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0	https://www.slideshare.net/slideshow/5207871-2/250042359	5207871 (2) \| PPT Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Cookie Policy Download free for 30 days Sign in UploadLanguage (EN)Support BusinessMobileSocial MediaMarketingTechnologyArt & PhotosCareerDesignEducationPresentations & Public SpeakingGovernment & NonprofitHealthcareInternetLawLeadership & ManagementAutomotiveEngineeringSoftwareRecruiting & HRRetailSalesServicesScienceSmall Business & EntrepreneurshipFoodEnvironmentEconomy & FinanceData & AnalyticsInvestor RelationsSportsSpiritualNews & PoliticsTravelSelf ImprovementReal EstateEntertainment & HumorHealth & MedicineDevices & HardwareLifestyle Change Language Language English Español Português Français Deutsche Cancel Save Submit search EN Uploaded byAmerWagdyGerges1 PPT, PDF 562 views 5207871 (2) AI-enhanced description The document discusses accounting for leases. It begins by explaining the nature of lease transactions and the advantages of leasing. It then describes the accounting criteria and procedures for classifying leases as capital or operating from the perspective of the lessee. Specifically, it outlines the four criteria that must be met for a lease to be classified as a capital lease by the lessee. The document also provides examples of accounting entries for capital and operating leases. Health & Medicine◦ Read more 0 Save Share Embed Download Download to read offline 1 / 87 2 / 87 3 / 87 4 / 87 5 / 87 6 / 87 7 / 87 8 / 87 9 / 87 10 / 87 11 / 87 12 / 87 13 / 87 14 / 87 15 / 87 16 / 87 17 / 87 18 / 87 19 / 87 20 / 87 21 / 87 22 / 87 23 / 87 24 / 87 25 / 87 26 / 87 27 / 87 28 / 87 29 / 87 30 / 87 31 / 87 32 / 87 33 / 87 34 / 87 35 / 87 36 / 87 37 / 87 38 / 87 39 / 87 40 / 87 41 / 87 42 / 87 43 / 87 44 / 87 45 / 87 46 / 87 47 / 87 48 / 87 49 / 87 50 / 87 51 / 87 52 / 87 53 / 87 54 / 87 55 / 87 56 / 87 57 / 87 58 / 87 59 / 87 60 / 87 61 / 87 62 / 87 63 / 87 64 / 87 65 / 87 66 / 87 67 / 87 68 / 87 69 / 87 70 / 87 71 / 87 72 / 87 73 / 87 74 / 87 75 / 87 76 / 87 77 / 87 78 / 87 79 / 87 80 / 87 81 / 87 82 / 87 83 / 87 84 / 87 85 / 87 86 / 87 87 / 87 ![Image 111: Chapter 21-24 E21-1: Journal entries for Adams through Jan. 1, 2012. LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Depreciation Expense 8,313 Accumulated Depreciation—Capital Leases 8,313 ($41,565 ÷ 5 = $8,313) Interest Expense 3,160 Interest Payable 3,160 ($41,565 – $9,968) X .10] 12/31/11 ]( Ad Recommended PPT Accounting Chapter 10.ppt bymayada34 PPTX Ch11 byESHETIE MEKONENE AMARE PPT Ch13 byhelal682621 PPTX Capital Structure bypremarhea PPTX Chapter3 bykim rae KI PPTX Ias 37 byESHETIE MEKONENE AMARE PDF PDF Kieso Intermediate Accounting chapter 12 bySafa444074 PPTX Cash Flow Statement PPT byWe Learn - A Continuous Learning Forum from Welingkar's Distance Learning Program. 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Explain thenature, economic substance, and advantages of lease transactions. 2. Describe the accounting criteria and procedures for capitalizing leases by the lessee. 3. Contrast the operating and capitalization methods of recording leases. 4. Identify the classifications of leases for the lessor. 5. Describe the lessor’s accounting for direct-financing leases. 6. Identify special features of lease arrangements that cause unique accounting problems. 7. Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. 8. Describe the lessor’s accounting for sales-type leases. 9. List the disclosure requirements for leases. Learning Objectives Chapter 21-4 Leasing Environment Who are players? Advantages of leasing Conceptual natureof a lease Accounting by Lessee Accounting by Lessor Special Accounting Problems Capitalization criteria Accounting differences Capital lease method Operating method Comparison Residual values Sales-type leases Bargain- purchase option Initial direct costs Current versus noncurrent Disclosure Unresolved problems Economics of leasing Classification Direct-financing method Operating method Accounting for Leases Chapter 21-5 Largest group ofleased equipment involves: Information technology Transportation (trucks, aircraft, rail) Construction Agriculture LO 1 Explain the nature, economic substance, and advantages of lease transactions. A lease is a contractual agreement between a lessor and a lessee, that gives the lessee the right to use specific property, owned by the lessor, for a specified period of time. The Leasing Environment Chapter 21-6 Three general categories: Banks. Captiveleasing companies. Independents. LO 1 Explain the nature, economic substance, and advantages of lease transactions. Who Are the Players? The Leasing Environment Chapter 21-7 1. 100% Financingat Fixed Rates. 2. Protection Against Obsolescence. 3. Flexibility. 4. Less Costly Financing. 5. Tax Advantages. 6. Off-Balance-Sheet Financing. The Leasing Environment LO 1 Explain the nature, economic substance, and advantages of lease transactions. Advantages of Leasing Chapter 21-8 Capitalize a leasethat transfers substantially all of the benefits and risks of property ownership, provided the lease is noncancelable. Leases that do not transfer substantially all the benefits and risks of ownership are operating leases. The Leasing Environment LO 1 Explain the nature, economic substance, and advantages of lease transactions. Conceptual Nature of a Lease Chapter 21-9 Operating Lease CapitalLease Journal Entry: Rent expense xxx Cash xxx Journal Entry: Leased equipment xxx Lease liability xxx The issue of how to report leases is the case of substance versus form. Although technically legal title may not pass, the benefits from the use of the property do. A lease that transfers substantially all of the benefits and risks of property ownership should be capitalized (only noncancellable leases may be capitalized). The Leasing Environment LO 1 Explain the nature, economic substance, and advantages of lease transactions. Chapter 21-10 If the lesseecapitalizes a lease, the lessee records an asset and a liability generally equal to the present value of the rental payments. Records depreciation on the leased asset. Treats the lease payments as consisting of interest and principal. Accounting by the Lessee LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Typical Journal Entries for Capitalized Lease Illustration 21-2 Chapter 21-11 To record alease as a capital lease, the lease must be noncancelable. One or more of four criteria must be met: 1. Transfers ownership to the lessee. 2. Contains a bargain-purchase option. 3. Lease term is equal to or greater than 75 percent of the estimated economic life of the leased property. 4. The present value of the minimum lease payments (excluding executory costs) equals or exceeds 90 percent of the fair value of the leased property. Accounting by the Lessee LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Chapter 21-12 LO 2 Describethe accounting criteria and procedures for capitalizing leases by the lessee. Lease Agreement Leases that DO NOT meet any of the four criteria are accounted for as Operating Leases. Accounting by the Lessee Illustration 21-4 Chapter 21-13 Capitalization Criteria LO 2Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Transfer of Ownership Test Not controversial and easily implemented. Bargain-Purchase Option Test At the inception of the lease, the difference between the option price and the expected fair market value must be large enough to make exercise of the option reasonably assured. Chapter 21-14 Capitalization Criteria LO 2Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Economic Life Test (75% Test) Lease term is generally considered to be the fixed, noncancelable term of the lease. Bargain-renewal option can extend this period. At the inception of the lease, the difference between the renewal rental and the expected fair rental must be great enough to make exercise of the option to renew reasonably assured. Chapter 21-15 Recovery of InvestmentTest (90% Test) LO 2 Accounting by the Lessee Minimum Lease Payments:  Minimum rental payment  Guaranteed residual value  Penalty for failure to renew  Bargain-purchase option Executory Costs:  Insurance  Maintenance  Taxes Exclude from PV of Minimum Lease Payment Calculation Capitalization Criteria Chapter 21-16 Accounting by theLessee Discount Rate Lessee computes the present value of the minimum lease payments using its incremental borrowing rate, with one exception.  If the lessee knows the implicit interest rate computed by the lessor and it is less than the lessee’s incremental borrowing rate, then lessee must use the lessor’s rate. Recovery of Investment Test (90% Test) Capitalization Criteria LO 2 Chapter 21-17 Asset and LiabilityRecorded at the lower of: 1. present value of the minimum lease payments (excluding executory costs) or 2. fair-market value of the leased asset. Asset and Liability Accounted for Differently LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Chapter 21-18 LO 2 Describethe accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Depreciation Period If lease transfers ownership, depreciate asset over the economic life of the asset. If lease does not transfer ownership, depreciate over the term of the lease. Asset and Liability Accounted for Differently Chapter 21-19 LO 2 Describethe accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Effective-Interest Method The effective-interest method is used to allocate each lease payment between principal and interest. Asset and Liability Accounted for Differently Depreciation Concept Depreciation and the discharge of the obligation are independent accounting processes. Chapter 21-20 E21-1 (Capital Leasewith Unguaranteed Residual Value): On January 1, 2011, Adams Corporation signed a 5-year noncancelable lease for a machine. The terms of the lease called for Adams to make annual payments of $9,968 at the beginning of each year, starting January 1, 2011. The machine has an estimated useful life of 6 years and a $5,000 unguaranteed residual value. Adams uses the straight-line method of depreciation for all of its plant assets. Adams’s incremental borrowing rate is 10%, and the Lessor’s implicit rate is unknown. LO 2 Accounting by the Lessee Instructions (a) What type of lease is this? Explain. (b) Compute the present value of the minimum lease payments. (c) Prepare all necessary journal entries for Adams for this lease through January 1, 2012. Chapter 21-21 E21-1: What typeof lease is this? Explain. LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Capitalization Criteria: 1. Transfer of ownership 2. Bargain purchase option 3. Lease term => 75% of economic life of leased property 4. Present value of minimum lease payments => 90% of FMV of property NO NO Lease term 5 yrs. Economic life 6 yrs. YES 83.3% FMV of leased property is unknown. Capital Lease, #3 Chapter 21-22 E21-1: Compute presentvalue of the minimum lease payments. LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Payment $ 9,968 Present value factor (i=10%,n=5) 4.16986 PV of minimum lease payments $41,565 Leased Machine Under Capital Leases 41,565 Lease Liability 41,565 Lease Liability 9,968 Cash 9,968 1/1/11 Journal Entries: Chapter 21-23 E21-1: Lease AmortizationSchedule LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee 10% Lease Interest Reduction Lease Date Payment Expense in Liability Liability 1/1/11 41,565 $ 1/1/11 9,968 $ 9,968 $ 31,597 12/31/11 9,968 3,160 6,808 24,789 12/31/12 9,968 2,479 7,489 17,300 12/31/13 9,968 1,730 8,238 9,062 12/31/14 9,968 906 9,062 0 Chapter 21-24 E21-1: Journal entriesfor Adams through Jan. 1, 2012. LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Depreciation Expense 8,313 Accumulated Depreciation—Capital Leases 8,313 ($41,565 ÷ 5 = $8,313) Interest Expense 3,160 Interest Payable 3,160 ($41,565 – $9,968) X .10] 12/31/11 Chapter 21-25 E21-1: Journal entriesfor Adams through Jan. 1, 2012. LO 2 Describe the accounting criteria and procedures for capitalizing leases by the lessee. Accounting by the Lessee Lease Liability 6,808 Interest Payable 3,160 Cash 9,968 1/1/12 Chapter 21-26 LO 3Contrast the operating and capitalization methods of recording leases. Accounting by the Lessee Operating Method The lessee assigns rent to the periods benefiting from the use of the asset and ignores, in the accounting, any commitments to make future payments. Illustration: Assume Adams accounts for it as an operating lease. Adams records this payment on January 1, 2011, as follows. Rent Expense 9,968 Cash 9,968 Chapter 21-27 E21-1: Comparison ofCapital Lease with Operating Lease LO 3 Contrast the operating and capitalization methods of recording leases. Accounting by the Lessee E21-1 Capital Lease Operating Depreciation Interest Lease Date Expense Expense Total Expense Diff. 2011 8,313 $ 3,160 $ 11,473 $ 9,968 $ 1,505 $ 2012 8,313 2,479 10,792 9,968 824 2013 8,313 1,730 10,043 9,968 75 2014 8,313 906 9,219 9,968 (749) 2015 8,313 8,313 9,968 (1,655) 41,565 $ 8,275 $ 49,840 $ 49,840 $ 0 Chapter 21-28 1. Interest Revenue. 2.Tax Incentives. 3. High Residual Value. Accounting by the Lessor Benefits to the Lessor LO 4 Identify the classifications of leases for the lessor. Chapter 21-29 A lessor determinesthe amount of the rental, based on the rate of return needed to justify leasing the asset. If a residual value is involved (whether guaranteed or not), the company would not have to recover as much from the lease payments Economics of Leasing Accounting by the Lessor LO 4 Identify the classifications of leases for the lessor. Chapter 21-30 E21-10 (Computation ofRental): Fieval Leasing Company signs an agreement on January 1, 2010, to lease equipment to Reid Company. The following information relates to this agreement. 1. The term of the noncancelable lease is 6 years with no renewal option. The equipment has an estimated economic life of 6 years. 2. The cost of the asset to the lessor is $343,000. The fair value of the asset at January 1, 2010, is $343,000. 3. The asset will revert to the lessor at the end of the lease term at which time the asset is expected to have a residual value of $61,071, none of which is guaranteed. 4. The agreement requires annual rental payments, beg. Jan. 1, 2010. 5. Collectibility of the lease payments is reasonably predictable. There are no important uncertainties surrounding the amount of costs yet to be incurred by the lessor. Accounting by the Lessor LO 4 Identify the classifications of leases for the lessor. Chapter 21-31 Accounting by theLessor LO 4 Identify the classifications of leases for the lessor. Residual value 61,071 $ PV of single sum (i=10%, n=6) 0.56447 PV of residual value 34,473 $ Fair market value of leased equipment 343,000 $ Present value of residual value (34,473) Amount to be recovered through lease payment 308,527 PV factor of annunity due (i=10%, n=6) 4.79079 Annual payment required 64,400 $ E21-10 (Computation of Rental): Assuming the lessor desires a 10% rate of return on its investment, calculate the amount of the annual rental payment required. ÷ x - Chapter 21-32 a. Operating leases. b.Direct-financing leases. c. Sales-type leases. Classification of Leases by the Lessor Accounting by the Lessor LO 4 Identify the classifications of leases for the lessor. Chapter 21-33 Classification of Leasesby the Lessor Accounting by the Lessor LO 4 Identify the classifications of leases for the lessor. A sales-type lease involves a manufacturer’s or dealer’s profit, and a direct-financing lease does not. Illustration 21-10 Chapter 21-34 Classification of Leasesby the Lessor Accounting by the Lessor LO 4 Identify the classifications of leases for the lessor. A lessor may classify a lease as an operating lease but the lessee may classify the same lease as a capital lease. Illustration 21-11 Chapter 21-35 In substance thefinancing of an asset purchase by the lessee. Direct-Financing Method (Lessor) Accounting by the Lessor LO 5 Describe the lessor’s accounting for direct-financing leases. Chapter 21-36 Accounting by theLessor E21-10: Prepare an amortization schedule that would be suitable for the lessor. LO 5 Describe the lessor’s accounting for direct-financing leases. Chapter 21-37 Accounting by theLessor E21-10: Prepare all of the journal entries for the lessor for 2010 and 2011. LO 5 Describe the lessor’s accounting for direct-financing leases. 1/1/10 Lease Receivable 343,000 Equipment 343,000 1/1/10 Cash 64,400 Lease Receivable 64,400 12/31/10 Interest Receivable 27,860 Interest Revenue 27,860 Chapter 21-38 Accounting by theLessor E21-10: Prepare all of the journal entries for the lessor for 2010 and 2011. LO 5 Describe the lessor’s accounting for direct-financing leases. 1/1/11 Cash 64,400 Lease Receivable 36,540 Interest Receivable 27,860 12/31/11 Interest Receivable 24,206 Interest Revenue 24,206 Chapter 21-39 Records each rentalreceipt as rental revenue. Depreciates the leased asset in the normal manner. Operating Method (Lessor) Accounting by the Lessor LO 5 Describe the lessor’s accounting for direct-financing leases. Chapter 21-40 Illustration: Assume Fievalaccounts for the lease as an operating lease. It records the cash rental receipt as follows: Operating Method (Lessor) Accounting by the Lessor LO 5 Describe the lessor’s accounting for direct-financing leases. Cash 64,400 Rental Revenue 64,400 Depreciation is recorded as follows: Depreciation Expense 57,167 Accumulated Depreciation 57,167 $343,000 / 6 years = 57,167 Chapter 21-41 1. Residual values. 2.Sales-type leases (lessor). 3. Bargain-purchase options. 4. Initial direct costs. 5. Current versus noncurrent classification. 6. Disclosure. Special Accounting Problems LO 6 Identify special features of lease arrangements that cause unique accounting problems. Chapter 21-42 Meaning of ResidualValue - Estimated fair value of the leased asset at the end of the lease term. Guaranteed Residual Value – Lessee agrees to make up any deficiency below a stated amount that the lessor realizes in residual value at the end of the lease term. Residual Values Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-43 Lessee Accounting forResidual Value The accounting consequence is that the minimum lease payments, include the guaranteed residual value but excludes the unguaranteed residual value. Residual Values Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-44 Illustration (Guaranteed ResidualValue – Lessee Accounting): Caterpillar Financial Services Corp. (a subsidiary of Caterpillar) and Sterling Construction Corp. sign a lease agreement dated January 1, 2011, that calls for Caterpillar to lease a front-end loader to Sterling beginning January 1, 2011. The terms and provisions of the lease agreement, and other pertinent data, are as follows. The term of the lease is five years. The lease agreement is noncancelable, requiring equal rental payments at the beginning of each year (annuity due basis). The loader has a fair value at the inception of the lease of $100,000, an estimated economic life of five years, and no residual value. Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-45 Illustration (Guaranteed ResidualValue – Lessee Accounting): Sterling pays all of the executory costs directly to third parties except for the property taxes of $2,000 per year, which is included as part of its annual payments to Caterpillar. The lease contains no renewal options. The loader reverts to Caterpillar at the termination of the lease. Sterling’s incremental borrowing rate is 11 percent per year. Sterling depreciates on a straight-line basis. Caterpillar sets the annual rental to earn a rate of return on its investment of 10 percent per year; Sterling knows this fact. Caterpillar estimates a residual value of $5,000 a the end of the lease. Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-46 Illustration (Guaranteed ResidualValue – Lessee Accounting): Caterpillar would compute the amount of the lease payments as follows: Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Illustration 21-16 NOTE: For the Lessee, the minimum lease payment includes the guaranteed residual value but excludes the unguaranteed residual value. Solution on notes page Chapter 21-47 Illustration 21-17 Illustration (GuaranteedResidual Value – Lessee Accounting): Computation of Lessee’s capitalized amount Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Solution on notes page Chapter 21-48 Illustration (Guaranteed ResidualValue – Lessee Accounting): Computation of Lease Amortization Schedule Illustration 21-18 Special Accounting Problems LO 7 Chapter 21-49 Illustration (Guaranteed ResidualValue – Lessee Accounting): At the end of the lease term, before the lessee transfers the asset to Caterpillar, the lease asset and liability accounts have the following balances. Illustration 21-19 Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-50 Illustration (Guaranteed ResidualValue – Lessee Accounting): Assume that Sterling depreciated the leased asset down to its residual value of $5,000 but that the fair market value of the residual value at December 31, 2015, was $3,000. Sterling would make the following journal entry. Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Loss on Capital Lease 2,000.00 Interest Expense (or Interest Payable) 454.76 Lease Liability 4,545.24 Accumulated Depreciation—Capital Leases 95,000.00 Leased Equipment under Capital Leases 100,000.00 Cash 2,000.00 Chapter 21-51 Illustration (Unguaranteed ResidualValue – Lessee Accounting): Assume the same facts as those above except that the $5,000 residual value is unguaranteed instead of guaranteed. Caterpillar would compute the amount of the lease payments as follows: Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Illustration 21-20 Solution on notes page Chapter 21-52 Illustration (Unguaranteed ResidualValue – Lessee Accounting): Computation of Lease Amortization Schedule Illustration 21-21 Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-53 Illustration (Unguaranteed ResidualValue – Lessee Accounting): At the end of the lease term, before Sterling transfers the asset to Caterpillar, the lease asset and liability accounts have the following balances. Illustration 21-22 Special Accounting Problems LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Chapter 21-54 Special Accounting Problems Illustration21-23 Comparative Entries, Lessee Company Chapter 21-55 Special Accounting Problems Illustration:Assume a direct-financing lease with a residual value (either guaranteed or unguaranteed) of $5,000. Caterpillar determines the payments as follows. LO 7 Describe the effect of residual values, guaranteed and unguaranteed, on lease accounting. Lessor Accounting for Residual Value The lessor works on the assumption that it will realize the residual value at the end of the lease term whether guaranteed or unguaranteed. Illustration 21-24 Chapter 21-56 Special Accounting Problems LessorAccounting for Residual Value Illustration: Lease Amortization Schedule, for Lessor— Guaranteed or Unguaranteed Residual Value Illustration 21-25 Chapter 21-57 LO 7 Describethe effect of residual values, guaranteed and unguaranteed, on lease accounting. Special Accounting Problems Lessor Accounting for Residual Value Illustration: Caterpillar would make the following entries for this direct-financing lease in the first year. Illustration 21-26 Solution on notes page Chapter 21-58 Primary difference betweena direct-financing lease and a sales-type lease is the manufacturer’s or dealer’s gross profit (or loss). Lessor records the sale price of the asset, the cost of goods sold and related inventory reduction, and the lease receivable. Difference in accounting for guaranteed and unguaranteed residual values. Sales-Type Leases (Lessor) Special Accounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Chapter 21-59 Sales-Type Leases (Lessor) SpecialAccounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Illustration: To illustrate a sales-type lease with a guaranteed residual value and with an unguaranteed residual value, assume the same facts as in the preceding direct- financing lease situation. The estimated residual value is $5,000 (the present value of which is $3,104.60), and the leased equipment has an $85,000 cost to the dealer, Caterpillar. Assume that the fair market value of the residual value is $3,000 at the end of the lease term. Chapter 21-60 Sales-Type Leases (Lessor) SpecialAccounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Illustration: Computation of Lease Amounts by Caterpillar Financial—Sales-Type Lease Illustration 21-28 Chapter 21-61 Sales-Type Leases (Lessor) SpecialAccounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Illustration: Caterpillar makes the following entries to record this transaction on January 1, 2011, and the receipt of the residual value at the end of the lease term. Illustration 21-29 Chapter 21-62 Sales-Type Leases (Lessor) SpecialAccounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Illustration: Caterpillar makes the following entries to record this transaction on January 1, 2011, and the receipt of the residual value at the end of the lease term. Illustration 21-29 (January 1, 2012) Chapter 21-63 Sales-Type Leases (Lessor) SpecialAccounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Illustration: Caterpillar makes the following entries to record this transaction on January 1, 2011, and the receipt of the residual value at the end of the lease term. Illustration 21-29 Chapter 21-64 Present value ofthe minimum lease payments must include the present value of the option. Only difference between the accounting treatment for a bargain-purchase option and a guaranteed residual value of identical amounts is in the computation of the annual depreciation. Bargain Purchase Option (Lessee) Special Accounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Chapter 21-65 The accounting forinitial direct costs: For operating leases, the lessor should defer initial direct costs. For sales-type leases, the lessor expenses the initial direct costs. For a direct-financing lease, the lessor adds initial direct costs to the net investment. Initial Direct Costs (Lessor) Special Accounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Chapter 21-66 FASB Statement No.13 does not indicate how to measure the current and noncurrent amounts. It requires that for the lessee the “obligations shall be separately identified on the balance sheet as obligations under capital leases and shall be subject to the same considerations as other obligations in classifying them with current and noncurrent liabilities in classified balance sheets.” Current versus Noncurrent Special Accounting Problems LO 8 Describe the lessor’s accounting for sales-type leases. Chapter 21-67 1. General descriptionof the nature of the lease. 2. Nature, timing and amount of cash inflows and outflows associated with leases, including payments for each of the five succeeding years. 3. Amount of lease revenues and expenses reported in the income statement each period. 4. Description and amounts of leased assets by major balance sheet classification and related liabilities. 5. Amounts receivable and unearned revenues under lease. Disclosing Lease Data Special Accounting Problems LO 9 List the disclosure requirements for leases. Chapter 21-68  Leasing wason the FASB’s initial agenda in 1973 and SFAS No. 13 was issued in 1976 (before the conceptual framework was developed). SFAS No. 13 has been the subject of more than 30 interpretations since its issuance.  The iGAAP leasing standard is IAS 17, first issued in 1982. This standard is the subject of only three interpretations. One reason for this small number of interpretations is that iGAAP does not specifically address a number of leasing transactions that are covered by U.S. GAAP. Examples include lease agreements for natural resources, sale-leasebacks, real estate leases, and leveraged leases. Chapter 21-69  Both U.S.GAAP and iGAAP share the same objective of recording leases by lessees and lessors according to their economic substance—that is, according to the definitions of assets and liabilities.  U.S. GAAP for leases in much more “rule-based” with specific bright-line criteria to determine if a lease arrangement transfers the risks and rewards of ownership; iGAAP is more general in its provisions. Chapter 21-70 LO 10 Chapter 21-71 LO 10Understand and apply lease accounting concepts to various lease arrangements. Solution on notes page Illustration 21A-2 Chapter 21-72 Solution on notes page Chapter 21-73 Illustration 21A-3 LO 10Understand and apply lease accounting concepts to various lease arrangements. Chapter 21-74 LO 10Understand and apply lease accounting concepts to various lease arrangements. Chapter 21-75 Illustration 21A-4 LO 10Understand and apply lease accounting concepts to various lease arrangements. Chapter 21-76 LO 10Understand and apply lease accounting concepts to various lease arrangements. Chapter 21-77 Illustration 21A-5 LO 10Understand and apply lease accounting concepts to various lease arrangements. Chapter 21-78 LO 11Describe the lessee’s accounting for sale-leaseback transactions. The term sale-leaseback describes a transaction in which the owner of the property (seller-lessee) sells the property to another and simultaneously leases it back from the new owner. Advantages: 1. May allow seller to refinance at lower rates. 2. May provide another source of working capital, particularly when liquidity is tight. 3. By selling the property, the seller-lessee may deduct the entire lease payment, which is not subject to alternative minimum tax considerations. Chapter 21-79 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Determining Asset Use To the extent the seller-lessee continues to use the asset after the sale, the sale-leaseback is really a form of financing.  Lessor should not recognize a gain or loss on the transaction. If the seller-lessee gives up the right to the use of the asset, the transaction is in substance a sale.  Gain or loss recognition is appropriate. Chapter 21-80 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Lessee If the lease meets one of the four criteria for treatment as a capital lease, the seller-lessee should  Account for the transaction as a sale and the lease as a capital lease.  Defer any profit or loss it experiences from the sale of the assets that are leased back under a capital lease.  Amortize profit over the lease term . Chapter 21-81 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Lessee If none of the capital lease criteria are satisfied, the seller-lessee accounts for the transaction as a sale and the lease as an operating lease.  Lessee defers such profit or loss and amortizes it in proportion to the rental payments over the period when it expects to use the assets. Exceptions:  Losses Recognized and Minor Leaseback Chapter 21-82 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Lessor If the lease meets one of the criteria in Group I and both of the criteria in Group II, the purchaser-lessor records the transaction as a purchase and a direct- financing lease. If the lease does not meet the criteria, the purchaser- lessor records the transaction as a purchase and an operating lease. Chapter 21-83 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Sale-Leaseback Example American Airlines on January 1, 2011, sells a used Boeing 757 having a carrying amount on its books of $75,500,000 to CitiCapital for $80,000,000. American immediately leases the aircraft back under the following conditions: 1. The term of the lease is 15 years, noncancelable, and requires equal rental payments of $10,487,443 at the beginning of each year. 2. The aircraft has a fair value of $80,000,000 on January 1, 2011, and an estimated economic life of 15 years. 3. American pays all executory costs. 4. American depreciates similar aircraft that it owns on a straight-line basis over 15 years. 5. The annual payments assure the lessor a 12 percent return. 6. American’s incremental borrowing rate is 12 percent. Chapter 21-84 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Sale-Leaseback Example This lease is a capital lease to American because  lease term exceeds 75 percent of the estimated life of the aircraft and  present value of the lease payments exceeds 90 percent of the fair value of the aircraft to CitiCapital. Assuming that collectibility of the lease payments is reasonably predictable and that no important uncertainties exist in relation to unreimbursable costs yet to be incurred by CitiCapital, it should classify this lease as a direct-financing lease. Chapter 21-85 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Sale-Leaseback Example Illustration 21B-1 Chapter 21-86 LO 11Describe the lessee’s accounting for sale-leaseback transactions. Sale-Leaseback Example Chapter 21-87 Copyright © 2009John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Copyright Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. 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1	https://math.fudan.edu.cn/_upload/article/files/8a/c7/b6bc60b64933acee041a351adda9/d0bb8522-a61d-4d6c-ab50-0051ba33a403.pdf	教案平面和直线教学内容平面和直线是几何学中最基本的研究对象，是一些向量空间和几何空间中某些对象的最基本原型。由于曲线在局部可以用它的切线来近似，曲面在局部可以用它的切平面来近似，所以平面和直线也是几何和分析中“以直代曲”的最基本元素。因此学习空间解析几何中处理平面与直线的方法非常重要，而且是必须要掌握的数学工具。在本节中主要讲解以下几方面的内容：（1）平面和直线的代数表示，即它们的方程的形式如何？以及如何用向量的运算方法来建立这些方程；（2）平面与平面，平面与直线、直线与直线，以及点与这些对象的位置关系，如距离、夹角等，以及如何利用代数方法来处理。教学思路和要求（1）首先讲解平面的方程、直线的方程、以及建立这些方程的方法。（2）讲解点与平面的距离、点与直线的距离公式的建立和运用。（3）讲解平面与平面，平面与直线、直线与直线的夹角概念与计算方法；（4）进一步，利用这些工具讲解一些具有某些特性的直线或平面的方程的具体计算方法。教学安排一．平面方程的几种形式在 3 R 中，给定了与平面垂直的方向和平面上的一个点，就可以唯一决定这个平面。与平面垂直的方向称为这个平面的法向量。设所求平面的法向量为 ) , , ( C B A n ，而且平面过点 0 P ( 0 0 0 , , z y x )。对平面上的任何一点 ) , , ( z y x P ，P 与 0 P 的连线依然在平面上，因而 ) , , ( 0 0 0 0 z z y y x x P P     与n 垂直，即 n P P 0 = 0。用分量表示，就是 0 ) ( ) ( ) ( 0 0 0       z z C y y B x x A ，这个关系式称为平面的点法式方程。记常数 ) ( 0 0 0 Cz By Ax D     ，则上述方程可以写成 0     D Cz By Ax ，这个关系式称为平面的一般方程。例6.2.1 求过原点 ) 0 , 0 , 0 ( O 和点 ) 2 , 3 , 6 (  P ，且与平面 8 2 4    z y x 垂直的平面方程。解记所求平面为。因为过原点 ) 0 , 0 , 0 ( O 和点 ) 2 , 3 , 6 (  P ，所以其法向量n应与 ) 2 , 3 , 6 (   OP 垂直。又垂直于平面 8 2 4    z y x ，所以n应与向量 z n P0 P  O y x 图6.2.1 ) 2 , 1 , 4 ( 1   n 垂直。故可取 k j i k j i n n 6 4 4 2 1 4 2 3 6 1         OP 。利用平面的点法式方程，所求平面方程为 0 6 4 4     z y x ，即 0 3 2 2    z y x 。确定平面的另一类条件是，不在一条直线上的3 个点唯一决定一张平面。设平面所过的3 个点为 0 P ( 0 0 0 , , z y x )， ) , , ( 1 1 1 1 z y x P ， ) , , ( 2 2 2 2 z y x P ，因此 1 0P P 和 2 0P P 与平面的法向量n 垂直，即可以取法向量 2 0 1 0 P P P P   n ，设P(x，y，z)是平面上的任何一点，则有 0 ) ( 0 2 0 1 0 0      P P P P P P P P n 。这称为平面的三点式方程。容易看出，它正好就是四点共面的条件 0 0 2 0 2 0 2 0 1 0 1 0 1 0 0 0           z z y y x x z z y y x x z z y y x x 。由行列式的性质，上式展开后一定是 0     D Cz By Ax 形式，因此实际计算时不必死记公式，可以将 0 P ， 1 P ， 2 P 的坐标直接代入平面的一般式方程，用待定系数法解出A，B，C，D。特别地，将平面所过的3 个点取为过坐标轴的点 0 P (a，0，0)， 1 P (0，b，0 )， 2 P (0，0，c )，代入三点式方程，就有 c a b a z y a x 0 0    = 0，展开整理后，得（1）当a、b、c 均不为0 时，平面方程为 1    c z b y a x ，它称为平面的截距式方程（见图6.2.2），其中a，b，c 依次称为该平面在x，y， z 轴上的截距。此时平面的法向为        c b a 1 , 1 , 1 n 。（2）当a，b，c 中只有一个为0 时，所决定的平面是坐标平面， . 0 ; 0 ; 0 0 0 0       z y x Oxy c Ozx b Oyz a z (0,0, c ) n O y (0,b,0) (a,0,0) x 图6.2.2 例6.2.2 求过点 ) 1 , 2 , 3 ( ， ) 0 , 1 , 0 ( ， ) 2 , 0 , 1 ( 的平面方程。解将三点的坐标代入平面的一般方程 0     D Cz By Ax ，得到关于A、B、C、D 的方程组                . 0 2 , 0 , 0 2 3 D C A D B D C B A 它的一组解为A= 3，B = -7，C= -2，D =7（此方程组有无穷多个解，只取一个解就可以了），于是所求的平面方程为 3x -7y -2z + 7 = 0。二．直线方程的几种形式与平面类似，要确定空间中的一条直线，主要条件也有两类。一类是确定直线的方向和直线上的一个点，另一类是确定直线上的两个点。设直线L 的方向向量为 ) , , ( n m l v ，它过点 0 P ( 0 0 0 , , z y x )。于是，直线L 上任何一点 ) , , ( z y x P 与 0 P 的连线与v 平行，即 v // 0P P （见图6.2.3），按分量写开，就是 n z z m y y l x x 0 0 0      ，它称为直线的对称式方程或点向式方程。注意，若l，m，n 中有等于0 的，例如，当 0  l ， 0 ,  n m 时，则应将上述方程理解为          . , 0 0 0 n z z m y y x x 当 0  m l ， 0  n 时，则应将上述方程理解为      . , 0 0 y y x x 例6.2.3 求过点 ) 4 , 1 , 2 (  ，方向向量为 ) 1 , 1 , 3 (  的直线方程。解直接代入直线的对称式方程，便得所求的直线方程为 1 4 1 1 3 2       z y x 。若给定了直线上的两个点 0 P ( 0 0 0 , , z y x )和 ) , , ( 1 1 1 1 z y x P ，则 1 0P P 的方向就是v 的方向向量（见图6.2.4），代入直线的对称式方程，即得到直线的两点式方程 z v P0 O P y x 图6.2.3 z v P1 P0 O y P x 图6.2.4 0 1 0 0 1 0 0 1 0 z z z z y y y y x x x x         。若在直线的对称式方程中记 t n z z m y y l x x       0 0 0 ，将等式写开，便得到            , , , 0 0 0 n t z z m t y y l t x x 它称为直线的参数方程，其中t 是参数。参数方程对于求解某些具体问题很有效。例6.2.4 求直线 2 4 1 3 1 2      z y x 与平面 0 6 2     z y x 的交点。解这就是求方程 2 4 1 3 1 2      z y x 与 0 6 2     z y x 的公共解。将直线方程写成参数方程            , 2 4 , 3 , 2 t z t y t x 其中t 是参数。代入平面的方程，便得到 2 (2+ t) + (3+ t) + (4 + 2 t) - 6 = 0，解得t = -1。代入直线的参数方程，得到 2 , 2 , 1    z y x 。即，交点为(1, 2, 2)。另外，如果给定了空间中两张互不平行的平面 1 ： 0 1 1 1 1     D z C y B x A 和 2 ： 0 2 2 2 2     D z C y B x A ，那么这两张平面交于一条直线。也就是说，这两个平面的联立方程            0 , 0 2 2 2 2 1 1 1 1 D z C y B x A D z C y B x A 同样表示一条直线，它称为直线的一般方程。直线的一般方程看起来不太直观，用起来有时也不太方便。由于 1 的法向量为 1 n ( 1 1 1 , , C B A )， 2 的法向量为 2 n ( 2 2 2 , , C B A )，由立体几何知识，直线的方向向量v 与 1 n 和 2 n 都垂直（见图6.2.5），因此，可以取   1 n v 2 n ，再从联立方程中求出一组解( 0 0 0 , , z y x )，也就是直线上的一个定点的坐标，这样就可以将它化成对称式方程了。例6.2.5 将直线的一般方程           0 4 2 , 0 1 2 z x z y x 化成对称式方程。解取直线的方向向量为 v n1 n2 2 1 图6.2.5    2 1 n n v 2 0 1 1 1 2  k j i = k j i  5 2 ，再在平面上任意取一个公共点，如令 0 0  x ，代入方程，         0 4 2 , 0 1 z z y 则可以解出 3 0   y ， 2 0   z 。于是，直线的对称式方程为 1 2 5 3 2       z y x 。例6.2.6 求过点 ) 2 , 0 , 0 ( 0  M ，与直线 1 L ： 1 2 3 4 1 z y x      相交，而且平行于平面 1 ： 0 1 2 3     z y x 的直线方程。解设所求直线为L 。其方向向量为 ) , , ( Z Y X v 。显然 ) 0 , 3 , 1 ( 1 M 是直线 1 L 上的点， ) 1 , 2 , 4 ( 1  v 是 1 L 的方向向量。由于L 过点 ) 2 , 0 , 0 ( 0  M ，且与直线 1 L 相交，因此向量 1 0M M ， 1 v 和v 共面，因此 0 ) ( 1 1 0    v v M M ，这就是说 0 1 2 4 ) 2 ( 0 0 3 0 1       Z Y X ，即 0 2    Z Y X 。又因为L 与平面 1 平行，所以 ) , , ( Z Y X v 与 1 的法向量 ) 2 , 1 , 3 (  n 垂直，因此 0  n v ，即 0 2 3    Z Y X 。联立上述各方程可解得 Z Y X 2 , 0   ，取 1  Z 得 2  Y 。因此直线L 的方程为 1 2 2 0    z y x ，即       . 0 , 4 2 x z y 本例也可先求过 0 M 且平行于 1 的平面 2 的方程，再求 2 与直线 1 L 的交点 1 M ，最后求出过 0 M 和 1 M 的直线方程。读者不妨自行计算。三．平面束空间直线L 的一般方程为            . 0 , 0 2 2 2 2 1 1 1 1 D z C y B x A D z C y B x A 对于任意一组不同时为零的常数  , ，方程 0 ) ( ) ( 2 2 2 2 1 1 1 1         D z C y B x A D z C y B x A   表示一张平面   。显然，满足L 的一般方程的点 ) , , ( z y x 一定满足平面   的方程，所以平面   通过直线L 。于是，对于不同的不同时为零的数对  , ， 0 ) ( ) ( 2 2 2 2 1 1 1 1         D z C y B x A D z C y B x A   就确定了一族通过L 的平面，它称为通过L 的平面的平面束，以上方程也称为通过L 的平面束方程。显然确定平面束中一张平面，只要确定与的比值，因此也常将通过L 的平面束方程写成 0 ) ( 2 2 2 2 1 1 1 1         D z C y B x A k D z C y B x A （注意这个束中不包含平面 0 2 2 2 2     D z C y B x A ），或 0 ) ( 2 2 2 2 1 1 1 1         D z C y B x A D z C y B x A k （注意这个束中不包含平面 0 1 1 1 1     D z C y B x A ）。另外，方程 0      Cz By Ax 确定一张平面  ，而当取不同值时，就得到一族相互平行的平面的方程。因此上式也称为平行平面束方程。例6.2.7 过点 ) 1 , 1 , 1 ( 和直线L ：            0 5 3 2 , 0 2 2 3 z y x z y x 的平面方程。解设所求通过L 的平面方程为 0 ) 5 3 2 ( 2 2 3         z y x k z y x 。它通过 ) 1 , 1 , 1 ( 点,所以将该点的坐标代入上式得 0 3 6  k 。所以 2  k 。于是，所求的平面方程为 0 ) 5 3 2 ( 2 2 2 3         z y x z y x ，即 0 8 8 5 5     z y x 。四．点到平面、直线的距离平面解析几何中讨论了某一平面上的点到直线的距离问题，现在我们将它推广到空间。先考虑点到平面的距离。设平面方程为 0     D Cz By Ax 。过空间一已知点 ) z , , ( y x P 作平面的垂线，显然，垂线的方向就是平面的法向量 ) , , ( C B A n 。取平面上一定点 0 P ( 0 0 0 , , z y x )，联结P 与 0 P ，则 ) z , , ( y x P 到平面的距离d 就是 P P 0 在n 方向的投影长度（见图6.2.6）。记平面的单位法向量 \|\| \|\| 0 n n n  ，由内积的定义，得 \| \| 0 0 n   P P d ， n P d P0 图6.2.6 用分量表示，就是 2 2 2 0 0 0 \| ) ( ) ( ) ( \| C B A z z C y y B x x A d         ，由于 ) ( 0 0 0 Cz By Ax D     ，因此 d = 2 2 2 \| \| C B A D Cz By Ax      ，这就是点 ) z , , ( y x P 到平面 0     D Cz By Ax 的距离的计算公式。当 ) z , , ( y x P 在该平面上时，显然有d = 0 。例6.2.8 求点 ) 1 , 1 , 2 (  与平面2x -3y -6z + 1 = 0 的距离。解这时 ) 1 , 1 , 2 (   P ， ) 6 , 3 , 2 ( ) , , (    C B A ，由点到平面距离的计算公式得 d = 2 2 2 ) 6 ( ) 3 ( 2 \| 1 ) 1 ( ) 6 ( 1 ) 3 ( 2 2 \|              = 7 8 。再考虑点到直线的距离。设直线L 的方程为 n z z m y y l x x 0 0 0      ，连接空间一已知点 ) z , , ( y x P 和直线L 上的点 ) , , ( 0 0 0 0 z y x P 。直线的方向向量为 ) , , ( n m l v ，因此单位方向向量为  0 v \|\| \|\| v v 。由图6.2.7 中可以看出，点P 到直线的距离d 是以 P P 0 和 0 v 为邻边的平行四边形的底边 0 v 上的高，由外积的几何意义，图6.2.7 中的平行四边形的面积 \|\| \|\| 0 v d S  \|\| \|\| 0 0 v   P P ，于是点 ) z , , ( y x P 到直线的距离公式为 \|\| \|\| 0 0 v   P P d 。例 6.2.9 求点 ) 3 , 2 , 5 (  与直线 3 2 1 1 1 z y x     的距离。解这时 ) 3 , 2 , 5 (   P ， ) 0 , 1 , 1 ( 0   P ，则 P P 0 = ) 3 , 1 , 4 ( ) 0 , 1 , 1 ( ) 3 , 2 , 5 (      。而 ) 3 , 2 , 1 (  v ，   \|\| \|\| 0 v v v 14 1 ) 3 , 2 , 1 ( ，因此   0 0 v P P 14 1 3 2 1 3 1 4  k j i 14 9   (1, 1, -1)，所以距离为 \|\|    \|\| \|\| 0 0 v P P d 14 3 9 。 P d P0 v0 图6.2.7 五．交角一．平面与平面的交角空间中两张平面的交角就是它们的法向量的交角（通常取 2 0    ）。因此，设两张平面的方程为 0 1 1 1 1     D z C y B x A 和 0 2 2 2 2     D z C y B x A ，那么它们的交角就是它们的法向量 ) , , ( 1 1 1 1 C B A n 和 ) , , ( 2 2 2 2 C B A n 的交角或补角，即  cos    \|\| \|\| \|\| \|\| \| \| 2 1 2 1 n n n n 2 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 \| \| C B A C B A C C B B A A        。特别地，（1）当 0 2 1 2 1 2 1    C C B B A A 时，这两张平面垂直；（2）当 2 1 2 1 2 1 2 1 D D C C B B A A    时，这两张平面平行；（3）当 2 1 2 1 2 1 2 1 D D C C B B A A    时，这两张平面重合。例6.2.10 求平面 0 3 2     z y x 和 0 1 2     z y x 的夹角。解此时 ) 2 , 1 , 1 ( 1   n ， ) 1 , 1 , 2 ( 2  n ，于是由平面与平面交角余弦计算公式得 2 1 1 1 2 2 ) 1 ( 1 \| 1 2 1 ) 1 ( 2 1 \| cos 2 2 2 2 2 2                ，因此，这两张平面的夹角为 3   。二．直线与直线的交角与平面与平面情况类似，空间中两条直线的交角就是它们的方向向量的交角 （通常取 2 0    ）。设两条直线的方程为 1 1 1 1 1 1 n z z m y y l x x      和 2 2 2 2 2 2 n z z m y y l x x      ，那么它们的交角（0≤≤2 ）满足 2 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 \| \| cos n m l n m l n n m m l l          。特别地，（1）当 0 2 1 2 1 2 1    n n m m l l 时，这两条直线垂直；（2）当 2 1 2 1 2 1 n n m m l l   时，这两条直线平行；（3）若 2 1 2 1 2 1 n n m m l l   ，且两条直线有一个公共点，则这两条直线重合。例6.2.11 求直线 1 1 4 2 1 1       z y x 和直线          0 2 2 , 0 1 z y y x 的夹角。解直线 1 1 4 2 1 1       z y x 的方向向量可取为 ) 1 , 4 , 1 ( 1   v ；直线          0 2 2 , 0 1 z y y x 的方向向量可取为 k j i k j i v       2 2 2 1 0 0 1 1 2 。利用直线与直线交角余弦计算公式得   cos 2 2 2 2 2 2 1 2 ) 2 ( 1 ) 4 ( 1 \| 1 1 2 ) 4 ( ) 2 ( 1 \|               = 2 1 ，因此，这两条直线的夹角为= 4 。三．平面与直线的交角直线与平面的交角是直线与它在平面上的垂直投影所夹的角（通常取 2 0    ，见图6.2.8）。设平面方程为 A x+ B y+ C z + D = 0，直线方程为 n z z m y y l x x 0 0 0      ，那么它们的交角（ 2 0    ）满足 2 2 2 2 2 2 \| \| \| cos \| sin n m l C B A Cn Bm l A            。特别地，（1）当 n C m B l A   时，平面与直线垂直；（2）当 0    Cn Bm Al 时，平面与直线平行；（3）若 0    Cn Bm Al ，且平面与直线有一个公共点，则直线属于平面。例6.2.12 求平面 0 3     z y x 和直线 4 3 1 2 3 2       z y x 的位置关系。解由于给定直线与平面的夹角的正弦为 0 ) 4 ( 1 3 1 1 1 \| ) 4 ( 1 1 1 3 1 \| sin 2 2 2 2 2 2                ， n v   图6.2.8 所以平面和直线平行。在直线 4 3 1 2 3 2       z y x 上任取一点，可以就取为 ) 3 , 2 , 2 (  ，它也满足平面方程 0 3     z y x ，所以 ) 3 , 2 , 2 (  也在平面上。因此，直线在平面上。六．习题 1，3．（1）、（3），4．（2），5．（1）、（3），6，7，9，11，13，15．（1）、（2）、（3）， 16。
2	https://www.scribd.com/document/704706131/Environmental-Engineering-by-Peavyrowe-Pdfdrive	Environmental Engineering by Peavyrowe Pdfdrive \| PDF \| Sewage Treatment \| Secondary Sector Of The Economy Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 5K views 214 pages Environmental Engineering by Peavyrowe Pdfdrive This document provides an overview of the McGraw-Hill textbook "Environmental Engineering" by Howard S. Peavy, Donald R. Rowe, and George Tchobanoglous. The textbook covers topics related to… Full description Uploaded by shakil1800151 AI-enhanced description Go to previous items Go to next items Download Save Save Environmental Engineering by Peavyrowe Pdfdrive For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Environmental Engineering by Peavyrowe Pdfdrive For Later You are on page 1/ 214 Search Fullscreen McGraw-Hill Series in Water Resources and Environmental Engineering Rolf Eliassen, Paul H. King, a nd Ray K. Linsley Consulting Editors Bailey and Ollis: Biochemical Engineering Fundamentals Bisho p: Marine Pollution and Its Control Biswa s: Model sfor Water Quality Managemeni Boc kra th: Environmental Law for Engineers. Scientisls, and Managers Bouw er: Gro {ndwater Hydrol ogy Canter: En vironmental Impact Assessment Chanlett: Environmental Pr ote ction Gaudy and Gaudy: Microbiolog yfor Environmental Scientists and Engineers Haimes: Hi erar ch ical Analysis of Water Resources Systems: Modelling and Optimization of Large-Scale System s Hall and Dracup: Wat er Resources Systems Engineering Linsley and Franzini: Water Resources Engineering Lins ley, Kohle r, an d P aulhus: Hydrology for Engineers Metcalf & Eddy, Inc.: Wastewater Engineering.' Collec t ion and Pumping of Wastewater Metcalf & Eddy, Inc.: Was tewater Engineering. Trealment, Disposal. Reuse Peavy, Rowe, and Tch oban oglou s: Environmental Engineering Ric h: Low-Maint enanc e, Mechan ically-Simpl e Wa ste water Treatment Systems Sawyer and McCarty : Chemistry for Environmental Engineering Steel and McGhe e: Water Supply and Sewerage Tcho bano glou s, The isen, a nd Eliassen: Solid Wastes, Engineering Principles and Management Issues ENVIRONMENTAL ENGINEERING Howard S. Peavy Professor of Civil Engineering Montana Scate University Donald R. Rowe Professor of Civil Engineering King Saud UniverSity Saudi Arabia George Tchobanoglous Professor of Civil Engineering Universi ty of Califo rnia, Davis McGraw-Hill Book C ompany New York St. Louis San Francisco Auckland Bogota H amburg London Madrid Mexico Montreal New Dehli Panama Paris Siio Paulo Singapore Sydney Tokyo To r onto adDownload to read ad-free ENVIRONMENTAL INTERNATIONAL EDITION 1985 Exclusive rights by McGraw-Hili Book Co., Singapor e for manufacture and export. This book ca nnot be r e-export ed from the country to which it is con s ign ed by McGraw -HilI. 40 39 38 37 36 35 34 33 32 31 15 14 13 12 11 CTP BJE Copyright © 1985 by Mc Graw-Hili, In c. All rights reserved. No part of this pub lica tio n m ay be re produ ced or dis tribut ed in any form or by any means, or stored in a data base or a retrieval sy stem, without the prior written perm ission of the publisher. This book was set in Tim es Roman. The editors were Kiran Verma and David A. Damstra. The production supe rvisor was Leroy A. Youn g. ... Libra./},.()f.c;:()1" 9.ress in Publication Data Peavy, Howard S. Env iro nmen t al e ngin ee rin g. (MC G raw-Hili se ries in water res our ces and enviro n mental engineering) I nclud es biblio graphi cal references and indexes. Environmental engin eeri ng. I. Rowe, Donald R. II. Tchobanoglou s, Geor ge . II I. Title. IV. Se ri e s. TD145.P43 1985 628 8 4- 3854 ISBN 0-07-04 91 34-8 When ordering this title use ISBN 0-07 -1 00231-6 Printed in Singapore CONTENTS Preface Intrn clu C i on I-I ·The Em 'i r on lll c nt 1-2 T he Im[lact 01 Il ulllam upon the Enviro nm ent 1-3 The Im[lact or th e En\ironm ent upon Hum ans 1- 4 I mpr o\ 'c ment of Environme ntal Qua li ty 1- 5 The Rolc of the Envir on men t al E ng in eer R eI- ercnces Part 1 Wat e r 2 Water Qua lit y: Def ini ti ons, C har a deri s tic s; and Persp ect ive s 2 -1 The Hydraul ic Cycle and Water Qual ity PH Y SICAL WATER- QUA LITY PARAME TER S 2-2 Su spended S,1 Iid , 2- 3 Turbi dit y 2-4 Color 2-5 Ta ste and O do l 2-6 T emper ature C HE M I CAL \ATE R -QUA LI TY PARAME TER S 2- 7 Che mi s tn of S"lutions' 2-8 Total Di ss oh cd Solid, 2-9 Alkali nit\ 2· 10 Har dness :2-1 1 Fl uo ri de 2· 12 Metals 2 -1 ] Org a nH :" 2- 14 Nutllc nh XI/ I 2 4 6 7 I I 12 1 4 15 17 1 8 20 22 23 23 28 3 1 3 (, 37 38 -1 4 adDownload to read ad-free vi CONTENTS BIOLOGICAL WATER-QU-\UTY PARAMETERS 2-15 Pathogens 2-16 Pathogen Indicators WATER QUALITY REQUIREMENTS 2-17 In-Stream Standards 2-18 Potable Water Standards 2-19 Wastewater Effluent Standards Discussion Topics and Problem s References . 3 Water Purification Processes in Natural Systems PHYSICAL PROCESSES 3-1 Dilution 3-2 Sedimentation and Resuspension 3-3 Filtration 3-4 Gas Transfer 3-5 Heat Transfer CHEMICAL PROCESSES 3-6 Chemical Conversions BIOCHEMICAL PROCESSES 3-7 Metabolic Processes 3-8 Microorganisms in Natural Wat er Systems RESPONSE OF STREM .. S TO BIODEGRADABLE. ORGANIC WASTE 3-9 Dissolved-Oxygen Balance 3-10 Dissolved-Oxygen Model 3-.11 Organic Discharge and Stream Ecology APPLICATION OF NATURAL .pROCESSES IN ENGINEERED SYSTEMS 3-12 Physical Processes 3- i3 Chemical Processes 3-14 Biological Processes Di s cu ss ion T o pics and Problem s Refere.nc es 4 Engineered Systems for Water Purification 4-1· Historical Overview of Water Treatment 4-2 Water-Treatment Processes 4-3 4-4 4 -5 4 -(, 4 -7 WA TER-TRFA 1M ENT PRO CES SE S. THEORY AND .APPLICATION Aeraiillil Solids Se paration \j)\ - \IJ) Settlin g Operatioll s Cllag ulati o ll S'\ - \G1 o ft e nin g 46 47 50 54 54 55 56 57 62 63 64 64 65 66 66 71 73 73 74 75 79 83 ....... .... ?;l. _ ...... . .. ... ..... . g5 94 104 105 107 109 110 11 3 In 1.1 1 151 ·4-8 Filtration '4-9 Di si nf ec ti on OTHER WATER-TREATMENT PROCESSES 4-10 Dissolved Solids Removal DiSCUSSIOn Topics and Problems References 5 Engineered Systems for Wastewater Treatment and Disposal 5 -1 Wastewater Characteristi cs 5-2 Effluent Standards 5-3 Terminology in Wastewater Treatment PRIMARY TREATMENT 5-4 Screening 5-5 Comminuting 5-6 Grit Removal 5-7 Flow Measurement 5-8 Primary Sedimentation SECONDARY TREATMENT 5-9 Growth and Food Utilization 5-10 Suspended-Culture System s 5·11 Activated Sludge 5-12 Ponds and Lagoons 5·13 Attached-Culture Systems 5-14 Secondary Clarification CONTENTS vii , t· 165 182 190 190 '104 . 207 208 211 212 2Pl 217 220 221 224 224 . Effiw:ots ..................................................... . 229 230 234 234 248 255 268 277 278 279 281 285 292 SLUDGE TREATMENT AND DISPOSAL 5-16 Sludge Characteristics 5-17 Sludge Thickening 5-18 Sludge Digestion 5-19 Sludge Disposal ADVANCED WASTEWATER TREATMENT 5 -20 Nutrient Removal 5- 21 Solids Removal WASTEWATER DISPOSAL AND REUSE 5-22 Wastewater Di s posal 5-23 Wastewater Reuse Discussion Topics and Problems References 6 Environmental Engineering Hydraulics Design WATER DISTRIBUTION SYSTEMS 6-1 Method s of Distributing Water 6-2 Di s tribut ion Re serv oir s 6 -3 Di s tribution Systems 294 295 :' 301 302 303 306 .314 322 324 324 324 326 331 adDownload to read ad-free -: - - """' '""""' -----.., v ii CONTENT S 6-4 6-5 <-6 6-8 6-9 6-10 6-11 < 6-12 6-13 6-14 6-15 6-16 6-17 6-18 6-19 6-20 6-21 6-22 6-23 6-24 6-25 6-26 6-27 Distribution System Components Capacity and Pres s ure Requirement s Design of Distribution Systems Hydraulic Analy sis of Di s tribution Systems Cross-Connections in Di s tr ib ution Systems Constructi on of Water Di s tribution Systems Pumping Re quired for Water Supply Syst ems WASTEWATER CO LLEC TION Types of Co llection Systems Types of Se wer s Collection System Appurten ances Basic Consideration s in th e De sign of Sewers Design of Sanitar y Sewers Preparation of Contact Drawings and Spe cif ica ti ons Construction of S ewe rs Maintenan ce of Sewers Design of Storm wa ter Sewers WATER AND WASTEWATER PUMPIN G Pump s Pump Drive Unit s Pump Application Te rminology and Usag e Pump Operating C haracte ristics and C ur ves Analysis of Pump S ys tem s Pump Stations for Water and Wa s tewater HYDRAULIC ANALYSIS OF WATER AN D WASTEWATER TREATMENT Treatment Plant De si gn Preparation of Hydraulic FFOfiks< <" < <" << " " <" 0< < -" , ---.. - .. ------.. - Discussion Topic s and Problem s References Part 2 . Air '7 Air Quality : Definitions, Characteristics , and Perspectives AIR POLLUTION -PAS T, PRESENT , AND FUTURE 7-1 Historical Overvi e>.' 