will cruise at V0/Ve ~ 1, a subsonic turbojet-powered aircraft is more likely to cruise at V0/Ve ~ 0.6. From Fig. 1.4 we see that the propeller-driven aircraft has a propulsive efficiency ηp ~ 1, and the value for the turbojet-powered aircraft is ηp = 0.7.
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Review article
A comprehensive review on countermeasures for CO emissions from ships
2020, Renewable and Sustainable Energy ReviewsHui Xing, ... Hua Chen
2.2 Propulsion efficiency
The propeller transforms the power delivered from the main engine via the shaft into thrust force to propel the ship. Total propulsive efficiency could be expressed as the product of hull efficiency, propeller open water efficiency, relative rotative efficiency and shaft efficiency . Typically, only two thirds of the delivered power from main engine is converted into thrust power to propel the hull. The main opportunities for propulsion efficiency improvements are in hull efficiency (typically 0.95–1.3), propeller open water efficiency (typically 0.55–0.70) and relative rotative efficiency (typically 0.98–1.07) , which correspondingly depend on high‐efficiency propellers, improving the wake distribution, and recovering rotational energy.
Researchers have shown enthusiasm for the design and development of high-efficiency propellers. Compared to conventional propellers, CLT propellers and Kappel propellers [45,46] draw most attention. The propeller operates in an inhomogeneous wake behind the ship and consequently loses some efficiency due to the disturbed flow field. A more homogeneous wake translates into better propeller efficiency. Wake equalising devices, such as ducted propellers, wake equalising ducts and nozzles aim to improve the wake distribution and reduce flow losses around the working propeller . Another group of power saving devices are aimed at recovering rotational energy from the water downstream of the propeller and converting it into thrust. A large number of devices have been proposed to recover some of the rotational energy of the water. These can be categorised into pre-swirl (upstream of the propeller) and post-swirl (downstream of the propeller) devices . Devices like PBCF , have been added to the propeller hubs to suppress the hub vortex. In recent years, CRP systems are widely regarded as an important option for recovering of rotational energy in the flow [49,50].
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Journal2020, Renewable and Sustainable Energy ReviewsHui Xing, ... Hua Chen
Related terms:
Energy Engineering
Vortex
Duct
Pressure Ratio
Airfoil
Bypass Ratio
Boundary Layer
Propulsion System
Reynolds' Number
Specific Fuel Consumption
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Rufei Huang
Rufei Huang
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Huan Xia
Huan Xia
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Ziyi Li
Ziyi Li
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Jingxian Deng
Jingxian Deng
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Tao Ye
Tao Ye
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Yadong Huang
Yadong Huang
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Yan Yang
Yan Yang
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1
Department of Cell Biology, Jinan University, Guangzhou 510632, China
2
Guangdong Province Key Laboratory of Bioengineering Medicine, Guangzhou 510632, China
3
National Engineering Research Center of Genetic Medicine, Guangzhou 510632, China
Authors to whom correspondence should be addressed.
†
These authors contributed equally to this work.
Int. J. Mol. Sci. 2024, 25(11), 5805;
Submission received: 7 April 2024
/
Revised: 1 May 2024
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Accepted: 22 May 2024
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Published: 26 May 2024
(This article belongs to the Special Issue Reproductive Endocrinology Research)
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Abstract
The testes serve as the primary source of androgens and the site of spermatogenesis, with their development and function governed by hormonal actions via endocrine and paracrine pathways. Male fertility hinges on the availability of testosterone, a cornerstone of spermatogenesis, while follicle-stimulating hormone (FSH) signaling is indispensable for the proliferation, differentiation, and proper functioning of Sertoli and germ cells. This review covers the research on how androgens, FSH, and other hormones support processes crucial for male fertility in the testis and reproductive tract. These hormones are regulated by the hypothalamic–pituitary–gonad (HPG) axis, which is either quiescent or activated at different stages of the life course, and the regulation of the axis is crucial for the development and normal function of the male reproductive system. Hormonal imbalances, whether due to genetic predispositions or environmental influences, leading to hypogonadism or hypergonadism, can precipitate reproductive disorders. Investigating the regulatory network and molecular mechanisms involved in testicular development and spermatogenesis is instrumental in developing new therapeutic methods, drugs, and male hormonal contraceptives.
Keywords:
hormones; spermatogenesis; testis; HPG axis; male reproduction
1. Introduction
The decline in male sperm count has been a common phenomenon worldwide. From 1973 to 2018, the mean sperm concentration and total sperm count among unselected men from all continents decreased by 51.6% and 62.3%, respectively [1,2]. This large and sustained decline is now considered a major public health issue, and the relationship between sperm count and infertility has received widespread attention. Infertility or reduced fertility can be attributed to endocrine diseases, testicular dysfunction, and poor lifestyle factors such as unhealthy diet and alcohol consumption. The endocrine system serves as the primary regulator of reproductive function, and a delicate hormonal balance and crosstalk are critical for testicular development and spermatogenesis.
The testes, being vital male reproductive organs, are responsible for the production of sperm and male sex hormones. Hormones play a vital role in regulating testicular development and function in males from fetal life through adulthood. Proper hormone regulation is essential for maintaining male reproductive health, which includes sexual maturation, germ cell production, and steroidogenesis. Imbalances in hormonal regulation associated with testicular disorders can result in a variety of health problems, such as infertility, sexual dysfunction, and testicular cancer. Therefore, having a thorough understanding of the hormonal regulation of testicular development and function is crucial for the accurate diagnosis and treatment of these disorders. This review aims to provide an overview of the hormones involved in the regulation of male reproduction, as well as recent advancements in related clinical and applied research.
2. Testis Physiology
The testes comprise multiple compartments, including the seminiferous tubules, interstitial cells (Leydig cells), and supporting cells (Sertoli cells). Testis physiology encompasses the intricate processes involved in sperm production, testosterone secretion, and hormonal regulation. It involves the coordinated functioning of various cells and hormones to maintain male reproductive health and fertility (Figure 1). Aside from sperm production and testosterone production, the testes also play a role in the regulation of sexual function. The testes produce other hormones, such as inhibin, which helps regulate follicle-stimulating hormone (FSH) levels, and estrogen, albeit in small amounts. Understanding testis physiology is crucial for diagnosing and treating male reproductive disorders and infertility. Any disruption in hormone production or spermatogenesis can lead to fertility issues, hormonal imbalances, and sexual dysfunctions.
Figure 1.
Schematic representation of the structure of the seminiferous tubules. There are somatic Sertoli cells and germ cells in the seminiferous tubules and Leydig cells, blood vessels, and immune cells in the interstitium. SSCs and Sertoli cells are attached to the basement membrane. In the process of spermatogenesis, germ cells (spermatogonia, pre-leptotene spermatocytes, pachytene spermatocytes, round spermatid, and elongating spermatid) move in the seminiferous epithelium and undergo meiosis until they are released to the seminiferous tubular fluid. There are four types of cell junctions in the seminiferous epithelium. The junctions between Sertoli cells form the BTB, which provides an immunologically privileged environment for spermatogenesis.
2.1. Testicular Structure
2.1.1. Seminiferous Tubule
Seminiferous tubules are home to germ cells, which are crucial for spermatogenesis, as well as Sertoli cells and peritubular myoid cells (PMCs). Sertoli cells offer support and nutrition to developing sperm, and PMCs surround the tubule’s external wall. The most optimal energy source for developing germ cells, the lactate molecule, is provided by Sertoli cells, which also furnish necessary growth factors and chemokines. The microenvironment created by neighboring Sertoli cells is vital for maintaining germ cell growth and initiating differentiation [3,4]. More specifically, Sertoli cells are indispensable for managing androgen production within the testis and encouraging the secretion of a range of bioactive peptides . These peptides facilitate communication between Sertoli cells and germ cells to support spermatogenesis and fertility [4,5]. PMCs play a role in regulating sperm and luminal fluid transport and secrete growth factors and an extracellular matrix to uphold the niche of spermatogonial stem cells (SSCs). For example, the glial cell-derived neurotrophic factor (GDNF) produced by PMCs is instrumental in the development of undifferentiated spermatogonia cells in vivo .
2.1.2. Interstitial Tissue
Interstitial tissue contains Leydig cells, which are responsible for testosterone production, as well as blood vessels, nerves, and connective tissue. The mammalian Leydig cell is the primary site for testosterone secretion, and this hormone is crucial for spermatogenesis. Additionally, two types of macrophages, peritubular macrophages, and interstitial macrophages, also contribute to testicular function . These macrophages differ in their location and function, but overall, they play a role in inducing spermatogonial proliferation and differentiation , providing a suitable niche for Leydig cells , and maintaining the immune privileges of the testis with the blood–testis barrier (BTB) .
2.1.3. Blood–Testis Barrier (BTB)
The BTB is a unique structure consisting the gap junctions, tight junctions (TJs), desmosomes, and ectoplasmic specializations between Sertoli cells (Figure 1). The tight junction proteins composing the BTB include the claudins (CLDNs), junctional adhesion molecules (JAMs), etc., and they bind to actin through the zonula occluden-1 (ZO-1), ZO-2, and ZO-3 [12,13]. The BTB is one of the strictest tissue barriers found in mammals and divides the seminiferous epithelium into the basal part and an adluminal part. This intricate barrier segregates the seminiferous tubules from the bloodstream, effectively restricting the diffusion of water, ions, electrolytes, paracrine factors, hormones, and other exogenous biomolecules through both the paracellular and transcellular pathways. Consequently, it creates a specialized microenvironment that is immune-privileged and conducive to germ cell development . The BTB is crucial for successful spermatogenesis as it helps safeguard developing sperm from autoimmune reactions. However, the presence of this barrier also poses a challenge for the development of drugs that target the testicles.
2.2. Testicular Function
2.2.1. Spermatogenesis
Spermatogenesis is a complex process that involves the development of male germ cells, known as spermatogonia, into fully mature spermatozoa, or sperm. This process occurs within the seminiferous tubules and consists of three main stages: the mitosis of spermatogonia, followed by the meiosis of spermatocytes, and finally the morphological transformation of spermatids into spermatozoa. During spermatogenesis, the germ cells undergo movement within the seminiferous epithelium, leading to the restructuring of the junctions between Sertoli cells and germ cells, as well as between Sertoli cells themselves at the BTB . After spermatogenesis is completed, mature sperm will be released from the basal part of the seminiferous epithelium into the seminiferous tubule lumen. Additionally, the premature shedding of germ cells from Sertoli cells indicated a failure of spermatogenesis . Sertoli cells play a critical role in spermatogenesis, as they provide a specialized environment for sperm cells. These cells contribute to the maintenance of the BTB and facilitate the proper development and release of mature spermatozoa. Understanding the intricate mechanisms involved in spermatogenesis is essential for elucidating the causes of male infertility and developing potential therapeutic interventions.
2.2.2. Testosterone Production
Testosterone is the primary androgen in the male body, and its levels directly affect spermatogenesis . Testosterone is primarily synthesized and secreted by Leydig cells in the testes through a series of enzymatic reactions. The regulation of testosterone levels is governed by the negative feedback mechanisms of the HPG axis, as well as local factors within the testes. The proper development of Leydig cells during puberty is essential for initiating spermatogenesis and promoting secondary sexual characteristics in males . Studies have shown that the transplantation of stem Leydig cells (SLCs) in Leydig cell-disrupted or aging models can help restore testosterone production, thereby accelerating meiosis and germ cell recovery [18,19].
In summary, testicular physiology comprises an intricate interrelationship involving various cell types, hormones, and signaling pathways, all of which play a crucial role in regulating spermatogenesis and hormone synthesis. These processes are fundamental for maintaining male fertility and overall reproductive health.
3. Hormone Regulation in Testicular Development and Function
The hypothalamic–pituitary–gonadal (HPG) axis is a vital system involved in hormonal regulation. It consists of the hypothalamus, which releases gonadotropin-releasing hormone (GnRH); the pituitary gland, which secretes LH and FSH; and the gonads (testes in males and ovaries in females). In males, GnRH from the hypothalamus stimulates the pituitary gland to release LH and FSH. LH acts on Leydig cells in the testes, stimulating the production of testosterone and INSL3. Testosterone plays a crucial role in the development of male secondary sexual characteristics, libido, and spermatogenesis. Insulin-like peptide 3 (INSL3) is involved in the descent of the testes during fetal development.
FSH acts on Sertoli cells in the testes, supporting spermatogenesis. Sertoli cells secrete inhibin, which inhibits the release of FSH from the pituitary gland, helping to regulate the levels of FSH (Figure 2). Sertoli cells also secrete anti-Müllerian hormone (AMH), which plays a role in the regression of female reproductive structures during male development. Imbalances in these hormones may lead to decreased fertility, infertility, or other reproductive/non-reproductive system diseases (Table 1).
Figure 2.
The HPG axis. The hypothalamus regulates the biosynthesis and secretion of pituitary hormones LH and FSH through GnRH. LH induces Leydig cells to secrete testosterone, which reduces GnRH and LH production through negative feedback. FSH acts on Sertoli cells to induce the secretion of inhibin, which in turn inhibits the production of FSH. Sertoli cells, under the combined action of testosterone and FSH, stimulate the proliferation and maturation of germ cells. Testosterone is also essential for promoting muscle and bone development. Furthermore, testosterone is metabolized by aromatase (CYP19A1) into 17β-estradiol (E2) which exerts inhibition on the hypothalamus. Normal spermatogenesis depends on the coordinated action of all these hormones.
Table 1.
Hormone secretory regions and target cells.
3.1. Testosterone
Testicular androgens, specifically testosterone and its metabolite, play a crucial role in male fertility and are essential for normal masculinization, testis function, and other androgenic targets such as muscle, fat, and bone. As with all steroid hormones, the initial precursor required for the production of androgen is cholesterol. The established pathway involves a series of enzymatic reactions that convert the steroid precursors into testosterone and dihydrotestosterone (DHT). A critical enzyme in this process is 17β-hydroxysteroid dehydrogenase type 3 (HSD17B3).
Testosterone exerts its effects by binding to the androgen receptor (AR). In the classical signaling pathway, AR in the cytoplasm undergoes conformational changes to androgen. Subsequently, it dissociates from chaperone proteins like heat shock proteins (HSPs), and the receptor–ligand complex enters the nucleus. Within the nucleus, the complex recruits coactivators or corepressors to bind with androgen response elements (AREs) in the promoter regions of targeted genes, thereby activating or inhibiting gene transcription (Figure 3). On the other hand, the non-classical testosterone signaling pathway becomes active when intratesticular testosterone levels are low, typically below 250 nM, and testosterone rapidly binds to the membrane AR and triggers a cascade reaction . This pathway predominantly functions by promoting the phosphorylation and activation of ERK1/2 and CREB . During this process, AR interacts with and activates Src kinase at the plasma membrane . Src activation leads to phosphorylation and activation of EGFR, which, in turn, activates MAPK cascades (Ras-Raf-MEK-ERK), resulting in the activation of transcription factors such as cAMP-response element-binding (CREB) proteins and a series of downstream effects [28,29] (Figure 3). Androgens regulate the expression of a series of genes through these two pathways, directly affecting the function of Sertoli cells or acting on germ cells via paracrine mechanisms to regulate sperm production.
Figure 3.
The classical and non-classical pathways of androgen between the two neighboring Sertoli cells allow the germ cells to continuously mature and move toward the lumen. The classic androgen pathway is shown in the Sertoli cells on the left. Androgen diffuses into the cytoplasm through the plasma membrane and interacts with HSP-binding AR, then the AR dissociates from HSP and localizes to the nucleus. In the nucleus, AR homodimers recruit co-regulators and bind to the ARE of the target gene to regulate their transcription. The classic pathways guarantee Sertoli cell maturation, BTB integrity, SSC differentiation, and spermatocyte meiosis. The non-classical androgen pathway is depicted within the Sertoli cells on the right. Androgen binds to AR on the cell membrane and subsequently interacts with Src. Src then activates EGFR, initiating a series of cascade reactions. Ultimately, the phosphorylated p90Rsk translocates to the nucleus and activates CREB to bind to the CRE of the target genes, thereby regulating their transcription. This process enables androgens to modulate the BTB integrity of Sertoli cells, the meiosis of spermatocytes, Sertoli cell–spermatid adhesion, spermatid development, and sperm release.
For Sertoli cells, AR signaling plays a role in regulating Sertoli cells to the cessation of proliferation and the initiation of differentiation. This is mainly through regulating Smad2/3 signaling , down-regulating the expression of AMH [31,32], and promoting the formation of cytoskeleton [33,34]. There is controversy over the effect of testosterone on Sertoli cell proliferation, and this effect appears to be species-specific and stage-specific [35,36,37]. Moreover, inhibiting the action of testicular hormones or mutations in the androgen receptor can lead to the increased permeability of the BTB [38,39,40]. Testosterone regulates the expression of TJ proteins (including the membrane-associated guanylate kinase (MAGUK) family, Claudins, and Occludins) [41,42,43,44], tissue-type plasminogen activator (tPA) (which is involved in BTB degradation) [45,46], basal ectoplasmic specialization (ES) proteins , and gap junction proteins (including Connexin 43) directly or indirectly to maintain the integrity of the BTB and support its renewal .
Also, the process of differentiation from SSCs to the eventual release of sperm is guaranteed by AR signaling. The effect of testosterone on male reproduction is also reflected in germ cells. After the differentiation of gonocytes into SSCs, they are influenced by the signaling network that triggers self-renewal and differentiation. Testosterone regulates the self-renewal and differentiation of SSC by modulating the expression of multiple genes, including Wnt5a , promyelocytic leukemia zinc finger (Plzf) , insulin-like growth factor 3 (Igf3) , and reproductive homeobox 5 (Rhox5) . What is more, testosterone plays a key role in spermatogenesis by participating in meiosis. The absence of testosterone stimulation leads to the failure of the transformation of round sperm cells in stages VII and VIII of the rat spermatogenic cycle during spermatogenesis [53,54]. On the one hand, the lack of AR signals leads to the dysfunction of the chromosome, mainly manifested in the tissues with double-strand breaks (DSBs) repair and chromosome synapsis . After androgen deprivation or the knockdown of AR in Sertoli cells, the expression and function of proteins essential for DSB repair, synaptonemal complex formation, spindle dynamics, and dynactin complex assembly are compromised, ultimately disrupting the process of meiosis [55,56]. On the other hand, testosterone prevents the apoptosis of germ cells during mitosis. Impaired AR signal transduction in Sertoli cells is caused by oxidative stress due to chromosomal dysfunction or breakage, resulting in cell apoptosis . Furthermore, reduced testosterone levels elevate ubiquitin carboxyl-terminal hydrolase L1 (UCHL1) expression in spermatogonia, thereby promoting the ubiquitination of the apoptosis factor p53 . Testosterone also promotes the phagocytic clearance of apoptotic germ cells by Sertoli cells, possibly by increasing the expression of miR-471-5p and Elmo1 [59,60,61]. The apoptosis of numerous germ cells may trigger the inner meiotic checkpoint, ultimately resulting in the termination of meiosis in the prophase I .
Additionally, the role of androgen signaling also includes maintaining the Sertoli cell–spermatid adhesion to prevent spermatid shedding from the epithelium and regulating the release of sperm. Low testosterone levels in rat testicles may lead to the loss of stage VIII and late spermatids [62,63]. Testosterone promotes the attachment of spermatids to Sertoli cells through the non-classical pathway . The activation of Src and FAK is essential to support sperm cell adhesion and sperm release, and both can be phosphorylated during non-classical androgen signaling activation [54,63,64].
Impaired fetal androgen action can lead to disorders of sexual differentiation , including the common human congenital disorder testicular dysgenesis syndrome (TDS). TDS is associated with various conditions such as cryptorchidism and hypospadias in male newborns [66,67], impaired spermatogenesis, and testicular germ cell cancer (TGCC) in young adult males . Interestingly, the role of androgens in establishing normal reproductive tract development and the masculinization of anogenital distance (AGD) is limited to a specific developmental window, known as the “masculinization programming window” (MPW). In rats, this window is between E15.5 and E18.5, while in humans, it is postulated to be between 8 and 14 weeks of gestation .
3.2. FSH
In the past century, it has been acknowledged that FSH plays a vital role in maintaining the normal production of germ cells in males . FSH is a glycoprotein composed of an α-subunit and a β-subunit. The α subunit is shared by TSH, LH, and HCG, while the FSHβ-subunit is unique to FSH . The FSH receptor (FSHR), responsible for binding FSH, is exclusively expressed on the cell membrane of Sertoli cells. FSH-induced signal transduction is mediated by FSHR, and its function reliant on interactions with numerous intracellular effectors.
The action of FSH mainly relies on the cAMP/PKA signaling pathway. FSH promotes Sertoli cell proliferation via the PI3K/Akt pathway and enhances the expression of cell-derived Myc (c-Myc) and type D1 cyclin (cyclin D1) through the cAMP-dependent pathway during fetal development and early postpartum stages [72,73,74]. The suppression of FSH in neonates reduces the number of final Sertoli cells by approximately 40% , which in turn affects the quantity of sperm. Furthermore, FSH targets functional factors and transcription factors through the cAMP/PKA pathway to affect Sertoli cell differentiation and apoptosis [76,77]. For example, transcription factors including Krüppel-like factor 4 (KLF4) , nuclear factor (NF)-κB , and activator protein-1 (AP-1) are involved in the regulation of Sertoli cell differentiation by FSH.
The cAMP/PKA signaling pathway plays a pivotal role in the regulation of FSH on the maintenance of the spermatogonia pool and the differentiation and apoptosis of spermatogonia. FSH stimulates the Sertoli cells to secrete GDNF and fibroblast growth factor 2 (FGF2), which are crucial for the self-renewal of spermatogonia stem cells (SSCs) and the proliferation of undifferentiated spermatogonia, as indicated by studies [81,82,83,84,85]. Furthermore, signaling enhances SSC differentiation through the activation of key factors such as stem cell factor (SCF), steel factor (SLF), bone morphogenetic protein-4 (BMP4), and insulin-like growth factor 3 (IGF3) [51,86,87]. This intricate hormonal interplay ensures the delicate balance between spermatogonia maintenance and differentiation, which is essential for testicular development and function.
FSH is also essential for meiosis. It controls DNA synthesis and meiotic chromosome dynamics through the regulation of activin A, inhibin B, IL-6, and nociception, thus promoting spermatocytes to enter meiosis [88,89,90]. In the early stage of meiosis, FSH protects spermatocytes from apoptosis by inducing the expression of galectin-3 in Sertoli cells and inhibiting the activation of transcription factor AP-1 [91,92]. The knockout of FSHR and FSH β in mice resulted in a decrease in the number of Sertoli cells, spermatogonia, and spermatocytes [93,94].
FSH signaling is intricately linked to testicular development and function, with pathways such as the ERK/MAPK, calcium, and phospholipase A2 pathways identified in FSH-mediated regulation [72,76,95]. Female transgenic mice lacking FSH or its receptor (FSHR) exhibit infertility, while their male counterparts, despite remaining fertile, display reduced testicular size and germ cell count [96,97,98]. Similarly, men with congenital FSH deficiency often face infertility, typically attributed to FSHβ gene mutations [99,100].
In a clinical setting, patients who have undergone hypophysectomy for pituitary tumors may still exhibit autonomous spermatogenesis due to the ligand-independent, constitutive activation of FSHR . This phenomenon underscores the potential for FSH signaling to sustain spermatogenesis independently of pituitary gonadotropins. Oduwole et al.’s research on transgenic mice with a constitutively active FSHR mutant form further supports this notion, as it demonstrated the partial restoration of fertility and the resumption of sperm production despite the absence of LH receptors . These findings suggest that robust FSHR activity can facilitate spermatogenesis without the need for testosterone, challenging the notion that testosterone is an absolute prerequisite for this process . This highlights the pivotal role of FSHR activation in the maintenance of sperm production, offering new insights into the hormonal regulation of testicular function.
3.3. Inhibin B
Inhibin B, along with inhibin A, are peptides produced by Sertoli cells within the testes, playing a crucial role in the endocrine feedback regulation of FSH at the pituitary level. Predominant in primates, inhibin B is more abundant in adult serum compared to inhibin A. A significant surge in inhibin B levels is observed postnatally, coinciding with the proliferation of Sertoli cells, followed by a decline until puberty, when FSH stimulation leads to a resurgence . Therefore, inhibin B is more appropriate for the early assessment of testicular function during these developmental stages [105,106]. Additionally, inhibin B is a vital biomarker for spermatogenesis and an indicator of Sertoli cell functionality, closely associated with the size of the Sertoli cell population and testicular volume [107,108]. In cases of oligospermia, there is a strong correlation between inhibin B and FSH levels with sperm concentration, and elevated FSH (>7.8 IU/I) and reduced inhibin B (<92 pg/mL) levels are indicative of a compromise [109,110,111]. This association underscores the importance of inhibin B in the diagnosis and monitoring of male fertility and testicular health.
3.4. Activin A
Activin is a dimer glycoprotein belonging to the transforming growth factor β (TGF-β) superfamily. It is composed of two β subunits, which can combine with the α subunit to form inhibin . Unlike inhibin, activin stimulates the pituitary gland to produce FSH. Activin signals are activated through interaction with specific receptors that belong to a serine/threonine kinase family and activate intracellular Smad proteins .
Activin A plays multiple roles in the testis. It serves as an important factor in regulating the proliferation of Sertoli cells and gonocytes, which are crucial for the establishment of a balance between Sertoli and germ cells at the beginning of spermatogenesis . The absence of activin A leads to a decrease in the proliferation of fetal Sertoli cells , while some gonocytes may bypass the quiescent state and multiply in number . Furthermore, activin A also influences the development and function of the epididymis. Transgenic mice that have been engineered to overexpress the activin antagonist follistatin (FST) in the testis exhibit fluid accumulation and sperm stasis . This result suggested that the blockade of activin signaling in males resulted in impaired testicular duct function, degenerative lesions, and impaired fertility, suggesting that the selective inhibitors of activin signaling could potentially be used for the development of male contraceptives without affecting androgen synthesis and actions. Additionally, the knockout of the activin/inhibin A subunit gene (INHBA) resulted in an abnormal morphology of Wolffian tubes, specifically a failure to develop the characteristic coiling in the epithelium . Recent studies have also revealed that activin A plays a role in regulating androgen biosynthesis in fetal testis by acting on Sertoli cells . Activin A deficiency can also lead to disturbances in the male reproductive system.
3.5. Anti-Müllerian Hormone (AMH)
Anti-Müllerian Hormone (AMH) serves as a marker for immature Sertoli cells , which are instrumental in the regression of Müllerian ducts in male fetuses, thus preventing the development of the uterus and fallopian tubes . AMH secretion by Sertoli cells remains elevated throughout fetal life and into postnatal development until the onset of puberty [121,122]. As puberty progresses, Sertoli cells transition from a proliferative, immature state to a mature, quiescent one, leading to a marked decrease in AMH levels . This reduction in AMH is linked to the maturation of Sertoli cells and inversely correlates with circulating androgen levels .
Serum AMH levels can be utilized as an indicator of testicular presence. A study revealed that serum AMH levels averaged around 48.2 ng/mL in children with normal testes, 11.5 ng/mL in children with abnormal testes, and only 0.7 ng/mL in cases of testicular absence . The AMH assay demonstrates high sensitivity, up to 92%, in detecting the absence of testicular tissue . Consequently, AMH levels are invaluable in differential diagnosis, particularly for conditions such as bilateral cryptorchidism and anorchidism in boys with nonpalpable gonads [119,120,124]. This biomarker provides a reliable tool for assessing testicular function and development, offering insights into the hormonal regulation of testicular health.
3.6. Insulin-like Factor 3 (INSL3)
INSL3 belongs to the insulin–relaxin family and is primarily produced in Leydig cells in human males . The expression of INSL3 is directly dependent on the number and differentiation status of Leydig cells and is an ideal biomarker for Leydig cells . During fetal development, INSL3 expression starts after gonadal sex determination and acts by binding to the RelaXin-like Family Peptide receptor 2 (RXFP2), which is expressed by gubernacular ligament mesenchymal cells that connect the testis to the inguinal wall . Their crucial role is reflected in the process of the descent of the fetal testes . During the first phase of testicular descent, INSL3 mediates the outgrowth of the gubernaculum, thereby retaining the fetal testis in the inguinal region [127,128]. In the second inguinal–scrotal phase, the testis descends into the scrotum through the inguinal canal, a process dependent on testosterone that also appears to involve the INSL3/RXFP2 signal . In INSL3 gene knockout mice, the ligaments did not develop, and the testes seemed to move loosely within the abdominal cavity, which leads to cryptorchidism .
Additionally, INSL3 has been found to inhibit germ cell apoptosis by binding to leucine-rich repeat-containing G protein-coupled receptor 8 (LGR8), which is expressed in germ cells [130,131,132]. Furthermore, studies have shown that higher serum INSL3 concentrations in men treated with hormonal male contraceptive regimens are associated with persistent sperm production . These findings collectively indicate a strong association between INSL3 and spermatogenesis.
3.7. Estrogen
In recent years, research has uncovered the presence of estrogen, a hormone traditionally associated with females, in male reproductive processes. One of the key enzymes involved in estrogen production, known as CYP19A1 or aromatase, plays a crucial role in the conversion of androstenedione and testosterone into estrone and estradiol. This enzyme has been found to be expressed in Leydig cells, germ cells, and epididymal sperm . When it comes to male reproduction, estrogen has direct effects on the development of male reproductive organs and the process of spermatogenesis. Studies on males with estrogen receptor-α (ERα) gene knockout have shown that the efferent ductule epithelium, which connects the testis to the initial segment of the epididymis, fails to absorb fluid properly [134,135]. This leads to backpressure atrophy in the testes and damages the process of spermatogenesis. Similar effects have also been observed in pubertal male mice that were treated with anti-estrogen compounds [136,137]. On the other hand, mice lacking the aromatase enzyme experience post-meiotic defects around 18 weeks of age, including increased apoptosis and reduced fertility, which can be improved by supplementing their diet with phytoestrogens [138,139,140].
However, adult male rats exposed to a diet high in phytoestrogens also experienced increased germ cell apoptosis and disruptions in spermatogenesis . The involvement of estrogen in the development of prostatic hyperplasia has also been demonstrated . Neonatal exposure to high levels of estrogen may permanently alter prostate development and differentiation . This effect may be attributed to the potential reprogramming of prostatic stem and progenitor cells as a result of early estrogen exposure, leading to changes in their proliferation status . Despite the recognized detrimental impact of estrogen on male fertility over the years, the appropriate expression of estrogen still plays a significant role in male reproductive capability and should not be overlooked.
3.8. Prolactin (PRL)
PRL is a peptide hormone secreted by lactotroph cells of the anterior pituitary gland, which plays a crucial role in female reproductive physiology. The release of PRL is regulated by various factors within the HPG axis, and it, in turn, regulates gonadal function by regulating the number of LH receptors on testicular Leydig cells and the release of pituitary gonadotropins . In immature male rats, the prolonged suppression of PRL can inhibit the process of spermatocyte–spermatid conversion, alter Leydig cell morphology, and increase serum LH levels. These effects can be mitigated by administering exogenous PRL . Furthermore, mice lacking the PRL gene showed no impaired fertility, but there were reductions in dopamine, LH, and FSH levels . Conversely, male hyperprolactinemia (HPRL), often resulting from a pituitary tumor, is associated with erectile dysfunction (ED), decreased libido, and dysfunction in orgasm or ejaculatory ability [147,148]. Although studies have shown the effects of impaired PRL function on male reproduction, sexuality, and metabolism, the role of prolactin and its receptors in males is still unclear [144,149].
3.9. Oxytocin (OT)
OT is a neurohormone stored and secreted by the posterior pituitary gland and has been found to be present in the male reproductive tract . Research has shown that OT promotes spermiation and sperm transfer, potentially by regulating steroidogenesis and the contractility of the seminiferous tubule [151,152]. Additionally, OT has been found to stimulate the secretion of testosterone and GnRH . Interestingly, treatment with OT has been found to increase the expression of proliferating cell nuclear antigen (PCNA) and B cell lymphoma 2 (Bcl-2) proteins in the testis, suggesting that OT may play an important role in the proliferation, survival, and death of germ cells . Although the molecular mechanism of the effect of OT on testicular activity is not clear, further exploration of the function and mechanism of OT in the testis is also of significant importance.
4. The HPG Axis in Hormonal Regulation
The HPG axis is the center of human reproduction, and its activation depends on the secretion of GnRH by the hypothalamus follows a pulsatile pattern. This pulsatile GnRH release was recently found to be caused by a group of kisspeptin neurons in the arcuate nucleus (ARC) of the hypothalamus . The frequency and amplitude modulation of these pulses control the release of LH and FSH, which triggers key events in the reproductive system. The HPG axis is active during critical stages of life, including fetal development (midgestational fetus), infancy (minipuberty), and puberty. At each stage, the activation of the HPG axis serves different biological functions related to growth, sexual maturation, and reproductive capability (Table 2).
Table 2.
The characteristics and function of hormones in different life stages.
4.1. HPG Axis in Fetal Life
During embryonic development, specifically around 42 days of gestation, neurons that secrete GnRH migrate from the epithelium of the medial olfactory pit. They travel along nerve fibers that are rich in neural cell adhesion molecule (N-CAM) to reach the fetal hypothalamus . GnRH expression begins in the first trimester, while levels of LH and FSH can be detected in the fetal serum and pituitary gland, starting at approximately 12 weeks of gestation in the second trimester (Figure 4). Towards the end of gestation, both LH and FSH levels decrease and become very low at term [155,160]. This decrease may be attributed to the inhibitory effects of high levels of placental estrogen on HPG axis activity at the end of pregnancy . At the eighth week of pregnancy, fetal testes begin to secrete T, and by 10–20 weeks, testosterone levels reach those typically found in adults. However, testosterone levels decline towards the end of pregnancy. During this stage, the testosterone stimulation of the sensory branch of the genitofemoral nerve causes rhythmic contractions of the gubernaculum that pull the testicles into the scrotum as the first process of testosterone regulation during development . Additionally, AMH is produced in the eighth week of pregnancy and causes the regression of the Müllerian ducts, which prevents the formation of the uterus and fallopian tubes . The attainment of proper masculinization after birth hinges on the adequate development of testicular somatic and germ cells during the fetal period, particularly during a pivotal phase termed the masculinization programming window. This critical developmental interval sets the stage for future reproductive health in adult males, and disruptions during this period can have lasting consequences. Consequently, many reproductive health issues encountered in adulthood may trace their roots back to fetal development [69,164]. Understanding the intricate processes and regulatory mechanisms at play during this window is essential for the prevention and management of male reproductive disorders.
Figure 4.
Developmental physiology of the testis and serum levels of gonadotropins, testosterone, inhibin B, and AMH from the fetal stage to puberty. The activation of the male HPG axis occurs in three distinct waves. The first wave commences during the fetal period, where testosterone expression prompts the differentiation of internal genitalia and the masculinization of the external genitalia. Additionally, AMH facilitates the regression of Müllerian ducts. This process reaches its peak in mid-pregnancy and decreases at birth. Following birth, there is a brief reactivation of the HPG axis during minipuberty. This second wave is characterized by an increased secretion of LH and FSH, leading to a testosterone peak comparable to that of puberty. Subsequently, during infancy and childhood, levels of gonadotropins and testosterone decrease, whereas levels of AMH and inhibin B continue to rise. Testicular volume experiences a slight increase during fetal and childhood stages, primarily due to the proliferation of Sertoli cells. It is not until puberty that the HPG axis is reactivated. During puberty, testosterone stimulates Sertoli cell maturation and initiates spermatogenesis, accompanied by a significant increase in testicular volume. Concurrently, testosterone inhibits AMH levels, and FSH stimulation elevates inhibin B levels.
4.2. HPG Axis in Minipuberty
In boys, the HPG axis becomes active again after birth. Correspondingly, other hormones also undergo changes throughout life under the influence of the HPG axis. Gonadotropin levels remain high for the first three months of life but start to decrease by the sixth month. During this time, testosterone levels reach their peak at one to three months of age [165,166]. This stage is referred to as minipuberty, and it plays a crucial role in the development of the fetus’s penis and testicles. After minipuberty, the HPG axis becomes inactive until puberty . During minipuberty, gonadotropins induced by GnRH stimulate Leydig cells to actively release androgen, while Sertoli cells secrete high levels of AMH. A study using a stereological approach suggests that the increase in Sertoli cell numbers primarily occurs during this period and puberty . At this time, germ cells go through mitosis but do not enter meiosis . Although Sertoli cells and germ cells remain immature before the onset of puberty, the size and potential for sperm production in adult rat testes are largely determined by the proliferation of testicular cells during fetal and neonatal development .
4.3. HPG Axis in Puberty
In mammals, the testicles contain Sertoli cells and undifferentiated spermatogonia from infancy until puberty, when spermatogonia begin to differentiate and undergo meiosis. Before the onset of puberty, the HPG axis remains relatively dormant during childhood, with GnRH, LH, and FSH being released in a pulsatile and nocturnal pattern and maintained at very low levels [171,172]. During puberty, the signal transduction of kisspeptin and neurokinin B (NKB) to GnRH neurons is further increased, resulting in gonadotropin secretion and gonadal maturation [171,173]. At this point, LH stimulates the differentiation of Leydig cells and produces testosterone. The combined effect of testosterone and FSH stimulation on Sertoli cells eventually leads to the initiation of spermatogenesis . At the same time, there is a significant increase in the volume of the testis and the diameter of seminiferous tubules due to germ cell proliferation, which is often the first indicator of puberty . Testosterone promotes the maturation of Sertoli cells and also reduces AMH production. The decrease in AMH levels is closely linked to an increase in inhibin B during early puberty . In the early stage of testicular maturation, the concentration of inhibin B is positively correlated with the level of FSH, while in the middle to late stage of adolescence, it shows an inverse relationship with FSH and slightly decreases [176,177] (Figure 4). However, the concentration of inhibin B is significantly correlated with the entire adolescent testicular volume , while the serum concentration of INSL3 gradually increases throughout puberty, and this increase is dependent on LH . This intricate hormonal interplay is pivotal in orchestrating the developmental changes in the testes during puberty.
4.4. HPG Axis in Adulthood
Male fertility peaks at puberty, when the reproductive system is fully developed, but the regulation of the HPG axis remains important in adulthood. In adulthood, the normal androgen levels maintained by the HPG axis are not only the basis for spermatogenesis, sexual function, and libido but also crucial for maintaining muscle mass, bone health, and erythropoiesis [179,180,181]. Consequently, disruptions in the HPG axis can have implications beyond reproductive health, potentially affecting the skeletal and cardiovascular systems as well [180,181].
5. Hormone Disorders and Male Hormonal Contraceptive
5.1. Hypogonadism
Male infertility is characterized by abnormal sperm parameters and can be divided into three categories: testicular dysfunction (possibly associated with primary hypogonadism), hypothalamic–pituitary diseases (resulting in secondary hypogonadism), and the obstruction of semen outflow (obstructive azoospermia, OA) . About 40% of infertile couples are affected by male hypogonadism .
Male hypogonadism is generally defined as low circulating testosterone levels, with signs and symptoms of testosterone deficiency. Serum testosterone levels are the primary indicator for detecting hypogonadism in adult males. Generally, repeated measurements of total testosterone concentration in the morning serum < 9.7–10.4 nmol/L and free testosterone < 0.17–0.31 nmol/L can be used as a reference value [184,185]. With decreased serum concentrations of total testosterone or free testosterone level, elevated LH and FSH levels should raise suspicion for a diagnosis of hypergonadotropic hypogonadism (primary hypogonadism), and low or inappropriately normal LH and FSH levels suggest hypogonadotropic hypogonadism (secondary hypogonadism) [186,187].
Primary hypogonadism is a failure of testicular testosterone production manifested by a significant decrease in testosterone concentrations and an increase in gonadotropin concentrations. Generally, primary hypogonadism can be caused by certain hereditary disorders (Klinefelter syndrome and other rare chromosomal abnormalities), infections, testicular injuries (torsion and trauma), orchitis, drugs, or congenital anorchidism . Secondary hypogonadotropin refers to the failure of gonadotropin secretion in the central nervous system, mainly caused by hypothalamic and pituitary disorders (including prolactinoma and other pituitary tumors), resulting in decreased spermatogenesis, low serum testosterone levels, and low or inappropriate normal LH and FSH levels [185,187]. Hemochromatosis, morbid obesity, hyperprolactinemia, medications (i.e., glucocorticoids and opioids), traumas, eating disorders, anabolic steroid abuse, or other genetic disorders (Prader–Willi syndrome and Kallmann’s syndrome) are also factors leading to secondary hypogonadism [188,189].
5.2. Male Hormonal Contraceptive
Although there are many methods of contraception, the unintended pregnancy rate in the United States remains at about 45%, and many people are reluctant to use long-acting reversible contraceptives (the IUD and the implant) despite their failure rate of only 1% [190,191]. There are not many reversible contraceptive methods available to men, including the male condom and withdrawal, with failure rates of 13% and 20%, respectively . Developing new contraceptives for men is challenging, and interfering with spermatogenesis through hormonal approaches is a hot topic of research. It is well known that intratesticular concentrations of testosterone (ITT) are 50 to 100 times higher than those in the peripheral circulation , which is thought to be essential for spermatogenesis and has been proposed as a new way of developing hormonal male contraceptives [193,194]. The use of exogenous steroids alone or in combination with progestins or GnRH agonists or antagonists can inhibit testicular testosterone production by the feedback inhibition of the HPG axis while also maintaining adequate serum levels to ensure other androgen-related functions [195,196].
However, oral testosterone is cleared too rapidly, hampering the search for safe and effective oral androgens . Therefore, there is a need for the development of new male hormonal contraceptives that are convenient, effective, reversible, and affordable. Testosterone undecanoate (TU), a testosterone ester created by the fatty acid esterification of testosterone, has been approved for use in several countries. It has been proven in clinical trials to be a highly reversible, effective, and safe male contraceptive, with a total efficacy was about 95% . 7α-methyl-19- nortestosterone (MENT, an androgenic–anabolic steroid) has been proven to be more effective in inhibiting pituitary gonadotropin than testosterone and the potential of its subcutaneous implants is being investigated [198,199]. Dimethandrostenolone undecanoate (DMAU) and 11β-methyl-19-norT (11β-MNT) are structurally similar and are currently under clinical investigation . Both of them bind avidly to AR and have proved to be safe and well tolerated without serious side effects and reversibly inhibit the hypothalamic–pituitary–testicular axis [201,202]. In addition, randomized, double-blind clinical trials demonstrated that a combination of testosterone and nestorone (NES, a nonandrogenic progestin) transdermal gel inhibited spermatogenesis, with 88.5% of men using the NES + T gel daily inhibiting sperm concentration to 1 million/mL or lower with minimal adverse effects .
Hormonal male contraception has been shown to be effective in clinical trials, but a novel male contraceptive will mitigate a myriad of biopsychosocial risks by male users and their partners . There must be further testing in long-term studies to determine whether these male hormones contraceptives are safe and effective in inhibiting sperm production. Realistically, long-term trials on a sufficient number of couples can take years before the product can be brought to market.
5.3. Research and Future Perspective
Although research began decades ago, our understanding of male hormone regulation is continually evolving with new discoveries. A recent study revealed that in LH receptor knockout mice, the resumption of spermatogenesis and sexual function required only approximately 5 nmol/L of circulating testosterone and intratesticular testosterone (ITT) concentration, suggesting that elevated ITT levels may be an inherent characteristic of the testis as a site of testosterone production . This finding challenges the traditional view of hormone dynamics in male fertility. Additionally, hormones play a crucial role in sexual function and spermatogenesis, and while there has been extensive research on the targets and functions of hormonal regulation, the endeavor to simulate these processes in vitro presents significant challenges. The production of sperm in vitro is a complex process that requires the accurate simulation of the physiological and hormonal signals within the natural environment of spermatogonial stem cells (SSCs). These cells are essential for spermatogenesis and necessitate a precise combination of hormones, growth factors, and other signals for successful development . However, the complexity of both endogenous and exogenous factors that influence aging and related diseases , as well as the challenge of developing high-throughput assays to evaluate the impacts of endocrine-disrupting chemicals , makes the task of mimicking these signals in vitro particularly daunting. Therefore, engineering reproductive tissues outside the body faces challenges in replicating hormonal fluctuations and re-establishing the intricate physiological interactions found in natural settings . This complexity is not unique to reproductive biology; for instance, the in vivo intestinal stem cell compartment also illustrates the challenges in capturing physiological interactions within tissues . Nevertheless, the endocrinology of the male reproductive system is particularly critical as it regulates the synthesis of sperm and other male reproductive entities . The regulation is intertwined with the endocrine hormones produced within the male reproductive system, highlighting the close relationship between hormonal action and spermatogenesis .
Approaches such as testicular tissue or SSC transplantation, and in vitro spermatogenesis, are emerging as vital strategies to restore fertility in individuals suffering from conditions like cancer or azoospermia . Promising results have been shown in mice, where neonatal fresh and cryopreserved mouse testicular tissue fragments were able to undergo spermatogenesis and produce viable spermatozoa [213,214]. However, achieving similar success in primates has remained elusive. One of the core difficulties lies in our limited understanding of the germ cell niche—its structural and nutritional requirements—and the complex regulatory mechanisms governing spermatogenesis. Hormonal regulation within this process is notably challenging to dissect due to the interconnected nature of the endocrine system, where hormones often interact with or influence other testis-acting hormones in largely synergistic ways.
In conclusion, in vitro sperm production is a highly intricate venture, contingent upon the replication of precise physiological and hormonal signals inherent to the natural developmental niche. Acknowledging and addressing the challenges in reproducing these signals is critical for advancing our capabilities in fertility restoration and for deepening our understanding of the elaborate interplay of physiological systems involved in sperm production. Future research should employ high-resolution, multidimensional detection and analytical methods to delve into the downstream regulatory networks and molecular underpinnings of male hormone regulation. Such studies could elucidate the dynamic effects of hormones during development, potentially leading to significant advancements in the diagnosis, classification, and treatment of male infertility. This could also pave the way for the discovery of a broader range of effective therapeutic strategies and medications.
6. Conclusions
The male reproductive system’s development and function are intricately linked to complex hormonal interactions, with the HPG axis’s pulsatile hormone release playing a key role in male fertility. Early diagnosis and intervention are vital for addressing hormonal disorders that can impact reproductive health and well-being. Understanding the complex regulatory networks and molecular mechanisms in testicular development and spermatogenesis is key to creating innovative treatments and contraceptives. Ongoing research is vital for advancing contraceptive options and enhancing overall reproductive health.
Author Contributions
Conceptualization, Y.Y.; writing—original draft preparation, L.L. and W.L.; writing—review and editing, Z.W., R.H., and H.X.; visualization, J.D., Z.L., and T.Y.; supervision, Y.H. and Y.Y.; funding acquisition, Y.H. and Y.Y. All authors have read and agreed to the published version of the manuscript.
Funding
This research was funded by the National Natural Science Foundation of China (No. U22A20277, 32170865, and 82071634); the Natural Science Foundation of Guangdong Province (No. 2022A1515012178); the Guangzhou Key R&D Program (No. 202103030003); and the Guangdong Key Areas R&D Program (No. 2022B1111080007).
Conflicts of Interest
The authors declare no conflicts of interest.
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Figure 1.
Schematic representation of the structure of the seminiferous tubules. There are somatic Sertoli cells and germ cells in the seminiferous tubules and Leydig cells, blood vessels, and immune cells in the interstitium. SSCs and Sertoli cells are attached to the basement membrane. In the process of spermatogenesis, germ cells (spermatogonia, pre-leptotene spermatocytes, pachytene spermatocytes, round spermatid, and elongating spermatid) move in the seminiferous epithelium and undergo meiosis until they are released to the seminiferous tubular fluid. There are four types of cell junctions in the seminiferous epithelium. The junctions between Sertoli cells form the BTB, which provides an immunologically privileged environment for spermatogenesis.
Figure 2.
The HPG axis. The hypothalamus regulates the biosynthesis and secretion of pituitary hormones LH and FSH through GnRH. LH induces Leydig cells to secrete testosterone, which reduces GnRH and LH production through negative feedback. FSH acts on Sertoli cells to induce the secretion of inhibin, which in turn inhibits the production of FSH. Sertoli cells, under the combined action of testosterone and FSH, stimulate the proliferation and maturation of germ cells. Testosterone is also essential for promoting muscle and bone development. Furthermore, testosterone is metabolized by aromatase (CYP19A1) into 17β-estradiol (E2) which exerts inhibition on the hypothalamus. Normal spermatogenesis depends on the coordinated action of all these hormones.
Figure 3.
The classical and non-classical pathways of androgen between the two neighboring Sertoli cells allow the germ cells to continuously mature and move toward the lumen. The classic androgen pathway is shown in the Sertoli cells on the left. Androgen diffuses into the cytoplasm through the plasma membrane and interacts with HSP-binding AR, then the AR dissociates from HSP and localizes to the nucleus. In the nucleus, AR homodimers recruit co-regulators and bind to the ARE of the target gene to regulate their transcription. The classic pathways guarantee Sertoli cell maturation, BTB integrity, SSC differentiation, and spermatocyte meiosis. The non-classical androgen pathway is depicted within the Sertoli cells on the right. Androgen binds to AR on the cell membrane and subsequently interacts with Src. Src then activates EGFR, initiating a series of cascade reactions. Ultimately, the phosphorylated p90Rsk translocates to the nucleus and activates CREB to bind to the CRE of the target genes, thereby regulating their transcription. This process enables androgens to modulate the BTB integrity of Sertoli cells, the meiosis of spermatocytes, Sertoli cell–spermatid adhesion, spermatid development, and sperm release.
Figure 4.
Developmental physiology of the testis and serum levels of gonadotropins, testosterone, inhibin B, and AMH from the fetal stage to puberty. The activation of the male HPG axis occurs in three distinct waves. The first wave commences during the fetal period, where testosterone expression prompts the differentiation of internal genitalia and the masculinization of the external genitalia. Additionally, AMH facilitates the regression of Müllerian ducts. This process reaches its peak in mid-pregnancy and decreases at birth. Following birth, there is a brief reactivation of the HPG axis during minipuberty. This second wave is characterized by an increased secretion of LH and FSH, leading to a testosterone peak comparable to that of puberty. Subsequently, during infancy and childhood, levels of gonadotropins and testosterone decrease, whereas levels of AMH and inhibin B continue to rise. Testicular volume experiences a slight increase during fetal and childhood stages, primarily due to the proliferation of Sertoli cells. It is not until puberty that the HPG axis is reactivated. During puberty, testosterone stimulates Sertoli cell maturation and initiates spermatogenesis, accompanied by a significant increase in testicular volume. Concurrently, testosterone inhibits AMH levels, and FSH stimulation elevates inhibin B levels.
Table 1.
Hormone secretory regions and target cells.
| Hormones | Secretory Regions | Target Cells | Disordered Diseases |
--- --- |
| GnRH | hypothalamic neurosecretory cells | pituitary gonadotrophs | precocious puberty, hypergonadism, Kallmann syndrome, oligospermia |
| LH | pituitary gonadotrophs | Leydig cells | hypogonadism |
| FSH | pituitary gonadotrophs | Sertoli cells, peritubular myoid cells, Spermatogoniums | hypogonadism |
| T | Leydig cells, Sertoli cells | germ cells, Sertoli cells, PMCs | TDS, micropenis [21,22] |
| AMH | Sertoli cells | Müllerian ducts mesenchymal cells | persistent Müllerian duct syndrome (PMDS) |
| Inhibin B | Sertoli cells | pituitary gonadotrophs | spermatogenesis disorder |
| INSL3 | Leydig cells | Leydig cells | cryptorchidism |
Table 2.
The characteristics and function of hormones in different life stages.
| Hormones | Fetal Life | Minipuberty | Puberty |
--- --- |
| GnRH | serum LH and FSH levels in the second trimester are independent of GnRH, and then GnRH gradually controls the release of LH and FSH | stimulates Sertoli cells to secrete inhibin B and AMH, and Leydig cells to produce INSL3 | increases gradually, triggering the secretion of LH and FSH |
| LH | replaces HCG to promote the secretion of testosterone by Leydig cells | stimulates Leydig cells to release testosterone | stimulated the differentiation of Leydig cells and their ability to produce testosterone |
| FSH | stimulates Sertoli cell proliferation and increases AMH and inhibin B | stimulates Sertoli cell proliferation and increases AMH and inhibin B | stimulates the proliferation of immature Sertoli cells and spermatogonia |
| T | induces the differentiation and development of the mesonephric duct into seminal vesicles, epididymis, and spermaduct | promotes the conversion of germ cells into spermatogonia | initiation of spermatogenesis |
| AMH | causes fallopian tube regression in men, preventing the formation of the uterus and fallopian tubes | as a diagnostic indicator of male fertility-related disorders | as a diagnostic indicator of male fertility-related disorders |
| Inhibin B | regulates FSH secretion and acts as a marker for Sertoli cell function | regulates FSH secretion and acts as a marker for Sertoli cell function | inhibit FSH secretion and markers of sperm production in men |
| INSL3 | the regulation of intra-abdominal testicular descent by regulating the growth and differentiation of the gubernaculum | as an accurate measure of Leydig cell functional capacity | as an accurate measure of Leydig cell functional capacity |
| | |
--- |
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Li, L.; Lin, W.; Wang, Z.; Huang, R.; Xia, H.; Li, Z.; Deng, J.; Ye, T.; Huang, Y.; Yang, Y.
Hormone Regulation in Testicular Development and Function. Int. J. Mol. Sci. 2024, 25, 5805.
AMA Style
Li L, Lin W, Wang Z, Huang R, Xia H, Li Z, Deng J, Ye T, Huang Y, Yang Y.
Hormone Regulation in Testicular Development and Function. International Journal of Molecular Sciences. 2024; 25(11):5805.
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Li, Lu, Wanqing Lin, Zhaoyang Wang, Rufei Huang, Huan Xia, Ziyi Li, Jingxian Deng, Tao Ye, Yadong Huang, and Yan Yang.
2024. "Hormone Regulation in Testicular Development and Function" International Journal of Molecular Sciences 25, no. 11: 5805.
APA Style
Li, L., Lin, W., Wang, Z., Huang, R., Xia, H., Li, Z., Deng, J., Ye, T., Huang, Y., & Yang, Y.
(2024). Hormone Regulation in Testicular Development and Function. International Journal of Molecular Sciences, 25(11), 5805.
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Areas of Right Triangles
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Alignments to Content Standards: 6.G.A.1
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Task
Explain why the right triangle shown has an area of exactly 20 square units.
The "legs" of a right triangle are the two sides that form the right angle. If one leg of a right triangle is 5 units long, explain what else would have to be true about the right triangle in order for its area to be 30 square units.
Here are leg measurements for more right triangles. What is the area of each?
6 and 3
12 and (4 {1\over2})
3 and 7
6.5 and 9
Explain in words how you can find the area of a right triangle when you know the lengths of its legs.
Let a represent the length of one leg of a right triangle and b represent the length of the other leg of the right triangle. Write a mathematical expression for the area of the right triangle in terms of a and b.
IM Commentary
This task is intended to help build understanding as students work toward deriving a general formula for the area of any triangle. The purpose of this task is for students to use what they know about area and express regularity in repeated reasoning (MP.8) to generate a formula for area of a right triangle.
This task is intended to be used before students know that the area of a triangle can be expressed with bh/2. So, in part (c), they should be reasoning using geometric methods (surround and subtract, composing and decomposing, counting unit squares, etc.). They are not expected to use a formula in part (c), because they are not expected to know a formula yet.
It would be appropriate to make grid paper, straight edges, and even scissors available to students who want to think about area in terms of counting square units. In part (c), this tactic will become more difficult, nudging students to think more abstractly.
Students thinking very concretely about the number of unit squares may struggle with parts of (c). For these students, it may be helpful to assist drawing pictures of all four part (c) triangles with leg lengths labeled. Even though it is difficult to count unit squares, students can still reason that the right triangle is half of an associated rectangle, and compute half the area of the rectangle. Generalizing this process will help them respond to parts (d) and (e).
Still other students may simply need support multiplying fractions.
Solution
Answers may vary. Techniques include "completing" the rectangle with sides of length 4 and 10 and arguing that half of its area must be 20 square units, chopping off and relocating a chunk of the right triangle to create a rectangle whose area is 20 square units, and "completing" individual unit squares to create a total of 20 unit squares. Here are some illustrations associated with these methods:
The other leg would have to measure 12 units in length. A 12 by 5 rectangle would have an area of 60 square units. Slicing this rectangle along a diagonal would create right triangles with legs of 12 and 5 with areas of 30 square units.
9
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(21\over2)or (10{1\over2})or 10.5
29.25 or 117/4
Answers may vary. Examples: Multiply the lengths of the legs together, and then divide by 2. One half of the product of the lengths of the legs.
(ab\over2), ({1\over2}ab) , (a\times b \div 2) , or equivalent
Areas of Right Triangles
Explain why the right triangle shown has an area of exactly 20 square units.
The "legs" of a right triangle are the two sides that form the right angle. If one leg of a right triangle is 5 units long, explain what else would have to be true about the right triangle in order for its area to be 30 square units.
Here are leg measurements for more right triangles. What is the area of each?
6 and 3
12 and (4 {1\over2})
3 and 7
6.5 and 9
Explain in words how you can find the area of a right triangle when you know the lengths of its legs.
Let a represent the length of one leg of a right triangle and b represent the length of the other leg of the right triangle. Write a mathematical expression for the area of the right triangle in terms of a and b.
Print Task
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An Introduction to Bayesian Reasoning and Methods
Chapter 9 Considering Prior Distributions
One of the most commonly asked questions when one first encounters Bayesian statistics is “how do we choose a prior?” While there is never one “perfect” prior in any situation, we’ll discuss in this chapter some issues to consider when choosing a prior. But first, here are a few big picture ideas to keep in mind.
Bayesian inference is based on the posterior distribution, not the prior. Therefore, the posterior requires much more attention than the prior.
The prior is only one part of the Bayesian model. The likelihood is the other part. And there is the data that is used to fit the model. Choice of prior is just one of many modeling assumptions that should be evaluated and checked.
In many situations, the posterior distribution is not too sensitive to reasonable changes in prior. In these situations, the important question isn’t “what is the prior?” but rather “is there a prior at all”? That is, are you adopting a Bayesian approach, treating parameters as random variables, and quantifying uncertainty about parameters with probability distributions?
One criticism of Bayesian statistics in general and priors in particular is that they are subjective. However, any statistical analysis is inherently subjective, filled with many assumptions and decisions along the way. Except in the simplest situations, if you ask five statisticians how to approach a particular problem, you will likely get five different answers. Priors and Bayesian data analysis are no more inherently subjective than any of the myriad other assumptions made in statistical analysis.
Subjectivity is OK, and often beneficial. Choosing a subjective prior allows us to explicitly incorporate a wealth of past experience into our analysis.
Example 9.1 Xiomara claims that she can predict which way a coin flip will land. Rogelio claims that he can taste the difference between Coke and Pepsi.
Before reading further, stop to consider: whose claim - Xiomara’s or Rogelio’s - is initially more convincing? Or are you equally convinced? Why? To put it another way, whose claim are you initially more skeptical of? Or are you equally skeptical? To put it one more way, whose claim would require more data to convince you?17
To test Xiomara’s claim, you flip a fair coin 10 times, and she correctly predicts the result of 9 of the 10 flips. (You can assume the coin is fair, the flips are independent, and there is no funny business in data collection.)
To test Rogelio’s claim, you give him a blind taste test of 10 cups, flipping a coin for each cup to determine whether to serve Coke or Pespi. Rogelio correctly identifies 9 of the 10 cups. (You can assume the coin is fair, the flips are independent, and there is no funny business in data collection.)
Let (\theta_X) be the probability that Xiomara correctly guesses the result of a fair coin flip. Let (\theta_R) be the probability that Rogelio correctly guesses the soda (Coke or Pepsi) in a randomly selected cup.
How might a frequentist address this situation? What would the conclusion be?
Consider a Bayesian approach. Describe, in general terms, your prior distributions for the two parameters. How do they compare? How would this impact your conclusions?
\iffalse{} Solution. to Example 9.1
For Xiomara, a frequentist might conduct a hypothesis test of the null hypothesis (H_0:\theta_X = 0.5) versus the alternative hypothesis: (H_a:\theta_X > 0.5). The p-value would be about 0.01, the probability of observing at least 9 out of 10 successes from a Binomial distribution with parameters 10 and 0.5 (1 - pbinom(8, 10, 0.5)). Rogelio’s set up would be similar and would yield the same p-value. So a strict frequentist would be equally convinced of the two claims.
Prior to observing data, we are probably more skeptical of Xiomara’s claim than Rogelio’s. Since coin flips are unpredictable, we would have a strong prior belief that (\theta_X) is close to 0.5 (what it would be if she were just guessing). Our prior for (\theta_X) would have a mean of 0.5 and a small prior SD, to reflect that only values close to 0.5 seem plausible. Therefore, it would require a lot of evidence to sway our prior beliefs. On the other hand, we might be familiar with people who can tell the difference between Coke and Pepsi; maybe we even can ourselves. Our prior for (\theta_R) would have a smaller prior SD than that of (\theta_X) to allow for a wider range of plausible values. We might even have a prior mean for (\theta_R) above 0.5 if we have experience with a lot of people who can tell the difference between Coke and Pepsi. Given the sample data, our posterior probability that (\theta_R>0.5) would be larger than the posterior probability that (\theta_X > 0.5), and we would be more convinced by Rogelio’s claim than by Xiomara’s.
Even if a prior does not represent strong prior beliefs, just having a prior distribution at all allows for Bayesian analysis. Remember, both Bayesian and frequentist are valid approaches to statistical analyses, each with advantages and disadvantages. That said, there are some issues with frequentist approaches that incorporating a prior distribution and adopting a Bayesian approach alleviates. (To be fair, an upcoming investigation will address some disadvantages of the Bayesian approach compared with the frequentist approach.)
Example 9.2 Tamika is a basketball player who throughout her career has had a probability of 0.5 of making any three point attempt. However, her coach is afraid that her three point shooting has gotten worse. To check this, the coach has Tamika shoot a series of three pointers; she makes 7 out of 24. Does the coach have evidence that Tamika has gotten worse?
Let (\theta) be the probability that Tamika successfully makes any three point attempt. Assume attempts are independent.
Prior to collecting data, the coach decides that he’ll have convincing evidence that Tamika has gotten worse if the p-value is less than 0.025. Suppose the coach told Tamika to shoot 24 attempts and then stop and count the number of successful attempts. Use software to compute the p-value. Is the coach convinced that Tamika has gotten worse?
Prior to collecting data, the coach decides that he’ll have convincing evidence that Tamika has gotten worse if the p-value is less than 0.025. Suppose the coach told Tamika to shoot until she makes 7 three pointers and then stop and count the number of total attempts. Use software to compute the p-value. Is the coach convinced that Tamika has gotten worse? (Hint: the total number of attempts has a Negative Binomial distribution.)
Now suppose the coach takes a Bayesian approach and assumes a Beta((\alpha), (\beta)) prior distribution for (\theta). Suppose the coach told Tamika to shoot 24 attempts and then stop and count the number of successful attempts. Identify the likelihood function and the posterior distribution of (\theta).
Now suppose the coach takes a Bayesian approach and assumes a Beta((\alpha), (\beta)) prior distribution for (\theta). Suppose the coach told Tamika to shoot until she makes 7 three pointers and then stop and count the number of total attempts. Identify the likelihood function and the posterior distribution of (\theta).
Compare the Bayesian and frequentist approaches in this example. Does the “strength of the evidence” depend on how the data were collected?
\iffalse{} Solution. to Example 9.2
The null hypothesis is (H_0:\theta = 0.5) and the alternative hypothesis is (H_a:\theta < 0.5). If the null hypothesis is true and Tamika has not gotten worse, then (Y), the number of successful attempts, has a Binomial(24, 0.5) distribution. The p-value is (P(Y \le 7) = 0.032) from pbinom(7, 24, 0.5). Using a strict threshold of 0.025, the coach has NOT been convinced that Tamika has gotten worse.
The null hypothesis is (H_0:\theta = 0.5) and the alternative hypothesis is (H_a:\theta < 0.5). If the null hypothesis is true and Tamika has not gotten worse, then (N), the number of total attempts required to achieve 7 successful attempts, has a Negative Binomial(7, 0.5) distribution. The p-value is (P(N \ge 24) = 0.017) from 1 - pnbinom(23 - 7, 7, 0.5). (In R, nbinom only counts the total number of failures, not the total number of trials.) Using a strict threshold of 0.025, the coach has been convinced that Tamika has gotten worse.
The data is (Y), the number of successful attempts in 24 attempts, which follows a Binomial(24, (\theta)) distribution. The likelihood is (P(Y=7|\theta)) [ f(y=7|\theta) = \binom{24}{7}\theta^7(1-\theta)^{17} \propto \theta^{7}(1-\theta)^{17}, \qquad 0 < \theta < 1. ] The posterior distribution is the Beta((\alpha + 7), (\beta + 17)) distribution.
The data is (N), the number of total attempts required to achieve 7 successful attempts, which follows a Negative Binomial(7, (\theta)) distribution. The likelihood is (P(N=24|\theta)) [ f(n = 24|\theta) = \binom{24 - 1}{7 - 1}\theta^7(1-\theta)^{17} \propto \theta^{7}(1-\theta)^{17}, \qquad 0 < \theta < 1. ] (The (\binom{24 - 1}{7 - 1}) follows from the fact that the last attempt has to be success.) Note that the shape of the likelihood as a function of (\theta) is the same as in the previous part. Therefore, the posterior distribution is the Beta((\alpha + 7), (\beta + 17)) distribution.
Even though both frequentist scenario involves 7 successes in 24 attempts, the p-value measuring the strength of the evidence to reject the null hypothesis differed depending on how the data were collected. Using a strict cutoff of 0.025 led the coach to reject the null hypothesis in one scenario but not the other. However, the Bayesian analysis is the same in either scenario since the posterior distributions were the same. For the Bayesian analysis, all that mattered about the data was that there were 7 successes in 24 attempts.
Bayesian data analysis treats parameters as random variables with probability distributions. The prior distribution quantifies the researcher’s uncertainty about parameters before observing data. Some issues to consider when choosing a prior include, in no particular order:
The researcher’s prior beliefs! A prior distribution is part of a statistical model, and should be consistent with knowledge about the underlying scientific problem. Researchers are often experts with a wealth of past experience that can be explicitly incorporated into the analysis via the prior distribution. Such a prior is called an informative or weakly informative prior.
A regularizing prior. A prior which, when tuned properly, reduces overfitting or “overreacting” to the data.
Noninformative prior a.k.a., (reference, vague, flat prior). A prior is sought that plays a minimal role in inference so that “the data can speak for itself”.
Mathematical convenience. The prior is chosen so that computation of the posterior is simplified, as in the case of conjugate priors.
Interpretation. The posterior is a compromise between the data and prior. Some priors allow for easy interpretation of the relative contributions of data and prior to the posterior. For example, think of the “prior successes and prior failures” interpretation in the Beta-Binomial model.
Prior based on past data. Bayesian updating can be viewed as an iterative process. The posterior distribution obtained from one round of data collection can inform the prior distribution for another round.
For those initially skeptical of prior distributions at all, the strategy of always choosing an noninformative or flat prior might be appealing. Flat priors are common, but are rarely ever the best choices from a modeling perspective. Just like you would not want to assume a Normal distribution for the likelihood in every problem, you would not to use a flat prior in every problem.
Furthermore, there are some subtle issues that arise when attempting to choose a noninformative prior.
Example 9.3 Suppose we want to estimate (\theta), the population proportion of Cal Poly students who wore socks at any point yesterday.
What are the possible values for (\theta)? What prior distribution might you consider a noninformative prior distribution?
You might choose a Uniform(0, 1) prior, a.k.a., a Beta(1, 1) prior. Recall how we interpreted the parameters (\alpha) and (\beta) in the Beta-Binomial model. Does the Beta(1, 1) distribution represent “no prior information”?
Suppose in a sample of 20 students, 4 wore socks yesterday. How would you estimate (\theta) with a single number based only on the data?
Assume a Beta(1, 1) prior and the 4/20 sample data. Identify the posterior distribution. Recall that one Bayesian point estimate of (\theta) is the posterior mean. Find the posterior mean of (\theta). Does this estimate let the “data speak entirely for itself”?
How could you change (\alpha) and (\beta) in the Beta distribution prior to represent no prior information? Sketch the prior. Do you see any potential problems?
Assume a Beta(0, 0) prior for (\theta) and the 4/20 sample data. Identify the posterior distribution. Find the posterior mode of (\theta). Does this estimate let the “data speak entirely for itself”?
Now suppose the parameter you want to estimate is the odds that a student wore socks yesterday, (\phi=\frac{\theta}{1-\theta}). What are the possible values of (\phi)? What might a non-informative prior look like? Is this a proper prior?
Assume a Beta(1, 1) prior for (\theta). Use simulation to approximate the prior distribution of the odds (\phi). Would you say this is a noninformative prior for (\phi)?
\iffalse{} Solution. to Example 9.3
(\theta) takes values in (0, 1). We might assume a flat prior on (0, 1), that is a Uniform(0, 1) prior.
We interpreted (\alpha) as “prior successes” and (\beta) as “prior failures”. So a Beta(1, 1) is in some some equivalent to a “prior sample size” of 2. Certainly not a lot of prior information, but it’s not “no prior information” either.
The sample proportion, 4/20 = 0.2.
With a Beta(1, 1) prior and the 4/20 sample data, the posterior distribution is Beta(5, 17). The posterior mean of (\theta) is 5/22 = 0.227. The posterior mean is a weighted average of the prior mean and the sample proportion: 0.227 = (0.5)(2/22) + (0.2)(20/22). The “noninformative” prior does have influence; the data does not “speak entirely for itself”.
If (\alpha+\beta) represents “prior sample size”, we could try a Beta(0, 0) prior. Unfortunately, such a probability distribution does not actually exist. For a Beta distribution, the parameters (\alpha) and (\beta) have to be strictly positive in order to have a valid pdf. The Beta(0, 0) density would be proportional to [ \pi(\theta) \propto \theta^{-1}(1-\theta)^{-1}, \qquad 0 < \theta <1. ] However, this is not a valid pdf since (\int_0^1 \theta^{-1}(1-\theta)^{-1}d\theta = \infty), so there is no constant that can normalize it to integrate to 1. Even so, here is a plot of the “density”.
Would you say this is a “noninformative” prior? It seems to concentrate almost all prior “density” near 0 and 1.
6. Beta(0, 0) is an “improper” prior. It’s not a proper prior distribution, but it can lead to a proper posterior distribution. The likelihood is (f(y=4|\theta) \propto \theta^4 (1-\theta)^{16}, 0 < \theta < 1). If we assume the prior is (\pi(\theta)\propto\theta^{-1}(1-\theta)^{-1}, 0 < \theta <1), then the posterior is [ \pi(\theta|y = 4) \propto \left(\theta^{-1}(1-\theta)^{-1}\right)\left(\theta^4 (1-\theta)^{16}\right) = \theta^{4 - 1} (1-\theta)^{16 - 1}, \qquad 0 <\theta < 1 ] That is, the posterior distribution is the Beta(4, 16) distribution. The posterior mean is 4/20=0.2, the sample proportion. Hoever, the posterior mode is (\frac{4- 1}{4 + 16 -2}= \frac{3}{18} = 0.167). So the posterior mode does not let the “data speak entirely for itself”.
7. If (\theta=0) then (\phi=0); if (\theta=1) then (\phi = \infty). So (\phi) takes values in ((0, \infty)). We might choose a flat prior on ((0,\infty)), (\pi(\phi) \propto 1, \phi > 0). However, this would be an improper prior.
8. Simulate a value of (\theta) from a Beta(1, 1) distribution, compute (\phi = \frac{\theta}{1-\theta}), and repeat many times. The simulation results are below. (The distribution is extremely skewed to the right, so we’re only plotting values in (0, 50).)
theta = rbeta(1000000, 1, 1) = rbeta 1000000 1 1odds = theta / (1 - theta) =/ 1 -hist(odds[odds<50], breaks = 100, xlab = "odds", freq = FALSE, hist< 50breaks = 100xlab = "odds"freq = FALSE ylab = "density",ylab = "density" main = "Prior distribution of odds if prior distribution of probability is Uniform(0, 1)")main ="Prior distribution of odds if prior distribution of probability is Uniform(0, 1)"
Even though the prior for (\theta) was flat, the prior for a transformation of (\theta) is not.
An improper prior distribution is a prior distribution that does not integrate to 1, so is not a proper probability density. However, an improper proper often results in a proper posterior distribution. Thus, improper prior distributions are sometimes used in practice.
Flat priors are common choices in some situations, but are rarely ever the best choices from a modeling perspective. Furthermore, flat priors are generally not preserved under transformations of parameters. So a prior that is flat under one parametrization of the problem will generally not be flat under another. For example, when trying to estimate a population SD (\sigma), assuming a flat prior for (\sigma) will result in a non-flat prior for the population variance (\sigma^2), and vice versa.
Example 9.4 Suppse that (\theta) represents the population proportion of adults who have a particular rare disease.
Explain why you might not want to use a flat Uniform(0, 1) prior for (\theta).
Assume a Uniform(0, 1) prior. Suppose you will test (n=100) suspected cases. Use simulation to approximate the prior predictive distribution of the number in the sample who have the disease. Does this seem reasonable?
Assume a Uniform(0, 1) prior. Suppose that in (n=100) suspected cases, none actually has the disease. Find and interpret the posterior median. Does this seem reasonable?
\iffalse{} Solution. to Example 9.4
We know it’s a rare disease! We want to concentrate most of our prior probability for (\theta) near 0.
If the disease is rare, we might not expect any actual cases in a sample of 100, maybe 1 or 2. However, the prior predictive distribution says that any value between 0 and 100 actual cases is equally likely! This seems very unreasonable given that the disease is rare.
theta_sim = runif(10000) = runif 10000y_sim = rbinom(10000, 100, theta_sim) = rbinom 10000 100hist(y_sim, hist xlab = "Simulated number of successes",xlab = "Simulated number of successes" main = "Prior predictive distribution")main = "Prior predictive distribution"
3. The posterior distribution is the Beta(1, 101) distribution. The posterior median is 0.007 (qbeta(0.5, 1, 101)). Based on a sample of 100 suspected cases with no actual cases, there is a posterior probability of 50% that more than 0.7% of people have the disease. A rate of 7 actual cases in 1000 is not a very rare disease, and we think there’s a 50% chance that the rate is even greater than this? Again, this does not seem very reasonable based on our knowledge that the disease is rare.
Prior predictive distributions can be used to check the reasonableness of a prior for a given situation before observing sample data. Do the simulated samples seem consistent with what you might expect of the data based on your background knowledge of the situation? If not, another prior might be more reasonable.
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THE GREEN-TAO THEOREM ON ARITHMETIC PROGRESSIONS IN THE PRIMES: AN ERGODIC POINT OF VIEW BRYNA KRA Abstract. A long standing and almost folkloric conjecture is that the primes contain arbitrarily long arithmetic progressions. Until recently, the only progress on this conjecture was due to van der Corput, who showed in 1939 that there are infinitely many triples of primes in arithmetic progression. In an amazing fusion of methods from analytic number theory and ergodic theory, Ben Green and Terence Tao showed that for any positive integer k, there exist infinitely many arithmetic progressions of length k consisting only of prime numbers. This is an introduction to some of the ideas in the proof, concentrating on the connections to ergodic theory.
1. Background For hundreds of years, mathematicians have made conjectures about patterns in the primes: one of the simplest to state is that the primes contain arbitrarily long arithmetic progressions. It is not clear exactly when this conjecture was first formalized, but as early as 1770 Lagrange and Waring studied the problem of how large the common difference of an arithmetic progression of k primes must be. A natural extension of this question is to ask if the prime numbers contain arbitrarily long arithmetic progressions.
Support for a positive answer to this question is provided by the following simple heuristic. The Prime Number Theorem states that the number of prime numbers less than the integer N is asymptotically N/ log N. It follows that the density of primes around a positive large x ∈R is about 1/ log x.
Thus if we model the sequence of prime numbers in {1, . . . , N} by choosing integers at random, with an integer in {1, . . . , N} being chosen with probability 1/ log N, then there ought to be approximately N 2/ logk N progressions of length k consisting of prime numbers less than or equal to N. Unfortunately, one cannot This article is an expanded version of a lecture presented January 7, 2005 at the AMS Special Session on Current Events, Joint Mathematics Meeting, Atlanta, GA.
The author was partially supported by NSF grant DMS-0244994.
1 2 BRYNA KRA hope to use this sort of argument to show that the primes contain any particular pattern, since the primes are far from being randomly distributed: 2 is the only even prime, 3 is the only prime congruent to 0 mod 3, and so on.
In 1923, Hardy and Littlewood made a very general conjecture (the k-tuple conjecture) about patterns and their distribution in the primes: if a1, . . . , ak and b1, . . . , bk are nonnegative integers such that P(x) = Qk i=1(aix + bi) is not identically 0 modulo any prime p, then there are infinitely many integers n such that {ain + bi : 1 ≤i ≤k} consists only of primes. This conjecture includes the twin prime con-jecture (there exist infinitely many primes p such that p + 2 is also prime) as a special case and it trivially implies that the primes contain arbitrarily long arithmetic progressions. Moreover it implies that the number of k-term arithmetic progressions in the primes bounded by N is asymptotically ckN 2/ logk N for a certain explicit value of ck.
There are numerous related conjectures about the existence of arith-metic progressions in certain subsets of the integers. For example, the famous conjecture of Erd¨ os and Tur´ an states that if A = {a1 < a2 < . . .} is an infinite sequence of integers with P i 1/ai = ∞, then A contains arbitrarily long arithmetic progressions. A corollary would be that the primes contain arbitrarily long arithmetic progressions.
The first major progress on arithmetic progressions in the primes was made by van der Corput , who proved in 1939 that the primes contain infinitely many arithmetic progressions of length 3. Further progress was only made in 1981, when Heath-Brown showed that there are infinitely many arithmetic progressions of length 4 consisting of three primes and an almost prime, meaning either a prime or a product of two primes. In a slightly different direction are the elegant results of Balog (, ) on patterns in the primes. For example, he shows that for any positive integer k, there exist infinitely many k-tuples of distinct primes p1 < p2 < . . . < pk such that (pi + pj)/2 is prime for all i, j ∈{1, . . . , k}. For k = 2 this implies, in particular, that the primes contain infinitely many arithmetic progressions of length 3.
Computational mathematicians have also worked on the problem of finding long arithmetic progressions in the primes. In 1995, Moran, Pritchard and Thyssen found a progression of length 22 in the primes. This record was finally broken in 2004, when Frind, Jobling and Underwood found a progression of length 23 starting with the prime 56211383760397 and with common difference 44546738095860.
In 2004, Ben Green and Terence Tao announced a major break-through, with a proof of the general case: THE GREEN-TAO THEOREM 3 Theorem 1.1 (Green and Tao ). For every integer k ≥1, the prime numbers contain an arithmetic progression of length k.
They also extract a bound on how far out in the primes one must go in order to guarantee finding an arithmetic progression of length k, showing that there is a k-term arithmetic progression of primes all of whose entries are bounded by 22222222(100k) .
This bound is considered far from optimal; standard heuristics in num-ber theory, plus a little calculation, lead to the conjecture that there is an arithmetic progression of length k in the primes all of whose entries are bounded by k! + 1.
Green and Tao prove a stronger statement than that given in The-orem 1.1. They show that not only do the primes contain arbitrarily long arithmetic progressions, but so does any sufficiently dense subset of the primes: Theorem 1.2 (Green and Tao ). If A is a subset of prime numbers with lim sup N→∞ 1 π(N) |A ∩{1, . . . , N}| > 0 , where π(N) is the number of primes in {1, . . . , N}, then for every in-teger k ≥1, A contains an arithmetic progression of length k.
For k = 3, this was proved by Green .
The theorem of Green and Tao is a beautiful result answering an old conjecture that has attracted much work. Perhaps even more im-pressive is the fusion of methods and results from number theory, er-godic theory, harmonic analysis, discrete geometry, and combinatorics used in its proof. The starting point for Green and Tao’s proof is the celebrated theorem of Szemer´ edi : a set of integers with positive upper density1 contains arbitrarily long arithmetic progressions. One of the main ideas is to generalize this, showing that a dense subset of a sufficiently pseudorandom collection (see Section 7 for the precise defi-nition) of the integers contains arbitrarily long arithmetic progressions.
There are three major ingredients in the proof. The first is Szemer´ edi’s Theorem itself. Since the primes do not have positive upper density, 1The upper density d∗(A) of a subset A of the integers is defined to be d∗(A) := lim sup N→∞ |A ∩{1, . . . , N}|/N .
4 BRYNA KRA Szemer´ edi’s Theorem can not be directly applied and the second major ingredient in Green and Tao’s proof is a certain transference principle that allows one to use Szemer´ edi’s Theorem in a more general set-ting. The last major ingredient is the use of specific properties of the primes and their distribution, based on recent work of Goldston and Yildirim , showing that this generalized Szemer´ edi Theorem applies to the primes.
It is impossible to give a complete proof of their theorem in this limited space, nor even to do justice to the main ideas. Our goal is to outline the main ingredients and focus on the relation between their work and recent parallel advances in ergodic theory. The interaction between combinatorial number theory and ergodic theory began with Furstenberg’s proof of Szemer´ edi’s Theorem (see Section 3) and has led to many new results. Until the present, this interaction has mainly taken the form of using ergodic theory to prove statements in combinatorial number theory, such as Szemer´ edi’s Theorem, its gener-alizations (including a multidimensional version and a polynomial version ), and the density Hales-Jewett Theorem . Green and Tao’s work opens a new chapter in this interaction, with ergodic the-oretic proof techniques being adapted for use in a number theoretic setting.
Acknowledgments: I appreciate the many helpful comments I re-ceived while preparing this manuscript, including those from Keith Burns, Nikos Frantzikinakis, Bernard Host, Margaret Symington, Ter-ence Tao, and Mate Wierdl, and especially those from Andrew Granville.
2. Szemer´ edi’s Theorem Substituting the set of all integers for the set of primes in Theo-rem 1.2, one obtains Szemer´ edi’s Theorem.
We state an equivalent finite version of this theorem: Theorem 2.1 (Finite Szemer´ edi ). Let 0 < δ ≤1 be a real number and let k ≥1 be an integer. There exists N0(δ, k) such that if N > N0(δ, k) and A ⊂{1, . . . , N} with |A| ≥δN, then A contains an arithmetic progression of length k.
It is clear that this version implies the first version of Szemer´ edi’s Theorem, and an easy argument gives the converse implication.
Szemer´ edi’s original proof in 1975 was combinatorial in nature.
Shortly thereafter, Furstenberg developed the surprising relation be-tween combinatorics and ergodic theory, proving Szemer´ edi’s Theo-rem via a multiple recurrence theorem (see Section 3). More recently, THE GREEN-TAO THEOREM 5 Gowers gave a new proof of Szemer´ edi’s Theorem using harmonic analysis, vastly improving the known bounds for N0(δ, k) in the finite version. Although the various proofs, Szemer´ edi’s, Furstenberg’s, and Gowers’, seem to use very different methods, they have several features in common. In each, a key idea is the dichotomy in the underlying space (whether it be a subset of the integers, a measure space, or the finite group Z/NZ) between randomness and structure.
One then has to analyze the structured part of the space to understand the intersection of a set with itself along arithmetic progressions. We start by further discussing Furstenberg’s proof of Szemer´ edi’s Theorem using ergodic theory.
3. Szemer´ edi’s Theorem and ergodic theory Furstenberg proved the multiple ergodic theorem: Theorem 3.1 (Multiple Recurrence ). Let (X, X, µ, T) be a mea-sure preserving probability system2 and let k ≥1 be an integer. For any set E ∈X with µ(E) > 0, (3.1) lim inf N→∞ 1 N N−1 X n=0 µ E ∩T −nE ∩T −2nE ∩. . . ∩T −(k−1)nE > 0.
An obvious corollary is: Corollary 3.2. Let (X, X, µ, T) be a measure preserving probability system and let k ≥1 be an integer. For any set E ∈X with µ(E) > 0, there exists an integer n ≥1 such that µ E ∩T −nE ∩T −2nE ∩. . . ∩T −(k−1)nE > 0 .
Furstenberg then made the beautiful connection to combinatorics, showing that regularity properties of integers with positive upper den-sity correspond to multiple recurrence results: Theorem 3.3 (Correspondence Principle ). Assume that A is a subset of integers with positive upper density. There exist a measure preserving probability system (X, X, µ, T) and a measurable set E ∈X 2By a measure preserving probability system, we mean a quadruple (X, X, µ, T), where X is a set, X denotes a σ-algebra on X (meaning an algebra X of subsets of X that is closed under countable unions), µ is a probability measure on (X, X), and T : X →X is a measurable map such that µ(A) = µ(T −1A) for all A ∈X.
Usually, we assume that X is a metrizable compact set and X is its Borel σ-algebra, meaning the σ-algebra generated by the open sets. In particular, for the spaces under consideration, L2(X, X, µ) is separable. We always denote the σ-algebra by the calligraphic version of the letter used for the space.
6 BRYNA KRA with µ(E) = d∗(A) such that for all integers k ≥1 and all integers m1, . . . , mk−1 ≥1, d∗ A∩(A+m1)∩. . .∩(A+mk−1) ≥µ E ∩T −m1E ∩. . .∩T −mk−1E .
Taking m1 = n, m2 = 2n, . . . , mk−1 = (k−1)n, Szemer´ edi’s Theorem follows from Corollary 3.2.
Furstenberg’s proof relies on a compactness argument, making it dif-ficult to extract any explicit bounds in the finite version of Szemer´ edi’s Theorem. On the other hand, Theorem 3.1 and its proof gave rise to a new area in ergodic theory, called “ergodic Ramsey theory.” Ergodic theoretic proofs have lead to many other results in combinatorics, such as the multidimensional Szemer´ edi Theorem and the polynomial Szemer´ edi Theorem , and some of these generalizations have yet to be attained by other methods. More recent developments in ergodic Ramsey theory closely parallel ideas in Green and Tao’s work; we re-turn to this in Section 5.
To prove Theorem 3.1, Furstenberg showed that in any measure pre-serving system, one of two distinct phenomena occurs to make the measure of this intersection positive. The first is weak mixing,3 when for any set E, µ(E ∩T −nE) is approximately µ(E)2 for most choices of the integer n. Then it can be shown that µ(E ∩T −nE ∩T −2nE ∩. . . ∩T −(k−1)nE) is approximately µ(E)k for most choices of n, which is clearly positive when E is a set of positive measure. The opposite situation is rigidity, when for appropriately chosen n, T n is very close to the identity. Then T jnE is very close to E and µ(E ∩T −nE ∩T −2nE ∩. . . ∩T −(k−1)nE) is very close to µ(E), again giving positive intersection for a set E of positive measure. One then has to show that the average along arith-metic progressions for any function can be decomposed into two pieces, one which exhibits a generalized weak mixing property and another that exhibits a generalized rigidity property. One of the difficulties lies in proving a structure theorem for the latter situation, showing that 3The system (X, X, µ, T) is weak mixing if for all A, B ∈X, lim N→∞ 1 N N−1 X n=0 µ T −nA ∩B −µ(A)µ(B) = 0 .
THE GREEN-TAO THEOREM 7 this portion of the system can be reduced to a finite series of com-pact extensions of a one point system (a Furstenberg tower) and then proving a recurrence statement for this tower.
4. Gowers norms in combinatorics In his proof of Szemer´ edi’s Theorem, Gowers defined certain norms, now referred to as Gowers (uniformity) norms, that capture behavior along arithmetic progressions. We start with a description of this key idea, explaining it in the combinatorial setup in this section and in the ergodic version in the next section. To define the norms, we introduce some notation.
For a positive integer N, let ZN := Z/NZ.
If f : ZN →C is a function, let E (f(x) | x ∈ZN) denote the average value of f on ZN: E (f(x) | x ∈ZN) = 1 N X x∈ZN f(x) .
We also use a higher dimensional version of the expectation. For ex-ample, by E(f(x, y) | x, y ∈ZN), we mean iteration of the one variable expectation: E E(f(x, y) | x ∈ZN) | y ∈ZN .
Taking f to be the indicator function of a set E whose average on ZN is at least δ, in this new terminology Szemer´ edi’s Theorem becomes: Theorem 4.1 (Reformulated Szemer´ edi). Let 0 < δ ≤1 be a real number and let k ≥2 be an integer.
If N is sufficiently large and f : ZN →R is a function with 0 ≤f(x) ≤1 for all x ∈ZN and E (f(x) | x ∈ZN) ≥δ, then (4.1) E f(x)f(x + r) . . . f(x + (k −1)r) | x, r ∈ZN ≥c(k, δ) for some constant c(k, δ) > 0 which does not depend either on f or on N.
At first glance, this appears to be a stronger version than the original statement of Szemer´ edi’s Theorem, showing not only the existence of a single arithmetic progression but of some positive multiple of N 2 pro-gressions. However, using some combinatorial trickery one can quickly show that the two versions are equivalent.
The average of Equation 4.1 is similar to the average in Equation 3.1, with the former being over ZN and the latter over Z. These averages along arithmetic progressions are controlled by certain norms and we now make this idea precise. The definition of the norms is motivated by a variation on the classic van der Corput difference theorem: 8 BRYNA KRA Lemma 4.2 (van der Corput Lemma for ZN). If f : ZN →C is a function, then |E(f(x) | x ∈ZN)|2 = E(f(x)f(x + h) | x, h ∈ZN) .
Since each of these expectations is a finite sum, the proof of the lemma is immediate by expanding both sides and using a change of variable.
The dth-Gowers (uniformity) norm ∥f∥Ud of a function f : ZN →C is defined inductively. Set ∥f∥U1 := |E(f(x) | x ∈ZN)| Thus for d = 1, ∥f∥Ud is only a seminorm.4 For d ≥2, we mimic successive uses of the van der Corput Lemma and define (4.2) ∥f∥Ud := E ffh 2d−1 Ud−1 | h ∈ZN 1/2d , where fh(x) = f(x + h). By definition, ∥f∥Ud is non-negative for d = 1 and therefore also for all higher d. Furthermore, Equation (4.2) shows that the dth-Gowers norm is shift invariant, meaning that ∥f(x)∥Ud = ∥f(x + h)∥Ud for any h ∈ZN. To justify the notation and terminology, we need to check that the Gowers norms are actually norms.
It follows immediately from the definitions and a change of variable that (4.3) ∥f∥U1 = E(f(x)f(x + h) | x, h ∈ZN) 1/2 .
The second Gowers norm can also be expressed in familiar terms. Using the Fourier expansion of f and computing, we have that ∥f∥U2 = X ξ∈ZN ˆ f(ξ) 4 !1/4 , where ˆ f denotes the Fourier transform of f. It follows that for d = 2, ∥f∥Ud is nondegenerate and so it is a norm.
For higher d, the situation is more complicated. To see that ∥f∥Ud is a norm for d ≥2, we give an equivalent characterization of the dth-Gowers norm as a certain average over a d-dimensional cube. This also allows us to express the definition of the norm in a closed form. We first need to introduce some more notation.
4A seminorm on a vector space V is a non-negative real valued function such that ∥f + g∥≤∥f∥+ ∥g∥and ∥cf∥= |c| · ∥f∥for all f, g ∈V and all scalars c.
Thus unlike a norm, one may have ∥f∥= 0 for some f ̸= 0.
THE GREEN-TAO THEOREM 9 We consider {0, 1}d as the set of vertices of the d-dimensional Eu-clidean cube, meaning it consists of points ω = (ω1, . . . , ωd) with ωj ∈{0, 1} for j = 1, . . . , d. For ω ∈{0, 1}d, define |ω| = ω1 + . . . + ωd and if ω ∈{0, 1}d and h = (h1, . . . , hd) ∈Zd N, we define ω · h := ω1h1 + . . . + ωdhd. Then if f : ZN →C is a complex valued function, it follows by inductively applying the definition in (4.2) that (4.4) ∥f∥Ud := E Y ω∈{0,1}d C|ω|f(x + ω · h) | x ∈ZN, h ∈Zd N 1/2d , where C is the conjugation operator Cf(x) := f(x). Thus the Gowers norms can be viewed as an average over the cube {0, 1}d.
By repeated applications of the Cauchy-Schwarz Inequality and the definitions of the norms, one obtains the Gowers Cauchy-Schwarz In-equality for 2d functions fω : ZN →C: (4.5) E Y ω∈{0,1}d C|ω|fω(x + ω · h) | x ∈ZN, h ∈Zd N ≤ Y ω∈{0,1}d ∥fω∥Ud .
From this, one can show that ∥f∥Ud is subadditive and so is a seminorm.
Furthermore, using the Gowers Cauchy-Schwarz Inequality, one has the chain of inequalities (4.6) ∥f∥U1 ≤∥f∥U2 ≤. . . ≤∥f∥L∞.
Since ∥f∥Ud is nondegenerate for d = 2, Inequality (4.6) implies that ∥f∥Ud is nondegenerate for all higher d, giving that the dth- Gowers norm is actually a norm for d ≥2.
We can also rewrite the Gowers norms in notation that is closer in spirit to the ergodic theoretic setup. Consider ZN endowed with the transformation T(x) = x + 1 mod N and the uniform measure m assigning weight 1/N to each element of ZN. Then the definition of Equation (4.4) becomes: (4.7) ∥f∥Ud = Z Y ω∈{0,1}d C|ω|f(T ω·hx) dm(x)dm(h1) . . . dm(hd) 1/2d .
These norms are used by Gowers (as well as by Host and Kra and more recently by Green and Tao and by Tao ) to control the average along arithmetic progressions, meaning the quantity in Equation (4.1). This type of control can be viewed as a generalized 10 BRYNA KRA version of the von Neumann Ergodic Theorem, which states that the average of a bounded function on a finite measure space converges in mean to its integral. We formalize this control, as stated by Tao : Theorem 4.3 (Generalized von Neumann Theorem ). Let k ≥2 be an integer, N be a prime number, and f0, . . . , fk−1 : ZN →C be functions with ∥f0∥∞, . . . , ∥fk−1∥infty ≤1. Then E f0(x)f1(x + n) . . . fk−1(x + (k −1)n) | x, n ∈ZN ≤ min 0≤j≤k−1 ∥fj∥Uk−1 .
The proof of this theorem is based on an induction argument, using the Cauchy-Schwarz Inequality and the van der Corput Lemma for ZN (Lemma 4.2).
To prove Szemer´ edi’s Theorem, Gowers studies the indicator function 1A of a set A ⊂ZN. As in Furstenberg’s proof, there are two distinct phenomena to consider.
If ∥1A −|A|/N∥Uk−1 is small, then a constant function is substituted for 1A and the average along arithmetic progressions is large. If ∥1A −|A|/N∥Uk−1 is large, then the restriction of 1A to some sufficiently large subset of ZN has many useful arithmetic properties and the average in Equation (4.1) is once again large. As in Furstenberg’s proof, a structure theorem is needed to analyze the second case: here the structure is a nested sequence of arithmetic progressions. The difficulty in this proof lies in showing that a usable version of the dichotomy between large and small always occurs.
5. Gowers (semi-)norms in ergodic theory Furstenberg’s proof of Theorem 3.1 left open the question of the existence of the limit in the left hand side of Equation (3.1). In , we show that this lim inf is actually a limit: Theorem 5.1 (Multiple Convergence ). Assume that (X, X, µ, T) is a measure preserving probability system, k ≥1 is an integer, and f1, f2, . . . , fk are bounded functions on X. Then the averages (5.1) 1 N N−1 X n=0 f1(T nx)f2(T 2nx) . . . fk(T knx) converge in L2(µ) as N →∞.
The existence of the limit for k = 1 is von Neumann’s ergodic theo-rem, existence for k = 2 was proven by Furstenberg , and for k = 3 was proven with a technical assumption by Conze and Lesigne and THE GREEN-TAO THEOREM 11 Furstenberg and Weiss , and in general by Host and Kra . More recently, Ziegler has an alternate approach for all k.
The first step in proving Theorem 5.1 is showing that instead of taking the average in the system (X, X, µ, T), it suffices to consider the average over some (ostensibly simpler) system (Y, Y, ν, S). This amounts to proving a generalized von Neumann Theorem, as in Gowers’ proof. This idea is implicit in Furstenberg’s proof of Szemer´ edi’s Theorem and made explicit in the proof of Theorem 5.1.
In , we introduced seminorms that generalize the Gowers norms; although the language is quite different, the form of the definition can be taken to closely resemble that of the Gowers norms. We consider a general probability measure preserving space (X, X, µ) with an invert-ible measurable, measure preserving transformation T : X →X on it.
For a function f ∈L∞(µ), we define (compare with Equation (4.2)) ∥f∥U1 := Z f(x) dµ(x) and inductively we define the dth-seminorm by (5.2) ∥f∥2d Ud := lim N→∞ 1 N N−1 X n=0 ∥fT nf∥2d−1 Ud−1 .
To recover the Gowers norms, we take the space ZN with the trans-formation x 7→x + 1 mod N and the uniform measure assigning each element of ZN weight 1/N.
Once again, there is an alternate presentation, analogous to that of Equation (4.7), as the integral with respect to a certain measure and this second presentation makes many properties of ∥f∥Ud more transparent. We need some notation to define this measure. (A reader not interested in the technical definition of this measure can omit this alternate presentation.) Assume that (X, X, µ) is a probability space.
If f ∈L1(µ) and Y ⊂X is a sub-σ-algebra, then the conditional expectation of f on Y is the Y-measurable function E(f | Y) such that Z A f dµ = Z A E(f | Y) dµ for all A ∈Y.
Fix an ergodic5 measure preserving probability system (X, X, µ, T).
Define X[d] = X2d and write points of X[d] as x = (xω : ω ∈{0, 1}d).
5The system (X, X, µ, T) is ergodic if the only sets A ∈X with µ(T −1A) = µ(A) have measure 0 or 1. Every system has a decomposition into ergodic components and so we can assume that the system being studied in Theorem 5.1 is ergodic.
12 BRYNA KRA Let T [d] = T × T × . . . × T taken 2d times. There is a natural identi-fication between X[d+1] and X[d] × X[d], with a point x ∈X[d+1] being identified with (x′, x′′) ∈X[d] × X[d], where x′ω = xω,0 and x′′ω = xω,1 for each ω ∈{0, 1}d.
For each integer d ≥0, we inductively define a T [d]-invariant measure µ[d] on X[d]. Define µ := µ. Assume that µ[d] is defined for some d ≥0. Let I[d] denote the T [d]-invariant σ-algebra of (X[d], µ[d], T [d]).
Using the natural identification of X[d+1] with X[d] × X[d], define the measure preserving (probability) system (X[d+1], µ[d+1], T [d+1]) to be the relatively independent joining of (X[d], µ[d], T [d]) with itself over I[d]: this means that the measure µ[d+1] is the measure such that for all bounded functions F ′ and F ′′ on X[d], we have Z X[d+1] F ′(x′)F ′′(x′′) dµd+1 = Z X[d] E(F ′ | I[d])E(F ′′ | I[d]) dµ[d] .
The measure µ[d+1] is invariant under T [d+1] and the two natural projections on X[d] are each µ[d]. Using induction, this gives that each of the 2d natural projections of µ[d] on X is equal to µ. Thus for a bounded function f on X, the integral Z X[d] Y ω∈{0,1}d C|ω|f(xω) dµd is real and nonnegative, where as before Cf(x) := f(x). An alternate definition of the seminorms is: (5.3) ∥f∥Ud = Z X[d] Y ω∈{0,1}d C|ω|f(xω) dµd 1/2d .
To show that these are seminorms, one proceeds in the same manner as in the combinatorial setup, deriving a version of the Cauchy-Schwarz Inequality (analogous to Equation (4.5)) and using it to show subad-ditivity.
Positivity follows immediately from definition (5.2).
From the definition of these measures and the Ergodic Theorem, we obtain that this second definition is equivalent to the first definition given in Equation (5.2).
The definition of Equation (5.3) can once again be viewed as an aver-age over the cube {0, 1}d. A convergence theorem for general averages along cubes is also proved in .
The first step in proving Theorem 5.1 is showing that the aver-ages along arithmetic progressions are once again controlled by the d-seminorms, meaning an analog of Theorem 4.3: THE GREEN-TAO THEOREM 13 Theorem 5.2 (Generalized von Neumann, revisited ). As-sume that (X, X, µ, T) is an ergodic measure preserving probability system. Let k ≥2 be an integer and assume that f1, . . . , fk are bounded functions on X with ∥f1∥∞, . . . ∥fk∥∞≤1. Then lim sup N→∞ 1 N N−1 X n=0 f1(T nx)f2(T 2nx) . . . fk(T knx) 2 ≤min 1≤j≤k(j∥fj∥Uk) .
The added factor of j which appears on the right hand side of this bound and not in Theorem 4.3 is due to the change in underlying space. In Theorem 4.3, we assumed that N is prime; in this case, for any integer j that is not a multiple of N, the map n 7→jn is onto in ZN, and this is not the case in Z. As for the earlier Generalized von Neumann Theorem, Theorem 5.2 is proved using induction, the Cauchy-Schwarz Inequality and a van der Corput lemma. This time we need a Hilbert space variation of this lemma: Lemma 5.3 (van der Corput Lemma, revisited ). Assume that H is a Hilbert space with inner product ⟨, ⟩and norm ∥· ∥, and that ξn, n ≥0, is a sequence in H with ∥ξn∥≤1 for all n. Then lim sup N→∞ 1 N N−1 X n=0 ξn 2 ≤lim sup H→∞ 1 H H−1 X h=0 lim sup N→∞ 1 N N−1 X n=0 ⟨ξn+h, ξn⟩ .
By Theorem 5.2, one can consider an average along arithmetic pro-gressions on an appropriate factor, rather than the whole space. We make this notion more precise.
For a measure preserving system (X, X, µ, T), the word factor is used with two different but equivalent meanings. First, it is a T-invariant σ-algebra of X. (Strictly speaking, this is a sub-σ-algebra, but through-out we omit the use of the word “sub”.) Secondly, if (Y, Y, ν, S) is a measure preserving system, a map π: X →Y is a factor map if π maps µ to ν and S ◦π = π ◦T. Then Y is said to be a factor of X and the two definitions coincide up to the identification of Y with π−1(Y). For f ∈L1(µ), we view E(f | Y) as a function on X and let E(f | Y ) denote the function on Y defined by E(f | Y )◦π = E(f | Y). It is characterized by Z Y E(f | Y )(y) · g(y) dν(y) = Z X f(x) · g(π(x)) dµ(x) for all g ∈L∞(µ).
The seminorms are used to define factors of the system (X, X, µ, T).
One presentation of these factors is by defining their orthogonal com-plements: for d ≥1, define Zd−1 to be the σ-algebra of X such that for 14 BRYNA KRA f ∈L∞(µ): ∥f∥Ud = 0 if and only if E(f | Zd−1) = 0 .
Thus a bounded function f is measurable with respect to Zd−1 if and only if R fgdµ = 0 for all functions g ∈L∞(µ) with ∥g∥Ud−1 = 0. This motivates an equivalent definition of the factors Zd with respect to a dual norm. Namely, defining the dual norm ∥f∥(Ud)∗by (5.4) ∥f∥(Ud)∗:= sup g∈L∞(µ) Z X fg dµ: ∥g∥Ud ≤1 , we have that the space of functions with finite (U d)∗norms is a dense subset (in L2) of the bounded functions that are measurable with re-spect to Zd−1.
Letting Zj denote the factor associated to the σ-algebra Zj, we have that Z0 is the trivial factor and Z1 is the Kronecker factor, meaning the σ-algebra which is spanned by the eigenfunctions of T. Furthermore, the sequence of factors is increasing (compare with Equation (4.6)): Z0 ←Z1 ←Z2 ←. . . ←X and if T is weakly mixing, then Zd is the trivial factor for all d.
Theorem 5.2 states that the factor Zd−1 is a characteristic factor for the average of Equation (5.1), meaning that the limit behavior of the averages in L2(µ) remains unchanged when each function is replaced by its conditional expectation on this factor. Thus it suffices to prove convergence when one of the factors Zd is substituted for the original system. For a progression of length k, this amounts to decomposing a bounded function f = g+h with g = f −E(f | Zk−1). The function g is the uniform component and has zero k−1 seminorm and so contributes zero to the average along arithmetic progressions. The second term h is the anti-uniform component and belongs to the algebra of functions measurable with respect to the factor Zk−1 and must be analyzed via a structure theorem for the characteristic factors. This decomposition of an arbitrary bounded function into uniform and anti-uniform compo-nents is unique. In the combinatorial setting, a similar decomposition (see Section 6) can only be carried out approximately. Ergodic theory is more precise than combinatorics in describing the second component of this decomposition.
When the description of a characteristic factor is “simple”, one has a better chance of proving convergence in this factor. For the given de-composition, the description of the characteristic factor is as an inverse THE GREEN-TAO THEOREM 15 limit of nilsystems, meaning that it can be approximated arbitrarily well by a rotation on a homogeneous space of a nilpotent Lie group.6 6. Quantitative ergodic theory Tao gave a new proof of Szemer´ edi’s Theorem, along the lines of Furstenberg’s original proof, but proving it in the finite system ZN rather than for an arbitrary measure space. This allows him to ex-tract explicit bounds for N0(δ, k) in the finite version (Theorem 2.1), although the bounds are nowhere near as good as those obtained by Gowers .
Once again, a generalized von Neumann Theorem (analogous to The-orems 4.3 and 5.2) is used to start the proof. Then, as in the ergodic setup, an arbitrary bounded function f on ZN is decomposed into pieces, each of which can be analyzed. This time the decomposition is into a term with small Gowers norm and a structured component, with the wrinkle that one also has to deal with a small error term.
The first term corresponds to a uniform component f −E(f | Z) for a well chosen σ-algebra Z (similar to the use of a characteristic factor in the ergodic setup) which has small Gowers norm and makes a small contribution to the average in Equation (4.1). Since the space being used is ZN, the σ-algebra Z is nothing more than a finite partition of ZN: elements of a σ-algebra are unions of elements (also called atoms) of some partition of ZN and a function is measurable with respect to this σ-algebra if it is constant on each element of the partition. The second term is the conditional expectation of f relative to Z, meaning that it is the function measurable with respect to Z defined by E(f|Z)(x) = E(1A(x)f(x) | x ∈ZN) E(1A(x) | x ∈ZN) , where A is the atom of Z containing x. This component is analyzed using a form of recurrence similar to that needed for a Furstenberg tower.
The second component of the decomposition, called the anti-uniform functions by Tao, is essentially dual to the uniform component where 6If G is a k-step nilpotent Lie group and Γ is a discrete cocompact subgroup, then a ∈G naturally acts on G/Γ by left translation by Ta(xΓ) = (ax)Γ. The Haar measure µ is the unique Borel probability measure on G/Γ that is invariant under this action of G by left translations. For a fixed element a ∈G, the system (G/Γ, G/Γ, Ta, µ) is a k-step nilsystem. The structure theorem in states that the factor Zk−1, which is a characteristic factor for the average along an arithmetic progressions of length k, is an inverse limit of such (k −1)-step nilsystems.
16 BRYNA KRA the anti-uniform (dual) norm ∥g∥(Ud)∗is defined by (compare with Equation (5.4)) ∥g∥(Ud)∗:= sup f : ZN→C {|⟨f, g⟩| : ∥f∥Ud ≤1} .
The contribution of this term to the average is bounded from below by van der Waerden’s Theorem,7 with the idea being that these functions lie in a sufficiently compact space so that a finite coloring argument can be used. Applying this idea to a function with positive expectation, the average along arithmetic progressions is positive.
This proof follows Furstenberg’s proof closely. One advantage is the elimination of the compactness argument, leading to explicit bounds on the size of the set needed to guarantee the existence of a progression of length k. The structure theorem corresponds to the tower of compact extensions used by Furstenberg and does not need an understanding of the precise structure of the chosen σ-algebra, such as the nilsystems in the structure theorem of Host and Kra. However, a more precise understanding of this structure should clarify the apparent link between the anti-uniform functions of level k appearing in Tao’s proof and the k-step nilsystems used to prove Theorem 5.1.
7. Arithmetic progressions in the primes Green and Tao continue in this vein to prove the existence of arith-metic progressions in the primes. The starting point is clear: study the averages of Equation (4.1) for the indicator function of the primes.
However, since the primes have density 0, any function that is 0 other than on the primes can not be bounded without its average on ZN be-coming arbitrarily small as N tends to infinity. Since such a function cannot be bounded independently of N, Szemer´ edi’s Theorem can not be applied directly.
Instead, Green and Tao begin with the closely related von Mangoldt function Λ(n), where Λ(n) = ( log p if n = pm for some m ∈N and a prime p 0 otherwise , and make use of the fact that this function is more natural analytically than Λ.
Although the von Mangoldt function is supported on the primes and their powers, the powers are sparsely enough distributed so 7Van der Waerden’s Theorem states that if the integers are partitioned into finitely many pieces, then one of these pieces contains arbitrarily long arith-metic progressions.
This theorem motivated Erd¨ os and Tur´ an to conjecture Szemer´ edi’s Theorem.
THE GREEN-TAO THEOREM 17 that they only contribute a small error term in the calculations. This function has had many uses in number theory; for example, the unique factorization theorem for integers is equivalent to the statement log n = X d|n Λ(d) for all positive integers n , and the Prime Number Theorem is equivalent to the statement 1 N X 1≤n≤N Λ(n) = 1 + o(1) .
(Throughout, by o(1), we mean a quantity that tends to 0 as N →∞, and when this quantity depends on other constants, we include them as subscripts on o.) The function Λ mostly avoids giving weight to arithmetic progres-sions congruent to a mod q when a and q are not relatively prime. Such arithmetic progressions are more dense when q has many small prime factors, making Λ too irregularly distributed for their purposes. There-fore Green and Tao are forced to modify Λ, quotienting out the small primes and make it more evenly distributed over all congruence classes.
They then majorize the modified function by something pseudorandom and much of the work is carried out for pseudorandom functions. The precise definition and modification is given below, but the idea is that the values of a pseudorandom function should be distributed so that using any statistic to measure the values, one gets approximately the same measurement as that arising from a random set of the same den-sity.
The goal then becomes to extend Szemer´ edi’s Theorem, showing that not only does a dense subset of the integers contain arbitrarily long arithmetic progressions, but a dense subset of a pseudorandom collec-tion of integers also contains arbitrarily long arithmetic progressions.
Green and Tao do this by “transferring” Szemer´ edi’s Theorem to a more general setting: the hypothesis in Theorem 4.1 that f : ZN →R satisfies 0 ≤f(x) ≤1 is replaced by f being bounded by a more general function ν : ZN →R+ with certain useful properties. (More precisely, for each N ∈N we have a function ν = νN : ZN →R+.) The function ν : ZN →R+ is assumed to be a measure,8 meaning that E(ν(x) | x ∈ZN) = 1+o(1), and ν is also assumed to be pseudorandom.
They show: 8As noted by Green and Tao, the name measure is a misnomer, and ν should more accurately be called a density relative to the uniform measure on ZN.