7-2 Global Impl i ca tions of Air P ollut i on 7-3 Units of Measur em ent 7-4 Sources of poJlutan ts CLASSIFICATION OF POLLUTA N TS 7-5 Particul ates 7-6 H y droc a rb ons 7-7 Carbon Monoxide 7 -8 Oxides o f Sulfur 333 335 33 7 338 346 347 348 348 3 49 3 49 3 53 3 54 360 369 371 37 1 37 1 372 372 375 378 384 3')0 397 397 397 ,--399, 406 412 417 41 8 418 420 424 426 429 431 442 445 44 9 7-9 Ox ides of N itr ogen 7-10 Phot ochemi cal Oxidant s 7 -1 1 I nd oo r Air Pollut i on 8 8-1 go: i)-} 8 -4 8-5 8-6 8-7 8-8 8-9 8 -1 0 8 -11 A IR- QUA LIT Y MA N AGEMENT C ONCEPTS Discussion To pics a nd Problems References Me t eoro logy and Natural Pur ifi cat i on Pr ocesses EL EME NT AL PR OPERT IES Of THE ATMOSP H ERE Scales of i\ l otill n Heat Press ur e Wlncl MOis tur e Rclative H umidit y IN FLUE N CE OF i\ I LTEOROLOG IC AL PHENOMENA ON AIR QUALITY Lapse Rat es and DI '> persion Press ur e Sy s tems and Dispersion Wind s and Dispersio n M OiS tur e and Dispersion Mo d el ing EFFECTS or AIR P OLLUT ION ON METEOROLOGICAL COND ITIONS 8-12 Change s on the M esos cale and Mic r os cale __ . __ 13 _ _ 0 .11 _ he_ M aC .f Qs eale Discussion Topics a nd P robl ems Referenc es 9 Engineered S ys tems for Air Pollution Control 9-1 Atm o sph eric Cle a nsing Processes 9-2 Approa ches to Co ntaminant Cont rol CON TR OL DEVICES rOR PARTICULATE CONTAM INA NTS 9 -3 Gra vit ati onal Se l1 ing Ch a mb ers 9-4 Ce ntr ifu gal 9 -5 Wct Colkctnr , '-)- (, Fa hr ic Fi lt er, (BaglHllls e Fi lte rs) 9-7 Electr os tat ic PrccipiL lt () fS ( ESP )' CONT I WL DEV I CES FOR GASEOUS CON T AM I NANTS ') -8 Ad so rpti(lll 9-9 ,\ O Sorptll )Jl ,-) · 10 Conlicns "tloll 9- 11 (' omo u '>1lt)n , \utt .lllIl1 t l\r C ONTENTS ix 455 461 463 464 477 480 483 483 484 486 491 493 495 495 495 496 498 498 499· 499 507 508 509 510 5 12 5 14 5 14 516 518 520 5 23 5 28 53 3 53 6 5 40 540 545 557 559 56J adDownload to read ad-free r CONTENTS xi x CONTENTS , 11-10 Dete rmination of Vehicle and Labor Requirements 607 Discu ssion Topic s and Problems 565 II-II Collection Routes 615 References 567 TRANSFER AND TRANSPORT 618 r 11-12 Transfer Sta tio n s 620 11-13 Location of Tr ansfer Station s 622 ... Part 3 Solid Waste 11-14 Transfe r Means and Method s 622 PROCESSING TECHNIQUES 626 t 10 Solid Waste: Defin itio ns, Char acteri stics, and r I-IS Mechan ical Volume Reducti on 627 Perspectives 573 11 - 16 Thermal Volume Redu c ti on 627 TYPES OF SOLID WASTES 573 11. - 17 Manual Co mp onent Se par a ti on 627 r 10-1 Municipal Wastes 574 ULTIMATE DISPOSAL 628 10-2 Indu strial Wa stes 574 11 -18 Landfi ll ing with Solid Wa stes 628 10- 3 Hazard ous Waste s 5 75 11-19 De sign and Operation of Landfills 638 SOUR C ES OF SOLID WASTES 5 75 11- 20 Landf arming 646 10-4 Sour ces of Municipal Wastes 575 11-21 Deep-Well Injection 647 10-5 Source s of Hazardou s Wastes 576 Di scussion Topic s and Pro bl ems 648 Refer ence s 652 PROPERTIES OF SOLID WASTES 576 10-6 Ph ysical C ompo sition 576 10- 7 Chemical Co mpo sition 582 10-8 Chan ges in Compo sition 5 88 . 12 Engineered Systems for Resou rce and SOLID-WASTE MANAGEMENT AN O VERVIEW 588 Energy Recovery 653 10-9 Materials Flow in S ociet y 5 88 10-10 Reduction in Raw Mat e ri als Usage 58 9 PRO C ESSING TECHNIQUES 653 10-11 Re ducti on in Sol id-W as te Qu antit ie s 590 12- i Mechanical Size Alter ati on 65.4 10-12 Reuse of Solid-Wa ste Mat erials 5 90 12 -2 Mechanical Compon ent Separation 656 10-13 .. I'vlil.t el: iil.l. s . .... 5 91 12-3 Ma gne ti c and Ele ctromec ha ni cal Sep ara tion 656-·-.. 10-14 En ergy Recovery 592 12-4 Drying and Dewatering 657 10-15 Day-to-Day Solid- Wa ste Managem ent 592 MA TERIALS-RECOVER Y SYSTEMS 657 Discu ssion Topics and Pro blems 5 92 Materials Speci ficati on s 657 Refer ences 59 3 1 2- 5 12 -6 Proce ssing and Rec ov er y Sy" stems 657 )2 -7 System De sign Layout 659 11 Engineered Systems for Solid-Waste Management 5 94 RE C OVERY OF BIOLOGICAL CONVERSION P·RODUCTS 659 II-I Functional Elements 5 94 12-8 Compo sting (Aerobic Conv ersion) 660 SOLID WASTE GENERATION 594 17.-9 Ana erobic Digestion .663 11-2 Typical Gene ration Rat es 595 RECOVERY OF THERMAL CONVERSION PRODUCTS 665 11-3 Es timat ion of Solid-Wa ste Qu antit ies 598 12 -10 Co mbu stion of Wa ste Mat erials 665 11-4 Fact ors Th at Affect Genera ti on Rat cs 5 98 12 -11 In cine ration with Heat Re cove ry 670 ON-SITE HANDLING. STORAG E. 12-12 Use of R efus e-D e ri v ed Fuel s (RDF) 671 t\ N D PRO CESS IN G 598 1 2- 13 Ga sification 671 11 -5 On -Site Handlin g 59 9 Pyrol ys is 672 11 -6 On- S it e Storage 599 1 2- 14 RE C OVERY OF ENE RGY FROM CO N VERSION PRODUCTS 672 11 -7 On -S it e Process in g of So lid Wastes 6 01 1 2- I 5 Ener gy -Reco ve ry S ys tems 673 C OLLE C TIO N OF SOLID \Vt\ ST ES 6 01 674 11 -8 Co ll ec ti on Services 6 01 1 2- 16 Efficiency Fact or s 675 11 -9 Types of Co ll ec ti o ll Syst cm, 1i05 1 2- 17 Dete rrnin ati on of t.ner g} Output and EtJi cicncy adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free 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3	https://www.youtube.com/watch?v=rBmLMUVNlEo	Cross Price Affect and Effect of Substitute Goods \| Microeconomics Consumer Equilibrium and Demand Scholarslearning Classes 30200 subscribers 38 likes Description 2466 views Posted: 24 Mar 2017 Class 12 Microeconomics-Consumer Equilibrium and Demand by Parul Madan Transcript: in this video we will understand the cross right effect now what is the cross price effect see the effect of change in the price of related goods on the quantity demanded of a commodity the new types are related with number one India substitute good what I am constitute so your substitute goods are those who that can be replaced for one another means your tea or coffee this little rf4 mine water these are into water gram right deep adversity could you can help you to lead on right people who do not happy what is in power to purchase week's loss is costlier than the low quality diet at all since these 24 K T right right so in these these are we no substitutes you will substitute we turn right into the substitute T or coffee you can substitute this layer of equipment to these are no particles these can be replaced for one another the shaking hands in your complimentary me you require the one hood to complete the other good one good complement or complete the other word let's take an example of this pink is been your lap top and motherhood your full octave are an engine so in order to date this would be required this is good complemented without this at work and animal vacuum about eight paint anonymous have to do without motherhood lack of cannot even capture so these are your compliment Rickles in this video mean understand detective substitute words or be quantity demanded in the next you another Santa complimentary bananas if that's not the quantity demanded substitute goods and it effect so let us take an example of tea and coffee here are some stupid we will be the effect of quantity demanded of tea do no change in the Kaiser call see e is rupees 10 per 100 okay now the price of coffee initial was rupees 10 but now you can some factors has increased to 315 per standard car when Li try the coffee was Ruby ten per hundred now this quantity demanded of tea was less than 150 units for one country like now the price of coffee has increased so if this type of copies increase the quantity demanded of casa might have decreased so where those people have shifted they might have shifted some coffee to tea so ultimately quantity demanded off he would have increased let's schedule $300 not plot this on the graph this is legal right this is the quantity we are taking the quantity of TN right Roxy T remains constant that is till 2 feet of our clamp as Wendy try the formula 10 repeat before Sigma of T 1 then take it on 50 100 150 200 250 300 115 300 the initial demand 50 was now the democracy has increased this is increased due to increase in the price of the substitute saddle pocket so this is how the substitute words that is coffee effective see through a quantity demanded of tea so the increase in this price of one commodity increases the quantity demanded of other commodity that they are being subsidy see we understood that with the rise in the price of the top to bottom topography how does it affect tea it's finally the market increases now maybe I understand it will decrease in the price of the substitute is coffee let's say the coffee prices decrease to reach 8 100 ground the users of P will shift there a man to copy the this quantity demanded reduces Clemson 102 degrees rise and decreasing beside us coffee let's see effect to graphically familiar quantity of tea this is the friend of t see the price of steel constant what is the affecting its demand the price of the its substitute right so the price of the party constant 1,500 150 when you find a copy 110 the quantity demanded for 200 now the coffee prices we do so people have shifted for me to coffee so ultimately reducing the demand of T this is the initiative Markinson and the price of the scale now there is no him the shifted marker there is decrease in the quantity month right so this is the effect of degrees in defense of the city to your commodity
4	https://www.reddit.com/r/AskPhysics/comments/1h3agfa/is_earths_rotation_slowing_down_gradually/	Is earths rotation slowing down gradually? : r/AskPhysics Skip to main contentIs earths rotation slowing down gradually? : r/AskPhysics Open menu Open navigationGo to Reddit Home r/AskPhysics A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to AskPhysics r/AskPhysics r/AskPhysics 1.5M Members Online •9 mo. ago PuddingWise3116 Is earths rotation slowing down gradually? I understand that the motion is what prevents the earth from falling directly into the sun. I also understand that Earth moves through a vacuum, so there is very little resistance. Are there there any other forces at play? I am also wondering whether electromagnetism affects earths kinetic energy in any way since moving electric charges (I perhaps naively assume earths electric charge isn't exactly 0) should lose energy. EDIT: I want to clarify that by rotation, I meant orbit around the sun. I apologise for not being clearer about that. Read more Share Related Answers Section Related Answers Understanding Earth's rotation and its forces Factors affecting Earth's rotation speed How does time dilation affect GPS satellites? Explaining black holes in simple terms How does gravity affect light? New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of November 30, 2024 Reddit reReddit: Top posts of November 2024 Reddit reReddit: Top posts of 2024 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
5	https://www.longdom.org/articles/qpochhammer-symbol-a-versatile-mathematical-tool-with-farreaching-applications-101786.html	q-Pochhammer Symbol: A Versatile Mathematical Tool with Far-Reach Home Journals A-Z JournalsBrowse By Subject Guidelines & Policies Editorial PoliciesOnline SubmissionInstructions to AuthorsPoliciesPublication ethics & malpractice statementReviewersTerms and Conditions AdvertisingConferencesContact us Powered by Translate Mathematica Eterna Open Access ISSN: 1314-3344 Journal HomeEditorial Board Instructions for Authors Instructions for AuthorsPublication ethics & malpractice statement Submit ManuscriptAims and ScopeArticles in processArchiveContact Awards Nomination25+ Million Readerbase Google Scholar citation report Citations : 1892 Mathematica Eterna received 1892 citations as per Google Scholar report Indexed In Google Scholar Gdansk University of Technology, Ministry Points 5 View More » Useful Links Covid-19 Journal Articles IssuesAim and ScopePeer Review ProcessOther CommentsAdvertisingCitations ReportIndexing and ArchivingTable of ContentsSubmit PaperTrack Your PaperFunded Work Share This Page ##### 25+ Million Website Visitors Journal Flyer Open Access Journals Agri and AquacultureBiochemistryBioinformatics & Systems BiologyBusiness & ManagementChemistryClinical SciencesEngineeringFood & NutritionGeneral ScienceGenetics & Molecular BiologyImmunology & MicrobiologyMedical SciencesNeuroscience & PsychologyNursing & Health CarePharmaceutical Sciences Commentary Article - (2023)Volume 13, Issue 2 View PDFDownload PDF q-Pochhammer Symbol: A Versatile Mathematical Tool with Far-Reaching Applications Stepen MarcCorrespondence: Stepen Marc, Department of Mathematics, University of Orléans, Orléans, France, Email: Author info » Description The q-pochhammer symbol, also known as the q-shifted factorial or the q-rising factorial, is a mathematical concept that finds its applications in diverse areas such as number theory, combinatorics, and quantum algebra. the symbol is named after the german mathematician h. pochhammer, who made significant contributions to the field. the q-pochhammer symbol generalizes the conventional pochhammer symbol, introducing a parameter q that adds an additional level of complexity and versatility to its applications. The q-pochhammer symbol is denoted as (a;q)n, where a and q are parameters, and n is a non-negative integer. (a;q)n = (1-a)(1-aq)(1-aq^2)...(1-aq^(n-1)) Where a and q are parameters, and n is a non-negative integer. It is important to note that the value of n determines the number of factors in the product. The q-pochhammer symbol arises in the theory of q-series, which is a generalization of power series where the terms are multiplied by certain factors involving the parameter q . these series have proven to be fundamental in the study of various mathematical objects, such as partitions, partitions with restricted parts, and mock theta functions. One of the key properties of the q-pochhammer symbol is its connection to basic hyper geometric series. A basic hyper geometric series, also known as a q-series, is a series that generalizes the classical hyper geometric series. The q-pochhammer symbol can be used to express basic hyper geometric series in a compact and elegant form. The q-pochhammer symbol also exhibits many interesting properties and identities. one such property is the q-binomial theorem, which states that for any complex number a and parameter q : (1-a)^{-1} = (a;q){_infty} Where (a;q){_infty} is the infinite product representation of the q-pochhammer symbol. This result provides a powerful tool for manipulating q-series and has applications in various branches of mathematics, such as number theory and combinatorics. Furthermore, the q-pochhammer symbol satisfies a q-analog of the addition theorem for binomial coefficients. this q-analog, known as the q-vandermonde identity, states that for any complex numbers a ,b , and parameter q : (a+b;q)n= _sum_{k= 0}^n(a;q)k(b;q){n-k}(q;q) _k Where (q;q) _k is the q-pochhammer symbol with a-1 and b=q^k . The q-pochhammer symbol also plays a fundamental role in the theory of quantum groups and quantum algebra. Quantum groups are noncommutative analogs of lie groups and lie algebras, and they arise naturally in the study of certain physical phenomena, such as quantum mechanics and statistical mechanics. the q-pochhammer symbol appears in the representation theory of quantum groups and provides a framework for constructing and classifying irreducible representations. The q-pochhammer symbol is a powerful mathematical tool with applications in diverse areas of mathematics, including number theory, combinatorics, and quantum algebra. Its connection to basic hyper geometric series, properties such as the q-binomial theorem and the q-vandermonde identity, and its role in the theory of quantum groups highlight its importance in various mathematical disciplines. The study of the q-pochhammer symbol continues to be an active area of research, with ongoing investigations into its properties and applications, ensuring its relevance in the future development of mathematics. Author Info Stepen Marc Department of Mathematics, University of Orléans, Orléans, France Citation: Marc S (2023) q-Pochhammer Symbol: A Versatile Mathematical Tool with Far-Reaching Applications. Math Eterna.13: 183. Received:22-May-2023, Manuscript No. ME- 23-24472; Editor assigned:25-May-2023, Pre QC No. ME- 23-24472 (PQ); Reviewed:09-Jun-2023, QC No. ME- 23-24472; Revised:16-Jun-2023, Manuscript No. ME- 23-24472 (R); Published: 23-Jun-2023 , DOI: 10.35248/1314-3344.23.13.183 Copyright:© 2023 Marc S. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited Content Links Tools Feedback Careers Privacy Policy Terms & Conditions Authors, Reviewers & Editors Contact Longdom Longdom Group SA Avenue Roger Vandendriessche, 18, 1150 Brussels, Belgium Phone: +442038085340 Email:info@longdom.org Connect FacebookLinkedinTwitterInstagram Copyright © 2025 Longdom Publishing. Top Original text Rate this translation Your feedback will be used to help improve Google Translate
6	https://www.nature.com/articles/s41415-022-4280-0	Osteoradionecrosis of the jaw: a window of opportunity \| British Dental Journal Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles Reviews & Analysis News & Comment Current issue Collections Follow us on Facebook Follow us on Twitter Subscribe Sign up for alerts RSS feed About the journal About the journal Journal Information BDJ Portfolio Call for Papers Open access publishing About the Editors About the BDA CPD BDJ Books For Advertisers Jobs Contact Publish with us Publish with us For Authors For Referees Language editing services Open access funding Submit manuscript Subscribe Sign up for alerts RSS feed nature british dental journal research insight article Research Insight Published: 13 May 2022 Osteoradionecrosis of the jaw: a window of opportunity Sven Butzin1 British Dental Journalvolume 232,page 628 (2022)Cite this article 400 Accesses 1 Citations Metrics details Access through your institution Buy or subscribe An audit into the timing of dental extractions pre-head and neck radiotherapy and the prevalence of osteoradionecrosis. Br Dent J 2022; To investigate the relative timings of dental extractions compared to the RCS guidelines, Ward et al. conducted an audit including 154 patients who received RTX for HNC at Queen Alexandra Hospital in Portsmouth between July 2016 and August 2017. The results revealed that only 125 (81.2%) patients were dentally assessed before the commencement of RTX; nevertheless 30.5% more compared to 2017. Patients who were not dentally assessed were mainly edentulous patients or those undergoing palliative treatment. In total, 102 patients (81.6%) required dental extractions which were carried out within a mean time of 5.7 days. Only one patient in the investigated cohort breached the recommended ten-day healing period due to urgency of treatment. In fact, 77.5% of cases requiring extractions were allowed more than 21 days for healing, and overall, only two patients developed ORNJ (1.3% ORNJ rate); one despite a healing time of 23 days and the other following extractions during RTX after the patient had initially refused pre-RTX dental treatment. This is a preview of subscription content, access via your institution Relevant articles Open Access articles citing this article. Valero’s host index is useful in predicting radiation-induced trismus and osteoradionecrosis of the jaw risks in locally advanced nasopharyngeal carcinoma patients Erkan Topkan , Efsun Somay …Ugur Selek BMC Cancer Open Access 12 July 2023 Access options Access through your institution Subscribe to this journal Receive 24 print issues and online access $259.00 per year only $10.79 per issue Learn more Buy this article Purchase on SpringerLink Instant access to full article PDF Buy now Prices may be subject to local taxes which are calculated during checkout Additional access options: Log in Learn about institutional subscriptions Read our FAQs Contact customer support Author information Authors and Affiliations Final Year Dental Student and Medical Doctor, University of Central Lancashire, UK Sven Butzin Authors 1. Sven ButzinView author publications Search author on:PubMedGoogle Scholar Rights and permissions Reprints and permissions About this article Cite this article Butzin, S. Osteoradionecrosis of the jaw: a window of opportunity. Br Dent J232, 628 (2022). Download citation Published: 13 May 2022 Issue Date: 13 May 2022 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative This article is cited by Valero’s host index is useful in predicting radiation-induced trismus and osteoradionecrosis of the jaw risks in locally advanced nasopharyngeal carcinoma patients Erkan Topkan Efsun Somay Ugur Selek BMC Cancer (2023) Access through your institution Buy or subscribe Associated content Collection Research Insights Sections Relevant articles Author information Rights and permissions About this article This article is cited by Advertisement British Dental Journal (Br Dent J) ISSN 1476-5373 (online) ISSN 0007-0610 (print) nature.com sitemap About Nature Portfolio About us Press releases Press office Contact us Discover content Journals A-Z Articles by subject protocols.io Nature Index Publishing policies Nature portfolio policies Open access Author & Researcher services Reprints & permissions Research data Language editing Scientific editing Nature Masterclasses Research Solutions Libraries & institutions Librarian service & tools Librarian portal Open research Recommend to library Advertising & partnerships Advertising Partnerships & Services Media kits Branded content Professional development Nature Awards Nature Careers Nature Conferences Regional websites Nature Africa Nature China Nature India Nature Japan Nature Middle East Privacy Policy Use of cookies Your privacy choices/Manage cookies Legal notice Accessibility statement Terms & Conditions Your US state privacy rights © 2025 Springer Nature Limited
7	https://artofproblemsolving.com/wiki/index.php/Summation?srsltid=AfmBOooZZZ88EH_7Pc8GcypHBKSe-SCFLipvLp5mKV1_15Cum7L3Lk29	Art of Problem Solving Summation - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Summation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Summation A summation is the sum of a number of terms (addends). Summations are often written using sigma notation . Contents [hide] 1 Definition 2 Identities 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Definition For , and a set (or any other algebraic structure), . Here refers to the index of summation, is the lower bound, and is the upper bound. As an example, . Note that if , then the sum is . Quite often, sigma notation is used in a slightly different format to denote certain sums. For example, refers to a cyclic sum, and refers to all subsets which are in . Usually, the bottom of the sigma contains a logical condition, as in , where the sum only includes the terms which divide into . Identities , and in general , and in general Or See PaperMath’s sums if you want to look deeper into these identities Problems Introductory Evaluate the following sums: Intermediate The nine horizontal and nine vertical lines on an checkerboard form rectangles, of which are squares. The number can be written in the form where and are relatively prime positive integers. Find (Source) Olympiad See Also Cyclic sum Symmetric sum This article is a stub. Help us out by expanding it. Retrieved from " Categories: Definition Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
8	https://math.stackexchange.com/questions/340744/what-do-the-symbols-d-dx-and-dy-dx-mean	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams What do the symbols d/dx and dy/dx mean? Ask Question Asked Modified 3 years, 9 months ago Viewed 188k times 31 $\begingroup$ Okay this may sound stupid but I need a little help... What do $\Large \frac{d}{dx}$ and $\Large \frac{dy}{dx}$ mean? I need a thorough explanation. Thanks. calculus notation Share edited Jun 27, 2019 at 16:30 Asaf Karagila♦ 407k4848 gold badges646646 silver badges1.1k1.1k bronze badges asked Mar 25, 2013 at 15:42 KamiKami 47511 gold badge44 silver badges55 bronze badges $\endgroup$ 1 3 $\begingroup$ This is not precalculus, it is calculus. These symbols are derivatives. Are you familiar with derivatives?\ $\endgroup$ Alex Becker – Alex Becker 2013-03-25 15:44:00 +00:00 Commented Mar 25, 2013 at 15:44 Add a comment \| 5 Answers 5 Reset to default 28 $\begingroup$ The symbol $$ \frac{dy}{dx} $$ means the derivative of $y$ with respect to $x$. If $y = f(x)$ is a function of $x$, then the symbol is defined as $$ \frac{dy}{dx} = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}. $$ and this is is (again) called the derivative of $y$ or the derivative of $f$. Note that it again is a function of $x$ in this case. Note that we do not here define this as $dy$ divided by $dx$. On their own $dy$ and $dx$ don't have any meaning (here). We take $\frac{dy}{dx}$ as a symbol on its own that can't be slit up into parts. The symbol $$ \frac{d}{dx} $$ you can consider as an operator. You can apply this operator to a (differentiable) function. And you get a new function. So if $f$ is a (differentiable) function that it makes sense to "apply" $\frac{d}{dx}$ to $f$ and write $$ \frac{d}{dx}f $$ If you write $y = f(x)$, then this is the same as $$ \frac{d}{dx}y = \frac{dy}{dx}. $$ Share edited Mar 25, 2013 at 15:55 answered Mar 25, 2013 at 15:49 ThomasThomas 44.6k1313 gold badges7676 silver badges139139 bronze badges $\endgroup$ 5 10 $\begingroup$ The confusion often arises from the fact that many writers call $dy/dx$ a "symbol" as if it were atomic, but then later start doing algebra with it. This leads to the question, "well, then what is $dy$ really?" $\endgroup$ Fixee – Fixee 2013-03-25 15:55:55 +00:00 Commented Mar 25, 2013 at 15:55 2 $\begingroup$ @Fixee: That is right. I often see this confusion which is why I always emphasize that $\frac{dy}{dx}$ is just a symbol. It is not a fraction. $\endgroup$ Thomas – Thomas 2013-03-25 15:57:40 +00:00 Commented Mar 25, 2013 at 15:57 $\begingroup$ @Thomas It's kinda a fraction. It's the limit of a fraction. $\endgroup$ Akiva Weinberger – Akiva Weinberger 2015-05-01 21:35:01 +00:00 Commented May 1, 2015 at 21:35 $\begingroup$ (Or the "standard part" of a fraction, if you're doing nonstandard analysis...) $\endgroup$ Akiva Weinberger – Akiva Weinberger 2015-05-01 21:35:50 +00:00 Commented May 1, 2015 at 21:35 $\begingroup$ @Fixee See my answer. It is nevertheless CAN be defined as a fraction of two functions, rather than an atomic object. $\endgroup$ porton – porton 2015-05-01 21:59:02 +00:00 Commented May 1, 2015 at 21:59 Add a comment \| 8 $\begingroup$ $\frac{d}{dx}$ means differentiate with respect to $x$. $\frac{dy}{dx}$ means differentiate $y$ with respect to $x$. Do you have any concrete examples for which you need to calculate these two? It would probably make it more easy to grasp for you if I could explain it in a few examples. Share edited Jun 27, 2019 at 4:53 nmasanta 9,6902525 gold badges2929 silver badges5858 bronze badges answered Mar 25, 2013 at 15:44 dreamerdreamer 3,4591010 gold badges5252 silver badges7979 bronze badges $\endgroup$ 1 $\begingroup$ Maybe calculating the flow rate of a tap based on how the container gets filled? $\endgroup$ The Empty String Photographer – The Empty String Photographer 2023-03-23 07:57:46 +00:00 Commented Mar 23, 2023 at 7:57 Add a comment \| 3 $\begingroup$ $d f$ means the differential of function $f$. By definition $(df)(x) = \lambda t\in\mathbb{R}:f'(x)\cdot t$. In other words, differential is the linear function (of an additional variable denoted $t$ here) whose tangent is the derivative of $f$. $d$ alone means the differential operator (a function of argument $f$). Exercise: Show that $\frac{df}{dx}=f'$. Share edited May 1, 2015 at 21:30 answered May 1, 2015 at 21:12 portonporton 5,23511 gold badge2020 silver badges4949 bronze badges $\endgroup$ 2 $\begingroup$ Note that $\frac{F}{G}$ is defined for two functions $F$ and $G$ as $\frac{F}{G}(x)=\frac{F(x)}{G(x)}$. Thus $\frac{df}{dx}$ makes sense. $\endgroup$ porton – porton 2015-05-01 21:16:52 +00:00 Commented May 1, 2015 at 21:16 $\begingroup$ It is described in en.wikipedia.org/wiki/Differential_of_a_function $\endgroup$ porton – porton 2015-05-02 19:33:53 +00:00 Commented May 2, 2015 at 19:33 Add a comment \| 1 $\begingroup$ I like to look at it this way: $dx$ and $dy$ are just representations of change in accordance to either $x$ or $y$ axis. If you take the the symbol for derivative $$\frac{dy}{dx}$$ and compare it to the formula for the slope: $$\frac{f(x_1) - f(x_2)}{x_1 - x_2}$$ we can clearly see that $dy$ and $dx$ depict change in $y$ and change in $x$ respectively. Share answered Dec 28, 2021 at 15:02 StanleyStanley 5399 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ If $y=f(x)$ i.e., where $y$ is the equation ( the dependent variable) and $x$ is the independent variable. Meaning $x$ changes $y$. Now $\frac{dy}{dx}$ means differentiate the equation $y$ in respect to $x$. $\frac{d}{dx}$ means differentiate in respect to $x$. Same way $\log x$ means find the natural logarithm of $x$, $\frac{d}{dx} x$ means find the derivative of $x$. N.B. In an equation $k= h²+5 $, $\frac{dk}{dh}$ means differentiate the equation $k$ in respect to $h$. Its not always $\frac{dy}{dx}$. I hope you understand Share edited Jun 27, 2019 at 4:40 nmasanta 9,6902525 gold badges2929 silver badges5858 bronze badges answered Jun 27, 2019 at 3:36 aisien osasumwenaisien osasumwen 11 $\endgroup$ 2 $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. $\endgroup$ dantopa – dantopa 2019-06-27 03:57:18 +00:00 Commented Jun 27, 2019 at 3:57 $\begingroup$ See the answer given by Thomas $\endgroup$ nmasanta – nmasanta 2019-06-27 04:52:47 +00:00 Commented Jun 27, 2019 at 4:52 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus notation See similar questions with these tags. 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9	https://famous-trials.com/animalrights/2595-the-case-against-animal-personhood	Published Time: Sun, 07 Sep 2025 15:52:37 GMT The Case AGAINST Animal Personhood Home The Case AGAINST Animal Personhood Details The Case AGAINST Recognizing Animals as "Persons" for Legal Purposes The Case AGAINST Animal Rights Several noted legal scholars have challenged the notion that certain animals should be recognized as “persons” entitled to at least some basic rights. Among the critics and skeptics are Judge Richard A. Posner who worries that listening “too raptly to the siren song of animal rights” might harm the interests not only of humans, but animals as well. [“Animal Rights”, 110 Yale L.J. (2000)]. Richard Epstein sees it as folly to argue other species should ever have legal rights approaching those we grant to humans. As Epstein puts it, “The blunt point is that we have, and will continue to have, different moral obligations to our own conspecifics than we do to chimps or members of any other species.” [Richard A. Epstein, "Animals as Objects, or Subjects, of Rights"(2002)]. Law professor Richard L. Cupp (Are all animal rights opponents named Richard?) argues that granting rights to animals would open the “floodgates” of litigation and that a far better focus for those who care for animals would be to seek more legislative protection against mistreatment of animals. [“Animals as More than ‘Mere Things,’ but Still Property: A Call for Continuing Evolution of the Animal Welfare Paradigm,” 82 Cinn. L. Rev (2016)]. The most compelling arguments AGAINST extending rights to animal are summarized below. 1.Cognitive capacity is neither a necessary nor a sufficient condition for having legally enforceable rights. Richard Posner observes that cognitive capacity is “a precondition of some rights, such as the right to vote.” But, he argues, we give rights to some humans with extremely little cognitive function. Posner writes: “Most of us would think it downright offensive to give greater rights to monkeys than to retarded people, upon a showing that the monkey has a greater cognitive capacity than a profoundly retarded human being, unless perhaps the human being has no brain function at all above the autonomic level, that is, is in a vegetative state.” (RP, 532) Why should cognitive capacity be the basis for personhood?—there really is no good reason. Posner points out: “Most people would think it distinctly odd to proportion animal rights to animal intelligence, as Wise wishes to do, implying that dolphins, parrots, and ravens are entitled to more legal protection than horses (or most monkeys), and perhaps that the laws forbidding cruelty to animals should be limited to the most intelligent animals, inviting the crack ‘They don't have syntax, so we can eat them.’” (RP, 532) Posner also asks, if cognitive capacity is the basis for determining personhood, “What about computers?” Posner points out: “Someday, perhaps soon, there will be computers that have as many ‘neurons’ as chimpanzees, and the ‘neurons’ will be ‘wired’ similarly. Such computers may well be conscious. This will be a problem for Wise, for whom the essence of equality under law is that individuals with similar cognitive capacities should be treated alike regardless of their species.” (RP, 531) Posner also notes: “Nature is not valued by environmentalists for its mental attributes, and so the environmentalist is unlikely to want to give special protection to chimpanzees, dolphins, and other highly intelligent animals.” (RP, 535-36) Richard Epstein argues that cognitive capacity has never been the essential basis for granting human rights: “The movement for equal rights of all human beings must take into account the fact that all people do not have anything remotely like the same cognitive abilities.” He notes that lack of cognitive capacity has been a justification for granting extra protection to certain humans, not less: “What fate lies for adult human beings whose mental disabilities in fact preclude them from taking advantage of many of the rights they are afforded? Our standard position is to give them extra protection, not to exterminate them, and to do so because they are human beings, entitled to protection as such. It follows therefore we should resist any effort to bootstrap legal rights for animals on the change in legal rights of women and slaves. There is no next logical step to restore parity between animals on the one hand and women and slaves on the other.” (RE, 17) Epstein, like Posner, sees no good reason for using cognitive capacity as the measure of rights. He quotes the utilitarian philosopher Jeremy Bentham: “The question is not, Can they reason? Nor, Can they talk? but, Can they suffer?” But Epstein sees legal efforts directed towards reducing animal suffering as fraught with another set of problems: “Our intervention to prevent suffering is, however, usually confined to questions of how human beings ought to interact with animals, and there the problems are difficult enough.” (RE, 36) STEVEN WISE'S RESPONSE:We humans don't like to be in prison. Imprisonment destroys your liberty, autonomy, and self-determination, your ability to go where you want to go and do what we want to do and with whom. These are basic human attributes. If you lack legal rights there is nothing you can do when someone else deprives you of your liberty. You become their slave. Thus judges and legislators value your right to autonomy and self-determination; they value your liberty so much and have for so long that they developed something called "habeas corpus," which is Latin for "you have the body." Habeas corpus is so important that it has long been referred to as "The Great Writ." Its single purpose is to test the legality of a "person's" imprisonment. By "person" the law does not mean "human being," but any entity that has the capacity for any legal right. That can be a human being. But it can be many other entities, too. The important thing is that if you are not a "person" you are just a "thing," unable to have any legal rights. Personhood is not connected to biology. In some countries, rivers, national parks, a religious idol, a religion's holy books, even the Amazon rainforest have been designated as "persons" with certain legal rights. On the other hand, at times humans such as fetuses, slaves, and women were not allowed to use habeas corpus because the courts believed they weren't "persons."Until recently all nonhuman animals were automatically seen as the "things" they have been for two thousand years. There was nothing they could do when a human deprived them of their liberty or other things they cared about. That is finally changing. The first nonhuman animals are starting to be recognized as "persons" in some countries. Some US states have enacted Pet Trust Statutes by which some or all nonhuman animals are given the legal rights of trust beneficiaries. The Nonhuman Rights Project began showing one way to establish the personhood and right to liberty of at least those nonhuman animals who are autonomous and self-determining when it started filing test cases in 2013 demanding that their nonhuman animal clients - so far chimpanzees and elephants - be given the right to bodily liberty.The NhRP brings in the world's greatest experts to explain to the judge how extraordinarily cognitive complex and autonomous its clients actually are. It never argues that autonomy and self-determination are required for legal personhood, just that those qualities should qualify their holders for the right to bodily liberty that is protected by habeas corpus, whoever they might be. Judge Richard A. Posner 2. There is no principled way to draw a line between animals that should be recognized as “persons” (at least for habeas corpus purposes) and those that should not be given such recognition—and judges need to draw lines. Animal rights advocates such as Steven Wise argue that chimpanzees (and bonobos and other primates) should be recognized as “persons” entitled to their freedom. He would say the same thing about elephants and whales and dolphins. And, no doubt, he believes that petitions for habeas corpus should be available to an intelligent bird species such as the African Grey Parrot. The octopus is a highly intelligent (albeit possessed with an almost alien sort of intelligence) creature—maybe an octopus should be a person? And then what about border collies or pigs? You see the problem. No judge would order a border collie freed from its owner and no judge would liberate all pigs on an Iowa hog farm. Nor would many, if any, judges be ready to order parrots freed from their owners and sent to an aviary, no matter how wonderful that change might be for the parrot. Richard Posner puts the problem this way: “Wise wants judges, in good common-law fashion, to move step by step, and for the first step simply to declare that chimpanzees have legal rights. But judges asked to step onto a new path of doctrinal growth want to have some idea of where the path leads, even if it would be unreasonable to insist that the destination be clearly seen. Wise gives them no idea.” (RP, 532) Epstein makes essentially the same point: “[If] the line between humans and chimps is no longer decisive, then some other line has to be. Perhaps it is the line between chimps and great apes, or between both and horses and cows, or between horses and cows and snails and fish. Which of these lines are decisive and why?” (RE, 22) Richard Cupp worries that the lack of an obvious place to draw a line will lead to excessive litigation: “If the legal wall between animals and humans is broken through, [that will open] a floodgate of expansive litigation without a meaningful standard for determining how many of the billions of animals in the world are intelligent enough to merit personhood.” (Cupp brief, 22) Of course, even if there might be disagreement precisely where the line should be drawn, we instinctively understand that some animals deserve more legal protections than others. Epstein acknowledges that fact: “The more animals look and act like human beings, the greater the level of protection that we as humans are willing to afford them. Rights of bodily integrity do not have much of a future for mosquitoes. Second, the higher the species ranks on own tree of life, the stronger the justifications that must be advanced in order to harm members of that species. Cost aside, we would be wholly inappropriate to think that we should capture or breed chimpanzees for food, whatever our views on their use for medical experimentation.” (RE, 26) STEVEN WISE'S RESPONSE:The English poet, Alexander Pope, wrote in his 1734 “An Essay On Man” that “Whatever is, is right.” Law professor Christopher Stone noted in his 1972 book, “Should Trees Have Standing” that “(t)he fact is that each time there is a movement to confer rights onto some new 'entity,' the proposal is bound to soundodd or frightening, or laughable,” while Justice Anthony Kennedy of the United States Supreme Court reminded us in 2003 that “times can blind us to certain truths and later generations can see that laws once thought necessary and proper in fact serve only to oppress.” Not long ago in Western law, only white people had rights., not blacks or indigenous peoples. Only men had rights, not women. Only adults had rights, not children. The claim that those lines needed to be redrawn, that blacks should have rights or women or children, sounded “odd or frightening or laughable” to many. Each claim was stoutly resisted for decades, sometimes centuries, by those blind to the true meaning of such fundamental values that we nearly all agree with as liberty and equality by their times. Today many, including Posner, Epstein, and Cupp, are similarly blind to the claim that a nonhuman can have any rights by their times. Our times. The irony is they argue that any new line must be shown to be decisive, oblivious that the ancient line that exists between humans and the rest of the universe was based upon outdated and disproven biology and superseded moral values that long ago rendered it arbitrary and unjust. 3. Once we recognize some rights for animals, courts will be drawn into endless cases turning on whether animals should be given some other right—and some rights, almost everyone agrees, should be rights for humans only. Epstein makes this argument: “The natural cognitive and emotional limitations of animals, even the higher animals, preclude any creation of full parity [with humans]. What animal can be given the right to contract? To testify in court? To vote? To participate in political deliberation? To worship?” (RE, 18) Epstein also argues that granting rights to animals will raise questions as to whether we have a legal obligation to intervene in nature to protect them. As he puts it: “Do we train the lion to lie down with the lamb, or do we let the lion consume the lamb in order to maintain his traditional folkways? Do we ask chimpanzees to forgo eating monkeys. It is odd to intervene in nature to forestall some deadly encounters, especially if our enforced nonaggression could lead to extermination of predator species. But, if animals have rights, then how do we avoid making these second-tier judgments?” (RE, 19) Posner also argues that starting down the path towards animal rights will lead to a judicial quagmire, thicket, or worse: “But what is meant by liberating animals and giving them the rights of human beings of the same cognitive capacity? Does an animal's right to life place a duty on human beings to protect animals from being killed by other animals? Is capacity to feel pain sufficient cognitive capacity to entitle an animal to at least the most elementary human rights? What kinds of habitats must we create and maintain for all the rights-bearing animals in the United States? Does human convenience have any weight in deciding what rights an animal has? Can common-law courts actually work out a satisfactory regime of animal rights without the aid of legislatures? When human rights and animal rights collide, do human rights have priority, and if so, why? And what is to be done when animal rights collide with each other, as they do with laws that by protecting wolves endanger sheep? Must entire species of animals be ‘segregated’ from each other and from human beings, and, if so, what does ‘separate but equal’ mean in this context? May we ‘discriminate’ against animals, and if so, how much? Do species have "rights," or just individual animals, and if the latter, does this mean that according special legal protection for endangered species is a denial of equal protection? Is domestication a form of enslavement? Wise does not try to answer any of these questions. He is asking judges to set sail on an uncharted sea without a compass.” (RP, 533) STEVEN WISE'S RESPONSE: These are all “straw man” arguments in that no one actually makes them or is likely to make them. Yet they are raised to prevent nonhuman animals from possessing legal rights against those humans who actually harm them. No one asks whether human fetuses should have any rights – they certainly have many rights now - because the law would then have to grapple with whether they should testify in court or have the right to vote or the right to contract or participate in political deliberation or have the right to worship. No one asks whether children should have any rights and for the same reasons. Humans have no legal rights to be protected from being attacked by nonhuman animals or hit by lightning or hurricanes or having trees fall on them. The rights of nonhuman animals, like the rights of humans, are rights against other humans and no one claims differently. Moreover the claim that, as judges would have to grapple with new legal claims, an entity should have no rights at all, is unprecedented. And every time lawyers advocate for a new human right those who will be unable to exploit other humans try to block it. Who believes that humans should not take steps toward preventing climate change or environmental damages just because later steps might have more serious economic impacts on those who are causing climate change or ruining the environment or which involve more complex questions? Finally, the only job judges have is to decide. Their compass, as always, is the fundamental values that have long guided judges in deciding cases involving humans. Richard Epstein 4. Many animals benefit from being treated as property and might actually fare less well if treated as “persons” entitled to their freedom. Another way of framing this argument is to ask: “Who’s to say animals want to be liberated?” Organizations like the Nonhuman Rights Project might say that they speak for the animals, but obviously there are some difficulties when it comes to determining the precise wishes of an animal. It is easy to imagine that in some cases freeing an animal from what appears to be a sub-optimal situation might lead to an even worse situation. Richard Epstein puts the argument this way: “Animals that are left to their own devices may have no masters; nor do they have any peace. Life in the wild leaves them exposed to the elements; to attacks by other animals; to the inability to find food or shelter; to accidental injury; and to disease. The expected life of animals in the wild need not be solitary, poor, nasty, brutish and short. But it is often rugged, and rarely placid and untroubled. . . . Because they use and value animals, owners will spend resources for their protection. Veterinary medicine may not be at the level of human medicine, but it is only a generation or so behind. When it comes to medical care, it’s better to be a sick cat in a middle-class United States household than a sick peasant in a third-world country. Private ownership of many pets gives them access to food and shelter (and sometimes clothing) which creates long lives of ease and comfort.” (RE, 10) STEVEN WISE'S RESPONSE:Humans often need to persuade themselves and others that the entities we enslave for our selfish purposes actually benefit from their enslavement. In one 18 th century debate in the English House of Lords, one Lord claimed that slaves in the Caribbean lived in a state of humble happiness. Another said slaves appeared so happy to that he often wished to be a slave himself. In the 19 th century United States Vice-President John C. Calhoun said “(n)ever before has the black races of Central Africa, from the dawn of civilization to the present day attained a condition so civilized and so improved, not only physically, but morally and intellectually.” After hearing evidence in a 2012 case alleging that elephants in the Los Angeles Zoo were being treated badly, a Los Angeles judge noted that the senior elephant keeper engaged in an “anthropomorphic fantasy that the elephants are happy to see her and live their lives in captivity” and observed that “(c)aptivity is a terrible existence for any intelligent self-aware species, which the undisputed evidence shows elephants are. To believe otherwise, as some high-ranking employees appear to believe, is delusional.” 5. If we don’t limit personhood to human beings, but grant personhood to other species based on their cognitive capacities, the net result might be that the law will give less respect to human rights and human dignity. Richard Posner makes this argument: “It is that if we fail to maintain a bright line between animals and human beings, we may end up by treating human beings as badly as we treat animals, rather than treating animals as well as we treat (or aspire to treat) human beings. Equation is a reflexive relation. If chimpanzees equal human infants, human infants equal chimpanzees.” (RP, 535) Richard Cupp makes this argument a central part of his amicus brief against extending rights to animals: “The pervasive view that all humans have distinctive and intrinsic human dignity regardless of their capabilities may have cultural, religious, or even instinctual foundations. Morally autonomous humans have unique natural bonds with other humans who have cognitive impairments, and thus denying rights to them also harms the interests of society—we are all in a community together. Infants’ primary identities are as humans, and adults with severe cognitive impairments’ primary identities are as humans who are other humans’ parents, siblings, children or spouse. However, good intentions sometimes create disastrous results. There should be deep concern that over a long horizon, allowing animal legal personhood based on cognitive abilities could unintentionally lead to gradual erosion of protections for these especially vulnerable humans. The sky would not immediately fall if courts started treating chimpanzees as persons. . . .But, over time, both the courts and society might be tempted not only to view the most intelligent animals more like we now view humans but also to view the least intelligent humans more like we now view animals. (Cupp brief, 14-18). Cupp concludes: “Deciding chimpanzees are legal persons based on the cognitive abilities we have seen in them may open a door that swings in both directions regarding rights for humans as well as for animals, and later generations may well wish we had kept it closed." (Cupp brief, 21) STEVEN WISE'S RESPONSE:Enormous concern is expressed that granting any legal rights to any nonhuman animal “may”, “could,” or “might” cause us humans to treat other humans as badly as we treat nonhuman animals, though no reasons are given for why that could happen, other than another “may” and a “sometimes,” with some pop psychology thrown in, and no claim is made that such a thing has ever happened. That is because it never has happened. When the question arose as to whether women should have the vote no one argued that if they got the vote some men might end up losing their right to vote vote. When the abolition of black chattel slavery arose no one claimed that some white people could end up being enslaved. And nowhere in this thicket of great unlikelihoods can one locate any concern about changing the admittedly bad way we humans actually treat nonhuman animals. And that’s a certainty. Richard Cupp 6. When push comes to shove, we have to favor our own species and cannot give anything like equal rights to another species. Our first moral obligation is to our own species. Richard Epstein raises the issue thusly: “Suppose that we manufacture limited supplies of a new pill that is a cure for some disease that is ravaging both human and chimpanzee populations. There is not enough to go around for both man and beast. Is there some kind of affirmative duty to assist chimps to the same extent that we assist other human beings? I should be stunned if any real world scenario would ever produce any result other than humans first, chimps second.” Epstein concludes: “The blunt point is that we have, and will continue to have, different moral obligations to our own conspecifics than we do to chimps or members of any other species.” (RE, 23) Gary Francione asks whether “we cannot prefer human over animals in situations of true emergencies or conflicts.” In his book, Your Child or the Dog? Francione asks how an individual should choose between saving the life of his or her child or his or her dog if both are trapped inside a burning house. Epstein argues the right answer is obvious: “The child, darn it, even if the child is unrelated and the dog is one’s own.” (RE, 23) Epstein agrees that line-drawing is uncomfortable, even though necessary: “The root of our discontent is that in the end we have to separate ourselves from (the rest of) nature from which we evolved. Unhappily but insistently, the ‘collective’ we is prepared to do just that. Such is our lot, and perhaps our desire, as human beings.” (RE, 27) Richard Posner believes that the special status humans accord themselves is not based on empirical evidence or logic. Rather, he says, it is based on “a moral intuition deeper than any reason that could be given for it and impervious to any reason that you or anyone could give against it.” (Posner debate with Peter Singer, 2001). STEVEN WISE'S RESPONSE:This sort of line-drawing by which powerful line-drawers accord themselves a special status that carries with it the entitlement to enslave others has been made many times throughout history, Egyptians of Hebrews, Greeks of barbarians, Romans of Greeks, Christians of indigenous peoples and Muslims, Muslims of Christians, and whites of blacks, the last exemplified perhaps most famously, and execrably, in Chief Justice Taney’s opinion in the Dred Scott case. The slave-holding Taney, referring both to the legal status of black people in the 1780s and in the present, wrote that blacks “had for more than a century before been regarded as beings of an inferior order, and altogether unfit to associate with the white race, either in social or political relations; and so far inferior, that they had no rights which the white man was bound to respect; and that the negro might justly and lawfully be reduced to slavery for his benefit … This opinion was at that time fixed and universal in the civilized portion of the white race. It was regarded as an axiom in morals as well as in politics, which no one thought of disputing, or supposed to be open to dispute; and men in every grade and position in society daily and habitually acted upon it in their private pursuits, as well as in matters of public concern, without doubting for a moment the correctness of this opinion … a perpetual and impassable barrier was intended to be erected between the white race and the one which they had reduced to slavery, and governed as subjects with absolute and despotic power, and which they then looked upon as so far below them in the scale of created beings …” Don E. Fehrenbacher noted in his Pulitzer Prize winning book, The Dred Scott Case,“Taney’s statement was to place Negroes … on the same level, legally, as domestic animals.” (p. 349). It is time that nonhuman animals themselves are no longer placed on the same level as nonhuman animals. 7. The concept of rights is essentially human and can have force and applicability only in the human moral world. The philosopher Carl Cohen puts the argument this way: “Animals cannot be the bearers of rights because the concept of right is essentially human; it is rooted in the human moral world and has force and applicability only within that world.” (Carl Cohen & Tom Regan, The Animal Rights Deba te. 2001, p. 30). Richard Cupp argues that because chimps should not be legally accountable, they also should not be given legal rights: “The pertinent question is not whether chimpanzees possess anything that could be characterized as a sense of responsibility, but rather whether they possess a sufficient level of moral agency to be justly held legally accountable as well as to possess legal rights under our human legal system. When, in 2012, an adult chimpanzee at the Los Angeles Zoo beat a three-month-old baby chimpanzee in the head until the baby died, doubtless no authorities seriously contemplated charging the perpetrator in criminal court." (Cupp brief, 5--6) STEVEN WISE'S RESPONSE:In 2018 New York Court of Appeals Judge Eugene Fahey, the only high court judge in the United States who has opined to date on whether nonhuman animals should have any legal rights, wrote in a habeas corpus case brought on behalf of a chimpanzee, that the lower appellate court had “reasoned that chimpanzees are not persons because they lack ‘the capacity or ability ... to bear legal duties, or to be held legally accountable for their actions. (The Nonhuman Rights Project and) law professors Laurence H. Tribe, Justin Marceau, and Samuel Wiseman question this assumption. Even if it is correct, however, that nonhuman animals cannot bear duties, the same is true of human infants or comatose human adults, yet no one would suppose that it is improper to seek a writ of habeas corpus on behalf of one’s infant child or a parent suffering from dementia. In short, being a ‘moral agent’ who can freely choose to act as morality requires is not a necessary condition of being a ‘moral patient’ who can be wronged and may have the right to redress wrongs. The (lower appellate court’s) conclusion that a chimpanzee cannot be considered a ‘person’ and is not entitled to habeas relief is in fact based on nothing more than the premise that a chimpanzee is not a member of the human species … The better approach … is to ask not whether a chimpanzee fits the definition of a person or whether a chimpanzee has the same rights and duties as a human being, but instead whether he or she has the right to liberty … Moreover, the answer to that question will depend on our assessment of the intrinsic nature of chimpanzees as a species. The record before us … contains unrebutted evidence, in the form of affidavits from eminent primatologists, that chimpanzees have advanced cognitive abilities and demonstrate autonomy by self–Initiating intentional, adequately informed actions, free of controlling influences.” 8. Most all of what animal rights proponents seek to accomplish can be accomplished better by seeking additional legislative protection against animal cruelty. Posner argues that seeking legislative protection is “more conservative, methodologically as well as politically, but possibly more efficacious, than rights-mongering.” He contends we should “extend, and move vigorously to enforce, laws designed to prevent gratuitous cruelty to animals. We should be able to agree without help from philosophers and constitutional theorists that gratuitous cruelty is bad.” (RP, 539) Posner writes: “Wise is another deer frozen in the headlights of Brown v. Board of Education.” Wise overlooks better approaches to securing animal welfare through legislative action or redefining property rights. Posner suggests we should simply “forbid treating chimpanzees, or any other animals with whom we sympathize, cruelly.” Animal rights, as Posner sees it, is “just an impediment to clear thought as well as a provocation in some legal and philosophical quarters.” (RP, 540) Richard Cupp agrees that the focus should be on legislative change to protect animal welfare. He approves of the language of New York’s Third Department, which refused to extend personhood to Tommy in the Lavery appeal: “Our rejection of a rights paradigm for animals does not, however, leave them defenseless,” noting that advocates for animals remained "free to importune the Legislature to extend further legal protections to chimpanzees.” (Cupp brief, 25). STEVEN WISE'S RESPONSE:Centuries of legal history have demonstrated that the only way in which a human’s most fundamental interests can be protected is through fundamental legal rights. That is why such international human rights documents as the Universal Declaration of Human Rights and the International Convention on Civil and Political Rights require that every individual be deemed a “person.” I invite any humans who claim that it would better for nonhuman animals to forgo all legal rights and be protected solely by welfare statutes to give up all their legal rights. That way even your most fundamental interests will be protected solely by human welfare statutes. As with animal welfare statutes, you too will be unable to enforce the statutes, their enforcement will be left to the complete discretion of a public prosecutor, and any prosecution will lead at best to the violator being convicted of a crime, with no remedy coming to you.. If you are unhappy you can always “importune the legislature for further legal protections” that don’t amount to legal rights. Not interested? I thought not. HomeTrial Account Other Resources Animal Rights on Trial: An Account Animals and Humans: A Brief History Animal Rights (Personhood) Trials: A Chronology The Case FOR Animal Personhood The Case AGAINST Animal Personhood Comparing Species: Autonomy, Intelligence, and Morality First Petition for Habeas Corpus (Tommy, a chimpanzee vs Patrick Lavery) What Chimps Can Do: The Cognitive Capacities of Chimpanzees Transcript: Hearing on Personhood for Herecules and Leo, Two Chimps NhRP Petition for for Writ of Habeas Corpus for Three Elephants (2017) What Elephants Can Do: The Cognitive Capacities of Elephants Transcript: Hearing on Personhood for Happy, an Elephant The James Sommersett Case (1772) Animal Rights Debate: Peter Singer vs Richard Posner Philosopher Tom Regan on Animal Rights Are All Species Created Equal? Animal Rights on Trial: Bibliography and Links Donate to Famous-Trials.com:With your help, Famous-Trials.com can expand and update its library of landmark cases and, at the same time, support the next generation of legal minds from UMKC School of Law. Donate Now Powered by UMKC School of Law Copyright © 1995 - 2025 Professor Douglas O. Linder To Top
10	https://www.gauthmath.com/solution/1803684607673350/What-is-the-probability-of-rolling-an-even-number-on-a-fair-six-sided-die-A-1-2-	Question Solution Respuesta: D. No, ya que sólo podemos elegir entre las condiciones disponibles y el rango de 1 a 5 no es una de ellas. contact@gauthmath.com
11	https://math.stackexchange.com/questions/4824606/circles-formed-by-points-of-common-tangency-of-two-circles	geometry - Circles formed by points of common tangency of two circles - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Circles formed by points of common tangency of two circles Ask Question Asked 1 year, 9 months ago Modified1 year, 9 months ago Viewed 457 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I was playing around in Geogebra with circles and their common tangents. For any two random circles (yellow), I noticed that the four points of tangency of direct common tangents form a circle, and so do the four points of tangency of the transverse common tangents. The two circles (blue) are concentric, and their common centre M also happens to be the midpoint of the line segment joining the centres of the original two circles. My question: how do I find the radii of the concentric circles, just using the radii of the original circles and the distance between them? (Also, are there any more interesting properties of these circles, which have been documented somewhere?) geometry euclidean-geometry circles tangent-line Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Dec 12, 2023 at 8:41 algorhythmalgorhythm asked Dec 11, 2023 at 17:12 algorhythmalgorhythm 77 7 7 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. The well-known geometric construction of direct common tangent, which is shown in the diagram below, can be used to easily determine the dimeter of the mentioned circle that passes through the four points of tangency. To construct the direct common tangents, we need to draw two auxiliary circles. One of those circles has the radius R−r R−r and the center C 1 C 1. The other is drawn with the radius d 2 d 2 and has the mod point of C 1 C 2 C 1 C 2 as its center. One of their points of intersection is Q Q. Draw the segment C 1 Q C 1 Q and then extend it so that it cuts the circle Ω 1 Ω 1 at P 1 P 1 as shown in the diagram. We also draw a line segment joining Q Q and C 2 C 2. To get the direct common tangent, draw a line parallel to Q C 2 Q C 2 through P 1 P 1 to through the circle Ω 2 Ω 2 at P 2 P 2. As per the construction, we can state that, C 1 Q=R−r,(0)(0)C 1 Q=R−r, P 1 P 2=Q C 2 and(1)(1)P 1 P 2=Q C 2 and ∡C 2 Q C 1=90 o.(2)(2)∡C 2 Q C 1=90 o. Now, mark N N as the midpoint of P 1 P 2 P 1 P 2. Since M M is the midpoint of C 1 C 2 C 1 C 2, M N M N is parallel to both C 1 P 1 C 1 P 1 and C 2 P 2 C 2 P 2. Therefore, we have, M N=1 2(C 1 P 1+C 2 P 2)=1 2(R+r).(3)(3)M N=1 2(C 1 P 1+C 2 P 2)=1 2(R+r). Because ∡C 2 Q C 1=90 o∡C 2 Q C 1=90 o, the triangle C 1 C 2 Q C 1 C 2 Q is right-angled. Using (0), we apply Pythagoras theorem to this triangle to obtain, Q C 2 2=C 1 C 2 2−C 1 Q 2=d 2−(R−r)2.(4)(4)Q C 2 2=C 1 C 2 2−C 1 Q 2=d 2−(R−r)2. Since Q C 2 P 2 P 1 Q C 2 P 2 P 1 is a rectangle, using (4), we shall write, P 1 P 2=Q C 2=d 2−(R−r)2−−−−−−−−−−−√.P 1 P 2=Q C 2=d 2−(R−r)2. N N is the midpoint of P 1 P 2 P 1 P 2. Therefore, P 1 N 2=1 4 P 1 P 2 2=1 4(d 2−(R−r))2.(5)(5)P 1 N 2=1 4 P 1 P 2 2=1 4(d 2−(R−r))2. Finally, using (1) and (5), apply Pythagoras theorem to the right-angled triangle M N P 1 M N P 1 to express M P 1 M P 1 in terms of the diameters of the two circles Ω 1 Ω 1 and Ω 2 Ω 2, and the distance d d between their centers. M P 2 1=M N 2+P 1 N 2==1 4(R+r)2+1 4(d 2−(R−r))2 M P 1 2=M N 2+P 1 N 2==1 4(R+r)2+1 4(d 2−(R−r))2 a=M P 1=1 2 d 2+4 R r−−−−−−−−√a=M P 1=1 2 d 2+4 R r Added at OP's request - Transverse Common Tangent: Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 12, 2023 at 18:50 answered Dec 12, 2023 at 13:17 YNKYNK 4,650 1 1 gold badge 12 12 silver badges 18 18 bronze badges 1 (+1) Thanks! Is it possible to also determine the radius of the circle formed by the transverse common tangency points?algorhythm –algorhythm 2023-12-12 13:49:53 +00:00 Commented Dec 12, 2023 at 13:49 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let C 1=(x 1,y 1)C 1=(x 1,y 1) and C 2=(x 2,y 2)C 2=(x 2,y 2). Call the points of tangency on one of the direct tangents A 1 A 1 and A 2 A 2. Then because each of the segments A 1 C 1 A 1 C 1 (of length r 1 r 1) and A 2 C 2 A 2 C 2 (of length r 2 r 2) is perpendicular to that tangent, they are parallel. So if O O is the intersection of the two direct tangents, then △O A 1 C 1△O A 1 C 1 and △O A 2 C 2△O A 2 C 2 are similar right triangles. The former has one leg r 1 r 1 and hypotenuse h h, And the latter has one leg r 2 r 2 and hypotenuse h+(x 1−x 2)2+(y 1−y 2)2−−−−−−−−−−−−−−−−−−√.h+(x 1−x 2)2+(y 1−y 2)2. Solving for h h will allow you to find ∠C 1 C 2 A 2∠C 1 C 2 A 2, and thus the coordinates of A 2 A 2, which, of course, allows you to calculate the length of M A 2 M A 2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 11, 2023 at 17:54 Paul TanenbaumPaul Tanenbaum 2,621 9 9 silver badges 19 19 bronze badges 1 (+1). Is it possible to find the radius using a more Euclidean approach? Like without finding the coordinates.algorhythm –algorhythm 2023-12-11 18:09:39 +00:00 Commented Dec 11, 2023 at 18:09 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. We use this property that the distances of points of intersection of the circle , passing the tangency points, with the line connecting the centers of two circles EN (in figure points P and O) from the centers of two circles are equal( in figure AP=CO=0.41). In particular case where the projection of point of tangency H on EN is coincided with reflect of point P about point A (the center of biger circle) we have AC, A H=r A A H=r A and C F=r C C F=r C as known of problem and we can write: ∠Q C A=∠P′H A∠Q C A=∠P′H A ⇒A P=P′A A H=r A=r A−r C A C⇒A P=P′A A H=r A=r A−r C A C in this way A P=C O A P=C O can be foud and the diameter of the circle passing the tangency points is: D=A C+2 A P D=A C+2 A P Update: to calculate the radius we can also use the power of E to the circle it's radius is required: E F×E H=E S 2 E F×E H=E S 2 t a n H E A=t a n Q C A=r A−r C C Q t a n H E A=t a n Q C A=r A−r C C Q s i n Q C A=r A−r C A C s i n Q C A=r A−r C A C These relations give measure of Q C=F H Q C=F H ∠F E C=∠Q C A∠F E C=∠Q C A E H=r c t a n F E C E H=r c t a n F E C E H=E F+F H E H=E F+F H E S 2=E F×E H E S 2=E F×E H E A=r A t a n H E A E A=r A t a n H E A E C=r c t a n H E A E C=r c t a n H E A E R=E A+E C 2 E R=E A+E C 2 Finally: r=R S=E R 2−E S 2−−−−−−−−−−√r=R S=E R 2−E S 2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 12, 2023 at 14:40 answered Dec 12, 2023 at 7:52 siroussirous 13.1k 3 3 gold badges 16 16 silver badges 25 25 bronze badges Add a comment\| You must log in to answer this question. 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12	https://medium.com/data-science/welchs-t-test-the-reliable-way-to-compare-2-population-means-with-unequal-variances-bbf7a62967bc	Sitemap Open in app Sign in Sign in ## TDS Archive · An archive of data science, data analytics, data engineering, machine learning, and artificial intelligence writing from the former Towards Data Science Medium publication. Welch’s t-Test: The Reliable Way to Compare 2 Population Means with Unequal Variances Discover why Welch’s t-Test is the go-to method for accurate statistical comparison, even when variances differ. Vito Rihaldijiran 8 min readJun 14, 2024 Part 1: Background In the first semester of my postgrad, I had the opportunity to take the course STAT7055: Introductory Statistics for Business and Finance. Throughout the course, I definitely felt a bit exhausted at times, but the amount of knowledge I gained about the application of various statistical methods in different situations was truly priceless. During the 8th week of lectures, something really interesting caught my attention, specifically the concept of Hypothesis Testing when comparing two populations. I found it fascinating to learn about how the approach differs based on whether the samples are independent or paired, as well as what to do when we know or don’t know the population variance of the two populations, along with how to conduct hypothesis testing for two proportions. However, there is one aspect that wasn’t covered in the material, and it keeps me wondering how to tackle this particular scenario, which is performing Hypothesis Testing from two population means when the variances are unequal, known as the Welch t-Test. To grasp the concept of how the Welch t-Test is applied, we can explore a dataset for the example case. Each stage of this process involves utilizing the dataset from real-world data. Part 2: The Dataset The dataset I’m using contains real-world data on World Agricultural Supply and Demand Estimates (WASDE) that are regularly updated. The WASDE dataset is put together by the World Agricultural Outlook Board (WAOB). It is a monthly report that provides annual predictions for various global regions and the United States when it comes to wheat, rice, coarse grains, oilseeds, and cotton. Furthermore, the dataset also covers forecasts for sugar, meat, poultry, eggs, and milk in the United States. It is sourced from the Nasdaq website, and you are welcome to access it for free here: WASDE dataset. There are 3 datasets, but I only use the first one, which is the Supply and Demand Data. Column definitions can be seen here: I am going to use two different samples from specific regions, commodities, and items to simplify the testing process. Additionally, we will be using the R Programming Language for the end-to-end procedure. Now let’s do a proper data preparation: ``` library(dplyr)# Read and preprocess the dataframewasde_data <- read.csv("wasde_data.csv") %>% select(-min_value, -max_value, -year, -period) %>% filter(item == "Production", commodity == "Wheat")# Filter data for Argentina and Australiawasde_argentina <- wasde_data %>% filter(region == "Argentina") %>% arrange(desc(report_month))wasde_oz <- wasde_data %>% filter(region == "Australia") %>% arrange(desc(report_month))() Read and preprocess the dataframe<-("wasde_data.csv")%>%(-, -, -, -)%>%(== "Production", == "Wheat") Filter data for Argentina and Australia<-%>%(== "Argentina")%>%(())<-%>%(== "Australia")%>%(()) ``` I divided two samples into two different regions, namely Argentina and Australia. And the focus is production in wheat commodities. Now we’re set. But wait.. Before delving further into the application of the Welch t-Test, I can’t help but wonder why it is necessary to test whether the two population variances are equal or not. Part 3: Testing Equality of Variances When conducting hypothesis testing to compare two population means without knowledge of the population variances, it’s crucial to confirm the equality of variances in order to select the appropriate statistical test. If the variances turn out to be the same, we opt for the pooled variance t-test; otherwise, we can use Welch’s t-test. This important step guarantees the precision of the outcomes, since using an incorrect test could result in wrong conclusions due to higher risks of Type I and Type II errors. By checking for equality in variances, we make sure that the hypothesis testing process relies on accurate assumptions, ultimately leading to more dependable and valid conclusions. Then how do we test the two population variances? We have to generate two hypotheses as below: The rule of thumb is very simple: If the test statistic falls into rejection region, then Reject H0 or Null Hypothesis. Otherwise, we Fail to Reject H0 or Null Hypothesis. We can set the hypotheses like this: ``` Hypotheses: Variance Comparisonh0_variance <- "Population variance of Wheat production in Argentina equals that in Australia"h1_variance <- "Population variance of Wheat production in Argentina differs from that in Australia" Hypotheses: Variance Comparison<- "Population variance of Wheat production in Argentina equals that in Australia"<- "Population variance of Wheat production in Argentina differs from that in Australia" ``` Now we should do the test statistic. But how do we get this test statistic? we use F-Test. An F-test is any statistical test used to compare the variances of two samples or the ratio of variances between multiple samples. The test statistic, random variable F, is used to determine if the tested data has an F-distribution under the true null hypothesis, and true customary assumptions about the error term. we can generate the test statistic value with dividing two sample variances like this: and the rejection region is: where n is the sample size and alpha is significance level. so when the F value falls into either of these rejection region, we reject null hypothesis. but.. the trick is: The labeling of sample 1 and sample 2 is actually random, so let’s make sure to place the larger sample variance on top every time. This way, our F-statistic will consistently be greater than 1, and we just need to refer to the upper cut-off to reject H0 at significance level α whenever. Get Vito Rihaldijiran’s stories in your inbox Join Medium for free to get updates from this writer. we can do this by: ``` Calculate sample variancessample_var_argentina <- var(wasde_argentina$value)sample_var_oz <- var(wasde_oz$value)# Calculate F calculated valuef_calculated <- sample_var_argentina / sample_var_oz Calculate sample variances<-($)<-($) Calculate F calculated value<-/ ``` we’ll use 5% significance level (0.05), so the decision rule is: ``` Define significance level and degrees of freedomalpha <- 0.05alpha_half <- alpha / 2n1 <- nrow(wasde_argentina)n2 <- nrow(wasde_oz)df1 <- n1 - 1df2 <- n2 - 1# Calculate critical F valuesf_value_lower <- qf(alpha_half, df1, df2)f_value_upper <- qf(1 - alpha_half, df1, df2)# Variance comparison resultif (f_calculated > f_value_lower & f_calculated < f_value_upper) { cat("Fail to Reject H0: ", h0_variance, "\n") equal_variances <- TRUE} else { cat("Reject H0: ", h1_variance, "\n") equal_variances <- FALSE} Define significance level and degrees of freedom<-<-/ 2<-()<-()<- - 1<- - 1 Calculate critical F values<-(,,)<-(1 -,,) Variance comparison result if(>&<){("Fail to Reject H0: ",, "\n")<- TRUE} else{("Reject H0: ",, "\n")<- FALSE} ``` the result is we reject Null Hypothesis at significance level of 5%, in other words, from this test we believe the population variances from the two populations are not equal. Now we know why we should use Welch t-Test instead of Pooled Variance t-Test. Part 4: The main course, Welch t-Test The Welch t-test, also called Welch’s unequal variances t-test, is a statistical method used for comparing the means of two separate samples. Instead of assuming equal variances like the standard pooled variance t-test, the Welch t-test is more robust as it does not make this assumption. This adjustment in degrees of freedom leads to a more precise evaluation of the difference between the two sample means. By not assuming equal variances, the Welch t-test offers a more dependable outcome when working with real-world data where this assumption may not be true. It is preferred for its adaptability and dependability, ensuring that conclusions drawn from statistical analyses remain valid even if the equal variances assumption is not met. The test statistic formula is: where: and the Degree of Freedom can be defined like this: The rejection region for the Welch t-test depends on the chosen significance level and whether the test is one-tailed or two-tailed. Two-tailed test: The null hypothesis is rejected if the absolute value of the test statistic \|t\| is greater than the critical value from the t-distribution with ν degrees of freedom at α/2. ∣t∣>tα/2,ν One-tailed test: The null hypothesis is rejected if the test statistic t is greater than the critical value from the t-distribution with ν degrees of freedom at α for an upper-tailed test, or if t is less than the negative critical value for a lower-tailed test. Upper-tailed test: t > tα,ν Lower-tailed test: t < −tα,ν So let’s do one example with One-tailed Welch t-Test. lets generate the hypotheses: h0_mean <- "Population mean of Wheat production in Argentina equals that in Australia"h1_mean <- "Population mean of Wheat production in Argentina is greater than that in Australia"<- "Population mean of Wheat production in Argentina equals that in Australia"<- "Population mean of Wheat production in Argentina is greater than that in Australia" this is a Upper Tailed Test, so the rejection region is: t > tα,ν and by using the formula given above, and by using same significance level (0.05): ``` Calculate sample meanssample_mean_argentina <- mean(wasde_argentina$value)sample_mean_oz <- mean(wasde_oz$value)# Welch's t-test (unequal variances)s1 <- sample_var_argentinas2 <- sample_var_ozt_calculated <- (sample_mean_argentina - sample_mean_oz) / sqrt(s1/n1 + s2/n2)df <- (s1/n1 + s2/n2)^2 / ((s1^2/(n1^2 (n1-1))) + (s2^2/(n2^2 (n2-1))))t_value <- qt(1 - alpha, df)# Mean comparison resultif (t_calculated > t_value) { cat("Reject H0: ", h1_mean, "\n")} else { cat("Fail to Reject H0: ", h0_mean, "\n")} Calculate sample means<-($)<-($) Welch's t-test (unequal variances)<-<-<-(-)/ sqrt(/ +/)<-(/ +/)^ 2/((^ 2/(^ 2(- 1))) +(^ 2/(^ 2(- 1))))<-(1 -,) Mean comparison result if(>){("Reject H0: ",, "\n")} else{("Fail to Reject H0: ",, "\n")} ``` the result is we Fail to Reject H0 at significance level of 5%, then Population mean of Wheat production in Argentina equals that in Australia. That’s how to conduct Welch t-Test. Now your turn. Happy experimenting! Part 5: Conclusion When comparing two population means during hypothesis testing, it is really important to start by checking if the variances are equal. This initial step is crucial as it helps in deciding which statistical test to use, guaranteeing precise and dependable outcomes. If it turns out that the variances are indeed equal, you can go ahead and apply the standard t-test with pooled variances. However, in cases where the variances are not equal, it is recommended to go with Welch’s t-test. Welch’s t-test provides a strong solution for comparing means when the assumption of equal variances does not hold true. By adjusting the degrees of freedom to accommodate for the uneven variances, Welch’s t-test gives a more precise and dependable evaluation of the statistical importance of the difference between two sample means. This adaptability makes it a popular choice in various practical situations where sample sizes and variances can vary significantly. In conclusion, checking for equality of variances and utilizing Welch’s t-test when needed ensures the accuracy of hypothesis testing. This approach reduces the chances of Type I and Type II errors, resulting in more reliable conclusions. By selecting the appropriate test based on the equality of variances, we can confidently analyze the findings and make well-informed decisions grounded on empirical evidence. Resources Nasdaq Data Link. (n.d.). Nasdaq Data Link. Retrieved June 14, 2024, from F-Test in Statistics. (2024, June 14). F-Test in Statistics. Two Independent Samples Unequal Variance (Welch’s Test) — ENV710 Statistics Review Website. (n.d.). Two Independent Samples Unequal Variance (Welch’s Test) — ENV710 Statistics Review Website. 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13	https://home.csulb.edu/~woollett/mbe4solve.pdf	Maxima by Example: Ch.4: Solving Equations ∗ Edwin L. Woollett January 29, 2009 Contents 4 Solving Equations 3 4.1 One Equation or Expression: Symbolic Solution or Roots . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4.1.1 The Maxima Function solve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4.1.2 solve with Expressions or Functions & the multiplicities List . . . . . . . . . . . . . . . . . . . . 4 4.1.3 General Quadratic Equation or Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 4.1.4 Checking Solutions with subst or ev and a Do Loop . . . . . . . . . . . . . . . . . . . . . . . 6 4.1.5 The One Argument Form of solve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 4.1.6 Using disp, display, and print . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 4.1.7 Checking Solutions using map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 4.1.8 Psuedo-PostFix Code: %% . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 4.1.9 Using an Expression Rather than a Function with Solve . . . . . . . . . . . . . . . . . . . . . . . 9 4.1.10 Escape Speed from the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 4.1.11 Cubic Equation or Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.1.12 Trigonometric Equation or Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.1.13 Equation or Expression Containing Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . 15 4.2 One Equation Numerical Solutions: allroots, realroots, nd root . . . . . . . . . . . . . . . . . . . . . . 16 4.2.1 Comparison of realroots with allroots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.2.2 Intersection Points of Two Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.2.3 Transcendental Equations and Roots: nd root . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2.4 nd root: Quote that Function! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 4.2.5 newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4.3 Two or More Equations: Symbolic and Numerical Solutions . . . . . . . . . . . . . . . . . . . . . . . . 28 4.3.1 Numerical or Symbolic Linear Equations with solve or linsolve . . . . . . . . . . . . . . . . . . 28 4.3.2 Matrix Methods for Linear Equation Sets: linsolve by lu . . . . . . . . . . . . . . . . . . . . . . 29 4.3.3 Symbolic Linear Equation Solutions: Matrix Methods . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.4 Multiple Solutions from Multiple Right Hand Sides . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.3.5 Three Linear Equation Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.3.6 Surpressing rat Messages: ratprint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.3.7 Non-Linear Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 4.3.8 General Sets of Nonlinear Equations: eliminate, mnewton . . . . . . . . . . . . . . . . . . . . . 37 4.3.9 Intersections of Two Circles: implicit plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.3.10 Using Draw for Implicit Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.3.11 Another Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.3.12 Error Messages and Do It Yourself Mnewton . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.3.13 Automated Code for mymnewton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 ∗This version uses Maxima 5.17.1. This is a live document. Check for the latest version of these notes. Send comments and suggestions to woollett@charter.net 1 COPYING AND DISTRIBUTION POLICY This document is part of a series of notes titled Maxima by Example and is made available via the author's webpage to aid new users of the Maxima com-puter algebra system. NON-PROFIT PRINTING AND DISTRIBUTION IS PERMITTED. You may make copies of this document and distribute them to others as long as you charge no more than the costs of printing. These notes (with some modications) will be published in book form eventually via Lulu.com in an ar-rangement which will continue to allow unlimited free download of the pdf les as well as the option of ordering a low cost paperbound version of these notes. 2 4 Solving Equations Maxima has several functions which can be used for solving sets of algebraic equations and for nding the roots of an expression. These are described in the Maxima manual, Sec. 21, and listed under Contents under Equations. This chapter gives examples of the following Maxima functions: • solve solves a system of simultaneous linear or nonlinear polynomial equations for the specied vari-able(s) and returns a list of the solutions. • linsolve solves a system of simultaneous linear equations for the specied variables and returns a list of the solutions. • nd root uses a combination of binary search and Newton-Raphson methods for univariate functions and will nd a root when provided with an interval containing at least one root. • allroots nds all the real and complex roots of a real univariate polynomial. • realroots nds all of the real roots of a univariate polynomial within a specied tolerance. • eliminate eliminates variables from a set of equations or expressions. • linsolve by lu solves a system of linear algebraic equations by the matrix method known as LU decom-position, and provides a Maxima method to work with a set of linear equations in terms of the matrix of coefcients. • newton, naive univariate Newton-Raphson, and mnewton, multivariate Newton-Raphson, can deal with nonlinear function(s). We also encourage the use of two dimensional plots to approximately locate solutions. This chapter does not yet include Solving Recurrence Relations, and Solving One Hundred Equations. 4.1 One Equation or Expression: Symbolic Solution or Roots 4.1.1 The Maxima Function solve Maxima's ability to solve equations is limited, but progress is being made in this area. The Maxima man-ual has an extensive entry for the important function solve, which you can view in Maxima with the input ? solve (no semicolon) followed by (Enter), or the equivalent command: describe(solve)$. The in-put example(solve)$ will show you the manual examples without the manual syntax material. We will present some examples of the use of solve and not try to cover everything. solve tries to nd exact solutions. If solve(f(x),x) cannot nd an exact solution, solve tries to return a simplied version of the original problem. Sometimes the simplied version can be useful: (%i1) f(x); (%o1) f(x) (%i2) solve( f(x)2-1 , x ); (%o2) [f(x) = - 1, f(x) = 1] Since Maxima's idea of what is simpler may not agree with your own, often the returned version is of no use. 3 The Maxima manual solve syntax discussion relevant to solving one equation is: Function: solve(expr, x) Function: solve(expr) Solves the algebraic equation expr for the variable x and returns a list of solution equations in x. If expr is not an equation, the equation expr = 0 is assumed in its place. x may be a function (e.g. f(x)), or other non-atomic expression except a sum or product. x may be omitted if expr contains only one variable. expr may be a rational expression, and may contain trigonometric functions, exponentials, etc. breakup if false will cause solve to express the solutions of cubic or quartic equations as single expressions rather than as made up of several common subexpressions which is the default. multiplicities will be set to a list of the multiplicities of the individual solutions returned by solve, realroots, or allroots. Try apropos (solve) for the switches which affect solve. describe may then by used on the individual switch names if their purpose is not clear. It is important to recognise that the rst argument to solve is either an equation such as f(x) = g(x) (or h(x) = 0), or simply h(x); in the latter case, solve understands that you mean the equation h(x) = 0, and the problem is to nd the roots of h(x), ie., values of x such that the equation h(x) = 0 is satised. Here we follow the manual suggestion about using apropos and describe: (%i1) apropos(solve); (%o1) [solve, solvedecomposes, solveexplicit, solvefactors, solvenullwarn, solveradcan, solvetrigwarn, solve_inconsistent_error] (%i2) describe(solveradcan)$ -- Option variable: solveradcan Default value: false' Whensolveradcan' is true',solve' calls radcan' which makessolve' slower but will allow certain problems containing exponentials and logarithms to be solved. (%i3) describe(solvetrigwarn)$ -- Option variable: solvetrigwarn Default value: true' Whensolvetrigwarn' is true',solve' may print a message saying that it is using inverse trigonometric functions to solve the equation, and thereby losing solutions. 4.1.2 solve with Expressions or Functions & the multiplicities List Let's start with a simple example where the expected answers are obvious and check the behavior of solve. In particular we want to check solve's behavior with both an expression and a function (dened via :=). We also want to check how the system list multiplicities is created and maintained. We include the use of realroots and allroots in this comparison, even though we will not have to use these latter two functions for a while. (%i1) multiplicities; (%o1) not_set_yet (%i2) ex1 : x2 - 2x + 1; 2 (%o2) x - 2 x + 1 4 (%i3) factor(ex1); 2 (%o3) (x - 1) (%i4) g(x) := x2 - 2x + 1$ (%i5) g(y); 2 (%o5) y - 2 y + 1 (%i6) solve(ex1); (%o6) [x = 1] (%i7) multiplicities; (%o7) (%i8) solve(g(y)); (%o8) [y = 1] (%i9) multiplicities; (%o9) (%i10) realroots(ex1); (%o10) [x = 1] (%i11) multiplicities; (%o11) (%i12) allroots(ex1); (%o12) [x = 1.0, x = 1.0] (%i13) multiplicities; (%o13) We see that we can use either an expression or a function with solve, and you can check that this also applies to realroots and allroots. It is not clear from our use of allroots above how allroots affects multiplicities, although, as we will see later, the manual does not assert any connection, and we would not expect there to be a connection because allroots returns multiple roots explicitly in %o12. Just to make sure, let's restart Maxima and use only allroots: (%i1) multiplicities; (%o1) not_set_yet (%i2) allroots(x2 - 2x + 1); (%o2) [x = 1.0, x = 1.0] (%i3) multiplicities; (%o3) not_set_yet As we expected, allroots does not affect multiplicities; only solve and realroots set its value. 4.1.3 General Quadratic Equation or Function To get our feet wet, lets turn on the machinery with a general quadratic equation or expression. There are some differences if you employ an expression rather than a function dened with :=. Each method has some advantages and some disadvantages. Let's rst use the function argument, rather than an expression argument. We will later show how the calculation is different if an expression is used. We will step through the process of verifying the solutions and end up with a do loop which will check all the solutions. We will use a function f(x) which depends parametrically on (a,b,c) as the rst argument to solve, and rst see what happens if we don't identify the unknown: how smart is Maxima?? (%i1) f(x) := ax2 + bx + c$ 5 (%i2) f(y); 2 (%o2) a y + b y + c (%i3) sol : solve( f(x) ); More unknowns than equations - solve' Unknowns given : [a, x, b, c] Equations given: 2 [a x + b x + c] -- an error. To debug this try debugmode(true); We see that Maxima cannot read our mind! We must tell Maxima which of the four symbols is to be considered the unknown. From Maxima's point of view (actually the point of view of the person who wrote the code), one equation cannot determine four unknowns, so we must supply the information about which of the four variables is to be considered the unknown. (%i4) sol : solve( f(x),x ); 2 2 sqrt(b - 4 a c) + b sqrt(b - 4 a c) - b (%o4) [x = - --------------------, x = --------------------] 2 a 2 a We see that solve returns the expected list of two possible symbolic solutions. 4.1.4 Checking Solutions with subst or ev and a Do Loop Let's check the rst solution: (%i5) s1 : sol; 2 sqrt(b - 4 a c) + b (%o5) x = - --------------------2 a Now we can use the subst( x = x1, f(x) ) form of the subst function syntax. (%i6) r1 : subst(s1, f(x) ); 2 2 2 (sqrt(b - 4 a c) + b) b (sqrt(b - 4 a c) + b) (%o6) ----------------------- - ------------------------ + c 4 a 2 a (%i7) expand(r1); (%o7) 0 Now that we understand what steps lead to the desired 0, we automate the process using a do loop: (%i8) for i:1 thru 2 do disp( expand( subst( sol[i], f(x) ) ) )$ 0 0 For each of the two solutions (for x) found by Maxima, the given expression evaluates to zero, verifying the roots of the expression. 6 Since the result (here) of using ev( f(x), sol[i]) is the same as using subst( sol[i], f(x) ), we can use ev instead: (%i9) for i:1 thru 2 do disp( expand( ev(f(x), sol[i]) ))$ 0 0 4.1.5 The One Argument Form of solve The simple one-argument form of solve can be used if all but one of the symbols in the expression is already bound. (%i10) solve(3x -2); 2 (%o10) [x = -] 3 (%i11) (a:1, b:2, c:3)$ (%i12) [a,b,c]; (%o12) [1, 2, 3] (%i13) solve(ax2 + bx + c); (%o13) [x = - sqrt(2) %i - 1, x = sqrt(2) %i - 1] (%i14) [a,b,c] : [4,5,6]; (%o14) [4, 5, 6] (%i15) solve(ax2 + bx + c); sqrt(71) %i + 5 sqrt(71) %i - 5 (%o15) [x = - ---------------, x = ---------------] 8 8 (%i16) [a,b,c] : [4,5,6]; (%o16) [4, 5, 6] 4.1.6 Using disp, display, and print We have seen above examples of using disp, which can be used to print out the values of symbols or text, and display, which can be used to print out the name of the symbol and its value in the form of an equation: x = value. Here is the do loop check of the roots of the quadratic found above using print instead of disp. However, we need to be careful, because we are using a function f(x) rather than an expression. We have just assigned the values of a, b, and c, and we want f(x) to have arbitrary values of these parameters. (%i17) [a,b,c]; (%o17) [4, 5, 6] (%i18) f(x); 2 (%o18) 4 x + 5 x + 6 (%i19) kill(a,b,c); (%o19) done (%i20) [a,b,c]; (%o20) [a, b, c] (%i21) f(x); 2 (%o21) a x + b x + c 7 (%i22) sol; 2 2 sqrt(b - 4 a c) + b sqrt(b - 4 a c) - b (%o22) [x = - --------------------, x = --------------------] 2 a 2 a (%i23) for i:1 thru 2 do print("expr = ", expand( subst(sol[i],f(x) ) ) )$ expr = 0 expr = 0 Here we use disp to display a title for the do loop: (%i24) ( disp("check roots"), for i thru 2 do print("expr = ", expand( subst( sol[i],f(x) ) ) ) )$ check roots expr = 0 expr = 0 The only tricky thing about this kind of code is getting the parentheses to balance. Note that that expand(...) is inside print, so the syntax used is do print(... ), ie., a one job do . The outside parentheses allow the syntax ( job1, job2 ). Note also that the default start of the do loop index is 1, so we can use an abbreviated syntax that does not have the i:1 beginning. 4.1.7 Checking Solutions using map One advantage of using a function f(x) dened via := as the rst argument to solve is that it is fairly easy to check the roots by using the map function. We want to use the syntax map( f, solnlist}, where solnlist is a list of the roots (not a list of replacement rules). To get the solution list we can again use map with the syntax map(rhs, sol}. (%i25) solnlist : map( rhs, sol ); 2 2 sqrt(b - 4 a c) + b sqrt(b - 4 a c) - b (%o25) [- --------------------, --------------------] 2 a 2 a (%i26) map( f, solnlist ); 2 2 2 (sqrt(b - 4 a c) + b) b (sqrt(b - 4 a c) + b) (%o26) [----------------------- - ------------------------ + c, 4 a 2 a 2 2 2 (sqrt(b - 4 a c) - b) b (sqrt(b - 4 a c) - b) ----------------------- + ------------------------ + c] 4 a 2 a (%i27) expand(%); (%o27) [0, 0] (%i28) expand( map(f, map(rhs, sol) ) ); (%o28) [0, 0] The last input %i27 shows a compact method which avoids having to name the solnlist and which also avoids having to look at the intermediate output. When you see someone's example written in a compact form like this, you should realize that the someone probably tried out the progression of steps one step at a time (just like we did) to see the correct route, and once the path to the result has been found, reduced the result to 8 the minimum number of steps and names. Often, one does not know in advance which progression of steps will succeed, and one must experiment before nding the true path. You should take apart the compact code, by reading from the inside out (ie., from right to left), and also try getting the result one step at a time to get comfortable with the method and notation being used. 4.1.8 Psuedo-PostFix Code: %% An alternative psuedo-postx (ppf) notation can be used which allows one to read the line from left to right, following the logical succession of procedures being used. Although this ppf notation costs more in keystrokes (an extra pair of outside parentheses, extra commas, and entry of double percent signs %%), the resulting code is usually easier for beginners to follow, and it is easier to mentally balance parentheses as well. As an example, the previous double map check of the roots can be carried out as: (%i29) ( map(rhs,sol), map(f,%%), expand(%%) ); (%o29) [0, 0] Note the beginning and ending parentheses for the whole line of input, with the syntax: ( job1, job2(%%), job3(%%),... ). The system variable %% has the manual description (in part): System variable: %% In compound statements, namely block, lambda, or (s_1, ..., s_n), %% is the value of the previ-ous statement. 