18 BRYNA KRA Theorem 7.1 (Transference Theorem ). Let 0 < δ ≤1 be a real number and let k ≥2 be an integer. If N is sufficiently large, ν : ZN → R+ is a k-pseudorandom measure, and f : ZN →R is function with 0 ≤f(x) ≤ν(x) for all x ∈ZN and E(f(x) | x ∈ZN) ≥δ, then (7.1) E f(x)f(x+r) . . . f(x+(k−1)r) | x, r ∈ZN ≥c(k, δ)−ok,δ(1) , where the constant c(k, δ) is the same as that in Theorem 4.1.
Other than the bounds on f, the only additional modification caused by bounding f by a pseudorandom measure instead of the constant function 1 is the introduction of the error term ok,δ(1), which tends to 0 as N →∞. The dependence of this error is only on k and δ.
Before giving an indication of the proof of Theorem 7.1, we make the notion of a pseudorandom measure more precise. (A reader not interested in the technical details can skip the next few paragraphs.) The measure ν : ZN →R+ is said to be k-pseudorandom if ν satisfies a k-linear forms condition and a k-correlation condition.
To define the linear forms condition, fix k, the length of the arith-metic progression and assume that N is prime and larger than k. As-sume that we have m linear forms ψi, 1 ≤i ≤m, with m ≤k · 2k−1 and t variables with t ≤3k −4. (The exact values of these constants are not important for the proof; the importance lies in showing that a particular choice of a pseudorandom function satisfies these conditions.
For this, it only matters that the values depend on nothing but k.) Let L = (Lij) be an m × t matrix, whose entries are rational numbers with numerator and denominator bounded in absolute value by k. By choice of N, we can view the entries of L as elements of ZN (recall that N is prime). Assume further that each of the t columns of L are not identically zero and that the columns are pairwise independent. Let ψi(x) = bi + Pt j=1 Lijxj denote the m linear forms, where x ∈Zt N and bi ∈ZN for 1 ≤i ≤m. The measure ν : ZN →R+ is said to satisfy the (m, t, L)-linear forms condition if E ν (ψ1(x)) . . . ν(ψm(x)) | x ∈Zt N = 1 + om,t,L(1) , where the dependence on N is assumed to be uniform in the choice of the linear forms ψi and so in particular uniform in the choice of the bi. The case m = 1 with ψ(x) = x corresponds to the measure ν with E(ν) = 1 + o(1), and this is the bound used in the Reformulated Szemer´ edi Theorem (Theorem 4.1). For higher m, the values of the measure ν evaluated on linear forms up to a certain complexity are, on average, independent. If there were no restriction on the complexity, the measure would be close to the ergodic theoretic notion of weak THE GREEN-TAO THEOREM 19 mixing, meaning its values along any distinct linear forms would be, on average, independent.
We now define the correlation condition. The measure ν : ZN →R+ is said to satisfy a 2k−1-correlation condition if for each m with 1 ≤ m ≤2k−1, there exists a weight function τ = τm : ZN →R+ with E(τ q | z ∈ZN) ≤C(m, q) for a constant C(m, q), for all 1 ≤q < ∞, and that E ν(x + h1)ν(x + h2) . . . ν(x + hm) | x ∈ZN ≤ X 1≤i k be a prime number, and assume that ν : ZN →R+ is a k-pseudorandom measure. Then ∥ν −1∥Ud = o(1) for all 1 ≤d ≤k −1.
The broad outline of the proof of Theorem 7.1 is similar to that of Tao’s proof of Szemer´ edi’s Theorem sketched in the last section, but both the technical details and the combination of ideas from seemingly unrelated areas of mathematics make it a significantly more ambitious undertaking. The innovation is the reduction of Theorem 7.1 to Sze-mer´ edi’s Theorem. The key argument, again, is a structure theorem, 20 BRYNA KRA but this time not only is there an error term in the decomposition, but the decomposition is only valid on most of the space. Green and Tao show: Theorem 7.3 (Decomposition Theorem ). Let k ≥2 be an integer, let 0 < ϵ ≪1 be a small parameter, and let N = N(ϵ) be sufficiently large.
Assume that ν : ZN →R+ is a k-pseudorandom measure and that f ∈L1(ZN) is a function satisfying 0 ≤f(x) ≤ν(x) for all x ∈ZN. Then there exists a σ-algebra Z and an exceptional set Ω∈Z with E(ν(x)1Ω(x) | x ∈ZN) = oϵ(1) such that ∥1ΩCE(ν −1 | Z)∥L∞= oϵ(1) and ∥1ΩC(f −E(f | Z))∥Uk−1 ≤ϵ1/2k , where ΩC denotes the complement of Ω.
This means that outside a small subset Ωof ZN, a function f that is bounded by a pseudorandom measure can be decomposed into a sum of a uniform function g and an anti-uniform function h, plus a small error term. The function g has small Gowers norm and corresponds to f −E(f | Z) in the ergodic theoretic setup, while the non-negative function h is bounded and corresponds to E(f | Z). Other than the er-ror terms, this parallels the ergodic theoretic decomposition associated to a characteristic factor described in Section 5 and the decomposition used by Tao described in Section 6.
The proof of the Decomposition Theorem follows a complicated it-erative procedure, designed to produce the σ-algebra Z. Starting with the trivial σ-algebra B = {∅, ZN}, if the function f −E(f|B) has small U k−1 norm, the algorithm terminates. If not, an appropriate addition is made to the σ-algebra B, taking care to increase E(f|B), yet keep it uniformly bounded. This process is repeated until f −E(f|B) is sufficiently small and the algorithm terminates.
The next ingredient in the proof of Theorem 7.1 is analogous to the generalized von Neumann Theorem; it gives a way to control the contri-bution of the Gowers uniform portion in the decomposition, meaning a way to bound the contribution coming from a function with small Gow-ers norm. Once again, the bound on the functions changes: instead of being bounded by the constant 1, the functions are now bounded point-wise by 1 plus a pseudorandom measure.
Theorem 7.4 (Pseudorandom Generalized von Neumann The-orem ). Let k ≥2 be an integer, let N be a prime number, and THE GREEN-TAO THEOREM 21 assume that ν : ZN →R+ is a k-pseudorandom measure. Assume that f0, . . . , fk−1 ∈L1(ZN) are functions such that |fj(x)| ≤ν(x) + 1 for all x ∈ZN, 0 ≤j ≤k −1 .
Then E f0(x)f1(x + n) . . . fk−1(x + (k −1)n) | x, n ∈ZN = O min 0≤j≤k−1 ∥fj∥Uk−1 + o(1) .
We are now ready to outline the proof of Theorem 7.1, still gloss-ing over many technical details. We fix a function f that is bounded by a pseudorandom measure and that has positive expectation on ZN.
Using the Decomposition Theorem, the expectation on the left hand side of Equation (7.1), which is the average along arithmetic progres-sions, is larger than the same expectation with 1ΩCf substituted for f, where Ωis some small set. Ignoring the error term, we now use the decomposition of this new function into g + h, where g is the Gowers uniform portion and h is some bounded function. Much like the idea of a characteristic factor in ergodic theory, we now want to discard the Gowers uniform portion g and replace our function by h. Making the substitution g + h for 1ΩCf, the expectation on the left hand side of Equation (7.1) can be expanded as a sum of 2k expectations of the form E(f0(x)f1(x + n) . . . fk−1(x + (k −1)n | x, n ∈ZN) , where each fi is equal either to g or to h. All terms but one contain an occurrence of g in it and each term containing a g is small by the Pseudorandom Generalized von Neumann Theorem. We are left only with a single term making a large contribution to the expectation, which is the only term with nothing but occurrences of the function h.
The good news is that now this function h is bounded and so the usual Szemer´ edi Theorem applies. Furthermore, f and h have approximately the same expectation, and in particular the expectation on ZN of h is also positive.
Thus by Szemer´ edi’s Theorem, the expectation in Equation (7.1) with f replaced by h is positive. Therefore, the same result holds for f.
Lastly we give an indication of the choice of the function f and measure ν needed to use the Transference Theorem for the primes.
The function f is a variation on the von Mangoldt function, cut off at a certain point, in order to make a function that is (vaguely speak-ing) supported on primes of magnitude log N. Unfortunately, it does not suffice to simply use a multiple of this function for ν, since as we 22 BRYNA KRA noted earlier, the primes, and therefore any multiple of the von Man-goldt function, are not uniformly distributed across all residue classes, whereas a pseudorandom function is. Instead, the measure ν is taken to have its support (again, vaguely speaking) on numbers n such that all the prime factors of n −1 are greater than some integer R. One can view this measure as approximately log R times the characteristic function of such numbers.
In the third century B.C., the scholar Eratosthenes came up with a simple algorithm for listing all the prime numbers up to a given N, referred to as the sieve of Eratosthenes. Given a list of the numbers between 1 and N, starting with 2, erase all multiples of 2 up to N, other thatn 2 itself.
Call the remaining set P2.
Returning to the beginning, take the first number greater than 2 and erase all of its multiples up to N, again other than the number itself. In general, the level R almost primes PR(N) are defined to be the set of all numbers between 1 and N that contain no nontrivial factors less than or equal to R. Thus if R = √ N, we have that P√ N(N) consists exactly of the prime numbers up to N. Mertens proved that the size |PR(N)| is approximately cN/ log R for some positive constant c. Combining this with the estimate from the Prime Number Theorem that the number of primes up to N is approximately N/ log N, we have that the density of primes in the almost primes PR(N) is about a multiple of log R/ log N.
Therefore if R is a small power of N, then the primes have positive density in the level R almost primes. This motivates the function and measure Green and Tao use. For completeness, we give the technical definitions.
Let W be the product of the primes up to some ω = ω(N), where ω(N) tends to infinity sufficiently slowly so that ˜ Λ has mean one (ω(N) = log log N suffices).
Taking place of the indicator function of the primes is the modified von Mangoldt function ˜ Λ, defined by ˜ Λ(n) = ( φ(W) W log(Wn + 1) when Wn + 1 is prime 0 otherwise .
Thus the function ˜ Λ is supported on the set QW = {n ∈Z: Wn + 1 is prime} .
It suffices to find an arithmetic progression in QW, since if {x, x + n, . . . , x + (k −1)n} is an arithmetic progression in QW, then {Wx + 1, W(x + n) + 1, . . . , W(x + (k −1)n) + 1} THE GREEN-TAO THEOREM 23 is an arithmetic progression in the primes with common difference Wn.
This modification is needed because the primes bounded by x are not uniformly spread out in arithmetic progressions. For example, there is only one prime congruent to 0 mod 2, while there are approximately x/ log x congruent to 1 mod 2. Furthermore, if a and q are relatively prime integers, the number of primes in the arithmetic progression a mod q up to x is approximately x log x · 1 φ(q). If one considers integers n with n ≡a mod q and for which Wn+1 is prime, then there are none only when q and Wa + 1 are not relatively prime and this can only happen when q and W are relatively prime. This means that q has no small prime factors and the set QW is more uniformly distributed among the arithmetic progressions.
Green and Tao do not show directly that ˜ Λ is k-pseudorandom, but instead majorize it by a measure ν whose values are more uniformly distributed and then are able to show that ν is k-pseudorandom. Be-fore defining the measure ν, we need one more variation on the von Mangoldt function. The truncated von Mangoldt function is defined to be ΛR(n) = X d|n,d≤R µ(d) log(R/d) , where µ is the M¨ obius function.9 The restriction d ≤R is needed to guarantee that the log remains positive. This is a cut offversion of the von Mangoldt function, since if R > n then ΛR(n) = Λ(n). This useful approximation to Λ(n) has been widely used in analytic number theory, notably by Selberg to give strong upper bounds for the number of primes predicted by an application of the Hardy-Littlewood k-tuple Conjecture. More recently, Goldston and Yildirim use it in their work on gaps in the primes.
Fix R = N k−12−k−4. (One can think of this choice of R as N ϵ for some ϵ < 1.) The measure ν : ZN →R+ is defined for 0 ≤n < N to be ν(n) = ( φ(W) W (ΛR(Wn+1))2 log R for N/(2k(k + 4)!) ≤n ≤2N/(2k(k + 4)!) 1 otherwise , 9The M¨ obius function µ(n) is defined by µ(1) = 1, µ(n) = 0 if n is not a square free integer and µ(n) = (−1)r if n is a square free integer and has r distinct prime factors.
24 BRYNA KRA where {0, 1, . . . , N−1} is naturally identified with ZN and φ denotes the Euler totient function.10 We make a few comments on this definition.
The bounds on n are designed to avoid counting progressions that are not progressions Z because they wrap around 0 in ZN, and the factors φ(w)/w guarantee that one has the correct average as N →∞. Since we are interested in majorizing ˜ Λ, the function ΛR is evaluated at Wn + 1.
One can quickly verify that this choice of ν majorizes the modified von Mangoldt function ˜ Λ.
The last major step is verifying that ν is k-pseudorandom. This relies on techniques from analytic number theory, using and extending recent results of Goldston and Yildirim on finding small gaps between primes. An alternate approach to this portion of the proof is given by Tao in , using only elementary properties of the primes and basic properties of the Riemann ζ function.
8. Further directions At this time, Green and Tao’s Theorem seems out of the reach of ergodic theory.
All combinatorial number theorems that have been proved using ergodic theory rely on some variant of Furstenberg’s Cor-respondence Principle, which only applies to sets of integers with posi-tive upper density. However, the many similarities between Green and Tao’s approach and proofs in ergodic theory suggest that a connection exists. The ultimate goal would be to use translations of the proof techniques of Green and Tao to obtain new convergence results in er-godic theory; in particular one may be able to use ergodic theory to show the existence of some patterns in certain subsets of density zero.
Tao’s proof of Szemer´ edi’s Theorem removes the compactness argument needed in Furstenberg’s proof and replaces it by a lengthy induction. This induction only needs finitely many steps, but the num-ber of steps is not explicitly known. A better understanding of the structure theorem used would probably improve the bounds that Tao extracts with this method. It seems that finding the exact link between the anti-uniform functions of level k and the k-step nilsystems found in the work of Host and Kra would clarify the connections between the two fields and probably lead to new and interesting developments.
A natural question arises from these considerations. Bergelson and Leibman used ergodic theory to establish a polynomial Szemer´ edi 10The Euler totient function φ(n) is defined to be the number of positive integers less than or equal to n that are relatively prime to n, with 1 being counted as relatively prime to all numbers.
THE GREEN-TAO THEOREM 25 type theorem 11 and perhaps it is possible to carry out a similar pro-gram to that of Green and Tao for this situation. Namely, prove a transference polynomial Szemer´ edi Theorem and show that not only do subsets of the integers with positive upper density contain poly-nomial patterns, but also dense subsets of pseudorandom sets contain polynomial patterns. This would prove, for example, that there exist infinitely many triples (p, k, n) of integers with p, n ≥1 and k > 1 such that p, p+n, p+n2, . . . , p+nk consists only of prime numbers. Finding polynomial patterns in the primes seems to have the added difficulty of lifting the result from ZN to Z.
One can also hope to use some variation of Green and Tao’s method to study the existence of other patterns in the primes, such as pairs of primes p, p + 2 (the twin prime conjecture) or pairs of primes p, 2p + 1.
Closer in spirit to arithmetic progressions, one might look for infinitely many pairs (p, d) with p a prime and d a positive integer such that p, p + 2d, p + 4d, 2p + d are all prime.
References A. Balog. The prime k-tuplets conjecture on average. Analytic number the-ory (Allerton Parl, IL., 1989), 47–75, Progr. Math., 85, Birkh¨ auser Boston, 1990.
A. Balog. Linear equations in primes. Mathematika, 39 (1992), 367–378.
V. Bergelson. Weakly mixing PET. Erg. Th. & Dyn. Sys., 7 (1987), 337– 349.
V. Bergelson and A. Leibman. Polynomial extensions of van der Waerden’s and Szemer´ edi’s theorems. J. Amer. Math. Soc., 9 (1996), 725–753.
J. -P. Conze and E. Lesigne. Sur un th´ eor` eme ergodique pour des mesures diagonales. C. R. Acad. Sci. Paris, S´ erie I, 306 (1988), 491–493.
P. Erd¨ os and P. Tur´ an. On some sequences of integers. J. Lond. Math. Soc., 11 (1936), 261-264.
11More precisely, Bergelson and Leibman’s Theorem generalizes Furstenberg’s Multiple Ergodic Theorem (Theorem 3.1). They show that if (X, X, µ, T) is an invertible measure preserving probability system, k ≥1 is an integer, p1, . . . , pk are polynomials taking integer values on the integers with p1(0) = . . . = pk(0) = 0, and A ∈X with µ(A) > 0, then lim inf N→∞ 1 N N−1 X n=0 µ T p1(n)A ∩. . . ∩T pk(n)A > 0 .
As for arithmetic progressions, the limits of the related averages 1 N N−1 X n=0 f1(T p1(n)x) . . . fk(T pk(n)x) for bounded functions f1, . . . , fk are known to exist in L2(µ) (see and ).
26 BRYNA KRA M. Frind, P. Jobling and P. Underwood. 23 primes in arithmetic progres-sion. Available at H. Furstenberg. Ergodic behavior of diagonal measures and a theorem of Szemer´ edi on arithmetic progressions. J. d’Analyse Math., 31 (1977), 204– 256.
H. Furstenberg and Y. Katznelson. An ergodic Szemer´ edi theorem for com-muting transformations. J. d’Analyse Math., 34 (1979), 275–291.
H. Furstenberg and Y. Katznelson. A density version of the Hales-Jewett Theorem. J. d’Analyse Math., 57 (1991), 64–119.
H.
Furstenberg and B.
Weiss.
A mean ergodic theorem for 1 N Pn n=1 f(T nx)g(T n2x). In Convergence in Ergodic Theory and Probabil-ity, Walter de Gruyter & Co, New York, 1996, 193–227.
T. Gowers. A new proof of Szemer´ edi’s Theorem. GAFA, 11 (2001), 465– 588.
A. Granville. Personal communication.
B. Green. Roth’s Theorem in the primes. To appear, Ann. Math.
B. Green and T. Tao. The primes contain arbitrarily long arithmetic pro-gressions. To appear, Ann. Math.
B. Green and T. Tao. A bound for progressions of length k in the primes.
Preprint.
D. Goldston and C. Y. Yildirim. Small gaps between primes, I. Preprint.
G. H. Hardy and J. E. Littlewood. Some problems of “partitio numerorum” III: on the expression of a number as a sum of primes. Acta Math., 44 (1923), 1–70.
D. R. Heath-Brown. Three primes and an almost prime in arithmetic pro-gression. J. Lond. Math. Soc. (2), 23 (1981), 396–414.
B. Host and B. Kra. Convergence of Conze-Lesigne Averages. Erg. Th. & Dyn. Sys., 21 (2001), 493–509.
B. Host and B. Kra. Nonconventional ergodic averages and nilmanifolds.
Ann. Math.161 (2005), 397–488.
B. Host and B. Kra. Convergence of polynomial ergodic averages. Isr. J.
Math.149 (2005), 1–19.
A. Leibman. Convergence of multiple ergodic averages along polynomials of several variables. Isr. J. Math.146, 303–316.
F. Mertens. Ein Beitrag zur analytischen Zahlentheorie. Journal f¨ ur Math., 78 (1874), 46–62.
A. Moran, P. Pritchard and A. Thyssen. Twenty-two primes in arithmetic progression. Math. Comp., 64 (1995), 1337–1339.
A. Selberg. The general sieve method and its place in prime number theory.
Proc. ICM, vol 1, Cambridge (1950), 286–292.
E. Szemer´ edi. On sets of integers containing no k elements in arithmetic progression. Acta Arith., 27 (1975), 299-345.
T. Tao. A quantitative ergodic theory proof of Szemer´ edi’s theorem.
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T. Tao. A remark on Goldston-Yildirim correlation estimates. Preprint.
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van der Corput.
¨ Uber Summen von Primzahlen und Primzahlquadraten. Math. Ann., 116 (1939), 1–50.
THE GREEN-TAO THEOREM 27 B. L. van der Waerden. Beweis einer Baudetschen Vermutung. Nieuw Arch.
Wisk., 15 (1927), 212–216.
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Department of Mathematics, Northwestern University, 2033 Sheri-dan Road, Evanston, IL 60208-2730 E-mail address: kra@math.northwestern.edu
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Scientific Notation Probl...
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What are the common difficulties students face when learning about scientific notation?
Marc W.
Programmer (Of FORTRAN, COBOL, PL/I, BAL, & More) (1975–present)
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Author has 7.1K answers and 4.6M answer views
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5y
1: Converting a number to Scientific Notation:
Some examples:
6020000 is that 6.02x10^6 or 6.02x10^7 or .602x10^8 or something else?
0.0000206 is that 206x10^-4 or 2.06 x 10^-4 or 2.06 x10^-5
2: Conversion from Scientific Notation:
What is 5.137x10^-3?
What about 6.136x10^4?
3: Using and keeping track of significant figures:
For example:
6.037x10^0 x 1.87325x10^5 = 1.13033770 x 10^6
(4 sig fig ……. 6 sig fig …………. 8 sig fig)
However since we only have at LEAST 4 significant figures the answer actually is
Gilbert Doan
Master (Unattempted) in Mathematics, San Jose State University (Graduated 2018)
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Author has 29.5K answers and 16.7M answer views
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5y
I do not think scientific notation is going to be hard for someone who is good in math. The students are going to have the hardest and most rewarding years in building their basic theory of science or general science. That will allow them to reason about the structures, objects, and relationships under the theory, which are evaluated based upon observation and experiment. You want the students to be very good at the scientific units, and working with units of measurement in all their data. That gives them a lot more precision and fundamental units to work with, which allow them to analyze and
I do not think scientific notation is going to be hard for someone who is good in math. The students are going to have the hardest and most rewarding years in building their basic theory of science or general science. That will allow them to reason about the structures, objects, and relationships under the theory, which are evaluated based upon observation and experiment. You want the students to be very good at the scientific units, and working with units of measurement in all their data. That gives them a lot more precision and fundamental units to work with, which allow them to analyze and identify the properties of something.
Assistant
Bot
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1y
Students often encounter several difficulties when learning about scientific notation. Here are some common challenges:
Understanding the Concept
:
Students may struggle to grasp why scientific notation is used and how it simplifies working with very large or very small numbers.
Conversion Between Forms
:
Converting numbers from standard form to scientific notation and vice versa can be confusing. Students may forget the rules for moving the decimal point or adjusting the exponent.
Exponents
:
The concept of exponents can be challenging, especially when students need to understand po
Students often encounter several difficulties when learning about scientific notation. Here are some common challenges:
Understanding the Concept
:
Students may struggle to grasp why scientific notation is used and how it simplifies working with very large or very small numbers.
Conversion Between Forms
:
Converting numbers from standard form to scientific notation and vice versa can be confusing. Students may forget the rules for moving the decimal point or adjusting the exponent.
Exponents
:
The concept of exponents can be challenging, especially when students need to understand positive and negative exponents, as well as how to perform arithmetic operations with them.
Multiplication and Division
:
Students often find it difficult to multiply or divide numbers in scientific notation, particularly when it comes to handling the coefficients and exponents separately.
Precision and Significant Figures
:
Understanding how to maintain the correct number of significant figures during conversions and calculations can lead to mistakes.
Application in Context
:
Applying scientific notation in real-world contexts (like scientific measurements) can be daunting, as students may struggle to see its relevance.
Misunderstanding Place Value
:
Students may confuse the place value of digits in scientific notation, particularly when transitioning from standard notation.
Practice and Familiarity
:
Lack of practice can lead to discomfort with scientific notation, making it harder for students to feel confident in using it.
Addressing these difficulties typically involves providing clear explanations, using visual aids, offering plenty of practice problems, and contextualizing scientific notation in real-world scenarios.
Related questions
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What are effective ways to teach scientific notation to students who struggle with it?
What are the applications of scientific notation?
How do I write 0.024 in scientific notation?
Jonathan E. Segal
Author has 436 answers and 328.5K answer views
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3y
Related
What are the biggest challenges that most will encounter learning math?
In California, there’s a law called AB705 at the “community” college level. By Fall 2022, there will be no more remedial Math classes offered. At my school, there will be a transfer level Algebra class. We must use a College Algebra Textbook. That will be the lowest level. People are going to be thrown into classes like Statistics who many not be ready for a class like that. If students are not prepared to take a transfer level class, the “Community” College has now shut the door on them. That’ll be a huge challenge in learning Math for some students.
For Math Majors, there is a class called “I
In California, there’s a law called AB705 at the “community” college level. By Fall 2022, there will be no more remedial Math classes offered. At my school, there will be a transfer level Algebra class. We must use a College Algebra Textbook. That will be the lowest level. People are going to be thrown into classes like Statistics who many not be ready for a class like that. If students are not prepared to take a transfer level class, the “Community” College has now shut the door on them. That’ll be a huge challenge in learning Math for some students.
For Math Majors, there is a class called “Introduction to Proofs”. That was not an easy class at first. It’s first introduction in how to do proof writing. Some people take to that skill easier than others. It’s very challenging at first. Some people also have a hard time with applications.
There are different challenges depending on the class you are taking.
Ann Cusano
Lives in Reading, PA
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Author has 6K answers and 5.6M answer views
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2y
Related
What are some common challenges students face with note taking?
I cannot speak for all students but my biggest challenge was getting all the information. I was pretty good at listening and writing but sometimes I would only get half an idea down and needed to move on to the next thing only to discover that evening that my notes were not complete. Some professors spoke quickly and did not like to be interrupted to repeat something for an individual. Fortunately in college my best friend was in almost all of my classes so we could compare notes. Otherwise I would have been missing crucial information. I believe all profs should have their lectures available
I cannot speak for all students but my biggest challenge was getting all the information. I was pretty good at listening and writing but sometimes I would only get half an idea down and needed to move on to the next thing only to discover that evening that my notes were not complete. Some professors spoke quickly and did not like to be interrupted to repeat something for an individual. Fortunately in college my best friend was in almost all of my classes so we could compare notes. Otherwise I would have been missing crucial information. I believe all profs should have their lectures available online for students to go back and listen to at home to be sure their notes are complete.
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Lucas Curtis
Science teacher (2001–present)
·
Author has 7.4K answers and 22.1M answer views
·
Updated Feb 20
Related
What are common mistakes students make when calculating empirical formulas?
It’s been a minute since I’ve taught chemistry, but here are a few mistakes I remember seeing rather frequently:
Not understanding how percentages work.
This is most prevalent when the question says “A compound is 60% carbon, 13.3% hydrogen, and
the remainder is oxygen
.” Lots of students will ignore the oxygen altogether if they don’t see a number in front of it.
Forgetting to convert grams to moles, or else converting incorrectly.
The most egregious error is
multiplying
by atomic weight rather than
dividing
, a mistake they wouldn’t make if they’d
just use dimensional analysis like you taught th
It’s been a minute since I’ve taught chemistry, but here are a few mistakes I remember seeing rather frequently:
Not understanding how percentages work.
This is most prevalent when the question says “A compound is 60% carbon, 13.3% hydrogen, and
the remainder is oxygen
.” Lots of students will ignore the oxygen altogether if they don’t see a number in front of it.
Forgetting to convert grams to moles, or else converting incorrectly.
The most egregious error is
multiplying
by atomic weight rather than
dividing
, a mistake they wouldn’t make if they’d
just use dimensional analysis like you taught them!
Rounding too soon.
Sometimes students forget that you have to divide the mole quantities by the lowest number of moles, to get a ratio. This is especially likely to happen if the mole quantities are already
kinda
close to whole numbers.
Rounding too soon, part 2.
Even after dividing by the lowest number of moles, you may wind up with a number that is half or a third of an integer. In that case, you have to multiply all the numbers by 2 or 3, respectively, to get a set of integers or near-integers.
Then
you can round.
Putting the wrong subscripts on the wrong elements.
This isn’t a conceptual mistake so much as an error of carelessness. After all that work to find the subscripts, students often write
[math]\rm C_4HO_{10}[/math]
rather than
[math]\rm C_4H_{10}O[/math]
.
Related questions
What is the scientific notation of 0.0000014 for eight students?
What are the difficulties faced by blind students?
What is scientific notation and why is it important for students learning science in school or college/university level courses?
How do you deal with difficult learners?
Why do students face difficulties in learning and speaking English?
Richard Gibson
Former O-6 at U.S. Air Force (1972–2002)
·
Author has 3.1K answers and 519.2K answer views
·
3y
Related
How do you learn scientific notation?
It’s a fairly simple process.
Take a number. Any number. For instance, 123456789.123456789
Move the decimal point however many spaces left or right you must to have only one digit to the left of it. For instance, 1.23456789123456789
Count the number of spaces you moved the decimal point and write that number after appending the letter E to your modified number. If you moved the decimal point to the LEFT, it’s a positive number. If you moved it to the RIGHT, it’s a negative number. For instance, 1.23456789123456789E8
That’s it.
Of course, if you’re given a number in scientific notation and want to k
It’s a fairly simple process.
Take a number. Any number. For instance, 123456789.123456789
Move the decimal point however many spaces left or right you must to have only one digit to the left of it. For instance, 1.23456789123456789
Count the number of spaces you moved the decimal point and write that number after appending the letter E to your modified number. If you moved the decimal point to the LEFT, it’s a positive number. If you moved it to the RIGHT, it’s a negative number. For instance, 1.23456789123456789E8
That’s it.
Of course, if you’re given a number in scientific notation and want to know what it looks like in long form, just do the reverse: take the number to the right of the E and move the decimal point that many spaces (it’s a reverse operation, so a negative number means you move the decimal point to the LEFT, nd a positive number means you move it to the RIGHT). Again, that’s all there is to it.
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Carter McClung
B.S. in UPSC General Studies & Mathematics, The University of Texas at Dallas (Graduated 2006)
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.6K answers and 6.5M answer views
·
Updated Apr 30
Related
What is scientific notation?
It's a notation used by scientists to reduce headaches.
About how many atoms are in your body?
About 7,000,000,000,000,000,000,000,000,000.
About how much do these atoms weigh?
About .000000000000000000000000001 g
Can you see how doing math with these numbers would be a headache?
We simplify this a bit by getting rid of the mass of trailing zeros on big numbers and leading zeros after the decimal on a small number by abusing a property of base 10 numbers.
70 can be rewritten as [math]7 10[/math], and 700 can be rewritten as [math]7 100[/math].
In the same way .7 can be rewritten as [math]7 10^{-1}[/math]and .07 can be rewritten as [math][/math]
It's a notation used by scientists to reduce headaches.
About how many atoms are in your body?
About 7,000,000,000,000,000,000,000,000,000.
About how much do these atoms weigh?
About .000000000000000000000000001 g
Can you see how doing math with these numbers would be a headache?
We simplify this a bit by getting rid of the mass of trailing zeros on big numbers and leading zeros after the decimal on a small number by abusing a property of base 10 numbers.
70 can be rewritten as [math]7 10[/math], and 700 can be rewritten as [math]7 100[/math].
In the same way .7 can be rewritten as [math]7 10^{-1}[/math]and .07 can be rewritten as [math]7 10^{-2}[/math].
Using this, we can say you have about [math]7 10^{27}[/math] atoms in your body, and they weigh about [math]110^{-26}[/math]g.
Scientific notation serves a second purpose beyond making numbers more manageable. It also shows significant digits to help describe how precise your measurements are so we don't assume a precision in calculations that we don't actually have.
For example, early measurements of the speed of light yielded 299,710 km/s. However this last zero isn't a significant digit (the measurement was plus or minus about 30 km/s) so we would write it as [math]2.9971 10^ 5[/math]km/s. If it were, we would write it as [math]2.99710 10^ 5[/math]. While 299,710 was ambiguous about the number of significant digits, the numbers in scientific notation aren't.
Weston Ding
B.S. in Mathematics Major, University of California, Los Angeles (Graduated 2015)
·
Author has 298 answers and 170.4K answer views
·
2y
Related
Can you explain scientific notation in easy English for a student who struggles with math?
Firstly, you have to solidify your knowledge of exponents. To do that you have to make sure you absolutely understand multiplication. You also need to solidify your knowledge decimals and integers. Math is way too dependent on previously learned content so you just cannot forget it. After this is solidified, I can explain.
Scientific notation is a method for expression very big and very small numbers (close to 0). It makes comparison easier and reduces the burden of multiplying and dividing large or small numbers and writing many 0s in computation. It also allows for easier (and sometimes neces
Firstly, you have to solidify your knowledge of exponents. To do that you have to make sure you absolutely understand multiplication. You also need to solidify your knowledge decimals and integers. Math is way too dependent on previously learned content so you just cannot forget it. After this is solidified, I can explain.
Scientific notation is a method for expression very big and very small numbers (close to 0). It makes comparison easier and reduces the burden of multiplying and dividing large or small numbers and writing many 0s in computation. It also allows for easier (and sometimes necessary) expression or significant digits which are necessary for verifying accuracy of computations (common and necessary in economic and science labs and measurements).
Scientific notation always expresses a number as a product of two numbers. The first number is between 1 and 10 not inclusive of 10) and the second number is a power of 10.
Here are some examples. The number 5,000,000 is written as 2 x 10^6. The number 73,000 is 7.3 x 10^4. The number 0.0000029 is 2.9 x 10^-6.
I can explain in more detail eventually but maybe these examples allow you to see what is going on.
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Ross Andersen
Former Metrologist - Director of Weights and Measures (1976–2010)
·
3y
Related
Why is it called "scientific" notation instead of just "notation"?
Because it has to be differentiated from other forms of notation. It needs a unique name. That’s the one that was chosen.
Up until the Hindus discovered zero, there were two kinds of notation, cardinal numbers for counting and ordinal numbers for ordering. That’s all we had and that’s all you will find in human languages. Think about counting: one, two, three, four, etc, and ordering: first, second, third, fourth, etc. We had words and numbers like 1, 2, 3, 4, 17 and 1st, 2nd, 3rd, 4th, 17th. Those are notations.
What is critical about cardinal and ordinal numbers is that they are exact. A count
Because it has to be differentiated from other forms of notation. It needs a unique name. That’s the one that was chosen.
Up until the Hindus discovered zero, there were two kinds of notation, cardinal numbers for counting and ordinal numbers for ordering. That’s all we had and that’s all you will find in human languages. Think about counting: one, two, three, four, etc, and ordering: first, second, third, fourth, etc. We had words and numbers like 1, 2, 3, 4, 17 and 1st, 2nd, 3rd, 4th, 17th. Those are notations.
What is critical about cardinal and ordinal numbers is that they are exact. A count of 452 is exactly 452. Third place is exactly the third place, following first and second. Let’s ignore ties at this point. It turns out some instances in the real world do lend themselves to exact numbers. Think, a dozen eggs, 42 sheep, and similarly with ordering. However, many instances in the real world don’t end up being exact.
Fractions such as 22/7. i.e., twenty two sevenths, were an attempt to describe the inexact world using exact numbers. Fractions are made up of two numbers, but it appears we articulate them as one cardinal number, 22, and one ordinal number, 7. Confused? Just ask any school child, Are fractions easy? I think most people just accept the inherent confusion without understanding why it’s confusing.
The division of 22 by 7 can be described as grouping the 22 items into orders of seven. You get three orders of seven with one item left over. The Greeks mathematicians, like Pythagoras, learned to do math this way and were very adept at it. They did so because cardinal and ordinal numbers were all they had. This continued until people finally grasped how zero could transform our world.
I chose the fraction 22/7 as it is an approximation of pi. Yet pi is a transcendental number that defies any exact definition. Now enter the world of scientific notation. Most things we measure in the real world also defy exact definition. Science (Knowledge) is pursued by measuring a “real” world that is inexact. We need a notation that expresses the “extent” of our inexact knowledge. Scientific notation does that through what are called significant digits. The more significant digits, the more knowledge.
Here’s an example. The sun is expressed as being 150,000,000 km from earth. That’s not exact. The distance ranges from about 147,000,000 km to 152,000,000 km. In scientific notation we would correctly write the sun is 1.50 x 10^8 km. In words that is one point five two times ten to the eighth power. What it communicates that the knowledge we have about the distance to the sun is only good to a few 10 million kilometers. All the zeros after the first comma are place holders and do not reflect real knowledge.
Because cardinal and ordinal numbers are exact but much of the real world is inexact, we encourage confusion when we try to describe the inexact world in fractions of exact numbers.
Scientific notation uses an inexact number to describe an inexact world. It’s biggest plus, when used correctly, is that the inexact number also communicates the level of inexactness. A synonym for scientific notation is “real numbers.”
I’ve just scratched the surface here. I’d suggest doing more research on real numbers and significant digits. There is plenty of good stuff on the internet. Also notice that there are even more kinds of notation, like imaginary numbers. Want to get really confused?
Physics, Math, Systems Engineer, Educator
·
Upvoted by
Peter Voke
, Ph.D. Physics, London (1976) · Author has 22.3K answers and 17.5M answer views
·
3y
Related
What is a nanometer in scientific notation?
A nanometre (US nanometer) is abbreviated as the 2 letters “nm”, with both letters in lower case.
1 nm = 10⁻⁹ m
10⁹ nm = 1 m
A nanometre (US nanometer) is abbreviated as the 2 letters “nm”, with both letters in lower case.
1 nm = 10⁻⁹ m
10⁹ nm = 1 m
Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR
·
Author has 5.6K answers and 10.9M answer views
·
Updated 3y
Related
Can you define what scientific notation is?
‘Scientific notation’ is the way of expressing the absolute value of numbers with very large or very small absolute value as one decimal number greater than 1 and smaller than 10, multiplied by the appropriate integer power of 10.
A number N is expressed in scientific notation if
|N| = n x 10^k
where 1 < n < 10 and k is an integer.
Some examples:
‘Scientific notation’ is the way of expressing the absolute value of numbers with very large or very small absolute value as one decimal number greater than 1 and smaller than 10, multiplied by the appropriate integer power of 10.
A number N is expressed in scientific notation if
|N| = n x 10^k
where 1 < n < 10 and k is an integer.
Some examples:
tired and retired, writer and re-writer (2018–present)
·
Author has 731 answers and 990.9K answer views
·
2y
Related
What are some common challenges students face with note taking?
When they begin, they tend to take down dictation, writing everything, rather then the salient points. I did exercises in note taking as a teacher because it was so vital in my advanced college and graduate courses. We practiced book briefing (writing out a “so what?” next to every paragraph or two in homework that they would then refer to in discussion.
More valuable students found was helping them to create an internal dialogue with one’s self while note-taking through symbols. A ? in the margin reminded the student later that they were not sure they understood the point and should follow up.
When they begin, they tend to take down dictation, writing everything, rather then the salient points. I did exercises in note taking as a teacher because it was so vital in my advanced college and graduate courses. We practiced book briefing (writing out a “so what?” next to every paragraph or two in homework that they would then refer to in discussion.
More valuable students found was helping them to create an internal dialogue with one’s self while note-taking through symbols. A ? in the margin reminded the student later that they were not sure they understood the point and should follow up. A CF next to something meant it should be researched to verify it is correct. A ! was something the teacher emphasized as important. Writing things in [ ] signified that this was not from the notes but internally generated as analysis, like [this seems similar to unit 2]. There are others, but you get the idea in general that notetaking is not mindless recording but an active process of synthesizing while recording information.
Related questions
What are the practical uses of scientific notation? Why is scientific notation so important in modern-day society?
How do I convert scientific notation to standard notation?
What are effective ways to teach scientific notation to students who struggle with it?
What are the applications of scientific notation?
How do I write 0.024 in scientific notation?
What is the scientific notation of 0.0000014 for eight students?
What are the difficulties faced by blind students?
What is scientific notation and why is it important for students learning science in school or college/university level courses?
How do you deal with difficult learners?
Why do students face difficulties in learning and speaking English?
What are the difficulties facing students in making a poem?
Why do students face difficulty in cracking NEET?
What are some of the difficulties that students face when learning in school? What are some of the difficulties that students face when learning outside of school?
Why do the Northeast students face many problems and difficulties?
What problems do students face in learning?
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WE ALL MAKE MISTAKES COMMON ERRORS IN FIRST YEAR CALCULUS (AND HOW TO AVOID MAKING THEM) Mark F. Schilling Professor of Mathematics California State University Northridge 1 INTRODUCTION: WE ALL MAKE MISTAKES The subject of calculus is over three hundred years old. Over the course of its history a general agreement has developed among mathematical educators about what topics constitute the core material of the subject. Thus today, while there are a large number of textbooks for first year calculus on the market, most look very much the same. z Not only are the included topics nearly the same from book to book, the techniques, examples and methods of solution also tend to be quite similar. Three centuries of experience have produced modern calculus textbooks in which subject matter is presented in an orderly, systematic way, and for each new topic, representative examples are provided and then worked out using a polished and ideal approach. Unfortunately beginning students, no matter how bright, fully comprehend the subject of calculus and solve problems precisely in the ways indicated in the text without making lots of errors (both conceptual and technical) along the way. It would be unreasonable to expect otherwise--the subject is too full of challenging new ideas for the student. In fact, Newton, Leibniz and the other mathematical giants who developed and refined calculus made plenty of mistakes of their own--often ones that would be quite embarrassing by today's standards. The authors of your text, as well as your professor, undoubtedly made lots of their own "dumb" mistakes as they first learned the subject. zzzz Your calculus text probably does not show these mistakes--it is quite full enough already--but nearly all students make at least some of them. Certain specific types of errors occur over and over, from student to student, year after year. Your instructor may discuss some of these mistakes, but time constraints often prevent an in-depth discussion of most of them. The purpose of this manual is to illustrate many of these common errors. Most of the problems presented are taken directly from student exam papers. Usually the student wound up with an incorrect answer to the problem, but in some cases the student merely got stuck and stopped with an incomplete solution due to an approach which appeared to make the problem more complicated than it really was. In the sections to follow, many of the most frequently made mistakes are illustrated in typical problems. The logical or technical flaw underlying each mistake is examined, the erroneous solution is contrasted with a correct one, and tips may be given for how to avoid the error in similar problems. In a sense, then, this manual illustrates how not to do calculus, with the hope that by anticipating certain common errors you will make fewer of them. It can also serve as a reference--if you have made a certain mistake on a problem or set of problems, you may find a careful analysis of a similar problem in the appropriate section of this manual. How To Use This Manual: You can use this manual in two ways. First, as you learn each main topic of calculus, read through the problems in the section of the manual that pertain to that topic to see what the most common mistakes are. You might wish to challenge yourself to find the error in each problem before reading on. Second, when you make a mistake on a homework or exam problem, look for a discussion of that error in this manual (you may find the Index useful for this) so that you fully understand the error and the correct alternative approach. 2 SECTION 1: ALGEBRAIC ERRORS Although calculus is a demanding subject with countless places to make errors along the way, most errors made by students learning calculus are algebraic mistakes. To see whether you are vulnerable to such mistakes, try the following quiz: zzzzzzzzzzzzz ALGEBRA QUIZ Which of the following calculations are correct? 1. 1 3 10 5 5 15 50 3 2 4 + − = + − t t t t t t for t ∫ 0 2. 4 3 5 4 1 4 3 ) 5 4 ( ) 1 4 3 ( 4 3 3 4 2 3 2 3 4 + = + + − − = + + − − x x x x x x x 3. ) tan( ) cos( ) sin( 2 3 θ = θ θ 4. 3 2 7 2 3 7 2 2 3 + = + x x x x 5. 10 4 2 10 4 2 − + = − + x x x x 6. ) 3 / tan( 2 1 2 3 2 7 2 ) 3 / tan( 3 7 2 2 3 π − + = / π − / + / x x x x 7. ( ) 2 2 1 4 4 − − = = ⋅ u u u u 8. n n x x + = 2 2) ( 9. 2 7 4 3 10 3 10 t t t t + − = + − 10. 5 25 25 1 1 2 2 > ⇒ > ⇒ > x x x 11. f(x) = 5x3, g(x) = 8x2 + 4x => f(x)g(x) = 5x3 ⋅ 8x2 + 4x = 40x5 + 4x 3 OK, ready? Only the first solution is correct. Did you miss any of the errors in the others? Here's why each solution after the first is wrong: #2: This is just like writing 4 3 4 3 2 2 + = + (which implies that 5 = 7). #3: When two trigonometric functions have different arguments (the inside parts--here, θ3 and θ), their ratio, product, sum, etc. cannot be combined into a single trigonometric expression. No simplification is possible in this problem. #4: A fractional expression cannot be broken up term by term unless the denominator is a single term. For instance, if we flip the problem upside down then splitting x x x 2 3 7 3 + as 2 3 2 7 2 + x is valid. A fractional expression cannot be separated into two or more expressions when the numerator is a single term, however. #5: 2x and −10 do not belong to the fraction. #6: Cancellation can only be done when every term contains the common factor. Since the term tan(π/3) does not have any x factor, the indicated cancellation is incorrect. (We can correctly simplify as follows, however: x x x x x x x x x 2 ) 3 / tan( 2 3 7 2 ) 3 / tan( 2 3 7 2 ) 3 / tan( 3 7 2 2 3 3 π − + = π − / / + = π − + / . #7: We must use subtraction of powers here: 2 7 2 1 2 8 2 1 4 4 u u u u u = = = − − . #8: When a power is raised to another power, those powers are multiplied, not added: n n x x 2 2) ( = . #9: The sign of the last term is wrong. This is easy to see if we add parentheses and break up the fractional expression: 2 7 2 7 4 4 3 10 3 10 3 10 3 10 t t t t t t t t t − − = + − = + − = + − . #10: There are two mistakes here. First, inverting both sides of an inequality reverses which side is larger. And second, the square root of x2 is |x|, not x. So the correct solution is 5 25 25 1 1 2 2 < ⇒ < ⇒ > x x x . #11: There are missing parentheses in the second expression from the right. As a result, the 5x3 factor did not get distributed onto the 4x. We should have: f(x)g(x) = 5x3 ⋅ (8x2 + 4x) = 40x5 + 20x4 . 4 SUMMARY OF COMMON ALGEBRAIC ERRORS: The most common algebraic mistakes arise from: • illegal simplifications of square root and other functions (see #2, #3) • splitting a fractional expression when the denominator is a sum of terms (see #4) • misreading fractional notation (see #5) • partial cancellation in a fractional expression (see #6) • mistakes when combining powers (see #7, #8) • failure to distribute minus signs and multiplying constants, usually due to missing parentheses (see #9, #11) 5 SECTION 2: ERRORS USING A GRAPHING CALCULATOR Many errors are possible when using a graphing calculator for calculus, including inputting incorrect values (unlike with pencil and paper there is no written record to check), omitting needed parentheses, and so forth. Several other types of errors that frequently arise are illustrated below. GRAPHING ERRORS Problem #1: Graph the function y = x2/3 . Solution: The Error: The domain of this function is all x, but the graph does not show any values for x < 0. This is because the calculator rounds the exponent to something like .666666666667, and the function x.666666666667 does not take on values for x < 0. (Can you see why?) A Correct Solution: Your calculator may have a special button for "cube root" that you could employ. However, this problem can occur for many fractional powers. A better approach is to simply use what you (should) already know about the graphs of elementary functions to check whether the graph is complete. If not, it is easy to draw the full graph when making a copy on paper if the function is even or odd. The function y = x2/3 is even (since (−x)2/3 = x2/3), so to extend the graph to the left of the y-axis, draw the mirror image of the part already shown on the calculator screen. Thus the correct graph is zzzzzz zzzzzzz 6 If the function had been odd (as with, say, y = x5/3) then to extend the graph to the left of the y-axis, use the inverted mirror image of the part to the right of the y-axis: Problem #2: Graph the function y = 3 3 5 x − . Solution: The Error: The given function is undefined at x = 3 and therefore has a discontinuity (gap) there. Your calculator, by default, assumes that any function you graph is continuous. As the graph is traced out from left to right on the screen, each point computed and plotted is joined to the previously plotted one. This gives an incorrect graph for discontinuous functions. A Correct Solution: When the graph has a long nearly vertical portion connecting two very different graph heights, this may indicate that the graph has an asymptote there. The nearly vertical portion should be omitted when copying the graph onto paper by hand: zzzz 7 Problem #3: Determine whether the graph of f(x) = (16 − x2) / (4 − x) is continuous. zzzz Solution: The graph below clearly indicates that the function is continuous. The Error: There is a "removable discontinuity" (a hole) in the graph at x = 4. Since the hole is infinitely small it cannot be seen in the graph. A Correct Solution: Note that at x = 4 the function is undefined since the denominator is zero, thus it must be discontinuous there. (The reason that the graph looks linear is that f(x) can be simplified as follows: f(x) = (4 + x)(4 − x)/(4 − x) = 4 + x for x ≠ 4.) zzzzz Problem #4: Determine whether the graph of f(x) = x + |x|/ x + 500 is continuous. Z zzzz Solution: The graph obtained when the default window ( [−10,10] × [−10,10] on TI calculators) is used is a blank screen. In order to see anything you must zoom out several times. Eventually you will see a graph something like this: The graph appears unbroken, hence the function is continuous. The Error: There is a break in the graph, but it cannot be seen at the scale used. The function is not continuous. Placing the cursor at the function's y-intercept (0, 500) and zooming in several times reveals a gap in the function: 8 But how would you know to do that? A Correct Solution: The correct approach is to study the function analytically: Since |x| = x for x > 0 and |x| = −x for x < 0, we can rewrite f(x) as < + > + = .
0 for 499 ; 0 for 501 ) ( x x x x x f This indicates that for x < 0 the function is linear; after x = 0 there is an upward shift of two units, creating a "jump discontinuity". An even simpler analytical solution is to note, as in the previous problem, that f(x) is undefined at x = 0 due to the x in the denominator of the term |x|/ x. Thus the graph is not continuous at x = 0. zzzzzz Problem #5: Consider the function f(x) = (x − 3)3 e−x. Determine the x-intervals for which this function is (i) positive; (ii) negative. Solution: Using a graphing calculator we find that the function is never positive, and is negative for 0 < x < 2. The Error: The default graph window ( [−10,10] x [−10,10] on TI calculators) does not reveal all the features of the function. Changing the y range to [−200,200] shows that the function is negative for values of x that are less than 0 as well: 9 Changing the y range to [−.5,.5] shows that the function also takes on positive values: zzzz A Correct Solution: Using a graph is not the best approach here. Studying the form of f(x), we can get the desired information quite easily. Since e−x is always positive, the sign of f(x) is determined by the sign of (x − 3)3. Thus clearly f(x) is positive for all x > 3, negative for all x < 3. COMPUTATION ERRORS Problem #6: Find all solutions of the equation x xe− = 0. Solution: The TI-85 SOLVER finds x = 2464.14388… as the only solution. The Error: This value of x does not satisfy the given equation. The calculator's equation solver operates by searching for a value of x for which the given expression is either equal to or extremely close to the specified value (here, 0). The value of 2464.14388…e−2464.14388… is so close to zero that the calculator regards x = 2464.14388… as a solution. A Correct Solution: Looking at the equation analytically is both easier and faster than using the calculator--and is capable of giving the correct solution! Note that e−x is always 10 positive, thus the only possible solution is x = 0. Moral: Never turn to your calculator as your first option! Problem #7: Compute the value of h h sin for h = 0.1, 0.01, 0.001. Solution: h 0.1 0.01 0.001 h h / ) (sin 0.017453284 0.017453292 0.017453293 The Error: The solver's calculator was set to degree mode, not radian mode. You should generally keep your calculator in radian mode for calculus. A Correct Solution: h 0.1 0.01 0.001 h h / ) (sin 0.9983 0.999983 0.99999983 Problem #8: Use your calculator to determine the value of 8 4 0 ) cos( 1 lim x x x − → . Solution: Plugging in values of x near zero gives the following table: x 0.03 0.02 0.01 0.001 −0.03 −0.02 −0.01 −0.001 8 4 / )) cos( 1 ( x x − .5030 .3906 0 0 .5030 .3906 0 0 It is apparent that the value of the indicated limit is 0. The Error: The tabled values are incorrect due to limitations on calculator precision. All calculators handle only a limited number of digits for any computation. Thus, for example, for the case x = 0.01 the calculator obtains cos(0.014) = 1 (try it!), hence computes (1 − cos(0.014))/0.018 = 0. Although the exact actual value of cos(0.014), 0.999999999999999950…, is very close to 1, it is different enough to make 8 4 / )) cos( 1 ( x x − very different from the tabled value of 0. There is also another important but less obvious error here--not in the solution but in the question itself: A calculator can often give good evidence for what a limit may be but can never determine it with certainty. The same remark applies to the finding of a limit by graphical means. Only analytical (algebraic) methods can establish the value of a limit with absolute certitude. 11 A Correct Solution: It is really not possible to solve the problem above with a calculator in a reasonable simple way. From analytical methods (one way is to use L'Hopital's Rule) it can be shown that the correct value of the limit is 1/2. The point of the given problem is to alert you to the dangers of calculator errors due to lack of precision. SUMMARY OF COMMON CALCULATOR ERRORS: Common errors during calculator use occur from: • trusting the graphical or numerical results displayed on your calculator without carefully thinking about the problem analytically (see #1−#6) • using a graph window that is poorly scaled or centered. Zooming in or out can often help. In some cases it may be necessary to use scales for the horizontal and vertical axes that are very different from each other (see #4, #5) • inaccuracy due to roundoff error, since your calculator only uses a limited number of digits in its computations (see #1, #2, #6, #8) • missing solutions when using your calculator's equation solver (see #6) • forgetting to keep your calculator in radian mode (see #7) 12 SECTION 3: THE DERIVATIVE CONCEPT LIMITS Problem #1: Find x x x 0 lim → if it exists, or explain why it does not exist. Solution: The limit does not exist, because the denominator becomes 0, making the expression undefined. The Error: It is not true that if the denominator takes the value 0 at the limit point then the limit cannot exist. More generally, it is not correct to say that limx→af(x) does not exist merely because f(a) is undefined. This is a critical fact about limits! In fact, most of the cases where you will be expected to find a limit will be of this type--that is, plugging in the limit point will give an undefined value. Here are some typical examples: zzz 3 21 10 lim 2 3 − + − → x x x x 2 1 0 lim x e x − →+ θ θ → θ sin lim 0 In every case, the limit exists even though the expression is undefined when the limit value for x or θ is substituted. A Correct Solution: Because absolute value is involved, we must consider the two cases x > 0 and x < 0. For x > 0, we have .