4.1.9 Using an Expression Rather than a Function with Solve Let's rework the general quadratic equation solution, including the checks of the solutions, using an expression ex rather than a function f(x) dened using :=. (%i1) ex : ax2 + bx + c$ (%i2) sol : solve( ex, x ); 2 2 sqrt(b - 4 a c) + b sqrt(b - 4 a c) - b (%o2) [x = - --------------------, x = --------------------] 2 a 2 a (%i3) s1 : sol; 2 sqrt(b - 4 a c) + b (%o3) x = - --------------------2 a (%i4) r1 : subst(s1, ex ); 2 2 2 (sqrt(b - 4 a c) + b) b (sqrt(b - 4 a c) + b) (%o4) ----------------------- - ------------------------ + c 4 a 2 a (%i5) expand(r1); (%o5) 0 (%i6) for i:1 thru 2 do disp( expand( subst( sol[i], ex ) ) )$ 0 0 (We could have also used ev instead of subst.) 9 Thus far, the methods have been similar. If we now bind the values of (a,b,c), as we did in the middle of our solutions using f(x), what is the difference? (%i7) [a,b,c] : [1,2,3]$ (%i8) [a,b,c]; (%o8) [1, 2, 3] (%i9) ex; 2 (%o9) a x + b x + c We see that the symbol ex remains bound to the same general expression. The symbol ex retains its original binding. We can make use of the values given to (a,b,c) with the expression ex by using two single quotes, which forces an extra evaluation of the expression ex by the Maxima engine, and which then makes use of the extra information about (a,b,c). (%i10) ''ex; 2 (%o10) x + 2 x + 3 (%i11) ex; 2 (%o11) a x + b x + c Forcing the extra evaluation in %10 does not change the binding of ex. Now let's try to check the solutions using map, as we did before. To use map we need a function, rather than an expression to map on a solution list. Let's try to dene such a function f(x) using the expression ex. (%i12) f(x); (%o12) f(x) (%i13) f(x) := ex; (%o13) f(x) := ex (%i14) f(y); 2 (%o14) a x + b x + c (%i15) f(x) := ''ex; 2 (%o15) f(x) := a x + b x + c (%i16) f(y); 2 (%o16) y + 2 y + 3 (%i17) kill(a,b,c); (%o17) done (%i18) f(y); 2 (%o18) a y + b y + c (%i19) solnlist : map(rhs,sol); 2 2 sqrt(b - 4 a c) + b sqrt(b - 4 a c) - b (%o19) [- --------------------, --------------------] 2 a 2 a 10 (%i20) map(f,solnlist); 2 2 2 (sqrt(b - 4 a c) + b) b (sqrt(b - 4 a c) + b) (%o20) [----------------------- - ------------------------ + c, 4 a 2 a 2 2 2 (sqrt(b - 4 a c) - b) b (sqrt(b - 4 a c) - b) ----------------------- + ------------------------ + c] 4 a 2 a (%i21) expand(%); (%o21) [0, 0] Output %14 showed that the syntax f(x) := ex did not succeed in dening the function we need. The input f(x) := ''ex suceeded in getting a true function of x, but now the function f(x) automatically makes use of the current binding of (a,b,c), so we had to kill those values to get a function with arbitrary values of (a,b,c). Having the function in hand, we again used the map function twice to check the solutions. Now that we have discovered the true path, we can restart Maxima and present the method as: (%i1) ex : ax2 + bx + c$ (%i2) sol : solve( ex, x ); 2 2 sqrt(b - 4 a c) + b sqrt(b - 4 a c) - b (%o2) [x = - --------------------, x = --------------------] 2 a 2 a (%i3) f(x) := ''ex$ (%i4) expand ( map(f, map(rhs, sol) ) ); (%o4) [0, 0] We can also use the (generally safer) syntax define( f(x),ex ); to obtain a true function of x: (%i5) define( f(x),ex ); 2 (%o5) f(x) := a x + b x + c (%i6) f(y); 2 (%o6) a y + b y + c (%i7) expand ( map(f, map(rhs, sol) ) ); (%o7) [0, 0] We can also use the unnamed, anonymous function lambda to avoid introducing needless names, like f: (%i8) expand( map(lambda([x],''ex), map(rhs,sol) ) ); (%o8) [0, 0] 4.1.10 Escape Speed from the Earth In this section we solve a physics problem which involves a simple quadratic equation. It is so simple that do-ing it on paper is faster than doing it with Maxima. In fact, once you understand the plan of the calculation, you can come up with the nal formula for the escape speed in your head. However, we will present practical details of setup and evaluation which can be used with more messy problems, when you might want to use Maxima. Let's use conservation of mechanical energy (kinetic plus potential) to rst calculate the initial radial speed a rocket must have near the surface of the earth to achieve a nal required radial speed far from the earth (far 11 enough away so we can neglect earth's gravitational pull). Let the mass of the rocket be m, the mass of the earth be M, the radius of the earth be R, a general radial distance from the center of the earth be r >= R, a general radial rocket speed be v, the maximum speed of the rocket near the surface of the earth be v0, and the nal radial speed of the rocket (as r becomes innite) be vf. At a general distance r from the center of the earth the rocket has kinetic energy ke = mv2/2, and gravitational energy pe = -GMm/r, where G is the gravitational constant: (G = 6.673 10(-11) newtonmeter2/kg2). (%i1) energy : mv2/2 - GMm/r; 2 m v m G M (%o1) ---- - -----2 r The initial energy e0 corresponds to the energy the rocket has achieved at the moment of maximum radial speed: this will occur at a radius r slightly larger than the radius of the earth R, but negligible error to the required lift-off speed v0 will be made by ignoring this difference in radius. (You can justify this as a good approximation by getting the answer when including this small difference, and comparing the percent difference in the answers.) (%i2) e0 : energy,v=v0,r=R; 2 m v0 m G M (%o2) ----- - -----2 R As the rocket rises, r becomes larger, and the magnitude of the gravitational energy becomes smaller. The nal energy efinal will be the energy when the gravitational energy is so small that we can ignore it; in practice this will occur when the magnitude of the gravitational energy is much smaller than the magnitude of the inital gravitational energy. The radial outward speed of the rocket then remains a constant value vf. (%i3) efinal : limit(energy,r,inf),v=vf; 2 m vf (%o3) -----2 If we neglect the loss of mechanical energy due to friction in leaving the earth's atmosphere, and also neglect other tiny effects like the gravitational interaction between the moon and the rocket, the sun and the rocket, etc, then we can approximately say that the total mechanical energy (as we have dened it) of the rocket is a constant, once chemical energy used to increase the rocket's speed is no longer a factor (which occurs at the moment of maximum radial speed). We can then get one equation by approximately equating the mechanical energy of the rocket just after achiev-ing maximum speed to the mechanical energy of the rocket when r is so large that we can ignore the instanta-neous gravitational energy contribution. (%i4) v0soln : solve(efinal = e0,v0); 2 G M 2 2 G M 2 (%o4) [v0 = - sqrt(----- + vf ), v0 = sqrt(----- + vf )] R R 12 (%i5) v0soln : v0soln; 2 G M 2 (%o5) v0 = sqrt(----- + vf ) R (%i6) v0; (%o6) v0 (%i7) v0 : rhs( v0soln ); 2 G M 2 (%o7) sqrt(----- + vf ) R This provides the required lift-off speed near the surface of the earth to achieve a given nal radial speed vf. We now want to nd the escape speed, the minimum value of the lift-off speed which will allow the rocket to escape the gravitational pull of the earth. Any rocket which has a radial speed when r is effectively innite will succeed in escaping, no matter how small that radial speed is. The limiting initial speed is then gotten by taking the limit of v0 as vf goes to zero. (%i8) vescape : ev( v0, vf = 0 ); G M (%o8) sqrt(2) sqrt(---) R Uing the mass of the earth, M = 5.974 10(24) kg, and the radius of the earth, R = 6.378 10(6) meters, we get an escape speed 11,181 m/s = 11.18 km/s. (%i09) ev( vescape, [G=6.673e-11,M=5.974e24,R=6.378e6] ); (%o09) 7905.892670345354 sqrt(2) (%i10) float(%); (%o10) 11180.62063706845 We have rounded the answer to four signicant gures, since that is the accuracy of the Earth data and gravita-tional constant we have used. If we need to use a set of physical or numerical constants throughout a session, we can dene a list of equali-ties, say clist, and use as follows: (%i11) clist : [G=6.673e-11,M=5.974e24,R=6.378e6]; (%o11) [G = 6.6729999999999999E-11, M = 5.9740000000000004E+24, R = 6378000.0] (%i12) ev( vescape, clist, float ); (%o12) 11180.62063706845 Note that adding the option variable oat as a switch (equivalent to float:true), gets the sqrt(2) in oating point. Looking at all those digits is unnecessary; set fpprintprec to something reasonable (this only affects the numbers presented on the screen, not the accuracy of the calculation): (%i13) fpprintprec:8$ (%i14) ev( vescape, clist, float ); (%o14) 11180.621 (%i15) clist; (%o15) [G = 6.673E-11, M = 5.974E+24, R = 6378000.0] 13 4.1.11 Cubic Equation or Expression Here is an example of using solve to solve a cubic equation, or, in the alternative language, nd the roots of a cubic expression. After checking the roots via the map function, we assign the values of the roots to the symbols(x1,x2,x3). The cubic expression we choose is especially simple, with no arbitrary parameters, so we can use the one argument form of solve. (%i1) ex : x3 + x2 + x$ (%i2) sol : solve(ex); sqrt(3) %i + 1 sqrt(3) %i - 1 (%o2) [x = - --------------, x = --------------, x = 0] 2 2 (%i3) define( f(x), ex )$ (%i4) expand ( map(f, map(rhs, sol) ) ); (%o4) [0, 0, 0] (%i5) [x1,x2,x3] : map(rhs,sol); sqrt(3) %i + 1 sqrt(3) %i - 1 (%o5) [- --------------, --------------, 0] 2 2 (%i6) x1; sqrt(3) %i + 1 (%o6) - --------------2 4.1.12 Trigonometric Equation or Expression Here is an exact solution using solve: (%i1) [fpprintprec:8,display2d:false]$ (%i2) ex : sin(x)2 -2sin(x) -3$ (%i3) sol : solve(ex);solve' is using arc-trig functions to get a solution. Some solutions will be lost. (%o3) [x = asin(3),x = -%pi/2] (%i4) define( f(x), ex )$ (%i5) expand ( map(f, map(rhs, sol) ) ); (%o5) [0,0] (%i6) numroots : float( map(rhs, sol) ); (%o6) [1.5707963-1.7627472%i,-1.5707963] The rst solution returned is the angle (in radians) whose sin is 3. For real x, sin(x) lies in the range -1 <= sin(x) <= 1 . Thus we have found one real root. But we have been warned that some solutions will be lost. Because the given expression is a polynomial in sin(x), we can use realroots: (%i7) rr : realroots(ex); (%o7) [sin(x) = -1,sin(x) = 3] However, by realroots, realroots means that the numbers [-1,3] are real! 14 We can of course take the output of realroots and let solve go to work. (%i8) map(solve, rr); solve' is using arc-trig functions to get a solution. Some solutions will be lost.solve' is using arc-trig functions to get a solution. Some solutions will be lost. (%o8) We know that the numerical value of the expression ex3 repeats when x is replaced by x + 2%pi, so there are an innite number of real roots, related to −π/2 by adding or subtracting 2nπ, where n is an integer. We can make a simple plot of our expression to see the periodic behavior and the approximate location of the real roots. -4 -2 0 2 4 -6 -4 -2 0 2 4 6 x 0.0 sin(x)2-2sin(x)-3 Figure 1: plot of ex3 We used the plot2d code: (%i18) plot2d([0.0,ex3],[x,-6,6],[y,-5,5] )$ 4.1.13 Equation or Expression Containing Logarithmic Functions Here is an example submitted to the Maxima mailing list and a method of solution provided by Maxima developer Stavros Macrakis. The problem is to nd the roots of the following expression ex: (%i1) [fpprintprec:8,display2d:false,ratprint:false]$ (%i2) ex : log(0.25(2x+5)) - 0.5log(5x - 5)$ We rst try solve, with the option variable solveradcan set equal to true. Remember that the syntax func, optvar; is equivalent to func, optvar:true;. (%i3) sol : solve(ex,x),solveradcan; (%o3) [log((2x+5)/4) = log(5x-5)/2] 15 We see that solve tried to nd a simpler form which it returned in %o2. The Maxima function fullratsimp has the manual description Function: fullratsimp(expr) fullratsimp repeatedly applies ratsimp followed by non-rational simplication to an expression until no further change occurs, and returns the result. When non-rational expressions are involved, one call to ratsimp followed as is usual by non-rational (general) simplication may not be sufcient to return a simplied result. Sometimes, more than one such call may be necessary. fullratsimp makes this process convenient. The effect of fullratsimp in our case results in the decimals being replaced by exact fractions. (%i4) ex : fullratsimp(ex); (%o4) (2log((2x+5)/4)-log(5x-5))/2 The logarithms can be combined using the Maxima function logcontract. This function was discussed in Chapter 2, Sect. 2.3.5. A partial description is: Function: logcontract(expr) Recursively scans the expression expr, transforming subexpressions of the form a1log(b1) + a2log(b2) + c into log(ratsimp(b1a1 b2a2)) + c (%i1) 2(alog(x) + 2alog(y))$ (%i2) logcontract(%); 2 4 (%o2) a log(x y ) Here is the application to our problem: (%i5) ex : logcontract(ex); (%o5) -log((80x-80)/(4x2+20x+25))/2 Having combined the logarithms, we try out solve on this expression: (%i6) sol : solve(ex ); (%o6) [x = -(2sqrt(30)-15)/2,x = (2sqrt(30)+15)/2] We have a successful exact solution, but solve (in its present incarnation) needed some help. We now use the map method to check the roots. (%i7) define( f(x), ex )$ (%i8) expand( map(f, map(rhs,sol) ) ); (%o8) [0,0] 4.2 One Equation Numerical Solutions: allroots, realroots, nd root We have already tried out the Maxima functions realroots and allroots. The most important restriction for both of these numerical methods is that the expression or equation be a polynomial, as the manual explains: Function: realroots(eqn, bound) Function: realroots(expr) Function: realroots(eqn) Computes rational approximations of the real roots of the polynomial expr or polynomial equation eqn of one variable, to within a tolerance of bound. Coefcients of expr or eqn must be literal numbers; symbol constants such as %pi are rejected. realroots assigns the multiplicities of the roots it nds to the global variable multiplicities. 16 realroots constructs a Sturm sequence to bracket each root, and then applies bisection to rene the approximations. All coefcients are converted to rational equivalents before searching for roots, and com-putations are carried out by exact rational arithmetic. Even if some coefcients are oating-point numbers, the results are rational (unless coerced to oats by the float or numer ags). When bound is less than 1, all integer roots are found exactly. When bound is unspecied, it is assumed equal to the global variable rootsepsilon (default: 10(-7)). When the global variable programmode is true (default: true), realroots returns a list of the form [x = , x = , ...]. When programmode is false, realroots creates interme-diate expression labels %t1, %t2, ..., assigns the results to them, and returns the list of labels. Here are the startup values of the option variables just mentioned: (%i1) fpprintprec:8$ (%i2) [multiplicities,rootsepsilon,programmode]; (%o2) [not_set_yet, 1.0E-7, true] The function allroots also accepts only polynomials, and nds numerical approximations to both real and complex roots: Function: allroots(expr) Function: allroots(eqn) Computes numerical approximations of the real and complex roots of the polynomial expr or polynomial equation eqn of one variable. The ag polyfactor when true causes allroots to factor the polynomial over the real numbers if the polynomial is real, or over the complex numbers, if the polynomial is complex (default setting of polyfactor is false). allroots may give inaccurate results in case of multiple roots. If the polynomial is real, allroots (%ip)) may yield more accurate approximations than allroots(p), as allroots invokes a different algorithm in that case. allroots rejects non-polynomials. It requires that the numerator after rat'ing should be a polynomial, and it requires that the denominator be at most a complex number. As a result of this, allroots will always return an equivalent (but factored) expression, if polyfactor is true. Here we test the default value of polyfactor: (%i3) polyfactor; (%o3) false 4.2.1 Comparison of realroots with allroots Let's nd the real and complex roots of a fth order polynomial which solve cannot solve, doesn't factor, and use both realroots and allroots. (%i4) ex : x5 + x4 -4x3 +2x2 -3x -7$ (%i5) define( fex(x), ex )$ We rst use realroots to nd the three real roots of the given polynomial, and substitute the roots back into the expression to see how close to zero we get. (%i6) rr : float( map(rhs, realroots(ex,1e-20) ) ); (%o6) [- 2.7446324, - 0.880858, 1.7964505] (%i7) frr : map( fex, rr ); (%o7) [0.0, - 4.4408921E-16, 0.0] 17 Next we nd numerical approximations to the three real roots and the two (complex-conjugate) roots of the given polynomial, using allroots( ex ) and substitute the obtained roots back into the expression to see how close to zero we get. (%i8) ar1 : map(rhs, allroots( ex ) ); (%o8) [1.1999598 %i + 0.41452, 0.41452 - 1.1999598 %i, - 0.880858, 1.7964505, - 2.7446324] (%i9) far1 : expand( map( fex, ar1 ) ); (%o9) [1.12132525E-14 %i + 4.4408921E-16, 4.4408921E-16 - 1.12132525E-14 %i, - 1.13242749E-14, 2.48689958E-14, - 2.84217094E-14] Finally, we repeat the process for the syntax allroots( %i ex ). (%i10) ar2 : map(rhs, allroots( %iex ) ); (%o10) [1.1999598 %i + 0.41452, - 3.60716392E-17 %i - 0.880858, 0.41452 - 1.1999598 %i, 6.20555942E-15 %i + 1.7964505, - 6.54444294E-16 %i - 2.7446324] (%i11) far2 : expand( map( fex, ar2 ) ); (%o11) [1.55431223E-15 %i - 1.77635684E-15, 5.61204464E-16 %i - 4.4408921E-16, 1.61204383E-13 %i + 2.26041408E-13, 2.52718112E-13 %i - 1.91846539E-13, - 6.32553289E-14 %i - 3.97903932E-13] (%i12) far2 - far1; (%o12) [- 9.65894031E-15 %i - 2.22044605E-15, 1.1774457E-14 %i - 8.8817842E-16, 1.61204383E-13 %i + 2.37365683E-13, 2.52718112E-13 %i - 2.16715534E-13, - 6.32553289E-14 %i - 3.69482223E-13] The three real roots of the given fth order polynomial are found more accurately by realroots than by either version of allroots. We see that the three real roots of this fth order polynomial are found more accurately by the syntax allroots(expr) (which was used to get ar1), than by the syntax allroots(%iexpr), used to get ar2. We also see that the syntax allroots(%iexpr) introduced a tiny complex piece to the dominant real part. The two extra complex roots found by the rst syntax (ar1) are the complex conjugate of each other. The two extra complex roots found by the alternative syntax (ar2) are also the complex conjugate of each other to within the default arithmetic accuracy being used. 4.2.2 Intersection Points of Two Polynomials Where do the two curves h(x) = x3 −8 x2 + 19 x −12 and k(x) = 1 2 x2 −x −1 8 intersect? We want approximate numerical values. We can plot the two functions together and use the cursor to read off the values of x for which the curves cross, and we can also nd the roots numerically. We rst dene the curves as expressions depending on x, then dene the difference of the expressions (rx) to work with using allroots rst to see if all the (three) roots are real, and then using realroots just for fun, and then checking the solutions. If we are just going to compare the results with a plot, we don't need any great accuracy, so we will use the default realroots precision. (%i1) fpprintprec : 8$ (%i2) hx : x3 - 8x2 + 19x - 12$ (%i3) kx : x2/2 -x -1/8$ (%i4) rx : hx - kx; 2 3 17 x 95 (%o4) x - ----- + 20 x - --2 8 18 (%i5) factor(rx); 3 2 8 x - 68 x + 160 x - 95 (%o5) -------------------------8 (%i6) define( fr(x), rx )$ (%i7) allroots(rx); (%o7) [x = 0.904363, x = 2.6608754, x = 4.9347613] (%i8) rr : float( realroots(rx) ); (%o8) [x = 0.904363, x = 2.6608754, x = 4.9347613] (%i9) rr : map( rhs, rr); (%o9) [0.904363, 2.6608754, 4.9347613] (%i10) map(fr, rr); (%o10) [2.04101367E-8, - 8.4406409E-8, 5.30320676E-8] We see that the numerical solutions are zeros of the cubic function to within the numerical accuracy realroots is using. Just out of curiosity, what about exact solutions of this cubic polynomial? Use of solve will generate a complicated looking expression involving roots and %i. Let's set display2d to false so the output doesn't take up so much room on the screen. (%i11) display2d : false$ (%i12) sx : solve(rx); (%o12) [x = (-sqrt(3)%i/2-1/2)(3-(3/2)sqrt(16585)%i/16+151/432)(1/3) +49(sqrt(3)%i/2-1/2)/(36(3-(3/2)sqrt(16585)%i/16+151/432) (1/3))+17/6, etc, etc ] (%i13) sx1 : map(rhs, sx); (%o13) [(-sqrt(3)%i/2-1/2)(3-(3/2)sqrt(16585)%i/16+151/432)(1/3) +49(sqrt(3)%i/2-1/2)/(36(3-(3/2)sqrt(16585)%i/16+151/432) (1/3))+17/6, etc, etc ] The list sx1 holds the exact roots of the cubic polynomial which solve found. We see that the form returned has explicit factors of %i. We already know that the roots of this polynomial are purely real. How can we get the exact roots into a form where it is obvious that the roots are real? The Maxima expert Alexey Beshenov (via the Maxima mailing list) suggested using rectform, followed by trigsimp. Using rectform gets rid of the factors of %i, and trigsimp does some trig simplication. (%i14) sx2 : rectform(sx1); (%o14) [49(3sqrt(3)sin(atan(4323-(3/2)sqrt(16585)/(15116))/3)/7 -3cos(atan(4323-(3/2)sqrt(16585)/(15116))/3)/7)/36 +7sqrt(3)sin(atan(4323-(3/2)sqrt(16585)/(15116))/3)/12 +%i(-7sin(atan(4323-(3/2)sqrt(16585)/(15116))/3)/12 +49(3sin(atan(4323-(3/2)sqrt(16585)/(15116))/3)/7 +3sqrt(3)cos(atan(4323-(3/2)sqrt(16585)/(15116))/3)/7)/36 -7sqrt(3)cos(atan(4323-(3/2)sqrt(16585)/(15116))/3)/12) -7cos(atan(4323-(3/2)sqrt(16585)/(15116))/3)/12+17/6, etc,etc ] (%i15) sx3 : trigsimp(sx2); (%o15) [ (7sqrt(3)sin(atan(9sqrt(16585)/(151sqrt(3)))/3) -7cos(atan(9sqrt(16585)/(151sqrt(3)))/3)+17)/6, -(7sqrt(3)sin(atan(9sqrt(16585)/(151sqrt(3)))/3) +7cos(atan(9sqrt(16585)/(151sqrt(3)))/3)-17)/6, (14cos(atan(9sqrt(16585)/(151sqrt(3)))/3)+17)/6 ] 19 The quantity sx3 is a list of the simplied exact roots of the cubic. Using oat we ask for the numerical values: (%i16) sx4 : float(sx3); (%o16) [2.6608754,0.904363,4.9347613] We see that the numerical values agree, although the order of the roots is different. Next we enquire whether or not the exact roots, when substituted back into the cubic, result in exact zeroes. Mapping the cubic onto the list of roots doesn't automatically simplify to a list of three zeroes, as we would like, although applying oat suggests the analytic roots are correct. The combination trigsimp( expand( [fr(root1), fr(root2), fr(root3)] ) ) still does not provide the alge-braic and trig simplication needed, but we nally get [0.0, 0.0, 0.0] when applying oat. (%i17) float( map(fr, sx3) ); (%o17) [-3.55271368E-15,1.88737914E-15,-1.42108547E-14] (%i18) float( expand( map(fr, sx3) ) ); (%o18) [6.66133815E-16,-2.44249065E-15,0.0] (%i19) float( trigsimp( expand( map(fr, sx3) ) ) ); (%o19) [0.0,0.0,0.0] Let's next plot the three functions, using the expressions hx, kx, and rx (the difference). -15 -10 -5 0 5 10 0 1 2 3 4 5 x hx kx rx=hx - kx Figure 2: Intersection Points are Zeroes of rx Here is code you can use to make something close to the above plot. (%i20) plot2d([hx,kx,rx],[x,0,5], [style, [lines,2,1], [lines,2,2], [lines,2,0] ], [legend, "hx", "kx", "rx=hx - rx"], [gnuplot_preamble, " set xzeroaxis lw 2 "])$ When you place the cursor over the places on the x axis where the expression rx is zero, you can read off the coordinates in the lower left corner of the plot window. The x coordinate (the rst) is the desired root. 20 4.2.3 Transcendental Equations and Roots: nd root A transcendental equation is an equation containing a transcendental function. Examples of such equations are x = ex and x = sin(x). The logarithm and the exponential function are examples of transcendental functions. We will include the trigonometric functions, i.e., sine, cosine, tangent, cotangent, secant, and cosecant in this category of functions. (A function which is not transcendental is said to be algebraic. Examples of algebraic functions are rational functions and the square root function.) To nd the roots of transcendental expressions, for example, we can rst make a plot of the expression, and then use nd root knowing roughly where nd root should start looking. The Maxima manual provides a lot of details, beginning with: Function: nd root(expr, x, a, b) Function: nd root(f, a, b) Option variable: nd root error Option variable: nd root abs Option variable: nd root rel Finds a root of the expression expr or the function f over the closed interval [a, b]. The expression expr may be an equation, in which case find_root seeks a root of lhs(expr) - rhs(expr). Given that Maxima can evaluate expr or f over [a, b] and that expr or f is continuous, find_root is guaranteed to nd the root, or one of the roots if there is more than one. Let's nd a root of the equation x = cos(x) . If we make a simple plot of the function x −cos(x), we see that there is one root somewhere between x = 0 and x = 1. -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0 0.2 0.4 0.6 0.8 1 plot of x - cos(x) Figure 3: Plot of x - cos(x) We can use either an expression or function as the entry to the rst slot of nd root. I nd that the most common mistake I make with nd root is to leave out the underline between nd and root, in which case, I simply get back the unevaluated ndroot(ex, x, 0, 1), since Maxima has no knowledge of ndroot (unless I create a homemade function with that name). 21 We can make a plot, nd the root in a variety of ways using nd root, and verify the accuracy of the root as follows: (%i1) fpprintprec:8$ (%i2) plot2d( x - cos(x), [ x, 0, 1 ], [style, [lines, 4, 1] ], [xlabel," plot of x - cos(x) "], [gnuplot_preamble, "set nokey; set xzeroaxis lw 2 "] )$ (%i3) find_root( x - cos(x),x, 0, 1); (%o3) 0.739085 (%i4) ex : x - cos(x)$ (%i5) [find_root( ex, x, 0, 1),find_root( ex, 0, 1)]; (%o5) [0.739085, 0.739085] (%i6) define( f(x), ex )$ (%i7) [find_root(f(x), x, 0, 1), find_root(f(x), 0, 1), find_root(f, 0, 1), find_root(f, x, 0, 1)]; (%o7) [0.739085, 0.739085, 0.739085, 0.739085] (%i8) ev(ex, x = first(%) ); (%o8) 0.0 As a second example, we will nd the roots of the function f(x) = cos(x/π) e−(x/4)2 −sin(x3/2) −5/4 Here is the plot of f(x). -2.5 -2 -1.5 -1 -0.5 0 0.5 0 1 2 3 4 5 plot of f(x) Figure 4: Plot of f(x) This plot shows roots near the points [x = 2.53, x = 2.97]. 22 Here is our session making a plot and using nd root: (%i1) fpprintprec:8$ (%i2) f(x):= cos(x/%pi)exp(-(x/4)2) - sin(x(3/2)) - 5/4$ (%i3) plot2d( f(x),[x,0,5], [style, [lines,4,1] ], [xlabel," plot of f(x) "],[ylabel," "], [gnuplot_preamble, "set nokey; set xzeroaxis lw 2 "] )$ (%i4) [find_root(f,2.5,2.6), find_root(f, x, 2.5, 2.6), find_root(f(x),x,2.5,2.6), find_root(f(y),y,2.5,2.6)]; (%o4) [2.5410501, 2.5410501, 2.5410501, 2.5410501] (%i5) [x1 : find_root(f,2.5,2.6),x2 : find_root(f, 2.9, 3.0 )]; (%o5) [2.5410501, 2.9746034] (%i6) float( map(f, [x1,x2] ) ); (%o6) [3.33066907E-16, 2.77555756E-16] We see that the numerical accuracy of nd root using default behavior is the normal accuracy of Maxima arithmetic. 4.2.4 nd root: Quote that Function! The Maxima function nd root is an unusual function. The source code which governs the behavior of nd root has been purposely designed to allow uses like the following: (%i1) fpprintprec:8$ (%i2) find_root(diff(cos(2x)sin(3x)/(1+x2),x), x, 0, 0.5); (%o2) 0.321455 nd root evaluates the derivative and parks the resulting expression in the code function fr(x),say, to be used to probe where the function changes sign, as the code executes a bisection search for a root in the range called for. The code rst checks that the sign of the function fr(x) has opposite signs at the end points of the given range. Next the code makes small steps in one direction, checking at each step if a sign change has occurred. As soon as a sign change has occurred, the code backs up one step, cuts the size of the step in half, say, and starts stepping again, looking for that sign change again. This is a brute force method which will nd that root if there is one in the given interval. Of course the user can always evaluate the derivative and submit the resulting expression to nd root to nd the same answer, as in: (%i3) ex : trigsimp( diff(cos(2x)sin(3x)/(1+x2),x) ); 2 (%o3) - (((2 x + 2) sin(2 x) + 2 x cos(2 x)) sin(3 x) 2 4 2 + (- 3 x - 3) cos(2 x) cos(3 x))/(x + 2 x + 1) (%i4) plot2d([0.0,ex],[x,-3,3])$ (%i5) find_root(ex,x,0,0.5); (%o5) 0.321455 23 If we assign the delayed derivative to an expression ex1, we can then use an argument ev(ex1,diff) to nd root as in: (%i6) ex1 : 'diff(cos(2x)sin(3x)/(1+x2),x); d cos(2 x) sin(3 x) (%o6) -- (-----------------) dx 2 x + 1 (%i7) find_root(ev(ex1,diff),x,0,0.5); (%o7) 0.321455 If we assign the delayed derivative to a function g(x), and assign ev(g(x),diff) to another function k(x), we can use nd root as in: (%i8) g(x) := 'diff(cos(2x)sin(3x)/(1+x2),x); d cos(2 x) sin(3 x) (%o8) g(x) := -- (-----------------) dx 2 1 + x (%i9) k(x) := ev(g(x),diff); (%o9) k(x) := ev(g(x), diff) (%i10) find_root(k(x),x,0,0.5); (%o10) 0.321455 or just use the syntax: (%i11) find_root( ev(g(x),diff),x,0,0.5 ); (%o11) 0.321455 However, the following syntax which leaves out the variable name produces an error message: (%i12) find_root( k, 0, 0.5 ); Non-variable 2nd argument to diff: 0.0 #0: g(x=0.0) #1: k(x=0.0) -- an error. To debug this try debugmode(true); In the above cases we want nd root to use the default initial evaluation of the rst slot argument before going to work looking for the root. The important thing to stress is that the nd root source code, by default, is designed to evaluate the rst slot expression before beginning the checking of the sign of the resulting function (after evaluation) at the end points and proceeding with the bisection search. If the user wants to use (for some reason) a function f(x) in the rst slot of nd root and wishes to prevent the initial evaluation of the rst slot expression, the user should use the syntax '(f(x)) for the rst slot entry, rather than f(x). However, the code is loosely written so that most of the time you can get correct results without using the single quote syntax '(f(x)). All of the following examples give the correct answer for this simple function. (%i13) f(x) := x - cos(x)$ (%i14) [find_root( f, 0, 1), find_root( 'f, 0, 1), find_root( '(f), 0, 1), find_root( f(x), x, 0, 1), find_root( '( f(x)), 'x, 0, 1),find_root( '( f(x)), x, 0, 1), find_root( 'f(x), x, 0, 1)]; (%o14) [0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085] 24 However, one can arrive at specialized homemade functions which require the strict syntax (a quoted func-tion entry in the rst slot) to behave correctly. Suppose we need to nd the numerical roots of a function dened by an integral. The following is a toy model which uses such a function in a case where we know the answer ahead of time. Instead of directly looking for the roots of the function f(x) = (x2 −5), we look for the roots of the function R x √ 5 2 y d y. (%i1) fpprintprec : 8$ (%i2) ex : integrate(2'y,'y,sqrt(5),x); 2 x 5 (%o2) 2 (-- - -) 2 2 (%i3) ex : expand(ex); 2 (%o3) x - 5 (%i4) define( f(x), ex ); 2 (%o4) f(x) := x - 5 (%i5) solve( ex ); (%o5) [x = - sqrt(5), x = sqrt(5)] (%i6) rr : float( map(rhs,%) ); (%o6) [- 2.236068, 2.236068] (%i7) map(f,rr); (%o7) [8.8817842E-16, 8.8817842E-16] (%i8) find_root(f,0,4); (%o8) 2.236068 Let's concentrate on nding the root 2.23606797749979 using the indirect route. Dene a function g(x) in terms of integrate: (%i9) g(x) := block([numer,keepfloat,y], numer:true,keepfloat:true, integrate(2y,y,sqrt(5),x) )$ (%i10) map(g, [1,2,3]); (%o10) [- 4.0, - 1.0, 4.0] (%i11) map(f, [1,2,3]); (%o11) [- 4, - 1, 4] (%i12) [find_root( g, 1, 4), find_root( g(x),x,1,4), find_root( '(g(x)),'x,1,4 ), find_root( 'g(x),x,1,4 )]; (%o12) [2.236068, 2.236068, 2.236068, 2.236068] We see that we have no problems with getting the function g(x) to work with nd root. However, if we replace integrate with quad qags, we nd problems. First let's show the numerical integration routine quad qags at work by itself: (%i13) quad_qags(2'y,'y,sqrt(5),2); (%o13) [- 1.0, 1.11076513E-14, 21, 0] (%i14) g(2.0); (%o14) - 1.0 The quad qags function returns a list of four items, the rst being the numerical value of the integral, the second being an estimate of the error of the answer calculated, the third being the number of function evaluations required, and the last an error code. The returned error code 0 means no problems were encountered, and we will write a function which ignores the error code returned, although in real life we would always want to report an error code value which was not 0. 25 Here we dene h(x) which employs the function quad qags to carry out the numerical itegration. (%i15) h(x) := block([numer,keepfloat,y,qlist], numer:true,keepfloat:true, qlist : quad_qags(2y,y,sqrt(5),x), qlist )$ (%i16) map(h,[1,2,3]); (%o16) [- 4.0, - 1.0, 4.0] (%i17) map(g,[1,2,3]); (%o17) [- 4.0, - 1.0, 4.0] The function h(x) does the same job as g(x), but uses quad qags instead of integrate. Now let's use h(x) in the Maxima function nd root. (%i18) find_root( h(x),x,1,4); function has same sign at endpoints [f(1.0) = - 5.0, f(4.0) = - 5.0] -- an error. To debug this try debugmode(true); We see that the syntax find_root( h(x),x,1,4) produced an error due to the way nd root evaluates the rst slot. Somehow nd root assigned -5.0 to the internal function fr(x) used to look for the root, and in the rst steps of that root location, checking for a difference in sign of fr(x) and the range end points, found the value -5.0 at both ends. In effect, the code was working on the problem find_root( -5.0, x, 1, 4). The following methods succeed. (%i19) [find_root( h, 1, 4),find_root( '(h(x)),'x,1,4 ), find_root( '(h(x)),x,1,4 ),find_root( 'h(x),x,1,4 )]; (%o19) [2.236068, 2.236068, 2.236068, 2.236068] 4.2.5 newton The Maxima function newton can also be used for numerical solutions of a single equation. The Maxima manual describes the syntax as: Function: newton(expr, x, x_0, eps) Returns an approximate solution of expr = 0 by Newton's method, considering expr to be a function of one variable, x. The search begins with x = x_0 and proceeds until abs(expr) < eps (with expr evaluated at the current value of x). newton allows undened variables to appear in expr, so long as the termination test abs(expr) < eps evaluates to true or false. Thus it is not necessary that expr evaluate to a number. load(newton1) loads this function. The two examples in the manual are instructive: (%i1) fpprintprec:8$ (%i2) load (newton1); (%o2) C:/PROGRA1/MAXIMA3.0/share/maxima/5.14.0/share/numeric/newton1.mac (%i3) newton (cos (u), u, 1, 1/100); (%o3) 1.5706753 (%i4) ev (cos (u), u = %); (%o4) 1.21049633E-4 (%i5) assume (a > 0); (%o5) [a > 0] 26 (%i6) newton (x2 - a2, x, a/2, a2/100); (%o6) 1.0003049 a (%i7) ev (x2 - a2, x = %); 2 (%o7) 6.09849048E-4 a Of course both of these examples are found exactly by solve: (%i8) solve( cos(x) ); solve' is using arc-trig functions to get a solution. Some solutions will be lost. %pi (%o8) [x = ---] 2 (%i9) float(%); (%o9) [x = 1.5707963] (%i10) solve( x2 - a2,x ); (%o10) [x = - a, x = a] I nd the source code (Windows XP) in the folder c:\Program Files\Maxima-5.14.0\share\maxima\5.14.0\share\numeric. Here is the code in the le newton1.mac: newton(exp,var,x0,eps):= block([xn,s,numer], numer:true, s:diff(exp,var), xn:x0, loop, if abs(subst(xn,var,exp))<eps then return(xn), xn:xn-subst(xn,var,exp)/subst(xn,var,s), go(loop) )$ We see that the code implements the Newton-Raphson algorithm. Given a function f(x) and an initial guess xg which can be assigned to, say, xi, a closer approximation to the value of x for which f(x) = 0 is generated by xi+1 = xi −f(xi) f ′(xi). The method depends on being able to evaluate the derivative of f(x) which appears in the denominator. Steven Koonin's (edited) comments (Computational Physics: Fortran Version, Steven E. Koonin and Dawn Meredith, WestView Press, 1998, Ch. 1, Sec.3) are cautionary: When the function f(x) is badly behaved near its root (e.g., there is an inection point near the root) or when there are several roots, the automatic Newton-Raphson method can fail to converge at all or converge to the wrong answer if the initial guess for the root is poor. 27 4.3 Two or More Equations: Symbolic and Numerical Solutions For sets of equations, we can use solve with the syntax: Function: solve([eqn_1, ..., eqn_n], [x_1, ..., x_n]) solve ([eqn_1, ..., eqn_n], [x_1, ..., x_n]) solves a system of simultaneous (linear or non-linear) polynomial equations by calling linsolve or algsys and returns a list of the solution lists in the variables. In the case of linsolve this list would contain a single list of solutions. This form of solve takes two lists as arguments. The rst list represents the equations to be solved; the second list is a list of the unknowns to be determined. If the total number of variables in the equations is equal to the number of equations, the second argument-list may be omitted. 4.3.1 Numerical or Symbolic Linear Equations with solve or linsolve The Maxima functions solve, linsolve, and linsolve by lu can be used for linear equations. Linear equations containing symbolic coefcients can be solved by solve and linsolve. For example the pair of equations ax + by = c, dx + ey = f . Here we solve for the values of (x,y) which simultaneously satisfy these two equations and check the solutions. (%i1) eqns : [ax + by = c, dx + ey = f]; (%o1) [b y + a x = c, e y + d x = f] (%i2) solve(eqns,[x,y]); c e - b f c d - a f (%o2) b d - a e b d - a e (%i3) soln : linsolve(eqns,[x,y]); c e - b f c d - a f (%o3) [x = - ---------, y = ---------] b d - a e b d - a e (%i4) (ev(eqns, soln), ratexpand(%%) ); (%o4) [c = c, f = f] Note the presence of the determinant of the coefcient matrix in the denominator of the solutions. A simple numerical (rather than symbolic) two equation example: (%i1) eqns : [3x-y=4,x+y=2]; (%o1) [3 x - y = 4, y + x = 2] (%i2) solns : solve(eqns,[x,y]); 3 1 (%o2) 2 2 (%i3) soln : solns; 3 1 (%o3) [x = -, y = -] 2 2 (%i4) for i thru 2 do disp( ev( eqns[i],soln ) )$ 4 = 4 2 = 2 28 Using linsolve instead returns a list, rather than a list of a list. (%i5) linsolve(eqns,[x,y]); 3 1 (%o5) [x = -, y = -] 2 2 4.3.2 Matrix Methods for Linear Equation Sets: linsolve by lu The present version (5.14) of the Maxima manual does not have an index entry for the function linsolve by lu. These notes include only the simplest of many interesting examples described in two mailing list responses by the creator of the linear algebra package, Barton Willis (Dept. of Mathematics, Univ. Nebraska at Kearney), dated Oct. 27, 2007 and Nov. 21, 2007. If we re-cast the two equation problem we have just been solving in the form of a matrix equation A . xcol = bcol, we need to construct the square matrix A so that matrix multiplication by the column vector xcol results in a column vector whose rows contain the left hand sides of the equations. The column vector bcol rows hold the right hand sides of the equations. Our notation below ( as xycol and xylist ) is only natural for a two equation problem (ie., a 2 x 2 matrix): you can invent your own notation to suit your problem. If the argument of the function matrix is a simple list, the result is a row vector (a special case of a matrix object). We can then take the transpose of the row matrix to obtain a column matrix, such as xcol below. A direct denition of a two element column matrix is matrix([x],[y]), which is probably faster than transpose(matrix([x,y])). To reduce the amount of space taken up by the default matrix output, we can set display2d:false. (%i6) m : matrix( [3,-1],[1,1] ); [ 3 - 1 ] (%o6) [ ] [ 1 1 ] (%i7) display2d:false$ (%i8) m; (%o8) matrix([3,-1],[1,1]) (%i9) xcol : matrix([x],[y]); (%o9) matrix([x],[y]) (%i10) m . xcol; (%o10) matrix([3x-y],[y+x]) The period allows non-commutative multiplication of matrices. The linear algebra package function lin-solve by lu allows us to specify the column vector bcol as a simple list: (%i11) b : [4,2]; (%o11) [4,2] (%i12) linsolve_by_lu(m,b); (%o12) [matrix([3/2],[1/2]),false] (%i13) xycol : first(%); (%o13) matrix([3/2],[1/2]) (%i14) m . xycol - b; (%o14) matrix(,) (%i15) xylist : makelist( xycol[i,1],i,1,length(xycol) ); (%o15) [3/2,1/2] (%i16) xyrules : map("=",[x,y],xylist); (%o16) [x = 3/2,y = 1/2] 29 The matrix argument needs to be a square matrix. The output of linsolve by lu is a list: the rst element is the solution column vector which for this two dimensional problem we have called xycol. We check the solution in input %i14. The output %o14 is a column vector each of whose elements is zero; such a column vector is ordinarily replaced by the number 0. One should always check solutions when using computer algebra software, since the are occasional bugs in the algorithms used. The second list element returned by linsolve by lu is false, which should always be the value returned when the calculation uses non-oating point numbers as we have here. If oating point numbers are used, the second element should be either false or a lower bound to the matrix condition number. (We show an example later.) We have shown how to convert the returned xycol matrix object into an ordinary list, and how to then construct a list of replacement rules (as would be returned by linsolve) which could then be used for other purposes. The use of makelist is the recommended way to use parts of matrix objects in lists. However, here is a simple method which avoids makelist: (%i17) flatten( args( xycol)); (%o17) [3/2,1/2] but makelist should normally be the weapon of choice, since the method is foolproof and can be extended to many exotic ways of using the various elements of a matrix. The Maxima function linsolve by lu allows the second argument to be either a list (as in the example above) or a column matrix, as we show here. (%i18) bcol : matrix(,)$ (%i19) linsolve_by_lu(m,bcol); (%o19) [matrix([3/2],[1/2]),false] (%i20) m . first(%) - bcol; (%o20) matrix(,) 4.3.3 Symbolic Linear Equation Solutions: Matrix Methods Here we use linsolve by lu for the pair of equations ax + by = c, dx + ey = f, seeking the values of (x,y) which simultaneously satisfy these two equations and checking the solutions. (%i1) display2d:false$ (%i2) m : matrix( [a,b], [d,e] )$ (%i3) bcol : matrix( [c], [f] )$ (%i4) ls : linsolve_by_lu(m,bcol); (%o4) [matrix([(c-b(f-cd/a)/(e-bd/a))/a],[(f-cd/a)/(e-bd/a)]),false] (%i5) xycol : ratsimp( first(ls) ); (%o5) matrix([-(bf-ce)/(ae-bd)],[(af-cd)/(ae-bd)]) (%i6) ( m . xycol - bcol, ratsimp(%%) ); (%o6) matrix(,) (%i7) (display2d:true,xycol); [ b f - c e ] [ - --------- ] [ a e - b d ] (%o7) [ ] [ a f - c d ] [ ---------] [ a e - b d ] (%i8) determinant(m); (%o8) a e - b d 30 We see the value of the determinant of the coefcient matrix m in the denominator of the solution. 4.3.4 Multiple Solutions from Multiple Right Hand Sides Next we show how one can solve for multiple solutions (with one call to linsolve by lu) corresponding to a number of different right hand sides. We will turn back on the default matrix display, and re-dene the rst (right hand side) column vector as b1col, and the corresponding solution x1col. (%i21) display2d:true$ (%i22) b1col : matrix( , ); [ 4 ] (%o22) [ ] [ 2 ] (%i23) x1col : first( linsolve_by_lu(m,b1col) ); [ 3 ] [ - ] [ 2 ] (%o23) [ ] [ 1 ] [ - ] [ 2 ] (%i24) b2col : matrix( , ); [ 3 ] (%o24) [ ] [ 1 ] (%i25) x2col : first( linsolve_by_lu(m, b2col) ); [ 1 ] (%o25) [ ] [ 0 ] (%i26) bmat : matrix( [4,3], [2,1] ); [ 4 3 ] (%o26) [ ] [ 2 1 ] (%i27) linsolve_by_lu( m, bmat ); [ 3 ] [ -1 ] [ 2 ] (%o27) [[ ], false] [ 1 ] [ -0 ] [ 2 ] (%i28) xsolns : first(%); [ 3 ] [ -1 ] [ 2 ] (%o28) [ ] [ 1 ] [ -0 ] [ 2 ] 31 (%i29) m . xsolns - bmat; [ 0 0 ] (%o29) [ ] [ 0 0 ] (%i30) x1col : col(xsolns,1); [ 3 ] [ - ] [ 2 ] (%o30) [ ] [ 1 ] [ - ] [ 2 ] (%i31) x2col : col(xsolns,2); [ 1 ] (%o31) [ ] [ 0 ] In input %i26 we dene the 2 x 2 matrix bmat whose rst column is b1col and whose second column is b2col. Using bmat as the second argument to linsolve by lu results in a return list whose rst element (what we call xsolns) is a 2 x 2 matrix whose rst column is x1col (the solution vector corresponding to b1col) and whose second column is x2col (the solution vector corresonding to b2col). In input %i29 we check both solutions simultaneously. The result is a matrix with every element equal to zero, which would ordinarily be replaced by the number 0. In input %i30 we extract x1col using the col function. 4.3.5 Three Linear Equation Example We next consider a simple three linear equation example. Although solve does not require the list [x,y,z] in this problem, if you leave it out, the solution list returned will be in an order determined by the peculiarities of the code, rather than by you. By including the variable list as [x,y,z], you are forcing the output list to be in the same order. (%i1) eqns : [2x - 3y + 4z = 2, 3x - 2y + z = 0, x + y - z = 1]$ (%i2) display2d:false$ (%i3) solns : solve( eqns,[x,y,z] ); (%o3) (%i4) soln : solns; (%o4) [x = 7/10,y = 9/5,z = 3/2] (%i5) for i thru 3 do disp( ev(eqns[i],soln) )$ 2 = 2 0 = 0 1 = 1 The Maxima function linsolve has the same syntax as solve (for a set of equations) but you cannot leave out the list of unknowns. (%i6) linsolve(eqns,[x,y,z]); (%o6) [x = 7/10,y = 9/5,z = 3/2] Notice that linsolve returns a list, while solve returns a list of lists. 32 Next we use linsolve by lu on this three equation problem. Using the laws of matrix multiplication, we reverse engineer the 3 × 3 matrix m and the three element column vector bcol which provide an equivalent problem in matrix form. (%i7) m : matrix( [2,-3,4],[3,-2,1],[1,1,-1] )$ (%i8) xcol : matrix( [x],[y],[z] )$ (%i9) m . xcol; (%o9) matrix([4z-3y+2x],[z-2y+3x],[-z+y+x]) (%i10) bcol : matrix( ,, )$ (%i11) linsolve_by_lu(m,bcol); (%o11) [matrix([7/10],[9/5],[3/2]),false] (%i12) xyzcol : first(%); (%o12) matrix([7/10],[9/5],[3/2]) (%i13) m . xyzcol - bcol; (%o13) matrix(,,) (%i14) xyzlist : makelist( xyzcol[i,1],i,1,length(xyzcol) ); (%o14) [7/10,9/5,3/2] (%i15) xyzrules : map("=",[x,y,z],xyzlist); (%o15) [x = 7/10,y = 9/5,z = 3/2] Both linsolve and solve can handle an equation list which is actually an expression list in which it is understood that the required equations are generated by setting each expression to zero. (%i16) exs : [2x - 3y + 4z - 2, 3x - 2y + z , x + y - z - 1]; (%o16) [4 z - 3 y + 2 x - 2, z - 2 y + 3 x, - z + y + x - 1] (%i17) linsolve(exs,[x,y,z]); 7 9 3 (%o17) [x = --, y = -, z = -] 10 5 2 (%i18) solve(exs); 3 9 7 (%o18) 2 5 10 The Maxima manual presents a linear equation example in which there is an undened parameter a. (%i1) eqns : [x + z = y,2ax - y = 2a2,y - 2z = 2]$ (%i2) solns : linsolve(eqns,[x,y,z] ); (%o2) [x = a + 1, y = 2 a, z = a - 1] (%i3) solve(eqns,[x,y,z]); (%o3) (%i4) for i thru 3 do ( e: expand( ev(eqns[i],solns) ),disp(lhs(e) - rhs(e)) )$ 0 0 0 (%i5) e; (%o5) 2 = 2 (%i6) [kill(e),e]; (%o6) [done, e] The code in input %i4 binds an equation to the symbol e which can be removed with kill. 33 We can avoid introducing a global binding to a symbol by using %%, which allows use of the previous result inside a piece of code. (%i7) for i thru 3 do ( expand( ev(eqns[i],solns) ),disp(lhs(%%) - rhs(%%)) )$ 0 0 0 Note the syntax used here: do ( job1, job2 ) . Let's try out linsolve by lu on this three linear (in the unknowns (x,y,z)) equation problem which involves the unbound parameter a. (%i8) display2d:false$ (%i9) m : matrix([1,-1,1],[2a,-1,0],[0,1,-2] )$ (%i10) xcol : matrix( [x],[y],[z] )$ (%i11) m . xcol; (%o11) matrix([z-y+x],[2ax-y],[y-2z]) (%i12) bcol : matrix( ,[2a2], )$ (%i13) soln : linsolve_by_lu(m,bcol)$ (%i14) xyzcol : ( first(soln), ratsimp(%%) ); (%o14) matrix([a+1],[2a],[a-1]) (%i15) ratsimp( m . xyzcol - bcol); (%o15) matrix(,,) In input %i15 we check the solution, and the result is a three element column vector, all of whose elements are zero. Such a column matrix is ordinarily replaced by the number 0. 4.3.6 Surpressing rat Messages: ratprint If we start with two linear equations with decimal coefcients, solve (and the other methods) converts the deci-mals to ratios of integers, and tries to nd an exact solution. You can avoid seeing all the rational replacements done by setting the option variable ratprint to false. Despite the assertion, in the manual section on rat, that keepfloat if true prevents oating point numbers from being converted to rational numbers. setting keepoat to true here does not stop solve from converting decimal numbers to ratios of integers. (%i1) [keepfloat,ratprint]; (%o1) [false, true] (%i2) display2d:false$ (%i3) fpprintprec:8$ (%i4) eqns : [0.2x + 0.3y = 3.3,0.1x - 0.8y = 6.6]$ (%i5) solns : solve(eqns, [x,y]);rat' replaced -3.3 by -33/10 = -3.3 rat' replaced 0.2 by 1/5 = 0.2rat' replaced 0.3 by 3/10 = 0.3 rat' replaced -6.6 by -33/5 = -6.6rat' replaced 0.1 by 1/10 = 0.1 rat' replaced -0.8 by -4/5 = -0.8 (%o5) (%i6) linsolve(eqns,[x,y]);rat' replaced -3.3 by -33/10 = -3.3 34 rat' replaced 0.2 by 1/5 = 0.2rat' replaced 0.3 by 3/10 = 0.3 rat' replaced -6.6 by -33/5 = -6.6rat' replaced 0.1 by 1/10 = 0.1 rat' replaced -0.8 by -4/5 = -0.8 (%o6) [x = 462/19,y = -99/19] (%i7) m : matrix([0.2,0.3],[0.1,-0.8] )$ (%i8) bcol : matrix( [3.3], [6.6] )$ (%i9) linsolve_by_lu(m,bcol);rat' replaced 0.2 by 1/5 = 0.2 rat' replaced 0.2 by 1/5 = 0.2rat' replaced 0.2 by 1/5 = 0.2 rat' replaced 0.95 by 19/20 = 0.95 (%o9) [matrix([24.315789],[-5.2105263]),false] (%i10) ratprint:false$ (%i11) solns : solve(eqns, [x,y]); (%o11) (%i12) linsolve(eqns,[x,y]); (%o12) [x = 462/19,y = -99/19] (%i13) linsolve_by_lu(m,bcol); (%o13) [matrix([24.315789],[-5.2105263]),false] (%i14) float(solns); (%o14) Matrix methods for sets of linear equations can be solved using IEEE double oats (as well as big oats) by including an optional method specication 'floatfield after the input column vector (or matrix of input column vectors). (%i15) linsolve_by_lu(m,bcol, 'floatfield); (%o15) [matrix([24.315789],[-5.2105263]),8.8815789] In this example the lower bound of the matrix condition number appears as the second element of the returned list. 4.3.7 Non-Linear Polynomial Equations Here is an example of using solve to nd a pair of exact solutions of a pair of equations, one equation being linear, the other quadratic. The pair of solutions represent the two intersections of the unit circle with the line y = −x/3. (%i1) fpprintprec:8$ (%i2) eqns : [x2 + y2 = 1, x + 3y = 0]$ (%i3) solns : solve(eqns,[x,y]); 3 1 (%o3) sqrt(10) sqrt(2) sqrt(5) 35 (%i4) solns : rootscontract(solns); 3 1 3 1 (%o4) sqrt(10) sqrt(10) sqrt(10) sqrt(10) (%i5) for i thru 2 do for j thru 2 do ( ev(eqns[i],solns[j]), disp(lhs(%%)-rhs(%%)) )$ 0 0 0 0 (%i6) float(solns); (%o6) The pair of solutions reect the symmetry of the given set of equations, which remain invariant under the transformation x →−y, y →−x. A set of two nonlinear polynomial equations with four solutions generated by solve is one of the examples in the Maxima manual. One of the solutions is exact, one is a real inexact solution, and the other two solutions are inexact complex solutions. (%i1) fpprintprec:8$ (%i2) eqns : [4x2 - y2 = 12, xy - x = 2]$ (%i3) solns : solve( eqns, [x,y] ); (%o3) (%i4) for i thru 2 do for j thru length(solns) do ( expand( ev(eqns[i],solns[j]) ), abs( lhs(%%) - rhs(%%) ), disp(%%) )$ 0 2.36036653E-15 2.36036653E-15 1.13954405E-6 0 0.0 0.0 9.38499825E-8 To get real numbers from the complex solutions, we used the abs function, which calculates the absolute value of a complex number. Note the syntax used to check the solutions: do ( job1, job2, job3). 