1 ) 1 ( lim lim lim 0 0 0 = = = + + + → → → x x x x x x x For x < 0, we have ) 1 ( lim lim lim 0 0 0 − = − = − − − → → → x x x x x x x = −1. Since the left-hand and right-hand limits are not equal, the overall limit doesn't exist--but not for the reason stated above! zzz z Problem #2: Find the following limits if they exist. If a limit does not exist explain why: (a) ) 5 ( lim 2 3 x x x − ∞ → (b) − + ∞ → x x x x 4 lim 2 Solution: For problem (a), ) 5 ( lim 2 3 x x x − ∞ → = 3 lim x x ∞ → − 2 5 lim x x ∞ → = ∞ − ∞ = 0. zzz Similarly for problem (b), we have − + ∞ → x x x x 4 lim 2 = x x x 4 lim 2 + ∞ → − x x ∞ → lim = ∞ − ∞ = 0. 13 The Error: It is never correct to subtract ∞ from ∞. The expression ∞ − ∞ is an indeterminate form. That means one can not say what its value is in a given problem. It could equal 0, 1, −37, or any other number--or its value could be undefined. A Correct Solution: (a) ) 5 ( lim 2 3 x x x − ∞ → = ) 5 ( lim 2 − ∞ → x x x . Since both x2 and x − 5 grow towards infinity as x → ∞, so does x3 − 5x2. Thus the given limit does not exist (or we might write ) 5 ( lim 2 3 x x x − ∞ → = ∞). (b) This problem is more difficult, and we will not provide a solution here. However, by evaluating the expression x x x − + 4 2 for large values of x or by graphing it you will see that the limit appears to be 2, which in fact is true. Thus in this problem, ∞ − ∞ = 2, whereas in problem (a), ∞ − ∞ = ∞. Problem #3: Find 3 27 6 lim 2 3 − − + → x x x x . Solution: Plugging in 3 gives 3 3 27 ) 3 ( 6 32 − − + = 3 3 27 27 − − = 0 0 = 1. The Error: 0/0 is another indeterminate form. Since plug-in gave an indeterminate form, we must find another method. A Correct Solution: 3 27 6 lim 2 3 − − + → x x x x = 3 ) 9 )( 3 ( lim 3 − + − → x x x x = ) 9 ( lim 3 + → x x = 3 + 9 = 12. Remark: There are many other indeterminate forms besides 0/0 and ∞ − ∞. For example, the following statements are all incorrect: 1 = ∞ ∞ 0 0 = ∞ ⋅ 1 1 = ∞ 1 0 = ∞ 1 00 = 0 00 = The expressions on the left side of each equation are indeterminate forms; their values must be found by other means. Problem #4: Find the derivative of f(x) = x2 + 3 by the difference quotient formula zzzzzz f′ (x) = h x f h x f h ) ( ) ( lim 0 − + → . 14 Solution: f′ (x) = h x h x h 3 3 lim 2 2 0 + − + + → = h h h 6 lim 0 + → = h h h 6 lim 0 + → = 6 . The Error: There are several errors here. First and foremost, when evaluating f(x + h), substitute x + h for x wherever x appears in the function--don't just tack it on after f(x). Here, f(x+h) = (x+h)2 + 3. The other errors are an important pair of missing parentheses and improper canceling: A Correct Solution: f′ (x) = h x h x h ) 3 ( 3 ) ( lim 2 2 0 + − + + → = h x h xh x h 3 3 2 lim 2 2 2 0 − − + + + → = h h xh h 2 0 2 lim + → = h h h x h ) 2 ( lim 0 + → = ) 2 ( lim 0 h x h + → = 2x. INTERPRETATION OF THE DERIVATIVE Problem #5: The temperature T (in degrees Celsius) at an agricultural research station t hours after sunrise is described by the function T = f(t). What is the practical meaning of the statement f′ (5) = −4? Solution: "The temperature five hours after sunrise is four degrees below zero." The Error: The equation f′ (5) = −4 is a statement about the derivative--i.e., the rate of change--of temperature. The solution above refers to the temperature value, and represents the equation f(5) = −4, not f′ (5) = −4. A Correct Solution: "After five hours the temperature is decreasing at four degrees per hour." Note that the word decreasing comes from the negative sign, and that the units of the derivative, degrees per hour, are "units of the dependent variable" per "units of the independent variable" (think of the notation dy/dx to remind yourself of this fact). Problem #6: A distance runner is running a marathon. The total number of calories she burns in the first t minutes is given by the function f(t). Give the practical meaning of the statement f′ (30) = 5. Solution: After running for 30 minutes, the runner burns five calories per minute for each additional minute she runs. 15 The Error: The derivative is a statement about the instantaneous rate of change--that is, about what is happening right when 30 minutes have been run. It does not tell us what happens for every minute after that. A Correct Solution: After running for 30 minutes, the runner is burning calories at a rate of five calories per minute. SUMMARY OF COMMON LIMIT AND DERIVATIVE CONCEPT ERRORS The most common errors involving limits and the derivative concept arise from: zz zz • thinking that if a function is undefined at a point x = c, then the limit cannot exist there (see #1) • treating indeterminate forms (such as ∞ − ∞, 0 ⋅ ∞ and 1∞) as having values (see #2, #3) • evaluating f(x + h) as f(x) + h in difference quotient problems (see #4) • confusing the meaning of the derivative in word problems with the function itself (see #5) • failing to interpret the derivative as representing instantaneous rate of change (see #6) 16 SECTION 4: TECHNIQUES OF DIFFERENTIATION Problem #1: Find the derivative of f(x) = x−4 . Solution: By the power rule for derivatives, f′ (x) = −4x−3 . The Error: This is a simple but very common error when using the power rule on an expression with a negative exponent. The power rule says to subtract 1 from the exponent, but here the solver has added 1. A Correct Solution: f′ (x) = −4x−4−1 = −4x−5 . Problem #2: Find the derivative of f (z) = z z 1 2 + . Solution: f(z) = 2 / 1 2 1 z z + => f′ (z) = 2 / 1 ) ( 0 2 2 / 1 − + z z = 2 z(2 z1/2) = 4 z3/2 . The Error: The derivative of a quotient is not equal to the derivative of the numerator divided by the derivative of the denominator. A Correct Solution: To differentiate a quotient you can either use the quotient rule or try to change the function into a different form before differentiating. Here, writing f(z) = 2 / 1 2 1 z z + = 2 / 1 2 / 1 2 1 z z z + = z3/2 + z−1/2 gives f′ (z) = 2 / 3 2 / 1 2 1 2 3 − − z z . Problem #3: Find the derivative of f(x) = 3 1 π + + x x + e . Solution: f′ (x) = 2 3 ln 2 1 π + + x x + e. The Error: There are three errors here. Can you find them all? If not, check the solution again before reading on. Error #1: ln x is the antiderivative of x 1 , not its derivative. The derivative of x 1 is 2 1 x − . 17 Errors #2 & 3: π3 = (3.14159…)3 and e = 2.71828… are constants. Thus their derivatives are 0. Always remember: Letters do not necessarily represent variables--often they are constants instead. A Correct Solution: f′ (x) = 2 1 2 1 x x − . Problem #4: Find the derivative of f(x) = (x3 − 4x2 + 13x + 10) sin x. Solution: f′ (x) = (3x2 − 8x + 13) cos x. The Error: This should be a fairly easy error to spot. The solver forgot to use the product rule. Remember: The derivative of a product of is not equal to the product of the two individual derivatives. A Correct Solution: f′ (x) = (3x2 − 8x + 13) sin x + (x3 − 4x2 + 13x + 10) cos x. Problem #5: Find the derivative of f(x) = (2x3 + x2 + 4x)e−x . Solution: Expanding gives f(x) = 2x3e−x + x2e−x + 4xe−x , then using the product rule gives f′ (x) = [6x2e−x + 2x3(−e−x)] + [2xe−x + x2(−e−x)] + [4e−x + 4x(−e−x)] = 6x2e−x − 2x3e−x + 2xe−x − x2e−x + 4e−x − 4xe−x = (6x2 − 2x3 + 2x − x2 + 4 − 4x) e−x = (−2x3 + 5x2 − 2x + 4) e−x . The Error: This solution is completely correct! However, the problem was solved very inefficiently by unnecessarily expanding the function at the very first step. A Correct (Better) Solution: Use the product rule immediately to obtain f′ (x) much more easily: f′ (x) = (6x2 + 2x + 4)e−x + (2x3 + x2 + 4x)(−e−x) = (6x2 + 2x + 4 − 2x3 − x2 − 4x)e−x = (−2x3 + 5x2 − 2x + 4) e−x . Remark: The first solution is so lengthy that many students would make a sign error or other error somewhere. And on a test, this approach would use up valuable time. When considering whether to take a certain step on a problem (such as clearing parentheses), always think first about whether it is helpful or necessary. 18 Problem #6: Find the derivative of f(x) = 7 3 sin 2 − x x . Solution: f′ (x) = x x 6 cos . The Error: This problem is similar to #2. The solver forgot to use the quotient rule. Again, the derivative of a quotient is not equal to the quotient of the derivatives. A Correct Solution: f′ (x) = 2 2 2 ) 7 3 ( sin 6 cos ) 7 3 ( − − − x x x x x . Problem #7: Find the derivative of f(x) = 8 2 5 3 + − x x . Solution: The quotient rule gives f′ (x) = 2 ) 8 2 ( 2 ) 5 3 ( ) 8 2 ( 3 + ⋅ − − + ⋅ x x x . Canceling 2x + 8 from the top and bottom gives 8 2 2 ) 5 3 ( 3 + ⋅ − − x x = 8 2 3 13 + − x x . The Error: The first step is correct, but then the solver cancelled improperly. (See #6 on the Algebra Quiz). A Correct Solution: f′ (x) = 2 ) 8 2 ( 2 ) 5 3 ( ) 8 2 ( 3 + ⋅ − − + ⋅ x x x = 2 ) 8 2 ( 10 6 24 6 + + − + x x x = 2 ) 8 2 ( 34 + x . Problem #8: Find dx dz for the function 2 3 + π = x z . Solution: The quotient rule gives 2 ) 2 ( 1 3 ) 2 ( 3 + π ⋅ − + π ⋅ = x dx dz = 2 ) 2 ( 3 6 3 + π − + π x . The Error: There are two errors. The main one is that π is a constant. Thus the derivative of π + 2 in the numerator of the solution should be 0, not 1. The other mistake is that it is not necessary to use the quotient rule at all. 19 A Correct Solution: Rewriting the function as x z 2 3 + π = easily gives dx dz = 2 3 + π . Problem #9: Find the derivative of f(x) = 2 / 3 2 1 x x x − + . Solution: By the quotient rule, f′ (x) = 2 2 / 3 2 / 1 2 2 / 3 2 / 1 ) ( ) 2 / 3 ( ) 1 ( ) ) 2 / 1 ( 2 ( x x x x x x x ⋅ − + − + − = … (etc.) The Error: The solver's approach results in a very messy algebra problem (although it could eventually give the correct answer if the solver made no algebraic mistakes along the way). It is not necessary to use the quotient rule here. A Correct (Better) Solution: Simplifying gives f(x) = 2 / 3 2 1 x x x − + = 2 / 3 2 / 3 2 / 1 2 / 3 2 1 x x x x x − + = 2 / 3 1 2 / 1 − −− + x x x . Then easily, zzz f′ (x) = 2 / 5 2 2 / 1 2 3 2 1 − − − + − x x x . Problem #10: Find the derivative of f(x) = xx. Solution: By the power rule for derivatives, f′ (x) = x ⋅ xx−1 = xx. The Error: The power rule can only be used when the base is a variable and the exponent is a constant. Here, the exponent is a variable. A Correct Solution: Rewrite the function in base e, as follows: f(x) = xx = (eln x)x = ex ln x. Then use the chain and product rules to obtain: zzzz f′ (x) = ex ln x (1 ⋅ ln x + x ⋅ 1/x) = ex ln x (ln x + 1) = xx (ln x + 1). Problem #11: Find the derivative of f(x) = sin(x2). Solution: f′ (x) = cos(2x) . 20 The Error: For a composite function, it is not correct to replace the inside and outside functions with their derivatives. The chain rule must be used instead. A Correct Solution: f′ (x) = cos(x2) ⋅ 2x = 2x cos(x2) . Problem #12: Find the derivative of f(x) = ln[(7x + 1)(x2 + 2)] . Solution: Using the chain and product rules, f′ (x) = ) 2 ( ) 1 7 ( ) 2 )( 7 ( 2)] 1)( [(7 1 2 2 x x x x x + + + ⋅ + + = ) 2 ( ) 1 7 ( 2) 1)( (7 ) 2 ( 7 2 2 x x x x x + + + + + = x x x 2 14 1 7 7 2 + + + . The Error: The solver has used many parentheses, but has omitted a crucial pair in the first step (See #11 on the Algebra Quiz): A Correct Solution: f′ (x) = 2) 1)( (7 1 2 + + x x ⋅ [ ) 2 ( ) 1 7 ( ) 2 ( 7 2 x x x + + + ] outer derivative × inner derivative = ) 14 2 21 ( 2) 1)( (7 1 2 2 + + ⋅ + + x x x x = 2) 1)( (7 14 2 21 2 2 + + + + x x x x . Problem #13: Find the derivative of f(x) = arcsin 2x. Solution: Using the formula for the derivative of arcsin x along with the chain rule, f′ (x) = 2 2 1 1 2 ⋅ −x = 2 2 1 2 x − . The Error: This is another case of a missing set of parentheses. Since 2x plays the role of x in the arcsine derivative formula, we should have: A Correct Solution: f′ (x) = 2 ) 2 ( 1 1 2 ⋅ − x = 2 4 1 2 x − . 21 Problem #14: Find the derivative of f(x) = ln(cos(4x)). Solution: By the chain rule, f′ (x) = )) 4 sin( ( ) 4 cos( 1 x x − ⋅ = ) 4 tan( ) 4 cos( ) 4 sin( x x x − = − . zz The Error: This is a doubly nested composite function. The chain rule requires that the derivatives of all three components (outer = ln, middle = cos, inner = 4x) be included. zzzz A Correct Solution: f′ (x) = 4 )) 4 sin( ( ) 4 cos( 1 ⋅ − ⋅ x x = ) 4 tan( 4 ) 4 cos( ) 4 sin( 4 x x x − = − . Problem #15: Find the second derivative of f(x) = 10x4 + 20x2. Solution: 10x4 + 20x2 = 40x3 + 40x = 120x2 + 40 . The Error: The answer is correct, but the solution is not, because the expressions given are not equal to each other. Make sure to use an equal sign only when expressions are equal, not merely to show the next step of your calculation: A Correct Solution: f(x) = 10x4 + 20x2 ⇒ f′ (x) = 40x3 + 40x ⇒ f′′ (x) = 120x2 + 40 . Problem #16: For the equation sin(x + y) = 4y, find dy/dx by implicit differentiation. zzz Solution: dx dy y x dx dy 4 ) cos( = + = … (etc.) The Error: There are two errors here. First, do not start an implicit differentiation problem by writing = dx dy … . Second, the problem requires that the chain rule be used for the left-hand side: A Correct Solution: dx dy dx dy y x 4 ) 1 )( cos( = + + ⇒ dx dy dx dy y x y x 4 ) cos( ) cos( = + + + ⇒ dx dy y x y x )) cos( 4 ( ) cos( + − = + ⇒ ) cos( 4 ) cos( y x y x dx dy + − + = . 22 Problem #17: The equation eyx = x + y defines a curve that passes through the point (0,1). Find the slope of the curve at that point. Solution: Use implicit differentiation, differentiate both sides with respect to x to get yeyx = 1 + dy/dx => dy/dx = yeyx − 1. Plugging in (0,1) gives dy/dx = 1 ⋅ e1⋅0 −1 = 0. zzzzz The Error: The derivative of eyx requires both the chain rule and the product rule, since y is a function of x. The solver was thinking of the y in the exponent of eyx as a constant. z A Correct Solution: Correctly differentiate both sides with respect to x to get eyx ⋅ ⋅ + ⋅ 1 y x dx dy = dx dy + 1 (Make sure to include the parentheses!) Now solve for dy/dx and plug in (0,1), or better yet, first plug in (0,1) and then find dy/dx: e1⋅0 ⋅ ⋅ + ⋅ 1 1 0 dx dy = dx dy + 1 => 1 = dx dy + 1 => dx dy = 0. SUMMARY OF COMMON DIFFERENTIATION ERRORS: Most common differentiation errors are due to: • computing the derivative of a product as a product of derivatives, the derivative of a quotient as a quotient of derivatives, or the derivative of a composite function as the corresponding composite function of the derivatives (see #2, #4, #6, #11) • treating constants as variables or variables as constants (see #3, #8, #10, #17) . • forgetting part of the derivative for multiply nested composite functions (see #14) • trying to solve implicit differentiation problems by starting with = dx dy (see #16) • algebraic mistakes and/or general carelessness, especially with parentheses (see #1, #3, #7, #12, #13) • not realizing the algebraic approach that allows for the simplest solution (not really an error, but doing a problem a hard way often leads to errors and takes more time) (see #2, #5, #8, #9) 23 SECTION 5: APPLICATIONS OF THE DERIVATIVE Problem #1: The graph of f′ is shown below. Find the intervals on which f is: (a) increasing (b) decreasing (c) concave up (d) concave down Also determine where f takes its minimum value. Solution: (a) & (b): Increasing for 1.5 < x < 2.5, decreasing for x < 1.5 and x > 2.5. ( (c) & (d): Concave up for x < 2, concave down for x > 2. Minimum value at x = 1.5. The Error: The student is thinking that the graph is of f, but it is a graph of f′ . zzzzzzzzzz A Correct Solution: (a) & (b): f is increasing where f′ is positive, namely for x < 1 and 2 < x < 3. For 1 < x < 2, f′ is negative so f is decreasing there. (c) & (d): Concavity depends on f′′, which is the slope of f′ . f is concave up where the slope of f′ is positive, namely for 1.5 < x < 2.5; f is concave down where the slope of f′ is negative, for x < 1.5 and x > 2.5. Determining where f has its minimum is trickier. Note that f′ is positive and large for x < 1. That tells us that f rises steeply on that interval. For x > 1 the value of f′ remains close to 0 so f is nearly horizontal there. Thus the minimum value of f is at x = 0. Problem #2: Find all inflection points of the function f(x) = 3x5 − 20x4 . Solution: We find inflection points by setting f′′ (x) = 0: f′ (x) = 15x4 − 80x3 => f′′ (x) = 60x3 − 240x2 = 60x2(x − 4) = 0 => x = 0, 4. Thus the function has inflection points at x = 0 and 4. The Error: The calculations above are correct. But not all points where f′′ (x) = 0 are necessarily inflection points. An inflection point is a point where the concavity of f changes. Thus the sign of f′′ (x) must change at that point. Here, the sign of f′′ (x) is negative on both sides of x = 0 near x = 0 (since 60x2 is positive and x − 4 is negative for x near 0). 1 2 3 24 A Correct Solution: Compute f′ ′ (x) = 60x2(x − 4) as above, find the zeros x = 0 and 4, then determine the sign of f′′ in between the zeros (one way is to plug other values of x into f′′ (x)). Then make a number line plot showing the sign of f′′ as x varies: f′′ : − 0 − 4 + x We conclude that f(x) has only one inflection point, at x = 4. Be sure to check your answer by graphing f(x). Problem #3: Use the graph of f′′ shown below to determine how many inflection points f has: Solution: There are three inflection points, marked below: The Error: This is another example of thinking of the given plot as the graph of f. But the graph shows f′′, not f. Since inflection points are points where the sign of f′′ changes, there is only one such point, where the graph crosses the x-axis. Problem#4: Find 2 ln lim x x x ∞ → . 25 Solution: L'Hopital's Rule gives 2 ln lim x x x ∞ → = 2 2 2 ) ( 2 ln ) / 1 ( lim x x x x x x ⋅ − ⋅ ∞ → = 4 ln 2 lim x x x x x − ∞ → = 3 ln 2 1 lim x x x − ∞ → = ??? (The solver is stuck because this expression is similar to but more complicated than the original problem.) The Error: The solver used the quotient rule rather than L'Hopital's Rule. A Correct Solution: Using L'Hopital's Rule correctly gives 2 ln lim x x x ∞ → = x x x 2 / 1 lim ∞ → = 2 2 1 lim x x ∞ → = 0. Problem #5: Find the following limits: (a) 1 5 2 sin lim 0 − → x x x ; (b) x x x x 2 ln lim ∞ → . Solution: Since both problems involve quotients, we can use L'Hopital's Rule for each: For (a), differentiating top and bottom gives 5 2 cos 2 lim 0 x x→ = 5 0 cos 2⋅ = 2/5 . For (b), differentiating top and bottom gives x x x x 2 ) 2 (ln / 1 lim ⋅ ∞ → = x x 2 ) 2 (ln 1 lim ∞ → = 0 since the denominator blows up. The Error--(a): L'Hopital's Rule can only be used on expressions that have the indeterminate form ∞ /∞ or 0/0. But plugging in 0 gives 1 ) 0 ( 5 ) 0 2 sin( − ⋅ = 1 0 − , so L'Hopital's Rule does not apply. A Correct Solution: 1 5 2 sin lim 0 − → x x x = 1 ) 0 ( 5 ) 0 2 sin( − ⋅ = 1 0 − = 0 . The Error--(b): This expression does have the indeterminate form ∞ /∞, so it is correct to use L'Hopital's Rule. However, the derivative of the numerator requires the product rule. A Correct Solution: x x x x 2 ln lim ∞ → = x x x x x 2 ) 2 (ln / 1 ln 1 lim ⋅ + ⋅ ∞ → = x x x 2 ) 2 (ln 1 ln lim + ∞ → . As this is still of the form ∞ /∞, we must apply L'Hopital's Rule again: 26 … = x x x 2 ) 2 (ln / 1 lim 2 ∞ → = x x x 2 ) 2 (ln 1 lim 2 ∞ → = 0, since the denominator blows up . Problem #6: Find the tangent line to f(x) = sin x near x = π/6. Solution: The formula for the tangent line is y = f(a) + f′ (a)(x − a) . We have f′ (x) = cos x, a = π/6, f(a) = sin (π/6) = 1/2, so the tangent line is y = 1/2 + cos x(x − π/6). The Error: Note that the solver's answer cannot be right because it is not even a linear function! The solver forgot that to obtain f′ (a) one must plug a = π/6 into f′ (x) = cos x. This is an extremely common error on tangent line problems. A Correct Solution: With f′ (a) = cos(π/6) = 3/3 replacing cos x in the solver's answer above, we have y = 1/2 + ( 3/3)(x − π/6), or y = .577x + .198 . SUMMARY OF COMMON DERIVATIVE APPLICATION ERRORS: Many errors involving applications of the derivative result from: • not noticing what a given graph refers to (f, f′ , f′ ′ , …) (see #1, #3) • assuming that all points where f′′ (x) = 0 are necessarily inflection points (see #2) • improper application of L'Hopital's Rule (see #4, #5) • using the derivative function rather than its numerical value as the slope of a tangent line (see #6) 27 SECTION 6: INTEGRATION Problem #1: Evaluate the integral ∫ + dx x x ) 4 3 ( 2 . Solution: ∫ + dx x x ) 4 3 ( 2 = x3 + 2x2 . The Error: This is a minor error, but an error nonetheless. In fact it is probably the most common integration error. Remember that an indefinite integral yields an entire family of solutions, not just one: A Correct Solution: ∫ + dx x x ) 4 3 ( 2 = x3 + 2x2 + C . Problem #2: Find the antiderivative of sin t . Solution: cos t + C . The Error: This simple error is made very often. The derivative of sin t is cos t, but the antiderivative of sin t is − cos t . A Correct Solution: − cos t + C . Remark: Also be careful when giving the antiderivative of cos t, which is sin t, not −sin t. Problem #3: Evaluate the integral ∫ dt 8 . Solution: ∫ dt 8 = 8x + C . The Error: The differential dt was ignored. The answer should involve t, not x. zzzzzzz A Correct Solution: ∫ dt 8 = 8t + C . Problem #4: Find the antiderivative of x5 + 2π. Solution: ∫ π + dx x ) 2 ( 5 = x6/6 + π2 + C . 28 The Error: 2π is a constant. Do not treat it as you would 2x. A Correct Solution: ∫ π + dx x ) 2 ( 5 = x6/6 + 2πx + C . Problem #5: Evaluate the integral ∫ − + dx x x ) 1 )( 2 ( . Solution: ∫ − + dx x x ) 1 )( 2 ( = C x x x C x x x x + − − = + − + 4 2 2 ) 2 )( 2 2 ( 4 3 2 2 2 . The Error: The antiderivative of a product of two expressions is not equal to the product of the antiderivatives of the two expressions (just as the derivative of a product is not the product of the individual derivatives--that's why we need the product rule). Other examples of this error are: ∫ + ≠ C x x dx x x sin 3 1 cos 3 2 ; ∫ ∫ + ≠ + ⋅ = + C x x dx x x dx x x arctan 1 1 2 1 2 2 2 2 . Problem #6: Evaluate the integral ∫ + dx x 1 5 . Solution: ∫ + dx x 1 5 = C x x + + + 2 5 2 . The Error: The power rule for integrals (∫ − ≠ + + = + 1 , 1 1 n C n x dx x n n ) cannot be applied here because the variable is in the exponent, not in the base. A Correct Solution: ∫ + dx x 1 5 = C C dx dx x x x x + = + ⋅ = = ⋅ + ∫ ∫ 5 ln 5 5 ln 5 5 5 5 5 5 1 (using the formula C a a dx a x x + = ∫ ln ). Problem #7: Find the derivative of f(x) = ∫ + x dt t 0 4 3 . Solution A: Let 3 2 1 4 4 4 ) 3 ( 2 1 , 3 t t dt dw t w ⋅ + = + = − , …? Solution B: f′ (x) = 3 4 + t . 29 The Error (Solution A): It is not necessary to find an explicit antiderivative in this problem. A Correct Solution: Simply use the Fundamental Theorem of Calculus, which tells us that f(x) is an antiderivative of 3 4 + x . Thus f′ (x) is the derivative of an antiderivative of 3 4 + x and therefore equals 3 4 + x . The Error (Solution B): Although the solver knew to use the Fundamental Theorem of Calculus, note that the answer looks wrong: f′ (x) must surely be a function of x, not of t. The variable t in the integrand is known as a dummy variable, and is not present in the evaluated integral. The answer, 3 4 + x , would be exactly the same if the problem had been given as, say, f(x) = ∫ + x du u 0 4 3 . Problem #8: Evaluate the integral ∫ − dx x 6 1 . Solution: ∫ − dx x 6 1 = ) 6 ln( − x + C . The Error: This is an example of one of the very most common integration errors made: The integration formula ∫ dx x 1 = ln x + C does not allow one to write dx ∫expression 1 = ln(expression) + C in general. A Correct Solution: Use the method of substitution: Let w = x − 6. Then dw = dx and we can write ∫ − dx x 6 1 = ∫ dw w 1 = ∫ − dw w 2 / 1 = 2w1/2 + C = 6 2 − x + C . Problem #9: Evaluate the integral ∫ + dx x x 2 4 2 . Solution A: ∫ + dx x x 2 4 2 = 2 3 4 3 / x x x + + C . 30 Solution B: Let w = x2 + 4. Then dw = 2x and ∫ + dx x x 2 4 2 = ∫dw w = etc. The Error: Solution A is wrong because it is not correct to integrate the numerator and denominator of a quotient separately, just as we do not differentiate a quotient by differentiating its the numerator and denominator separately (we use the quotient rule for derivatives). Unfortunately there is no quotient rule for integrals, so various approaches must be taken for integrals containing quotients. Solution B is wrong because it is never right to have dw in the denominator. Therefore the substitution w = x2 + 4, though tempting, cannot be used here. A Correct Solution: Change the form of the integrand first: ∫ + dx x x 2 4 2 = ∫ + dx x x x 2 4 2 2 = ∫ + dx x x 2 2 = x2/4 + 2ln |x| + C . Problem #10: Evaluate the integral ∫− − 3 3 2 9 dx x . Solution: ∫− − 3 3 2 9 dx x = ∫− − 3 3 2 / 1 2) 9 ( dx x = 3 3 2 / 3 2 2 / 3 ) 9 ( − −x = 2 / 3 ) ) 3 ( 9 ( 2 / 3 ) 3 9 ( 2 / 3 2 2 / 3 2 − − − − = 0. The Error: It is incorrect to apply the power rule for integrals here. To see that this is wrong, note that the derivative of 2 / 3 ) 9 ( 2 / 3 2 x − is not 2 9 x − , because the chain rule comes into play: 2 2 / 1 2 2 / 1 2 2 / 3 2 9 ) 9 ( 2 ) 2 ( 2 / 3 ) 9 )( 2 / 3 ( 2 / 3 ) 9 ( x x x x x x dx d − ≠ − − = − ⋅ − = − . In general, when the integrand is a composite function with the outer function being a power, the power rule for integrals should not be applied. This is because the derivative of the inner function is not accounted for in this approach. Here are several more examples of this error: ∫ + ≠ C x dx x 4 3 sin 4 1 sin C e dx e x x + + ≠ + ∫ 3 2 ) 1 ( 3 1 ) 1 ( 31 C x x dx x x + + − ≠ + − − ∫ 2 4 3 4 ) 2 ( 2 1 ) 2 ( ∫ + ≠ C x dx x 3 2 ln 3 1 ln A Correct Solution: The best way to evaluate the integral ∫− − 3 3 2 9 dx x is to recognize what it represents geometrically. The graph of the function 2 9 x y − = is shown below. Notice that just by making this graph you can immediately tell that the answer of 0 above cannot be right since the area under a curve lying above the x-axis must be positive. zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz To get the right answer, note that 2 9 x y − = is the upper half of a semicircle of radius 3, thus the integral given above is the area of that semicircle, π(3)2/2 = 9π/2. -3 3 (You could also solve the integral with the rationalizing substitution x = 3 sin t if you have learned that method.) Problem #11: Evaluate the integral ∫ + − 5 4 2 4 121 22 dx x x . Solution: ∫ + − 5 4 2 4 121 22 dx x x = ∫ + − 5 4 2 / 1 2 4 ) 121 22 ( dx x x = 5 4 2 / 3 3 5 121 3 22 5 3 2 + − x x x = … = 93.87. The Error: When integrating a composite function, do not integrate both the inside and outside functions. A Correct Solution: ∫ + − 5 4 2 4 121 22 dx x x = ∫ − 5 4 2 2 ) 11 ( dx x = ∫ − 5 4 2 | 11 | dx x = ∫ − 5 4 2 ) 11 ( dx x (since x2 − 11 > 0 for 4 ≤ x ≤ 5) = 5 4 3 11 3 − x x = 9.33. Problem #12: Evaluate the integral ∫ + dx x x cos 1 sin . 32 Solution: Let w = 1 + cos x. Then dw = sin x dx, and we obtain ∫ + dx x x cos 1 sin = ∫ + + = + = C x C w w dw ) cos 1 ln( ln . The Error: There are two careless errors here. First, the derivative of cos x is − sin x, so dw is not right. Getting the sign wrong with either the derivative or antiderivative of sin x or cos x is a very common mistake, so be careful! Second, the derivative of ln x is 1/x, but the antiderivative of 1/x is ln|x|, not ln x. A Correct Solution: Let w = 1 + cos x. Then dw = − sin x dx, and we get∫ + dx x x cos 1 sin = ∫ + + − = + − = − C x C w w dw cos 1 ln ln . Problem #13: Evaluate the integral ∫ + 4 2 x dx . Solution: Let dx x dw x w 2 Then .
4 2 = + = => dx = dw/2x, so ∫ + 4 2 x dx ∫ = w x dw 2 / ∫ = w dw x 2 1 = C w x + = ln 2 1 = C x x + + = 4 ln 2 1 2 . The Error: There are two mistakes here: Error #1: Pulling out x 2 1 = from the integral is not an allowable step. . Only constants can be pulled outside an integral, not variables. When performing a substitution, you should convert all instances of the original variable (here, x) into the substitution variable (here, w) and leave them inside the integral. Unfortunately, while that is the correct technique, if we try it above we obtain zzzzz ∫ w x dw 2 / ∫ − = w w dw 4 2 / (x was replaced by solving 4 2 + = x w for x) ∫ − = 4 2 w w dw , which is a worse integral than the one we started with. And that zzzzzz brings us to the second error: Error #2: The substitution w = denominator should not be used when the integrand has the form 1/quadratic. 33 A Correct Solution: Recall that ∫ +1 2 x dx = arctan x. We can convert the given integral to this form as follows: ∫ + 4 2 x dx = ∫ + ) 1 4 / ( 4 2 x dx = ∫ + ) 1 ) 2 / ( ( 4 2 x dx ; now let w = x/2, => dw = dx/2, to obtain ∫ + ) 1 ( 4 2 2 w dw = ∫ +1 2 1 2 w dw = C x C w + = + 2 arctan 2 1 arctan 2 1 . Remark: Letting 4 2 + = x w would have worked if the integral had contained an x in the numerator to accommodate dw = 2xdx. SUMMARY OF COMMON INTEGRATION ERRORS: Typical integration errors are due to: • forgetting to include + C in indefinite integral problems (see #1) • treating constants as variables or variables as constants (see #4, #6, #13) . • failing to use the Fundamental Theorem of Calculus when x appears in the limits of integration (see #7) • computing the integral of a product as a product of integrals or the integral of a quotient as a quotient of integrals (see #5, #9) • integrating a composite function by integrating the inside and outside functions (see #11) • writing dx ∫expression 1 = ln(expression) + C when the expression is something other than x (see #8) • using the power rule for dx n ∫ ) expression ( when the integrand is a composite function with the outer function being a power (see #10) • using a substitution that doesn't work (see #13) • general carelessness, especially involving the antiderivatives of sin t and cos t (see #3, #4, #12) 34 SECTION 7: ERRORS IN THE SECOND SEMESTER While this manual is focused on common errors that arise initially in first semester calculus, there are several mistakes that occur often enough in the second semester that they are well worth mentioning. What follows should not be regarded as a comprehensive list, but rather as a collection of a few of the most frequently made mistakes. Problem #1: Evaluate the integral ∫− 1 1 4 1 dx x . Solution: ∫− 1 1 4 1 dx x = ∫− − 1 1 4 dx x = 1 1 3 3 − − − x = 3 2 3 1 3 1 − = − − − − . The Error: Notice that the integrand is always positive, so the definite integral cannot possibly come out negative. The error is failing to recognize that this an improper integral, since 1/x4 blows up at x = 0. A Correct Solution: ∫− 1 1 4 1 dx x = ∫ 1 0 4 1 dx x + ∫− 0 1 4 1 dx x . Now ∫ 1 0 4 1 dx x = ∫ + → 1 4 0 1 lim a a dx x = 1 3 0 3 lim a a x − − →+ = − − − + → 3 0 3 1 3 1 lim a a . Since a is in the denominator and approaching 0, the limit does not exist. Hence the integral ∫ 1 0 4 1 dx x diverges, and therefore the original integral diverges as well. zzzzzzzzz Problem #2: The region bounded by the lines x = 0 and x = 1 and the curves y = x and y = ex (see next page) is revolved around the x-axis. Set up an integral that represents the volume of the resulting solid. 35 0 1 2 Solution: A thin vertical slice at location x has width ∆x and height ex − x. Revolving the slice thus gives a "washer" having a volume of π(ex − x)2∆x, so the desired integral is ∫ − ( π 1 0 2 ) dx x e x . The Error: The washer's volume is actually π(ex)2∆x − πx2∆x (think of the "disk" obtained by revolving y = ex minus the "hole" obtained by revolving y = x). A Correct Solution: ∫ − π 1 0 2 2 ] ) ( [ dx x e x . Problem #3: Find the second degree Taylor polynomial that approximates the function f(x) = cos(x + 2π) near x = 0. Solution: The first two derivatives of f are f′ (x) = − sin(x + 2π), f′′ (x) = − cos(x + 2π). Substituting these into the Taylor polynomial formula f(a) + f′ (a) (x−a) + 2 1 f′′ (a) (x−a)2 and using a = 0 produces cos(x + 2π) − sin(x + 2π)(x − 0) − 2 1 cos(x + 2π)(x − 0)2 = cos(x + 2π) − sin(x + 2π)x − 2 1 cos(x + 2π)x2 . The Error: The solver used the function f(x) and the derivative functions f′ (x), f′′ (x) rather than their values f(a), f′ (a), f′′ (a) in the Taylor polynomial formula. Note that the solver's answer is not even a polynomial, as it has trigonometric factors. A Correct Solution: Using f(a) = f(0) = cos(0 + 2π) = 1, f′ (a) = f′ (0) = − sin(0 + 2π) = 0 and f′′ (a) = f′′ (0) = − cos(0 + 2π) = −1, we obtain the correct Taylor polynomial 1 + 0(x − 0) + 2 1 (−1)(x − 0)2 = 1 − x2/2 . y = ex y = x x = 0 x = 1 36 Problem #4: Determine whether the series ∑ ∞ = + 1 2 n n n converges or diverges. Solution: n n n 2 lim + ∞ → = + ∞ → n n 2 1 lim = 1 + 0 = 1. The series converges to 1. The Error: The solver has confused the convergence/divergence of the sequence of terms in the series with the convergence/divergence of the series itself. A Correct Solution: Each term in the series is greater than 1, thus the series is greater than the series 1 + 1 + 1 + … , which diverges to infinity. Hence the given series diverges to infinity also. Problem #5: Determine whether the series ∑ ∞ = + 1 2 2 n n n converges or diverges. Solution: 2 2 lim n n n + ∞ → = 2 2 2 2 / / 2 / lim n n n n n n + ∞ → = 1 / 2 / 1 lim 2 n n n + ∞ → = 1 0 0 + = 0. Since the terms of the series tend to 0, the series converges. The Error: The solver is confused about the following fact (sometimes known as the divergence test): If the sequence of terms in a series does not tend to 0, the series diverges. But this does not mean that if the terms do converge to 0, then the series converges. We cannot tell what a series does just from knowing that its terms converge to 0. For example, the series ∑ 2 1 n converges while the series ∑n 1 diverges, although the two sequences of terms 2 1 n and n 1 both converge to 0. A Correct Solution: The series ∑ ∞ = + 1 2 2 n n n is larger, term-by-term, than the harmonic series ∑ ∞ =1 1 n n since for every n, n n n n n 1 2 2 2 = > + . Since the harmonic series is known to diverge, so does the given series, by comparison. Problem #6: Determine whether the series ∑ ∞ = − 1 3 2 1 2 n n n converges or diverges. 37 Solution: ∑ ∞ = − 1 3 2 1 2 n n n < ∑ ∞ =1 3 2 2 n n n (term-by-term) = ∑ ∞ =1 1 2 n n Since the harmonic series ∑ ∞ =1 1 n n diverges, so does the given series, by comparison. zzzzzz The Error: In order to use comparison to show that a series diverges, you must show that it is greater than another series that is known to diverge. In the solution above, the solver showed that the given series was smaller than a known divergent series. That tells nothing about whether the given series converges or diverges. A Correct Solution: ∑ ∞ = − 1 3 2 1 2 n n n = ∑ ∞ = − + 1 3 2 2 1 n n n n > ∑ ∞ =1 3 2 n n n (term-by-term) = ∑ ∞ =1 1 n n Since the given series is greater than the harmonic series, which diverges, so does the given series by comparison. Problem #7: Determine whether the series ∑ ∞ =10 3 ) (ln 1 n n n converges or diverges. Solution: ∑ ∞ =10 3 ) (ln 1 n n n < ∑ ∞ =10 2 1 n n (term-by-term) Sine the series on the right is a convergent p-series (p > 1), the series on the left converges by comparison. The Error: The inequality is not true: (ln n)3 grows very slowly as n → ∞, and after a while remains less than n from then on. Thus eventually, 3 ) (ln 1 n n > n n⋅ 1 = 2 1 n and so the comparison given above is not valid. A Correct Solution: We can compare the given series to the improper integral∫ ∞ 10 3 ) (ln 1 dx x x . Using the substitution w = ln x, we can show that this integral converges. By comparison, so does the series. 38 SUMMARY OF COMMON ERRORS IN SECOND SEMESTER CALCULUS Common mistakes in second semester calculus include: • failing to notice that an improper integral is improper (see #1) • subtracting before squaring, rather than after squaring, when finding the volume of a solid of revolution that has a hole (see #2) • forgetting to evaluate each derivative at x = a in Taylor polynomial problems (see #3) • concluding that an infinite series converges to a certain value when its terms converge to that value (see #4) • concluding that a series converges because its terms tend to 0 (see #5) • using infinite series comparisons the wrong way around (see #6, #7) 39 INDEX A algebraic mistakes, 19, 22 D derivative chain rule, 19, 20, 21, 22, 30 composite function, 20, 21, 22 graph, 23, 24, 26 implicit differentiation, 21, 22 interpretation, 14, 15 power rule, 16, 19 product rule, 17, 22, 25, 28 quotient rule, 16, 18, 19, 22 variables vs. constants, 17, 18, 19, 22 G graphing errors continuity, 6, 7 fractional powers, 5 roundoff error, 10, 11 SOLVER, 9, 11 wrong mode, 10, 11 wrong view, 8, 11 I indefinite integral, 27, 33 infinite series, 36, 37, 38 comparison, 36, 37, 38 divergence test, 36 inflection points, 23, 24, 26 integral antiderivative of a product, 28 Fundamental Theorem of Calculus, 29, 33 improper, 34, 37, 38 method of substitution, 29, 30, 31, 32, 33, 37 of a composite function, 30, 31, 33 of a quotient, 16, 18, 19, 22, 25, 30, 33 power rule for integrals, 28, 30, 33 variables vs. constants, 28, 32, 33 L limits, 10, 11, 12, 13, 15, 25, 33, 34 difference quotient, 13, 15 indeterminate form, 13, 15, 25 L'Hopital's Rule, 11, 25, 26 P parentheses, 3, 4, 5, 14, 17, 20, 22 S solid of revolution, 34, 38 T tangent line, 26 Taylor polynomial, 35, 38
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ResourcesAops Wiki 2016 AMC 12A Problems/Problem 23
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2016 AMC 12A Problems/Problem 23
Contents
[hide]
1 Problem
2 Solution
3 Solution 1: Super WLOG
4 Solution 2: Conditional Probability
5 Solution 3: Calculus
6 Solution 4: Geometry
7 Solution 5: More Calculus
7.1 Solution 5.1 (better explanation)
8 Solution 7: More WLOG, Complementary Probability
9 Solution 8: 3D geometry
10 Solution 9(Fastest solution if you have no time): Stick Solution
11 Video Solution by Punxsutawney Phil
12 Video Solution by The Art of Problem Solving
13 See Also
Problem
Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
Solution
Solution 1: Super WLOG
Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be and . WLOG assume is the largest. Then, , meaning the solution is , as shown in the graph below.
Solution 2: Conditional Probability
WLOG, let the largest of the three numbers drawn be . Then the other two numbers are drawn uniformly and independently from the interval . The probability that their sum is greater than is
Solution 3: Calculus
When , consider two cases:
1) , then
2), then
is the same. Thus the answer is .
Solution 4: Geometry
The probability of this occurring is the volume of the corresponding region within a cube, where each point corresponds to a choice of values for each of and . The region where, WLOG, side is too long, , is a pyramid with a base of area and height , so its volume is . Accounting for the corresponding cases in and multiplies our answer by , so we have excluded a total volume of from the space of possible probabilities. Subtracting this from leaves us with a final answer of .
Solution 5: More Calculus
The probability of this occurring is the volume of the corresponding region within a cube, where each point corresponds to a choice of values for each of and . We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when , which has area or when or , which have an area of Integrating this expression from 0 to 1 in the form
![Image 55: $\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}$
Solution 5.1 (better explanation)
This problem is going to require some geometric probability, so let's dive right in. Take three integers , , and . Then for the triangle inequality to hold, the following inequalities must be true
Now, it would be really easy if these equations only had two variables instead of , because then we could graph it in a -dimensional plane instead of a -dimensional cube. So, we assume is a constant. We will deal with it later.
Now, since we are graphing, we should probably write these equations in terms of so they are in slope-intercept form and are easier to graph.
Now, note that all solutions are in a square in the -plane because , .
I recommend drawing the following figure to get an idea of what is going on. The first line is a line with a negative slope that cuts off a triangle with side length of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a triangle with side length off the top left corner of the square. The third line also has a positive slope and cuts a triangle with side length off the bottom right corner of the square.
Note: All triangles are because the lines have slopes of or .
Using the and signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to
Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.
Except, now, we have one problem. is still a variable. But, we want to be a constant. Well, what if we just took the area over every possible value of ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!
To find the area of the triangles, our plan of attack will be
Find the area in terms of -> take the integral from to of the expression for the area (this will cover every possible value of
The area of the triangles is
.
The integral from to is
The total area of all the possible unit squares is quite obviously
Thus, the area not enclosed by the triangles is , and the total area of the square is . Thus, the desired probability is
~Extremelysupercooldude
Solution 7: More WLOG, Complementary Probability
The triangle inequality simplifies to considering only one case: . Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) is the largest, so on average (now equal to becomes a degenerate case with probability , so we no longer need to consider it). We now want , so imagine choosing at once rather than independently. But we know that is between and . The complement is thus: . But keep in mind that we choose each and randomly and independently, so if there are ways to choose together, there are ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if , then we only count this once, but in reality: we have two cases , and ; similar reasoning also generalizes to non-integral values). The complement is then actually . Therefore, our desired probability is given by
Solution 8: 3D geometry
We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines and We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length and the other has 3 sides of length and 3 sides of length The volume of this region is . Hence our solution is
Solution 9(Fastest solution if you have no time): Stick Solution
Consider a stick of length . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is probability that the second cut is on the opposite side.
-this isn't necessarily true: what if the cuts are very very close to the edges? -happypi31415
-Think about it - Judokid
-alanisawesome2018
Video Solution by Punxsutawney Phil
Video Solution by The Art of Problem Solving
AMBRIGGS
See Also
2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22Followed by
Problem 24
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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The following 48 pages are in this category, out of 48 total. This list may not reflect recent changes.
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https://www.youtube.com/watch?v=dw_ujbbN8ys
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How to Find Local Maximum and Minimum Values of a Rational Function | Calculus | Glass of Numbers
Glass of Numbers
11100 subscribers
253 likes
Description
26847 views
Posted: 19 Mar 2020
In this video, we use the First Derivative Test to find the local maximum and minimum values of a rational function.
How to find the local extreme values? We first need to find critical numbers of the function. Then plot the critical numbers on the number line (x-axis) to construct a sign analysis chart. This chart will tell us on what interval the function is increasing or decreasing. We can next apply the First Derivative Test to find the local maximum and minimum values.
Note that because we are finding the max and min values, we will finding the y-values of the function. Do not confuse this with the location of the extrema, which are the x-values.
See here how to find the local extreme values for a polynomial function:
See here how to deal with a trigonometric function with infinitely many local max and min:
See here how to find the local extreme values for a sum of exponential function:
📺 Subscribe to my channel for Math Learning!
Please help me by subscribing to my YouTube channel. I want to use my free time to make a lot of math videos to help others and to make education available to the world. Your sharing and subscribing will help this channel be reached by more people. Thank you!
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Calculus #LocalExtrema #CurveSketching
32 comments
Transcript:
hello everyone this is Wilson here we are going to find the local maximum and minimum values of this rational function and we are gonna use the first derivative test to find the local maximum minimum values okay so let's get started first we need to find the derivative so that we can find the critical numbers so when we are doing the derivative we are going to use the quotient rule for this so f prime right so let's start taking the derivative here we are going to so we are gonna differentiate the numerator which is going to be 3 so let me just highlight the differentiation here so we are gonna get 3 here and then what do we get in the we are going to copy from the denominator right so that's going to be X square plus 4 so let's know that we are doing quotient rule so we have minus and then the next step it's going to be what it's going to be just copying the Liberator and then we are going to differentiate the denominator which is going to be 2x you use the wrong color for this one but actually it doesn't really matter ok now what about the denominator for the denominator we are going to square the X square plus 4 so that's our derivative and then we can simplify by combining like terms right this you doing distributing the 3 and then combine like terms so we are going to get what do we get here we have 3x square and then that's 6x square so when we subtract them we are going to get lated 3x square so we are going to get laid' 3x square and then plus now there is only a there is only one turn left that we need to write down here that's 3 times the 4 which is 12 so now that's the numerator and we actually don't need to do anything to the denominator okay and then it would actually be a good idea to factor out legging 3 from the numerator which would give negative 3 times X square what is that that's minus 4 ok oh yeah so we can simply just write down just make a note right here - just to indicate that that's what we are doing right yeah so comes from here it comes from The Liberator ok now you know that we can factor this x squared minus 4 further right so we can factor it as continue to factor this would be what leg of the 3 and then X plus 2 X minus 2 okay so our factored form for the first derivative would be like the 3x plus 2 times X minus 2 and then we do not really touch the denominator so we'll just leave it x squared plus 4 and then that's square ok so that's the first derivative next step is that we need to find the critical numbers and so to find the critical numbers we are going to do this let's just write down the critical numbers right here so to find critical numbers we have two cases okay so one of the cases one of the cases is that we have F prime is equal to zero now for F prime equal to zero this is the fraction here what we want is that in order for a fraction to be equal to zero we want the numerator to be zero but the denominator is not zero at the same time so that means we can set this later three and then X plus two times X minus two equal to zero because that's not the same as the denominator right so the denominator will not be be zero when the numerator is equal to zero so what happened is that we are getting what do we get here we are getting this blue expression right so we have negative 3 times X plus 2 and then X minus 2 it's equal to zero okay so and then that actually tells you that the two critical numbers that we are getting is x equals negative 2 and then x equals positive 2 okay so that's case one right the second case is that f prime does not exist so we want to find X values that when we plug into the F prime it will be it will cause a problem and in this case that's when the denominator is equal to zero for a fraction so this X square plus four square is equal to zero now when we are considering only X real numbers X then we are actually not going to be able to get a real solution for this equation so there are actually no values for this case so the only two critical numbers that we have would be negative two and then positive two okay so the next step is the sign analysis chart so what we are going to do is that we are going to set up the chart right here so let's do that so here we have the sign Alyce's chart then this is actually the number line right the x axis and then we are going to label the leg of the two on there and then we maybe the positive 2 on there make sure that we label the smaller number on the left side right so even if we got to here and then lega to here we still need to label the leg that you on the left side at the llama line okay so those two critical numbers will break the lumber line into three intervals and then we are going to just pick some numbers right now okay so we are going to pick what we need to pick here between negative 2 and Q it's easy we can pick the 0 right and then on the right side of the 2 we can pick 1,000 and then on the left side of the lega - we can pick negative 1,000 so now if we are plugging 0 into our derivative then we can just put that into the factored form so that will be easier for us to figure out the sign so when we plug the 0 in there that's negative that's okay so let's just write those things down right so we are going to get the form to be what is that that's negative ok and then when we plug the 0 into this 0 plus 2 that's positive and then when we plug the 0 in here 0 minus 2 that's going to be negative and then now for the denominator because we have a square on the outside that whole quantity this is always positive right so we do not really need to worry about the bottom part right here we only need to worry about the top part to determine the sign because the bottom it's always positive so we do not really need to worry about any more so in this case we just count how many number of negative signs that we have here we have two negative signs and because that's an even number like the signs we are going to get a positive quantity for this interval here so that's positive here and so what that means is that the function is increasing from negative two to two okay so now let's do let's just keep doing the checking with the other two intervals with the 1000 year so for the 1000 when we plug in the 1000 year and this leg the 3 is always late right so we don't really need to just do any calculation and when we put the 1000 in here 1000 plus 2 is positive and then 1000 minus 2 is also positive so now because we have only one negative sign then this result would be negative so if the function is decreasing from two to infinity okay same thing here with the leg at 1,000 we are going to just do the check then so the leg of the three is always negative so we don't need to worry about it here like the 1000 that would make it a negative here and then negative 1000 minus two that will also make it look at it and because we have three legged signs right here then our result will also be a leg deep lumber when we multiply those two numbers together and so what does that mean that means we can say that the function is decreasing from negative infinity to negative two and then it's increasing from negative two to two and then it's going to be increase decreasing after the two right from two to infinity so we can actually write out all the information here we can write down the increasing in opposed right here so for the increasing intervals what is that that's actually just like it - two - all right so like the two - two and then for the decreasing intervals then we are getting what we are going to get for this interval here it's going to be a negative infinity all right negative infinity to negative two and the other one is going to be what from 2 to a positive infinity so we have 2 to a positive infinity ok now using the first derivative test we can actually see what which one is the local maximum which one is a local minimum okay so for the leg of the tube because the function changes from decreasing to increasing then we can see that the curve looks like this it first goes down and then after like the two it goes up so we have a function that looks like this so that's a local minimum and then for the tube the function changes from increasing to decreasing then we can see that that's increasing first and then decreasing after so that will actually give us the what local maximum okay so now we just need to find the values by plugging those two critical numbers back into the original function and then we can get the values okay so let's write down our results so for a local maximum value right we have what let me see so we have we have F okay so local maximum is at 2 so we gotta plug in the 2 in here and then it's going to be 3 times 2 over what is that that's 2 square plus 4 okay so let's just plug that number in there so we get the 2 and the 2 here 2 here then you do the calculation and then if we do the calculation that's going to be what's 6 6 over 8 right so that's 3 over 4 so that's our local maximum values just just one value right I think that's yes so that's just one local maximum value and then the other one the other one is a local minimum value so is what same thing here do some plugging in and calculation basic calculation right there so it's going to be like either - like the - like the two here and then if you do all that calculation you are going to get legs 3 over 4 so those are the values that refound okay and if we won the point right if we won the point we can also write down the point and so the point for this one it's going to be what 2 3 over 4 make sure you put the x value first and then the point for the local minimum it's going to be what leg the two and then lay the 3 over 4 okay yeah so that's it for this problem and then if you like this video please subscribe my channel and then give me some support thank you for watching this video I will see you next
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Math 1010 on-line - Roots and Radicals
Department of Mathematics --- College of Science --- University of Utah
Mathematics 1010 online
Roots and Radicals
Roots and Radicals deserve their own chapter and homework because they occur frequently in applications.
Let be a natural number, and let be a real number. The -th root of is a number that satisfies The number is denoted by
For example, since , and since .
The symbol is called the radical symbol, and an expression involving it is called a radical (expression).
If then is thesquare root of and the number is usually omitted. For example,
If , then is the cube root of . For example, the cube root of is , and that of is .
If is even and is positive then there are two -th roots of , each being the negative of the other. For example, since there are two square roots of . In that case by convention the symbol means the positive -th root of , and it is called the principal (-th) root of .
If is negative and is odd then there is just one -th root, and it is negative also. For example,
At this stage we do not know of an -th root if is even and is negative. This leads to the subject of complex numbers which we will take up later in the course.
Radicals are just special cases of powers, and you can simplify much of your thinking by keeping this fact in mind:
It follows immediately from that observation and the properties of powers that
Solving Radical Equations
An equation involving radicals is called a radical equation (naturally). To solve it you simply apply our generalprinciple:
To solve an equation figure out what bothers you and then do the same thing on both sides of the equation to get rid of it.
To get rid of a radical you take it to a power that will change the rational exponent to a natural number. This will work if the radical is on one side of the equation by itself.
Let's look at a few simple examples:
Suppose
We proceed as follows:
Here is a slightly more complicated problem:
We obtain
Our last example shows how to get rid of more than one radical:
To get rid of the square roots we isolate them and square one at a time:
In each case, we check our answer by substituting it in the original equation. For example, in the last equation we obtain:
Later in the course we will consider more complicated cases of radical equations.
Numerical Values
The radicals in the above examples were all natural numbers. This is due only to a judicious choice of examples. Frequently the roots occurring in applications are irrational numbers with decimal expansions that never repeat or terminate. The following table lists approximations of a few specific radicals.
Some Radicals (Approximately)
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Archimedean Tilings and Egyptian Fractions | Azimuth
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Archimedean Tilings and Egyptian Fractions
Ever since I was a kid, I’ve loved Archimedean tilings of the plane: that is, tilings by regular polygons where all the edge lengths are the same and every vertex looks alike. Here’s my favorite:
There are also 11 others, two of which are mirror images of each other. But how do we know this? How do we list them all and be sure we haven’t left any out?
The interior angle of a regular -sided polygon is obviously
since it’s a bit less than 180 degrees, or and how much?— well, times a full turn, or But these ‘s are getting annoying: it’s easier to say ‘a full turn’ than write Then we can say the interior angle is
times a full turn.
Now suppose we have an Archimedean tiling where polygons meet: one with sides, one with sides, and so on up to one with sides. Their interior angles must add up to a full turn. So, we have
or
or
So: to get an Archimedean tiling you need n whole numbers whose reciprocals add up to one less than n/2.
Looking for numbers like this is a weird little math puzzle. The Egyptians liked writing numbers as sums of reciprocals, so they might have enjoyed this game if they’d known it. The tiling I showed you comes from this solution:
since it has 3 polygons meeting at each vertex: a 4-sided one, a 6-sided one and a 12-sided one.
Here’s another solution:
It gives us this tiling:
Hmm, now I think this one is my favorite, because my eye sees it as a bunch of linked 12-sided polygons, sort of like chain mail. Different tilings make my eyes move over them in different ways, and this one has a very pleasant effect.
Here’s another solution:
This gives two Archimedean tilings that are mirror images of each other!