36 4.3.8 General Sets of Nonlinear Equations: eliminate, mnewton Solving systems of nonlinear equations is much more difcult than solving one nonlinear equation. A wider variety of behavior is possible: determining the existence and number of solutions or even a good starting guess is more complicated. There is no method which can guarantee convergence to the desired solution. The computing labor increases rapidly with the number of dimensions of the problem. 4.3.9 Intersections of Two Circles: implicit plot Given two circles, we seek the intersections points. We rst write down the dening equations of the two circles, and look visually for points (x,y) which simultaneously lie on each circle. We use implicit plot for this visual search. (%i1) [eq1 : x2 + y2 = 1,eq2 : (x-2)2 + (y-2)2 = 4]$ (%i2) eqns : [eq1,eq2]$ (%i3) load(implicit_plot); (%o3) C:/PROGRA1/MAXIMA3.0/share/maxima/5.14.0/share/contrib/implicit_plot.lisp (%i4) implicit_plot(eqns,[x,-6,6],[y,-6,6],[nticks,200], [gnuplot_preamble, "set zeroaxis" ])$ We are not taking enough care with the x and y ranges to make the circles circular, but we can use the cursor to read off approximate intersections points: (x,y) = (0.26,0.98), (0.96,0.30). However, the dening equations are invariant under the symmetry transformation x ↔y, so the solution pairs must also respect this symmetry. We next eliminate y between the two equations and use solve to nd accurate values for the x's. Since we know that both solutions have positive values for y, we enforce this condition on equation 1. (%i5) solve(eq1,y); 2 2 (%o5) [y = - sqrt(1 - x ), y = sqrt(1 - x )] (%i6) ysoln : second(%); 2 (%o6) y = sqrt(1 - x ) (%i7) eliminate(eqns,[y]); 2 (%o7) [32 x - 40 x + 9] (%i8) xex : solve(first(%)); sqrt(7) - 5 sqrt(7) + 5 (%o8) [x = - -----------, x = -----------] 8 8 (%i9) (fpprintprec:8, xex : float(xex) ); (%o9) [x = 0.294281, x = 0.955719] (%i10) [x1soln : first(xex), x2soln : second(xex) ]$ (%i11) [ev(%o7,x1soln), ev(%o7,x2soln)]; (%o11) (%i12) y1soln : ev(ysoln,x1soln); (%o12) y = 0.955719 (%i13) y2soln : ev(ysoln,x2soln); (%o13) y = 0.294281 (%i14) [soln1:[x1soln,y1soln],soln2:[x2soln,y2soln] ]$ (%i15) [ev(eqns,soln1), ev(eqns,soln2) ]; (%o15) (%i16) [soln1,soln2]; (%o16) 37 We have solutions (%o16) which respect the symmetry of the equations. The solutions have been numerically checked in input %i15. -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2 3 4 5 6 Intersections of Two Circles Figure 5: two circles 4.3.10 Using Draw for Implicit Plots The gure above was created using the draw package with the following code in a separate work le implic-itplot1.mac. The code the the le is / file implicitplot1.mac / / need load(implicit_plot); to use this code / disp(" doplot() ")$ doplot() := block([ x,y, eq1, eq2, y1:-2, y2:4,r, x1 ,x2 ], r : 1.56, x1 : ry1, x2 : ry2, eq1 : x2 + y2 = 1, eq2 : (x-2)2 + (y-2)2 = 4, draw2d( grid = true, line_type = solid, line_width = 3, color = blue, implicit(eq1, x, x1,x2, y, y1,y2), color = red, implicit(eq2, x, x1,x2, y, y1,y2), color = black, point_type = filled_circle, point_size = 2, 38 points( [ [0.294281,0.955719], [ 0.955719, 0.294281] ] ), title = "Intersections of Two Circles" , terminal = 'eps , file_name = "c:/work2/mycircle2" ) )$ Here is record of use: (%i1) load(draw); (%o1) C:/PROGRA1/MAXIMA3.0/share/maxima/5.14.0/share/draw/draw.lisp (%i2) load(implicitplot1); doplot() (%o2) c:/work2/implicitplot1.mac (%i3) doplot(); (%o3) [gr2d(implicit, implicit, points)] Note that no actual plot was drawn on the screen, since an eps le was created mycircle2.eps in my work folder c:\work2 for use in my latex le. To use this code to get a gure on your screen, you would remove the last two lines: terminal = 'eps , file_name = "c:/work2/mycircle2" and, very important!, also remove the comma at the end of two circles. 4.3.11 Another Example We next work through Example 6.5.1 (page 149), in Numerical Methods for Unconstrained Optimization and Nonlinear Equations by J. E. Dennis, Jr. and Robert B. Schnabel (Prentice-Hall, 1983). (%i1) eq1 : x2 + y2 = 2; 2 2 (%o1) y + x = 2 (%i2) eq2 : exp(x - 1) + y3 = 2; 3 x - 1 (%o2) y + %e = 2 We will concentrate on the solution (x,y) = (1,1). We can eliminate y and plot the resulting function of x to visually locate the x solutions. (%i3) eliminate([eq1,eq2],[y]); 2 x x + 1 2 6 2 4 2 2 2 (%o3) [%e - 4 %e + %e x - 6 %e x + 12 %e x - 4 %e ] (%i4) ex : first(%); 2 x x + 1 2 6 2 4 2 2 2 (%o4) %e - 4 %e + %e x - 6 %e x + 12 %e x - 4 %e (%i5) plot2d([0.0,ex],[x,-5,5] )$ (%i6) plot2d([0.0,ex],[x,0,2] )$ (%i7) x1 : find_root(ex,x,0.5,1.5); (%o7) 1.0 (%i8) yeqn : ev(eq1,x = x1); 2 (%o8) y + 1.0 = 2 (%i9) solve(yeqn);rat' replaced -1.0 by -1/1 = -1.0 (%o9) [y = - 1, y = 1] 39 (%i10) ysol : second(%); (%o10) y = 1 (%i11) soln1 : [x = x1, ysol]; (%o11) [x = 1.0, y = 1] (%i12) ev(eq1,soln1); (%o12) 2.0 = 2 (%i13) ev(eq2,soln1); (%o13) 2.0 = 2 One solution is then %o11 and the solution has been checked. -1.5 -1 -0.5 0 0.5 1 1.5 -3 -2 -1 0 1 2 3 Y X dennis and schnabel example Figure 6: Dennis and Schnabel Example Let's try mnewton on this problem. Wikipedia, under Newton's Method, has the introduction: In numerical analysis, Newton's method (also known as the Newton Raphson method or the Newton Fourier method) is perhaps the best known method for nding successively better approximations to the zeros (or roots) of a real valued function. Newton's method can often converge remarkably quickly, especially if the iteration begins sufciently near the desired root. Just how near sufciently near needs to be and just how quickly remarkably quickly can be depends on the problem, as is discussed in detail below. Unfortunately, far from the desired root, Newton's method can easily lead an unwary user astray, and astray with little warning. Such users are advised to heed the advice of Press, et. al. (1992), who suggest embedding Newton's method in a routine that also detects possible convergence failures. Newton's method can also be used to nd a minimum or maximum of such a function, by nding a zero in the function's rst derivative, see Newton's method as an optimization algorithm. The algorithm is rst in the class of Householder's methods, succeeded by Halley's method. The Maxima manual entry for mnewton begins: Function: mnewton(FuncList,VarList,GuessList) Multiple nonlinear functions solution using the Newton method. FuncList is the list of functions to solve, VarList is the list of variable names, and GuessList is the list of initial approximations. The solution is returned in the same format that solve() returns. If the solution isn't found, [] is returned. This function is controlled by global variables newtonepsilon and newtonmaxiter. 40 Here are the Maxima manual examples of mnewton: (%i1) load(mnewton); (%o1) C:/PROGRA1/MAXIMA3.0/share/maxima/5.14.0/share/contrib/mnewton.mac (%i2) mnewton([2aa-5],[a],); (%o2) (%i3) mnewton([23u-v/u-5, u+2v-4], [u, v], [2, 2]); (%o3) (%i4) mnewton([x1+3log(x1)-x22, 2x12-x1x2-5x1+1], [x1, x2], [5, 5]); (%o4) In the above examples, mnewton is presented with a list of expressions. Here we use mnewton on the Dennis and Schnabel problem we solved earlier using eliminate and nd root. We rewrite the equations as expressions here. (%i1) fpprintprec:8$ (%i2) load(mnewton); (%o2) C:/PROGRA1/MAXIMA3.0/share/maxima/5.14.0/share/contrib/mnewton.mac (%i3) exs : [x2 + y2 -2, exp(x-1)+y3-2]$ (%i4) mn(x0,y0) := mnewton(exs,[x,y],[x0,y0] )$ (%i5) mn(1.1,0.9); (%o5) (%i6) mn(1.2,0.8); (%o6) (%i7) mn(1.3,0.7); (%o7) (%i8) mn(1.4,0.6); (%o8) (%i9) mn(1.5,0.5); (%o9) (%i10) mn(1.5,0.6); (%o10) (%i11) mn(1.7,0.6); (%o11) (%i12) mn(1.9,0.6); (%o12) (%i13) mn(1.9,0.7); (%o13) (%i14) mn(2,0.7); (%o14) (%i15) mn(0.8,1.1); (%o15) (%i16) mn(0.5,1.2); (%o16) (%i17) mn(0.1,1.2); (%o17) We have to be close enough to nd the (1,1) root with mnewton. Note that Maxima's function mnewton can't nd the desired root starting with (x0,y0) = (2,0.5) as the textbook example does with eventual success. 41 4.3.12 Error Messages and Do It Yourself Mnewton Let's explore the Newton Raphson method using an easy example which solve has no trouble with. As a by-product, we show that mnewton can deal with a list of equations, rather than a list of expressions. After getting an error message from Maxima's mnewton, we work this same problem by hand, using matrix methods. A good reference is Chapter 5, Nonlinear Equations, of the text Scientic Computing: An Introductory Survey (2nd ed.), by Michael T. Heath. ( see webpage: (%i18) eqns : [x+y=3,x2+y2=9]$ (%i19) mn(x0,y0) := mnewton(eqns,[x,y],[x0,y0] )$ (%i20) solve(eqns,[x,y]); (%o20) (%i21) mn(1,2); (%o21) (%i22) mn(-1,4); (%o22) (%i23) mn(-2,5); (%o23) (%i24) mn(0,0); Maxima encountered a Lisp error: Error in FUNCALL [or a callee]: Zero divisor. Automatically continuing. To reenable the Lisp debugger set debugger-hook to nil. (%i25) mn(0,1); (%o25) (%i26) mn(2,0); (%o26) The zero divisor message from the Maxima code for mnewton probably means that the starting point (x,y) = (0,0) resulted in an attempted division by 0. To explore what kind of problems can arise, we implement a naive (ie., a strategy-less) iteration algorithm as presented in Heath's text. We present the al-gorithm rst in terms of a 2 × 2 matrix and a two element column vector which are functions of the scalars (x,y). We then convert the algorithm to a form which works with two element column vectors v and b. We let the column vector b hold the elements x and y and we seek b such that the equation f(b) = 0, where f is a column vector which denes the problem. If b is an approximate solution of this equation, then bbetter = b + s, where s is the solution of the matrix equation j(b) . s = -f(b), and the 2 × 2 matrix j(b) is the jacobian matrix: j[1,1] : diff( f[1,1],x ), j[1,2] : diff( f[1,1],y ), and j[2,1] : diff( f[2,1],x), j[2,2] : diff( f[2,1],y ). (%i1) (fpprintprec:8,ratprint:false)$ (%i2) g : matrix( [x + y -3],[x2 + y2 -9] ); [ y + x - 3 ] (%o2) [ ] [ 2 2 ] [ y + x - 9 ] 42 (%i3) gv : ev(g, x=v[1,1], y=v[2,1] ); [ v + v - 3 ] [ 2, 1 1, 1 ] (%o3) [ ] [ 2 2 ] [ v + v - 9 ] [ 2, 1 1, 1 ] (%i4) define(f(v), gv); [ v + v - 3 ] [ 2, 1 1, 1 ] (%o4) f(v) := [ ] [ 2 2 ] [ v + v - 9 ] [ 2, 1 1, 1 ] (%i5) b : matrix(,); [ 1 ] (%o5) [ ] [ 2 ] (%i6) f(b); [ 0 ] (%o6) [ ] [ - 4 ] (%i7) (r1 : g[1,1], r2 : g[2,1] )$ (%i8) h : matrix( [diff(r1,x), diff(r1,y)], [diff(r2,x), diff(r2,y) ] ); [ 1 1 ] (%o8) [ ] [ 2 x 2 y ] (%i9) hv : ev(h, x=v[1,1], y=v[2,1] ); [ 1 1 ] (%o9) [ ] [ 2 v 2 v ] [ 1, 1 2, 1 ] (%i10) define( j(v), hv ); [ 1 1 ] (%o10) j(v) := [ ] [ 2 v 2 v ] [ 1, 1 2, 1 ] (%i11) j(b); [ 1 1 ] (%o11) [ ] [ 2 4 ] (%i12) ls : linsolve_by_lu(j(b),-f(b) ); [ - 2 ] (%o12) [[ ], false] [ 2 ] (%i13) s : first(ls) ; [ - 2 ] (%o13) [ ] [ 2 ] 43 (%i14) b : b + s; [ - 1 ] (%o14) [ ] [ 4 ] (%i15) b : b + first( linsolve_by_lu(j(b),-f(b) ) ); [ 1 ] [ - - ] [ 5 ] (%o15) [ ] [ 16 ] [ --] [ 5 ] (%i16) b : b + first( linsolve_by_lu(j(b),-f(b) ) ); [ 1 ] [ - -- ] [ 85 ] (%o16) [ ] [ 256 ] [ ---] [ 85 ] (%i17) b : b + first( linsolve_by_lu(j(b),-f(b) ) ); [ 1 ] [ - ----- ] [ 21845 ] (%o17) [ ] [ 65536 ] [ -----] [ 21845 ] (%i18) b : b + float( first( linsolve_by_lu(j(b),-f(b) ) ) ); [ - 6.98491931E-10 ] (%o18) [ ] [ 3.0 ] (%i19) f(b); [ 0.0 ] (%o19) [ ] [ 4.19095159E-9 ] Starting with the guess (x0,y0) = (1,2), this iteration process has converged to the approximate solu-tion given by %o18, which we check as an approximate solution in input %i19. Now let's start with the dangerous guess: (x0,y0) = (0,0). (%i20) b : matrix( , ); [ 0 ] (%o20) [ ] [ 0 ] (%i21) f(b); [ - 3 ] (%o21) [ ] [ - 9 ] (%i22) j(b); [ 1 1 ] (%o22) [ ] [ 0 0 ] 44 (%i23) ls : linsolve_by_lu(j(b),-f(b) ); Division by 0 -- an error. To debug this try debugmode(true); (%i24) determinant(j(b)); (%o24) 0 Thus, a check of the non-vanishing of the determinant of the jacobian matrix would have kept us out of trouble. 4.3.13 Automated Code for mymnewton Writing code for an arbitrary number of dimensions is a suggested homework problem. Here we just assume the problem is two dimensional and assume the variables are called x and y. To check for the nonvanishing of the determinant of the jacobian, we ask if the absolute value is less than some very small number. Here is the code, written with notepad2 in a le mymnewton.mac, placed in my work directory c:\work2. You can download this le from the author's webpage, and experiment with it. You can reduce the size of the output on the screen by adding the line display2d:false, in the program, or outside the pro-gram in your work session. If you make changes to this code, add some extra debug printouts at rst like display(newval1,newval2), or print(" v1 = ",v1), to make sure you are on the right track. Once the program has been debugged, you can comment out the debug version in your work le, copy the whole code to a new section, remove the debug printouts, and use as your working version. / working version / / file: mymnewton.mac e. woollett, april, 08 / disp("working version mymnewton, assumes two dimensional problem only, syntax: mymnewton( exprlist, guesslist, numiter )$ exprlist should have the form: [expr1, expr2], to find (x,y) such that simultaneously expr1=0,expr2=0, expr1, expr2 should be explicit functions of x and y, guesslist should be in the form: [xguess,yguess], numiter = number of iterations ")$ mymnewton( exprlist, guesslist, numiter) := block([numer,ratprint,fpprintprec, small : 1.0e-30 , g,x,y,gv,h,hv,b,v,ls,d ] , local(f,j), numer:true, ratprint:false, fpprintprec:8, / g = col vec: row 1=expr 1, row 2=expr 2 depends on (x,y) / g : matrix( [ exprlist ],[ exprlist ] ), display(g), gv : ev(g, x=v[1,1], y=v[2,1] ), / v is generic col vector / 45 define( f(v), gv ), / h is jacobian matrix associated with col vec g(x,y) / h : matrix( [diff( g[1,1], x), diff( g[1,1], y)], [diff( g[2,1], x), diff(g[2,1], y) ] ), hv : ev(h, x=v[1,1], y=v[2,1] ), define( j(v), hv ), / b is col vec containing (x,y) values / b : matrix([ guesslist ],[ guesslist ]), / start iterations / for i:0 thru numiter do ( print(" "), print(" i = ",i," b = ",b," f(b) = ",f(b) ), if ( i = 0) then print(" starting values ") else print(" condition number = ", second(ls) ), / check jacobian determinant / d : ( determinant( j(b) ), float(%%) , abs(%%) ), if ( d < small ) then ( print(" abs(det(jacobian)) is ", d," program halt " ), return() ), / using 'floatfield arg gets condition number returned / ls : linsolve_by_lu(j(b),-f(b), 'floatfield ), / improved (hopefully) estimates of values of (x,y) which simultaneously satisfy expr1=0 and expr2=0 / b : b + first(ls) ) / end do / )$ / end block / 46 and here is a little output from this code: (%i1) load("mymnewton.mac"); working version mymnewton, assumes two dimensional problem only, syntax: mymnewton( exprlist, guesslist, numiter )$ exprlist should have the form: [expr1, expr2], to find (x,y) such that simultaneously expr1=0,expr2=0, expr1, expr2 should be explicit functions of x and y, guesslist should be in the form: [xguess,yguess], numiter = number of iterations (%o1) c:/work2/mymnewton.mac (%i2) mymnewton([x+y-3, x2+y2-9],[1,2], 5)$ [ y + x - 3 ] g = [ ] [ 2 2 ] [ y + x - 9 ] [ 1 ] [ 0 ] i = 0 b = [ ] f(b) = [ ] [ 2 ] [ - 4 ] starting values [ - 1.0 ] [ 0.0 ] i = 1 b = [ ] f(b) = [ ] [ 4.0 ] [ 8.0 ] condition number = 22.5 [ - 0.2 ] [ 0.0 ] i = 2 b = [ ] f(b) = [ ] [ 3.2 ] [ 1.28 ] condition number = 19.5 [ - 0.0117647 ] [ 0.0 ] i = 3 b = [ ] f(b) = [ ] [ 3.0117647 ] [ 0.0708651 ] condition number = 10.92 [ - 4.57770657E-5 ] [ 4.4408921E-16 ] i = 4 b = [ ] f(b) = [ ] [ 3.0000458 ] [ 2.74666585E-4 ] condition number = 7.212872 [ - 6.98491837E-10 ] [ 0.0 ] i = 5 b = [ ] f(b) = [ ] [ 3.0 ] [ 4.19095159E-9 ] condition number = 7.000824 47 Here is a test of the dangerous initial condition case: (%i3) mymnewton([x+y-3, x2+y2-9],[0,0], 5)$ [ y + x - 3 ] g = [ ] [ 2 2 ] [ y + x - 9 ] [ 0 ] [ - 3 ] i = 0 b = [ ] f(b) = [ ] [ 0 ] [ - 9 ] starting values abs(det(jacobian)) is 0.0 program halt It is left as a homework problem to incorporate a criterion for nding an accurate enough solution without providing the program a number of iterations, and allowing the program to hide the details of the iteration process, providing the user with an answer and an estimate of the likely error. 48
14	https://testbook.com/question-answer/successive-discounts-of-20-10-and-15-are-equiv--5d9b263df60d5d5c70b6da30	Get Started Exams SuperCoaching Test Series Skill Academy Pass Skill Academy Free Live Classes Free Live Tests & Quizzes Previous Year Papers Doubts Practice Refer & Earn All Exams Our Selections Careers English Home Quantitative Aptitude Profit and Loss Discount and MP Question Download Solution PDF Successive discounts of 20%, 10% and 15% are equivalent to a single discount of: This question was previously asked in SSC CGL Previous Paper 64 (Held On: 19 June 2019 Shift 3) Attempt Online View all SSC CGL Papers > 38.8% 43.5% 42.2% 44.5% Answer (Detailed Solution Below) Option 1 : 38.8% Free Tests View all Free tests > SSC CGL Tier 1 2025 Full Test - 01 4.6 Lakh Users 100 Questions 200 Marks 60 Mins Start Now Detailed Solution Download Solution PDF Given: The successive discount are 20%, 10% and 15%. Formula: Successive discount = X + Y - (XY)/100 Where, X = First discount Y = Second discount Calculation: Successive discount = X + Y - (XY)/100 ⇒ 20 + 10 - 200/100 ⇒ 28% Now, again Successive discount = X + Y - (XY)/100 X = 28%, Y = 15% ⇒ 28 + 15 - (28 × 15)/100 ⇒ 28 + 15 - 420/100 ⇒ 43 - 4.2% = 38.8% ∴ The single equivalent discount is 38.8%. Let the amount be 100. Successive discount on 100 = (100 - 80% of 90% of 85% of 100) ⇒ 100 - 61.2 = 38.8% Download Solution PDF Share on Whatsapp Latest SSC CGL Updates Last updated on Sep 24, 2025 -> The SSC CGL Tier 1 is being conducted from 12th September, 2025 onwards. 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India’s #1 Learning Platform Start Complete Exam Preparation Daily Live MasterClasses Practice Question Bank Mock Tests & Quizzes Get Started for Free Trusted by 7.6 Crore+ Students More Discount and MP Questions Q1.On a shirt of Rs. 500 a shop is giving successive discounts of 50%, 30% and 20%. What is the final selling price of the shirt? Q2.4 articles A,B,C,D are sold by the shopkeeper. The ratio of selling price of A to marked price of C is 2:3. Article B and article D is marked 33.33% above its cost price. Discount% that is offered on article D is same as that on article B. Selling price of A is 100% more than marked price of B. A discount of 10% is provided while selling article B. Sum of marked price of C and D is Rs 540, while sum of profit earned on article B and D is Rs 33. If profit earned on selling article A is 66.66% more than that by selling article B. Find discount % on article A if it is marked Rs 120 above its cost price. Q3.Difference between cost price of Pen and Copy is Rs. 200 (Pen < copy). Marked price of Pen is twice its CP, and it is sold at discount of 25%. While marked price of copy is Z% more than its cost price and it is sold after discount of Y% and thus earning profit of 20%. Total selling price of both articles together is Rs. 1320. Which of the following can be uniquely determined? I. Selling price of copy. II. Marked price of article Pen. III. Value of Y, if Z is 50. Q4.A shopkeeper sells an item for ₹1000, making a profit of 25%. The marked price of the item is ₹600 more than its cost price. What was the discount given on the item? Q5.The marked price of an article is Rs. 2600. If a discount of 38 percent is given, then what is the selling price of the article? Q6.The marked price of a car is Rs 550000. Rahul buys that car after two successive discounts of 10 percent and 30 percent. If he sells the car for Rs. 360000, then what is the profit percentage? [Give your answer correct to two places of decimal.] Q7.A shopkeeper sold an article at 20% discount and earned a profit of 4%. By what percent the marked price of the article more than the cost price? Q8.A store offers a 25% discount on all of its products. If a customer pays ₹6,000 for a product, what was the marked price of this product? Q9.A customer purchases a product for ₹550 after a 12% discount. What was the marked price? Q10.A supplier offers a trade discount of 15% on the list price of ₹10,000. Additionally, a cash discount of 5% is offered if payment is made within 10 days. What is the final price to be paid if the buyer avails both discounts? More Profit and Loss Questions Q1.On a shirt of Rs. 500 a shop is giving successive discounts of 50%, 30% and 20%. What is the final selling price of the shirt? Q2.A fruit vendor bought some oranges 3 for a rupee and equal number 2 for a rupee. If he wants to make a profit of 20%, what should be his selling price per dozen? Q3.The cost price of an item is ₹300. It is increased by 10% by the wholesaler and again increased by 10% by the retailer. What is the final price of the item? Q4.4 articles A,B,C,D are sold by the shopkeeper. The ratio of selling price of A to marked price of C is 2:3. Article B and article D is marked 33.33% above its cost price. Discount% that is offered on article D is same as that on article B. Selling price of A is 100% more than marked price of B. A discount of 10% is provided while selling article B. Sum of marked price of C and D is Rs 540, while sum of profit earned on article B and D is Rs 33. If profit earned on selling article A is 66.66% more than that by selling article B. Find discount % on article A if it is marked Rs 120 above its cost price. Q5.A man sold his horse for ₹19800 and gained 10%. The horse was bought for: Q6. On a particular day each of Anya, Ben and Carlos sold three varieties of notebooks from their respective shops. Coincidentally, each of them sold an identical number of notebooks of Variety X, an identical number of notebooks of Variety Y and an identical number of notebooks of Variety Z. However, the numbers of notebooks of each variety that were sold were all different from one another. Also, the three sellers sold each of the varieties at different profit (or loss) per unit. Assertion (A): It is possible that Anya sold each notebook of Variety X at a profit of Rs 3, each notebook of Variety Y at a loss of Rs 2, and each notebook of Variety Z at a loss of Rs 9 and made an overall profit of Rs 11; Ben sold each notebook of Variety X at a profit of Rs 4, each notebook of Variety Y at a loss of Rs 5, and each notebook of Variety Z at a loss of Rs 2 and made an overall profit of Rs 10; and Carlos sold each notebook of Variety X at a profit of Rs 2, each notebook of Variety Y at a prof Q7.Difference between cost price of Pen and Copy is Rs. 200 (Pen < copy). Marked price of Pen is twice its CP, and it is sold at discount of 25%. While marked price of copy is Z% more than its cost price and it is sold after discount of Y% and thus earning profit of 20%. Total selling price of both articles together is Rs. 1320. Which of the following can be uniquely determined? I. Selling price of copy. II. Marked price of article Pen. III. Value of Y, if Z is 50. Q8.A shopkeeper sells an item at its face value. If he sells it at 75% of his face value then he gains 5%. If he sells it at its face value, what is his profit percentage? Q9.The ratio of the time investment of A : B : C is 5 : 8 : 4 and the ratio of their amount is 8 : 6 : 3. Find the ratio of profit. Q10.X and Y invest in a business in the ratio 2 : 3. After donating 10% of the total profit to a school, X gets Rs. 900 as profit. Then Y’s profit is Crack SSC CGL Offline Mock Test with India's Super Teachers Ananya Singh Testbook Varun Awasthi Testbook Explore Supercoaching For FREE Suggested Test Series View All > SSC CGL Mock Test Series 2025 (Tier I & Tier II) 2174 Total Tests with 39 Free Tests Start Free Test Current Affairs (CA) 2025 Mega Pack for SSC/Railways/State Exam Mock Test 430 Total Tests with 1 Free Tests Start Free Test Suggested Exams SSC CGLSSC CPO SSC CGL Important Links More Quantitative Aptitude Questions Q1.Solve the given series below and answer the following question. Series: 125r, M, 120q, (100p + 4), (30q + 42), (8r + 4.8) Note: A) p and q are consecutive prime numbers where q > p. B) r = p + q C) r is a factor of 60 and greater than 10. Which of the series follows the same pattern as the above series? I) 2000, 1600, 1120, 672, 336, 134.4 II) 1000, 400, 240, 192, 192, 230.4 III) 1200, 1300, 1500, 1800, 2200, 2700 Q2.There have two equations given: Equation I. p2 − 10Mp + 6N = 0 Equation II. q2 + Mq − 5N=0 Where M + N = 39, 3 ≤ M ≤ 5, and N is a multiple of 3. Find the relationship between p and q. Q3.Two trains, Train A and Train B, have lengths of x meters and (x + 50) meters respectively. The ratio of the speed of Train A to Train B is 5 : 4. Train A crosses an electric pole in 12 seconds. When moving in the same direction, Train A completely overtakes Train B in ____ seconds. If Train B were to cross a platform of 200 meters, it would take ____ seconds. Which of the following values can fill the blanks in the same order? A: 130 and 35 B: 140 and 40 C: 150 and 45 Q4.A rectangular field has dimensions where one side is longer than the other. A person decides to walk directly across the field along the diagonal instead of walking along the two adjacent sides. By doing this, the person covers a distance that is shorter than the total walking path by exactly one-third of the length of the longer side. Determine the ratio of the length of the shorter side to the longer side of the field. Q5.In a Monty Hall game with 3 doors, the host does not know where the car is and opens a door at random. By chance, it reveals a goat. You chose a door initially. What is your probability of winning if you stick with your original choice? Q6.For the dataset: 10, 15, 20, 25, 30, 35, 40, 45, what is the value of the first quartile (Q1)? Q7.For the dataset: 5, 8, 12, 15, 18, 22, 25, 30, what is the value of the 3rd decile (D3)? Q8.What is the percentile rank of a score of 40 in a dataset where 60% of the data lies below 40? Q9.What does Karl Pearson’s Coefficient of Skewness measure? Q10.If the mean of a dataset is 50, the median is 48 and the standard deviation is 10, what is Karl Pearson’s coefficient of skewness? 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15	https://en.wikipedia.org/wiki/XO_sex-determination_system	XO sex-determination system - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Evolution 2 Parthenogenesis 3 See also 4 References XO sex-determination system [x] 13 languages العربية বাংলা Bosanski Ελληνικά فارسی Français Galego Magyar 日本語 Polski Português Svenska 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Biological system that determines the sex of offspring Heredity of sex chromosomes in XO sex determination Part of a series on Sex Biological terms Sexual dimorphism Sexual differentiation Feminization Virilization Sex-determination system XY XO ZW ZO Temperature-dependent Haplodiploidy Heterogametic sex Homogametic sex Sex chromosome X chromosome Y chromosome Sex chromosome anomalies Testis-determining factor Hermaphrodite Sequential hermaphroditism Simultaneous hermaphroditism Intersex (biology) Mating type Sexual reproduction Evolution of sexual reproduction Anisogamy Isogamy Germ cell Meiosis Gametogenesis Spermatogenesis Oogenesis Gamete spermatozoon ovum Fertilization External fertilization Internal fertilization Sexual selection Plant reproduction Fungal reproduction Sexual reproduction in animals Sexual intercourse Penile–vaginal intercourse Copulation Human reproduction Lordosis behavior Pelvic thrust Sexuality Plant sexuality Animal sexuality Human sexuality Mechanics Differentiation Activity Sexual system Dioecy Gonochorism Hermaphrodite Intersex v t e The XO sex-determination system (sometimes referred to as X0 sex-determination system) is a system that some species of insects, arachnids, and mammals (not including humans) use to determine the sex of offspring. In this system, there is only one sex chromosome, referred to as X. Males only have one X chromosome (XO), while females have two (XX). The letter O (sometimes a zero) signifies the lack of a Y chromosome. Maternal gametes always contain an X chromosome, so the sex of the animals' offspring depends on whether a sex chromosome is present in the male gamete. Its sperm normally contains either one X chromosome or no sex chromosomes at all. This system determines the sex of offspring among: Most arachnids with the exception of mites where a small majority are haplodiploid Almost all apterygote and Paleopteran insects (e.g., dragonflies, silverfish) Most exopterygote insects (e.g., grasshoppers, crickets, cockroaches) Some nematodes,crustaceans,gastropod molluscs, and bony fish, notably in the genus Ancistrus Several mammals, including: A few species of bat, including the hammer-headed bat,Buettikofer's epauletted fruit bat, Franquet's epauletted fruit bat, Peters's epauletted fruit bat, and Gambian epauletted fruit bat The Ryukyu spiny rat and Tokunoshima spiny rat In a variant of this system, most individuals have two sex chromosomes (XX) and are hermaphroditic, producing both eggs and sperm with which they can fertilize themselves, while rare individuals are male and have only one sex chromosome (XO). The model organismCaenorhabditis elegans—a nematode frequently used in biological research—is one such organism. Most spiders have a variation of the XO system in which males have two different X chromosomes (X 1 X 2 O), while females have a pair of X 1 chromosomes and a pair of X 2 chromosomes (X 1 X 1 X 2 X 2). Some spiders have more complex systems involving as many as 13 different X chromosomes. Some Drosophila species have XO males. These are thought to arise via the loss of the Y chromosome.[original research] In humans the XO designation attaches to individuals with Turner syndrome. Evolution [edit] XO sex determination can evolve from XY sex determination within about 2 million years.[clarification needed] It typically evolves due to Y-chromosome degeneration. As the Y-chromosome is not paired (though see pseudoautosomal region), it is susceptible to decay by Muller's ratchet. Similarly, the W chromosome in a ZW sex-determination system is susceptible to decay, resulting in a ZZ/ZO system. Parthenogenesis [edit] Parthenogenesis with XO sex-determination can occur by different mechanisms to produce either male or female offspring. See also [edit] Sex-determination system Sexual differentiation Haplodiploid sex-determination system XY sex-determination system ZO sex-determination system ZW sex-determination system Temperature-dependent sex determination X chromosome Y chromosome References [edit] ^ Jump up to: abcSember, Alexandr; Pappová, Michaela; Forman, Martin; Nguyen, Petr; Marec, František; Dalíková, Martina; Divišová, Klára; Doležálková-Kaštánková, Marie; Zrzavá, Magda; Sadílek, David; Hrubá, Barbora; Král, Jiří (24 July 2020). "Patterns of Sex Chromosome Differentiation in Spiders: Insights from Comparative Genomic Hybridisation". Genes. 11 (8): 849. doi:10.3390/genes11080849. PMC7466014. PMID32722348. ^ Jump up to: abcBull, James J.; Evolution of sex determining mechanisms; p. 17 ISBN0805304002 ^Bachtrog, Doris; Mank, Judith E.; Peichel, Catherine L.; Kirkpatrick, Mark; Otto, Sarah P.; Ashman, Tia-Lynn; Hahn, Matthew W.; Kitano, Jun; Mayrose, Itay; Ming, Ray; Perrin, Nicolas; Ross, Laura; Valenzuela, Nicole; Vamosi, Jana C. and The Tree of Sex Consortium; ‘Sex Determination: Why So Many Ways of Doing It?’; PLoS Biol12(7): e1001899 ^Thirot-Quiévreux, Catherine (2003). "Advances in Chromosomal Studies of Gastropod Molluscs". Journal of Molluscan Studies. 69 (3): 187–201. doi:10.1093/mollus/69.3.187. ^Devlin, R.H. and Y. Nagahama, 2002. ‘Sex determination and sex differentiation in fish: an overview of genetic, physiological, and environmental influences’; Aquaculture 208: 191–364. ^Anderson, Luís Alves; Oliveira, Claudio; Nirchio, Mauro; Granado, Ángel and Foresti, Fausto; ‘Karyotypic relationships among the tribes of Hypostominae (Siluriformes: Loricariidae) with description of XO sex chromosome system in a Neotropical fish species’; Genetica, vol. 128 (2006); pp. 1-9 ^Hsu, T. C.; Benirschke, Kurt (1977). "Hypsignathus monstrosus (Hammer-headed fruit bat)". An Atlas of Mammalian Chromosomes. pp.13–16. doi:10.1007/978-1-4615-6436-2_4. ISBN978-1-4684-7997-3. ^Denys, C.; Kadjo, B.; Missoup, A. D.; Monadjem, A.; Aniskine, V. (2013). "New records of bats (Mammalia: Chiroptera) and karyotypes from Guinean Mount Nimba (West Africa)". Italian Journal of Zoology. 80 (2): 279–290. doi:10.1080/11250003.2013.775367. hdl:2263/42399. S2CID55842692. ^Kobayashi, Tsuyoshi; Yamada, Fumio; Hashimoto, Takuma; Abe, Shintaro; Matsuda, Yoichi; Kuroiwa, Asato (2007). "Exceptional minute sex-specific region in the XO mammal, Ryukyu spiny rat". Chromosome Research. 15 (2): 175–187. doi:10.1007/s10577-006-1093-y. PMID17294259. S2CID6461447. ^Patterson, J. T.; Stone, W. S. (1952). Evolution in the Genus Drosophila. New York: Macmillan. ^Nei, Masatoshi (2013-05-02). Mutation-Driven Evolution. OUP Oxford. p.168. ISBN978-0-19-163781-0. ^Hales, Dinah F.; Alex C. C. Wilson; Mathew A. Sloane; Jean-Christophe Simon; Jean-François Legallic; Paul Sunnucks (2002). "Lack of Detectable Genetic Recombination on the X Chromosome During the Parthenogenetic Production of Female and Male Aphids". Genetics Research. 79 (3): 203–209. doi:10.1017/S0016672302005657. PMID12220127. \| hide 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 \| Retrieved from " Categories: Sex-determination systems Insect genetics Arachnid anatomy Hidden categories: Articles with short description Short description is different from Wikidata All articles that may contain original research Articles that may contain original research from August 2021 Wikipedia articles needing clarification from August 2024 This page was last edited on 6 March 2025, at 16:47(UTC). 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16	https://www.youtube.com/watch?v=H3QLFx7yH7Q	Vertical And Horizontal Line Distance mrmaisonet 68100 subscribers 303 likes Description 56934 views Posted: 18 Apr 2018 Learn how to recognize when 2 points on the coordinate plane will create a horizontal or a vertical line. Also see how to find the distance between the two points that form vertical or horizontal lines. This video does not go into the distance between points that from diagonal lines on the coordinate plane which would require the distance formula. 17 comments Transcript: Intro during this lesson I'm going to review how to find the distance between two points on the coordinate plane that form either a vertical or a horizontal line we are also going to learn how to recognize when a pair of points will form a vertical or a horizontal line let's look at the first pair of points positive 7 positive 8 and positive 7 positive 2 notice how each of the X values are the same for both points whenever you have two points with identical X values the two points will form a vertical line the second pair of points negative 5 positive 4 and negative 5 negative 5 also have identical X values meaning they will also form a vertical mind and looking at the third pair of points negative 6 negative 5 and positive 7 negative 5 we should notice that the Y values are identical when the Y values are Horizontal Line identical the two points will form a horizontal line let's plot each pair of points so we may examine this further the first point positive 7 positive 8 is located right here and positive 7 positive 2 is located right here as you can see both of these points have an x value of 7 and they form a vertical line now let's plot the second pair of points negative 5 positive 4 is located right here and negative 5 negative 5 is located right here once again the X values are identical meaning they form a vertical line now looking at the third pair of points we should notice that the Y values are the same when the Y values are the same the graph points will form a horizontal line let us plot the points just to verify this negative 6 negative 5 is located right here and positive 7 negative 5 is located here and as you can see the two points form a horizontal line so whenever you have two or more points with the same X values they will form a vertical mind and when you have two or more points with the same Y values they will form a horizontal line now if you recognize right away that the two points will form a horizontal line or a vertical line there is a simple way to find the distance between the two points what you can do is examine the values in your pair of points that are not the same and find the distance between those two values for example looking at the first pair of points the Y values in and two are not the same eight and two are six away from each other so the distance between the two points is six now of course we could just graph the points and physically count the distance between the points however it is convenient at times just to recognize when two points will form a vertical or a horizontal line and what its distance is by simply examining the two values that are different looking at the second pair of points we can see that the x's are the same so we must examine the Y values to determine the distance the Y values of these two points are positive 4 and negative 5 positive 4 and negative 5 are a distance of 9 away from each other so the distance between the two points is 9 now notice how in this example we had a negative value in a positive value when finding the distance between a negative value and a positive Absolute Values value you must add their absolute values together if you forgot what absolute value is it is simply the distance a number is away from 0 the absolute value of a number is always positive we can look at the coordinate plane to see why that is notice how this point is 4 units above 0 and this point right here is 5 units below zero because one number is below 0 and the other is above zero we must add their absolute values to find the distance between them now let's look at our third pair of points that formed a horizontal line the Y values are the same so we must examine the X values negative 6 and positive 7 remember to find the distance between a negative value and a positive value we must add their absolute values together the absolute value of negative 6 is 6 and the absolute value of positive 7 is 7 so basically one of the numbers is 6 below zero and the other is 7 above zero but if we add their absolute values together the total distance between those two points would be 13 so remember if you notice that two points have the same x value you should recognize that a vertical line will be formed and if you have the same Y values you can immediately recognize that the points would create a horizontal line you can then take the two values that are not the same from your two points and just find the distance between those two numbers and that will give you the distance between the two points [Music]
17	https://math.stackexchange.com/questions/2606084/find-common-ratio-with-two-terms	Skip to main content Find common ratio with two terms [closed] Ask Question Asked Modified 7 years, 7 months ago Viewed 10k times This question shows research effort; it is useful and clear -2 Save this question. Show activity on this post. I am trying to find common ratio but ... If b1=−2 and b8=−384, how can I find common ratio (q)? sequences-and-series geometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Jan 15, 2018 at 10:51 Air Conditioner 1,96233 gold badges1212 silver badges1919 bronze badges asked Jan 15, 2018 at 10:32 DonnaTellooDonnaTelloo 10111 silver badge33 bronze badges 1 1 Hint: b8b1=q7. Also, please try defining bn in advance. I assume it means term number. – For the love of maths Commented Jan 15, 2018 at 10:35 Add a comment \| 3 Answers 3 Reset to default This answer is useful 1 Save this answer. Show activity on this post. The fact that you're asking for a common ratio probably means that the sequence with terms bn form a geometric sequence but that's something you should mention. It also helps to show your own effort, ideas or relevant formulas you have seen. You probably know that bn=q⋅bn−1 and from this also bn=qn−1⋅b1. You know b1 and b8 so use this formula with n=8 to get an equation for q. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 15, 2018 at 10:35 StackTDStackTD 28.4k3838 silver badges6767 bronze badges 3 And if I know that b2 = 7 and b5 = 56 – DonnaTelloo Commented Jan 15, 2018 at 11:00 Does this formula help me? I think no. Am I wrong? – DonnaTelloo Commented Jan 15, 2018 at 11:01 Then you either work your way backwards and go via b1 (formula above), or you could think a bit longer/harder and come up with a direct relation between b2 and b5. It's not that hard: b3=q⋅b2 so b4=…. – StackTD Commented Jan 15, 2018 at 11:52 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Hint: bn=b0qn. Compute b8/b1. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 15, 2018 at 10:35 lhflhf 222k2020 gold badges254254 silver badges584584 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Hint: For a geometric progression with first term a and common ratio r, the nth term is given by: bn=arn−1 Here, b1=a=−2 and b8=−384. Can you take it from here? Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jan 15, 2018 at 10:38 answered Jan 15, 2018 at 10:35 user371838user371838 1 I think the OP would prefer bn to Tn. – For the love of maths Commented Jan 15, 2018 at 10:37 Add a comment \| Start asking to get answers Find the answer to your question by asking. 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18	https://www.cuemath.com/ncert-solutions/find-the-area-of-the-quadrilateral-whose-vertices-taken-in-order-are-4-2-3-5-3-2-and-2-3/	LearnPracticeDownload Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3) Solution: Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). Then, Area of a triangle is given by 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] ------ (1) Let the vertices of the quadrilateral be A (- 4, - 2), B (- 3, - 5), C (3, - 2), and D (2, 3) Join AC to form two triangles ∆ABC and ∆ACD. We know that, area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] By substituting the values of vertices, A, B, C in the formula. Area of ΔABC = 1/2 [(- 4){(- 5) - (- 2)} + (- 3){(- 2) - (- 2)} + 3{(- 2) - (- 5)}] = 1/2 (12 + 0 + 9) = 21/2 square units By substituting the values of vertices, A, C, D in the Equation (1), Area of ΔACD = 1/2 [(- 4){(- 2) - 3} + 3{(3) - (- 2)} + 2{(- 2) - (- 2)}] = 1/2 (20 + 15 + 0) = 35/2 square units Area of ABCD = Area of ΔABC + Area of ΔACD = (21/2 + 35/2) square units = 28 square units ☛ Check: NCERT Solutions for Class 10 Maths Chapter 7 Video Solution: Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3) NCERT Class 10 Maths Solutions - Chapter 7 Exercise 7.3 Question 4 Summary: The area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3) is 28 square units. ☛ Related Questions: Find the area of the triangle whose vertices are:(i) (2, 3), (- 1, 0), (2, - 4)(ii) (- 5, - 1), (3, - 5), (5, 2) In each of the following find the value of ‘k’, for which the points are collinear.(i) (7, - 2), (5, 1), (3, k)(ii) (8, 1), (k, - 4), (2, - 5) Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, - 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2) Math worksheets and visual curriculum FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Events MATH WORKSHEETS Kindergarten Worksheets 1st Grade Worksheets 2nd Grade Worksheets 3rd Grade Worksheets 4th Grade Worksheets 5th Grade Worksheets 6th Grade Worksheets 7th Grade Worksheets 8th Grade Worksheets 9th Grade Worksheets 10th Grade Worksheets Terms and ConditionsPrivacy Policy
19	https://www.youtube.com/watch?v=IZrUk3RGjrc	Gas is $4.50, your car gets 20 MPG – how far can you go with $175 in gas? TabletClass Math 854000 subscribers 79 likes Description 4043 views Posted: 11 Mar 2022 TabletClass Math: High gas prices, how to calculate how far your car will travel with a certain mpg. For more math help to include math lessons, practice problems and math tutorials check out my full math help program at Math Notes: Pre-Algebra Notes: Algebra Notes: Geometry Notes: Algebra 2 / Trig Notes: 31 comments Transcript: okay so what do you think of these high gas prices so uh it's definitely got a lot of people looking like this including myself i'm like wow you know four dollars fifty cents now uh depending on where you live uh like my family uh my sisters and whatnot they live in california southern california and they're paying close to this so you know their expression is more like so they got a little smoke coming through their ears so uh yes high gas prices uh certainly hasn't been this high i think maybe one of the time uh gas has been this high but uh anyways i know it's one uh one of many things that we are probably thinking about uh these days but my philosophy in life is listen when you're handing some lemons try to make lemonade so what i'm gonna do here is we're going to take these high gas prices and uh play around with them and make this into a math problem right so that's kind of the idea with this video and uh here is the problem so uh 4.50 this is the price per gallon that you would pay and let's say you have a vehicle and you get 20 miles per gallon on this vehicle now that's not the best it's not the worst but let's say that's the case and you have 175 dollars so if you have 175 dollars and every uh gas station to go to you pay 4.50 per gallon and you have this vehicle that gets 20 miles per gallon how far could you travel in miles okay so this is the problem kind of basic practical math now what i'd like to do in these videos is always encourage you to pause videos see if you can figure it out okay now you don't uh have to do it in my exact way but as long as you understand and you have good justification for your final answer that's really what counts but we're going to practice some math again we're going to take lemons with these high gas prices hopefully they come down here we'll see okay but um anyways we're going to take these lemons and turn them into lemonade and practice some very important math skills uh that you definitely need to know because this little uh miles per gallon right here this little thing like this this has to do with uh ratios and proportions so that's something very important that you need to understand in mathematics we're gonna get to this just one second but uh first let me quickly introduce myself my name is john i'm the founder of tablet class math i'm also a middle and high school math teacher and uh over several years i've constructed what i like to believe is one of the best online math health programs there is now that's a pretty bold statement but i really feel that way if you're interested you can check out my math help program by following the link in the description of this video but basically i have 100 plus different math courses ranging from pre-algebra algebra 1 geometry algebra 2. i'm going to be launching pre-calculus here very shortly i'm very excited about that but i also have many courses in the area of test preparation so if you're preparing for the ged high set maybe the task exam maybe the s-a-t-a-c-t uh alex exam clep exam accuplacer maybe a teacher's certification exam maybe the asvab all those courses or are those exams and many others have a good amount of math on them if you don't do them on a math section do not do well in the exam so let me help you out just go to my website again the link is in the description of this video and you can check out my full course catalog i should have uh the test you're preparing for now if i do not um have that test drop me a line and i'll help you out the best i can i also do a lot with independent learners like homeschoolers so if you homeschool been helping homeschoolers for 15 plus years have a great homeschool learning program and want to check that out then obviously help those of you that are just having a tough time in your current math course now if you're serious about wanting to learn math and ken if you're not serious and just disregard i'm going to tell you but if you're truly serious about wanting to get better at mathematics then you got to be serious about this and that is you have to take great math notes okay so over decades of teaching math those students who do the work on a daily basis and have fantastic math notes they always do very very well and the reverse is true those students who uh you know like to you know look at their cell phone talk to their friends uh do homework uh from another class while their math class all the things i did way back in the good old 1980s except for the cell phone part we didn't have those cell phones at least uh we had those gigantic ones that cost five thousand dollars and of course nobody that i knew had one of those but uh you know if you kind of blow off the work you need to do in math you're going to end up looking like this person at the end of the school year all right so again you have to ask yourself are you serious about wanting to uh you know improve in mathematics and learn well then you got to be serious about note-taking now in the meantime if you don't have great math notes you can use mine so i offer detail comprehensive math notes to include pre-algebra algebra and geometry algebra 2 and trigonometry you can find the links to those notes in the description of this video okay so let's get back to this problem and uh give it a shot okay so here's the deal uh you're gonna be paying four dollars and fifty cents at uh how many gas stations you need to go uh with this 175 dollars now you're probably gonna have to fill up it definitely at least uh one time okay let's assume you left with the full tank and your vehicle gets 20 miles per gallon all right so how far how many miles can you go with 175 bucks under this uh you know these conditions all right so let's get to it again a couple different ways you can approach this problem but i'm going to use um proportions because it's just good practice okay so let's start off with the gas okay so we know uh the gas is four dollars and fifty cents but really it's four dollars fifty cents per gallon right so that's really what that means so one gallon is four dollars and fifty cents so this per 4.50 per gallon this little like thing right there that's a division side so we can set up a nice proportion here okay now what i'm trying to figure out is how much um how many gallons okay can i get um with 175 dollars so i want to start i want to figure out how many gallons first so 175 bucks how many gallons of gas will i get well you could simply take that 175 right divided by 450 yes you can you can all right and you'll get 38.8 gallons but let's just practice uh proportions here okay so one gallon is uh uh per 4.50 cents okay so how many gallons per 175 dollars so notice here um what i have is what we call proportion this is going to be two equal fractions and when you set this up you have to put the same units of measure in the respective places in these fractions so here down in the denominator i have both dollars okay and then up in the numerator i have gallons okay you can't have gallons dollars and then dollars and gallons they have to be in the same position so if you uh do this okay we'll be able to determine let me kind of erase all this um how many gallons now of course i could take the 175 divided by 450 and so i know it's going to be 38.8 gallons but we're practicing math here again all right so we're just going to take these high gas prices and make all kinds of crazy math problems out of them so how do i solve this well i can just use the cross product all right this is one thing you need to know about proportions so it's going to be 1 times 75 which is a 75 and 450 times x right so now i end up with this basic algebra equation right here and to solve for x i need to divide both sides of the equation by four dollars and 50 cents or 450 and i get x is equal to 38.8 gallons okay so that's kind of the first uh part of this problem so with my 175 dollars i can purchase uh 38.8 gallons now i'm going to start off getting gallons uh first because i know my vehicle is rated uh in gallons right it gets 20 miles per gallon so i want to uh do i want to determine how many gallons i can get with this 175 bucks so i get 38.8 gallons and we're going to see how far we can go now with those gallons all right so let's get to this next part so remember um let me put this up here our 175 dollars was able to get us or purchase a purchase uh 38.8 gallons all right so that's the situation again we figured that out up there now let's talk about this vehicle right so 20 miles per uh gallon all right that's mpg uh pretty common um specification for vehicles and people now are probably uh paying attention uh pretty closely to their miles per gallon so obviously things like make sure you have good tire inflation and not you know driving like a a maniac all those things count right like not hitting the accelerator everyone's gonna try to optimize their miles per gallon but let's say this car gets a consistent 20 miles per gallon for this entire 38.8 gallons now let's go ahead and figure this out how far you can go so again we're going to set up a proportion here so 20 miles okay this right here this little fraction bar is per okay so 20 miles per 1 gallon that's what mpg means all right it's 20 miles this is per one gallon per gallon right a gallon so 20 miles per gallon is one gallon so it's 20 miles per one gallon so one gallon we get is 20 miles so 38 gallons okay 38.8 gallons will get us how many miles so we're setting up a proportion again a proportion is two equal fractions and notice again that my units of measure are in the same respective places so down here i have gallons in the denominator i have miles in the numerator so to solve this i simply just use the cross product now if you're quite um if you're a little rusty on proportions and ratios rates and proportions i have tons of videos on this in my algebra playlist pre-algebra and editor playlist of course i teach this extensively in those courses as well but effectively all i need to do is multiply across and i can solve for x so let's go ahead and do that now so x times 1 is just 1x or x that's what i'm looking for because that's how many miles and then 20 times 38.8 is right here so when you cross multiply that's what the cross product means you can do that when you have a proportion and so x is going to be 20 times 38.8 you end up with 700 or i said yeah 776 miles that's how uh many miles we can go with this vehicle that gets uh 20. that's a terrible looking car but you get the idea 20 miles per gallon and we have 38.8 gallons uh that we spent 175 dollars um at gas stations that were charging four dollars and fifty cents per gallon okay so that is the scenario now if you got this problem right and maybe you did it a different way but if you did get this right then i must in turn give you a happy face with a little mohawk right that's a happy expression uh you know that's how we kind of looked at least a lot of us did back in the 1980s i didn't have this uh hairstyle but a lot of my friends did we used a lot of hairspray spray back in the 1980s it was quite dangerous i actually kind of sported around with this little flat top you don't see those anymore but i look pretty cool at least i thought i did but anyways um you know you deserve a happy face we figured this thing out right so let's have fun with these high gas prices the best we can and try to do some uh mathematics so uh at least we're getting something out of it all right so that's basically it again um a couple things here you know very common to be uh seeing these ratios rates and proportions right like things like uh mpg miles per gallon this is you know you got to set up a ratio with these problems or a rate okay and then you got to use proportions so you know use this as a kind of you know as something to hopefully incentivize you to want to learn more about uh just basic mathematics and i would classify rates ratios and proportions that's good practical math to learn again you can um uh you know if you're quite you know like a little bit rusty on this to learn a ton of this stuff on my youtube channel but if this video was interesting and you enjoyed it to some degree please consider smashing that like button and again i hope that you take advantage of all my videos on my channel and become a subscriber i've been on youtube for 10 plus years i have over a thousand videos and uh you know i teach from basic to advanced math my goal is always to try to teach math in a clear and understandable way all right i want you to succeed in mathematics so if you like my teaching style you can learn a lot from all the videos that i post but my best math help will certainly be in my math help program okay so hopefully these gas prices come down we'll see but uh with that being said i definitely wish you all the best in your mathematics adventures thank you for your time and have a great day
20	https://commons.wikimedia.org/wiki/Category:Spanning_tree_protocol	Jump to content Search Category:Spanning tree protocol Help Category Discussion Read Edit View history Tools Actions Read Edit View history General What links here Related changes Permanent link Page information Get shortened URL Download QR code Print/export Create a book Download as PDF Printable version In other projects Wikipedia Wikidata item Appearance From Wikimedia Commons, the free media repository The article Spanning tree protocol on Wikipedia projects: (en) Spanning tree protocol (ca) Spanning tree (de) Spanning Tree Protocol (es) Spanning tree (eu) Spanning tree (fr) Spanning tree protocol (it) Spanning Tree Protocol (ja) スパニングツリープロトコル (ko) 신장 트리 프로토콜 (nl) Spanning Tree Protocol (pl) Protokół drzewa rozpinającego (pt) Spanning Tree Protocol (ru) STP (sv) Spanning tree (th) สแปนนิ่งทรี (yi) ספענינג טרי פראטאקאל +/− Spanning tree; 生成樹協定; Spanning Tree Protocol; STP; Spanning Tree Protocol; Spanning Tree Protocol; Spanning Tree Protocol; پروتکل درخت پوشا; Spanning Tree Protocol; 生成樹協議; Spanning tree protocol; STP; 生成樹協議; 生成树协议; 신장 트리 프로토콜; Spanning Tree Protocol; Spanning Tree Protocol; Spanning tree; Spanning tree protocol; Täispuuprotokoll; Spanning tree protocol; Spanning Tree Protocol; Savienotā koka protokols; Протокол разапињућег стабла; Spanning Tree Protocol; เกณฑ์วิธีต้นไม้แบบทอดข้าม; Spanning Tree Protocol; Spanning Tree Protocol; Spanning Tree Protocol; Protokół STP; スパニングツリープロトコル; Spanning tree protocol; פרוטוקול העץ הפורש; Spanning Tree Protocol; 生成树协议; STP; Spanning Tree Protocol; بروتوكول الشجرة المتفرعة; 生成树协议; Spanning Tree Protocol; канальный протокол, устраняющий образование петель в топологии произвольной сети Ethernet; protocol de la xarxa d'ordinador; protocole réseau de niveau 2; omrežni protokol za logično topologijo brez zank za omrežja Ethernet; network protocol that builds a loop-free logical topology for Ethernet networks; Ethernet-Netzwerkprotokoll für schleifenfreie Netz-Topologien; protokol v počítačových sítích; protocollo di comunicazione standard; Spanning Tree Protocol; Spanning Tree; Spanning tree; STP; ספענינג טרי פראטאקאל; Bridge Protocol Data Unit; Bpdu; STP; Rapid spanning tree; Spanning tree protocol; STP; Spanning Tree Protocol; Spanning tree; Protokół drzewa rozpinającego; Spanning-Tree Protocol; Spanning Tree Protocol; Spanning Tree Protocol; STP; MST; IEEE 802.1d; Spanning Tree Protocol; STP; Rapid Spanning Tree; Spanning Tree Protocol; Spanning-tree protocol; Protokol Pohon Rentangan; IEEE 802.1w Spanning Tree Protocol \| network protocol that builds a loop-free logical topology for Ethernet networks \| \| Upload media \| \| Wikipedia \| \| Instance of \| communication protocol graph algorithm \| \| Part of \| Ethernet \| \| Discoverer or inventor \| Radia Perlman (1984) \| \| Has part(s) \| Bridge Protocol Data Units \| \| Publication date \| 1985 \| \| Time of discovery or invention \| 1985 \| \| \| Authority file \| \| Q852555 GND ID: 7647999-7 \| \| Reasonator Scholia Wikidocumentaries PetScan statistics WikiMap Locator tool KML file Search depicted \| \| The Spanning Tree Protocol (STP) is a network protocol that ensures a loop-free topology for any bridged Ethernet local area network. Media in category "Spanning tree protocol" The following 21 files are in this category, out of 21 total. Bridge cascade STP limit.png 972 × 231; 24 KB Spanning tree protocol at work 1.svg 591 × 591; 36 KB Spanning tree protocol at work 2.svg 591 × 591; 48 KB Spanning tree protocol at work 3.svg 591 × 591; 63 KB Spanning tree protocol at work 4.svg 591 × 591; 64 KB Spanning tree protocol at work 5.svg 591 × 591; 86 KB Spanning tree protocol at work 6.svg 591 × 591; 83 KB Spanning tree topology.png 739 × 763; 35 KB Spanning trees.png 285 × 579; 11 KB STP designated port criterium A.svg 99 × 101; 193 KB STP designated port criterium B.svg 164 × 102; 311 KB STP designated port criterium C.svg 133 × 88; 223 KB STP geographical scale.svg 99 × 64; 68 KB STP loop problem.svg 250 × 215; 566 KB STP root bridge avoided positioning.png 370 × 234; 11 KB STP root bridge optimal positioning.png 309 × 222; 10 KB STP root port criterium A.svg 116 × 90; 168 KB STP root port criterium B.svg 163 × 99; 287 KB STP root port criterium C.svg 135 × 89; 224 KB STP root port criterium D.svg 137 × 95; 205 KB STP unidirectional links.png 512 × 156; 11 KB Retrieved from " Categories: Network protocols Spanning trees Non-topical/index: Uses of Wikidata Infobox Uses of Wikidata Infobox with no image Category:Spanning tree protocol Add topic
21	https://pmc.ncbi.nlm.nih.gov/articles/PMC1794423/	Photosynthesis: what color was its origin? - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Genome Biol . 2007 Jan 3;7(12):245. doi: 10.1186/gb-2006-7-12-245 Search in PMC Search in PubMed View in NLM Catalog Add to search Photosynthesis: what color was its origin? Jin Xiong Jin Xiong 1 Department of Biology, Texas A&M University, College Station, TX 77843, USA Find articles by Jin Xiong 1,✉ Author information Article notes Copyright and License information 1 Department of Biology, Texas A&M University, College Station, TX 77843, USA ✉ Corresponding author. Issue date 2006. Copyright © 2006 BioMed Central Ltd PMC Copyright notice PMCID: PMC1794423 PMID: 17210067 Short abstract Recent studies suggest several alternative evolutionary scenarios for the origin of photosynthesis. Abstract Recent studies using geological and molecular phylogenetic evidence suggest several alternative evolutionary scenarios for the origin of photosynthesis. The earliest photosynthetic group is variously thought to be heliobacteria, proteobacteria or a precursor of cyanobacteria, organisms whose photosynthetic pigments make them different colors. The origin of photosynthesis using tetrapyrrole compounds (such as chlorophylls) has long been one of the most complex and challenging issues in biology. Many schools of thought have emerged, each with its own assumptions and with evidence supporting a particular origin of photosynthesis. A number of recently published landmark papers [1-3] have contributed further to the debate. To get a better grasp of the important issue, one needs to first understand the distribution of extant photosynthetic groups and the types of photosynthetic apparatus within each group. Within the prokaryotic domain, there are five main groups of bacteria that perform tetrapyrrole-based photosynthesis. They are proteobacteria (also known as purple bacteria), heliobacteria, Chloroflexi (filamentous bacteria also known as green non-sulfur bacteria), Chlorobi (green sulfur bacteria) and cyanobacteria. Cyanobacteria are known evolutionary progenitors of chloroplasts in algae and plants which evolved at a later stage . Therefore, to understand the early evolution of photosynthesis, one needs to focus on the photosynthetic prokaryotes. The types of photosynthesis and photosynthetic apparatus in these groups of organisms also vary. Cyanobacteria contain two types of photosystems (type I, also known as Fe-S type, and type II, also known as quinone type) and carry out oxygen-evolving photosynthesis. The other four groups of photosynthetic bacteria contain only one type of photosystem and perform non-oxygen-evolving photosynthesis. Proteobacteria and Chloroflexi are known to contain a simplified type II photosystem whereas Chlorobi and heliobacteria contain a simplified type I photosystem only. In addition, the chlorophyll pigments contained in the photosystems of these bacterial groups also differ structurally with cyanobacteria having chlorophyll a, heliobacteria having bacteriochlorophyll g, and the other three groups synthesizing various bacteriochlorophylls from a to e. These pigments absorb light at different frequencies and thus have slightly different colors. The question of the original nature of the most ancestral photosynthetic apparatus can thus be metaphorically encapsulated by asking the color of the first photosynthetic apparatus. (It needs to be pointed out that the actual colors of the photosynthetic organisms are often masked by non-chlorophyll pigments such as carotenoids and phycobilins). Geological evidence on the origin of photosynthesis The advent of photosynthesis is known from geological studies to be a very ancient event. The earliest evidence for biological carbon fixation was dated to 3.8 billion years ago (or Giga-annum, Ga) [5,6] from the isotopic composition of sedimentary rocks. The best known fossil evidence of the earliest photosynthetic forms of life has been dated to 3.5 Ga and was found to contain filamentous cellular structures [7,8]. From their morphology alone, Schopf and co-workers [7,8] proposed that these structures were oxygen-evolving cyanobacteria. This finding is significant in that it suggests that photosynthetic organisms were the earliest forms of life on Earth and that oxygen-evolving photosynthesis started in the early Archean age . The finding remains controversial, however, because it is difficult to determine cell physiology on the basis of the shape of the structures in ancient microfossils. More recently, Brasier et al. challenged the early interpretation of the microfossils by suggesting (using data from electron microscopy, digital imaging and Raman spectroscopy) that the structures were in fact artifacts of amorphous graphite. However, emerging geochemical studies [11,12] seem to have reconfirmed the biogenic nature of the microfossils and thus reversed the conclusion of Brasier et al. . Despite the controversies surrounding the 3.5 Ga fossils, there are some other microfossils thought to be cyanobacteria that are more likely to be genuine, the oldest of which were dated to 2.6-2.7 Ga by more reliable biomarkers [13,14]. This timing is significant because it predates slightly the early rise of oxygen on Earth, which was about 2.3 Ga [15-17]. Recently, Tice and Lowe showed additional evidence of photosynthetic carbon fixation by filamentous microbial mats found in 3.4 Ga cherts (flint-like sedimentary rocks) in completely anoxic environments. Their geochemical analysis further ruled out the possibility that the primary electron donor for the carbon fixation could be H 2 O (as used by plants), Fe 2+ or H 2 S. Instead, the primary electron source for this type of photosynthesis is most likely to have been hydrogen, which was abundant in the atmosphere in the early Archean age. The result is consistent with the view that the early photosynthesis was most likely to have been carried out by anoxygenic photosynthetic bacteria rather than cyanobacteria. Molecular phylogenetic evidence on the evolution of photosynthesis Although the geological records provide the timing information for the evolutionary events, finding the sequence with which the five main groups of photosynthetic micro-organisms evolved from a common ancestor requires molecular phylogenetic analysis of the genetic components of extant photosynthetic organisms. The use of molecular sequences to discover this ordering has, however, so far generated even greater controversies than the study of microfossils. Various hypotheses have been proposed and various methodologies used in the course of reconstructing the early history of photosynthesis. Studies of whole organisms and genomes In the early days of molecular phylogenetics, bacterial relationships were usually resolved using 16S ribosomal RNA (rRNA) [18,19], which allowed classification and identification of the major bacterial groups. From the 16S rRNA phylogenetic trees, the evolutionary pathway of the five photosynthetic bacterial groups can be compiled, giving Chloroflexi as the earliest photosynthetic lineage, with heliobacteria as the second, followed by Chlorobi, cyanobacteria and proteobacteria, in that order (Figure 1a). Figure 1. Open in a new tab Four representative scenarios for the early evolution of the photosynthetic process among photosynthetic bacteria. (a)The branching order of photosynthetic organisms based on 16S rRNA [4,18,19]. (b)The branching order for photosynthetic organisms proposed by Gupta and coworkers using heat shock proteins; (c)the branching order that we derived from analysis of (bacterio)chlorophyll biosynthesis genes (yellow arrows represent the direction of lateral gene transfer for the type II reaction center; green arrows represent the direction of lateral gene transfer for the type I reaction center); and (d)the branching order evolutionary pathway derived from analysis of photosynthesis genes by Galperin and coworkers (black arrows represent the direction of lateral gene transfer for some photosynthesis genes). Gupta et al. used heat shock proteins (Hsp60 and Hsp70) as the molecular markers and relied heavily on conserved insertions and deletions (indels) in the sequence alignment to derive phylogenetic trees for the photosynthetic bacterial groups. The results led to the conclusion that the heliobacterial group was the most ancestral out of the photosynthetic groups and that the evolutionary pathway followed a linear order, with Chloroflexi branching second, then cyanobacteria, Chlorobi and proteobacteria in that order (Figure 1b). With the rapid accumulation of bacterial whole-genome sequence data, phylogenetic relationships are now more often studied at the whole-genome level to obtain a clearer picture of bacterial evolution. Raymond et al. analyzed one representative genome from each of the five photosynthetic taxa and discovered highly incongruent evolutionary patterns among the five genomes. They observed 15 possible tree topologies from the commonly shared proteins encoded by all five genomes. To resolve the evolutionary pattern for photosynthesis further, the authors then compiled a set of 'photosynthesis-specific' and 'photosynthesis-related' genes and performed phylogenetic analysis on each gene product, but they failed to reach a phylogenetic consensus. This confirms that bacterial genome evolution involves extensive lateral gene transfer, which also had a role in the development of the photosynthetic apparatus. Recently, Mulkidjanian et al. analyzed 15 cyanobacterial genomes and derived a set of genes commonly shared by all the genomes, in the form of a minimal cyanobacterial genome. The photosynthesis-related portion of the minimal gene set was found to be much larger than the gene set previously derived by the Blankenship group because many genes are specific to cyanobacteria. The more comprehensive nature of the cyanobacterial gene set prompted the conclusion that cyanobacteria were the most ancestral phototroph. As the conclusion was not drawn from accepted rooted phylogenies using bona fide photosynthesis genes found in all photosynthetic lineages, however, the logic behind this proposal seems weak. Studies using chlorophyll biosynthesis markers Because (bacterio)chlorophylls are integral components of the photosynthetic apparatus, enzymes involved in the biosynthesis of this pigment (encoded by the bch genes) could be used as specific indicators for the evolution of photosynthesis. The main advantage of this set of markers is their ubiquitous presence among all the photosynthetic bacterial groups. Most of the bch trees can be unambiguously rooted, because a composite tree can be constructed with a reliable outgroup from a different but homologous enzyme family. The analysis of the Bch enzymes has been instrumental in testing the long-standing Granick hypothesis , which states that biosynthetic pathways recapitulate their evolution: in a multi-step biosynthetic pathway, products produced in early steps would evolutionarily predate products produced in later steps. As a general guide to biochemical evolution, this hypothesis makes sense, but when it is used to reconstruct the evolutionary history of photosynthesis, it may generate erroneous conclusions. In the chlorophyll biosynthesis pathway, chlorophyll a biosynthesis requires shorter steps and appears before bacteriochlorophyll a. According to the Granick hypothesis, this would indicate that cyanobacteria (which contain chlorophyll a) predate anoxygenic photosynthetic bacteria (containing bacteriochlorophyll a) [24,25]. Therefore, by applying the Granick hypothesis, one would conclude that photosynthesis originated with cyanobacteria. This view agrees with that of Mulkidjanian et al. . But the molecular phylogenetic analysis of a number of enzymes involved in (bacterio)chlorophyll biosynthesis, performed by my group and others [26-28] using carefully selected outgroups for rooting the trees, indicates that the anoxygenic photosynthetic lineages are almost certain to be more deeply rooted than the oxygenic cyanobacterial and chloroplast lineages. Proteobacteria seem to be the earliest evolving among the anoxygenic lineages, suggesting that bacteriochlorophyll a predates chlorophyll a during evolution. A Bayesian analysis that we subsequently performed on the dataset delineated the sequence of evolution for (bacterio)cholorophyll biosynthesis (Figure 1b). In this scenario, the pigment biosynthesis genes were laterally transferred from proteobacteria to Chlorobi, from which the lineage bifurcated to Chloroflexi and cyanobacteria, which gave rise to heliobacteria (Figure 1c). This result seems to contradict the Granick hypothesis. A simple explanation for this paradox could be that gene loss of some of the bch genes occurred during the evolution of the genes in the cyanobacterial lineage, leading to a shortened biosynthesis pathway. Studies using reaction centers The reaction center is the core of the (bacterio)chlorophyll-containing protein complex where the primary electron transfer event takes place during photosynthesis. Because of their central importance, reaction center proteins have naturally become the focus of study for the evolutionary pathway of photosynthesis. However, the main difficulty of using the reaction center proteins as molecular markers is the extremely high divergence of the sequences between the two types of reaction center, making it difficult to derive an evolutionary scenario for all five photosynthetic bacterial groups. Currently, several hypotheses have been put forward to postulate the origin and developmental pathway of the photosynthetic apparatus. Generally, they fall within two schools of thought, the selective loss model and the fusion model. The selective loss model The selective loss model [24,25,30] postulates an ancestral photosynthetic organism, similar to oxygenic cyanobacteria, containing both types of reaction center. A subsequent loss of one of the reaction center types gave rise to a single reaction center found in extant anoxygenic photosynthetic bacteria. The model suggests that organisms like cyanobacteria were present in the prebiotic phase, when life first originated. This view can find initial support from the 3.5 Ga microfossil study by Schopf and coworkers [7,8]. The most recent support for the selective loss theory came from Mulkidjanian et al. , who believe that the enlarged photosynthesis core gene set suggests a cyanobacterial origin of photosynthesis. Given the compelling geological evidence that anoxygenic photosynthesis evolved before oxygenic photosynthesis, the authors offered a revised selective loss model in which a group termed 'procyanobacteria', which was largely similar to extant cyanobacteria but did not evolve oxygen, was the most ancient phototroph and that it subsequently spread photosynthesis genes to other anoxygenic photosynthetic bacterial groups by lateral gene transfer and large-scale gene loss (Figure 1d). The authors further postulated that procyanobacteria, as the photosynthetic progenitors, contained the type I reaction center only. This idea was based primarily on the geological evidence of Tice and Lowe that the 3.4 Ga phototroph performed hydrogen-based photosynthesis. According to Mulkidjanian et al. , only procyanobacteria were suitable for this type of photosynthesis. Chlorobi and heliobacteria were excluded from consideration because they do not contain the Calvin cycle (in which a six carbon sugar molecule is synthesized by fixing CO 2 and combining it with a five carbon molecule, 1,5-ribulose bisphosphate). Chloroflexi and proteobacteria were excluded because hydrogen is too reducing for the quinone-type reaction centers that they contain. Both arguments seem weak because green sulfur bacteria (Chlorobi) are known to fix CO 2 not through the Calvin cycle but through the reductive tricarboxylic acid (TCA) cycle (the traditional citric acid cycle running in reverse), which uses hydrogen or reduced sulfur compounds as electron donors. In addition, it is well established that Chlorobi, Chloroflexi and proteobacteria can use hydrogen as the sole electron donor and CO 2 the sole electron acceptor for photoautotrophic growth (that is, growth that solely depends on light and inorganic nutrients). In contrast, normal cyanobacterial cells are not capable of using hydrogen as the sole electron donor in photosynthesis. Though some specialized cyanobacterial cells such as heterocysts (nitrogen fixing cells with multi-layered cell walls) are capable of anoxic photosynthetic electron transfer using the type I photosystem only with hydrogen or sulfur compounds serving as electron donors, this special type of differentiated cells are considered a relatively recent evolutionary invention . The fusion model The fusion model [4,34] postulates that the type I and type II reaction centers could have been established independently in two different ancestral lineages (one in proteobacteria or Chloroflexi and the other in heliobacteria or Chlorobi) before being brought together into one lineage to produce the cyanobacterial dual photosystem. The model envisages the photosynthetic apparatus as having evolved from simple to complex, which seems more reasonable than the opposite scenario. A colleague and I proposed one version of the fusion model , in which the direction of reaction-center evolution was inferred from a Bayesian analysis. The most ancestral form of the reaction center was proposed to be a type II reaction center of proteobacterial origin. The subsequent divergence of the proteobacterial lineage into Chloroflexi and cyanobacteria gave rise to the extant type II reaction center in these two lineages. The type I reaction center, which was thought to be relatively late evolving, may have been formed through a fusion by the primordial type II reaction center and a light-harvesting antenna protein (which contains chlorophyll pigments that harvest light energy and transfer it to the reaction center). We proposed that the gene fusion event occurred in a heliobacterial lineage, resulting in an enlarged reaction center. This type I-like reaction center later diverged into those found in Chlorobi and cyanobacteria. The arrival of both types of reaction centers in cyanobacteria enabled the later establishment of a linear electron transfer between the two. Our proposed evolutionary scenario for the reaction centers is distinct from the evolutionary pathway for (bacterio)chlorophyll biosynthesis (Figure 1c), which adds an additional layer of lateral gene transfer relative to the 16S rRNA evolutionary pathway. In view of the lack of obvious sequence similarity between the two types of reaction centers, which makes it difficult to derive a common evolutionary tree for them, a new approach was adopted by the Blankenship group , based on both structure and sequence. Because of the known structural similarity of the two types of reaction centers, Blankenship and coworkers first aligned the conserved core structures of two reaction centers, which exposed the structurally corresponding residues. The structurally aligned residues were then used to construct a sequence alignment that was then used to build a unified phylogeny of the reaction centers. The reaction center trees were unrooted and thus did not allow direct inference of the most ancestral reaction center. If a midpoint rooting technique were used, however, the trees would suggest that the earliest reaction center was anoxygenic and probably a homodimeric complex. In conclusion, although no consensus for the evolutionary history of photosynthetic apparatus has yet emerged, it is widely accepted that it is a very complex process involving multi-layered lateral gene transfer . The lateral gene transfer events can seem so complex that the origin of photosynthesis could become an intractable issue. As a solution to the problem, instead of assuming that all genes are equally important in their ability to reveal the early evolutionary history of photosynthesis, we suggested focusing on a sub-process, (bacterio)chlorophyll biosynthesis, as the factor most likely to have determined the advent of photosynthesis. Along with the development of the most important elements of the photosynthetic apparatus, a functional apparatus could have been assembled through a multi-staged recruitment of reaction center proteins and antenna proteins, which could conceivably have had separate evolutionary histories and performed different functions before the recruitment. The recruitment process may have undergone several intermediate stages, producing products with various degrees of complexity. In essence, the precise picture of early evolution of photosynthesis still remains to be understood. To reveal the true color of the origin of photosynthesis will require years of painstaking biogeochemistry and molecular phylogenetic studies. Acknowledgments Acknowledgements I thank the Welch Foundation for support (grant no. A1589). References Tice MM, Lowe DR. Hydrogen-based carbon fixation in the earliest known photosynthetic organisms. Geology. 2006;34:37–40. doi: 10.1130/G22012.1. [DOI] [Google Scholar] Mulkidjanian AY, Koonin EV, Makarova KS, Mekhedov SL, Sorokin A, Wolf YI, Dufresne A, Partensky F, Burd H, Kaznadzey D, et al. The cyanobacterial genome core and the origin of photosynthesis. Proc Natl Acad Sci USA. 2006;103:13126–13131. doi: 10.1073/pnas.0605709103. [DOI] [PMC free article] [PubMed] [Google Scholar] Sadekar S, Raymond J, Blankenship RE. Conservation of distantly related membrane proteins: photosynthetic reaction centers share a common structural core. Mol Biol Evol. 2006;23:2001–2007. doi: 10.1093/molbev/msl079. [DOI] [PubMed] [Google Scholar] Blankenship RE. 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[DOI] [PubMed] [Google Scholar] Articles from Genome Biology are provided here courtesy of BMC ACTIONS View on publisher site PDF (93.8 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Short abstract Abstract Geological evidence on the origin of photosynthesis Molecular phylogenetic evidence on the evolution of photosynthesis Acknowledgments References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
22	https://courses.lumenlearning.com/wm-developmentalemporium/chapter/7-2-2-adding-and-subtracting-radicals/	Adding and Subtracting Radicals Learning Outcomes Identify radicals that can be added or subtracted Add and subtract radical expressions Adding and subtracting radicals is much like combining like terms with variables. We can add and subtract expressions with variables like this: 5x+3y−4x+7y=x+10y There are two keys to combining radicals by addition or subtraction: look at the index, and look at the radicand. If these are the same, then addition and subtraction are possible. If not, then you cannot combine the two radicals. Remember the index is the degree of the root and the radicand is the term or expression under the radical. In the diagram below, the index is n, and the radicand is 100. The radicand is placed under the root symbol and the index is placed outside the root symbol to the left: Index and radicand In the graphic below, the index of the expression 123√xy is 3 and the radicand is xy. Practice identifying radicals that are compatible for addition and subtraction by looking at the index and radicand of the roots in the following example. Example Identify the roots that have the same index and radicand. 10√6 −13√6 √25 12√6 123√25 −73√6 Show Solution Let’s start with 10√6. The index is 2 because no root was specified, and the radicand is 6. The only other radical that has the same index and radicand is 12√6. −13√6 has an index of 3, and a radicand of 6. The only other radical that has the same index and radicand is −73√6. √25 has an index of 2 and a radicand of 25. There are no other radicals in the list that have the same index and radicand. 12√6 has the same index and radicand as 10√6 123√25 has an index of 3 and a radicand of 25. There are no other radicals in the list that share these. −73√6 has the same index and radicand as −13√6 Making sense of a string of radicals may be difficult. One helpful tip is to think of radicals as variables, and treat them the same way. When you add and subtract variables, you look for like terms, which is the same thing you will do when you add and subtract radicals. Let’s use this concept to add some radicals. In this first example, both radicals have the same radicand and index. Example Add. 3√11+7√11 Show Solution The two radicals have the same index and radicand. This means you can combine them as you would combine the terms 3a+7a. 3√11 + 7√11 Answer 3√11+7√11=10√11 It may help to think of radical terms with words when you are adding and subtracting them. The last example could be read “three square roots of eleven plus 7 square roots of eleven”. This next example contains more addends, or terms that are being added together. Notice how you can combine like terms (radicals that have the same root and index) but you cannot combine unlike terms. Example Add. 5√2+√3+4√3+2√2 Show Solution Rearrange terms so that like radicals are next to each other. Then add. 5√2+2√2+√3+4√3 Answer 5√2+√3+4√3+2√2=7√2+5√3 Notice that the expression in the previous example is simplified even though it has two terms: 7√2 and 5√3. It would be a mistake to try to combine them further! (Some people make the mistake that 7√2+5√3=12√5. This is incorrect because√2 and √3 are not like radicals so they cannot be added.) Example Add. 3√x+123√xy+√x Show Solution Rearrange terms so that like radicals are next to each other. Then add. 3√x+√x+123√xy Answer 3√x+123√xy+√x=4√x+123√xy In the following video, we show more examples of how to identify and add like radicals. Try It Sometimes you may need to add and simplify the radical. If the radicals are different, try simplifying first—you may end up being able to combine the radicals at the end, as shown in these next two examples. Example Add and simplify. 23√40+3√135 Show Solution Simplify each radical by identifying perfect cubes. 23√8⋅5+3√27⋅523√(2)3⋅5+3√(3)3⋅523√(2)3⋅3√5+3√(3)3⋅3√5 Simplify. 2⋅2⋅3√5+3⋅3√5 Add. 43√5+33√5 Answer 23√40+3√135=73√5 Example Add and simplify. x3√xy4+y3√x4y Show Solution Simplify each radical by identifying perfect cubes. x3√x⋅y3⋅y+y3√x3⋅x⋅yx3√y3⋅3√xy+y3√x3⋅3√xyxy⋅3√xy+xy⋅3√xy Add like radicals. xy3√xy+xy3√xy Answer x3√xy4+y3√x4y=2xy3√xy The following video shows more examples of adding radicals that require simplification. Subtracting Radicals Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices (plural of index) must be the same for two (or more) radicals to be subtracted. In the examples that follow, subtraction has been rewritten as addition of the opposite. Example Subtract. 5√13−3√13 Show Solution The radicands and indices are the same, so these two radicals can be combined. 5√13−3√13 Answer 5√13−3√13=2√13 Example Subtract. 43√5a−3√3a−23√5a Show Solution Two of the radicals have the same index and radicand, so they can be combined. Rewrite the expression so that like radicals are next to each other. 43√5a−3√3a−23√5a43√5a−23√5a)−3√3a) Combine. Although the indices of 23√5a and −3√3a are the same, the radicands are not—so they cannot be combined. 23√5a−3√3a) Answer 43√5a−3√3a−23√5a=23√5a−3√3a In the video examples that follow, we show more examples of how to add and subtract radicals that don’t need to be simplified beforehand. Example Subtract and simplify. 54√a5b−a4√16ab, where a≥0 and b≥0 Show Solution Simplify each radical by identifying and pulling out powers of 4. 54√a4⋅a⋅b−a4√(2)4⋅a⋅b5⋅a4√a⋅b−a⋅24√a⋅b5a4√ab−2a4√ab The answer is 3a4√ab. In our last videos, we show more examples of subtracting radicals that require simplifying. Summary Combining radicals is possible when the index and the radicand of two or more radicals are the same. Radicals with the same index and radicand are known as like radicals. It is often helpful to treat radicals just as you would treat variables: like radicals can be added and subtracted in the same way that like variables can be added and subtracted. Sometimes, you will need to simplify a radical expression before it is possible to add or subtract like terms. Candela Citations CC licensed content, Original Screenshot: keys. Provided by: Lumen Learning. License: CC BY: Attribution Screenshot: Index and radicand. Provided by: Lumen Learning. License: CC BY: Attribution Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: License: CC BY: Attribution Ex: Add and Subtract Radicals - No Simplifying. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Ex: Add and Subtract Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Original Screenshot: keys. Provided by: Lumen Learning. License: CC BY: Attribution Screenshot: Index and radicand. Provided by: Lumen Learning. License: CC BY: Attribution Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: License: CC BY: Attribution Ex: Add and Subtract Radicals - No Simplifying. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Ex: Add and Subtract Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