Of course, whether you count these as two different Archimedean tilings or just one depends on what rules you choose. And by the way, people usually don’t say a tiling is Archimedean if all the polygons are the same, like this:
They instead say it’s regular. If modern mathematicians were inventing this subject, we’d say regular tilings are a special case of Archimedean tilings—but this math is all very old, and back then mathematicians treated special cases as not included in the general case. For example, the Greeks didn’t even consider the number 1 to be a number!
So here’s a fun puzzle: classify the Archimedean tilings! For starters, you need to find all ways to get whole numbers whose reciprocals add up to one less than . That sounds hard, but luckily it’s obvious that
since an equilateral triangle has the smallest interior angle, of any regular polygon, and you can only fit 6 of them around a vertex. If you think a bit, you’ll see this cuts the puzzle down to a finite search.
But you have to be careful, since there are some solutions that don’t give Archimedean tilings. As usual, the number 5 causes problems. We have
but there’s no way to tile the plane so that 2 regular pentagons and 1 regular decagon meet at each vertex! Kepler seems to have tried; here’s a picture from his book Harmonices Mundi:
It works beautifully at one vertex, but not for a tiling of the whole plane. To save the day he had to add some stars, and some of the decagons overlap! The Islamic tiling artists, and later Penrose, went further in this direction.
If you get stuck on this puzzle, you can find the answer here:
• Michal Krížek, Jakub Šolc, and Alena Šolcová, Is there a crystal lattice possessing five-fold symmetry?, AMS Notices59 (January 2012), 22-30.
• Combinations of regular polygons that can meet at a vertex, Wikipedia.
Not enough?
In short, all Archimedean tilings of the plane arise from finding whole numbers whose reciprocals sum to But what if the total is not enough? Don’t feel bad: you might still get a tiling of the hyperbolic plane. For example,
so you can’t tile the plane with 3 heptagons meeting at each corner… but you still get this tiling of the hyperbolic plane:
which happens to be related to a wonderful thing called Klein’s quartic curve.
You don’t always win… but sometimes you do, so the game is worth playing. For example,
so you have a chance at a tiling of the hyperbolic plane where five equilateral triangles and a square meet at each vertex. And in this case, you luck out:
For more beautiful pictures like these, see:
• Uniform tilings in hyperbolic plane, Wikipedia.
• Don Hatch, Hyperbolic tesselations.
Too much?
Similarly, if you’ve got reciprocals that add up to more than , you’ve got a chance at tiling the sphere. For example,
and in this case we luck out and get the snub dodecahedron. I thought it was rude to snub a dodecahedron, but apparently not:
These tilings of the sphere are technically called Archimedean solids and (if all the polygons are the same) Platonic solids. Of these, only the snub dodecahedron and the ‘snub cube’ are different from their mirror images.
Fancier stuff
In short, adding up reciprocals of whole numbers is related to Archimedean tilings of the plane, the sphere and the hyperbolic plane. But this is also how Egyptians would write fractions! In fact they even demanded that all the reciprocals be distinct, so instead of writing 2/3 as , they’d write
It’s a lousy system—doubtless this is why King Tut died so young. But forget about the restriction that the reciprocals be distinct: that’s silly. If you can show that for every the number can be written as for whole numbers you’ll be famous! So far people have ‘only’ shown it’s true for up to a hundred trillion:
• Erdös–Straus conjecture, Wikipedia.
So, see if you can do better! But if you’re into fancy math, a less stressful activity might be to read about Egyptian fractions, tilings and ADE classifications:
• John Baez, This Week’s Finds in Mathematical Physics (Week 182).
This only gets into ‘Platonic’ or ‘regular’ tilings, not the more general ‘Archimedean’ or ‘semiregular’ ones I’m talking about today—so the arithmetic works a bit differently.
For the special magic arising from
see:
• John Baez, 42.
In another direction, my colleague Julie Bergner has talked about how they Egyptian fractions show up in the study of ‘groupoid cardinality’:
• Julie Bergner, Groupoids and Egyptian fractions.
So, while nobody uses Egyptian fractions much anymore, they have a kind of eerie afterlife. For more on what the Egyptians actually did, try these:
• Ron Knott, Egyptian fractions.
• Egyptian fractions, Wikipedia.
Related
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26 Responses to Archimedean Tilings and Egyptian Fractions
jim stasheff says: 5 February, 2012 at 2:12 pm
Thanks for the colorful pix and even an animation.
Reply
2. Todd Trimblesays: 5 February, 2012 at 3:19 pm
Along these lines, one of my current favorite Archimedean solids is the truncated icosidodecahedron, which looks sort of like your first favorite at the top of the article. It’s got a pleasing mix of shapes and sizes, and it’s the Cayley graph of the Coxeter group (which is in turn the full group of isometries of the icosahedron).
It’s also called “the great rhombicosidodecahedron”. Don’t you love these quaint old-fashioned names? :-)
Reply
John Baezsays: 5 February, 2012 at 10:44 pm
I too love the great rhombicosidodecahedron! It’s great! Since I’m the only one who can include images in the comments here, it’s my duty to show everyone what we’re talking about:
As you hinted, it’s what we get when we replace the hexagons in the great rhombitrihexagonal tiling by pentagons:
So, here’s a question for you, Todd: is this tiling also the Cayley graph of some group? Maybe a group coming from an affine Coxeter diagram?
Don’t you love these quaint old-fashioned names? :-)
Yes, I think the name of a polytope should resemble its appearance, so if it has lots of faces of different shapes its name needs lots of shiny, sophisticated syllables.
The modern standard name for the ‘great rhombitrihexagonal tiling’ is apparently the ‘truncated trihexagonal tiling’. Conway has his own names for all these, and for him it’s the ‘truncated hexadeltille’. I like the idea of extended Schläfli symbols, which seek to reduce this fancy nomenclature to a logical numerical code, but even these seem to stagger under the sheer diversity of interesting polytopes. Hmm, Wikipedia says these gave way to Coxeter diagrams. That suggests I could look up the answer to my question to you—but I won’t!
Reply
Todd Trimblesays: 5 February, 2012 at 11:30 pm
My off-hand guess is yes, it’s the Cayley graph of the Coxeter group whose Coxeter matrix is 3 by 3 and whose off-diagonal entries are , , and . However, I’m drinking right now (it’s the Superbowl), and last time I drank and posted I said some things which were embarrassingly wrong. I should come back to this later, when I can look into this more closely.
Reply
John Baezsays: 6 February, 2012 at 8:59 am
It sounds like you’re guessing what I’d guess: you’re taking the Coxeter diagram here:
and jacking up the 5 by one to get this:
6
o---o---o
If that’s right, I feel like calling the result an ‘affine Coxeter diagram’, but I don’t think I’ve actually ever seen it. Hmm… oh yes, it’s called in this list on Wikipedia. They even discuss how these diagrams are related to ‘uniform tilings’ of the plane, which I’ve been calling Archimedean tilings.
Todd Trimblesays: 5 February, 2012 at 9:30 pm
Just found out that a ‘snub’ — also called an alternation — is a certain process for deriving polytopes. It’s also called ‘snubification’. :-)
Reply
4. Sammy Black says: 5 February, 2012 at 10:53 pm
I think that you have a typo in the formula where you introduce Kepler’s attempted tiling (5, 5, 10).
Reply
John Baezsays: 5 February, 2012 at 11:16 pm
You’re right! I’ll fix it now. Thanks!
Reply
chalico6says: 6 February, 2012 at 3:06 am
Great going John, Bravo!
I’m not a scientist, but I love new angles on things – gravity as an entropic by-product, amazing!
Reply
6. Mike Staysays: 7 February, 2012 at 5:41 am
Since hyperbolic space has an intrinsic length scale, if you make polygons very small, they begin to behave more like Euclidean ones.
If you made the heptagons in the (7,3) tiling of the hyperbolic plane a little larger, they wouldn’t quite reach all the way around, and would begin to leave a gap rather like pentagons do in Euclidean space. Is there a regular hyperbolic polyhedron whose sides are all heptagons?
Reply
John Baezsays: 7 February, 2012 at 9:22 am
I have no idea what you mean by a “regular hyperbolic polyhedron”!
If you hadn’t said more, I’d assume you meant a tiling of the hyperbolic plane by regular polygons—just as an ordinary regular polyhedron is a tiling of the 2-sphere by regular polygons. Spherical geometry and hyperbolic geometry are the two most symmetrical non-Euclidean 2d geometries.
And so, if someone asked me about a “regular hyperbolic polyedron whose sides are all heptagons”, I’d show them this:
But, given the rest of what you said, I’d guess you’re thinking about a tiling of the “hyperbolic 2-sphere” by regular heptagons. The problem is, there’s no such thing as the “hyperbolic 2-sphere”: thanks to the Gauss–Bonnet theorem, there’s no way to give the 2-sphere a Riemannian metric with everywhere negative curvature.
Take a n-holed torus, give it a Riemannian metric, and tile it with heptagons whose sides are geodesics and whose interior angles are the usual interior angles of a regular heptagon. Then n can’t be 0 or 1; it must be 2 or more. Again this is a spinoff of the Gauss–Bonnet theorem.
Luckily, you can give the 3-holed torus a metric with constant negative curvature, and tile it by 24 heptagons whose sides geodesics and whose interior angles are the usual interior angles of a regular heptagon. This gadget is the closest thing I have to a good answer to your question. This is called ‘Klein’s quartic curve’ and I have a nice webpage on it.
It’s hard to visualize this 3-holed torus because you can’t embed it in Euclidean 3d space in a metric-preserving way. But that webpage has lots of pictures. In particular, you can get Klein’s quartic curve by taking the hyperbolic plane and ‘curling it up’, identifying heptagons that have the same numbers here:
This in fact gives the most symmetrical Riemannian metric possible on a 3-holed torus! The symmetry group of the 3-holed torus with this metric has 336 elements. That number is 24 × 7 × 2. You’ve seen the numbers 24 and 7 in my discussion already; the factor of 2 comes from an orientation-reversing symmetry.
(Coincidentally, 336 is the number of hours in a fortnight).
Reply
Mike Staysays: 7 February, 2012 at 1:16 pm
The hyperbolic plane has a constant negative Gaussian curvature K. Define ; this is a length scale intrinsic to the space. A regular heptagon will have the sum of its exterior angles be greater than 360 degrees, but how much greater depends on the size of the heptagon. Only when the heptagon is the correct size will the exterior angle be 60 degrees so we can fit three of them together at a vertex to get the (7,3) tiling. If they’re large enough, the external angle becomes 90 degrees and we get the (7,4) tiling.
Between these two sizes, seven heptagons around a central one leave a gap, rather like five pentagons around a central one. In Euclidean 3-space we can curl up pentagons into a dodecahedron; we can do the same in hyperbolic 3-space for pentagons of the correct size whose external angles are 90 degrees—in fact, these tile hyperbolic 3-space and one appears in the Mathematica logo. Since these “hyperbolic dodecahedra” exist, I was wondering if one can use the gap between heptagons in the hyperbolic plane whose size is somewhere between the ones in the (7,3) and (7,4) tiling of the plane to curl them up into a polyhedron with regular heptagons as sides.
Reply
Mike Staysays: 7 February, 2012 at 3:57 pm
I guess I might as well start with hexagons rather than heptagons. Is it possible to have a hyperbolic polyhedron with three hexagons meeting at a vertex? The hyperbolic dodecahedron has dihedral angles of 90 degrees and tiles hyperbolic 3-space; every vertex has three orthogonal lines passing through it, and there are planar slices that result in the (5,4) tiling of the hyperbolic plane. What prevents a similar construction involving hexagons and the (6,4) plane?
Reply
John Baezsays: 7 February, 2012 at 11:21 pm
I still don’t really understand what you’re saying, so I’ll just blather at random.
I’m pretty sure it’s impossible to take a sphere and subdivide its surface into just hexagons, or indeed any mix of just hexagons, heptagons, octagons, nonagons and so on. You need some polygons with fewer sides! This should be a purely topological result, nothing to do with the geometry of the sphere or the polygons: their edges don’t need to be ‘straight’ in any sense. It should follow from Euler’s formula
where is the genus, or number of holes in our doughnut; here .
It’s easiest for me to explain this in the case where we imagine subdividing a sphere into heptagons with meeting at each vertex. You can easily generalize what I’m about to say.
In this special case we have
since each face has 7 edges but each edge is shared by 2 faces. We also have
since each vertex has 3 edges coming out of it, but each edge ends at 2 vertices.
So
and thus
so the problem is, there’s no way to obtain
with a positive number of faces. (Please don’t invent polyhedra with a negative number of faces to get around this, at least not today.)
So, we can’t ’tile’ a 2-sphere, no matter how irregularly, with heptagons meeting three at a vertex. Nor can we do a torus, since that would require . If we try a 2-holed torus we get
so maybe we can do that with 12 heptagons—I don’t know. The case I’m most familiar with is the 3-holed torus, which gives
or . That’s why we can tile a 3-holed torus with 24 heptagons meeting 3 at each vertex! Or dually we can tile the 3-holed torus with triangles, 7 meeting at each of 24 vertices, as shown in this picture by Greg Egan:
If you allow more than 3 heptagons to meet at some vertices we instead have
and I think you can show it’s still impossible to tile a sphere, which would require
I get the impression you’re trying to violate such laws of topology.
Reply
Mike Staysays: 8 February, 2012 at 12:29 am
That’s a great proof, thank you!
John Baezsays: 10 February, 2012 at 7:34 am
Thanks!
I noticed yesterday that we can easily tile a sphere with just two heptagons: take one to be the northern hemisphere, one to be the southern hemisphere, and divide the equator into 7 segments.
But in this case we only have two heptagons meeting at each vertex: that’s a bit degenerate.
So let me show that you can’t subdivide a sphere into just polygons with 6 or more sides, if at least 3 polygons meet at each vertex. By the reasoning above, if we could do this we’d have
So, expressing everything in terms of edges, we get
so
so
which contradicts
This argument also gives a bit more. For a torus we must have
There are lots of ways to subdivide a torus into hexagons that meet 3 at a vertex: just take a hexagonal lattice and curl it up. But if we try to stick in one or more polygons with more than 6 edges, or try to have more than 3 hexagons meet somewhere, the argument above shows we’re doomed: then we’ve tipped the situation over from
to
A similar calculation shows that if you subdivide a sphere into just hexagons and pentagons, with 3 meeting at each vertex, there have to be exactly 12 pentagons. People working on fullerenes know this well! The first interesting example is this:
but people go further, like this:
Forrest Sheldon says: 7 February, 2012 at 10:04 am
I read this blog quite often and have never had much to share but I think my friends post that I read moments before this fits this blog quite nicely…It may not be “Great”, but behold, the recycled rhombicosidodecahedron!
Reply
John Baezsays: 7 February, 2012 at 10:55 am
Nice, Forrest Sheldon! Here’s one photo from that blog post… but people here should read the whole thing:
Reply
Todd Trimblesays: 7 February, 2012 at 11:58 am
Awesome! These past few weeks I’ve had a craving for a bunch of Magnetix, precisely to construct such a thing… but you can’t buy them in stores anymore because of a recall a few years back.
Reply
Fourteenth Linkfestsays: 7 February, 2012 at 5:59 pm
[…] John Baez: Archimedean Tilings and Egyptian Fractions […]
Reply
9. Circle Limit III animation | Newton Excel Bach, not (just) an Excel Blogsays: 16 November, 2012 at 7:29 am
[…] Hyperbolic geometry is fun. And it’s much easier to create regular & semi-regular tessellations in the hyperbolic plane than in the Euclidean plane. John Baez discussed this topic a little while ago: Archimedean Tilings and Egyptian Fractions […]
Reply
10. 42 | Azimuthsays: 25 May, 2013 at 3:44 pm
It turns out that if you look for natural numbers that make
as close to 1 as possible, while still less than 1, the very best you can do is It comes within of equalling 1, and this has amazing consequences.
Reply
11. Marek14 says: 25 February, 2016 at 5:20 pm
Archimedean hyperbolic tilings get much more interesting with 4 or more edges per vertex. At this point, there is a possibility of having several different tilings which, nonetheless, have the same sequence of polygons around a vertex.
For example, a tiling with three hexagons and one triangle around each vertex has three different forms. With more complex vertex configurations, there can be many distinct tilings.
Reply
John Baezsays: 25 February, 2016 at 8:33 pm
Is there a place I can see pictures of these?
Reply
Marek14 says: 25 February, 2016 at 9:11 pm
Well, basics of the theory is here:
Some pictures are at and
Lately, I tried using HyperbolicApplet ( but it requires some image editing to get a proper colored picture.
Reply
Solve abc+bcd+cda+dab=abcd+3abc+bcd+cda+dab=abcd+3 in Nmathbb{N}. – Math Solutionsays: 10 March, 2022 at 4:40 pm
[…] in classifying tilings of the regular surfaces (the plane, sphere, and hyperbolic plane); see e.g. for a nice discussion of […]
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Pregnancy and isotretinoin therapy - PMC
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Pregnancy and isotretinoin therapy
June Seek Choi
June Seek Choi, MD
1 From the Motherisk Program (Choi, Koren, Nulman) Division of Clinical Pharmacology/Toxicology, Hospital for Sick Children, Toronto, Ont., and Cheil General Hospital and Women’s Healthcare Centre (Choi), Seoul, Republic of Korea.
Find articles by June Seek Choi
1, Gideon Koren
Gideon Koren, MD
1 From the Motherisk Program (Choi, Koren, Nulman) Division of Clinical Pharmacology/Toxicology, Hospital for Sick Children, Toronto, Ont., and Cheil General Hospital and Women’s Healthcare Centre (Choi), Seoul, Republic of Korea.
Find articles by Gideon Koren
1, Irena Nulman
Irena Nulman, MD
1 From the Motherisk Program (Choi, Koren, Nulman) Division of Clinical Pharmacology/Toxicology, Hospital for Sick Children, Toronto, Ont., and Cheil General Hospital and Women’s Healthcare Centre (Choi), Seoul, Republic of Korea.
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1,✉
Author information
Copyright and License information
1 From the Motherisk Program (Choi, Koren, Nulman) Division of Clinical Pharmacology/Toxicology, Hospital for Sick Children, Toronto, Ont., and Cheil General Hospital and Women’s Healthcare Centre (Choi), Seoul, Republic of Korea.
✉
Correspondence to: Irena Nulman, irena.nulman@sickkids.ca
© 1995-2013, Canadian Medical Association
PMC Copyright notice
PMCID: PMC3602257 PMID: 23296582
A 29-year-old woman was given isotretinoin for severe, recalcitrant acne. During the first 3 months of treatment, she was taking an oral contraceptive agent that she had started at the same time as the isotretinoin. After deciding she wanted to become pregnant, the patient stopped taking the contraceptive agent but continued to take isotretinoin for another 2 months.
When the patient became pregnant, she sought advice from the Motherisk program regarding the fetal risks of exposure to isotretinoin. With counselling, it became apparent that the patient had not understood the connection between isotretinoin and birth defects, and hence the need for continued effective contraception while taking the drug. After repeated counselling, the patient decided to terminate her pregnancy.
Discussion
Oral isotretinoin was introduced in the United States in 1982 and in Canada in 1983 for the treatment of severe recalcitrant acne. In the intervening decades, additional off-label indications have become established, including the treatment of gram-negative folliculitis, recalcitrant rosacea, pyoderma facial, generalized lichen planus, psoriasis, cutaneous lupus erythematosus and acne fulminans.
Although effective for treating severe acne, isotretinoin is a potent human teratogen. There is an estimated 20%–35% risk for congenital defects in infants exposed to the drug in utero,1 including craniofacial, cardiovascular, neurological and thymic malformations. About 30%–60% of children exposed to isotretinoin prenatally have been reported to show neurocognitive impairment, even in the absence of physical defects.1
The gold standard for preventing fetal exposure to isotretinoin relies on the following tenets: the woman’s awareness of the risks; ruling out pregnancy at the start of treatment; the use of 2 forms of contraception in parallel during treatment; monitoring for pregnancy throughout treatment; and avoiding pregnancy until the drug has been cleared from the body.2 Although these principles have been incorporated in all regulatory programs aimed at preventing fetal risk, repeated studies have shown failures in each step.
Causes of isotretinoin exposure in pregnancy
The most commonly reported reasons for pregnancy during isotretinoin therapy are unsuccessful attempts at abstaining from sexual intercourse, ineffective or inconsistent use of contraceptive agents or failure to detect pregnancy.2
Contraceptive failure is a leading reason for pregnancy during isotretinoin treatment. Despite acknowledging their familiarity with the risks of isotretinoin to the fetus, 7%–59% of women taking the drug do not use any form of contraception,3–5 and as many as 80% of women do not use 2 methods of birth control in parallel, as recommended.2,5 Our patient, for example, used only a single method of contraception. Improper use of contraceptives, in our experience, is the most common reason for unexpected pregnancies among women using isotretinoin and highlights the need for more comprehensive contraceptive counselling for all women starting treatment with this drug.2 Although most women know that taking isotretinoin confers a high risk of congenital defects if pregnancy occurs,3–5 they still choose to use it for the treatment of acne.
Many women taking isotretinoin do not monitor for pregnancy before and during treatment. Only 44%–67% of women report taking a pregnancy test before starting isotretinoin therapy, and only about 1 in 5 reports taking a repeated pregnancy test.6 Moreover, most women use over-the-counter pregnancy tests for human chorionic gonadotropic hormone. These tests have limited sensitivity and may result in false-negative results, particularly if they are done shortly after conception.7 Repeat testing may lead to earlier detection of pregnancy and subsequent earlier cessation of treatment or pregnancy termination.8
Variability in the pharmacokinetics of isotretinoin may also lead to exposure during pregnancy. Continued use of contraception for 1 month after the end of isotretinoin therapy has been advocated to provide an adequate safety margin to prevent fetal exposure, based on more than 5 elimination half-lives of the drug.9 However, there is evidence of greater variability in isotretinoin elimination half-life (from 5.3 h to 7 d); thus, 1 month may not allow clearing of the drug in all women.10 Indeed, there have been reports of isotretinoin-induced fetal malformations despite stopping treatment 1 month before pregnancy.11 A 3-month window may be more appropriate.
Preventing fetal exposure to isotretinoin
In the past 3 decades, several programs have been started in an attempt to prevent pregnancy during isotretinoin therapy. Although these programs have been successful in keeping pregnancy rates low relative to the general population of women of childbearing age who use contraception, they have all been ineffective in completely eliminating and preventing fetal exposure to the drug. In 1988, Roche and the US Food and Drug Administration (FDA) introduced the Retinoid Pregnancy Prevention Program. This program included explaining to women the risk to their unborn babies and obtaining their written agreement to use 2 methods of contraception in parallel, have monthly pregnancy tests and otherwise comply with the program. This program has been practised in Canada ever since. An evaluation of the program in 1995 showed that despite 99% of patients recalling being told to avoid pregnancy, 36% failed to receive a pregnancy test before starting treatment.12 An assessment in 2000 concluded that the program was unable to adequately prevent pregnancies during treatment, suggesting the need for improvement.13
In April 2002, the System to Manage Accutane-Related Teratogenicity (SMART) program began in the US to prevent women from starting isotretinoin therapy during pregnancy and to prevent conception during treatment. In addition to the steps included in the Retinoid Pregnancy Prevention Program, SMART called for 2 consecutive pregnancy tests with negative results before starting treatment and a voluntary registration system. The program did not include verification of compliance by physicians and pharmacists. Similar to the Retinoid Pregnancy Prevention Program, SMART has also been shown to be ineffective: 34% of women did not receive the required 2 pregnancy tests before starting treatment, and 54% of women of childbearing age did not use 2 methods of birth control in parallel during treatment.13
In 2006, the FDA implemented iPLEDGE, a program designed to provide more exhaustive guidelines for the safe prescription and use of isotretinoin. This program enforces monthly pregnancy tests, documentation of contraceptive use and patient education to reinforce key safety messages. Furthermore, all patients with prescriptions for isotretinoin and their pharmacists are registered in a single database.8 Seven years after its introduction, this program does not appear to have solved the issue of fetal exposure to isotretinoin. Shin and colleagues recently reported 29 fetal exposures out of a total of 9912 courses of treatment with the drug.14 The authors found no evidence that iPLEDGE significantly decreased the risk of fetal exposure compared with the SMART program. Health Canada’s Scientific Advisory Panel on isotretinoin has unanimously rejected the idea of a similar registry in Canada.15
In addition, we are facing a serious new challenge. After decades of the availability of only a single isotretinoin product (Accutane, Roche Inc.), there is now broad availability of multiple products, particularly on the Web. There are 232 types of generic and original isotretinoin made in 36 countries around the world, with 4 versions (Accutane, Accutane Roche, Clarus and Isotrex) currently available in Canada. Isotretinoin can be purchased online from different sources, without prescription or patient education. The amount of isotretinoin circulating in Canada is probably higher than estimated, and many users will not have received proper drug safety education.
Multilevel approaches
Preventing fetal exposure to isotretinoin requires intervention at multiple levels. In addition to the programs currently in place, other options for system intervention include banning online sales of isotretinoin, as well as mandatory online certification programs for the physicians and pharmacists providing the drug that include education about contraceptive methods and the drug’s therapeutic and fetotoxic effects. A similar program currently exists for health care providers prescribing methadone in Canada, which appears to function well16
It is the responsibility of the prescribing physician to counsel women and their sexual partners as to the teratogenic effects of isotretinoin. This advice should include appropriate information about contraceptive methods and how to properly perform pregnancy tests. However, in a survey of internists, only 65% of responding physicians appeared to have correct knowledge of published failure rates of contraceptive agents (Box 1).17 In that study, most of the physicians agreed there was a need for continuing medical education on counselling patients about potential teratogens and contraception.17 This education should precede patient education.17
Box 1: Failure rate (%) of common contraceptive methods under typical use.
Hormonal implant = 0.05
Combined injectable hormones = 0.2
Intrauterine device (IUD) with progesterone = 0.2
IUD (copper) = 0.8
Medroxyprogesterone acetate = 3
Spermicidal agents = 8
Combined contraceptive pill = 8
Male latex condom = 15
Diaphragm and spermicide = 16
Standard day method (calendar method) = 25
Source: General Practice Notebook, available at www.gpnotebook.co.uk/simplepage.cfm?ID=-1590362086
Physicians in clinical practice are typically very busy, with little time to educate patients about effective contraception; some patients may have limited ability to understand written or verbal instructions in English or French; and teenage patients may be hesitant to discuss their sexual activities with health care providers. Appropriate contraceptive counselling takes time and may not be effectively achieved during an interview lasting only a few minutes. Multidisciplinary involvement may be helpful, including the reinforcement of information (both written and verbal) by pharmacists and other health care providers.
Conclusion
The medical community has a major role to play in ensuring that babies are not affected by prescription drugs taken during pregnancy. The continued occurrence of fetal exposure to isotretinoin highlights the need for improved practices. These improvements will likely require intervention at multiple levels, including improved education for patients and health care providers, improved patient adherence with established safety protocols and proper control of isotretinoin’s distribution.
Key points
Isotretinoin is a known teratogen, but women taking this drug may fail to use effective contraception.
Current programs in Canada do not appear to be effective in preventing pregnancies during isotretinoin therapy.
Prescribing physicians should ensure that women of reproductive age who take isotretinoin understand the fetal risks of this drug and use effective contraception.
Preventing fetal exposure to isotretinoin will require system-level internventions in addition to those at the level of individual health care providers and patients.
Footnotes
Competing interests: None declared.
This article has been peer reviewed.
Contributors: June Seek Choi wrote the first draft of the paper, and revised and edited it for important intellectual content. Irena Nulman participated in conception and design and the paper, and the acquisition and interpretation of the data. Gideon Koren participated in the concept and design of the paper, its drafting and revision, and the interpretation of the data. All of the authors approved the final version submitted for publication.
References
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13.Brinker A, Kornegay C, Nourjah P. Trends in adherence to a revised risk-management program designed to decrease or eliminate isotretinoin-exposed pregnancies: evaluation of the Accutane SMART program. Arch Dermatol 2005;141:563–9 [DOI] [PubMed] [Google Scholar]
14.Shin J, Cheetham TC, Wong L, et al. The impact of the iPLEDGE program on isotretinoin fetal exposure in an integrated health care system. J Am Acad Dermatol 2011;65:1117–25 [DOI] [PubMed] [Google Scholar]
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https://ccom.ucsd.edu/~ctiee/notes/grad_n_curl.pdf
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1 Why Gradients Must Have Zero Curl M20E CT 1 Why must the gradient have zero curl?
The curious reader may have asked the question “Why must the gradient have zero curl?” The answer, given in our textbook and most others is, simply “equality of mixed partials” that is, when computing the curl of the gradient, every term cancels another out due to equality of mixed partials. That’s all well and good, analytically, and the fact that mixed partials are equal at all is a deep and important result on its own.1 But what does this mean geometrically?
Let’s recall what a gradient field ∇f actually is, for f : R2 →R (using 2D to assist in visualiza-tion), in terms of the scalar function f. It is a vector pointing in the direction of increase of f, pointing away from the level curves of f in the most direct manner possible, i.e. perpendicularly. But what are the level curve, anyway? The sets f(x, y) = c; this value of c is the height of the graph of f on that curve. Hence ∇f always points in the uphill direction of a graph. For example, with the function f(x, y) = x2 + y2 we have ∇f = (2x, 2y) (see pictures below, the right hand figure puts the level curves and gradient field together): -2 0 2 -2 0 2 0 2 4 6 8 -2 0 2 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 Now let’s take a look at our standard Vector Field With Nonzero curl, F(x, y) = (−y, x) (the curl of this guy is (0, 0, 2): 1In fact, a fellow by the name of Georg Friedrich Bernhard Riemann developed a generalization of calculus which one can apply to curved spaces, in which space is flat if and only if (the sutiably generalized notion of) mixed partials are equal.
2 1 WHY MUST THE GRADIENT HAVE ZERO CURL?
What’s wrong with this picture? If all those vectors always point in the direction of increase, then if we traverse a circle counterclockwise, f goes uphill all the way around in a circle. The nostalgic reader may remember the old gag parents tell their children “When I was your age, I had to walk to school uphill both ways,” and intuitively knew that this was of course impossible. For if we made such a trip and returned to our starting point, how could we be at a higher level? How can we be walking uphill the whole time? (Actually the fact that the parents’ claim is wrong is precisely the fact that the gravitational field is in fact a gradient field! This is not a loosely relevant insert just to make the reader laugh, it is a direct consequence of everything we are talking about right now). But nevertheless let’s examine the situation a little more closely: Suppose we could find f such that ∇f = F. Then looking at the “level curves” of such a function we would have something like the left hand figure below: -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -2 0 2 -2 0 2 0 0.5 1 1.5 2 -2 0 2 0 5 1 The straight level curves—here line segments from the origin to a point on the circle r = 3—would increase in value as we go counterclockwise. To get back to the same height we would have to make a sharp drop; the right hand figure shows a potential (ha ha) candidate for a function f such that F = ∇f—i.e., if there were an f such that ∇f = F, then it would have to look like the thing on the right above. But sharp drops are extremely bad from the standpoint of differential calculus. This is not 3 anyone’s idea of a differentiable function. Even if the sharp drop were approximated by some steeply sloping, yet differentiable function (this is always possible by using using what are called “smooth bump functions”), this would imply we sketched our vector field wrong. During the steep drop, the gradient field would suddenly swerve into the opposite direction and have a very large magnitude for a short time. But the field (−y, x) never does anything like that, so such a function could never correspond to (−y, x).
Right now the suspicious reader may be wondering, “Well, why can’t we just keep going up and up instead of having to make that drop?” Good question. Recall that we are speaking of the graph of the function f. A function can have only one value for each point in its domain, so its graph must pass the so-called “vertical line test”—a vertical line (line parallel to the z-axis) cannot intersect the graph at two distinct points.2 The astute reader may have recalled that there is a technique to try to compute the “inverse gra-dient” — given a vector field F we seek to construct an f such that ∇f = F. For example for F(x, y) = (2x, 2y), we note that if such an f exists, fx = ∂f/∂x = 2x implies f(x, y) = x2 + C(y) for some function C. C is a function of only y because we did a “partial integration” with respect to x and so we must have a “constant” of integration—but any function of only y is constant with respect to x. So we know partially (ha ha) what f must look like. Now we know fy = 2y and so C′(y) = 2y.
This means C(y) = y2 + K where K is now a genuine constant (because C was only a function of y).
Therefore any f such that ∇f = (2x, 2y) must have the form f(x, y) = x2 + y2 + K for any constant K.
Now let’s try it on F(x, y) = (−y, x) which we know isn’t the gradient of anything. Suppose such an f exists. Then fx = −y. Therefore f(x, y) = −yx + C(y) for some function C of y only. Then fy = x = −x + C′(y). This means C′(y) = 2x or that C(y) = 2xy + K for some constant K. But this is Very Bad, because we assumed that C was a function of y only, whereas this shows quite blatantly it is not a function of only y. But still, the stiff-necked reader may insist upon taking f(x, y) = −xy+2xy+K anyway. This gives f(x, y) = xy + K. But ∇f = (y, x) which is most certainly not (−y, x) so still no luck. And since its curl is not zero, further efforts along these lines will be thwarted.
2 The Converse Question The exceedingly inquisitive reader is likely to have asked the question: Does ∇× F = 0 imply F = ∇f for some f? The answer is a resounding sometimes. This is, in fact, a very deep question in which the most precise answer one can get requires graduate-level mathematics. Or, as the cinephilic reader may recall, mathematics that John Nash demanded of his vector calculus students. The scene is where John Nash throws a multivariable calculus text into the trash and writes up on the board the following: Given V = {F : R3 ∖X →R3 | ∇× F = 0} and W = {F | F = ∇g}, what is dim(V/W)? A full answer to this requires something called the de Rham cohomology. Asking whether or not the converse hold is equivalent to asking whether or not the answer to this question is 0. It turns out that for X = ∅, that is for vector fields F defined on all of R3, the converse is in fact true, and we can always find an f such that ∇f = F. For vector fields that omit one point (for example 2Actually, one can define graphs of functions like this, called Riemann surfaces, yes that same Riemann in the last footnote. But the usual methods of multivariable calculus cannot be used to handle these objects.
4 2 THE CONVERSE QUESTION vector fields not defined at 0, that is X in the above is the singleton {0}) this need not be the case. We can prove that the dimension is at least 1 by giving the following evil vector field: G(x, y) = −y x2 + y2 , x x2 + y2 !
= −y r2 , x r2 for (x, y) , 0 (where r = p x2 + y2). This vector field looks like our canonical example of a Vector Field With Nonzero Curl, (−y, x) except that the vectors grow in magnitude as they approach the origin, and it is left undefined at 0. By the same arguments above, this function is certainly not the gradient of anything as such a function would suffer the same problems as the (−y, x) field (which we will call F as before). But a calculation gives ∇× G = ∇× 1 r2 F = ∇(r−2) × F + r−2∇× F The second term in this sum is just (0, 0, 2r−2) since we know ∇× F = (0, 0, 2). Now ∇(r−2) = −2r−4r where r = (x, y) by a general formula (problem 30 of 4.4). What is r×F (physicists: no it is not torque, sorry =) actually well if F is a force-field and r is the radius vector, it really is a torque, but this is another story). (x, y, 0)×(−y, x, 0) = (0, 0, x2+y2) = (0, 0, r2). Therefore −2r−4(0, 0, r2) = (0, 0, −2r−2) which, when added to (0, 0, 2r−2) gives us 0. This answers the converse in the negative. Geometrically speaking, the structure of gradient fields vs. that of curl-less (irrotational) fields is related to the number of “holes” in the space; R2 minus the origin has one “hole.” To answer Nash’s question in this case, the span of this single function, a “nontrivial representative of V/W” means the space V/W is at least 1-dimensional. To prove that it is exactly 1-dimensional, however, is the hard part and that is what requires the de Rham cohomology (that sounds really cool, doesn’t it?). I am only barely just learning this material right now, whereas one Winter Quarter 5 years ago, I learned vector calculus.
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IB HL Vectors – IB Maths Resources from Intermathematics
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https://www.varsitytutors.com/practice/subjects/7th-grade-math/help/know-and-use-the-formulas-for-the-area-and-circumference-of-a-circle-ccss-math-content-7-g-b-4
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Know and Use the Formulas for the Area and Circumference of a Circle: CCSS.Math.Content.7.G.B.4
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7th Grade Math › Know and Use the Formulas for the Area and Circumference of a Circle: CCSS.Math.Content.7.G.B.4
If a circle has an area of , what is the circumference?
Explanation
For a circle, the formula for area is and the formula for circumference is , where is the radius and is the diameter.
Plug the known quantities into the area formula and solve for the radius:
Now plug this value into the circumference formula to solve:
If a circle has an area of , what is the circumference?
Explanation
For a circle, the formula for area is and the formula for circumference is , where is the radius and is the diameter.
Plug the known quantities into the area formula and solve for the radius:
Now plug this value into the circumference formula to solve:
If a circle has an area of , what is the circumference of the circle?
Explanation
The formula for the area of a circle is πr2. For this particular circle, the area is 81π, so 81π = πr2. Divide both sides by π and we are left with r2=81. Take the square root of both sides to find r=9. The formula for the circumference of the circle is 2πr = 2π(9) = 18π. The correct answer is 18π.
If a circle has an area of , what is the circumference of the circle?
Explanation
The formula for the area of a circle is πr2. For this particular circle, the area is 81π, so 81π = πr2. Divide both sides by π and we are left with r2=81. Take the square root of both sides to find r=9. The formula for the circumference of the circle is 2πr = 2π(9) = 18π. The correct answer is 18π.
The radius of a circle is 4 cm, what is the area?
Explanation
The area of a circle is found by: , where r is the radius.
.
The area of the circle is .
The radius of a circle is 4 cm, what is the area?
Explanation
The area of a circle is found by: , where r is the radius.
.
The area of the circle is .
The radius of a circle is . Give the area of the circle.
Explanation
The area of a circle can be calculated as , where is the radius of the circle, and is approximately .
What is the circumference of a circle with a radius of ?
(Round your answer to the nearest tenth.)
Explanation
The circumference is given by the formula:
where is the radius.
The radius of a circle is . Give the area of the circle.
Explanation
The area of a circle can be calculated as , where is the radius of the circle, and is approximately .
What is the circumference of a circle with a radius of ?
(Round your answer to the nearest tenth.)
Explanation
The circumference is given by the formula:
where is the radius.
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https://www.grc.nasa.gov/www/k-12/airplane/propth.html
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Propeller Thrust
+ Text Only Site + Non-Flash Version + Contact Glenn Most general aviation or private airplanes are powered by internal combustion engines which turn propellers to generate thrust. The details of how a propeller generates thrust is very complex, but we can still learn a few of the fundamentals using the simplified momentum theory presented here. Propeller Propulsion System On the slide, we show a schematic of a propeller propulsion system at the top and some of the equations that define how a propeller produces thrust at the bottom. The details of propeller propulsion are very complex because the propeller is like a rotating wing. Propellers usually have between 2 and 6 blades. The blades are usually long and thin, and a cut through the blade perpendicular to the long dimension will give an airfoil shape. Because the blades rotate, the tip moves faster than the hub. So to make the propeller efficient, the blades are usually twisted. The angle of attack of the airfoils at the tip is lower than at the hub because it is moving at a higher velocity than the hub. Of course, these variations make analyzing the airflow through the propeller a verydifficult task. Leaving the details to the aerodynamicists, let us assume that the spinning propeller acts like a disk through which the surrounding air passes (the yellow ellipse in the schematic). The engine, shown in white, turns the propeller and does work on the airflow. So there is an abrupt change in pressure across the propeller disk. (Mathematicians denote a change by the Greek symbol "delta" ( ). Across the propeller plane, the pressure changes by "delta p" (p). The propeller acts like a rotating wing. From airfoil theory, we know that the pressure over the top of a lifting wing is lower than the pressure below the wing. A spinning propeller sets up a pressure lower than free stream in front of the propeller and higher than free stream behind the propeller. Downstream of the disk the pressure eventually returns to free stream conditions. But at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli'sequation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine violates an assumption used to derive the equation. Simple Momentum Theory Turning to the math, the thrust F generated by the propeller disk is equal to the pressure jump delta p times the propeller disk area A: F = delta p A A units check shows that: force = (force/area) area We can use Bernoulli's equation to relate the pressure and velocity ahead of and behind the propeller disk, but not through the disk. Ahead of the disk the total pressure pt0 equals the static pressurep0 plus the dynamic pressure .5 r V0 ^2. pt0 = p0 + .5 r V0 ^2 where r is the air density and V0 is the velocity of the aircraft. Downstream of the disk, pte = p0 + .5 r Ve ^2 where pte is the downstream total pressure and Ve is the exit velocity. At the disk itself the pressure jumps delta p = pte - pt0 Therefore, at the disk, delta p = .5 r [Ve ^2 - V0 ^2] Substituting the values given by Bernoulli's equation into the thrust equation, we obtain F = .5 r A [Ve ^2 - V0 ^2] We still must determine the magnitude of the exit velocity. A propeller analysis based on the momentum equation provides this value. Note that this thrust is an ideal number that does not account for many losses that occur in practical, high speed propellers, like tip losses. The losses must be determined by a more detailed propeller theory, which is beyond the scope of these pages. The complex theory also provides the magnitude of the pressure jump for a given geometry. The simple momentum theory, however, provides a good first cut at the answer and could be used for a preliminary design. Activities: Guided Tours Propellers: Turboprops: Navigation .. Beginner's Guide Home Page + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility CertificationEditor: Nancy Hall NASA Official: Nancy Hall Last Updated: May 13 2021 + Contact Glenn
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Reference Notes for Lectures on Signed Graphs and Geometry by Thomas Zaslavsky Workshop on Signed Graphs, Set-Valuations and Geometry CMS Pala Campus, Pala, and Mary Matha Arts and Science College, Manathavady, Kerala1 16–19 August 2010 Contents Introduction 2 1.
Graphs 2 2.
Signed Graphs 4 2.1.
Balance and switching 4 2.2.
Deletion, contraction, and minors 6 2.3.
Frame circuits 7 2.4.
Closure and closed sets 8 3.
Geometry 10 3.1.
Vectors for edges 10 3.2.
The incidence matrix 11 3.3.
Arrangements of hyperplanes 11 3.4.
Matroid 12 4.
Coloring 12 4.1.
Chromatic polynomials 12 4.2.
Chromatic numbers 13 5.
Examples 14 5.1.
Full signed graphs 14 5.2.
All-positive signed graphs 14 5.3.
All-negative signed graphs 15 5.4.
Complete signed graphs 15 5.5.
Signed expansion graphs 15 Bibliography 16 1I deeply appreciate the generous labors of Sr. Dr. Germina K.A. and Professor A.M. Mathai in organizing the workshop at Pala and Mananthavady and inviting me to give a lecture series, and especially their kind care and their wish to show the beauties of Kerala to ‘a traveller from a distant land’.
1 2 Reference Notes on Signed Graphs and Geometry Introduction These notes and the lectures are a personal introduction to signed graphs, concentrating on the aspects that have been most persistently interesting to me. This is just a few corners of the theory; I am leaving out a great deal. The emphasis is on the way signed graphs arise naturally from geometry, especially from the geometry of the classical root systems.
The arrangement of the notes is topical, not historical. In the lectures I will talk about the historical development, but in the notes the purpose is to provide a printed reference for most of the definitions, theorems, and examples, and possibly some proofs.
The principal reference for most of the properties of signed graphs treated here is .
A simple introduction to the hyperplane geometry is [B, Zaslavsky (1981a)]. Many of my articles can be downloaded from my Web site, Bon voyage! Suffa yathra.
1. Graphs In these lectures all graphs are finite.
A graph is Γ = (V, E), where V := V (Γ) is the vertex set and E := E(Γ) is the edge set.
Notation: • n := |V |, called the order of Γ.
• V (e) is the multiset of vertices of the edge e.
• If S ⊆E, V (S) is the set of endpoints of edges in S.
Edges and edge sets: • There are four kinds of edge: A link has two distinct endpoints. A loop has two equal endpoints. A half edge has one endpoint. A loose edge has no endpoints. The set of loose edges of Γ is E0(Γ).
• An ordinary edge is a link or a loop. An ordinary graph is a graph in which every edge is ordinary. A link graph is a graph whose edges are links.
• The set of loose edges of Γ is E0(Γ). The set of ordinary edges of Γ is E∗:= E∗(Γ).
• Edges are parallel if they have the same endpoints. A simple graph is a link graph with no parallel edges.
• If S ⊆E, Sc := E \ S is its complement.
• E(X, Y ), where X, Y ⊆V , is the set of edges with one endpoint in X and the other in Y .
• A cut or cutset is an edge set E(X, Xc) that is nonempty.
The degree of a vertex v, d(v) := dΓ(v), is the number of edges of which v is an endpoint, but a loop counts twice. Γ is regular if every vertex has the same degree. If that degree is k, it is k-regular.
Walks, trails, paths, circles: • A walk is a sequence v0e1v1 · · · elvl where V (ei) = {vi−1, vi} and l ≥0. Its length is l. It may be written e1e2 · · · el or v0v1 · · · vl.
• A closed walk is a walk where l ≥1 and v0 = vl.
• A trail is a walk with no repeated edges.
• A path or open path is a trail with no repeated vertex, or the graph of such a trail (technically, the latter is a path graph), or the edge set of a path graph.
Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 3 • A closed path is a closed trail with no repeated vertex other than that v0 = vl. (A closed path is not a path.) • A circle (also called ‘cycle’, ‘polygon’, etc.) is the graph, or the edge set, of a closed path. Equivalently, it is a connected, regular graph with degree 2.
• C = C(Γ) is the class of all circles in Γ.
Examples: • Kn is the complete graph of order n. KX is the complete graph with vertex set X.
• Kc n is the edgeless graph of order n.
• Γc is the complement of Γ, if Γ is simple.
• Pl is a path of length l (as a graph or edge set).
• Cl is a circle of length l (as a graph or edge set).
• Kr,s is the complete bipartite graph with r left vertices and s right vertices. KX,Y is the complete bipartite graph with left vertex set X and right vertex set Y .
• The empty graph, ∅:= (∅, ∅), has no vertices and no edges. It is not connected.
Types of subgraph: In Γ, let X ⊆V and S ⊆E.
• A component (or connected component) of Γ is a maximal connected subgraph, ex-cluding loose edges. An isolated vertex is a component that has one vertex and no edges.
• c(Γ) is the number of components of Γ (excluding loose edges). c(S) is short for c(V, S).
• A spanning subgraph is Γ′ ⊆Γ such that V ′ = V .
• Γ|S := (V, S). This is a spanning subgraph.
• S:X := {e ∈S : ∅̸= V (e) ⊆X} = (E:X ∩S. We often write S:X as short for the subgraph (X, S:X).
• The induced subgraph Γ:X is the subgraph Γ:X := (X, E:X). An induced subgraph has no loose edges. We often write E:X as short for (X, E:X).
• Γ \ S := (V, E \ S).
• Γ \ X is the subgraph with V (Γ \ X) := V \ X and E(Γ \ X) := {e ∈E | V (e) ⊆V \ X}.
We say X is deleted from Γ. Γ \ X includes all loose edges, if there are any.
Vertices and vertex sets in Γ: Let X ⊆V .
• An isolated vertex is a vertex that has no incident edges; i.e., a vertex of degree 0.
• X is stable or independent if E:X = ∅.
• X is a clique if every pair of its vertices is adjacent.
• Xc denotes V \ X.
Graph structures and types: • A theta graph is the union of 3 internally disjoint paths that have the same endpoints.
• A block of Γ is a maximal subgraph without isolated vertices or loose edges, such that every pair of edges is in a circle together. The simplest kind of block is ({v}, {e}) where e is a loop or half edge at vertex v. A loose edge or isolated vertex is not in any block.
• Γ is inseparable if it has only one block or it is an isolated vertex.
• A cutpoint is a vertex that belongs to more than one block.
4 Reference Notes on Signed Graphs and Geometry Let T be a maximal forest in Γ. If e ∈E∗\ T, there is a unique circle Ce ⊆T ∪{e}. The fundamental system of circles for Γ, with respect to T, is the set of all circles Ce for e ∈E∗\T.
The set sum or symmetric difference of two sets A, B is denoted by A⊕B := (A\B)∪(B\A).
Proposition 1.1. Choose a maximal forest T. Every circle in Γ is the set sum of funda-mental circles with respect to T.
Proof. C = L e∈C\T CT(e).
□ 2. Signed Graphs A signed graph Σ = (Γ, σ) = (V, E, σ) is a graph Γ together with a function σ that assigns a sign, σ(e) ∈{+, −}, to each ordinary edge (link or loop) in Γ. A half or loose edge does not get a sign. Thus, the sign function (or signature) is σ : E∗→{+, −}. Notation: • |Σ| is the underlying graph Γ.
• E+ := σ−1(+) = {e ∈E : σ(e) = +}. The positive subgraph is Σ+ := (V, E+).
• E−:= σ−1(+) = {e ∈E : σ(e) = −}. The negative subgraph is Σ−:= (V, E−).
• +Γ := (Γ, +), an all-positive signed graph (every ordinary edge is +). e ∈E∗= E∗(Γ) becomes +e ∈+E = E(+Γ).
• −Γ := (Γ, −), an all-negative signed graph (every ordinary edge is −).
e ∈E∗ becomes −e ∈−E = E(−Γ).
• ±Γ := (+Γ) ∪(−Γ). E(±Γ) = ±E := (+E) ∪(−E). This is the signed expansion of Γ.
• Σ• := Σ with a half edge or negative loop attached to every vertex that does not have one. Σ• is called a full signed graph.
• Σ◦:= Σ with a negative loop attached to every vertex that does not have one.
• If ∆is a simple graph, then K∆:= (−∆)∪(+∆c), with underlying graph |K∆| = Kn.
This is a signed complete graph.
Equivalent notations for the sign group: {+, −}, {+1, −1}, or Z2 := {0, 1} modulo 2.
2.1. Balance and switching.
2.1.1. Balance.
Signs and balance: • The sign of a walk, σ(W), is the product of the signs of its edges, including repeated edges.
• The sign of an edge set, σ(S), is the product of the signs of its edges, without repetition.
• The sign of a circle, σ(C), is the same whether the circle is treated as a walk or as an edge set.
• The class of positive circles is B = B(Σ) := {C ∈C(|Σ|) : σ(C) = +}.
• Σ is balanced if it has no half edges and every circle in it is positive. Similarly, any subgraph or edge set is balanced if it has no half edges and every circle in it is positive.
• A circle is balanced if and only if it is positive. However, in general, a walk cannot be balanced because it is not a graph or edge set.
• b(Σ) is the number of components of Σ (omitting loose edges) that are balanced.
• πb(Σ) := {V (Σ′) : Σ′ is a balanced component of Σ}. Then b(Σ) = |πb(Σ)|.
Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 5 • V0(Σ) is the set of vertices of unbalanced components of Σ.
Formally, V0(Σ) := V \ S W∈πb(Σ) W.
• πb(S) is short for πb(Σ|S). V0(S) is short for V0(Σ|S).
Types of vertices and edges in Σ: • A balancing vertex is a vertex v such that Σ\v is balanced although Σ is unbalanced.
• A partial balancing edge is an edge e such that Σ \ e has more balanced components than does Σ.
• A total balancing edge is an edge e such that Σ \ e is balanced although Σ is not balanced. A total balancing edge is a partial balancing edge, but a partial balancing edge may not be a total balancing edge.
Proposition 2.1. An edge e is a partial balancing edge of Σ if and only if it is either (i) an isthmus between two components of Σ \ e, of which at least one is balanced, or (ii) a negative loop or half edge in a component Σ′ such that Σ′ \ e is balanced, or (iii) a link with endpoints v, w, such that every vw-path in Σ \ e has sign opposite to that of e.
Lemma 2.2. Σ is balanced if and only if every block is balanced.
A bipartition of a set X is an unordered pair {X1, X2} such that X1 ∪X2 = X and X1 ∩X2 = ∅. X1 or X2 could be empty.
Theorem 2.3 (Harary’s Balance Theorem ). Σ is balanced ⇐ ⇒it has no half edges and there is a bipartition V = V1 ∪ · V2 such that E−= E(V1, V2).
Corollary 2.4. −Γ is balanced if and only if Γ is bipartite.
2.1.2. Switching.
A switching function for Σ is a function ζ : V →{+, −}.
The switched signature is σζ(e) := ζ(v)σ(e)ζ(w), where e has endpoints v, w. The switched signed graph is Σζ := (|Σ|, σζ). We say Σ is switched by ζ. Note that Σζ = Σ−ζ.
If X ⊆V , switching Σ by X means reversing the signs of every edge in the cutset E(X, Xc).
The switched graph is ΣX. (This is the same as Σζ where ζ(v) := −if and only if v ∈Z.
Switching by ζ or X is the same operation, with different notation.) Note that ΣX = ΣXc.
Proposition 2.5.
(i) Switching leaves the signs of all circles unchanged. That is, B(Σζ) = B(Σ).
(ii) If |Σ1| = |Σ2| and B(Σ1) = B(Σ2), then there exists a switching function ζ such that Σ2 = Σζ 1.
Proof of (i). Let ζ be a switching function and let C = v0e0v1e1v2 · · · vn−1en−1v0 be a circle.
Then σζ(C) = ζ(v0)σ(e0)ζ(v1) ζ(v1)σ(e1)ζ(v2) . . .
ζ(vn−1)σ(en−1)ζ(v0) = σ(e0)σ(e1) · · · σ(en−1) = σ(C).
□ Proof of (ii). We may assume Σ1 is connected. Pick a spanning tree T and list the vertices in such a way that vi is always adjacent to a vertex in {v1, . . . , vi−1} (for i > 1). Let ti be the unique tree edge connecting vi to Σ:{v1, . . . , vi−1}.
6 Reference Notes on Signed Graphs and Geometry We define a switching function ζ: ζ(vi) = ( + if i = 1, σ1(ti)σ2(ti)ζ(vj) if i > 1, where vj is the endpoint of ti that is not vi.
Now it is easy to show that Σζ 1 = Σ2.
□ Signed graphs Σ1 and Σ2 are switching equivalent, written Σ1 ∼Σ2, if they have the same underlying graph and there exists a switching function ζ such that Σζ 1 = Σ2. The equivalence class of Σ, [Σ] := {Σ′ : Σ′ ∼Σ}, is called its switching class.
Proposition 2.6. Switching equivalence, ∼, is an equivalence relation on signatures of a given underlying graph.
Corollary 2.7. Σ is balanced if and only if it has no half edges and it is switching equivalent to +|Σ|.
Proof of Harary’s Balance Theorem. Σ has the form stated in the theorem ⇐ ⇒ it is (+|Σ|)V1 ⇐ ⇒it is a switching of +|Σ| ⇐ ⇒(by Proposition 2.5) it is balanced.
□ 2.2. Deletion, contraction, and minors.
R, S denote subsets of E. A component of S means a component of (V, S).
The deletion of S is the signed graph (V, Sc, σ|Sc).
2.2.1. Contracting an edge e.
If e is a positive link, delete e and identify its endpoints (this is how to contract a link in an unsigned graph); do not change any edge signs. If e is a negative link, switch Σ by a switching function ζ, chosen so e is positive link in Σζ; then contract e as a positive link.
Lemma 2.8. In a signed graph Σ any two contractions of a link e are switching equivalent.
The contraction of a link in a switching class is a well defined switching class.
To contract a positive loop or a loose edge e, just delete e.
If e is a negative loop or half edge and v is the vertex of e, delete v and e, but not any other edges. Any other edges at v lose their endpoint v. A loop or half edge at v becomes a loose edge. A link with endpoints v, w becomes a half edge at w.
2.2.2. Contracting an edge set S.
The edge set and vertex set of Σ/S are E(Σ/S) := E \ S, V (Σ/S) := πb(Σ|S) = πb(S).
For w ∈V , define This means we identify all the vertices of each balanced component so they become a single vertex. For e ∈E(Σ/S), the endpoints are given by the rule VΣ/S(e) = {W ∈πb(S) : w ∈VΣ(e) and w ∈W ∈πb(S)}.