23	https://www.scribd.com/document/418744956/Radicals-What-Rational-Numbers-Have-Rational-Square-Roots-Mathematics-Stack-Exchange	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Upload 0 ratings0% found this document useful (0 votes) 788 views7 pages Rational Numbers with Rational Roots Rational numbers have rational square roots when the numerator and denominator can both be expressed as perfect squares. Specifically: A rational number a/b has a rational square root if an… Uploaded by Frankie Liu You are on page 1/ 7 7/18/2019radicals - What rational numbers have rational square roots? - Mathematics Stack Exchange Mathematics Stack Exchange is a question and answer site for people studying math at any level andprofessionals in related fields. Join them; it only takes aminute: Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top What rational numbers have rational square roots? All rational numbers have the fraction form where a and b are integers(). ab ≠ 0 My question is: for what and does the fraction have rational square root? The simple answer would be when both are perfect squares, but if two perfect squares are multiplied by a commoninteger , the result may not be tw o perfect squares. Like: a bn 49 → 818 And intuitively, without factoring, and must qualify by some standard to have a rationalsquare root. = 8 b = 18 Once this is solved, can this be extended to any degree of roots? Like for what and does thefraction have rational th root? a bn elementary-number-theoryradicalsrationality-testingedited Aug 25 '15 at 12:19 punctured dusk 33088asked Jul 20 '13 at 16:44 137 1 – Sorry for the disturbing grammar mistake. 2. Sorry for this elementary question. user2386986Jul 20 '13 By using our site, you acknowledge that you have read and understand our , , and our .Cookie PolicyPrivacy PolicyTerms of Service Download to read ad-free 7/18/2019radicals - What rational numbers have rational square roots? - Mathematics Stack Exchange is a related question.math.stackexchange.com/questions/324724/… Baby DragonJul 20 '13 at 17:26 – is not in the "lowest terms", which means a , b ) = 1 KazJul 21 '13 at 5:563 – possible duplicate of is either an integer or an irrational number. a punctured dusk Aug 25 '15 at 12:20 – is a square rational iff is a square integer, as several people pointed out. The natural question is, is there asimilar test for cube rationals? Here it is: is a cube rational iff is a cube integer. ab ab ab Akiva Weinberger Nov10 '15 at 17:20 8 Answers 48 A nice generalization of the is that every rational number isuniquely represented as a product of primes raised to integer powers. For example:fundamental theorem of arithmetic 49 = 2 2 ∗ 3 −2 This is the natural generalization of factoring integers to rational numbers. Positive powers are partof the numerator, negative powers part of the denominator (since ). a − b = 1 a b When you take the th root, you divide each power by : n n n √ 2 p 2 ∗ 3 p 3 ∗ 5 p 5 . . . = 2 p 2 / n ∗ 3 p 3 / n ∗ 5 p 5 / n . . . For example: √ 49 = 2 2/2 ∗ 3 −2/2 = 23 In order for the powers to continue being integers when we divide (and thus the result a rationalnumber), they must be multiples of . In the case where is , that means the numerator anddenominator, in their reduced form, are squares. (And for , cubes, and so on...) n n 2 n = 3 In your example, when you multiply the numerator and denominator by the same number, they continue to be the same rational number, just represented differently. 2 ∗ 42 ∗ 9 = 2 2+1−1 ∗ 3 −2 = 2 2 ∗ 3 −2 You correctly recognize the important of factoring, though you don't really want to use it in youranswer. But the most natural way to test if the fraction produced by dividing by has a rational throot, is to factor and look at the powers. Or, equivalently, reduce the fraction and determine if the numerator and denominator are integers raised to the power of . a b na / bn edited Jul 20 '13 at 17:12answered Jul 20 '13 at 17:06 By using our site, you acknowledge that you have read and understand our , , and our .Cookie PolicyPrivacy PolicyTerms of Service Download to read ad-free 7/18/2019radicals - What rational numbers have rational square roots? - Mathematics Stack Exchange colah 602 47 3 – Good answer. They should teach this stuff in school. goblinJul 20 '13 at 17:11 – Very useful to switch to prime product representation. mvwJul 21 '13 at 11:06 7 We give a fairly formal statement and proof of the result described in the post. Let and be integers, with . Suppose that has a rational square root. Then thereexists an integer , and integers and , such that and , Theorem: a b b ≠ 0 ab e m n a = em 2 b = en 2 It is enough to prove the result for positive . For if is negative and has a square root,then we must have . Thus . If we know that there are integers , , such that and , then and . Proof: b b ab a ≤ 0 ab = \| a \|\| b \| e m n \| a \| = em 2 \| b \| = en 2 a = ( − e ) m 2 b = ( − e ) n 2 So suppose that , and . Let be the greatest common divisor of and . Then , and, for some relatively prime and . b 0 a ≥ 0 d a b a = da ∗ b = db ∗ a ∗ b ∗ It will be sufficient to prove that each of and is a perfect square. a ∗ b ∗ Since is a square, there exist relatively prime integers and such that . a ∗ b ∗ m n a ∗ b ∗ = ( mn ) 2 With some algebra we reach a ∗ n 2 = b ∗ m 2 . By Euclid's Lemma, since divides the product on the left, and is relatively prime to , we havethat divides . Also, because divides the expression on the right, and is relatively prime to , we have divides . Since is positive, we conclude that . Now it is easy to show that . b ∗ a ∗ b ∗ n 2 n 2 n 2 m 2 n 2 b ∗ b ∗ b ∗ = n 2 a ∗ = m 2 A similar theorem can be stated and proved for -th roots. k edited Jul 20 '13 at 17:28answered Jul 20 '13 at 17:22 André Nicolas 458k 36440827 6 Trivially, a rational number has a rational square root if and only if it's the square of some rationalnumber. As the other answers note, various other characterizations can be given, e.g. that has a rationalsquare root if and only if: ab By using our site, you acknowledge that you have read and understand our , , and our .Cookie PolicyPrivacy PolicyTerms of Service Download to read ad-free 7/18/2019radicals - What rational numbers have rational square roots? - Mathematics Stack Exchange and for some integers , and , a = ec b = ed c d e the numerator and denominator of the reduced form of are both squares, or ab the unique prime factorization of has only even exponents. ab Yet another "cute" characterization is that of an integer. I've sketched a proof of this below. has a rational square root if and only if is asquare ab ab To show that this rule works, first note that, if and , then .Showing the converse is a bit trickier, but it follows easily enough from prime factorization.Specifically, let where denotes the -th prime. Since is a square, we know that must be even for all ; from , we know that for all . Together, these imply that, if is odd for some , so is ,and the converse also holds. Now let where if and are odd, and let if they are even. Since and for all , divides both and . Furthermore, since and are both even for all , it follows that and are both squares, and so we can write and . Proof. a = ec 2 b = ed 2 ab = c 2 d 2 e 2 = ( cde ) 2 ab = x 2 = y = ∞ ∏ i =1 p y i i , a = ∞ ∏ i =1 p a i i and b = ∞ ∏ i =1 p b i i p i i y = x 2 y i i y = ab y i = a i + b i i a i i b i e = ∞ ∏ i =0 p e i i , e i = 1 a i b i e i = 0 e i ≤ a i e i ≤ b i i ea b a i − e i b i − e i ia / e = c 2 b / e = d 2 a = ec 2 b = ed 2 ◻ As Akiva Weinberger notes in the comments below, this "trick" can also be generalized to higherpowers. Specifically, has a rational -th root if and only if (or, equivalently, ) is an -thpower of an integer. Ps. ab n a n −1 b b n −1 a n edited Nov 10 '15 at 19:55answered Jul 20 '13 at 19:06 Ilmari Karonen 20.6k 25288 – "Yet another "cute" characterization is that ab has a rational square root if and only if ab is a square of aninteger." This is the answer I actually want, because it disregards whether a and b are in lowest term. But itfails to hold for higher degree of roots... user2386986Jul 20 '13 at 21:021 – Yes, it only works for square roots. Essentially, that's because the additive ring of integers modulo 2 has 2, and thus addition and subtraction modulo 2 are the exact same operation. The primefactorization of is given by adding the exponents of the factorizations of and , whereas that of is given by subtracting them. But for determining whether a number is a , we only need to look at the exponentsmodulo 2, and thus addition and subtraction give the same result.characteristic ab a b a / b square Ilmari KaronenJul 20 '13 at 21:131 "Alasthistrickdoesnoteneralizetohiherowers."Sureitdoesouustneedaslihtmodification. has a By using our site, you acknowledge that you have read and understand our , , and our .Cookie PolicyPrivacy PolicyTerms of Service Download to read ad-free Download to read ad-free Download to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Prove Square Root of 3 Is Irrational 0% (1) Prove Square Root of 3 Is Irrational 4 pages Rajesh Verma Math PDF 100% (2) Rajesh Verma Math PDF 1,728 pages Rational Numbers Properties No ratings yet Rational Numbers Properties 4 pages Prove Root 2 Is Irrational No ratings yet Prove Root 2 Is Irrational 4 pages 01-Rational and Irrational Numbers - Summary Note 100% (2) 01-Rational and Irrational Numbers - Summary Note 8 pages A Proof That The Square Root of 2 Is Irrational No ratings yet A Proof That The Square Root of 2 Is Irrational 29 pages Math 7 ADM Module 7 Edited 0% (1) Math 7 ADM Module 7 Edited 26 pages G7 Math Q1 Week 6 - Rational and Irrational Numbers No ratings yet G7 Math Q1 Week 6 - Rational and Irrational Numbers 19 pages Rational and Irrational Numbers No ratings yet Rational and Irrational Numbers 8 pages Irrational Numbers No ratings yet Irrational Numbers 14 pages Chapter 1 - Real Numbers Revision Notes No ratings yet Chapter 1 - Real Numbers Revision Notes 1 page Rational Numbers: Definition & Properties No ratings yet Rational Numbers: Definition & Properties 4 pages History of Rational Numbers 40% (5) History of Rational Numbers 4 pages Real Analysis HW1 No ratings yet Real Analysis HW1 4 pages Doulus SystemC Tutorial No ratings yet Doulus SystemC Tutorial 23 pages Notes To Accompany The Theorem of Pythagoras and Properties of No ratings yet Notes To Accompany The Theorem of Pythagoras and Properties of 4 pages Notes2 34t4 No ratings yet Notes2 34t4 9 pages The TLM-2.0 Standard: John Aynsley, Doulos No ratings yet The TLM-2.0 Standard: John Aynsley, Doulos 43 pages Merged 20240208 0717 No ratings yet Merged 20240208 0717 115 pages Real Numbers No ratings yet Real Numbers 30 pages 1-5 Roots and Irrational Numbers No ratings yet 1-5 Roots and Irrational Numbers 34 pages Ross Problem Set No ratings yet Ross Problem Set 12 pages Types of Rational Numbers No ratings yet Types of Rational Numbers 4 pages Year 9 Unit 1.1 Irrational Numbers No ratings yet Year 9 Unit 1.1 Irrational Numbers 23 pages The Real Numbers No ratings yet The Real Numbers 12 pages SystemC for Advanced System Design No ratings yet SystemC for Advanced System Design 25 pages Páginas de Dokumen - Pub - Mathematical-Analysis-Solution-Manual No ratings yet Páginas de Dokumen - Pub - Mathematical-Analysis-Solution-Manual 3 pages 2019 DVCon India Modern SystemC.v2 - 4.3 No ratings yet 2019 DVCon India Modern SystemC.v2 - 4.3 41 pages Principal Roots & Irrational Numbers Module No ratings yet Principal Roots & Irrational Numbers Module 25 pages Binary Operations 100% (1) Binary Operations 22 pages Smp08alg Na Te2 C13 L07 13 No ratings yet Smp08alg Na Te2 C13 L07 13 7 pages HW 3 Solutions No ratings yet HW 3 Solutions 7 pages En 0105 Serge Lang Basic Mathematics Answers 100% (2) En 0105 Serge Lang Basic Mathematics Answers 16 pages Math 7 Q1Wk6 ABCD Final No ratings yet Math 7 Q1Wk6 ABCD Final 8 pages 8 Order of Operations Pemdas No ratings yet 8 Order of Operations Pemdas 4 pages M1F Summary Notes No ratings yet M1F Summary Notes 11 pages BJU Press Alg. 1 Section 11.1 No ratings yet BJU Press Alg. 1 Section 11.1 26 pages Real Numbers & Calculus Basics No ratings yet Real Numbers & Calculus Basics 25 pages Determinant Practice Sheet 100% (1) Determinant Practice Sheet 13 pages LESSON 10 Principal Roots - Irrational Numbers-jdSinlao No ratings yet LESSON 10 Principal Roots - Irrational Numbers-jdSinlao 50 pages 8th Square and Square Root 8th Cube No ratings yet 8th Square and Square Root 8th Cube 3 pages Computer Number System Number System: I. 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24	https://lshtm-hds.github.io/Math-Refresher/cm_combinatorics.html	Combinatorics — Maths, Probability and Statistics Refresher Maths, Probability and Statistics Refresher =========================================== Maths, Probability and Statistics Refresher Mathematics Functions, exponential and logarithm rules Differentiation and integration Matrices Combinatorics Probability and statistics Probability, definitions and axioms Probability Theorems Discrete Random Variables Continuous Random Variables Correlation and covariance Distributions Quizzes Quiz 1: Powers and logs Quiz 2: Differentiation and integration Quiz 3: Matrices Quiz 4: Combinatorics Quiz 5: Probability Quiz 6: Probability theorems Quiz 7: Discrete Random Variables Quiz 8: Continuous Random Variables Quiz 9: Correlation and covariance Quiz 10: More complex questions Solutions Solution to quizzes Worked solutions - Quiz 1 Worked solutions - Quiz 2 Worked solutions - Quiz 3 Worked solutions - Quiz 4 Worked solutions - Quiz 5 Worked solutions - Quiz 6 Worked solutions - Quiz 7 Worked solutions - Quiz 8 Worked solutions - Quiz 9 Worked solutions - Quiz 10 Further resources Further material Powered by Jupyter Book .ipynb.pdf repositoryopen issue Binder Contents Permutations Permutations of n n objects Permutations of a subset of x x objects chosen from n n Combinations Combinations of x x objects chosen from n n Combinatorics¶ Combinatorics is the branch of maths concerned with counting. A permutation is an arrangement with regard to order, whereas in a combination the order does not matter. For example, the letters AB and BA are two different permutations, but the same combination of letters. Permutations¶ Permutations of n n objects¶ If we have 3 distinct objects, labelled A A, B B and C C, how many permutations are there? There are 6 permutations: A B C,A C B,B A C,B C A,C A B A B C,A C B,B A C,B C A,C A B and C B A C B A. More generally, suppose we have n n different objects. In how many ways can n n different objects be arranged? Each arrangement (i.e. respecting different orders) is a permutation. To obtain a general expression for the number of permutations, imagine there are n n empty boxes and n n items labelled 1 1, 2 2, …, n n, to put in the n n boxes. One way of putting the items in the boxes would be: \| Box 1 \| Box 2 \| . \| . \| . \| Box n n \| --- --- --- \| \| Item 1 \| Item 2 \| . \| . \| . \| Item n \| So, let’s start with the n n empty boxes and think about ways in which we can fill them. There are n n ways of filling the first box (we can pick item 1, item 2, …, or item n n). Once we have filled the first box, we have (n−1)(n−1) objects left and (n−1)(n−1) empty boxes left. There are (n−1)(n−1) ways of filling the second box, leaving (n−2)(n−2) objects to choose from. There are (n−2)(n−2) ways of filling the third box, leaving (n−3)(n−3) objects. … There is just one way of filling the (last) n n th box. For each of the n n ways of filling the first slot, there are n−1 n−1 ways of filling the second slot, i.e. there are n(n−1)n(n−1) ways of filling the first two slots. For each of these n(n−1)n(n−1) ways, there are n−2 n−2 ways to fill the third slot, etc. So there are n×(n−1)×(n−2)×...×3×2×1 n×(n−1)×(n−2)×...×3×2×1 ways of arranging the n n objects, i.e.the number of permutations of n n objects is equal to n×(n−1)×…×3×2×1=n!n×(n−1)×…×3×2×1=n! We denote this expression by n!n! (pronounced n n factorial). As a matter of convention we set 0!=1 0!=1. Permutations of a subset of x x objects chosen from n n¶ Now consider, how many possible permutations are there of x x objects if x x objects are selected from the total n n possible? Once again, we can think of the number of ways that we can fill x x boxes, as in the diagram below. \| Box 1 \| Box 2 \| . \| . \| . \| Box x x \| --- --- --- \| \| ? \| ? \| . \| . \| . \| ? \| So, let’s start with the x x empty boxes and think about ways in which we can fill them with the n n items. There are n n ways of filling the first box, leaving (n−1)(n−1) objects and (x−1)(x−1) boxes. There are (n−1)(n−1) ways of filling the second box, leaving (n−2)(n−2) objects to choose from. There are (n−2)(n−2) ways of filling the third box, leaving (n−3)(n−3) objects. … There are (n−x+1)(n−x+1) ways of filling the (last) x x th box. So there are n×(n−1)…×(n−x+1)n×(n−1)…×(n−x+1) ways of arranging (or permuting) x x objects from n n. We often write n P x n P x or n P x n P x to denote the number of permutations of x x objects from n n different objects. Including (n−x)(n−x−1)...×3×2×1(n−x)(n−x−1)...×3×2×1 in the numerator and denominator, we can express the number permutations of x x objects from n n as: n P x=n(n−1)(n−2)...(n−x+1)(n−x)(n−x−1)...×3×2×1(n−x)(n−x−1)...×3×2×1=n!(n−x)!n P x=n(n−1)(n−2)...(n−x+1)(n−x)(n−x−1)...×3×2×1(n−x)(n−x−1)...×3×2×1=n!(n−x)! We can think of the number of permutations of x x objects chosen from n n as the number of permutations of n n objects, divided by the number of of permutations of the (n−x)(n−x) objects that we don’t choose. Combinations¶ Combinations of x x objects chosen from n n¶ Suppose now we select x x objects chosen from n n, but we are not concerned with the order of the x x objects. Let k k be the number of ways of choosing x x from n n objects without regard to order, i.e. the number of combinations of x x objects from n n objects. From above, we know that for each of these combinations (of x x objects) there are x!x! permutations. Hence k=n!x!(n−x)!k=n!x!(n−x)! Intuitively, there are more permutations than combinations, since for every combination there are several permutations: n!(n−x)!>n!x!(n−x)!n!(n−x)!>n!x!(n−x)! We often write n C x n C x or the binomial coefficient (n x)(n x) to denote the number of combinations of x x objects from n n, n C x=(n x)=n!x!(n−x)!.n C x=(n x)=n!x!(n−x)!. MatricesProbability, definitions and axioms By MSc Health Data Science, LSHTM © Copyright 2021.
25	https://www.youtube.com/watch?v=iHNa1fhV7IA	Second derivatives (parametric functions) \| Advanced derivatives \| AP Calculus BC \| Khan Academy Khan Academy 9090000 subscribers 859 likes Description 106492 views Posted: 26 Jul 2016 Sal finds the second derivative of the function defined by the parametric equations x=3e__ and y=3__-1. Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? AP Calculus BC on Khan Academy: Learn AP Calculus BC - everything from AP Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP Test About Khan Academy: Khan Academy is a nonprofit with a mission to provide a free, world-class education for anyone, anywhere. We believe learners of all ages should have unlimited access to free educational content they can master at their own pace. We use intelligent software, deep data analytics and intuitive user interfaces to help students and teachers around the world. Our resources cover preschool through early college education, including math, biology, chemistry, physics, economics, finance, history, grammar and more. We offer free personalized SAT test prep in partnership with the test developer, the College Board. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide every year. For more information, visit www.khanacademy.org, join us on Facebook or follow us on Twitter at @khanacademy. And remember, you can learn anything. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy's AP Calculus BC channel: Subscribe to Khan Academy: 31 comments Transcript: so here we have a a set of parametric equations where X and Y are both defined in terms of T so if you input all the possible T's that you can into these functions and then plot the corresponding X and Y's for each for each T this would plot a curve in the XY plane what I want to do in this video is figure out the first derivative of y with respect to X and the second derivative of y with respect to X and in both cases it's going to be in terms of T so let's let's get to it so first let's find the first derivative of y with respect to X first derivative of y with respect to X and we've seen this before in other videos where this is going to be the derivative of y with respect to T over the derivative of x with respect to T and so this is going to be equal to well what is the derivative of y with respect to T Dy DT is equal to let's see the derivative of e to the 3T with respect to 3T is just e to the 3T and then the derivative of 3T with respect to T is going to be three so I could say times three like that or I could put that three out front and then the derivative of negative one well negative a constant doesn't change no matter what you do with to your T so that's just going to be zero so that's dydt so it's going to be equal to 3 e to the 3 T all of that over well what's the derivative of x with respect to T derivative of x with respect to T is equal two well we're going to have the three out front and so the derivative of e to the 2T with respect to 2T is going to be e to the 2T and then we're going to take the derivative of 2T with respect to T which is just two so this is going to be 6 e to the 2T 6 e to the 6 e to 2T and let's see we could simplify this a little bit I'll now go to a neutral color this is equal to so e to the so this is going to be 12 that's 3 over 6 e to the 3T minus 2T 3T minus 2T and I'm just using exponent properties right over here but three if I have three T's and I take away two of those t's I'm just going to have a t so this is just going to simplify to a t right over here so now that we know the we we've now figured out the the first derivative of y with respect to X in terms of T now how do we find the second derivative how do we find the second derivative of y with respect with respect to X now and I'll give you a hint we're going to use this same idea if you want to find the rate of change of something with respect to X you find the rate of change of that something with respect to T and divide it by the rate of change of X with respect to T so what this is going to be this is going to be we want to find the derivative of the first derivative with respect to T so let me write this down so we want to find we want to take the derivative with respect to T in the numerator of the first derivative which I will put in blue now of Dy DX all of that over all of that over the all of that over DX DX DT now I want you to if it didn't if it doesn't jump out at you why this might why this is the same thing that we did before I encourage you to pause the video and think about it think about what we did over here the first time when we find wanted to find the derivative of y with respect to to X we found the derivative of y with respect to T and then divided that by the derivative of x with respect to T here we want to find the derivative we want to find the second derivative of y with respect to X actually let me just write it down out here a little bit clearer what we really want to do is we want to find the derivative with respect let me write it this way when we wanted to find the derivative with respect to X of y that was equal to derivative of y with respect to T over the derivative of x with respect to T now we want to find the derivative with respect to X of the first derivative with respect to X and so everywhere we saw a y here replace that with the first derivative so this is going to be equal to in the numerator the derivative with respect to T of Dy DX notice this was derivative with respect to T of Y in fact let me write it that way just so you can see it so if I clear this out so if I clear that out we're going to get this is the derivative with respect to T of Y so hopefully you see before we had a y there now we have a dydx DX DT now this might seem really daunting and complicated except for the fact that these are actually fairly straight things to evaluate the of taking the derivative with respect to T of the first derivative well that's just taking the derivative with respect to T of this and this is pretty easy this is the derivative was just going to be/ 12 and the derivative with respect to T of e to the t is just e to the T and so that's going to be over the derivative of x with respect to T which we already saw is 6 e to the 2T 6 e to the 2 2 T and so we can write this we can write this as let's see2 / 6 is 1 over 12 and then e to the T minus 2T which is equal to we could write this as 11 12th e to the - t or we could write this as 1 over 12 e to the T and we're done
26	https://futuresetcamp.com/math/multiplying-and-dividing-negative-numbers	Cart 0 Overview Testimonials Why Future Set? Mission Statement Director's Bio Location Contact Math Resources Blog Cart 0 Math for Computer Science 6 \| Multiplying and dividing negative numbers Back to Table of Contents Previous Lesson Next Lesson Examples Practice Problems Introduction How are negative numbers multiplied and divided? Multiplying and dividing negative numbers can be summarized with the following rules: Multiplying a negative and a positive results in a negative Multiplying two negatives results in a positive Dividing a negative and a positive results in a negative Dividing two negatives results in a positive Examples Multiplying negative numbers Q: 3 × (-5) A: Multiplying 3 and -5 means adding 3 groups of -5. Since adding negatives results in a move to the left on the number line, 3 × (-5) is -15. Dividing negative numbers Q: -15 ÷ (-5) A: Dividing -15 and -5 means counting groups of -5 until we get to -15. Since adding negatives results in a move to the left on the number line, our quotient will be positive. Thus, -15 ÷ (-5) = 3. Practice \| Multiplying and Dividing Negative Numbers Click "Create Report" upon completion of any practice set to document your progress: Example Problem0 of 10 1Multiplying negative numbers 2Dividing negative numbers 3Multiplying and dividing negative numbers (no models) 4Mixed practice (all operations, no models) Previous Lesson Next Lesson
27	https://youglish.com/pronounce/reveal/english/us	Reveal \| 13192 pronunciations of Reveal in American English Toggle navigation Login Sign up Daily Lessons Submit Get your widget Donate! for English▼ • Arabic • Chinese • Dutch • English • French • German • Greek • Hebrew • Italian • Japanese • Korean • Polish • Portuguese • Romanian • Russian • Spanish • Swedish • Thai • Turkish • Ukrainian • Vietnamese • Sign Languages Say it! All US UK AUS CAN IE SCO NZ [x] All - [x] United States - [x] United Kingdom - [x] Australia - [x] Canada - [x] Ireland - [x] Scotland - [x] New Zealand Close How to pronounce reveal in American English (1 out of 13192): Speed: arrow_drop_down Normal arrow_drop_up vertical_align_top emoji_objects settings ▼ ▲ ↻ ↻ U × My favorite way to reveal I speak the language though ••• [Feedback] [Share] [Save] [Record] [YouTube] [G. TranslateB. TranslateDeepLReverso▼] Google Translate Bing Translate DeepL Reverso Definition: Click on any word below to get its definition: my favorite way to reveal i speak the language though Discover more YouGlish Get professional language services ESL textbooks Purchase English audio lessons Find English speaking partner Accent reduction software Find online English courses Bilingual dictionaries Online English classes Translation services Nearby words: Having trouble pronouncing 'reveal' ? Learn how to pronounce one of the nearby words below: review revelation revolution revenue revealed reverse reviews reviewed revolutionary reveals revenues reviewing revealing reversed revival revenge revisit revolutions revised reverend revision reverence revelations revolt revise revolver reversal rev reviewers Discover more Cultural immersion experiences Phrase pronunciation guide Speech therapy services English grammar workbooks Accent reduction training English vocabulary flashcards Pronunciation practice software High-quality headphones Travel guides to English-speaking countries English language dictionaries Phonetic: When you begin to speak English, it's essential to get used to the common sounds of the language, and the best way to do this is to check out the phonetics. Below is the UK transcription for 'reveal': Modern IPA: rɪvɪ́jl Traditional IPA: rɪˈviːl 2 syllables: "ri" + "VEEL" Test your pronunciation on words that have sound similarities with 'reveal': revealed reveals reveille ravel arrival revealing revel revell revile revolt rival arrivals drivel prevail ravi ravioli removal repeal reveled reveler reveller revels Discover more English language tutor Cultural immersion experiences Online English pronunciation courses English listening practice English grammar workbooks Language learning community Real-world speech examples Language learning games High-quality headphones Speech therapy services Tips to improve your English pronunciation: Here are a few tips that should help you perfect your pronunciation of 'reveal': Sound it Out: Break down the word 'reveal' into its individual sounds "ri" + "veel". Say these sounds out loud, exaggerating them at first. Practice until you can consistently produce them clearly. Self-Record & Review: Record yourself saying 'reveal' in sentences. Listen back to identify areas for improvement. YouTube Pronunciation Guides: Search YouTube for how to pronounce 'reveal' in English. Pick Your Accent: Mixing multiple accents can be confusing, so pick one accent (US or UK) and stick to it for smoother learning. Here are a few tips to level up your english pronunciation: Mimic the Experts: Immerse yourself in English by listening to audiobooks, podcasts, or movies with subtitles. Try shadowing—listen to a short sentence and repeat it immediately, mimicking the intonation and pronunciation. Become Your Own Pronunciation Coach: Record yourself speaking English and listen back. Identify areas for improvement, focusing on clarity, word stress, and intonation. Train Your Ear with Minimal Pairs: Practice minimal pairs (words that differ by only one sound, like ship vs. sheep) to improve your ability to distinguish between similar sounds. Explore Online Resources: Websites & apps offer targeted pronunciation exercises. Explore YouTube channels dedicated to pronunciation, like Rachel's English and English with James for additional pronunciation practice and learning. Discover more English grammar guide English vocabulary flashcards Language learning platform Virtual language tutors Daily pronunciation lessons English dictionary Voice recording devices Language learning community English language dictionaries Language learning apps YouGlish for: Arabic Chinese Dutch English French German Greek Hebrew Italian Japanese Korean Polish Portuguese Romanian Russian Spanish Swedish Thai Turkish Ukrainian Vietnamese Sign Languages Choose your language: English Français Español Italiano Português Deutsch العربية HOME ABOUT CONTACT PRIVACY & TERMS SETTINGS API BROWSE CONTRIBUTE ×Close ■Definitions■Synonyms■Usages■Translations Translate to : Close More
28	https://www.thoughtco.com/the-slope-of-the-aggregate-demand-curve-1146834	What Is the Slope of the Aggregate Demand Curve? Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Social Sciences› Economics› Supply & Demand› The Slope of the Aggregate Demand Curve Print UpperCut Images / UpperCut Images / Getty Images Social Sciences Economics Supply & Demand U.S. Economy Employment Psychology Sociology Archaeology Ergonomics By Jodi Beggs Jodi Beggs Economics Expert Ph.D., Business Economics, Harvard University M.A., Economics, Harvard University B.S., Massachusetts Institute of Technology Jodi Beggs, Ph.D., is an economist and data scientist. She teaches economics at Harvard and serves as a subject-matter expert for media outlets including Reuters, BBC, and Slate. Learn about ourEditorial Process Updated on February 03, 2019 Close Students learn in microeconomics that the demand curve for a good, which shows the relationship between the price of a good and the quantity of the good that consumers demand- i.e. are willing, ready, and able to purchase- has a negative slope. This negative slope reflects the observation that people demand more of almost all goods when they gets cheaper and vice versa.This is known as the law of demand. The Aggregate Demand Curve in Macroeconomics In contrast, the aggregate demand curve used in macroeconomics shows the relationship between the overall (i.e. average) price level in an economy, usually represented by the GDP Deflator, and the total amount of all goods demanded in an economy. Note that "goods" in this context technically refers to both goods and services. Specifically, the aggregate demand curve shows real GDP, which, in equilibrium, represents both total output and total income in an economy, on its horizontal axis. Technically, in the context of aggregate demand, the Y on the horizontal axis represents aggregate expenditure.As it turns out, the aggregate demand curve also slopes downwards, giving a similar negative relationship between price and quantity that exists with the demand curve for a single good. The reason that the aggregate demand curve has a negative slope, however, is quite different. In a lot of cases, people consume less of a particular good when its price increases because they have an incentive to substitute away to other goods that have become relatively less expensive as a result of the price increase. On an aggregate level, however, this is somewhat difficult to do- though not totally impossible, since consumers can substitute away to imported goods in some situations. Therefore, the aggregate demand curve must slope downwards for different reasons. In fact, there are three reasons why the aggregate demand curve exhibits this pattern: the wealth effect, the interest-rate effect, and the exchange-rate effect. The Wealth Effect When the overall price level in an economy decreases, consumers' purchasing power increases, since every dollar they have goes further than it used to. On a practical level, this increase in purchasing power is similar to an increase in wealth, so it shouldn't be surprising that an increase in purchasing power makes consumers want to consume more. Since consumption is a component of GDP (and therefore a component of aggregate demand), this increase in purchasing power caused by a reduction in the price level leads to an increase in aggregate demand. Conversely, an increase in the overall price level decreases the purchasing power of consumers, making them feel less wealthy, and therefore decreases the number of goods that consumers want to purchase, leading to a decrease in aggregate demand. The Interest-Rate Effect While it is true that lower prices encourage consumers to increase their consumption, it is often the case that case that this increase in the number of goods purchased still leaves consumers with more money left over than they had before. This leftover money is then saved and lent out to companies and households for investment purposes. The market for "loanable funds" responds to the forces of supply and demand just like any other market, and the "price" of loanable funds is the real interest rate. Therefore, the increase in consumer saving results in an increase in the supply of loanable funds, which decreases the real interest rate and increases the level of investment in the economy. Since investment is a category of GDP (and therefore a component of aggregate demand), a decrease in the price level leads to an increase in aggregate demand. Conversely, an increase in the overall price level tends to decrease the amount that consumers save, which lowers the supply of savings, raises the real interest rate, and lowers the quantity of investment. This decrease in investment leads to a decrease in aggregate demand. The Exchange-Rate Effect Since net exports (i.e. the difference between exports and imports in an economy) is a component of GDP (and therefore aggregate demand), it's important to think about the effect that a change in the overall price level has on the levels of imports and exports. In order to examine the effect of price changes on imports and exports, however, we need to understand the impact of an absolute change in the price level on relative prices between different countries. When the overall price level in an economy decreases, the interest rate in that economy tends to decline, as explained above. This decline in the interest rate makes saving via domestic assets look less attractive compared to saving via assets in other countries, so demand for foreign assets increases. In order to purchase these foreign assets, people need to exchange their dollars (if the U.S. is the home country, of course) for foreign currency. Like most other assets, the price of currency (i.e. the exchange rate) is determined by the forces of supply and demand, and an increase in demand for foreign currency increases the price of foreign currency. This makes domestic currency relatively cheaper (i.e. the domestic currency depreciates), meaning that the decrease in the price level not only reduces prices in an absolute sense but also reduces prices relative to the exchange-rate adjusted price levels of other countries. This decrease in the relative price level makes domestic goods cheaper than they were before for foreign consumers. The currency depreciation also makes imports more expensive for domestic consumers than they were before. Not surprisingly, then, a decrease in the domestic price level increases the number of exports and decreases the number of imports, resulting in an increase in net exports. Because net exports is a category of GDP (and therefore a component of aggregate demand), a decrease in the price level leads to an increase in aggregate demand. Conversely, an increase in the overall price level will increase interest rates, causing foreign investors to demand more domestic assets and, by extension, increase the demand for dollars. This increase in demand for dollars makes dollars more expensive (and foreign currency less expensive), which discourages exports and encourages imports. This decreases net exports and, as a result, decreases aggregate demand. Cite this Article Format mlaapachicago Your Citation Beggs, Jodi. "The Slope of the Aggregate Demand Curve." ThoughtCo, Apr. 5, 2023, thoughtco.com/the-slope-of-the-aggregate-demand-curve-1146834.Beggs, Jodi. (2023, April 5). The Slope of the Aggregate Demand Curve. Retrieved from Beggs, Jodi. "The Slope of the Aggregate Demand Curve." ThoughtCo. 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29	https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/x786f2022:polar-spherical-cylindrical-coordinates/a/triple-integrals-in-cylindrical-coordinates	Triple integrals in cylindrical coordinates (article) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Multivariable calculus Course: Multivariable calculus>Unit 4 Lesson 9: Polar, spherical, and cylindrical coordinates Double integrals in polar Triple integrals in cylindrical coordinates Triple integrals in spherical coordinates Integrals in spherical and cylindrical coordinates Math> Multivariable calculus> Integrating multivariable functions> Polar, spherical, and cylindrical coordinates © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Triple integrals in cylindrical coordinates Google Classroom Microsoft Teams How to perform a triple integral when your function and bounds are expressed in cylindrical coordinates. Background Triple integrals Double integrals in polar coordinates Need a quick refresher of cylindrical coordinates? Cylindrical coordinates are essentially the same as polar coordinates in two-dimensions, just with a z‍-coordinate thrown in to make it three-dimensional. Each point in space is described with three coordinates: r‍ θ‍ z‍ As with any coordinate system, these three numbers give instructions for how to get to a given point starting at the origin. Start by drawing a line of length r‍ away from the origin, along the x‍-axis. Rotate that line by θ‍ radians on the x y‍-plane, counterclockwise away from the x‍-axis, fixing one end at the origin. From the tip of the line, move a distance z‍ in the third dimension, perpendicular to the x y‍-plane. Converting between cylindrical coordinates and cartesian coordinates is the same as converting to polar coordinates in two dimensions. x 2+y 2=r 2‍ tan⁡(θ)=y x‍ The only new coordinate to three dimensions, z‍, remains unchanged as you convert back and forth. Note, many authors will use the coordinates (ρ,ϕ,z)‍ instead of (r,θ,z)‍, but I have opted to use (r,θ,z)‍ to emphasize the connection with polar coordinates. What we're building to The main thing to remember about triple integrals in cylindrical coordinates is that d V‍, representing a tiny bit of volume, is expanded as d V=r d θ d r d z‍ (Don't forget to include the r‍) Using cylindrical coordinates can greatly simplify a triple integral when the region you are integrating over has some kind of rotational symmetry about the z‍-axis. The one rule When performing double integrals in polar coordinates, the one key thing to remember is how to expand the tiny unit of area d A‍ in terms of d r‍ and d θ‍ ∬R f(r,θ)d A=∬R f(r,θ)r d θ d r‍ Note that the variable r‍ is part of this expansion. Expanding the tiny unit of volume d V‍ in a triple integral over cylindrical coordinates is basically the same, except that now we have a d z‍ term: ∭R f(r,θ,z)d V=∭R f(r,θ,z)r d θ d r d z‍ Remember, the reason this little r‍ shows up for polar coordinates is that a tiny "rectangle" cut by radial and circular lines has side lengths r d θ‍ and d r‍. 1‍2‍−1‍−2‍1‍2‍−1‍−2‍y‍x‍d r‍r d θ‍r‍ The key thing to remember here is that θ‍ is not a unit of length, so d θ‍ does not represent a tiny length in the same way that d r‍ and d z‍ do. It measures radians, which need to be multiplied by the distance r‍ from the origin to become a length. Thinking more deeply about this expansion Think of your little unit of volume d V‍ as a block with the following properties. One pair of faces represent constant values of θ‍. One pair of faces represent constant values of r‍ (these will be slightly curved, as if hugging a cylinder). One pair of faces represent constant values of z‍ (these are the flat top and bottom). When you are integrating over a three-dimensional region R‍, you can think of it as being composed of infinitely many infinitely small blocks like this. When they are small enough, the curved edge will basically be a straight line, so you can treat it as a rectangular prism. As the size of the block approaches zero, each edge will represent an infinitesimal change in one variable. One edge will have length d r‍, representing a small change in r‍ while θ‍ and z‍ stay constant. One edge will have length d z‍, representing a small change in z‍ while r‍ and θ‍ stay constant. Another edge will represent a tiny change in θ‍ while r‍ and z‍ remain constant. The length of this edge is not d θ‍, since θ‍ is not a unit of length. Its length is r d θ‍ where r‍ is the distance from the box to the origin. In total, the volume of this box is d V=(r d θ)(d r)(d z)‍. Example 1: Volume of a sphere Problem: Find the volume of a sphere with radius 1‍ using a triple integral in cylindrical coordinates. First of all, to make our lives easy, let's place the center of the sphere on the origin. Next, I'll give the sphere a name, S‍, and write the abstract triple integral to find its volume. ∭S d V=∭S r d θ d r d z‍ As always, the hard part is putting bounds on the integral. However, doing this with cylindrical coordinates is much easier than it would be for cartesian coordinates. In particular, r‍ and θ‍ will just live within the unit disc, which is very natural to describe in polar coordinates: 1‍−1‍1‍−1‍y‍x‍r‍θ‍ Concept check: Which of the following sets of bounds for r‍ and θ‍ should we use to integrate over the unit disc? Choose 1 answer: Choose 1 answer: (Choice A) 0≤r≤1‍ 0≤θ≤2 π‍ A 0≤r≤1‍ 0≤θ≤2 π‍ (Choice B) −1≤r≤1‍ 0≤θ≤2 π‍ B −1≤r≤1‍ 0≤θ≤2 π‍ (Choice C) 0≤r≤1‍ 0≤θ r≤2 π‍ C 0≤r≤1‍ 0≤θ r≤2 π‍ Check Answer The first choice is correct: 0≤r≤1‍ 0≤θ≤2 π‍ Within the unit disc, the radius varies between 0‍ and 1‍, and the angle θ‍ runs along its full range from 0‍ to 2 π‍. Since the bounds of z‍ will depend on the value of r‍, we let the innermost integral handle z‍, while the outer two integrals take care of r‍ and θ‍. Writing down what we have so far, we get ∫0 2 π∫0 1∫??r d z d r d θ‍ Remember, it's important to make sure the order of the differential terms d z‍, d r‍ and d θ‍ matches up with the appropriate integral. This next question is a little trickier. Concept check: For a given value of r‍, which of the following shows the right range of values for z‍? Choose 1 answer: Choose 1 answer: (Choice A) −1−r 2≤z≤1−r 2‍ A −1−r 2≤z≤1−r 2‍ (Choice B) −1+r 2≤z≤1+r 2‍ B −1+r 2≤z≤1+r 2‍ (Choice C) −r 2≤z≤r 2‍ C −r 2≤z≤r 2‍ (Choice D) −r≤z≤r‍ D −r≤z≤r‍ Check Answer The first choice is correct. −1−r 2≤z≤1−r 2‍ In cylindrical coordinates, the distance from a point (r,θ,z)‍ to the origin is r 2+z 2‍ To see why, relate this to cartesian coordinates using the fact that r 2=x 2+y 2‍ r 2+z 2=x 2+y 2+z 2‍ Therefore, the unit sphere could be defined as all points such that r 2+z 2=1 2‍. Solving for z‍, we get its upper and lower bounds if it is to remain inside the sphere. r 2+z 2=1 z 2=1−r 2 z=±1−r 2‍ Applying this bound to our innermost integral, we get something that can be worked out. Concept check: Solve this triple integral. ∫0 2 π∫0 1∫−1−r 2 1−r 2 r d z d r d θ=‍ Check Answer ∫0 2 π∫0 1∫−1−r 2 1−r 2 r d z⏟Integrate with respect to z d r d θ=∫0 2 π∫0 1[r z]z=−1−r 2 z=1−r 2 d r d θ=∫0 2 π∫0 1 r(1−r 2−(−1−r 2))d r d θ=∫0 2 π∫0 1 2 r 1−r 2 d r d θ‍ Using the inverse chain rule, we can quickly scribble down some notes to see what this antiderivative is: ∫2 r 1−r 2 d r=2 r(2 3(1−r 2)3/2)⏟Antiderivative of something 1−2 r⏞Divide by d d r(1−r 2)=−2 3(1−r 2)3/2‍ Continuing on with our integral then, =∫0 2 π∫0 1 2 r 1−r 2 d r d θ=∫0 2 π[−2 3(1−r 2)3/2]r=0 r=1 d θ=∫0 2 π(−2 3(1−(1)2)3/2−(−2 3(1−(0)2)3/2))d θ=∫0 2 π(−2 3(0)−−2 3(1))d θ=∫0 2 π 2 3 d θ=4 3 π‍ And with that, you just found the volume of a unit sphere! Moreover, this tool is powerful enough to do more than just find the volume of the sphere. For example, you could integrate a three-variable function f(r,θ,z)‍ inside the sphere, ∫0 2 π∫0 1∫−1−r 2 1−r 2 f(r,θ,z)r d z d r d θ‍ The hard part of finding the bounds is no different, but the computation of the integrals (done by either you or a computer) will change. Example 2: Integrating over a pie slice For this example, we will integrate over a region which looks kind of like a slanted pie slice: Khan Academy video wrapper See video transcript In a problem, this region might be described to you using the following list of properties: x≥0‍ y≥0‍ z≥0‍ y≤x‍ x 2+y 2≤4‍ z≤y x‍ This time, we will not just be finding the volume of this region. Instead, our task is to integrate the following three-variable function: f(x,y,z)=z−x 2−y 2‍ This might seem out of place in an article about integrating in cylindrical coordinates, since everything here is given in cartesian coordinates. Indeed, you could setup the triple integral using cartesian coordinates if you wanted. However, there's one key fact suggesting that our lives can be made dramatically easier by converting to cylindrical coordinates first: The expression x 2+y 2‍ shows up in the function f‍, as well as in the description of the bounds. This suggests some rotational symmetry around the z‍-axis, which cylindrical coordinates are well-suited for. For example, look at the range for our x‍ and y‍ values: x≥0‍ y≥0‍ y≤x‍ x 2+y 2≤4‍ 1‍2‍−1‍−2‍1‍2‍−1‍−2‍y‍x‍ Describing this with a pair of integrals over d x‍ and d y‍ is a real pain. However, in polar coordinates, this becomes very simple: 0≤θ≤π 4‍ 0≤r≤2‍ This means the bounds on the integrals handling d θ‍ and d r‍ will be constants. You can't do better than that! What about the other criteria, such as z≤y x‍ Since converting to polar coordinates involves the property tan⁡(θ)=y x‍ The bounds on z‍ can be translated to 0≤z≤tan⁡(θ)‍ Putting this together, our triple integral looks like this: ∫0 π/4∫0 2∫0 tan⁡(θ)f d V‍ Notice how simple the bounds are. If you are up for a little pain, you can try finding the appropriate triple integral bounds in cartesian coordinates to see just how much uglier they are. We now write the function f‍ using polar coordinates. f(x,y,z)=z−x 2−y 2⇓f(r,θ,z)=z−r 2‍ And of course, we incorporate the main takeaway of this article, which is how to write d V‍ in polar coordinates: d V=r d θ d r d z‍ Putting this all together, we get our triple integral in its final solvable state. More practice: Solve this integral ∫0 π/4∫0 2∫0 tan⁡(θ)(z−r 2)r d z d r d θ=‍ Check Answer Again, in practice, the computation of integrals like this one are taken care of by programs performing numerical integration, or even computer algebra systems. By hand, here's what it looks like: ∫0 π/4∫0 2∫0 tan⁡(θ)(z−r 2)r d z d r d θ=∫0 π/4∫0 2∫0 tan⁡(θ)(z r−r 3)d z d r d θ=∫0 π/4∫0 2[z 2 2 r−r 3 z]0 tan⁡(θ)d r d θ=∫0 π/4∫0 2(tan 2⁡(θ)2 r−r 3 tan⁡(θ))d r d θ=∫0 π/4[r 2 tan 2⁡(θ)4−r 4 4 tan⁡(θ)]r=0 r=2 d θ=∫0 π/4((2)2 tan 2⁡(θ)4−(2)4 4 tan⁡(θ))d θ=∫0 π/4(tan 2⁡(θ)−4 tan⁡(θ))d θ‍ Here, I had to look things up. The antiderivative of tan 2⁡θ‍ is tan⁡(θ)−θ‍ The antiderivative of tan⁡(θ)‍ is −ln⁡(\|cos⁡(θ)\|)‍ Continuing on our merry way then: =∫0 π/4(tan 2⁡(θ)−4 tan⁡(θ))d θ=[tan⁡(θ)−θ−4(−ln⁡(cos⁡(θ)))]0 π/4=[tan⁡(θ)−θ+4 ln⁡(cos⁡(θ))]0 π/4=(tan⁡(π/4)−π/4+4 ln⁡(cos⁡(π/4)))−(tan⁡(0)−0+4 ln⁡(cos⁡(0)))=(1−π 4+4 ln⁡(2 2))−(0−0+4 ln⁡(1))=1−π 4+4 ln⁡(2 2)‍ Summary The main thing to remember about triple integrals in cylindrical coordinates is that d V‍, representing a tiny bit of volume, is expanded as d V=r d θ d r d z‍ (Don't forget to include the r‍) Using cylindrical coordinates can greatly simplify a triple integral when the region you are integrating over has some kind of rotational symmetry about the z‍-axis. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Iron Programming 5 years ago Posted 5 years ago. Direct link to Iron Programming's post “Howdy, Could someone ple...” more Howdy, Could someone please elaborate on how/when to use the Jacobian matrix with a "change" of variables or bounds. To be honest I didn't find sufficient learning material on "changing bounds" or "transformation" of variables. In the previous skill "Practice: Double integrals in polar", we need to use the Jacobian determinant to "scale" our transformation. I saw the videos KA had on the Jacobian determinant, but I'm afraid I was quite there when it came to using this for a transformation of variables. Is there a good external (non KA) source you'd recommend for me? Ever since KA made the skills for Multivariable Calculus I've been reinvigorated to learn some more math & I've been making a lot of notes but I've been slightly lost on some of the new skills. Regardless, I've been loving the new skills, so thanks KA! Answer Button navigates to signup page •Comment Button navigates to signup page (10 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer lakern 5 years ago Posted 5 years ago. Direct link to lakern's post “Glad you like the new ski...” more Glad you like the new skills! Paul's Online Math Notes has some good material to learn about change of variables. Check out this link for example: Also Professor Leonard on YouTube has comprehensive material on multivariable calculus. Check out his video on change of variable: Apologies that KA has no videos yet on change of variables. Maybe the hints can guide you a bit, but if you're stuck that's understandable. I hope there'll be videos soon! Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more 26cperezserna 2 years ago Posted 2 years ago. Direct link to 26cperezserna's post “Can you Add A Video for t...” more Can you Add A Video for this Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jonathan 9 years ago Posted 9 years ago. Direct link to Jonathan's post “In the xy diagram for pie...” more In the xy diagram for pie slice, why doesn't the shaded area include points under the X-axis? For example, wouldn't the point (1,-1) satisfy the conditions? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer samweegman 8 years ago Posted 8 years ago. Direct link to samweegman's post “Why can't you leave it as...” more Why can't you leave it as tan^2(theta) and tan(theta) Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer 𝜏 Is Better Than 𝝅 5 years ago Posted 5 years ago. Direct link to 𝜏 Is Better Than 𝝅's post “In example 2 where the ar...” more In example 2 where the article says: " If you are up for a little pain, you can try finding the appropriate triple integral bounds in Cartesian coordinates to see just how much uglier they are", I've tried to do this, but I just can't seem to find the corresponding triple integral in Cartesian coordinates. For instance, if you do it with respect to dz, then dx, then dy, the bounds (I thought) should be, for the respective integrals: [0,y/x], [y,sqrt(4-y^2)], [0,sqrt(2)], but when I check this with a computer, the two triple integrals do not spit out the same value. Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Up next: article Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. 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30	https://web.ma.utexas.edu/users/m408m/Display12-4-5.shtml	Applications M408M Learning Module Pages Main page ### Chapter 10: Parametric Equations and Polar Coordinates### Chapter 12: Vectors and the Geometry of Space#### Learning module LM 12.1: 3-dimensional rectangular coordinates:#### Learning module LM 12.2: Vectors:#### Learning module LM 12.3: Dot products:#### Learning module LM 12.4: Cross products:Areas in the plane Definition Computing cross products Applications #### Learning module LM 12.5: Equations of Lines and Planes:#### Learning module LM 12.6: Surfaces: ### Chapter 13: Vector Functions ### Chapter 14: Partial Derivatives ### Chapter 15: Multiple IntegralsApplications ============ Cross Products p4 =============== Geometric Application 1: when vectors u u, v v are adjacent sides of the parallelogram shown to right, then the height of the parallelogram is ∥v∥sin θ‖v‖sin⁡θ, so its area=base×height=∥u∥∥v∥sin θ=∥u×v∥.area=base×height=‖u‖‖v‖sin⁡θ=‖u×v‖. Thus the length of u×v u×v is the area of the parallelogram whose sides are u u and v v. This should come as no surprise. If we have two vectors in the x x-y y plane, a=a 1 i+a 2 j a=a 1 i+a 2 j and b=b 1 i+b 2 j b=b 1 i+b 2 j, then a×b=(a 1 b 2−a 2 b 1)k a×b=(a 1 b 2−a 2 b 1)k, and we already know that a 1 b 2−a 2 b 1 a 1 b 2−a 2 b 1 is the area of a parallelogram in the plane. Geometric Application 2: the vectors a a, b b and c c shown respectively in blue, red and green to the right form adjacent edges of a parallelepiped. Now by 1., the base has area =∥a×b∥=‖a×b‖, while its height=comp a×b(c)=∥c⋅(a×b)∥∥a×b∥.height=comp a×b(c)=‖c⋅(a×b)‖‖a×b‖. Thus the parallelepiped has volume=(area of base)×height=∥c⋅(a×b)∥.volume=(area of base)×height=‖c⋅(a×b)‖. Remember that c⋅(a×b)c⋅(a×b) is the determinant of a 3×3 3×3 matrix whose rows are c c, a a and b b. 3×3 3×3 determinants give volume the same way that 2×2 2×2 determinants give area. Mechanics Application 1: when we push down on a bike pedal we exert a force, a vector, on the bike pedal. The objective is to turn the chain wheel. In the figure to the right only the component ∥sin θ F∥‖sin⁡θ F‖ of F F will have any effect. Torque as defined by the cross product τ=r×F,τ=r×F, is a measure of the turning force exerted as the cyclist pushes on the pedal. Note that the torque is perpendicular to the plane of rotation. In most examples, the direction of the torque is along the axis of rotation. Example:compute the volume of the parallelepiped with adjacent edges O P¯¯¯¯¯¯¯¯,O Q¯¯¯¯¯¯¯¯O P¯,O Q¯ and O R¯¯¯¯¯¯¯¯O R¯ determined by vertices P(1,−1,−2),Q(1,1,−1),R(1,3,4),P(1,−1,−2),Q(1,1,−1),R(1,3,4), where O O is the origin. Solution: the parallelepiped is determined by a=O P−→−=⟨1,−1,−2⟩,a=O P→=⟨1,−1,−2⟩, b=O P−→−=⟨1,1,−1⟩,b=O P→=⟨1,1,−1⟩, c=O P−→−=⟨1,3,4⟩.c=O P→=⟨1,3,4⟩.So its volume is given by scalar triple product a⋅(b×c)=∣∣∣∣1 1 1−1 1 3−2−1 4∣∣∣∣a⋅(b×c)=\|1−1−2 1 1−1 1 3 4\| =∣∣∣1 3−1 4∣∣∣−(−1)∣∣∣1 1−1 4∣∣∣−2∣∣∣1 1 1 3∣∣∣.=\|1−1 3 4\|−(−1)\|1−1 1 4\|−2\|1 1 1 3\|. Consequently, the parallelepiped has volume=8 units.volume=8 units. Two very useful properties of the (scalar) triple product follow immediately from its interpretation as the volume of a parallelepiped: u⋅(u×v)=0,v⋅(u×v)=0,u⋅(u×v)=0,v⋅(u×v)=0, because in both cases the parallelepiped collapses to a parallelogram, and so has zero volume.