(If e is a loop at w in Σ, then w is a repeated vertex in VΣ(e); if w ∈W ∈πb(S), then W is a repeated vertex in VΣ/S(e) so e is a loop in Σ/S. If w ∈VΣ(e)∩V0(S), then w disappears from VΣ/S(e). To define the signature of Σ/S, first switch Σ to Σζ so every balanced component of S is all positive. Then σΣ/S(e) := σζ(e).
Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 7 Lemma 2.9. (a) Given Σ a signed graph and S ⊆E(Σ), all contractions Σ/S (by different choices of switching Σ) are switching equivalent. Any switching of one contraction Σ/S is another contraction and any contraction Σζ/S of a switching of Σ is a contraction of Σ.
(b) If |Σ1| = |Σ2|, S ⊆E is balanced in both Σ1 and Σ2, and Σ1/S and Σ2/S are switching equivalent, then Σ1 and Σ2 are switching equivalent.
(c) For e ∈E, [Σ/e] (defined in Section 2.2.1) and [Σ/{e}] are essentially the same switching class.
Part (a) means that the switching class [Σ/S] is uniquely defined, even though the signed graph Σ/S is not unique.
2.2.3. Minors.
A minor of Σ is any contraction of any subgraph.
Theorem 2.10. Given a signed graph Σ, the result of any sequence of deletions and con-tractions of edge and vertex sets of Σ is a minor of Σ. In other words, a minor of a minor is a minor.
Proof. See [4, Proposition 4.2].
□ 2.3. Frame circuits.
A frame circuit of Σ is a subgraph, or edge set, that is either a positive circle or a loose edge, or a pair of negative circles that intersect in precisely one vertex and no edges (this is a tight handcuffcircuit), or a pair of disjoint negative circles together with a minimal path that connects them (this is a loose handcuffcircuit). We regard a tight handcuffcircuit as having a connecting path of length 0 (it is the common vertex of two the circles).
Proposition 2.11. Σ contains a loose handcuffcircuit if and only if there is a component of Σ that contains two disjoint negative circles.
Proposition 2.12. Let e ∈E. If Σ contains a handcuffcircuit C such that e ∈C, then e is in an unbalanced component Σ′ of Σ and e is not a partial balancing edge. If e is in an unbalanced component Σ′ of Σ and e is not a partial balancing edge, then Σ contains a frame circuit C such that e ∈C.
Proof. If there exists C as stated in the proposition, then the component Σ′ that contains e also contains C so it is unbalanced.
As C \ e is unbalanced, Σ′ \ e has no balanced component (this requires checking cases), so e is not a partial balancing edge (this requires checking definitions).
Conversely, suppose e is not a partial balancing edge and it is in an unbalanced component Σ′. Since Σ′ \ e is unbalanced, it has a negative circle C1. If e is an unbalanced edge at v, there is a path P in Σ′ from v to C1; then C = C1 ∪P ∪e.
If e is a balanced edge, it is a link with endpoints v, w. If it is an isthmus, then Σ′ \ e has two components, both unbalanced (by Proposition 2.1), so C is a negative circle in each of those components, together with a connecting path (which must contain e). If e is not an isthmus, it lies in a circle C′. If C′ is positive, let C = C′. But suppose C′ is negative; then there are three subcases, depending on how many points of intersection C′ has with C1. If there are no such points, take a minimal path P connecting C′ to C1 and let C = C1∪P ∪C′.
If there is just one such point, C = C1 ∪C′. If there are two or more such points, take P to be a maximal path in C′ that contains e and is internally disjoint from C1. Then P ∪C1 is 8 Reference Notes on Signed Graphs and Geometry a theta graph in which C1 is negative; hence one of the two circles containing P is positive, and this is the circuit C.
□ Theorem 2.13 ([B, Slilaty (2007a)]). Σ has no two vertex-disjoint negative circles if and only if one or more of the following is true: (1) Σ is balanced, (2) Σ has a balancing vertex, (3) Σ embeds in the projective plane, or (4) Σ is one of a few exceptional cases.
We will not discuss projective planarity, which is a large topic in itself (see [B, Zaslavsky (1993a)], [B, Archdeacon and Debowsky (2005a)]).
2.4. Closure and closed sets.
The balance-closure of an edge set R is bcl(R) := R ∪{e ∈Rc : ∃a positive circle C ⊆R ∪e such that e ∈C} ∪E0(Σ).
The closure of an edge set S is clos(S) := E:V0(S) ∪ k [ i=1 bcl(Si) ∪E0(Σ), where S1, . . . , Sk are the balanced components of S.
An edge set is closed if it equals its own closure: clos S = S. We write Lat Σ := {S ⊆E : S is closed}.
When partially ordered by set inclusion, Lat Σ is a lattice.
Note that a half edge and a negative loop are equivalent in everything that concerns closure or circuits.
The usual closure operator in a graph Γ is the same as closure in +Γ. In that case, since +Γ is balanced, a frame circuit is simply a positive circle (or a loose edge).
Lemma 2.14. bcl(R) is balanced if and only if R is balanced. Furthermore, bcl(bcl R) = bcl(R) = clos(R).
Lemma 2.15. For an edge set S, πb(clos S) = πb(bcl S) = πb(S) and V0(clos S) = V0(S).
Let E be any set; its power set P(E) is the class of all subsets of E. A function J : P(E) → P(E) is an (abstract) closure operator on E if it has the three properties (C1) J(S) ⊇S for every S ⊆E (increase).
(C2) R ⊆S = ⇒J(R) ⊆J(S) (isotonicity).
(C3) J(J(S)) = J(S) (idempotence).
Theorem 2.16. The operator clos on subsets of E(Σ) is an abstract closure operator.
Proof. The definition makes clear that clos is increasing and isotonic. What remains to be proved is that clos(clos(S)) = clos(S).
Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 9 Let πb(S) = {B1, . . . , Bk}; thus, S:Bi is balanced. Using Lemma 2.15, clos(clos S) = E:V0(clos S) ∪ k [ i=1 bcl (clos S):Bi = E:V0(S) ∪ k [ i=1 bcl (bcl S):Bi = E:V0(S) ∪ k [ i=1 bcl(S:Bi) = clos S.
□ Theorem 2.17. For S ⊆E, clos S = S ∪{e / ∈S : ∃a frame circuit C such that e ∈C ⊆S ∪e}.
Proof. We treat a half edge as if it were a negative loop.
Necessity. We want to prove that if e ∈clos S, then a frame circuit C exists. There are three cases, depending on where the endpoints of e are located.
Case 0. A trivial case is where e is a loose edge. Then e ∈clos S and C = {e}.
Case 1. Suppose e has its endpoints within one component, S′. Then there is a circle C′ in S′ ∪e that contains e. If C′ is positive it is our circuit C. (This includes the case of a positive loop e, where C = {e}.) If S′ is balanced, then e ∈bcl S′ so there exists a positive circle C′. If S′ is unbalanced and e is not a partial balancing edge, then C exists by Proposition 2.11.
Suppose S′ is unbalanced and e is a partial balancing edge. Then Proposition 2.1 tells us that e cannot be in clos S (this requires checking the three cases of that proposition).
Case 2. Suppose e has endpoints in two different components, S′ and S′′. For e to be in the closure, it must be in E:V0. Hence, S′ and S′′ are unbalanced. Each of them contains a negative circle, C′ and C′′ respectively, and there is a connecting path P in S ∪e which contains e. Then C′ ∪P ∪C′′ is the desired circuit.
Sufficiency. Assuming a circuit C exists, we want to prove that e ∈clos S. Again there are three cases, this time depending on C and its relationship with e.
Case 0. C is balanced. Then e ∈bcl S ⊆clos S.
Case 1. C is unbalanced and e is not in the connecting path. Let C1, C2 be the two negative circles and P the connecting path of C, and assume e ∈C1. Since C \ e is connected, it lies in one component S′ of S. Thus, C2 ⊆S′, whence S′ is unbalanced. It follows that e ∈E:V0 ⊆clos S.
Case 2. C is unbalanced and e is in the connecting path. With notation as in Case 1, now C \ e has two components, one containing C1 and the other containing C2. The components of S that contain C1 and C2 are unbalanced. (There may be one such component or two, depending on whether C1 and C2 are connected by a path in S.) Therefore, e has both endpoints in V0, so again, e ∈E:V0 ⊆clos S.
□ 10 Reference Notes on Signed Graphs and Geometry 3. Geometry In this section we write the vertex set as V = {v1, v2, . . . , vn}. F denotes any field. The most important field will be the real numbers R. Other important fields are F2, the 2-element field of arithmetic modulo 2, and F3, the 3-element field of arithmetic modulo 3.
3.1. Vectors for edges.
We have a signed graph Σ of order n. For each edge e there is a vector x(e) ∈Fn, whose definition is, for the four types of edge: i j 0 .
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0 ±1 0 .
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0 ∓σ(e) 0 .
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0 a link e:vivj, i 0 .
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0 ±1 ∓σ(e) 0 .
.
.
0 a loop e at vi, i 0 .
.
.
0 ±1 0 .
.
.
0 a half edge e at vi, 0 .
.
.
0 .
.
.
0 a loose edge.
These vectors are well defined only up to sign, i.e., the negative of x(e) is another possible choice of x(e). We make an arbitrary choice x(e) for each edge e. The choice does not affect the linear dependence properties.
For a set S ⊆E, define x(S) := {x(e) : e ∈S}.
Theorem 3.1. Let S be an edge set in Σ and consider the corresponding vector set x(S) in the vector space Fn over a field F.
(1) When char F ̸= 2, x(S) is linearly dependent if and only if S contains a frame circuit.
(2) When char F = 2, x(S) is linearly dependent if and only if S contains a circle or a loose edge.
The proof is not short.
Corollary 3.2. The minimal linearly dependent subsets of x(E) are the sets x(C) where C is a frame circuit in Σ.
The proofs of the next results are short. Define a set S ⊆E(Σ) to be independent if x(S) is linearly independent.
Corollary 3.3. A set S ⊆E(Σ) is independent if and only if it does not contain a frame circuit.
The vector subspace generated by a set X ⊆Fn is denoted by ⟨X⟩. We write LF(Σ) := {⟨X⟩: X ⊆x(E)}.
When partially ordered by set inclusion, LF(Σ) is a lattice.
Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 11 Corollary 3.4. For S ⊆E(Σ), x(E) ∩⟨x(S)⟩= x(clos S). Thus, LR(Σ) ∼ = Lat Σ.
The rank of S ⊆E is defined to be rk S := n −b(S).
Theorem 3.5. Let S ⊆E. Then dim⟨x(S)⟩= rk S.
Proof. The proof is simplest when expressed in terms of the frame matroid (Section 3.4), so I omit it; see [4, Theorem 8B.1 and following remarks]. The essence of the proof is using Corollary 3.3 to compare the minimum number of edges required to generate S by closure in Σ to the minimum number of vectors x(e) required to generate ⟨x(S)⟩.
□ 3.2. The incidence matrix.
The incidence matrix H(Σ) (read ‘Eta of Sigma’) is a V × E matrix (thus, it has n rows and m columns where m := |E|) in which the column corresponding to edge e is the column vector x(e).
Theorem 3.6. Let S ⊆E. The rank of H(Σ|S) is rk S.
Proof. The column rank is the dimension of the span of the columns corresponding to S, which is the span of x(S). Apply Theorem 3.5.
□ 3.3. Arrangements of hyperplanes.
An arrangement of hyperplanes in Rn, H = {h1, h2, . . . , hm}, is a finite set of hyper-planes. The complement is Rn \ Sm k=1 hk . A region of H is a connected component of the complement. We write r(H) := the number of regions. The intersection lattice is the family L(H) of all subspaces that are intersections of hyperplanes in H, partially ordered by s ≤t ⇐ ⇒t ⊆s (reverse inclusion). The characteristic polynomial of H is (3.1) pH(λ) := X S⊆H (−1)|S|λdim S, where dim S := dim T hk∈S hk .
Theorem 3.7 ([3, Theorem A]). We have r(H) = (−1)npH(−1).
A signed graph Σ, with edge set {e1, e2, . . . , em}, gives rise to a hyperplane arrangement H[Σ] := {h1, h2, . . . , hm} where hk has the equation xj = σ(ek)xi, if ek is a link or loop with endpoints vi, vj, xi = 0, if ek is a half edge or a negative loop at vi, 0 = 0, if ek is a loose edge or a positive loop.
(The last equation has the solution set Rn, so it is not truly a hyperplane, but I allow it under the name ‘degenerate hyperplane’.) The hyperplane hk is the solution set of the equation x(ek) · x = 0; i.e., hk = {x ∈Rn : x(ek) · x = 0}.
(· is the usual inner product or ‘dot product’.) Lemma 3.8. Let S = {hi1, . . . , hil} be the subset of H[Σ] that corresponds to the edge set S = {ei1, . . . , eil}. Then dim S = b(S).
12 Reference Notes on Signed Graphs and Geometry Proof. Apply vector space duality to Theorem 3.5.
□ Theorem 3.9. L(H[Σ]) is isomorphic to LR(Σ) and Lat Σ.
Proof. The isomorphism between L(H[Σ]) and LR(Σ) is standard vector-space duality. The isomorphism LR(Σ) ∼ = Lat Σ is in Corollary 3.4.
□ 3.4. Matroid.
The frame matroid G(Σ) is an abstract way of describing all the previous characteristics of a signed graph: linearly dependent edge sets, minimal dependencies, rank, closure, and closed sets. See [4, Section 5] for more information. For matroid theory, consult [Oxley].
4. Coloring We color a signed graph from a color set Λk := {±1, ±2, . . . , ±k} ∪{0} or a zero-free color set Λ∗ k := Λk{0} = {±1, ±2, . . . , ±k}.
A k-coloration (or k-coloring) of Σ is a function γ : V →Λk. A coloration is zero free if it does not use the color 0. Coloring comes from and [B, Zaslavsky (1982c)].
A coloration γ is proper if it satisfies all the properties ( γ(vj) ̸= σ(e)γ(vi) for a link or loop e with endpoints vi, vj, γ(vi) ̸= 0 for a half edge e with endpoint vi, and there are no loose edges. (These are the negations of the equations of the hyperplanes hk.) 4.1. Chromatic polynomials.
There are two chromatic polynomials of a signed graph. For aninteger k ≥0, define χΣ(2k + 1) := the number of proper k-colorations, and χ∗ Σ(2k) := the number of proper zero-free k-colorations.
Theorem 4.1. The chromatic polynomials have the properties of (i) Unitarity: χ∅(2k + 1) = 1 = χ∗ ∅(2k) for all k ≥0.
(ii) Switching Invariance: If Σ ∼Σ′, then χΣ(2k +1) = χΣ′(2k +1) and χ∗ Σ(2k) = χ∗ Σ′(2k).
(iii) Multiplicativity: If Σ is the disjoint union of Σ1 and Σ2, then χΣ(2k + 1) = χΣ1(2k + 1)χΣ2(2k + 1) and χ∗ Σ(2k) = χ∗ Σ1(2k)χ∗ Σ2(2k).
(iv) Deletion-Contraction: If e is a link, a positive loop, or a loose edge, χΣ(2k + 1) = χΣ\e(2k + 1) −χΣ/e(2k + 1) and χ∗ Σ(2k) = χ∗ Σ\e(2k) −χ∗ Σ/e(2k).
Outline of Proof. The hard part is the deletion-contraction property. The proof is similar to the usual proof for ordinary graphs: count proper colorations of Σ\e. If e is a link, switch so it is positive. Then a proper coloration of Σ\e give unequal colors to the endpoints of e, and is a proper coloration of Σ, or it gives the same color to the endpoints, and it corresponds to a proper coloration of Σ/e. If e is a half edge or a negative loop, the two cases are when the endpoint gets a nonzero color or is colored 0.
□ Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 13 Theorem 4.2. χΣ(λ) is a polynomial function of λ = 2k + 1 > 0; specifically, (4.1) χΣ(λ) = X S⊆E (−1)|S|λb(S).
Also, χ∗ Σ(λ) is a polynomial function of λ = 2k ≥0. Specifically, (4.2) χ∗ Σ(λ) = X S⊆E:balanced (−1)|S|λb(S).
Proof. Apply Theorem 4.1 and induction on n.
□ Therefore, we can extend the range of λ to all of R. In particular, we can evaluate χΣ(−1).
This lets us draw an important connection between the geometry and coloring of a signed graph.
Theorem 4.3. χΣ(λ) = pHΣ.
Proof. Compare the summation expressions, (4.1) and (3.1), for the two polynomials, and note that by Lemma 3.8 b(S) = dim S if S ⊆H[Σ] corresponds to the edge set S.
□ Corollary 4.4. The number of regions of H[Σ] equals (−1)nχΣ(−1).
To compute the chromatic polynomial it is often easiest to get the zero-free polynomial first and use Theorem 4.5 (Zero-Free Expansion Identity). The chromatic and zero-free chromatic poly-nomials are related by χΣ(λ) = X W⊆V : stable χ∗ Σ\W(λ −1).
Proof. Let λ = 2k+1. For each proper k-coloration γ there is a set W := {v ∈V : γ(v) = 0}, which must be stable. The restricted coloration γ|V \W is a zero-free, proper k-coloration of Σ \ W. This construction is reversible.
□ 4.2. Chromatic numbers.
The chromatic number of Σ is χ(Σ) := min{k : ∃a proper k-coloration}, and the zero-free chromatic number is χ∗(Σ) := min{k : ∃a zero-free proper k-coloration}.
Thus, χ(Σ) = min{k ≥0 : χΣ(2k + 1) ̸= 0} and χ∗(Σ) = min{k ≥0 : χ∗ Σ(2k) ̸= 0}.
Almost any question about the behavior of chromatic numbers of signed graphs is open.
What I know is in [B, Zaslavsky (1984a)], where I studied complete signed graphs with very large or very small zero-free chromatic number.
14 Reference Notes on Signed Graphs and Geometry 5. Examples The standard basis vectors of Rn are b1 = (1, 0, . . . , 0), b2, . . . , bn.
5.1. Full signed graphs.
In this example Σ is a signed graph with no half or loose edges or negative loops, Σ• is Σ with a half edge at every vertex, and Σ◦is Σ with a negative loop at every vertex. Whether a half edge or negative loop is added makes little difference, because each is an unbalanced edge. Write fi for the unbalanced edge added to vi.
• Graph theory: The balanced subgraphs in Σ• are the same as those of Σ.
• Closed sets: An edge set in Σ• is closed if and only if it consists of the induced edge set E(Σ•):W together with a balanced, closed subset of E(Σ):W c, for some vertex set W ⊆V . Σ◦is similar.
• Vectors: x(E(Σ•)) is x(E(Σ)) together with the unit basis vectors bi of Rn. x(E(Σ◦)) is x(E(Σ)) together with the vectors 2bi.
• Hyperplane arrangement: H[Σ•] = H[Σ◦], and they equal H[Σ] together with all the coordinate hyperplanes xi = 0.
• Chromatic polynomials: χΣ•(λ) = χΣ◦(λ), and both = χΣ•(λ) = χ∗ Σ(λ −1) by Theo-rem 4.5, since the only stable set is W = ∅.
χ∗ Σ•(λ) = χ∗ Σ◦(λ) = χ∗ Σ(λ).
• Chromatic numbers: χ(Σ•) = χ(Σ◦) = χ∗(Σ) since the unbalanced edges prevent the use of color 0.
5.2. All-positive signed graphs.
Assume Γ is a graph with no unbalanced edges and no loose edges. +Γ has almost exactly the same properties as its underlying graph.
• Graph theory: Every subgraph is balanced; b(S) = c(S) for all S ⊆E.
• Closed sets: S is closed ⇐ ⇒ every edge with endpoints connected by S is in S.
Closure in +Γ is identical to the usual closure in Γ, and the closed sets in +Γ are the same as in Γ.
• Vectors: If e has endpoints vi, vj, then x(e) = ±(bj −bi). All x(e) ∈the subspace x1 + · · · + xn = 0.
If Γ = Kn and one takes both signs, the set of vectors is the classical root system An−1. Thus, x(E) for any graph is a subset of An−1.
• Incidence matrix: M(+Γ) is the ‘oriented incidence matrix’ of Γ.
• Hyperplane arrangement: If ek has endpoints vi, vj, then hk has equation xi = xj.
All hk ⊇the line x1 = · · · = xn.
Take Γ = Kn; then H[+Kn] = An−1, the hyperplane arrangement dual to An−1.
• Chromatic polynomials: χ+Γ(λ) = χ∗ +Γ(λ) = χΓ(λ), the chromatic polynomial of Γ.
• Chromatic numbers: χ(+Γ) = ⌊χ(Γ)/2⌋and χ∗(+Γ) = ⌈χ(Γ)/2⌉.
The full graph +Γ• is very much like Γ+v0 := Γ with an extra vertex v0 which is adjacent to all of V by edges e0i between v0 and vi. Define α : E(+Γ•) →E(Γ + v0) by α(e) := e if e ∈E(Γ) and α(fi) := e0i.
• Graph theory: S is balanced if and only if it does not contain an unbalanced edge fi.
• Closed sets: S is closed ⇐ ⇒α(S) is closed in Γ + v0.
• Chromatic polynomials: χ+Γ(λ) = χ∗ +Γ(λ −1) = χΓ(λ −1).
• Chromatic numbers: χ(+Γ•) = χ∗(+Γ•) = ⌈χ(Γ)/2⌉.
Thomas Zaslavsky | Pala–Manathavady 16–19 August 2010 15 5.3. All-negative signed graphs. Assume Γ is a graph with no unbalanced edges. −Γ is very interesting.
• Graph theory: A subgraph is balanced ⇐ ⇒it is bipartite. b−Γ(S) = the number of bipartite components of S (including isolated vertices).
• Closed sets: S is closed if the union of its non-bipartite components is an induced subgraph.
• Vectors: If e has endpoints vi, vj, then x(e) = bi + bj (or its negative).
• Incidence matrix: M(−Γ) is the ‘unoriented incidence matrix’ of Γ.
• Hyperplane arrangement: hk has equation xi +xj = 0 if ek has endpoints vi, vj. Also, r(H[−Γ]) = P F∈Lat Γ |χΓ/F(−1 2)|.
• Chromatic polynomials: χ∗ −Γ(λ) = P F∈Lat Γ χΓ/F(1 2λ) [B, Zaslavsky (1982c), Theo-rem 5.2]. χ−Γ(λ) has not seemed interesting.
• Chromatic numbers: χ∗(−Γ) = the largest size of a matching in the complement of a contraction of Γ [B, Zaslavsky (1982c), page 299]. χ(−Γ) has not seemed interesting.
5.4. Complete signed graphs.
The signed expansions ±Kn, called the complete signed link graph, and ±K• n, called the complete signed graph, have very simple properties.
• Closed sets: Lat(±K• n) ∼ = the lattice of signed partial partitions of V [B, Dowling (1973b)].
• Vectors: x(E(±Kn)) = {±(bj −bi), ±(bj + bi) : i ̸= j} where we take either + or − for each vector. x(E(±K• n)) = {±(bj −bi), ±(bj + bi) : i ̸= j} ∪{±bi} (if fi is a half edge; but ±2bi if fi is a negative loop) where we take either + or −for each vector.
If we take both signs, we get the classical root systems Dn := {±(bj−bi), ±(bj+bi) : i ̸= j} (where we take both + and −signs) from ±Kn, and Bn := Dn ∪{±bi} and Cn := Dn ∪{±2bi} from ±K• n (the former if all fi are half edges, the latter if they are negative loops).
• Hyperplane arrangement: H[±K• n] = Bn = Cn and H[±Kn] = Dn, the duals of Bn, Cn, and Dn. The numbers of regions are 2nn! and 2n−1n!, respectively.
• Chromatic polynomials: χ±K• n(λ) = (λ −1)(λ −3) · · · (λ −2n + 1), χ±Kn(λ) = (λ −1)(λ −3) · · · (λ −2n + 3) · (λ −n + 1), and χ∗ ±Kn(λ) = χ∗ ±K• n(λ) = λ(λ −2) · · · (λ −2n + 2).
• Chromatic numbers: χ(±K• n) = χ∗(±K• n) = χ∗(±Kn) = n and χ(±Kn) = n −1.
5.5. Signed expansion graphs. The properties of ±Γ and ±Γ• are closely related to those of Γ.
• Graph theory: Each balanced set S ⊆E(Γ) gives 2b(S) balanced subsets of E(±Γ).
• Hyperplane arrangement: r(H[±Γ•]) = 2n(−1)nχΓ(−1) = 2n|χΓ(−1)| and r(H[±Γ]) = X W⊆V : stable in Γ (−2)n−|W||χΓ\W(−1)|.
• Chromatic polynomials: χ±Γ•(λ) = 2nχΓ(1 2(λ −1)), χ∗ ±Γ(λ) = 2nχΓ(1 2λ), and χ±Γ(λ) = X W⊆V : stable in Γ 2n−|W|χΓ\W(1 2(λ −1)).
• Chromatic numbers: χ(±Γ•) = χ∗(±Γ) = χ(Γ), the chromatic number of Γ, and χ(±Γ) = χ(Γ) −1.
16 Reference Notes on Signed Graphs and Geometry Bibliography References in the style [B, Name (yeara)] are to the author Name, item (yeara) in [B].
[B] Thomas Zaslavsky, A mathematical bibliography of signed and gain graphs and allied areas. Electronic J.
Combin., Dynamic Surveys in Combinatorics (1998), No. DS8 (electronic), vi + 151 pp. Current version (about 240 pp.) available at MR 2000m:05001a. Zbl 898.05001.
[G] Thomas Zaslavsky, Glossary of signed and gain graphs and allied areas. Electronic J. Combinatorics (1998), Dynamic Surveys in Combinatorics, #DS9 (electronic), 41 pp. MR 2000m:05001b. Zbl 898.05002.
F. Harary, On the notion of balance of a signed graph. Michigan Math. J. 2 (1953–54), 143–146 and addendum preceding p. 1. MR 16, 733h. Zbl 56, 421c (e: 056.42103).
J.G. Oxley, Matroid Theory. Oxford Univ. Press, Oxford, 1992. MR 94d:05033. Zbl 784.05002.
Thomas Zaslavsky, Facing Up to Arrangements: Face-Count Formulas for Partitions of Space by Hyper-planes. Mem. Amer. Math. Soc., No. 154 (= Vol. 1, Issue 1). American Mathematical Society, Providence, R.I., 1975. MR 50 #9603. Zbl 296.50010.
Thomas Zaslavsky, Signed graphs. Discrete Appl. Math. 4 (1982), 47–74. Erratum. Discrete Appl. Math.
5 (1983), 248. MR 84e:05095. Zbl 503.05060.
Thomas Zaslavsky, Signed graph coloring. Discrete Math. 39 (1982), 215–228. MR 84h:05050a. Zbl 487.05027.
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Difference Between Mutually Exclusive and Independent Events
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HomeKey Differences
Difference Between Mutually Exclusive and Independent Events
Difference Between Mutually Exclusive and Independent Events
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Overview
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Understanding the concepts of mutually exclusive and independent events is crucial in the field of probability and statistics. These terms are used to describe the relationship between events in a probability experiment. In this article, we will explore the key differences between mutually exclusive and independent events and discuss their significance in various contexts.
PointMutually Exclusive EventsIndependent Events
1 Events that cannot occur simultaneously or overlap Events that do not affect or influence each other
2 The occurrence of one event excludes the possibility of the other event happening The occurrence or non-occurrence of one event does not affect the probability of the other event happening
3 The events have no common outcomes or elements The events can have common outcomes or elements
4 The probability of the combined events happening is the sum of their individual probabilities The probability of one event happening is unaffected by the occurrence or non-occurrence of the other event
5 Examples of mutually exclusive events: flipping a coin and getting either heads or tails, rolling a die and getting an odd number or an even number Examples of independent events: flipping a coin and rolling a die, drawing cards from a deck with replacement
6 When calculating the probability of mutually exclusive events, you add their individual probabilities When calculating the probability of independent events, you multiply their individual probabilities
7 Mutually exclusive events have a zero probability of occurring together Independent events have a non-zero probability of occurring together
8 In a Venn diagram, the sets representing mutually exclusive events do not overlap In a Venn diagram, the sets representing independent events may overlap
9 The presence or absence of one event provides no information about the occurrence of the other event The presence or absence of one event does not provide any information about the occurrence of the other event
10 In probability theory, mutually exclusive events have a joint probability of zero In probability theory, independent events have a joint probability equal to the product of their individual probabilities
11 Mutually exclusive events involve a choice between two or more exclusive options Independent events involve separate occurrences or outcomes that are not influenced by each other
12 The occurrence of one event gives certainty about the non-occurrence of the other event The occurrence or non-occurrence of one event has no impact on the probability of the other event
13 Mutually exclusive events can be visualized as separate, non-overlapping regions in a probability distribution Independent events can be visualized as separate, overlapping regions in a probability distribution
14 The concept of mutually exclusive events applies to situations where only one outcome can happen at a time The concept of independent events applies to situations where the outcome of one event does not affect the outcome of the other event
15 In statistical analysis, mutually exclusive events can help determine the probability of specific outcomes or occurrences In statistical analysis, independent events can help assess the probability of combined outcomes or predict future events
16 Mutually exclusive events involve distinct choices, alternatives, or possibilities Independent events involve separate occurrences or actions that do not depend on each other
17 The occurrence of one event provides information about the non-occurrence of the other event The occurrence or non-occurrence of one event provides no information about the occurrence of the other event
18 Mutually exclusive events have no shared outcomes or dependencies Independent events can have shared outcomes or dependencies
19 The addition rule applies to mutually exclusive events to calculate the probability of their combination The multiplication rule applies to independent events to calculate the probability of their combination
20 The probability of at least one of the mutually exclusive events happening is the sum of their individual probabilities The probability of one event happening does not affect the probability of the other event happening
21 Mutually exclusive events can be thought of as mutually contradictory or incompatible options Independent events can be thought of as unrelated or unrelated occurrences
22 Mutually exclusive events have a negative relationship, meaning the occurrence of one event precludes the occurrence of the other event Independent events have no inherent relationship or dependency
23 The probability of mutually exclusive events happening simultaneously is always zero The probability of independent events happening simultaneously is the product of their individual probabilities
24 The concept of mutually exclusive events helps in decision-making, problem-solving, and statistical analysis The concept of independent events helps in probability calculations, statistical modeling, and risk assessment
25 Mutually exclusive events involve distinct possibilities or outcomes that cannot coexist Independent events involve separate possibilities or outcomes that can coexist without affecting each other
Understanding Mutually Exclusive Events
Mutually exclusive events are events that cannot occur simultaneously. When two events are mutually exclusive, the occurrence of one event excludes the possibility of the other event happening. For example, when rolling a fair six-sided die, the outcomes of "getting a 3" and "getting an odd number" are mutually exclusive because obtaining a 3 means it cannot be an odd number.
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Understanding Independent Events
Independent events are events where the occurrence or non-occurrence of one event does not affect the probability of the other event happening. In other words, the outcomes of independent events are not dependent on each other. For instance, flipping a fair coin twice, the result of the first flip does not impact the outcome of the second flip.
Key Differences Between Mutually Exclusive and Independent Events
While both mutually exclusive and independent events involve the relationship between events, they have distinct characteristics:
Mutually Exclusive Events
Definition: Mutually exclusive events cannot occur simultaneously.
Overlap: These events have no common outcomes.
Impact: The occurrence of one event affects the probability of the other event happening.
Symbol: Represented by the symbol "∩" (intersection).
Formula: P(A ∩ B) = 0, where A and B are mutually exclusive events.
Independent Events
Definition: Independent events do not affect each other's outcomes.
Overlap: Independent events can have common outcomes.
Impact: The occurrence of one event does not influence the probability of the other event happening.
Symbol: Represented by the symbol "∩" (intersection) with a line through it.
Formula: P(A ∩ B) = P(A) P(B), where A and B are independent events.
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Examples of Mutually Exclusive Events
To further illustrate the concept of mutually exclusive events, consider the following examples:
Selecting a card from a well-shuffled deck and getting a red card and a black card at the same time.
Choosing a random student from a classroom and selecting a student who is a freshman and a senior simultaneously.
Rolling a fair six-sided die and getting an odd number and an even number in the same roll.
In each of these examples, the events cannot occur simultaneously, making them mutually exclusive.
Examples of Independent Events
Let's explore some examples to better understand independent events:
Tossing a fair coin twice and getting heads on the first toss and tails on the second toss.
Drawing a card from a well-shuffled deck, returning it to the deck, and drawing another card. The outcomes of both draws are unrelated.
Rolling two fair dice and getting a 4 on the first die and a 6 on the second die.
In each of these examples, the occurrence of one event has no impact on the probability of the other event happening, indicating independence.
Importance of Distinguishing Between Mutually Exclusive and Independent Events
Understanding the distinction between mutually exclusive and independent events holds significant importance in various fields. Whether it's business, finance, or scientific research, recognizing the nature of the events involved is crucial for making informed decisions and accurate predictions. By correctly identifying whether events are mutually exclusive or independent, we can effectively assess probabilities and mitigate risks.
Calculating Probabilities for Mutually Exclusive and Independent Events
When it comes to calculating probabilities for mutually exclusive and independent events, different formulas and approaches are used.
For mutually exclusive events, the probability of their intersection is always zero. This means that if events A and B are mutually exclusive, the probability of both A and B occurring together is P(A ∩ B) = 0. To calculate the probability of either event A or event B occurring, we use the addition rule: P(A or B) = P(A) + P(B).
For independent events, the probability of their intersection is the product of their individual probabilities. If events A and B are independent, the probability of both A and B occurring is given by P(A ∩ B) = P(A) P(B). To calculate the probability of either event A or event B occurring, we use the addition rule as well: P(A or B) = P(A) + P(B) - P(A ∩ B).
Understanding these formulas and applying them correctly is essential for accurate probability calculations.
Applications in Real-Life Situations
The concepts of mutually exclusive and independent events find applications in various real-life situations. Here are a few examples:
Insurance: Assessing the probability of mutually exclusive events, such as accidents or damages, allows insurance companies to calculate premiums and provide coverage accordingly.
Market Research: Analyzing independent events, such as consumer preferences and buying behavior, helps businesses make informed decisions about product development, marketing strategies, and target audiences.
Genetics: Identifying mutually exclusive events in genetic traits helps scientists study inheritance patterns and understand the probability of certain genetic disorders or traits occurring in individuals or populations.
Sports: Analyzing independent events, such as the performance of individual players or teams, allows coaches and analysts to make strategic decisions based on probabilities and outcomes.
These are just a few examples of how the concepts of mutually exclusive and independent events have practical applications in various domains.
Common Misconceptions about Mutually Exclusive and Independent Events
Misconceptions about mutually exclusive and independent events can lead to errors in probability calculations and decision-making. Here are a few common misconceptions:
Assuming independence when events are mutually exclusive: It's important to recognize that mutually exclusive events cannot be independent. The occurrence of one event affects the probability of the other event happening.
Believing that independent events are always unrelated: Two events can be independent even if they have some connection. Independence refers to the lack of influence between events, not necessarily a lack of association.
Confusing mutually exclusive events with exhaustive events: Mutually exclusive events cannot occur together, but that doesn't mean they account for all possible outcomes. Mutually exclusive events can be part of a larger set of possibilities.
Considering independence based on past occurrences: Independence is a property of events, not past occurrences. Just because two events have been independent in the past does not guarantee their independence in the future.
Avoiding these misconceptions is crucial for a clear understanding of mutually exclusive and independent events and their implications.
Relationship between Mutually Exclusive and Dependent Events
While mutually exclusive and independent events are distinct concepts, it's important to note that dependent events can be mutually exclusive. Dependence refers to the influence between events, while mutual exclusivity refers to their simultaneous occurrence.
For example, consider the events "rain" and "sunshine" for a specific day. These events are mutually exclusive because it cannot rain and be sunny at the same time. However, their occurrence may depend on factors like weather patterns, seasons, or geographical location. If it rained yesterday, the probability of rain today may be higher, indicating a dependent relationship.
Understanding the relationship between these concepts allows for a more comprehensive analysis of probability and outcomes.
Tips for Identifying Mutually Exclusive and Independent Events
Distinguishing between mutually exclusive and independent events can sometimes be challenging. Here are a few tips to help identify the nature of events:
Read the problem carefully: Pay attention to the wording and context of the problem. Look for phrases like "at the same time," "either/or," or "independent of."
Analyze the outcomes: Consider whether the events share any common outcomes. If they have no outcomes in common, they are likely mutually exclusive.
Consider cause and effect: Reflect on whether the occurrence of one event influences the probability of the other event happening. If there is no impact, the events may be independent.
Apply formulas and calculations: Use the appropriate formulas to calculate probabilities and test the relationship between events. Comparing the joint probability to the product of individual probabilities can provide insights into independence or mutual exclusivity.
By employing these tips, you can effectively identify whether events are mutually exclusive or independent and make accurate probability assessments.
Summary of Differences and Similarities
In summary, mutually exclusive events cannot occur simultaneously and have no common outcomes, while independent events do not affect each other's outcomes and can have common outcomes. Both concepts play important roles in probability analysis and decision-making. Understanding their distinctions and applying the appropriate formulas enable us to calculate probabilities, assess risks, and make informed choices in various contexts.
Conclusion
Distinguishing between mutually exclusive and independent events is essential for accurate analysis and prediction in probability experiments. Mutually exclusive events cannot occur simultaneously, while independent events do not affect each other's outcomes. Understanding these concepts enhances our ability to interpret and calculate probabilities effectively.
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FAQs on Difference Between Mutually Exclusive and Independent Events
What is the difference between mutually exclusive and independent events?
Mutually exclusive events cannot occur simultaneously, while independent events are not influenced by each other and can occur separately or together.
Can mutually exclusive and independent events have different probabilities?
Yes, mutually exclusive events have a combined probability of 0, while independent events have probabilities that can be calculated based on their individual probabilities.
Are mutually exclusive and independent events related to each other?
No, mutually exclusive events are not independent, as the occurrence of one event affects the probability of the other event.
Can mutually exclusive and independent events be illustrated using a Venn diagram?
Yes, mutually exclusive events are represented as non-overlapping circles in a Venn diagram, while independent events are depicted as separate circles.
Are mutually exclusive and independent events applicable to both theoretical and real-life scenarios?
Yes, both concepts are applicable to both theoretical probability calculations and real-life situations involving events and outcomes.
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slow-paced
adjective as in slow
Weak matches
apathetic
crawling
creeping
dawdling
delaying
deliberate
dilatory
disinclined
dreamy
drowsy
easy
gradual
heavy
idle
imperceptible
inactive
indolent
inert
lackadaisical
laggard
lagging
leaden
leisurely
lethargic
listless
loitering
measured
moderate
negligent
passive
phlegmatic
plodding
ponderous
postponing
procrastinating
quiet
reluctant
remiss
slack
sleepy
slothful
slow-footed
slow-going
slow-moving
sluggish
snaillike
stagnant
supine
tardy
torpid
tortoiselike
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Related Words
Words related to slow-paced are not direct synonyms, but are associated with the word slow-paced. Browse related words to learn more about word associations.
slow-footed
adjectiveas in slow
apathetic
crawling
creeping
dawdling
delaying
deliberate
dilatory
disinclined
dreamy
drowsy
easy
gradual
heavy
idle
imperceptible
inactive
indolent
inert
lackadaisical
laggard
lagging
leaden
leisurely
lethargic
listless
loitering
measured
moderate
negligent
passive
phlegmatic
plodding
ponderous
postponing
procrastinating
quiet
reluctant
remiss
slack
sleepy
slothful
slow-going
slow-moving
slow-paced
sluggish
snaillike
stagnant
supine
tardy
torpid
tortoiselike
slow-going
adjectiveas in slow
apathetic
crawling
creeping
dawdling
delaying
deliberate
dilatory
disinclined
dreamy
drowsy
easy
gradual
heavy
idle
imperceptible
inactive
indolent
inert
lackadaisical
laggard
lagging
leaden
leisurely
lethargic
listless
loitering
measured
moderate
negligent
passive
phlegmatic
plodding
ponderous
postponing
procrastinating
quiet
reluctant
remiss
slack
sleepy
slothful
slow-footed
slow-moving
slow-paced
sluggish
snaillike
stagnant
supine
tardy
torpid
tortoiselike
From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group.
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290
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https://dept.math.lsa.umich.edu/~speyer/PolymathIdentities.pdf
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SOME SUMS OVER IRREDUCIBLE POLYNOMIALS DAVID E SPEYER Abstract. We prove a number of conjectures due to Dinesh Thakur concerning sums of the form P P h(P) where the sum is over monic irreducible polynomials P in Fq[T], the function h is a rational function and the sum is considered in the T −1-adic topology. As an example of the simplest of our results, in F2[T], the sum P P 1 P k+1 always converges to a rational function, and is 0 for k = 1.
1. Introduction The point of this document is to explain some identities experimentally found by Dinesh Thakur, involving sums over irreducible polynomials in finite fields. We begin by stating the simplest of these identities: Let P be the set of irreducible polynomials in F2[T]. Then X P∈P 1 P −1 = 0.
Here the sum must be interpreted as a sum of power series in T −1. For example, the first five summands are 1 T−1 = T −1+T −2+T −3+ · · · 1 (T+1)−1 = T −1 1 (T 2+T+1)−1 = T −2+T −3+ · · · 1 (T 3+T+1)−1 = T −3+ · · · 1 (T 3+T 2+1)−1 = T −3+ · · · .
As the reader can see, only finitely many terms contribute to the coefficient of each power of T −1, and the coefficient of T −j is 0 for each j.
We now introduce the notation necessary to state our general results. To aid the reader’s comprehension, we adopt the following conventions: Integers will always be denoted by lower case Roman letters (k, p, q . . . ); polynomials over finite fields will always be denoted by capital Roman letters (A, F, P . . . ), sets of such polynomials will always be denoted by calligraphic letters (A, P, R, . . . ), symmetric polynomials will be denoted by bold letters (ek, pk, . . . ). Of course, there will be other sorts of mathematical objects as well, which we trust the reader to accommodate as they occur.
Let p be a prime and q a power of p. Let Fq be the field with q elements.
Let R be the polynomial ring Fq[T]. Let K be the fraction field Fq(T) and 1 2 DAVID E SPEYER let b K be the T −1-adic completion of K. All infinite sums will be understood in the T −1-adic topology.
Let P be the set irreducible polynomials in R; let P1 be the set of monic irreducible polynomials. Here is our main result for the case p = 2.
Theorem 1.1. If p = 2 then, for any positive integer k ≡0 mod q −1, the sum X P∈P1 1 P k −1 is in K.
The reader may wonder what happens is we sum over all irreducible poly-nomials rather than monic ones; that is an easy corollary: Corollary 1.2. For any positive integer k, the sum X P∈P 1 P k −1 is in K.
Proof. We rewrite the sum as P P∈P1 P a∈F× q 1 (aP)k−1. The corollary then follows from the identity X a∈F× q 1 (aX)k −1 = 1 XLCM(k,q−1) −1 in Fq(U). To prove this identity, write 1 (aX)k −1 = ∞ X j=1 1/(aX)kj and recall that X a∈F× q am = ( 1 m ≡0 mod q −1 0 otherwise .
□ We now discuss the case of a general prime. Define the rational function Gp(U) by Gp(u) = (1 −U p) −(1 −U)p p(1 −U)p .
When p = 2, we have G2(U) = 2U+2U2 2(1−U)2 = U/(1−U), so G2(1/P) = 1/(P−1).
When p is odd, we have the following alternate expressions for Gp: Proposition 1.3. If p is odd, then, as rational functions in Fp(U), we have Gp(U) = Pp−1 j=1 Uj j (1 −U)p = X 0≤j<∞ j̸≡0 mod p Uj j .
SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 3 Proof. If p is odd, then (1−U p)−(1−U)p = Pp−1 j=1(−1)j−1 p j U j. (If p = 2, there is also a 2U 2 term.) We have (−1)j−1 p p j = (−1)j−1(p −1)(p −2) · · · (p −j + 1) 1 · · · 2 · · · (j −1)j ≡1 j mod p.
This proves the first equality, and the second is immediate.
□ Theorem 1.4. For any positive integer k ≡0 mod q −1, the sum X P∈P1 Gp(1/P k) is in K.
As we noted, G2(1/X) = 1/(X −1), so Theorem 1.4 implies Theorem 1.1.
Once again, we have a trivial variant where we sum over P: Corollary 1.5. For any positive integer k, the sum X P∈P Gp(1/P k) is in K.
Proof. If p = 2, we proved this in Corollary 1.2, so we may (and do) assume p is odd.
As in the proof of Corollary 1.2, we rewrite the sum as P P∈P1 P a∈F× q Gp(1/(aP)k).
We now need the identity X a∈F× q Gp((aU)k) = GCD(q −1, k)Gp(U LCM(q−1,k)) in Fq(U). To prove this identity, we use the formula Gp(U) = P j̸≡0 mod p U j/j and the identity X a∈F× q am = ( q −1 m ≡0 mod q −1 0 otherwise .
So X a∈F× q Gp(1/(aU)k) = X j̸≡0 mod p X a∈F× q 1 j(aU)kj = (q −1) X j̸≡0 mod p kj≡0 mod q−1 1 jU kj .
Putting kj = LCM(q −1, k)ℓ, this is (q −1) P ℓ̸≡0 mod p k LCM(q−1,k)ℓULCM(q−1,k)ℓ = k(q−1) LCM(q−1,k)Gp(U LCM(q−1,k)) = GCD(q −1, k)Gp(U LCM(q−1,k)) as required.
□ We also compute explicit values for the sum when k is not too large.
4 DAVID E SPEYER Theorem 1.6. Let k = (q−1)ℓ. If 1 ≤ℓ≤q/p, then P P∈P1 Gp(1/P k) = 0.
If q/p + 1 < ℓ≤2q/p, then X P∈P1 Gp(1/P k) = ℓ (T q −T)q+1 (T q2 −T q)(T q2 −1).
In principle, our methods are capable of computing P P∈P1 Gp(1/P k) for any k ≡0 mod q −1, but they become impractical beyond ℓ= 2q/p.
1.1. History of the problem. Dinesh Thakur experimentally discovered most of the relations described above in characteristic two, and suspected there should be similar results in odd chracteristic. Thakur published these computations them in a preprint entitled “Surprising symmetries in distri-bution of prime polynomials”, arXiv:1512.02685. At Thakur’s suggestion, Terence Tao promoted the problem in posts on his blog and on the Poly-math blog. I am grateful to Thakur for finding such an elegant problem and to Tao for bringing it to my attention. My thanks also to all who partici-pated in the discussion on the Polymath blog – Noam Elkies, Ian Finn, Ofir Gorodetsky, Jesse, Gil Kalai, David Lowry-Duda, Dustin G. Mixon, John Nicol, Partha Solapurkar, John Voight, Victor Wang, Qiaochu Yuan, Joshua Zelinsky.
2. The Carlitz exponential, and symmetric polynomials The main tool in our proofs is the theory of the Carlitz exponential. Put Di = (T qi −T)(T qi −T q)(T qi −T q2) · · · (T qi −T qi−1).
Define eC(Z) = ∞ X j=0 T qj Dj this sum is T −1-adically convergent for any Z ∈b K. We will make use of the product identity: eC(πZ) πZ = Y A∈R{0} 1 + Z A where π is the element of b K( q−1 √ T) such that the coefficients of Zq−1 match.
This identity should be thought of as similar to Euler’s identity: sin(πz) πz = Y a∈Z̸=0 1 + z a .
We introduce the notations A for the nonzero polynomials of R, and A1 for the monic polynomials.
Writing ek for the elementary symmetric function of degree k, this implies ek(1/A)A∈A = ( πk/Dj k = qj −1 0 otherwise SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 5 Since the ring of symmetric polynomials is generated by the ek, we deduce Proposition 2.1. If f is a homogenous symmetric polynomial of degree k, then f(1/A)A∈A is in πkK.
Here we note that f(1/A)A∈A is always defined, since only finitely many terms contribute to the coefficient of any particular power of T −1.
The above considers symmetric polynomials in {1/A}A∈A, but we would rather restrict to the case of A monic. To this end, we have Proposition 2.2.
eℓ(1/Aq−1)A∈A1 = ( (−1)ℓπℓ(q−1)/Dj ℓ= qj−1 q−1 0 otherwise Proof. Grouping together scalar multiples of the same polynomial in the Carlitz product identity, we have eC(πZ) πZ = Y A∈A1 1 −Zq−1 Aq−1 .
Equate coefficients of Zℓ(q−1) on both sides.
□ Corollary 2.3. If f is a homogenous symmetric polynomial of degree ℓ, then f(1/Aq−1)A∈A1 is in πℓ(q−1)K.
3. Proofs of rationality We now have enough background Theorem 1.4 and, hence, Theorem 1.1.
Throughout, let k ≡0 mod q −1.
Consider the symmetric polynomial gp(X1, X2, . . . , ) := 1 p X Xi p − X Xp i .
The polynomial gp has integer coefficients, so we may discuss plugging ele-ments of K into it.
Let C be the cyclic group of order p, and let C act on Ap 1 by rotating coordinates. Let ∆denote the small diagonal: ∆:= {(A, A, . . . , A)} ⊂Ap 1.
Then gp(1/Ak)A∈A1 = X (A1,...,Ap)∈(Ap 1\∆)/C 1 Ak 1Ak 2 · · · Ak p .
The sum is over cosets for the free action of C on Ap \ ∆.
Let Φ = {(A1, . . . , Ap) ∈Ap 1 : GCD(A1, . . . , Ap) = 1}.
Any (A1, . . . , Ap) ∈Ap 1 can be uniquely factored as Ai = DBi for some D ∈A1 and (B1, . . . , Bp) ∈Φ. So we can factor the above sum as gp(1/Ak)A∈A1 = X D∈A1 1 Dkp X (B1,...,Bp)∈(Φ{(1,...,1)})/C 1 Bk 1Bk 2 · · · Bk p .
6 DAVID E SPEYER Now, from Proposition 2.3, gp(1/Ak)A∈A, is in πpkK. Also from Proposi-tion 2.3, P D∈A1 1/Dkp is in πpkK, and a quick computation shows that this sum is 1 plus terms in t−1Fq, so it is not zero. We deduce that X (B1,...,Bp)∈(Φ(1,...,1))/C 1 Bk 1Bk 2 · · · Bk p ∈K.
For B ∈A1, let Ψ(B) be the set of p-tuples (B1, B2, . . . , Bp) for which GCD(B1, . . . , Bp) = 1 and Q Bi = B. Let ψ(B) = #Ψ(B). So we have shown that X B∈A1{1} ψ(B)/p Bk ∈K.
Here, to interpret the numerator, we must divide ψ(B) by p as integers and only then consider the quotient in Fp.
If B = P k1 1 P k2 2 · · · P kr r then there is an easy bijection between Ψ(B) and Ψ(P k1 1 ) × Ψ(P k2 2 ) × · · · × Ψ(P kr r ), so ψ(B) = Q ψ(P ki i ). If P is irreducible then ψ(P r) is divisible by p for any r > 0, since C acts freely on Ψ(P r).
So, if B is divisible by two different irreducible polynomials, then ψ(B) is divisible by p2. So we can rewrite the sum as X P∈P1 ∞ X r=1 ψ(P r)/p P rk .
We now compute ψ(P r); which is the number of p-tuples (P r1, . . . , P rp) with Q P ri = P r and GCD(P r1, . . . , P rp) = 1. In other words, we must count (r1, . . . , rp) ∈Zp ≥0 with P ri = r and min(r1, . . . , rp) = 0. The number of (r1, . . . , rp) ∈Zp ≥0 with P ri = r is the coefficient of U r in 1/(1 −U)p.
In order to impose min(r1, . . . , rp) = 0, we subtract offthe terms with min(r1, . . . , rp) > 0. These are in bijection with (s1, . . . , sp) ∈Zp ≥0 with p + P si = r. So ψ(pr) is the coefficient of U r in 1/(1 −U)p −U p/(1 −U)p.
In other words, P∞ r=0 ψ(P r)U r = (1 −U p)/(1 −U)p. So ∞ X r=1 ψ(P r) p U r = 1 p 1 −U p (1 −U)p −1 = Gp(U).
We deduce that P∞ r=1 ψ(P r)/p P rk = Gp(1/P k).
We have now shown that P P∈P1 Gp(1/P k) ∈K, as claimed. □ We record the specific formula we have proved: Proposition 3.1. Let k be a positive integer. Then X P∈P1 Gp(1/P k) = gp(1/Ak)A∈A1 P A∈A1 1/Apk We will rewrite this formula in various ways in Section 5. We remark that this formula is correct even if k is not divisible by q −1, although we have only shown the ratio is in K when k ≡0 mod q −1.
SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 7 4. Vanishing We will now prove the claim in Theorem 1.6 that the sum vanishes when k = (q −1)ℓfor 1 ≤ℓ≤q/p. From Proposition 3.1, it is equivalent to show that gp(1/Aℓ(q−1))A∈A1 = 0. To this end, we must explicitly write gp(1/Aℓ(q−1)) as a polynomial in the ek(1/Aq−1).
The variables λ or µ will always denote partitions, meaning weakly de-creasing sequences (λ1, λ2, . . . , λr) of positive integers; sums over λ or µ implicitly contain the condition that the summation variable is a partition.
We define eλ = Q s eλs. The symmetric polynomials eλ form an integer basis for the symmetric polynomials with integer coefficients.
Lemma 4.1. Write gp(Xℓ 1, Xℓ 2, . . . , ) = X |λ|=pℓ cλeλ(X1, X2, . . .) for some integers cλ. Then c11···1 = 0.
Proof. Note that e11···1 is in the only eλ with a nonzero coefficient of Xpℓ 1 .
The coefficient of Xpℓ 1 in gp(Xℓ 1, Xℓ 2, . . .) is clearly 0.
□ Now, suppose that ℓ≤q/p, so we have pℓ< q + 1. So any partition (λ1, . . . , λr) of pℓother than (1, 1, . . . , 1) contains a λi between 2 and q. By Lemma 2.2, em(1/Aq−1)A∈A1 = 0 for 2 ≤m ≤q, so eλ(1/Aq−1)A∈A1 = 0 whenever λ is a partition of pℓother than (1, 1, . . . , 1).
We deduce that gp(1/Aq−1)A∈A1 = 0 as desired. □ 5. Computations for small k In this section, we will discuss the computation of P P∈P1 1/P k for k ≡ 0 mod q −1 and, in particular, prove the remaining half of Theorem 1.6.
Our strategy is to combine Propositions 3.1 and 2.2.
We must compute gp(1/Aℓ(q−1))A∈A1 and P A∈A1 1/Apk. Note the latter is pℓ(1/Aq−1)A∈A1 p.
We write k = (q −1)ℓ.
Put gp(Xℓ 1, Xℓ 2, . . . ) = P |λ|=ℓp cλeλ(X1, X2, . . .) pℓ(X1, X2, . . . ) = P |µ|=ℓdµeµ(X1, X2, . . .).
Note that em(1/Aq−1)A∈A1 = 0 unless m is of the form (qj −1)/(q −1). So we only need to sum over partitions where all the parts of λ are of the form (qj −1)/(q −1).
From now on, we now impose that q/p + 1 ≤ℓ≤2q/p. So ℓ< q + 1. Any partition of ℓcannot contain any parts of size (qj −1)/(q − 1), for j ≥1.
Similarly, pℓ< 2q + 2, so a partition of pℓcan contain at most one part of size (q2 −1)/(q −1) = q + 1 and no parts of size (qj −1)/(q −1) for j ≥2. We deduce that the only terms which contribute to our final answer come from λ = (1, 1, . . . , 1) or λ = (q + 1, 1, 1, . . . , 1) when computing gp(1/Aℓ(q−1))A∈A1, and from µ = (1, 1, . . . , 1) in computing 8 DAVID E SPEYER pℓ(1/Aq−1)A∈A1 p. Moreover, from Lemma 4.1, the coefficient c(1,1,...,1) is zero.
We deduce that P P∈P1 1/P k = c(q+1,1pℓ−q−1) e(q+1,1pℓ−q−1)(1/Aq−1)A∈A1 (d1ℓe1ℓ(1/Aq−1)A∈A1) p = c(q+1,1pℓ−q−1) eq+1(1/Aq−1)A∈A1(e1(1/Aq−1)A∈A1) pℓ−q−1 d1ℓ(e1(1/Aq−1)A∈A1) pℓ = c(q+1,1pℓ−q−1) eq+1(1/Aq−1)A∈A1 d1ℓ(e1(1/Aq−1)A∈A1) q+1 .
Here 1r is shorthand for r parts equal to 1.
We now use Proposition 2.2. The powers of π and (−1) cancel to give X P∈P1 1/P k = c(q+1,1pℓ−q−1) d1ℓ Dq+1 1 D2 = c(q+1,1pℓ−q−1) d1ℓ (T q −T)q+1 (T q2 −T q)(T q2 −1).
To finish the computation, we must find cq+1,1ps−1 and d1ℓ. The latter is easy: Comparing coefficients of Xℓ 1 on both sides of pℓ(X1, X2, . . . ) = P |µ|=ℓdµeµ(X1, X2, . . .), we deduce that d1ℓ= 1.
To compute c(q+1,1pℓ−q−1), we begin with the formula gp(Xℓ 1, Xℓ 2, . . . ) = 1 p pℓ(X1, X2, . . . , )p −ppℓ(X1, X2, . . .) .
For brevity, we write f(X) to indicate that the inputs to a symmetric poly-nomial are (X1, X2, . . .). Note that we are working with symmetric polyno-mials with integer coefficients, so it makes sense to divide by p.
We rewrite the right hand side of the previous equation as 1 p e1(X)ℓ+ · · · p − e1(X)pℓ+ dq+1,1pℓ−q−1eq+1(X)e1(X)pℓ−q−1 + · · · .
Here the ellipses denote terms eλ where λ has some part not of the form (qj −1)/(q −1). We deduce that cq+1,1pℓ−q−1 = −1 pdq+1,1pℓ−q−1.
Now, observe the identity X j (−1)j−1pj(X)U j j = X i log(1 + XiU) = log Y i (1 + XiU) = log 1 + ∞ X m=1 em(X)U m !
.
The coefficient of U pℓon the left is (−1)pℓ pℓ ppℓ. Expanding the log on the right hand side as a Taylor series, only one term contributes to U pℓeq+1epℓ−q−1 1 .
SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 9 So we obtain (−1)pℓ−1 pℓ ppℓ(X) = (−1)pℓ−q−1 pℓ−q pℓ−q 1 eq+1(X)epℓ−q−1 1 (X) + · · · where the ellipses denote a sum of eλ other than eq+1(X)epℓ−q−1 1 (X). So dq+1,1pℓ−q−1 = (−1)qpℓand cq+1,1ps−1 = (−1)q−1ℓ. Plugging into our previ-ous formula, and using that (−1)q−1 ≡1 mod p, X P∈P1 1/P k = ℓ (T q −T)q+1 (T q2 −T q)(T q2 −1).
This concludes the proof of Theorem 1.6. □ We conclude by verifying one Thakur’s conjectures which goes beyond the range ℓ≤2q/p. Let p = q = 2. Thakur conjectures X P∈P1 1 P 3 −1 = 1 T 4 + T 2 .
We begin by computing p3(X) = e1(X)3 + 3e3(X)3 −3e2(X)e1(X) p3(X)2 = e1(X)6 + 6e1(X)3e3(X) + 9e3(X)2 + · · · .
Here and in the following equations, the ellipses denote eλ terms where λ contains a part other than 1 and 3. (Note that (23 −1)/(2 −1) = 7, too large to contribute to a symmetric polynomial of degree 6.) Similarly, p6(X) = e1(X)6 + 6e1(X)3e3(X) + 3e3(X)2 + · · · .
So g2(X3 1, X3 2, . . .) = 1 2 p3(X)2 −p6(X) = 3e3(X)2 + · · · and g2(1/A3)A∈A1 = (e3(1/A)A∈A1)2 = π6 D2 2 = π6 (T 4 −T 2)2(T 4 −T)2 .
(Recall that we are working modulo 2, so 3 = 1 = −1 and 2 = 0.) Similarly, p6(1/A)A∈A1 = (e1(1/A)A∈A1)6 + (e3(1/A)A∈A1)2 = π D1 6 + π3 D2 2 = π6 1 T 2−T 6 + 1 (T 4−T 2)(T 4−T) 2 We verify Thakur’s claim: X P∈P1 1 P 3 −1 = 1/((T 4 −T 2)2(T 4 −T)2) 1/(T 2 −T)6 + 1/((T 4 −T 2)2(T 4 −T)2) = 1 T 4 + T 2 .
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1-Propanol
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1-Propanol
Formula: C 3 H 8 O
Molecular weight: 60.0950
IUPAC Standard InChI:InChI=1S/C3H8O/c1-2-3-4/h4H,2-3H2,1H3 Copy
IUPAC Standard InChIKey:BDERNNFJNOPAEC-UHFFFAOYSA-N Copy
CAS Registry Number: 71-23-8
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Other names: Propyl alcohol; n-Propan-1-ol; n-Propanol; n-Propyl alcohol; Ethylcarbinol; Optal; Osmosol extra; Propanol; Propylic alcohol; 1-Propyl alcohol; n-C3H7OH; 1-Hydroxypropane; Propanol-1; Propan-1-ol; n-Propyl alkohol; Alcool propilico; Alcool propylique; Propanole; Propanolen; Propanoli; Propylowy alkohol; UN 1274; Propylan-propyl alcohol; NSC 30300; Alcohol, propyl
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n-propanol
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Notice: Except where noted, spectra from this collection were measured on dispersive instruments, often in carefully selected solvents, and hence may differ in detail from measurements on FTIR instruments or in other chemical environments. More information on the manner in which spectra in this collection were collected can be found here.
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| Owner | COBLENTZ SOCIETY Collection (C) 2018 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. |
| Origin | DOW CHEMICAL COMPANY |
| Source reference | COBLENTZ NO. 10490 |
| Date | 1965/10/15 |
| Name(s) | 1-propanol |
| State | GAS (SATURATED) |
| Instrument | DOW KBr FOREPRISM-GRATING |
| Instrument parameters | BLAZED AT 3.5, 12.0, 20.0 MICRON, CHANGED AT 5.0, 7.5, 14.9 MICRON |
| Path length | 12.5 CM CELL |
| Resolution | 4 |
| Sampling procedure | TRANSMISSION |
| Data processing | DIGITIZED BY NIST FROM HARD COPY (FROM TWO SEGMENTS) |
This IR spectrum is from the Coblentz Society's evaluated infrared reference spectra collection.
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Data from NIST Standard Reference Database 69: NIST Chemistry WebBook
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Back To Independent EventsBack
12.16
Independent Events
Written by:CK-12 |Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Jul 01, 2025
Disjoint and Overlapping Events
When you toss a coin to calculate a probability, it is impossible to get a head and a tail at the same time. If you get a heads then the probability of tails is 0. Similarly, when you roll a single die, it is impossible to get an odd number and an even number at the same time. The above events are disjointed, or not connected, that is they have no outcomes in common. Events that don't have any outcomes in common are disjoint events.
The Venn diagram for disjoint events shows no overlap between the two events.
[Figure 1]
Disjoint Events have no overlap
Disjoint events cannot happen at the same time. In other words, they are mutually exclusive. Put in formal terms: events A and B are disjoint if their intersection is an empty set:
P(A∩B)=∅
The events that are connected, that is they have some outcomes in common, are overlapping events.
The Venn diagram for overlapping events shows some overlap between the two events.
[Figure 2]
Overlapping Events have some outcomes that appear in multiple sets.
INTERACTIVE
Independent Events (Warm Up)
Warm Up:
You are playing a game of basketball, and you have just learned about disjoint events! The ball is passed to you and you have many choices of what you can do with it. Two of these choices are listed on the ball. Think about whether these events must happen independently or can happen simultaneously, and decide whether or not they are disjoint events.
Shoot the basketball into one of the "Disjoint Events" or "Non Disjoint Events" boxes by dragging it. Once the ball is in a valid field, a "Submit" button will appear. Click this button to reveal the correct answer and then click the subsequent "Next" button to reset the basketball.
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Determining Probability
Recall that the probability of an event is the chance of it happening. Probabilities can be written as fractions or decimals between 0 and 1, or as percents between 0% and 100%. T o find the probability of an event if all the outcomes in the event have an equal chance of occurring, find the number of outcomes in the event and divide by the number of outcomes in the sample space.
P(E)=#of outcomes in E#of outcomes in sample space
Consider a dartboard. To determine the probability of landing a dart in a green zone, find the number of green sectors and divide by the total number of sectors on the dartboard. Similarly, if each of the letters of the word 'PROBABILITY' is written on a card and all the cards are well-shuffled and placed face down. To determine the probability of drawing a vowel card, find the number of vowel cards and divide by the total number of cards.
A spinner is divided into 5 equal sectors of different colors. What is the probability of the pointer landing on the green, when the spinner is spun?
The probability of the pointer landing on the green is .
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
The sample space is S={H H,H T,T H,T T}, which contains four outcomes.
Let A be the event that heads comes up exactly once.
A={H T,T H}
Therefore:P(A)=2 4=1 2
To find the probability of a single event, all you need to do is count the number of outcomes in the event and the number of outcomes in the sample space. Probability calculations become more complex when you consider the combined probability of two or more events.
Two events are independent if one event occurring does not change the probability of the second event occurring. Two events are dependent if one event occurring causes the probability of the second event to go up or down. Two events are independent if the probability of A and B occurring together is the product of their individual probabilities:
P(A∩B)=P(A)P(B) if and only if A and B are independent events.
In some cases it is pretty clear whether or not two events are independent. In other cases, it is not at all obvious. You can always test if two events are independent by checking to see if the probabilities satisfy the relationship above.
CK-12 PLIX Interactive
D etermining Independence
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
The sample space is S={H H,H T,T H,T T}, which contains four outcomes. Let C be the event that the first coin is a heads. Let D be the event that the second coin is a tails.
List the outcomes in events C and D.
C={H H,H T}.D={H T,T T}. Note that the outcomes in the two events overlap. This does NOT mean that the events are not independent!
Take a guess at whether or not you think the two events are independent.
If you get heads on the first coin, that shouldn't have any effect on whether you get tails for the second coin. It makes sense that the events should be independent.
Find P(C) and P(D).
P(C)=2 4=1 2.P(D)=2 4=1 2.
1b) Find P(C∩D)- Are the two events independent?
C∩D is the event of getting heads first and tails second. C∩D={H T}.
P(C∩D)=1 4.P(C)P(D)=(1 2)(1 2)=1 4.
Because P(C∩D)=P(C)P(D), the events are independent.
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
The sample space is S={H H,H T,T H,T T}, which contains four outcomes. Let C be the event that the first coin is a heads. Let E be the event that both coins are heads.
List the outcomes in events C and E.
C={H H,H T}. E={H H}.
Take a guess at whether or not you think the two events are independent.
If you get heads on the first coin, then you are more likely to end up with two heads than if you didn't know anything about the first coin. It seems like the events should NOT be independent.
Find P(C) and P(E).
P(C)=2 4=1 2. P(E)=1 4.
2b) Find P(C∩E). Are the two events independent?
C∩E is the event of getting heads first and both heads. This is the same as the event of getting both heads, since if you got both heads then you definitely got heads first. C∩E={H H}.
P(C∩E)=1 4.P(C)P(E)=(1 2)(1 4)=1 8.
Because P(C∩E)≠P(C)P(E), the events are NOT independent.
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
The sample space is S={H H,H T,T H,T T}, which contains four outcomes. Let E be the event that both coins are heads. Let F be the event that both coins are tails.
List the outcomes in events E and F.
E={H H}. F={T T}.
Take a guess at whether or not you think the two events are independent.
If you get both heads, then you definitely didn't get both tails. It seems like the events should NOT be independent.
Find P(E) and P(F).
P(E)=1 4. P(F)=1 4.
3b) Find P(E∩F). Are the two events independent?
E∩F is the event of getting two heads and two tails. This is impossible to do because these two events are disjoint. E∩F={}.
P(E∩F)=0.P(E)P(F)=(1 4)(1 4)=1 16.
Because P(E∩F)≠P(E)P(F), the events are NOT independent.
P(G)=1 3 and P(H)=1 2. If P(G∩H)=1 4, are events G and H independent?
The events G and H are .
Examples
Example 1
Two events are disjoint if they have no outcomes in common. Consider two events A and B that are disjoint. Can you say whether or not events A and B are independent?
If A and B are disjoint, then A∩B={} and P(A∩B)=0.
For the two events to be independent, P(A)P(B)=P(A∩B). This means that P(A)P(B)=0. By the zero product property, the only way for P(A)P(B)=0 is if P(A)=0 or P(B)=0.
In other words, two disjoint events are independent if and only if the probability of at least one of the events is 0.
Example 2
Consider the experiment of tossing a coin and then rolling a die. Event A is getting a tails on the coin. Event B is getting an even number on the die. Are the two events independent? Justify your answer using probabilities.
The sample space for the experiment is S={H 1,H 2,H 3,H 4,H 5,H 6,T 1,T 2,T 3,T 4,T 5,T 6}. Next consider the outcomes in the events. A={T 1,T 2,T 3,T 4,T 5,T 6}. B={H 2,H 4,H 6,T 2,T 4,T 6}. A∩B={T 2,T 4,T 6}.
P(A)=6 12=1 2
P(B)=6 12=1 2
P(A)P(B)=(1 2)(1 2)=1 4
P(A∩B)=3 12=1 4
The events are independent because P(A)P(B)=P(A∩B).
Example 3
Consider the experiment of rolling a pair of dice. Event C is a sum that is even and event D is both numbers are greater than 4. Are the two events independent? Justify your answer using probabilities.
First find the sample space and the outcomes in each event:
S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}C={(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)}D={(5,5),(5,6),(6,5),(6,6)}C∩D={(5,5),(6,6)}
Next find the probabilities:
P(C)=18 36=1 2
P(D)=4 36=1 9
P(C)P(D)=(1 2)(1 9)=1 18
P(C∩D)=2 36=1 18
The events are independent because P(C)P(D)=P(C∩D).
Summary
Overlapping events are events that have outcomes in common.
Disjoint or mutually exclusive events cannot both occur in a single trial of a given experiment.
The probability of E is P(E)=# outcomes in E# outcomes in sample space
Two events are independent if one event occurring does not change the probability of the second event occurring.
P(A∩B)=P(A)P(B)
Dependent events are events where one outcome impacts the probability of the other.
Review
Consider the experiment of flipping a coin three times and recording the sequence of heads and tails. The sample space is S={H H H,H H T,H T H,T H H,H T T,T H T,T T H,T T T}, which contains eight outcomes. Let A be the event that exactly two coins are heads. Let B be the event that all coins are the same. Let C be the event that at least one coin is heads. Let D be the event that all coins are tails.
List the outcomes in each of the four events. Which of the two events are complements?
Find P(A),P(B),P(C),P(D).
Find P(A∩C). Are events A and C independent? Explain.
Find P(B∩D). Are events B and D independent? Explain.
Find P(B∩C). Are events B and C independent? Explain.
Create two new events related to this experiment that are independent. Justify why they are independent using probabilities.
Consider the experiment of drawing a card from a deck. The sample space is the 52 cards in a standard deck. Let A be the event that the card is red. Let B be the event that the card is a spade. Let C be the event that the card is a 4. Let D be the event that the card is a diamond.
Describe the outcomes in each of the four events.
Find P(A),P(B),P(C),P(D).
Find P(A∩B). Are events A and B independent? Explain.
Find P(B∩C). Are events B and C independent? Explain.
Find P(A∩D). Are events A and D independent? Explain.
P(A)=1 4 and P(B)=1 8. If P(A∩B)=1 16 are events A and B independent?
P(A)=1 4 and P(B)=1 8. If P(A∩B)=1 32 are events A and B independent?
What is the difference between disjoint and independent events?
Two events are disjoint, and both have nonzero probabilities. Can you say whether the events are independent or not?
Bag A contains 5 yellow, 6 blue, and 4 white marbles. Bag B contains 8 green, 5 black, and 4 red marbles. Create a probability problem that would show that is you pick two marbles from the bags, that these events would be independent?
Consider the outcomes for the tossing of 3 coins. Let A represent the outcomes where the first coin tossed is a head. Let B represent the outcomes where there is at least 2 heads tossed. Create a problem, using notation found in the concept, that would illustrate two independent events.
You and your friend are in separate lines for tickets to the concert. Describe the outcome(s) that would make these event disjoint. What about independent? Is it possible to be both disjoint and independent? Why or why not?
Review (Answers)
To see the answer key for this book, go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
Image Attributions
Back to Independent Events
| Image | Reference | Attributions |
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| | [Figure 1] | Credit:CK 12 Source:CK 12 License:CC BY-NC 3.0 |
| | [Figure 2] | Credit:CK 12 Source:CK 12 License:CC BY-NC 3.0 |
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Opacity - Simple English Wikipedia, the free encyclopedia
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294
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https://www.sfu.ca/~mdevos/notes/geom-sym/14_transpositions.pdf
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14 Transpositions Definition.
A cycle is a permutation A with the property that the cycle representation of A has exactly one cycle. For instance A = (a1a2 . . . ak). We call k the length of the cycle.
Note: It may seem that there is ambiguity about an expression such as (164)(29)(8735).
Is this one permutation with three cycles, or a product of the three cycles (164), (29), and (8735)? Fortunately, the permutation (164)(29)(8735) is equal to the product of the three cycles (164), (29), and (8735), so there is no trouble.
Definition.
A transposition is a cycle of length 2. So, in cycle notation, a transposition has the form (ab). Note that every transposition is its own inverse: (ab)(ab) = I.
Lemma 14.1. Every permutation can be expressed as a product of transpositions.
Proof. A quick check reveals that a cycle (a1, a2, . . . , ak) can be represented as follows: (a1a2a3 . . . ak) = (a1ak) . . . (a1a4)(a1a3)(a1a2) Since every permutation is a product of cycles, every permutation may be represented as a product of transpositions.
Example: Represent the permutation (13584)(2967) ∈S9 as a product of transpositions.
(13584)(2967) = (14)(18)(15)(13)(27)(26)(29) Note: Every permutation can be expressed as a product of transpositions in many (actually infinitely many) ways. For instance, the permutation (13584)(2967) from the above example can also be expressed in all of the following ways (13584)(2967) = (72)(38)(17)(28)(47) = (69)(64)(68)(12)(17)(15)(13)(56)(24) = (91)(95)(98)(94)(93)(92)(97)(96)(45)(83)(12) We have just represented a particular permutation as a product of 5, 9, and 11 transpositions— all odd numbers. This is not a coincidence! In fact for every permutation either all such expressions will all have an even number of terms, or all such expressions will have an odd number of terms. We next prove that this property holds for the identity.
2 Lemma 14.2. If T1, . . . , Tm are transpositions and T1T2 · · · Tm = I, then m is even.
Proof. We return to our original way of thinking about composition using a figure where each bijection in is represented by arrows from 1, 2, . . . , n to 1, 2, . . . , n as follows: 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 T1 Tm−1 Tm 1 2 3 4 5 T2 Now, if we start at the rightmost 1 and follow the arrows we go along a path, finally ending up at the leftmost 1 (since the product T1T2 · · · Tm is the identity). Let’s imagine this path as a string which we call strand 1. Similarly, for every 2 ≤i ≤n we have a strand starting and ending at i. The figure may be complicated since the strands may cross one another many times, but nevertheless, we can reason about these crossings. Make the following definitions: 1. Let ci,j be the number of times strand i and strand j cross.
2. Let c be the total number of times one strand crosses another.
3. Let tk be the number of times one strand crosses another in the transposition Tk.
For any two distinct strands, say i and j, it must be that ci,j is even, since these strands must cross an even number of times in order to end in the same positions in which they begin. It follows from this that the total number of crossings c must also be even.
Next let us think about the number of crossings contributed by a single transposition Tk.
Suppose (without loss) that Tk = (ab) where a < b. If b = a + 1 then the only strands crossing in Tk are strand a and a + 1 so tk = 1. If b = a + 2 then strands a and a + 2 cross in Tk but strand a + 1 also gets crossed by strand a and a + 2 for a total of 3 crossings. More generally, if b = a + p then strands a and a + p will cross one another, and both of these strands will cross all of the strands numbered a + 1, a + 2, . . . , a + p −1. This gives a total of 1 + 2(p −1) crossings, so tk = 1 + 2(p −1) is odd. The total number of crossings can also be counted by summing the contributions from each transposition, giving us the equation c = t1 + t2 + . . . + tm Now c is even, but each tk is odd, and it follows that m must be even, as desired.
3 Now we are ready to prove that this even/odd property holds for every permutation.
Theorem 14.3. For every permutation A ∈Sn, either every representation of A as a product of transpositions has an odd number of transpositions, or every such representation has an even number of transpositions.
Proof. Let T1, . . . , Tj and U1, . . . , Uk be transpositions satisfying A = T1T2 · · · Tj = U1U2 · · · Uk To prove the theorem it suffices to show that either j and k are both even, or j and k are both odd. For any product of permutations B = B1B2 · · · Bm−1Bm the inverse is always given by B−1 = B−1 m B−1 m−1 · · · B−1 2 B−1 1 . Since every transposition is its own inverse, we can therefore express A−1 as A−1 = UkUk−1 · · · U2U1 Now we have I = AA−1 = T1 · · · TjUk · · · U1 By the previous lemma we deduce that j + k is even, so either j and k are both even, or j and k are both odd, as desired.
Definition.
We call a permutation A ∈Sn even if it can be represented as a product of an even number of transpositions, and odd if it can be represented as a product of an odd number of transpositions. By Theorem 14.3 every permutation is either even and not odd, or is odd and not even. We call this characteristic (even or odd) the parity of A.
Note: If A is a cycle of length k, say A = (a1a2 . . . ak) then we can express A as A = (a1ak) . . . (a1a3)(a1a2). Therefore a cycle of even length is an odd permutation, and a cycle of odd length is an even permutation!
Observation 14.4. If A, B ∈Sn then the product AB satisfies AB is ( even odd ) if ( A, B are either both even or both odd one of A, B is even and the other is odd ) .
4 Proof. Express A and B as products of transpositions A = T1T2 · · · Tj B = U1U2 · · · Uk Now AB = T1T2 · · · TjU1U2 · · · Uk so AB is even if j + k is even and odd if j + k is odd.
The previous observation shows that parity of permutations acts just like the parity of integers: Adding two integers that are both even or both odd gives an even integer; adding two integers with one odd and the other even gives an odd integer.
Proposition 14.5. A permutation is ( even odd ) if, in cycle notation, there are an ( even odd ) number of cycles of even length.
Proof. This follows from the previous observation and the fact that a cycle of odd length is an even permutation while a cycle of even length is an odd one.
Example: The parity of the permutation (13)(94)(657)(28) is odd since this permutation has three cycles of even length.
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https://www.engineeringtoolbox.com/inductors-energy-stored-d_1391.html
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Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Inductors - Stored Energy
Energy stored in a magnetic field.
The energy stored in the magnetic field of an inductor can be calculated as
W = 1/2 L I2 (1)
W = energy stored (joules, J)
L = inductance (henrys, H)
I = current (amps, A)
Example - Energy Stored in an Inductor
The energy stored in an inductor with inductance 10 H with current 5 A can be calculated as
W = 1/2 (10 H) (5 A)2
= 125 J
Unit Converter
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296
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https://www.parabola.unsw.edu.au/sites/default/files/2024-03/vol39_no1_3.pdf
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Parabola Volume 39, Issue 1 (2003) Rearrangement Inequalities David Angell1 Let’s begin with a puzzle.
x 1 2 3 y xy total = You are to write the numbers 4, 5, 6 in the row marked “y”; multiply the two numbers in each column to obtain three products xy in the third row; add these products to obtain a total. The problem is to choose the order in which 4, 5, 6 are placed so as to obtain the largest possible total. Once you have done this, find the order which will give the smallest total.
Of course this is really a very easy puzzle. There are only six possible ways to write 4, 5 and 6 in the three slots; all you have to do is calculate the six totals and then pick out the largest and the smallest. You should have found that the maximum total of 32 is obtained by writing the numbers in increasing order, 4, 5, 6; while the minimum total, 28, is obtained by writing the numbers in decreasing order, 6, 5, 4.
This is an example of a rearrangement inequality. In order to write down a precise statement of this result, let’s first note that if we had written the x values in a different order, then we should have had to change the order of the ys correspondingly. In fact, it clearly makes no difference in what order we add the products to get the final total; so we may as well assume that the xs appear in increasing order.
Theorem Rearrangement inequalities for sums of products. Let x1, x2, . . . , xn be pos-itive real numbers in increasing order, that is, 0 < x1 ≤x2 ≤· · · ≤xn, and let y1, y2, . . . , yn be positive real numbers. Let z1, z2, . . . , zn be a rearrangement of the yj, and consider the sum x1z1 + x2z2 + · · · + xnzn .
(1) This sum has its maximum value when the zj are arranged in increasing order, and its minimum value when they are arranged in decreasing order.
Comment Observe carefully that we have not said that the maximum occurs only when the zj are in increasing order. If some of the xj are equal then there may be a different order of zj for which the sum is still maximal. We are only saying that, compared with the increasing order, there is no other order which actually gives a greater sum.
1David Angell is an Associate Lecturer at UNSW 1 Examples We shall give some examples first and prove the theorem later.
1. Consider two positive real numbers a and b, with a ≤b; take n = 2 and x1 = a, x2 = b and { y1, y2 } = { a, b }. Then ab + ba ≤x1z1 + x2z2 ≤a2 + b2 ; ignoring the middle term and dividing both sides by 2 yields ab ≤a2 + b2 2 , which is a version of the arithmetic mean–geometric mean inequality.
2. If a, b, c are positive then a + b c + c + a b + b + c a ≥6 .
Proof By symmetry we may assume that a ≤b ≤c; take x1 = a , x2 = b , x3 = c and y1 = 1 a , y2 = 1 b , y3 = 1 c .
Then y3 ≤y2 ≤y1, and so for any rearrangement z1, z2, z3 of y1, y2, y3 we have x1z1 + x2z2 + x3z3 ≥x1y1 + x2y2 + x3y3 = a a + b b + c c = 3 .
Making two particular choices for z1, z2, z3 gives a b + b c + c a ≥3 and a c + b a + c b ≥3 ; adding these and collecting terms with the same denominator proves the result claimed.
3. Let a1, a2, . . . , an be real numbers in increasing order; take x1 = a2 1 , . . . , xn = a2 n and y1 = 1 a1 , . . . , yn = 1 an .
Then (using only one half of the rearrangement inequality) x1z1 + · · · + xnzn ≥x1y1 + · · · + xnyn = a1 + · · · + an .
In particular, if we take n = 2002 and z1 = y2, z2 = y3, ..., z2001 = y2002, z2002 = y1 we obtain a2 1 a2 + a2 2 a3 + · · · + a2 2001 a2002 + a2 2002 a1 ≥a1 + a2 + · · · + a2001 + a2002 , which solves question 6 on the 2002 UNSW mathematics competition.
Instead of multiplying pairs of numbers and adding the products we could have done the reverse: added pairs of numbers and then multiplied the sums. A bit of experi-menting (perhaps with x = 1, 2, 3 and y = 4, 5, 6) suggests that the result is, in a sense, the opposite of the one we have seen above.
2 Theorem Rearrangement inequalities for products of sums. Let x1, x2, . . . , xn be pos-itive real numbers in increasing order, that is, 0 < x1 ≤x2 ≤· · · ≤xn, and let y1, y2, . . . , yn be positive real numbers. Let z1, z2, . . . , zn be a rearrangement of the yj, and consider the product (x1 + z1)(x2 + z2) · · · (xn + zn) .
This product has its maximum value when the zj are arranged in decreasing order, and its minimum value when they are arranged in increasing order.
Examples 1. Let x1 = 1, x2 = 2 and x3 = 3. If z1, z2 and z3 are 4, 5 and 6 (not necessarily in that order), then (1 + 4)(2 + 5)(3 + 6) ≤(x1 + z1)(x2 + z2)(x3 + z3) ≤(1 + 6)(2 + 5)(3 + 4) , that is, 315 ≤(1 + z1)(2 + z2)(3 + z3) ≤343 .
2. Let x1 = 12, x2 = 22, ..., xn = n2. If z1, z2, . . . , zn is any arrangement of 1, 2, . . . , n then (x1 + z1)(x2 + z2) · · · (xn + zn) ≥(x1 + 1)(x2 + 2) · · · (xn + n) .
That is, (12 + z1)(22 + z2) · · · (n2 + zn) ≥(12 + 1)(22 + 2) · · · (n2 + n) = (1 × 2) × (2 × 3) · · · n(n + 1) = (n + 1)(n!)2 .
Another way in which the above ideas can be varied is to consider sums of products of three or more numbers. The maximum result generalises in a fairly obvious way, but, curiously, the minimum result does not. For products of sums it is the minimum which generalises and the maximum which does not.
Theorem Extended rearrangement inequalities. Let x1, x2, . . . , xn be positive numbers in increasing order; let y11, y12, . . . , y1n and y21, y22, . . . , y2n and so on be m collections of n positive real numbers. Let z11, z12, . . . , z1n be an arrangement of y11, y12, . . . , y1n, let z21, z22, . . . , z2n be an arrangement of y21, y22, . . . , y2n and so forth. Then the sum of products (x1z11z21 · · · zm1) + (x2z12z22 · · · zm2) + · · · + (xnz1nz2n · · · zmn) takes its maximum value when zj1, zj2, . . . , zjn are in increasing order for each j. The product of sums (x1 + z11 + z21 · · · + zm1)(x2 + z12 + z22 · · · + zm2) · · · (xn + z1n + z2n · · · + zmn) has its minimum value when zj1, zj2, . . . , zjn are in increasing order for each j.
Examples 3 1. Suppose that x1 = 1, x2 = 2, x3 = 3 and { y11, y12, y13 } = { 4, 5, 6 } and { y21, y22, y23 } = { 1, 3, 5 }. The maximum value of (x1z11z21) + (x2z12z22) + (x3z13z23) is (1 × 4 × 1) + (2 × 5 × 3) + (3 × 6 × 5) = 124 , where the 4, 5, 6 and the 1, 3, 5 have been listed in increasing order. However if we list them both in decreasing order, 6, 5, 4 and 5, 3, 1, we obtain (1 × 6 × 5) + (2 × 5 × 3) + (3 × 4 × 1) = 72 , which is not the minimum value of the sum! In similar fashion, the minimum value of (x1 + z11 + z21)(x2 + z12 + z22)(x3 + z13 + z23) is (1 + 4 + 1)(2 + 5 + 3)(3 + 6 + 5) = 840 , but the maximum value is not (1 + 6 + 5)(2 + 5 + 3)(3 + 4 + 1) = 960 .
Exercise What is the minimum value of the sum of products? What is the maximum value of the product of sums?
2. Let θ = π/2n and consider the product of sums P = sin θ + cos θ + cot θ sin 2θ + cos 2θ + cot 2θ · · · · · · sin(n −1)θ + cos(n −1)θ + cot(n −1)θ .
Note first of all that 0 < θ < 2θ < · · · < (n −1)θ < 1 2π , and that on the interval 0 < x < 1 2π the sine function is increasing, while the cosine and cotangent functions are decreasing.
If we take xj = sin jθ then the xs are in increasing order, so to obtain the minimum product of sums we must arrange y1j = cos jθ and y2j = cot jθ in increasing order too. Therefore P ≥ sin θ + cos(n −1)θ + cot(n −1)θ sin 2θ + cos(n −2)θ + cot(n −2)θ · · · · · · sin(n −1)θ + cos θ + cot θ .
4 Since (n −k)θ = 1 2π −kθ we have cos(n −j)θ = sin jθ and cot(n −j)θ = tan jθ , so this inequality can be rewritten as P ≥ 2 sin θ + tan θ 2 sin 2θ + tan 2θ · · · · · · 2 sin(n −1)θ + tan(n −1)θ .
Proof of the rearrangement inequalities for sums of products We shall show that if the zj are not in increasing order, then there is an arrangement with fewer terms out of order, and for which the sum (1) is the same or larger. If the new arrangement still has terms which are not in increasing order we can repeat the process, again increasing the sum; eventually we shall reach a sequence which has no terms out of order. This will show that the sequence with all terms in increasing order gives the largest possible sum.
Suppose, then, that there is some k for which zk > zk+1. Exchange zk and zk+1 to get a new arrangement z′ 1, z′ 2, . . . , z′ n. That is, z′ k = zk+1 , z′ k+1 = zk and z′ j = zj if j ̸= k, k + 1.
For convenience we write S′ = x1z′ 1 + x2z′ 2 + · · · + xnz′ n and S = x1z1 + x2z2 + · · · + xnzn ; most of the products in these two sums are the same, and if we subtract them we get S′ −S = (xkz′ k + xk+1z′ k+1) −(xkzk + xk+1zk+1) = xkzk+1 + xk+1zk −xkzk −xk+1zk+1 = (xk+1 −xk)(zk −zk+1) ≥0 , because each of the bracketed terms is non–negative. Thus S′ ≥S, and z′ 1, z′ 2, . . . , z′ n has fewer terms out of order than z1, z2, . . . , zn since zk and zk+1 have been “corrected”.
As explained above, this proves the maximum inequality.
The result that the sum is minimal when the zj are in decreasing order may be proved by a very similar argument (exercise!), or may be deduced from what we have proved in the following way. Choose a real number M larger than all the yj, and apply the maximum result to the numbers M −y1, M −y2, . . . , M −yn. Each zj must be replaced by M −zj, and so the sum (1) is x1(M −z1) + x2(M −z2) + · · · + xn(M −zn) = (x1 + x2 + · · · + xn)M −(x1z1 + x2z2 + · · · + xnzn) .
Now the first sum on the right hand side is fixed, and so the minimum value of x1z1 + x2z2 + · · · + xnzn corresponds to the maximum value of the left hand side. From what we have just proved this occurs when M −z1, M −z2, . . . , M −zn are in increasing order, that is, when z1, z2, . . . , zn are in decreasing order.
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https://www.nice.org.uk/guidance/ng192/resources/2021-exceptional-surveillance-of-caesarean-birth-morbidly-adherent-placenta-nice-guideline-ng192-pdf-12003150630085
|
2021 exceptional
surveillance of caesarean
birth – morbidly adherent
placenta (NICE guideline
NG192)
Surveillance report
Published: 19 October 2021
www.nice.org.uk
© NICE 2023. All rights reserved. Subject to Notice of rights ( Contents
Surveillance decision .................................................................................................................. 3
Exceptional surveillance review summary ............................................................................................ 3
Information considered in this exceptional surveillance review ......................................................... 4
How we will handle this .......................................................................................................................... 6
Equalities .................................................................................................................................................. 7
Overall decision ....................................................................................................................................... 7
2021 exceptional surveillance of caesarean birth – morbidly adherent placenta (NICE guideline NG192)
© NICE 2023. All rights reserved. Subject to Notice of rights (
Page 2 of 7 Surveillance decision
We propose to update the section on morbidly adherent placenta in the NICE guideline on
caesarean birth . The term morbidly adherent placenta will be editorially amended to
placenta accreta spectrum (PAS) with committee input.
Exceptional surveillance review summary
Reason for considering this area
NICE received feedback that the section of the NICE guideline on morbidly adherent
placenta may be out of date with regards to terminology and that a number of the
recommendations within it need reviewing and may need updating in light of more recent
guidance from the Royal College of Obstetrics and Gynaecology (RCOG) on Placenta
Praevia and Placenta Accreta: Diagnosis and Management .
Methods
To review the impact of this enquiry on NICE guidance we took the following approach:
• Considered the information submitted by the stakeholder.
• Considered the evidence used to develop the NICE guideline on caesarian birth
related to the section on morbidly adherent placenta.
• Discussed the issues raised with the NICE clinical adviser.
• Considered feedback from topic experts.
It was concluded that full updated literature searches were not needed because the
information we obtained was enough to establish whether an amendment to the guideline
was needed.
For further information see ensuring that published guidelines are current and accurate in
developing NICE guidelines: the manual .
2021 exceptional surveillance of caesarean birth – morbidly adherent placenta (NICE guideline NG192)
© NICE 2023. All rights reserved. Subject to Notice of rights (
Page 3 of 7
Information considered in this exceptional
surveillance review
How the guideline was developed
The section on morbidly adherent placenta dates back to 2011 and includes 6
recommendations:
Recommendation 1.2.7: For women who have had a previous caesarean birth, offer colour-
flow Doppler ultrasound at 32 to 34 weeks as the first diagnostic test for morbidly
adherent placenta if low-lying placenta is confirmed.
Recommendation 1.2.8: If a colour-flow Doppler ultrasound scan result suggests morbidly
adherent placenta:
• discuss with the woman how MRI in addition to ultrasound can help diagnose morbidly
adherent placenta and clarify the degree of invasion, particularly with a posterior
placenta
• explain what to expect during an MRI procedure
• inform the woman that current experience suggests that MRI is safe, but that there is a
lack of evidence about any long-term risks to the baby
• Offer MRI if this is acceptable to the woman.
Recommendation 1.2.9: Discuss birth options (for example, timing of birth, operative
interventions including possibility of hysterectomy, need for blood transfusion) with a
woman suspected to have morbidly adherent placenta. This discussion should be carried
out by a consultant obstetrician, or with a consultant obstetrician present.
Recommendation 1.2.10: When performing a caesarean birth for a woman suspected to
have a morbidly adherent placenta, ensure that:
• a consultant obstetrician and a consultant anaesthetist are present in the operating
theatre
• a paediatric registrar, consultant, or equivalent, is present
2021 exceptional surveillance of caesarean birth – morbidly adherent placenta (NICE guideline NG192)
© NICE 2023. All rights reserved. Subject to Notice of rights (
Page 4 of 7
• a haematology registrar, consultant, or equivalent, is available for advice
• a critical care bed is available
• sufficient cross-matched blood and blood products are readily available.
Recommendation 1.2.11: Before performing a caesarean birth for women suspected to have
morbidly adherent placenta, the multidisciplinary team should agree which other
healthcare professionals need to be consulted or present, and the responsibilities of each
team member.
Recommendation 1.2.12: All hospitals should have a locally agreed protocol for managing
morbidly adherent placenta that sets out how these elements of care should be provided.
In the 2011 guideline the section on morbidly adherent placenta was fully updated from the
original 2004 version. The update focused on the diagnosis and management of morbidly
adherent placenta and included an economic analysis to compare different diagnostic
strategies for morbidly adherent placenta in praevia.
Previous surveillance
The 2017 surveillance review prompted an update of the guideline in 2020/21 . However,
the section on morbidly adherent placenta was not updated, as although during the
surveillance review experts highlighted this section as an important one, no new evidence
was identified that would impact on recommendations.
Five of the recommendations within the section on morbidly adherent placenta were
editorially refreshed for sense and style during the 2021 update of the guideline. There
was no update to the content of the recommendations or evidence base underpinning
them.
New intelligence
NICE received feedback that the section of the guideline on morbidly adherent placenta
may be out of date and explained that the terminology of morbidly adherent placenta is no
longer used and instead the term PAS is preferred. The feedback also highlighted that
recommendations 1.2.7 to 1.2.9 and 1.2.12 within this section may need reconsidering as
they differ from more recent guidance from RCOG on Placenta Praevia and Placenta
Accreta .
2021 exceptional surveillance of caesarean birth – morbidly adherent placenta (NICE guideline NG192)
© NICE 2023. All rights reserved. Subject to Notice of rights (
Page 5 of 7
In particular, the following suggestions were made:
• Recommendation 1.2.7 needs to consider the role of screening for PAS at the mid-
pregnancy routine scan.
• Recommendation 1.2.8 should reconsider the role of MRI in diagnosis of PAS.
• Recommendation 1.2.9 needs to consider the need for delivery in a specialist centre
with access to blood products and intensive care unit.
• Recommendation 1.2.12 should consider the role of nationally agreed protocols.
Expert feedback
We contacted 5 experts with backgrounds in obstetrics, operational department practice
and midwifery to determine if the term PAS had replaced morbidly adherent placenta. Two
experts replied and agreed that PAS is the preferred term, however 1 expert explained that
morbidly adherent placenta is still in use and that students may need to be aware of both
terms. As such, it was deemed that morbidly adherent placenta should be editorially
amended to PAS with committee input.
How we will handle this
We propose to update the section on morbidly adherent placenta in the NICE guideline on
caesarean birth taking into consideration the guidance from RCOG. There are several
aspects to this update. Firstly, the term morbidly adherent placenta needs to be integrated
or replaced with the newer term PAS but with committee consideration to ensure any
replacement is helpful to practice.
The evidence base identified in the RCOG guidance on MRI compared with colour-flow
Doppler ultrasound for diagnosing PAS needs reviewing as the RCOG guidance includes
new evidence that was not considered during NICE guideline development.
Issues around when to screen for PAS, the role of specialist centres for birth, and
nationally agreed protocols also need committee consideration. Furthermore, NICE's
guideline on antenatal care may need to be cross referred to, particularly as it provides
useful advice on screening during routine antenatal scans.
2021 exceptional surveillance of caesarean birth – morbidly adherent placenta (NICE guideline NG192)
© NICE 2023. All rights reserved. Subject to Notice of rights (
Page 6 of 7
Equalities
No equalities issues were identified.
Overall decision
We propose to update the section on morbidly adherent placenta in the NICE guideline on
caesarean birth . The term morbidly adherent placenta will be editorially amended to PAS
with committee input.
ISBN: 978-1-4731-4318-0
2021 exceptional surveillance of caesarean birth – morbidly adherent placenta (NICE guideline NG192)
© NICE 2023. All rights reserved. Subject to Notice of rights (
Page 7 of 7
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https://arxiv.org/pdf/1610.06094
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arXiv:1610.06094v3 [quant-ph] 8 Jun 2017
Perfect quantum state transfer using Hadamard diagonalizable graphs
Nathaniel Johnston a,b , Steve Kirkland c, Sarah Plosker b,c,d , Rebecca Storey d,Xiaohong Zhang c
aDepartment of Mathematics & Computer Science, Mount Allison University, Sackville, NB, Canada E4L 1E4
bDepartment of Mathematics & Statistics, University of Guelph, Guelph, ON, Canada N1G 2W1
cDepartment of Mathematics, University of Manitoba, Winnipeg, MB, Canada R3T 2N2
dDepartment of Mathematics & Computer Science, Brandon University, Brandon, MB, Canada R7A 6A9
Abstract
Quantum state transfer within a quantum computer can be achieved by using a net-work of qubits, and such a network can be modelled mathematically by a graph. Here, we focus on the corresponding Laplacian matrix, and those graphs for which the Laplacian can be diagonalized by a Hadamard matrix. We give a simple eigen-value characterization for when such a graph has perfect state transfer at time π/ 2;this characterization allows one to choose the correct eigenvalues to build graphs having perfect state transfer. We characterize the graphs that are diagonalizable by the standard Hadamard matrix, showing a direct relationship to cubelike graphs. We then give a number of constructions producing a wide variety of new graphs that exhibit perfect state transfer, and we consider several corollaries in the set-tings of both weighted and unweighted graphs, as well as how our results relate to the notion of pretty good state transfer. Finally, we give an optimality result, show-ing that among regular graphs of degree at most 4, the hypercube is the sparsest Hadamard diagonalizable connected unweighted graph with perfect state transfer.
Keywords: Laplacian matrix, Hadamard diagonalizable graph, quantum state transfer, cubelike graphs, double cover, perfect state transfer
2010 MSC: 05C50, 05C76, 15A18, 81P45
Email address: ploskers@brandonu.ca (Sarah Plosker)
Introduction
Accurate transmission of quantum states between processors and/or registers of a quantum computer is critical for short distance communication in a physical quantum computing scheme. Bose first proposed the use of spin chains to accomplish this task over a decade ago. Since then, much work has been done on
perfect state transfer (PST) , which accomplishes this task perfectly in the sense that the state read out by the receiver at some time t0 is, with probability equal to one, identical up to complex modulus to the input state of the sender at time t = 0 .Many families of graphs have been found to exhibit PST, including the join of a weighted two-vertex graph with any regular graph , Hamming graphs (see also [7, 12, 13]), a family of double-cone non-periodic graphs , and a family of integral circulant graphs (see also ). It is easy to see that the Cartesian product of two graphs having PST at the same time also has PST [1, Sec. 3.3]. Much work has also been done with respect to analyzing the sensitivity [10, 15, 16, 17, 20, 21, 23], or even correcting errors , of quantum spin systems. Signed graphs and graphs with arbitrary edge weights have also been considered (see and the references therein), due to the intriguing fact that certain graphs that do not exhibit PST when unsigned/unweighted can exhibit PST when signed or weighted properly. Three articles particularly relevant to the work herein are [7, 11], which characterized perfect state transfer in cubelike graphs, a family of graphs that are Hadamard diagonalizable, and , which shows that perfect state transfer occurs in graphs constructed in a manner similar to our merge operation, called the “ ⋉”operation. The general approach taken in the literature is to model a quantum spin system with an undirected connected graph, where the dynamics of the system are gov-erned by the Hamiltonian of the system: for XX dynamics the Hamiltonian is the adjacency matrix corresponding to the graph, and for Heisenberg ( XXX ) dynam-ics the Hamiltonian is the Laplacian matrix corresponding to the graph. In the case of XXX dynamics (on which we focus exclusively in this paper), there is more structure to work with since we know that the smallest eigenvalue of a Laplacian matrix is zero, with corresponding eigenvector 1 (the all-ones vector). We note in passing that in the case of regular graphs (which we deal with frequently in this paper), presence or absence of PST is identical under XX dynamics and XXX
dynamics. Our contribution to the theory of perfect state transfer is to characterize graphs that are diagonalizable by the standard Hadamard matrix, connecting this prop-erty with the notion of cubelike graphs, and to detail procedures for creating new 2graphs with PST. Our focus on graphs having a Hadamard diagonalizable Laplacian is not as restrictive as it might seem at first glance; Hadamard matrices are ubiquitous in quantum information theory, and because of the special structure of Hadamard matrices the corresponding graphs tend to exhibit a good deal of symmetry. As a result, many of the known graphs with PST are actually Hadamard diagonalizable, such as the hypercube. Furthermore, integer-weighted graphs with Hadamard di-agonalizable Laplacian are convenient to work with in our setting because they are known to be regular, with spectra consisting of even integers (see and The-orem 2 below); consequently the corresponding graph often exhibits PST between two of its vertices at time t0 = π/ 2 (see Theorem 3 for a more specific statement). In Section 2, we give a quick review of the graph theory and quantum state transfer definitions and tools that we will use. In Section 3, we give an eigen-value characterization connecting a graph being Hadamard diagonalizable and it having PST at time π/ 2 between two of its vertices. We further give a con-nection between diagonalizability by the standard Hadamard matrix and cubelike graphs, completely characterizing such graphs. In Section 4, we describe several ways to construct new Hadamard diagonalizable graphs from old ones, including our “merge” operation, a weighted variant of the “ ⋉” operation, which takes two Hadamard diagonalizable graphs as input, and produces a new (larger) Hadamard diagonalizable graph with PST as output under a wide variety of conditions. We also present several results demonstrating the usefulness of this operation and the types of graphs with PST that it can produce. In Section 5, we discuss how our results generalize to graphs with non-integer edge weights, which involves the notion of pretty good state transfer (PGST), and we close in Section 6 with some results concerning the optimality in terms of timing errors and manufacturing er-rors of Hadamard diagonalizable graphs.
Preliminaries
2.1. Graph Theory Basics
For a weighted undirected graph G on n vertices, its corresponding n × n
adjacency matrix A = ( ajk ) is defined by
ajk =
{
wj,k if j and k are adjacent
0 otherwise, where wj,k is the weight of the edge between vertices j and k. Its corresponding
n × n Laplacian matrix is defined by L = D − A, where D is the diagonal matrix 3of row sums of A, known as the degree matrix associated to G. Often wj,k in the above is taken to be 1 for all adjacent j, k , in which case the graph is said to be
unweighted . A signed graph is a graph for which the non-zero weights can be either ±1. A weighted graph is a graph for which there is no restriction on wj,k
(although the weights are typically taken to be in R, as they are in this paper). An unweighted graph G is regular if each of its vertices has the same number of neighbours, or, more specifically, k-regular if each of its vertices has exactly k
adjacent neighbours. The weighted analogue of a regular graph is a graph where the sum of all the weights of edges incident with a particular vertex is the same for all vertices. We will be interested in weighted graphs with this equal “weighted degree” property; for simplicity, we simply call this the degree of the graph. A graph is connected if there is a path (a sequence of edges connecting a sequence of vertices) between every pair of distinct vertices and complete if there is an edge between every pair of distinct vertices (the complete graph on n vertices is denoted
Kn). There are several different operations that can be performed to turn two graphs into a new (typically larger) graph. Specifically, given graphs G1 = ( V1, E 1) and
G2 = ( V2, E 2), where V1 and E1 are the set of all vertices and the set of all edges, respectively, in the graph G1 (and similarly for V2 and E2), then 1. The union of G1 and G2 is the graph G1 + G2 = ( V1 ∪ V2, E 1 ∪ E2);2. The join of G1 and G2 is the graph G1 ∨G2 = ( Gc
1
+Gc
2
)c where every vertex of G1 is connected to every vertex of G2, and all of the original edges of G1
and G2 are retained as well; 3. The Cartesian product of G1 and G2 is the graph G1G2 = ( V1 × V2, E 3)
where V1 × V2 is the cartesian product of the two original sets of vertices, and there is an edge in G1G2 between vertices (g1, g 2) and (h1, h 2) if and only if either (i) g1 = h1 and there is an edge between g2 and h2 in G2, or (ii) g2 = h2 and there is an edge between g1 and h1 in G1.One can also define the Cartesian product of weighted graphs G1 and G2 by defining (i) the weight of the edges between (g1, g 2) and (g1, h 2) in G1G2
to be the same as the weight between g2 and h2 in G2, and (ii) the weight of the edges between (g1, g 2) and (h1, g 2) in G1G2 to be the same as the weight between g1 and h1 in G1; and 4. If V1 = V2, let G1 ⋉ G2 be the graph defined by the adjacency matrix
A(G1 ⋉ G2) =
[A(G1) A(G2)
A(G2) A(G1)
]
, where A(·) is the adjacency matrix of the given graph. If the edge sets of G1 and G2 are disjoint, then G1 ⋉ G2 is a double cover of the graph with adjacency matrix A(G1) + A(G2).4We recall that a Hadamard matrix (or simply, a Hadamard) of order n is an
n × n matrix H with entries +1 and −1, such that HH T = nI . Let H1 =[1 11 −1
]
, H2 =
[H1 H1
H1 −H1
]
, . . . , H n =
[Hn−1 Hn−1
Hn−1 −Hn−1
]
. This construction gives the standard Hadamards of order 2n. The results herein may be of use in the physical setting because Hadamards are among the simplest non-trivial gates to implement in the lab (the standard n-qubit Hadamard with a scaling factor of
1/2n/ 2 is frequently used in quantum information theory). From the definition of a Hadamard matrix, it is clear that any two rows of H are orthogonal, and any two columns of H are also orthogonal. This property does not change if we permute rows or columns or if we multiply some rows or columns by −1. This leads to the simple but important observation that, given a Hadamard matrix, it is always possible to permute and sign its rows and columns so that all entries of the first row and all entries of the first column are all 1’s. A Hadamard matrix in this form is said to be normalized . Given a graph G on n vertices with corresponding Laplacian matrix L, if we can write L = 1
n
HΛHT for some Hadamard H and diagonal matrix Λ, then we say that G (or, that L) is Hadamard diagonalizable .If G is Hadamard diagonalizable by some Hadamard H, then G is also Hadamard diagonalizable by a corresponding normalized Hadamard [5, Lemma 4]. Thus, there is no loss of generality in assuming that a Hadamard diagonalizable graph is in fact diagonalized by a normalized Hadamard matrix. Note that “normalized” in this setting does not imply scaling H to satisfy ‖H‖ = 1 .
2.2. Perfect state transfer basics
A graph exhibits perfect state transfer (PST) at time t0 if p(t0) := |eTj eit 0Hek|2 =1 for some vertices j 6 = k and some time t0 > 0, where H is the Hamiltonian of the system (either the adjacency matrix A or the Laplacian matrix L, depending on the system’s dynamics). In other words, the graph has perfect state transfer if and only if eit 0Hek is a scalar multiple of ej (or, equivalently, if eit 0 Hej is a scalar multiple of ek). Typically we say that a graph has PST from vertex j to vertex k
if it exhibits PST for some vertices j and k and j < k .A slightly weaker property is that of pretty good state transfer (PGST) : a graph exhibits PGST (for some vertices j 6 = k) if for every ε > 0, there exists a time tε
such that p(tε) := |eTj eit εHek|2 ≥ 1 − ε.The following observation is well-known.
Remark 1. For a general integer-weighted graph G, assume that a is the greatest common divisor of all the edge weights of G and that L is the Laplacian matrix
5of G. Let G′ denote the integer-weighted graph with Laplacian 1/a L . Since
eitL = eita ( 1
aL)
for all t, we find that G has PST at π/ (2 a) if and only if G′ has PST at π/ 2. This allows us to identify more graphs having PST: for example, if G
has PST at π/ 2, and we are given the graph with Laplacian matrix 2L, we know that it has PST at π/ 4.2.3. Cubelike graphs
A large family of graphs, of which the hypercube is a member, is the family of cubelike graphs [7, 11]: Take a set C ⊂ Zd
2
= Z2 × · · · × Z2 (d times), where C does not contain the all-zeros vector. Construct the cubelike graph G(C)
with vertex set V = Zd
2
and two elements of V are adjacent if and only if their difference is in C. The set C is called the connection set of the graph G(C). The following result characterizes PST at π/ 2 for cubelike graphs.
Theorem 1. [7, Theorem 1], [11, Theorem 2.3] Let C be a subset of Zd
2
and let
σ be the sum of the elements of C. If σ 6 = 0 , then PST occurs in G(C) from j to
j + σ at time π/ 2. If σ = 0 , then G(C) is periodic with period π/ 2 (every vertex has perfect state transfer with itself at time t0 = π/ 2).
The code of G(C) is the row space of the d × | C| matrix M constructed by taking the elements of C as its columns. When the sum of the elements of C
is zero, it has been shown that if perfect state transfer occurs on a cubelike graph, then it must take place at time π/ 2D, where D is the greatest common divisor of the (Hamming) weights of the binary strings in the code.
Hadamard diagonalizable graphs with PST
The following theorem originally appeared in , restricted to the case of unweighted graphs. The version below allows for arbitrary integer edge weights. Although its proof is almost identical to its unweighted version, we include it here for completeness.
Theorem 2. [5, Theorem 5] If G is an integer-weighted graph that is Hadamard diagonalizable, then G is regular and all the eigenvalues of its Laplacian are even integers. Proof. Without loss of generality we assume that the Laplacian matrix for G is di-agonalized by a normalized Hadamard matrix; observe then that the first column of that Hadamard is the all-ones vector, and that it corresponds to the eigenvalue 60. Choose a non-zero eigenvalue λ of the Laplacian matrix L associated to G; the corresponding column of the Hadamard matrix that diagonalizes L is an eigenvec-tor corresponding to λ. One can split the graph G into two subgraphs, G1 and G2
(with Laplacians L1 and L2), corresponding to the n/ 2 entries of 1 and the n/ 2
entries of −1 of the eigenvector corresponding to λ. By applying a permutation similarity if necessary, we find that
[L1 + X1 −R
−RT L2 + X2
] [ 1
−1
]
= λ
[ 1
−1
]
,
for some matrices X1, X 2, and R. Necessarily X1, X 2 are diagonal, and note that we have X11 = R1 and X21 = RT 1.Since λ1 = L11 + X11 + R1 = 2 X11, and since G is integer-weighted, we deduce that λ is an even integer. Hence each eigenvalue of the Laplacian is an even integer. Next we show that G is regular. For concreteness, suppose that G has n ver-tices and that H is a normalized Hadamard matrix so that LH = HD for some di-agonal matrix D. Fix an index j between 1 and n, and let Sj be the diagonal matrix with diagonal entries ±1 such that eTj HS j = 1T . Observe that LHS j = HS j D,and that HS j is also a Hadamard matrix. Since the j-th row of HS j is the all-ones vector and the remaining rows are orthogonal to it, we deduce that HS j 1 = ne j .Consequently, eTj LHS j 1 = ne Tj Le j . On the one hand, we have eTj LHS j 1 =
eTj HS j D1 = 1T D1. Thus, for each j = 1 , . . . , n, e Tj Le j = 1
n
1T D1, so G is regular, as desired.
For an integer-weighted graph that is diagonalizable by some Hadamard ma-trix, we now give a precise characterization of its eigenvalues when it exhibits PST at time t0 = π/ 2. The proof applies a standard characterization of PST; see , for example.
Theorem 3. Let G be an integer-weighted graph that is Hadamard diagonaliz-able by a Hadamard of order n. Let H = ( huv ) be a corresponding normalized Hadamard. Denote the eigenvalues of the Laplacian matrix L corresponding to G
by λ1, · · · , λ n, so that LH ej = λj Hej , j = 1 , . . . , n . Then G has PST from vertex
j to vertex k at time t0 = π/ 2 if and only if for each ℓ = 1 , · · · , n , λℓ ≡ 1 − hjℓ hkℓ
mod 4 .Proof. Let Λ be the diagonal matrix of eigenvalues such that L = 1
n
HΛHT , and hence ei(π/ 2) L = 1
n
He i(π/ 2)Λ HT . By the definition of PST, it follows that G has 7PST from vertex j to vertex k at t0 = π/ 2 if and only if ei(π/ 2)Λ HT ej is a scalar multiple of HT ek. Since the first column of H is the all ones vector 1, i.e. an eigenvector of L corresponding to the eigenvalue 0, we know that the first entry of
ei(π/ 2)Λ HT ej is hj1 = 1 , and the first entry of HT ek = hk1 = 1 . Thus we deduce that not only is ei(π/ 2)Λ HT ej a scalar multiple of HT ek, but that the multiple must be 1, i.e., we have PST from vertex j to k at π/ 2 if and only if
ei(π/ 2)Λ HT ej = HT ek. (1) Note that
ei(π/ 2) λℓ =
{
1 if λℓ ≡ 0 mod 4
−1 if λℓ ≡ 2 mod 4 .
Consequently, (1) holds if and only if, for each ℓ = 1 , · · · , n , if hjℓ hkℓ = 1 then
λℓ ≡ 0 mod 4 , and if hjℓ hkℓ = −1 then λℓ ≡ 2 mod 4 . The conclusion follows.
It is worth noting that Theorem 3 already gives an extremely easy method for creating weighted Hadamard diagonalizable graphs exhibiting PST, since for any normalized Hadamard matrix H we can choose the eigenvalues in Λ to satisfy the required mod 4 equation, and then L = 1
n
HΛHT will necessarily be the Laplacian of some rational-weighted graph with PST at time t0 = π/ 2 (the graph will be integer-weighted provided n divides each edge weight in this construction). It is known that the adjacency matrix of any cubelike graph is diagonalized by the standard Hadamard matrix (see ). The following result provides the con-verse; in the proof, it will be convenient to denote the graph (possibly containing loops) with adjacency matrix A by Γ( A).
Lemma 1. Suppose that k ∈ N and that A is a symmetric (0 , 1) matrix that is diagonalizable by the standard Hadamard matrix of order 2k. Then 1. A has constant diagonal; 2. if A has zero diagonal then it is the adjacency matrix of a cubelike graph; 3. if A has all ones on the diagonal, then A − I is the adjacency matrix of a cubelike graph. Proof. We proceed by induction on k. For k = 1 , it is straightforward to see that the (0 , 1) symmetric matrices that are diagonalized by H1 =
[1 11 −1
]
are: 8[0 00 0
]
,
[1 00 1
]
,
[0 11 0
]
,
[1 11 1
]
. For these matrices, conclusions (1)–(3) follow readily. Suppose that the result holds for some k ∈ N and that A is of order 2k+1 . Write the standard Hadamard matrix of order 2k+1 as Hk+1 =
[Hk Hk
Hk −Hk
]
, where Hk is the standard Hadamard matrix of order 2k. Partition A accordingly as
[ A1 XXT A2
]
.Then there are diagonal matrices D1, D 2 such that
[Hk Hk
Hk −Hk
] [ A1 XXT A2
] [ Hk Hk
Hk −Hk
]
=
[D1 OO D2
]
.
Hence
[Hk(A1 + A2 + X + XT )Hk Hk(A1 − A2 − X + XT )Hk
Hk(A1 − A2 + X − XT )Hk Hk(A1 + A2 − X − XT )Hk
]
=
[D1 OO D2
]
.We deduce that A1 − A2 = X − XT ; since A1 − A2 is symmetric and X − XT is skew-symmetric, it must be the case that A1 = A2 and X = XT . Then Hk diag-onalizes both 2( A1 + X) and 2( A1 − X), and we conclude that Hk diagonalizes
A1 and diagonalizes X. In particular the induction hypothesis applies to A1 and
X. Thus A1 has constant diagonal, and hence so does A.Suppose that A has zero diagonal. Applying the induction hypothesis to A1,we find that Γ( A1) is cubelike. Let C1 denote its connection set. Applying the induction hypothesis to X, then either X has zero diagonal and so that Γ( X) is a cubelike graph with connection set C2, say, or Γ( X − I) is a cubelike graph with connection set ˜C2. Set C2 = ˜C2 ∪ { 0}.We label the vertices of the graph Γ( A) with vectors in Zk+1 2 in increasing order if considered as binary numbers. So the first 2k rows/columns of A are labelled as
[0
z
]
, where z ∈ Zk
2
, and the last 2k rows/columns of A are of la-belled as
[1
z
]
, where z ∈ Zk
2
. Now construct the following connection set:
C =
{[ 0
x
]
, x ∈ C1
}
∪
{[ 1
y
]
, y ∈ C2
}
. It follows that A is the adjacency matrix of the cubelike graph with connection set C.9If A has all ones on the diagonal we proceed as above with A − I.This establishes the induction steps for (1)–(3).
Corollary 1. Let G be an unweighted graph with Laplacian matrix L. Then L
is diagonalized by the standard Hadamard matrix if and only if G is a cubelike graph. Proof. If L is diagonalized by the standard Hadamard matrix, then in particular
G is regular by Theorem 2. Hence the adjacency matrix of G is diagonalized by the standard Hadamard matrix, so by Lemma 1, G is cubelike. Conversely, if
G is cubelike, it is regular and its adjacency matrix is diagonalized by the stan-dard Hadamard matrix. We now deduce that L is diagonalized by the standard Hadamard matrix.
Creation of new Hadamard diagonalizable graphs with PST
It is known that the union of a PST graph with itself still exhibits PST. Here, we show that for a graph G on n ≥ 4 vertices that is diagonalizable by some Hadamard matrix and that has PST at time π/ 2, both its complement and the join of G with itself are Hadamard diagonalizable and have PST at time t0 = π/ 2.
Proposition 1. Let G be an integer-weighted graph on n ≥ 4 vertices that is di-agonalizable by a Hadamard matrix H, and that has perfect state transfer from vertex j to vertex k at time t0 = π/ 2. Then its complement Gc is also diagonal-izable by H, and has the same PST pairs and PST time as G. Furthermore, the join G ∨ G of G with itself is diagonalizable by the Hadamard
[H HH −H
]
, and has PST from vertex j to vertex k at time t0 = π/ 2.Proof. Without loss of generality we can assume that H is a normalized Hadamard matrix. The result that Gc and G ∨ G are diagonalizable follows from Lemma 7 in . If we denote the eigenvalues of the Laplacian of G by λ1 = 0 , λ 2, · · · , λ n,then from Theorem 3 we know that for ℓ = 1 , · · · , n , λℓ ≡ 1 − hjℓ hkℓ mod 4 .Therefore the eigenvalues 0, n − λ2, . . . , n − λn of Gc satisfy (n − λℓ) ≡ − (1 −
hjℓ hkℓ ) ≡ 1 − hjℓ hkℓ mod 4 , since 1 − hjℓ hkℓ is either 0 or 2 mod 4 and n must be a multiple of 4 in order for a Hadamard of order n to exist. Again from Theorem 3, we then know that Gc has PST from vertex j to k at time π/ 2. Thus G ∨ G = ( Gc + Gc)c also has PST from vertex j to k at time π/ 2.
10 Note that we can also prove that Gc exhibits PST at t0 = π/ 2 by noticing that if λ is a nonzero Laplacian eigenvalue for G, then n − λ is a Laplacian eigenvalue for Gc with the same eigenvector. As n − λ ≡ λ mod 4 , the conclusion now follows from Theorem 3. We now introduce a modification of G1 ⋉ G2 that, much like G1 ⋉ G2, can be used to construct new graphs with PST from old ones. Suppose that G1 and G2 are two weighted graphs of order n, with Laplacians L1 = D1−A1 and L2 = D2 −A2,respectively. Then we define the merge of G1 and G2 with respect to the weights
w1 and w2 to be the graph G1 ⊙w1 w2 G2 with Laplacian
[w1L1 + w2D2 −w2A2
−w2A2 w1L1 + w2D2
]
.
In the case that w1 = w2 = 1 , we denote the merge simply by G1 ⊙ G2, and it recovers G1 ⋉ G2.
1
2 3
4
1
2 3
4
1
2 3
4
5
6 7
8
Figure 1: A depiction of two Hadamard diagonalizable graphs (left) and their merge (right). The new graph has two copies of the original vertex set, and there is now an edge (j, k )and
(n+j, n +k)if and only if G1(top left) had edge (j, k ), and there is an edge (j, n +k)if and only if G2(bottom left) had edge (j, k ).
Observe that if G1 and G2 are both diagonalizable by the same Hadamard matrix H, then G1 ⊙w1 w2 G2 is also Hadamard diagonalizable, by the matrix 11 [H HH −H
]
; this observation is what motivates our definition of the merge. While this operation is a bit less intuitive than the other ones we saw, it does have an in-terpretation in terms of the vertices and edges of the original graphs. Specifically, if G1 and G2 each have vertices labelled {1, . . . , n }, then G1 ⊙w1 w2 G2 has twice as many vertices, which we label {1, . . . , 2n}. Furthermore, if G1 has edge (j, k )
with weight wjk then G1 ⊙w1 w2 G2 has edges (j, k ) and (n + j, n + k), each with weight w1wjk . Similarly, if G2 has edge (j, k ) with weight wjk then G1 ⊙w1 w2 G2
has edge (j, n + k) and (k, n + j) with weight w2wjk . See Fig. 1 for an exam-ple in the unweighted case—the Laplacian matrices corresponding to G1, G2, and
G1 ⊙ G2 in the example are, respectively,
L1 =
2 −1 0 −1
−1 2 −1 00 −1 2 −1
−1 0 −1 2
, L2 =
2 −1 −1 0
−1 2 0 −1
−1 0 2 −10 −1 −1 2
, and
L3 =
4 −1 0 −1 0 −1 −1 0
−1 4 −1 0 −1 0 0 −10 −1 4 −1 −1 0 0 −1
−1 0 −1 4 0 −1 −1 00 −1 −1 0 4 −1 0 −1
−1 0 0 −1 −1 4 −1 0
−1 0 0 −1 0 −1 4 −10 −1 −1 0 −1 0 −1 4
.
We now describe an exact characterization of when the merge of two integer-weighted graphs which are diagonalizable by the same Hadamard matrix has PST at time t0 = π/ 2. This gives us a wide variety of new graphs with PST; in par-ticular, the merge operation produces perfect state transfer graphs in a variety of scenarios. We note that the result below can be proven by using techniques de-veloped in for the adjacency matrix. However for completeness, we give a separate proof that relies on Theorem 3.
Theorem 4. Suppose G1 and G2 are integer-weighted graphs on n vertices, both of which are diagonalizable by the same Hadamard matrix H. Fix w1, w 2 ∈ Z
and let L1 = d1I − A1, L2 = d2I − A2 be the Laplacian matrices for G1, G 2,respectively. Then G1 ⊙w1 w2 G2 has PST from vertex p to q, where p < q , at time
t0 = π/ 2 if and only if one of the following 8 conditions holds:
12 1. p, q ∈ { 1, . . . , n } and (a) w1 is odd, w2 is even, and G1 has PST from p to q at t0 = π/ 2, or (b) w1 and d2 are even, w2 is odd, and G2 has PST from p to q at t0 = π/ 2,or (c) w1 and w2 are odd, d2 is even, and the weighted graph with Laplacian
L1 + L2 has PST from p to q at t0 = π/ 2;2. p, q ∈ { n + 1 , . . . , 2n} and (a) w1 is odd, w2 is even, and G1 has PST from p − n to q − n at t0 = π/ 2,or (b) w1 and d2 are even, w2 is odd, and G2 has PST from p − n to q − n at
t0 = π/ 2, or (c) w1 and w2 are odd, d2 is even, and the weighted graph with Laplacian
L1 + L2 has PST from p − n to q − n at t0 = π/ 2;3. p ∈ { 1, . . . , n }, q ∈ { n + 1 , . . . , 2n} and (a) w1 is even, w2 and d2 are odd, and G2 has PST from p to q − n at
t0 = π/ 2, or (b) w1, w2, and d2 are all odd, and the weighted graph with Laplacian matrix L1 + L2 has PST from p to q − n at t0 = π/ 2.Proof. Without loss of generality we can assume that H is a normalized Hadamard matrix. Denote the diagonal matrices of eigenvalues for L1, L 2 by Λ1, Λ2, respec-tively, so that Lj = 1
n
HΛj HT , j = 1 , 2. Then the Laplacian of G1 ⊙w1 w2 G2 is
L3 =
[w1L1 + w2d2I −w2A2
−w2A2 w1L1 + w2d2I
]
. Further,
L3 = 1
2n
[H HH −H
] [ w1Λ1 + w2Λ2 00 w1Λ1 − w2Λ2 + 2 w2d2I
] [ H HH −H
]T
.
Denote the eigenvalues of L1, L 2 by λ(1)
ℓ
, λ (2)
ℓ
, ℓ = 1 , · · · , n , respectively. 1. Suppose that p, q ∈ { 1, · · · , n } and that the graph with Laplacian L3 has PST from p to q. Then for each ℓ = 1 , · · · , n , w1λ(1)
ℓ
+w2λ(2)
ℓ
≡ (1 −hpℓ hqℓ )mod 4 and w1λ(1)
ℓ
− w2λ(2)
ℓ
2 w2d2 ≡ (1 − hpℓ hqℓ ) mod 4 . In particular,
2w2d2 ≡ 0 mod 4 , i.e., w2d2 is even. Note that if w1 and w2 are both even, then hpℓ hqℓ = 1 for ℓ = 1 , · · · , n , which is impossible. If w1 is odd and w2 is even, then w1λ(1)
ℓ
w2λ(2)
ℓ
≡ λ(1)
ℓ
mod 4 , so that
λ(1)
ℓ
≡ (1 − hpℓ hqℓ ) mod 4 , ℓ = 1 , · · · , n . Hence G1 has PST from p to 13 q. Similarly, if w1 is even and w2 is odd, then necessarily d2 is even, and as above G2 has PST from p to q.If w1 and w2 are both odd, then necessarily d2 is even. Also w1λ(1)
ℓ
+
w2λ(2)
ℓ
≡ λ(1)
ℓ
λ(2)
ℓ
≡ (1 − hpℓ hqℓ ) mod 4 , ℓ = 1 , · · · , n . We deduce that the graph with Laplacian L1 + L2 has PST from p to q.2. If p, q ∈ { n + 1 , · · · , 2n} and the graph with Laplacian L3 has PST from p
to q, the conclusions (a), (b), and (c) follow analogously to Case 1 above. 3. Suppose that p ∈ { 1, · · · , n }, q ∈ { n + 1 , · · · , 2n} and that the graph with Laplacian L3 has PST from p to q. Set ˆq = q − n. Then for each
ℓ = 1 , · · · , n , we have
w1λ(1)
ℓ
w2λ(2)
ℓ
≡ (1 − hpℓ hˆqℓ ) mod 4 , and (2)
w1λ(1)
ℓ
− w2λ(2)
ℓ
2 w2d2 ≡ (1 + hpℓ hˆqℓ ) mod 4 . (3) Summing equations (2) and (3), we find that 2w1λ(1)
ℓ
2 w2d2 ≡ 2 mod 4
and hence 2w2d2 ≡ 2 mod 4 since all the eigenvalues of L1 are even inte-gers, and therefore w2d2 must be odd, i.e., w2 is odd and d2 is odd. We have the following two cases. If w1 is even, then (2) simplifies to λ(2)
ℓ
≡ (1 − hpℓ hˆqℓ ) mod 4 , ℓ =1, · · · , n , so for even w1, and odd w2 and d2, G2 has PST from p to ˆq.If w1 is odd, then (2) simplifies to λ(1)
ℓ
λ(2)
ℓ
≡ (1 − hpℓ hˆqℓ ) mod 4 ,
ℓ = 1 , · · · , n , which shows that the integer-weighted graph with Laplacian
L1 + L2 has PST from p to ˆq.The converses are straightforward.
Note that when both w1 and w2 are even, the graph G1 ⊙w1 w2 G2 does not have PST at time π/ 2. However, it might have PST at some other time. To see this, we decompose the two integer weights wj as wj = 2 rj .b j (for j = 1 , 2), where bj
are odd integers. Let r = min( r1, r 2). Then the PST property of the graph with Laplacian 1
2r
L3 at time π/ 2 can be determined according to Theorem 4. In the case that PST occurs, the graph G1 ⊙w1 w2 G2 would then have PST at time π/ 2r+1 .Also note that Theorem 4 is true for any graphs whose Laplacian eigenvalues are all even integers (including non integer-weighted graphs).
Remark 2. Assume that G1 and G2 are two graphs on 2m vertices for m ≥ 2
and that they are diagonalizable by the same Hadamard matrix. Suppose that G1
has PST from vertex p to vertex q, and G2 has all its eigenvalues being multiples
14 of 4 and that its degree d2 is odd (for example, a disjoint union of 2m−r copies of K2r for 2 ≤ r ≤ m). Then G1 ⊙ G2 has PST from p to q + 2 m according to Case 3(b) in Theorem 4. Similarly, G2 ⊙ G1 has PST from vertex p to q if d1 is even (Case 1(c)), and it has PST from vertex p to q + 2 m if d1 is odd (Case 3(b)).
The requirement that both graphs are diagonalizable by the same Hadamard matrix is necessary for Theorem 4 to hold. As a concrete example, let G1 be equal to K8 with a K3 removed, G2 be equal to the 3-cube, w1 = 2 and w2 = 1 (and
d2 = 3 ). Then G1 ⊙w1 w2 G2 is equal to
13 0 0 −2 −2 −2 −2 −2 0 −1 −1 0 −1 0 0 00 13 0 −2 −2 −2 −2 −2 −1 0 0 −1 0 −1 0 00 0 13 −2 −2 −2 −2 −2 −1 0 0 −1 0 0 −1 0
−2 −2 −2 17 −2 −2 −2 −2 0 −1 −1 0 0 0 0 −1
−2 −2 −2 −2 17 −2 −2 −2 −1 0 0 0 0 −1 −1 0
−2 −2 −2 −2 −2 17 −2 −2 0 −1 0 0 −1 0 0 −1
−2 −2 −2 −2 −2 −2 17 −2 0 0 −1 0 −1 0 0 −1
−2 −2 −2 −2 −2 −2 −2 17 0 0 0 −1 0 −1 −1 00 −1 −1 0 −1 0 0 0 13 0 0 −2 −2 −2 −2 −2
−1 0 0 −1 0 −1 0 0 0 13 0 −2 −2 −2 −2 −2
−1 0 0 −1 0 0 −1 0 0 0 13 −2 −2 −2 −2 −20 −1 −1 0 0 0 0 −1 −2 −2 −2 17 −2 −2 −2 −2
−1 0 0 0 0 −1 −1 0 −2 −2 −2 −2 17 −2 −2 −20 −1 0 0 −1 0 0 −1 −2 −2 −2 −2 −2 17 −2 −20 0 −1 0 −1 0 0 −1 −2 −2 −2 −2 −2 −2 17 −20 0 0 −1 0 −1 −1 0 −2 −2 −2 −2 −2 −2 −2 17
.
There is no PST at time π/ 2, though the parameters are set up so that they sat-isfy 3(a) of Theorem 4 (but not the hypothesis of both Laplacians being diagonal-ized by the same Hadamard). Thus, unlike a similar result [14, Theorem 5.2] for the “ ⋉” operation (which uses the adjacency matrices), graphs whose Laplacian matrices aren’t diagonalizable by the same Hadamard matrix do not necessarily satisfy the conclusion of the theorem. This may be due to the difference between Laplacian dynamics and adjacency dynamics. The following corollary to Theorem 4 provides an instance where the state-ment of the theorem simplifies considerably, and generalizes the known fact that the unweighted hypercube graph has PST.
Corollary 2. Suppose w1, w 2, . . . , w n are nonzero integers, exactly d of which are odd, and consider the weighted hypercube Cn := ( w1K2)(w2K2) · · · (wnK2).
15 For each vertex u of Cn, there is a vertex v at distance d from u such that there is perfect state transfer in Cn from u to v at time t0 = π/ 2.Proof. We prove the result by induction on n. For the base case, we simply note that it is straightforward to verify that the weighted 1-cube w1K2 has perfect state transfer at time t = π/ 2 if and only if w1 is an odd integer. For the inductive hypothesis, we use Theorem 4 with G1 = Cn (which we will assume has perfect state transfer at time t = π/ 2 from vertex j to k, which are a distance of d apart) and G2 is the graph on the same number of vertices where every vertex has a self-loop (of weight 1) and no other edges (note that this graph has perfect state transfer between any vertex and itself at any time). Then it is straightforward to verify that the graph G1 ⊙1 wn+1 G2 is exactly the weighted
(n + 1) -cube:
G1 ⊙1 wn+1 G2 = ( wn+1 K2)Cn = ( wn+1 K2)(w1K2)(w2K2) · · · (wnK2).
So condition 1(a) of Theorem 4 tells us that if wn+1 is even then G1 ⊙1 wn+1 G2 has perfect state transfer at time t0 = π/ 2 from vertex j to k (which still have a dis-tance of d from each other). On the other hand, if wn+1 is odd then condition 3(b) of Theorem 4 says that G1 ⊙1 wn+1 G2 has perfect state transfer at time t = π/ 2
from vertex j to k +2 n (which have a distance of d+1 from each other). By noting that the particular labelling of the weights is irrelevant (i.e., permute the indices of the weights so that G1 ⊙1 wn+1 G2 = ( w1K2)(w2K2) · · · (wn+1 K2) = Cn+1 ), this completes the inductive step and the proof.
Example 1. From Lemma 9 and Proposition 10 of , one can conclude that there is no unweighted graph of order 12 that is Hadamard diagonalizable and exhibits PST. However, it is easy to construct weighted graphs of this type. Let G1
16 be the graph whose Laplacian is
L1 = 1
3
18 0 −1 −1 −1 −3 −3 −3 −1 −3 −1 −10 18 −1 −1 −1 −3 −3 −3 −1 −3 −1 −1
−1 −1 18 −2 −2 0 −2 0 −2 −2 −4 −2
−1 −1 −2 18 −4 0 0 −2 −2 −2 −2 −2
−1 −1 −2 −4 18 −2 −2 0 −2 0 −2 −2
−3 −3 0 0 −2 18 −2 −2 0 −2 −2 −2
−3 −3 −2 0 −2 −2 18 −2 −2 −2 0 0
−3 −3 0 −2 0 −2 −2 18 −2 −2 −2 0
−1 −1 −2 −2 −2 0 −2 −2 18 0 −2 −4
−3 −3 −2 −2 0 −2 −2 −2 0 18 0 −2
−1 −1 −4 −2 −2 −2 0 −2 −2 0 18 −2
−1 −1 −2 −2 −2 −2 0 0 −4 −2 −2 18
Then one can easily verify that L1 is Hadamard diagonalizable by the order 12 Hadamard
H =
1 1 1 1 1 1 1 1 1 1 1 11 −1 1 −1 1 1 1 −1 −1 −1 1 −11 −1 −1 1 −1 1 1 1 −1 −1 −1 11 1 −1 −1 1 −1 1 1 1 −1 −1 −11 −1 1 −1 −1 1 −1 1 1 1 −1 −11 −1 −1 1 −1 −1 1 −1 1 1 1 −11 −1 −1 −1 1 −1 −1 1 −1 1 1 11 1 −1 −1 −1 1 −1 −1 1 −1 1 11 1 1 −1 −1 −1 1 −1 −1 1 −1 11 1 1 1 −1 −1 −1 1 −1 −1 1 −11 −1 1 1 1 −1 −1 −1 1 −1 −1 11 1 −1 1 1 1 −1 −1 −1 1 −1 −1
and that the (1 , 2) entry of ei(π/ 2) L1 is 1, thus showing that L1 exhibits PST between vertices 1 and 2 at time t0 = π/ 2. Let G2 = K12 , which we note is Hadamard diagonalizable by H but does not exhibit PST, and let w1 = 5 and w2 = 2 .Direct computation shows that all the eigenvalues of L1 are even integers. Hence Theorem 4 still applies here, and Case 1(a) of the theorem tells us that G1 ⊙5 2 G2
has PST from vertex 1 to vertex 2 at time t0 = π/ 2. One can indeed verify that
17 L3 =
[5L1 + 2 D2 −2A2
−2A2 5L1 + 2 D2
]
, where D2 = 11 I and A2 = J − I (where J is the all-ones matrix), is Hadamard diagonalizable by
[H HH −H
]
with eigenvalues (in the order determined by that diagonalization) equal to 0, 54 , 64 , 54 , 64 , 64 , 64 ,
54 , 54 , 54 , 44 , 54 , 44 , 50 , 60 , 50 , 60 , 60 , 60 , 50 , 50 , 50 , 40 , and 50 . Furthermore, by checking the (1 , 2) entry of ei(π/ 2) L3 we see that this graph exhibits PST between vertices 1 and 2 at time t0 = π/ 2.
Remark 3. For each k ≥ 3 and each d with k + 1 ≤ d ≤ 2k − 2, we can construct a graph that is d-regular, unweighted, connected, and non-bipartite on
2k vertices, that is diagonalizable by the standard Hadamard matrix and has PST at time t0 = π/ 2. This can be done with cubelike graphs by using Theorem 1. To ensure the cubelike graph is connected, we just need to make sure the connection set contains a basis of Zk
2
when considered as a vector space. Let us take the standard ordered basis: e1, · · · , ek. Assume they form the set T . For d = k + 1 ,take the connection set C = T ∪{ e1 +e2}. Then the induced subgraph on vertices
0, e1, e2, e1 + e2 is K4; hence the corresponding cubelike graph is not bipartite. Also note that the sum of the elements in C is not 0 for k ≥ 3. For d > k + 1 , we always keep C as a subset of the connection set S (|S| = d, 0 /∈ S). If the sum of all elements in S is not 0, then the cubelike graph G(S) is a desired graph. On the other hand, if the sum of all elements in S is 0, then we delete some element
c0 from the set S \ C. Denote the set S \ { c0} by S0 and we know the sum of all its elements is c0 6 = 0 (in Zk
2
, every element has itself as its inverse). Finally, we pick any element c1 ∈ Zk
2
\ (S ∪ { 0}) (this set has cardinality 2k − d − 1 > 0)and form a new set S1 = S0 ∪ { c1}. Then S1 has cardinality d and the sum of all its element is c = c0 + c1 6 = 0 . Hence there is PST from u to u + c at time π/ 2 in the connected (since S1 is a generating set of the group Zk
2
) nonbipartite cubelike graph G(S1).
This remark can be stated as follows. As a means of highlighting the utility of the merge operation, we present an alternate proof that constructs such graphs using the merge.
Theorem 5. Suppose that k ∈ N with k ≥ 3. For each d ∈ N with k + 1 ≤ d ≤
2k − 2, there is a connected, unweighted, non-bipartite graph that is (1) diagonalizable by the standard Hadamard matrix of order 2k,(2) d-regular, and
18 (3) has PST between distinct vertices at time t0 = π/ 2.Proof. For every integer k ≥ 3, it is easy to see that the complement of a perfect matching (disjoint union of 2k−1 copies of K2) is a (2 k − 2) -regular graph with the desired properties. So we just need to prove the result for k + 1 ≤ d ≤ 2k − 3.We proceed by induction on k. For k = 3 , it is straightforward to check that
(K2,2K2)c, (K2,2 + K2,2)c and (K2 + K2 + K2 + K2)c are 4-, 5-, and 6-regular graphs, respectively, having the described properties. Now suppose the result holds for some fixed k ≥ 3; that is, for each d with k + 1 ≤ d ≤ 2k − 2,we have a d-regular graph Gk,d on 2k vertices with the desired properties. To construct desired graphs on 2k+1 vertices, we split into two cases depending on the regularity d of the graph that we are trying to construct. Case 1: (k + 1) + 1 ≤ d ≤ 2k − 1. Let the Laplacian matrix for Gk,d −1 be
Lk,d −1. Consider the graph K2Gk,d −1. This graph has 2k+1 vertices and Laplacian
[Lk,d −1 + I −I
−I Lk,d −1 + I
]
. It is straightforward to see that this graph is d-regular, connected, non-bipartite (since Gk,d −1 is) and satis-fies (1) and (3). Case 2: k + 4 ≤ d ≤ 2k+1 − 3. For each 2 ≤ r ≤ k, let Gr be the disjoint union of 2k−r copies of K2r and let Lr denote its Laplacian matrix. Note that each eigenvalue of Lr is congruent to 0 (mod 4); further, with a natural labelling of the vertices, Lr is diagonalizable by the standard Hadamard matrix of order 2k.Fix d′ with k + 1 ≤ d′ ≤ 2k − 2 and let Ak,d ′ be the adjacency matrix of Gk,d ′ . Let G(r)
k,d ′
= Gr ⊙ Gk,d ′ be the graph on 2k+1 vertices whose Laplacian matrix is L(r)
k,d ′
=
[Lr + d′I −Ak,d ′
−Ak,d ′ Lr + d′I
]
. Then G(r)
k,d ′
is not bipartite (it has K4 as an induced subgraph), and it is Hadamard diag-onalizable by the standard Hadamard matrix. By Remark 2, it also has PST between a pair of distinct vertices. It is regular of degree d′ + 2 r − 1.Also in the notation of Theorem 4, −Λ2 + 2 d′I has positive diagonal en-tries (since Gk,d ′ is not bipartite) and so we deduce that the nullity of L(r)
k,d ′
is 1 and that G(r)
k,d ′
is connected. Thus G(r)
k,d ′
is a graph on 2k+1 vertices, satisfying the desired properties. We denote it as Gk+1 ,d ′+2 r −1.19 Thus we have produced the desired graphs whose degrees fall in the set
[k + 2 , 2k − 1] ∪
k
⋃
r=2
[k + 2 r, 2k + 2 r − 3] . (4) For k = 3 , this set covers the integers 5, 6, 7, 8, 9, 11, 12, 13. From Propo-sition 1 we know that if G = ( K2,2 + K2,2)c, our 5-regular graph on 8 vertices satisfying the desired properties, then the graph Gc ∨ Gc of order 16 is 10-regular and has the desired properties. So the result is also true for graphs on 24 vertices. For k ≥ 4 we have k + 4 ≤ 2k−1. Then for any r ≤ k − 1, we have k + 4 + 2r+1 ≤ 2k−1 + 2 r+1 = 2 k−1 + 2 r + 2 r ≤ 2k−1 + 2 k−1 + 2 r = 2 k + 2 r, which in turn implies that k + 2 r+1 ≤ 2k + 2 r − 3. It follows that the set (4) contains all of the integers in [k + 2 , 2k+1 − 3] .
Remark 4. Note that for 2k +1 ≤ d ≤ 2k+1 −2, we can also construct a d-regular graph with the desired properties on 2k+1 vertices using the join operation. From the induction hypothesis, we have a graph Gk,d with the desired properties for
k + 1 ≤ d ≤ 2k − 2. Now we use the result in Proposition 1: if G is a Hadamard diagonalizable graph on n ≥ 4 vertices and that G has PST at time π/ 2, then its complement also has PST at the same time, then we get a non-empty d-regular Hadamard diagonalizable PST graph G for each d such that 1 ≤ d ≤ 2k − 2.Then the graph G ∨ G has PST at time t0 = π/ 2 and is diagonalizable by the standard Hadamard matrix, whose regularity is 2k + d, ranging from 2k + 1 to
2k + 2 k − 2. Since G is not empty, G ∨ G has cycles of length 3, and therefore it is not bipartite.
PST for graphs with non-integer weights
We now consider some ways in which our results generalize to the case of Hadamard diagonalizable graphs with non-integer edge weights. In the case where all of the edge weights are rational, the idea is rather straightforward.
Proposition 2. Suppose the graph G1 with Laplacian L1 is a rational-weighted Hadamard diagonalizable graph, and let lcm be the least common multiple of the denominators of its edge weights, and gcd be the greatest common divisor of all the new integer edge weights lcm ·w(j, k ). Then G1 has PST at time t1 =
lcm
gcd
· π/ 2 if and only if the integer-weighted Hadamard diagonalizable graph G2
with Laplacian L2 = lcm
gcd
L1 has PST at time t0 = π/ 2 between the same pair of vertices.
20 Proof. The result follows simply from noticing that for each j and k we have
|eTj eit 0 L2 ek|2 = |eTj eit 0 lcm
gcd L1
ek|2 = |eTj eit 1L1 ek|2, (5) and G1 has PST between vertex j and vertex k at time t1 if and only the rightmost quantity in (5) equals 1, while G2 has PST at time t0 if and only if the leftmost quantity in (5) equals 1.
While we are not able to extend Proposition 2 to the case of irrational weights directly—in general such a graph may not exhibit PST at any time—it is true at least that the resulting graph has pretty good state transfer when exactly one of the two weights in G1 ⊙w1 w2 G2 is irrational. Before giving the theorem, we recall the following result about approximating an irrational real number with rational numbers.
Theorem 6 () . Let o denote the odd integers and e denote the even integers. Then for every real irrational number w, there are infinitely many relatively prime numbers u, v with [u, v ] in each of the three classes [o, e ], [e, o ], and [o, o ], such that the inequality |w − u/v | < 1/v 2 holds.
For the graph G1 ⊙w1 w2 G2, we say it has parameters [w1, w 2, d 2], where as in Theorem 4, d2 denotes the degree of G2. In particular, if w1, w 2, and d2 are all odd integers, we say the graph G1 ⊙w1 w2 G2 has type [o, o, o ]. We will denote the set of irrational numbers by Q.
Theorem 7. Assume that G1 and G2 are integer-weighted graphs on n vertices, both of which are diagonalizable by the same Hadamard matrix H. Let d2 be the degree of G2. Let L1 and L2 denote the Laplacian matrices of G1 and G2,respectively. Suppose that one of w1, w 2 is rational and the other is irrational, and suppose that p, q ∈ { 1, . . . , n }. Then the weighted graph G1 ⊙w1 w2 G2 has PGST as stated in the following cases. 1. Suppose that G1 has PST from p to q at time π/ 2. Then G1 ⊙w1 w2 G2 has PGST from p to q and from p + n to q + n.2. Suppose that G2 has PST from p to q at time π/ 2. If d2 is even, then G1 ⊙w1 w2 G2
has PGST from p to q and from p + n to q + n. If d2 is odd, then G1 ⊙w1 w2 G2
has PGST from p to q + n and from q to p + n.3. Suppose that the graph with Laplacian L1 + L2 has PST from p to q at time
π/ 2. If d2 is even, then G1 ⊙w1 w2 G2 has PGST from p to q and from p + n to
q + n. If d2 is odd, then G1 ⊙w1 w2 G2 has PGST from p to q + n and from q to
p + n.
21 Before proving this result, we note that it can alternatively be proved via Kro-necker’s theorem using the techniques of . However, this would require proving that vertices p and q are strongly cospectral, as well as some knowledge of eigen-values and eigenprojection matrices, so we instead give the following proof that is somewhat more self-contained.
Proof. As in the proof of Theorem 4, without loss of generality we assume that
H is a normalized Hadamard matrix. Assume w1 is rational and w2 is irrational (case 1). We denote the graph G1 ⊙w1 w2 G2 as G3, with corresponding Laplacian
L3. It suffices to consider w1 odd. Indeed, if w1 = a
b
, with a and b being relatively prime integers, assume a = 2 rk where r ∈ N and k odd, then eitL 3 = eit 2r
b(L3b/ 2r)
so that L3b/ 2r is the Laplacian of the graph G1 ⊙k w2b/ 2r G2 where k is odd and
w2b/ 2r is irrational. Note that if L3b/ 2r has PGST, then so does L3. Thus, for notational simplicity, we consider w1 odd. We approach w2 with fractions u/v such that |w2−u/v | < 1/v 2. For each such pair of u, v , we denote the graph G1 ⊙w1 u/v G2 as G4, and the graph G1 ⊙0 w2−u/v G2
as G5. In particular, the Laplacian of G3 is the sum of the Laplacian of G4 with the Laplacian of G5. Denote the Laplacian matrices of G4 and G5 as L4 and L5,respectively. Now consider the integer-weighted graph G′
4
= G1 ⊙vw 1 u G2, then its Laplacian is vL 4 and has parameters [vw 1, u, d 2].There are now a number of cases to consider. If [u, v ] is of type [o, e ] and d2 is even, the graph G′
4
is of type [e, o, e ]. From Theorem 4 we know, if G2 has PST from p to q at π/ 2, then G′
4
has PST at π/ 2 from p to q and from p + n to q + n
(Case 1(b), 2(b)). If [u, v ] is of type [o, e ] and d2 is odd, the graph G′
4
is of type
[e, o, o ]. From Theorem 4 we know that if G2 has PST at π/ 2 from p to q at π/ 2,then G′
4
has PST at π/ 2 from p to q + n and from q to p + n (Case 3(a)). If [u, v ] is of type [e, o ], then the graph G′
4
is of type [o, e, f ], where f denotes the parity of d2. From Theorem 4 we know that if G1 has PST from p to q at π/ 2,then G′
4
has PST at π/ 2 from p to q and from p + n to q + n (Case 1(a), 2(a)). If [u, v ] is of type [o, o ] and d2 is even, the graph G′
4
is of type [o, o, e ]. From Theorem 4 we know that if the graph with Laplacian L1 + L2 has PST from p
to q at π/ 2, then G′
4
has PST from p to q and from p + n to q + n (Case 1(c), 2(c)). If [u, v ] is of type [o, o ] and d2 is odd, the graph G′
4
is of type [o, o, o ]. From Theorem 4 we know that if the integer weighted graph with Laplacian L1 + L2
has PST from p to q at π/ 2, then G′
4
has PST from p to q + n and from q to p + n
(Case 3(b)). Similarly we can get the results when w1 is irrational and w2 is rational (case 2). (We can assume w2 is odd by way of a similar argument to w1 being odd in 22 case 1.) For all the above cases, G4 has PST at time t0 = vπ/ 2. Next, we recall the following result from [16, Theorem 4] (here we take the absorbed constant factor
t0 out): Suppose PST occurs for the graph with Laplacian matrix L and assume that ˆL = t0(L + L0) due to a small nonzero edge-weight perturbation L0. Then
1 − | eTj eit 0 (L+L0)ek|2 ≤ 2‖t0L0‖ + ‖t0L0‖2 − ‖ t0L0‖3. (6) Now, G4 is a graph with PST at time t0, and L3 = L4 + L5. Then the fidelity of state transfer of G3 between the corresponding pair of vertices satisfies
|eTj eit 0L3 ek|2 ≥ 1 − 2‖t0L5‖ − ‖ t0L5‖2 + ‖t0L5‖3
≥ 1 − 2cnπ/ (2 v) − (cnπ/ (2 v)) 2 + ( cnπ/ (2 v)) 3,
where c is the maximum edge weight in G2. Since there are infinitely such integers
v, the expression on the right hand side in the above inequality can be made as close to one as possible.
It is known (see ) that if there is perfect state transfer from vertex j to vertex k at time t0, and perfect state transfer from vertex j to vertex l at time
t1, then necessarily k = l. The following example, which is a straightforward consequence of Theorem 7, shows that the situation with respect to pretty good state transfer is markedly different. This is a potentially important application to routing—the task of choosing between several possible recipients of the state.
Example 2. Consider the unweighted graphs G1, G 2 with the following Laplacian matrices:
L1 =
3 −1 −1 0 −1 0 0 0
−1 3 0 −1 0 −1 0 0
−1 0 3 −1 0 0 −1 00 −1 −1 3 0 0 0 −1
−1 0 0 0 3 −1 −1 00 −1 0 0 −1 3 0 −10 0 −1 0 −1 0 3 −10 0 0 −1 0 −1 −1 3
23 which has PST at time π/ 2 for the pairs (1 , 8) , (2 , 7) , (3 , 6) , (4 , 5) , and
L2 =
3 0 −1 −1 −1 0 0 00 3 −1 −1 0 −1 0 0
−1 −1 3 0 0 0 −1 0
−1 −1 0 3 0 0 0 −1
−1 0 0 0 3 0 −1 −10 −1 0 0 0 3 −1 −10 0 −1 0 −1 −1 3 00 0 0 −1 −1 −1 0 3
which has PST at time π/ 2 for the pairs (1 , 6) , (2 , 5) , (3 , 8) , (4 , 7) . It turns out that
L1 + L2 has PST at time π/ 2 between the pairs (1 , 3) , (2 , 4) , (5 , 7) , (6 , 8) . From the above collection of cases, we find for example that if w1 ∈ Q and w2 ∈ Q (or
w1 ∈ Q and w2 ∈ Q), then G1 ⊙w1 w2 G2 has the intriguing property that there is PGST between the pairs (1 , 8) , (1 , 11) , (1 , 14) (among others).
Optimality
6.1. Timing errors
In , the authors analyse the sensitivity of the probability of state transfer in the presence of small perturbations. Bounds on the probability of state trans-fer with respect to timing errors and with respect to manufacturing errors were given in the most general setting where no information is known about the graph in question. Specifically, suppose that under XXX dynamics, a graph G on n
vertices has PST from vertex 1 to vertex 2 at time t0. Suppose further that there is a small perturbation so that the readout time is instead t0 + h, where |h| < π
λn
,and where λn is the largest eigenvalue of the corresponding Laplacian. Decom-pose the Laplacian as L = QΛQT , where Λ = diag ( λ1 = 0 , λ 2, . . . , λ n), with
0 < λ 2 ≤ · · · ≤ λn, and Q is an orthogonal matrix of corresponding eigenvectors. If q1 and q2 are the first and second columns of QT , respectively, then for some
θ ∈ R we have eiθ q1 = eit 0Λq2. Setting M = diag ( eihλ 1 , . . . , e ihλ n )eiθ , it follows that
p(t0) − p(t0 + h) = 1 − | qT
1
Mq1|2.
24 In the special case that G is Hadamard diagonalizable, we have Q = 1√n H, where
H is a Hadamard matrix, so we can say more. In that case,
|qT
1
Mq1| = 1
n
∣∣∣∣∣
n
∑
j=1
eihλ j
∣∣∣∣∣ . (7) This suggests that, in order to find a lower bound for |qT
1
Mq1| (and thus an upper bound for p(t0) − p(t0 + h)), the goal should be to make the numbers eihλ j as closely-spaced on the complex unit circle as possible. This agrees with the known fact that minimizing the spectral spread has the effecting of maximizing the bound for the fidelity of state transfer due to timing errors . Thus, this remark is not surprising but rather confirms the known rule while at the same time providing a more accurate bound on timing errors for Hadamard diagonalizable graphs.
6.2. Manufacturing errors: sparsity of graphs with PST
It is desirable to minimize the number of edges that need to be engineered in a graph (so as to minimize manufacturing errors), so one question of interest in the theory of perfect state transfer is how sparse a graph with perfect state transfer can be. Among the sparsest known graph with PST is the k-cube, which has 2k
vertices, each with degree k. We now show that if we restrict our attention to Hadamard diagonalizable unweighted graphs, then for k ≤ 4 the k-cube is indeed the sparsest connected graph with PST.
Theorem 8. Let G be a simple, connected, unweighted r-regular graph on n
vertices. Suppose further that G is Hadamard diagonalizable, has perfect state transfer at π/ 2, and that r ≤ 4. Then n ≤ 2r.Proof. If L is the Laplacian of G then the result follows by computing some quan-tities of the form Tr( Lk) (k ≥ 0 is an integer) in two different ways. First, if
λ1, . . . , λ n are the eigenvalues of L then Tr( Lk) = ∑nj=1 λkj , and we know by the Gershgorin circle theorem that 0 ≤ λj ≤ 2r for each j. On the other hand,
L = rI − A, where A is the adjacency matrix of G, so Tr( L) = rn − Tr( A) and
Tr( L2) = r2n − 2rTr( A) + Tr( A2). Since A is simple, we know that Tr( A) = 0
and it is straightforward to compute Tr( A2) = rn . Thus we have the following system of equations:
n
∑
j=1
λj = rn and
n
∑
j=1
λ2
j
= rn (r + 1) .
25 If we let cλ denote the number of eigenvalues of L equal to λ (with the con-vention that if λ is not an eigenvalue, then cλ = 0 ), then these equations tell us that
r
∑
j=1
(2 j)c2j = rn and
r
∑
j=1
(2 j)2c2j = rn (r + 1) . (8) If we add in the equation ∑rj=1 c2j = n − 1 (since one of the eigenvalues equals
0), then we have a system of 3 linear equations in the variables n, c 2, c 4, . . . , c 2r.If r ≤ 2 then it is straightforward to solve this system of equations to get n = 2 r.If r = 3 then by adding the equation c2 + c6 = c4 + 1 (since we know that half of L’s eigenvalues must belong to each equivalence class mod 4) we can similarly solve the system of equations to get n = 8 = 2 r.For the r = 4 case, we use the Equations (8) together with the equation
c2 + c6 = c4 + c8 + 1 (again, because the eigenvalues are split evenly between the mod 4 equivalence classes). These equations together can be reduced to the system of equations c2 = 3 n/ 8 − 2, c4 = 3 n/ 8, c6 = n/ 8 + 2 , and c8 = n/ 8 − 1.To reduce this system further and get a unique solution, we need to compute
Tr( L3) in two different ways (similar to at the start of the proof): Tr( L3) = ∑nj=1 λ3
j
= r3n − 3r2Tr( A) + 3 rTr( A2) − Tr( A3) = r3n + 3 r2n − Tr( A3). Since
Tr( A3) ≥ 0 we arrive at the inequality ∑nj=1 λ3
j
≤ r2n(r + 3) , which is equivalent to ∑rj=1 (2 j)3c2j ≤ r2n(r + 3) . Plugging in r = 4 then gives
8c2 + 64 c4 + 216 c6 + 512 c8 ≤ 112 n.
It is then straightforward to substitute the equations c2 = 3 n/ 8 − 2, c4 = 3 n/ 8,
c6 = n/ 8 + 2 , and c8 = n/ 8 −1 into this inequality to get n ≤ 2r = 16 , as desired.
It seems reasonable to believe that Theorem 8 could be generalized to arbitrary
r, but the method of proof that we used does not seem to generalize in a straight-forward way, as there are no more obvious equations or inequalities involving the
c2j ’s that we can use. For example, if we try to extend the proof of Theorem 8 to the r = 5 case, we might try computing Tr( L4) in two different ways. However, we then end up with an equation involving both −Tr( A3) and +Tr( A4), and it is not clear how to bound such a quantity. 26 7. Acknowledgements
The authors are grateful to an anonymous referee, whose constructive com-ments resulted in improvements to the paper. N.J., S.K., and S.P. were supported by NSERC Discovery Grant number RGPIN-2016-04003, RGPIN/6123-2014, and 1174582, respectively; N.J. was also supported by a Mount Allison Marjorie Young Bell Faculty Fund; R.S. was supported through a NSERC Undergraduate Student Research Award; X.Z. was supported by the University of Manitoba’s Faculty of Science and Faculty of Graduate Studies.
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内积和外积
在本文之前，特别说明一下翻译的相关问题。由于历史原因，数学学科和物理学科关于「inner product」和「outer product」两个词汇有着五花八门的翻译。
在物理学科，一般翻译成「标积」和「矢积」，表示运算的结果为标量和矢量。高中数学课本上「数量积」和「向量积」也采用了这种意译的办法。
在数学学科，通常也可以翻译成「内积」和「外积」，是两个名词的直译。「点乘」和「叉乘」是根据运算符号得来的俗称，这种俗称也很常见。
在「点乘」运算中，经常省略运算的点符号，在线性代数中更是会直接看作矩阵乘法，不写点符号。
内积
内积的概念 对于任意维数的向量都适用。
定义
内积有不同但等价的定义方法，下面介绍其中一些。
几何定义
在 维欧氏空间 下，已知两个向量 ，它们的夹角为 ，那么：
就是这两个向量的 内积，也叫 点积 或 数量积。其中称 为 在 方向上的投影。内积的几何意义即为：内积 等于 的模与 在 方向上的投影的乘积。
代数定义
在 维欧氏空间 下，已知两个向量 ，那么：
就是这两个向量的 内积，也叫 点积 或 数量积。内积的几何定义与代数定义在欧氏空间下是等价的，而后者更方便使用。
在不引起混淆的情况下，内积的点号可以省略不写。如果在向量的右上角有上角标 ，表示向量与自身内积的简写，即 向量模长的平方，省略模长记号。该上角标 不可以理解为向量的平方，这是因为，向量内积的结果为标量，不存在除了 以外任何个数的向量的内积。同理，向量模长平方的平方，不可以简写为上角标 ，而是必须将上角标 的结果视为一个整体，以此类推。
性质
可以发现，内积得到的结果是一个标量，其特别之处在于，它是关于两个向量分别都线性的双线性运算。具体而言，内积满足：
内积还满足交换律，即：
应用
下面介绍内积运算的一些常见应用。
判定两向量垂直：
即互相垂直的两个向量的内积，结果为 ；向量与零向量内积，结果为 。如果使用内积为零作为垂直的定义，则可以得出零向量与任何向量都垂直。
2. 判定两向量共线：
3. 计算向量的模：
4. 计算两向量的夹角：
二阶与三阶行列式
二阶与三阶行列式，可以作为行列式的较为简单的情形特殊定义。在微积分的最后一个部分场论部分，格林公式用到了二阶行列式，高斯公式用到了点乘，斯托克斯公式用到了三阶行列式。
二阶行列式可以视为四元函数，其定义为：
三阶行列式可以视为九元函数，其定义为：
一种特殊的记忆方法是采用「对角线法则」，对角线法则只适用于二阶与三阶行列式。
特别注意：四阶行列式展开后共有 24 项，并且副对角线一项的符号为正。如果强行应用三阶行列式的「对角线法则」，不仅项数不够，副对角线一项的符号也不正确，因此三阶行列式的「对角线法则」不适用于更高阶的行列式，更高阶的行列式也不适合使用直接展开法计算。
外积
外积是 三维向量特有的运算。
在物理学中，三维向量为默认与空间位置相关的向量，一律采用粗体表示。然而，物理学中与相对论相关的四维向量不会采用粗体，而是使用特殊的记号与下标。
在线性代数中，所有的向量都会用粗体表示，并且由于麻烦，并且线性代数中大多为向量与矩阵的运算，很难造成歧义，在手写时可以省略向量记号不写。
定义
外积有不同但等价的定义方法，下面介绍其中一些。
几何定义
在三维欧氏空间 下，定义向量 的外积为一个向量，记为 ，其模与方向定义如下：
；
与 都垂直，且 的方向符合右手法则。
注意到外积的模，联想到三角形面积计算公式 ，可以发现外积的几何意义是： 是以 为邻边的平行四边形的面积。
代数定义
在三维欧氏空间 下，定义向量 的外积为一个向量 ，记作 ，其结果可以使用三阶行列式表示：
其中 表示朝向为坐标轴 的单位向量，并写在对应坐标处。展开得
性质
外积是关于两个向量分别都线性的双线性运算。具体而言，外积满足：
前两行性质亦可称为分配律，即外积对于向量加法满足乘法分配律。
2. 外积满足反交换律，即：
3. 根据上文内积与外积的几何定义：
可以写出恒等式：
4. 外积满足 Jacobi 恒等式：
应用
下面介绍外积运算的一些常见应用。
判定两向量是否共线：
即共线的两个三维向量的外积，结果为 ；三维向量与自身外积，结果为 ；三维向量与零向量外积，结果为 。若使用外积为零作为两向量共线的定义，则可以得出零向量与任何向量都共线。
2. 计算两向量张成的平行四边形面积：
二维向量的情形
对于二维向量，无法计算外积，但是仍然可以计算两向量张成的平行四边形面积：
记 ，将平面直角坐标系扩充为空间直角坐标系，原平面位于新坐标系的 平面，原本的坐标 和 变为 和 。
那么两个向量的外积为 ，因此平行四边形的面积为 ，可以视为二阶行列式运算结果的绝对值。
此时，根据右手法则和 坐标的符号，可以推断出 相对于 的方向，若在逆时针方向则 坐标为正值，反之为负值，简记为 顺负逆正。
混合积
与外积一样，向量的混合积是 三维向量特有的运算。
定义
设 是三维空间中的三个向量，则 称为三个向量 的混合积，记作 或 或 或 。混合积的绝对值 的几何意义表示以 为棱的平行六面体的体积。
向量的混合积可以使用三阶行列式表示：
性质
混合积关于三个向量都分别线性，具体而言，有：
混合积具有反对称性，交换两个向量的位置会使混合积变成其相反数，因此有：
据此还可以得到内积与外积有如下关系：
应用
向量的混合积有如下常见应用。
计算四面体 的体积：
判定 是否共面；
三个三维向量 共面的充分必要条件是 。
3. 判定 构成的坐标系的手性；
混合积 的符号是正还是负，取决于 与 形成的夹角是锐角还是钝角，即指向 与 张成平面的同侧还是异侧，这相当于 三个向量依序构成右手系还是左手系。具体而言：
等价于 依序构成左手系；
等价于 依序构成右手系。
二重外积
三维向量的混合积是内积与外积的混搭，具有轮换对称性。三维向量和三维向量的外积还是三维向量，那么外积的外积是否存在相关结论？
先证明一个引理。
证明：由右手定则， 与 和 都垂直，待证等式左端与 垂直，因此待证等式左端与 和 共面。
因此可以假设：
根据混合积的相关结论，上式两端同时对于 和 分别做内积，有：
由前文推出的恒等式：
可以解得：
证毕。
在上文的证明中提到， 与任意向量叉乘，得到的向量与 和 共面。接下来证明 二重外积 的结论：
上述共面性有助于二重外积结论的记忆。可见，上文的引理为二重外积的特殊情况。
证明：这里只需考虑三个向量均为非零且不共线的情况，其他特例为显然的。
三维向量 ， 和 不共面，因此可以假设：
所以有：
根据上文的引理有：
证毕。
本页面最近更新：2024/9/25 21:52:26，更新历史
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