31	https://www.frontiersin.org/journals/endocrinology/articles/10.3389/fendo.2025.1560981/full	Frontiers \| Infertility management in resistant ovary syndrome: a review Frontiers in Endocrinology About us About us Who we are Mission and values History Leadership Awards Impact and progress Frontiers' impact Our annual reports Publishing model How we publish Open access Peer review Research integrity Research Topics FAIR² Data Management Fee policy Services Societies National consortia Institutional partnerships Collaborators More from Frontiers Frontiers Forum Frontiers Planet Prize Press office Sustainability Career opportunities Contact us All journalsAll articlesSubmit your researchSearchLogin Frontiers in Endocrinology Sections Sections Adrenal Endocrinology Bone Research Cancer Endocrinology Cardiovascular Endocrinology Cellular Endocrinology Clinical Diabetes Developmental Endocrinology Diabetes: Molecular Mechanisms Endocrinology of Aging Experimental Endocrinology Gut Endocrinology Molecular and Structural Endocrinology Neuroendocrine Science Obesity Pediatric Endocrinology Pituitary Endocrinology Renal Endocrinology Reproduction Systems Endocrinology Thyroid Endocrinology Translational and Clinical Endocrinology ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? 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Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Submit your researchSearchLogin Your new experience awaits. Try the new design now and help us make it even better Switch to the new experience MINI REVIEW article Front. Endocrinol., 18 March 2025 Sec. Reproduction Volume 16 - 2025 \| Infertility management in resistant ovary syndrome: a review Hong Yi 1†Lin Chen2†Jinglei Zhang 3Yangxing Wen 3Xi Zheng 4Xiaoyan Chen5 1 Department of Reproductive Health, Shenzhen Baoan Women’s and Children’s Hospital, Shenzhen University, Shenzhen, China 2 Department of Obstetrics and Gynecology, Faculty of Medicine, The Chinese University of Hong Kong, Hong Kong, Hong Kong SAR, China 3 Reproductive Medicine Center, The First Affiliated Hospital, Sun Yat-sen University, Guangzhou, China 4 The First Clinical Medical College, Southern Medical University, Guangzhou, China 5 Department of Obstetrics and Gynecology, Maternal-Fetal Medicine Institute, Shenzhen Baoan Women’s and Children’s Hospital, Shenzhen University, Shenzhen, China Resistant ovary syndrome is a rare endocrinological disorder characterized by elevated serum gonadotropins and normal ovarian reserves. The leading causes of this condition include FSHR mutations, and autoimmune disorders. Due to follicle maturation defects and anovulation, these patients have menstrual disturbances and infertility. Endocrinological disorders can be treated by regular hormone replacement therapy. However, there is no consensus regarding the infertility issues and current treatment remains experimental with controlled ovarian stimulation (COS) and in vitro maturation. Herein, we conducted a review of the current literature, which concludes that: 1) patients with FSHR mutations had very poor COS outcomes; 2) follicle-stimulating hormone and luteinizing hormone levels poorly predict COS outcomes; 3) both recombinant and urinary gonadotropins may be effective in COS; 4) the dosage of exogenous gonadotropins is not the key to successful COS; 5) in vitro maturation is a feasible option for patients carrying FSHR mutations or unsuccessful COS cycles. Introduction Resistant ovary syndrome (ROS), originally described by Jones et al. in 1969 (1), is also known as ovarian insensitive syndrome and Savage syndrome. It is a rare reproductive disorder characterized by a seemingly conflicting picture of hypergonadotropic hypogonadism and normal ovarian reserves. Women with ROS generally have anovulation disorders because of ovarian hyposensitivity to endogenous gonadotropins (2), which further cause endocrinological disturbance, primary or secondary amenorrhea, and infertility. With normal ovarian reserves, ROS differs from premature ovarian insufficiency, which represents the loss of ovarian functions before the age of 40 due to the depletion of primordial follicles. However, ovarian follicle maturation is impaired due to various etiologies. Regular hormone replacement therapy can treat endocrinological and menstrual disorders, but parenthood planning remains a tricky problem. The core of fertility management is to resume follicle maturation and the current strategies include controlled ovarian stimulation (COS) (3–6), in vitro maturation (IVM) (7), and using donor oocytes (8). Yet, fertility management in ROS is still experimental due to its heterogenous etiology and low incidence. This review focuses on the pathogenesis and fertility management of ROS and aims to summarize the common rules of fertility management based on available evidence. The etiology of ROS The etiology of ROS remains largely unclear to date. In the initial report, Jones et al. (1) proposed two hypotheses: 1) there were biologically inactive follicle-stimulating hormone (FSH) molecules in serum; and 2) the follicle apparatus itself was defective and resistant to FSH stimulation. Current studies on the pathogenesis of ROS include follicle-stimulating hormone receptor (FSHR) mutations and autoimmune disorders. FSHR mutations An FSHR-inactivating mutation is the most studied mechanism in ROS. In 1995, Aittomaki et al. discovered the first FSHR mutation (c.556C>T, p.Ala189Val) in hypergonadotropic ovarian dysgenesis families (9). The mutation was located at the 7th exon of FSHR and caused a substitution of Ala by Val at the extracellular domain of the FSH receptor, which disrupted the cell surface targeting of the receptor and abolished cAMP production upon FSH stimulation despite the affinity with FSH remaining unaltered (9, 10). Thereafter, more than 200 mutations in FSHR have been discovered according to the CliniVar database, however, the clinical information was not documented in most cases. In total, 30 mutations were reported with detailed phenotypes. These mutations involved multiple domains of the FSH receptor, and most mutations have been confirmed to impair the functions of the receptor in in vitro assays. In Fshr knock-out mouse models, preantral follicles were present in the ovaries, indicating that preantral follicle development does depend not on FSH receptors. However, no follicle progressed to the antral stages, suggesting an FSH receptor-dependent mechanism in follicle maturation (11). Similarly, equivalent small follicles were observed in ovary sections between patients with and without FSHR mutations, however, no follicles beyond the antral stages were observed in patients with FSHR mutations. Autoimmune disorders In 1982, Chiauzzi et al. identified circulating immunoglobulins that inhibited FSH binding to its receptor in two patients with hypergonadotropic amenorrhea who were complicated with myasthenia gravis (12). This discovery was subsequently validated in a larger cohort consisting of 23 patients who were previously diagnosed with ROS, whereas the immunoglobulins were negative in the control group (13). Rogenhofer and colleagues reported a patient with ROS whose serum showed strong reactivity against human menopausal gonadotropins (hMG) but no reactivity to recombinant FSH (rFSH). Interestingly, the patient achieved pregnancy after controlled ovarian stimulation with a daily injection of hMG (225 IU/d) and rFSH (75IU/d) (5). Li et al. detected circulating autoimmune antibodies against FSH receptors in a patient with ROS who was successfully managed with exogenous gonadotropins and dexamethasone (4). In addition, Chitnis et al. isolated and purified an oligopeptide from human ovarian follicular fluid, which was demonstrated to inhibit the binding of FSH to ovarian granulosa cells in vitro and to induce atresia in developing follicles in rodent models (14). This oligopeptide may represent a potential contributing factor to gonadotropin resistance and ovarian resistance syndrome, highlighting its relevance in the pathophysiology of these conditions. Diagnosis of ROS ROS shares similar clinical manifestations with premature ovarian insufficiency (POI), including menstrual disturbance (oligomenorrhea or amenorrhea), hypergonadotropinemia, decreased circulating estradiol levels, and fertility issues. However, they differ in ovarian reserves as POI represents the absolute depletion of ovarian primordial follicles at early ages whereas ROS is characterized by normal ovarian reserves, which can be distinguished by serum anti-mullerian hormone (AMH) levels and antral follicle count under transvaginal ultrasound scans. In 1972, Van Campenhout et al. (2) proposed that the diagnosis of ROS must satisfy the following three criteria: 1) endogenous hypergonadotropinemia; 2) presence of normal ovarian follicles; and 3) hyposensitivity of the ovaries to excessive stimulation of exogenous human gonadotropins. However, there were many cases that satisfied the first two criteria but showed normal ovarian responsiveness to exogenous human gonadotropins (3–5). Currently, there is no consensus regarding the diagnosis criteria of ROS but all cases to date satisfied the first two criteria. Infertility management strategies in ROS Patients with ROS are primarily affected by anovulation disorder-associated menstrual disturbances, endocrinological disorders, and infertility. The first two symptoms are successfully managed with hormonal replacement therapy, whereas fertility management has been a tricky issue. Currently, there is no systematic strategy for ROS-related infertility due to heterogeneous etiology and low incidence. Potential strategies include controlled ovarian hyperstimulation (5), in vitro maturation (7), and using donor oocytes (8). However, these methods were attempted in separate cases with varying clinical outcomes. ROS patients who have undergone fertility management are summarized in Table 1, including 16 cases with FSHR mutations. COS and IVM were the leading strategies, and the common themes (or outcomes) of these reports are presented below. Table 1 Table 1. Summary of the application of controlled ovarian stimulation in resistant ovary syndrome. Patients with confirmed FSHR mutations had very poor COS outcomes To date, more than 30 inactivating mutations of FSHR with demonstrated phenotypes in women have been discovered (38, 39), most of which have been confirmed to undermine FSH receptor functions by in vitro assays, and there were 16 cases (cases 26–41) who received COS (Table 1) for infertility management. Excessive exogenous gonadotropins were administrated but dominant follicles were observed in only two patients (cases 30 and 38) (23, 30). Case 30 had a homozygous c.1540 T>C mutation of the FSHR and she developed ovarian hyperstimulation syndrome during ovarian stimulation. However, in vitro assays confirmed that the mutation enhanced the function of FSH receptors, indicating that it is an activating mutation. In addition, she sought assisted reproductive technology (ART) treatment due to tubal and male factors rather than ovulatory disorders (30). Similar findings were also observed in case 29, a patient with compound heterozygous c.662A>T, c.919G>A, and c.2039G>A mutations in FSHR who also developed ovarian hyperstimulation syndrome during ovarian stimulation (29), but no experimental evidence regarding the impact of these mutations on receptor functions was available. Interestingly, in case 38, a patient with a homozygous c.919A>G mutation in FSHR was unresponsive to exogenous FSH in a 7-day stimulation according to follicle size under transvaginal ultrasound scans and serum estradiol levels, but four mature eggs, along with four metaphase I (MI) oocytes and five germinal vesicles (GV), were retrieved after being triggered with hCG for IVM purposes (23). This result may be explained by residual FSHR function despite its mutations; however, no experimental evidence was available. Furthermore, there were many patients with FHSR mutations for whom ovarian stimulation was not performed because of their younger age or lack of pregnancy desire despite clinical and experimental evidence showing that mutations compromise FSHR functions (30, 33, 40–46). Among the cases where ovarian stimulation with exogeneous gonadotropins was effective, genetic tests regarding FSHR were negative (4, 18) or were not performed (5, 6, 25). It is worth noting the obviously divided ovarian responsiveness to endogenous and exogenous gonadotropins in some cases (cases 7, 8, 10, 16, 24, and 25; Table 1). These cases were normal or high ovarian responders when exogenous gonadotropin was administrated despite high endogenous gonadotropin levels (4–6, 19, 21). Therefore, abnormalities of the endogenous gonadotropins may be the cause, as proposed by Jones and Moraes in the initial report (1). In conclusion, for patients with homozygous and compound heterozygous FSHR inactivating mutations, a trial of controlled ovarian stimulation is of limited value. The prognostic value of FSH and luteinizing hormone levels for COS outcomes For those without an FHSR mutation or where genetic tests are not available, ovarian stimulation is still worth trying for pregnancy management. Many women have achieved pregnancies and live births with the use of appropriate ovarian stimulation. According to Huang et al., basal FSH levels were negatively associated with COS outcomes. The higher the FSH levels, the poorer the outcomes (47). However, the conclusion was limited by its sample size, which included only six patients, five of whom were carrying FSHR mutations. A downregulation protocol was the most popular protocol in patients with ROS, intended to improve ovarian responsiveness to FSH stimulation by inhibiting serum gonadotropin levels (48), and similar strategies included pretreatment with oral contraceptives (OC). However, it seems that FSH and luteinizing hormone (LH) levels (after pretreatment) were not correlated with COS outcomes in previous studies. In case 7 (Table 1), for example, FSH/LH levels before and after downregulation were as high as 70.3/7.6 IU/L, 62.5/1.7 IU/L, and 38.3/0.7 IU/L, respectively. However, 13 and 5 oocytes were obtained after ovarian stimulation with exogenous gonadotropins (19). Similar results were also observed in other cases (cases 17 and 24; Table 1) (6, 24). However, successful inhibition of serum gonadotropins does not necessarily lead to improved ovarian sensitivity to FSH stimulation and favorable oocyte yields, such as in cases 12, 13, and 16 (Table 1) (3, 4, 21). There were also successful COS cycles after successful FSH/LH inhibition and failed COS cycles after failed FSH/LH inhibition (Table 1). Unfortunately, the serum gonadotropin levels at the time of exogenous FSH stimulation were not documented in most cases, leading to a lack of data for further analysis. Given the complexities of the pathogenesis of ROS and insufficient etiology investigations (e.g. FSHR mutations), downregulation may be attempted when considering COS, whereas the failure to inhibit serum FSH and LH levels can then lead to ovarian stimulation. Recombinant and urinary gonadotropins can be equally effective in COS Among the cases successfully managed by COS, hMG (cases 7 and 17; Table 1), rFSH (cases 6 and 10; Table 1) and the combination of both (cases 8, 16, 24, and 25; Table 1) have been attempted, leading to a median (P25, P75) yield of 4.0 (2.0, 12.0) oocytes. In addition, hMG and rFSH were used in two individual cycles in case 7, and mature oocytes were obtained in both cycles (19). Interestingly, Rogenhofer et al. detected serum antibodies against hMG but not rFSH in a patient with ROS. Nevertheless, 11 oocytes were retrieved after stimulation with hMG (225 IU/d) and recombinant FSH (75 IU/d) (5). Given the complex mixture of hMG (49), the antibodies concerned may target other ingredients rather than FSH. Meanwhile, both recombinant hCG and urinary hCG had been proven to be effective if dominant follicles were observed after FSH stimulation. Therefore, both recombinant and urinary gonadotropins are effective in patients with resistant ovary syndrome. The dosage of exogenous gonadotropins is not the key to successful COS Excessive gonadotropins were prescribed in several cases (cases 3–5, 9, 12–14, 20, 23, 26-28, 31, 34, and 40; Table 1) (7, 16, 17, 21, 22, 25–28, 32, 36), but in vain. There were 19 successful COS cycles (defined by the retrieval of mature oocytes) and 33 failed COS cycles (Table 1), respectively. The median (P25, P75) stimulating duration was 13.00 (10.25, 15.50) vs. 15.00 (10.00, 17.00) (p = 0.529, by Mann Whitney test) days, respectively. Furthermore, the total gonadotropin (FSH) dosage per cycle was 3,375 (2,700, 4,800) vs. 4,500(3,000, 6,038) (p = 0.190, by Mann Whitney test) IU, respectively. The total dosages of FSH prescribed in the failed cycles were surprisingly higher than the successful cycles, though not statistically significant. In addition, several cycles were canceled due to a lack of follicle growth after a short period of ovarian stimulation, which led to the underestimation of the actual dosage of gonadotropins in this group. Therefore, simply increasing the dosage of gonadotropins does not necessarily improve COS outcomes, while investigations into etiologies may provide more clues for further management (e.g., genetic evaluations). For example, similar dosages of exogenous FSH [3,500 (1 700, 5 625)] were prescribed to those patients with FSHR mutations1 when compared with those successful cycles, but seldomly did it work. Given the complexities of ROS, simply increasing the dosage of gonadotropins in COS is not recommended. IVM is a feasible option for ROS In vitro maturation, as a method that supports immature GV-stage cumulus-oocyte complexes (COCs) from antral follicles to grow into the metaphase II (MII) stage, may serve as a final resort for fertility management in ROS (50). Several live births have been reported in patients with this condition who showed no response to conventional ovarian stimulation (7, 18, 21, 25, 32, 51). Currently, 12 patients have achieved live births via IVM, including 10 who were resistant to exogenous gonadotropin stimulation (Table 2). On average, 10 COCs were retrieved per IVM cycle, yielding an overall maturation rate of 41.51% per cycle and a live birth rate of 54.54% per patient (Table 2). In addition, there were serval patients who still had their embryos cryopreserved, and it is likely that the number of live births will increase (25). Eftekhar et al. also reported a series of cases managed with IVM, however, these patients seemed unlikely to have resistant ovary syndrome as their FSH levels were within the normal range (52). Overall, IVM is a feasible option for patients with FSHR mutations or unsuccessful COS cycles. Table 2 Table 2. In vitro maturation for patients with resistant ovary syndrome. Conclusions Patients with FSHR mutations had very poor COS outcomes, making it not worthy of a trial. For those without an FSHR mutation, COS is worth trying, with either recombinant or urinary gonadotropins at approximate dosages, whereas FSH and LH levels poorly predict COS outcomes. Furthermore, IVM is a feasible option for patients carrying FSHR mutations or with unsuccessful COS cycles. Author contributions HY: Conceptualization, Data curation, Formal analysis, Writing – original draft, Writing – review & editing. LC: Conceptualization, Data curation, Formal analysis, Writing – original draft, Writing – review & editing. JZ: Data curation, Formal analysis, Writing – original draft, Writing – review & editing. YW: Data curation, Formal analysis, Supervision, Writing– original draft, Writing – review & editing. XZ: Data curation, Writing – original draft, Writing – review & editing. XC: Conceptualization, Funding acquisition, Supervision, Writing – original draft, Writing – review & editing. Funding The author(s) declare that financial support was received for the research, authorship, and/or publication of this article. This study was funded by the Shenzhen Key Medical Discipline Construction Fund (Grant no. SZXK028), the Shenzhen Science and Technology Program (Grant no. JCYJ20210324141403009 and RCYX20210609104608036), and the Natural Science Funding of China (Grant no. 82201851). Conflict of interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Generative AI statement The author(s) declare that no Generative AI was used in the creation of this manuscript. Publisher’s note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. Footnotes ^ Case 30 was excluded from calculation due to the activating nature of the mutation. References 1. Jones GS, De-Moraes-Ruehsen M. A new syndrome of amenorrhae in association with hypergonadotropism and apparently normal ovarian follicular apparatus. Am J Obstet Gynecol. (1969) 104:597–600. doi:10.1016/S0002-9378(16)34255-7 PubMed Abstract \| Crossref Full Text \| Google Scholar 2. Van Campenhout J, Vauclair R, Maraghi K. Gonadotropin–resistant ovaries in primary amenorrhea. Obstet Gynecol. (1972) 40:6–12. Google Scholar 3. Mu Z, Song J, Yu Y, Sun Z. 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(2021) 19:569–74. doi:10.18502/ijrm.v19i6.9378 PubMed Abstract \| Crossref Full Text \| Google Scholar Keywords: resistant ovary syndrome, follicle-stimulating hormone receptor, gene mutation, controlled ovarian stimulation, in vitro maturation Citation: Yi H, Chen L, Zhang J, Wen Y, Zheng X and Chen X (2025) Infertility management in resistant ovary syndrome: a review. Front. Endocrinol. 16:1560981. doi: 10.3389/fendo.2025.1560981 Received: 15 January 2025; Accepted: 21 February 2025; Published: 18 March 2025. Edited by: Duan Xing, Southeast University, China Reviewed by: Jeremy Egbert, UCONN Health, United States Copyright © 2025 Yi, Chen, Zhang, Wen, Zheng and Chen. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Xiaoyan Chen, chenxiaoyan@cuhk.edu.hk †These authors have contributed equally to this work Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. 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32	https://web2.ph.utexas.edu/~vadim/Classes/2016s/BSL.pdf	MAGNETIC FIELDS OF ELECTRIC CURRENTS BIOT–SAVART–LAPLACE LAW AND ITS APPLICATIONS The magnetic ﬁeld of an inﬁnite straight wire carrying a steady current I is given by a simple formula B = µ0I 2π ˆ φ r (1) where µ0 is the fundamental constant of the MKSA system of units, µ0 = 4π · 10−7 T m/A, exactly, (2) r is the distance from the wire and ˆ φ is a unit vector in the circular direction around the wire in the plain ⊥to the wire. Speciﬁcally, if you are looking down the wire and the current ﬂows away from you, then circular direction ˆ φ of the magnetic ﬁeld is clockwise, while if the current ﬂows toward you, then the magnetic ﬁeld is counterclockwise. Here are the pictures of the magnetic ﬁeld lines for the two cases: The eﬀect of this magnetic ﬁeld on another long wire parallel to the ﬁrst wire is an attractive force if the currents in the two wires ﬂow in the same direction, and a repulsive force if the 1 currents ﬂow in the opposite directions, I1 I2 F F I1 I2 F F I1 I2 F F (3) Here is the graphical explanation of the force’s direction for the currents in the same direction: (4) The magnitude of the force between two wires per unit of wire’s length is F L = µ0 2π × I1I2 r (5) where r is the distance between the wires and µ0 is the vacuum permeability, a fundamental constant of the MKSA system of units set to µ0 = 4π · 10−7 T m/A, exactly. (6) In other words, the Ampere — the unit of electric current — is deﬁned such that two long parallel wires separated by 1 m distance and each carrying 1 A current are attracted or repelled with a force of 2 · 10−7 Newtons per meter of length. 2 In the Gaussian system of units, there is no µ0. Instead there are factors 1/c (where c is the speed of light in vacuum) all over the place, For example, the Lorentz Force on a particle in Gaussian units become F = q E + v c × B , (7) the magnetic force on a current-carrying wire F = 1 c I⃗ ℓ× B, (8) the magnetic ﬁeld of an inﬁnite straight wire B = 2I c ˆ φ r , (9) and the force between 2 parallel wires F L = 2 c2 × I1I2 s . (10) ⋆ ⋆ ⋆ The wires of geometries other than an inﬁnite straight line create magnetic ﬁelds much more complicated that (1). For a steady current in a wire of most general geometry, there is an integral formula known as the Biot–Savart–Laplace equation or Biot–Savart–Laplace Law: B(r) = µ0 4π Z wire I dr′ × r −r′ \|r −r′\|3 (11) in MKSA units, or B(r) = 1 c Z wire I dr′ × r −r′ \|r −r′\|3 (12) in Gaussian units. In these formulae, the r′ spans the wire and the dr′ = d⃗ ℓis the inﬁnitesimal 3 length vector along the wire in the direction of the current. The expression r −r′ \|r −r′\|3 = unit vector from r′ to r distance between r′ and r 2 (13) should be familiar to you from the Coulomb Law for the electric ﬁeld of a charge distribution, for example the electric ﬁeld of a charged wire E(r) = 1 4πǫ0 Z wire r −r′ \|r −r′\|3 λ dℓ. (14) But there is a crucial diﬀerence between the Biot–Savart–Laplace equation (11) and the Coulomb equation (14) — the cross product of Id⃗ ℓwith (13) in the BSL equation (11). In these notes, I shall explore the consequences of this vector product for several examples of wire geometries. Example#1: Infinite Long Wire For my ﬁrst example, let me reproduce eq. (1) for the magnetic ﬁeld of an inﬁnite straight wire from the Biot–Savart–Laplace Law. Let me use a coordinate system where the wire runs along the z axis with the current ﬂowing in the +ˆ z direction. Consequently, in the BSL equation B(r) = µ0 4π Z wire I dr′ × r −r′ \|r −r′\|3 (11) we have r′ = (0, 0, z′) for a variable z′ but ﬁxed x′ = y′ = 0, and I dr′ = +I dz′ ˆ z; on the other hand, the coordinates (x, y, z) of the point r where we measure the magnetic ﬁeld are completely general. Therefore, r −r′ = x ˆ x + y ˆ y + (z −z′)ˆ z, (15) I dz′ ˆ z × (r −r′) = I dz′ (y ˆ x −x ˆ y), (16) \|r −r′\|2 = x2 + y2 + (z −z′)2, (17) \|r −r′\|3 = x2 + y2 + (z −z′)23/2, (18) 4 and plugging all these formulae into eq. (11) gives us B(x, y, z) = Iµ0 4π (y ˆ x −x ˆ y) × +∞ Z −∞ dz′ x2 + y2 + (z −z′)23/2 . (19) To evaluate the integral here, let’s change the integration variable from z′ to α = arctan s z −z′ = ⇒ z′ = z − s tan α = z −s×ctan α where s = p x2 + y2. (20) Consequently, dz′ = + s dα sin2 α , (21) x2 + y2 + (z −z′)2 = s2 + s2 tan2 α = s2 sin2 α , (22) dz′ x2 + y2 + (z −z′)23/2 = s dα sin2 α × sin3 α s3 = sin α dα s2 = d(−cos α) x2 + y2 , (23) while the angle α runs from 0 for z′ →−∞to π for z′ →+∞. Hence, +∞ Z −∞ dz′ x2 + y2 + (z −z′)23/2 = π Z 0 d(−cos α) x2 + y2 = −cos(π) + cos(0) x2 + y2 = 2 x2 + y2 (24) and therefore B(x, y, z) = µ0I 2π y ˆ x −x ˆ y x2 + y2 ⟨ ⟨in Cartesian coordinates ⟩ ⟩ = µ0I 2π ˆ φ s ⟨ ⟨in cylindrical coordinates ⟩ ⟩, (25) in perfect agreement with eq. (1) for the inﬁnite straight wire. 5 Example#2: Circular Ring For the next example, consider a wire shaped into a circular ring of radius R. For simplicity, let me limit the calculation of the magnetic ﬁeld to the axis of the ring, otherwise we would have to deal with elliptic integrals. Let’s use the coordinate system where the ring lies in the xy plane while its symmetry axis is the z axis, thus x y z I r′ r (26) Along the circular wire, r′ = R cos φ ˆ x + R sin φ ˆ y, (27) dr′ = R(−sin φ ˆ x + cos φ ˆ y) dφ, (28) while the points r where we measure the magnetic ﬁeld are restricted to r = z ˆ z, hence r −r′ = −R cos φ ˆ x −R sin φ ˆ y + z ˆ z, (29) I dr′ × (r −r′) = IR dφ z cos φ ˆ x + z sin φ ˆ y + R(sin2 φ + cos2 φ = 1)ˆ z , (30) \|r −r′\|2 = R2 + z2, (31) \|r −r′\|3 = R2 + z23/2. (32) 6 Plugging all these formulae into the Biot–Savart–Laplace equation, we obtain B(0, 0, z) = µ0 4π Z wire I dr′ × r −r′ \|r −r′\|3 = µ0 4π IR R2 + z23/2 2π Z 0 dφ z cos φ ˆ x + z sin φ ˆ y + Rˆ z , (33) where the integral evaluates to 2π Z 0 dφ z cos φ ˆ x + z sin φ ˆ y + Rˆ z = z ˆ x × 2π Z 0 dφ cos φ + z ˆ y × 2π Z 0 dφ sin φ + Rˆ z × 2π Z 0 dφ = z ˆ x × 0 + z ˆ y × 0 + Rˆ z × 2π = 2πRˆ z. (34) Altogether, the magnetic ﬁled along the ring’s axis is B(0, 0, z) = µ0I 2 R2 (R2 + z2)3/2 ˆ z . (35) In particular, at the center of the ring, the magnetic ﬁeld is B(center) = µ0I 2R ˆ z. (36) Note: on the diagram (26), the current in the wire ﬂows counterclockwise; consequently, the magnetic ﬁeld (35) points up, in the +ˆ z direction. For a clockwise current, we would have an opposite sign of I dr′ and hence opposite direction of the magnetic ﬁeld — −ˆ z, i.e., down. This is an example of the right screw rule for the current loops: turn a right screw (almost all the screws are right) in the direction of the current in the loop, and the screw will move in the direction of the B ﬁeld. Equivalently, you may use the right hand rule: curl the ﬁngers of your right hand around the loop in the direction of the current, and your thumb will point the direction of the B ﬁeld. 7 Segments: In many cases, a wire is made of several segments. Each segment has a simple geometric shape — a piece of a straight line, or a circular arc — but the overall geometry can be quite elaborate. For example, consider a star made of 5 straight-line segments, (37) For a wire like this, the Biot–Savart–Laplace integral over the whole wire becomes a sum of integrals over the individual segments, B(r) = µ0 × I 4π × segments X i Z segment#i dr′ × (r −r′) \|r −r′\|3 . (38) Let’s work out the integrals here for the straight-line and the circular-arc segments, and than we shall see a few interesting combinations. 8 Example#3: Straight-Line Segment: Consider a wire segment which follows a straight line from point r′ 1 to point r′ 1. Let’s picture the triangle made by the two ends of this segment and by the point r where we measure the magnetic ﬁeld: ⃗ r′ 1 ⃗ r′ 2 ⃗ r ⃗ h α1 α2 O (39) Since the wire segment is straight, the inﬁnitesimal vector d⃗ ℓ= dr′ along the segment has a ﬁxed direction, same as r′ 2 −r′ 1. Consequently, the vector product in the numerator of the BSL integral remains constant along the whole segment, dr′ × (r −r′) = dr′ × (r −r′ O) −dr′ × (r′ −r′ O) ⟨ ⟨where the second term vanishes since dr′ ∥(r′ −r′ O) ∥(r′ 2 −r′ 1) ⟩ ⟩ ≡dr′ × (r −r′ O) = d⃗ ℓ×⃗ h, (40) where d⃗ ℓ= dr′ is the inﬁnitesimal length element along the straight segment, and ⃗ h is the height of the triangle (39). In other words, ⃗ h is the line to the point r where we measure the magnetic ﬁeld from the wire segment — or from the extrapolated straight line of the wire segment — in the direction ⊥to the segment. Note: if we measure the magnetic ﬁeld at a point r which happens to lie right on the extrapolated straight line of the wire segment, then ⃗ h = 0 and hence d⃗ ℓ× ⃗ h ≡0. Consequently, the whole BSL integral vanishes regardless of the denominator’s details, and the magnetic ﬁeld of the segment is zero. Thus, straight segments ‘pointing’ directly towards or directly away from r do not contribute to the magnetic ﬁeld at r. 9 For ⃗ h ̸= 0, the direction of the magnetic ﬁeld is the direction of the vector product d⃗ ℓ×⃗ h in the numerator of the BSL integral. This direction is ⊥to the wire and to the ⃗ h; in other words, the direction of B(r) is ⊥to the whole triangle (39). The speciﬁc perpendicular obtains from the right screw rule: If from your point of view, the current ﬂows in the clockwise direction around r — as it does on ﬁgure (39)— then take the perpendicular which points away from you. OOH, if you see the current ﬂows counterclockwise around r, then take the perpendicular which points towards you. Now that we know the direction of the magnetic ﬁeld, let’s ﬁnd its magnitude B = µ0I 4π × ℓ2 Z ℓ1 dℓ× h \|r′ −r\|3 (41) In this formula, ℓis the coordinate along the wire, and I take it’s origin ℓ= 0 to be the point O where the height ⃗ h of the triangle touches the wire or the extrapolated line of the wire. In terms of this ℓ, \|r′ −r\|2 = ℓ2 + h2 = ⇒ \|r′ −r\|3 = ℓ2 + h23/2 , (42) so the BSL integral (41) becomes B = µ0I 4π × ℓ2 Z ℓ1 h × dℓ ℓ2 + h23/2 . (43) To evaluate this integral, we proceed similarly to eqs. (20) through (24): we change the integration variable ℓto the angle α = arc ctan −ℓ h = ⇒ ℓ= −h × ctan α, (44) hence dℓ ℓ2 + h23/2 = d(−cos α) h2 (45) 10 and therefore ℓ2 Z ℓ1 h × dℓ ℓ2 + h23/2 = α2 Z α1 d(−cos α) h = cos α1 −cos α2 h (46) where the angles α1 and α2 are exactly as shown on the diagram (39). Altogether, the magnetic ﬁeld of a straight wire segment is B(r) = µ0 × I 4πh × (cos α1 −cos α2) × n (47) where h and the angles α1 and α2 are as shown on ﬁgure (39) and n is the unit vector ⊥to the whole triangle in the direction given by the right-screw rule. Note: in the limit of inﬁnitely long segment in both directions, α1 →0, α2 →π, hence cos α1 −cos α2 →2, and the magnetic ﬁeld of the segment agrees with the formula for an inﬁnite wire, B∞= µ0 × I 2πh . (48) Example#4: A Square Loop Consider a closed loop of wire in the shape of an a × a square: 45◦ 135◦ (49) Let’s calculate the magnetic ﬁeld at the center of the square (shown in blue). The square wire consists of 4 similar straight-line segments, so all we need is to evaluate eq. (47) for the magnetic ﬁeld due to each segment, and then total up the 4 segments’ contributions. For each segment, h = 1 2a, α1 = 45◦, α2 = 135◦, hence B1 segment = µ0 × I 4π(a/2) × cos 45◦−cos 135◦= √ 2 = √ 2 2π × µ0I a . Also, for each segment the triangle spanning the wire and the center of the square where we measure B lies in the plane of the square, so the direction of the magnetic ﬁeld due to 11 each segment is ⊥to the whole square. Speciﬁcally, the magnetic ﬁeld points into the page since in each segment the current ﬂows clockwise around the center. Thus, altogether, the magnetic ﬁeld points into the screen and its magnitude is Bwhole square = 4 × B1 segment = 4 √ 2 2π × µ0I a = 8 √ 2 π × µ0I perimeter = 4a . (50) Example#5: Symmetric N-sided Polygon In this example the wire also makes a complete loop, this time in the shape of symmetric N–sided polygon with side a, for example (51) Again, we focus on the magnetic ﬁeld at the center of the polygon, so by symmetry each segment of the wire contributes a similar B1 segment. All these contributions are directed ⊥ to the polygon, speciﬁcally into the screen, hence Bpolygon = N × B1 segment × n (52) where n is the unit vector pointing into the page. Now let’s draw a single segment of the wire and the triangle connecting it to the center point where the magnetic ﬁeld is measured: 2π N ⃗ h α1 α2 12 Simple geometry+trigonometry for this triangle gives us α1,2 = π 2 ∓π N , (53) cos α1 −cos α2 = 2 sin π N , (54) h = a 2 × ctan π N , (55) and therefore B1 segment = µ0I 4π × cos α1 −cos α2 h = µ0I 4π × 2 tan π N a × 2 sin π N . (56) Finally, combining all N segments, we ﬁnd the magnetic ﬁeld at the center of the polygon is B = N × B1 segment = N × µ0 × I πa × sin π N × tan π N = µ0 × I perimeter = Na × N2 π × sin π N × tan π N , (57) where in the last expression P = N × a is the polygon’s perimeter. To check this formula, we ﬁrst plug compare it for N = 4 with eq. (50) for the square loop and see that they indeed produce the same magnetic ﬁeld at the center. Second, let’s take a large N limit in which the polygon becomes a circular ring of perimeter P = 2πR. In this limit, lim N→∞ N2 π × sin π N × tan π N = π, (58) hence the magnetic ﬁeld at the center of the polygon becomes B = µ0 × I 2πR × π = µ0 × I 2 , (59) which indeed agrees with the magnetic ﬁeld (36) at the center of a circular ring. 13 Example#6: A Circular Arc As our ﬁnal example, let’s calculate the magnetic ﬁeld at the center of a circular arc. More generally, consider a wire comprised of a semicircle and two straight segments C ϕ (60) and calculate the magnetic ﬁeld at point C at the center of the circular arc. Note: besides being at the center of the arc, the point C happens to lie on the straight-line extrapolations of the straight segments of the wire. Consequently, the two straight segments do not contribute to the magnetic ﬁeld B(C) at that point. Thus, the entire ﬁeld at point C comes from the circular arc segment only. Let’s parametrize the arc segment by the angle φ from the point C; φ ranges from φ0 to φ0 + ϕ. In terms of φ, r′ = R cos φ ˆ x + R sin φ ˆ y, (61) r −r′ = −R cos φ ˆ x −R sin φ ˆ y, (62) dr′ = R(−sin φ ˆ x + cos φ ˆ y) dφ, (63) hence in the numerator of the Biot–Savart–Laplace integral dr′ × (r −r′) = R2 dφ 0ˆ x + 0ˆ y + (sin2 φ + cos2 φ)ˆ z = R2 dφˆ z Note: the direction of this vector product is always vertically Up, ⊥to the plane of the ring, so the magnetic ﬁeld’s direction is going to be vertically Up. 14 As to the denominator of the BSL formula, the whole circular arc is at constant distance \|r −r′\| ≡R from the ring’s center, so the denominator is a constant R3. Altogether, Z arc dr′ × (r −r′) \|r −r′\|3 = φ0+ϕ Z φ0 R2 ˆ z dφ R3 = ϕ R ˆ z , (64) so the magnetic ﬁeld at point C is B(C) = µ0I 2iR × ϕ 2π × ˆ z . (65) Note: the ϕ angle in this formula should be taken in radians. Magnetic Fields of Thick Conductors The Biot–Savart–Laplace formula B(r) = µ0 4π Z wire I dr′ × r −r′ \|r −r′\|3 (11) gives the magnetic ﬁeld of a steady current ﬂowing through a thin wire which may be approximated by an inﬁnitely thin line, straight or curved. For such a wire, it does not matter how the current is distributed across the wire’s cross-section, only the net current I enter the formula. But sometimes we have currents ﬂowing through the volume of a conductor which is too thick to be approximated as a line. For such conductors, Z wire I dr′ − − − − → becomes ZZZ conductor′s volume d3Vol ′ J(r′) (66) where J(r′) is the current density at the point r′. Consequently, the Biot–Savart–Laplace equation for such current densities generalizes to B(r) = µ0 4π ZZZ d3Vol J(r′) × r −r′ \|r −r′\|3 . (67) Likewise, for a steady current ﬂowing along a conducting surface with density K(r′), the 15 BSL equation becomes B(r) = µ0 4π ZZ d2A K(r′) × r −r′ \|r −r′\|3 . (68) Example#7: Current Flowing Along an Infinite Plane As our ﬁnal example, consider a uniform current K ﬂowing along an inﬁnite plane. Let’s choose our coordinates so that this plane is the xy plane and the current ﬂows in the +ˆ y direction. Then the magnetic ﬁeld at some generic point r = (x, y, z) is given by eq. (68), speciﬁcally, B(x, y, z) = µ0K 4π ZZ plane dx′ dy′ ˆ y × (r −r′) \|r −r′\|3 . (69) In the numerator inside the integral here, r −r′ = (x −x′)ˆ x + (y −y′)ˆ y + z ˆ z, (70) ˆ y × (r −r′) = −(x′ −x)ˆ z + z ˆ x, (71) while in the denominator \|r −r′\|3 = (x −x′)2 + (y −y′)2 + z23/2. (72) To simplify these expressions, let’s change the integration variables x′ and y′ to the polar coordinates (s, φ) centered at (x, y), thus x′ = x + s × cos φ, (73) y′ = y + s × sin φ, (74) dx′ dy′ = s ds dφ, (75) ˆ y × (r −r′) = −s cos φˆ z + z ˆ x, (76) \|r −r′\|3 = s2 + z23/2. (77) 16 Plugging all these formulae into eq. (69), we arrive at B(s, y, z) = µ0K 4π ∞ Z 0 ds s 2π Z 0 dφ −s cos φˆ z + z ˆ x (s2 + z2)3/2 . (78) Integrating over the polar angle φ, we immediately obtain 2π Z 0 dφ −s cos φˆ z + z ˆ x = 0 × ˆ z + 2πz × ˆ x, (79) hence B(x, y, z) = µ0K 2 × ˆ x × ∞ Z 0 zs ds (s2 + z2)3/2 , (80) hence the magnetic ﬁeld everywhere points in the ±ˆ x direction, depending on the sign of the remaining integral in this formula. To evaluate this integral, we change variables from s to t = s2 + z2, thus zs ds = z 2 dt, (z2 + s2)3/2 = t3/2, (81) hence ∞ Z 0 zs ds (s2 + z2)3/2 = z 2 ∞ Z z2 dt t3/2 = z 2 × −2 √ t +∞ z2 = z 2 × 0 −−2 √ z2 = z √ z2 = sign(z). (82) Altogether, we have B(x, y, z) = µ0K 2 × sign(z) ˆ x (83) Note: the magnetic of the current sheet is completely uniform above the sheet (z > 0) and likewise completely uniform below the sheet (z < 0), but jumps discontinuously across the sheet. Its magnitude is the same B = µ0K/2 above and below the sheet, but the directions are opposite: above the sheet, the magnetic ﬁeld points in the +ˆ x direction while above the sheet it points in the −ˆ x direction. Relative to the currents direction +ˆ y, the magnetic ﬁeld above the sheet points 90◦to the right of the current, while below the sheet it points 90◦to the left of the current. 17
33	https://www.pbslearningmedia.org/resource/mgbh.math.oa.pemdas/order-of-operations-pemdas/	Order of Operations: Flocabulary’s PEMDAS Lesson \| PBS LearningMedia FOR TEACHERS Is WMHT your local station?Yes No, change Sign in Sign up FOR TEACHERS Brought to you by New York PBS Stations Subjects Grades Student site Math at the Core: Middle School Share to Google ClassroomShare link with studentsBuild a lessonSocial share Favorite Order of Operations: Flocabulary’s PEMDAS Lesson Video Grades: 3-5, 6-8 Collection: Math at the Core: Middle School PreK-12 Resources for New School Routines To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video Video Player is loading. Play Video Play Seek back 10 seconds Mute Current Time 0:00 / Duration 3:46 Loaded: 4.38% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-3:46 1x Playback Rate 2x 1.75x 1.5x 1.25x 1x, selected 0.75x 0.5x 0.25x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off English Captions, selected CC1 Captions Audio Track default, selected Settings Auto 64 kbps 480p 360p, selected 234p 180p Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. About Standards Watch and listen to a recitation of the order of operations set to hip-hop music. This video focuses on PEMDAS as the acronym for order of operations and demonstrates the concept by walking through a problem in the correct order. This video was submitted through the Innovation Math Challenge, a contest open to professional and nonprofessional producers. Permitted use Stream Only Accessibility Caption Credits New York State Education Department Want to see state standards for this resource? Sign In Nationwide Common Core State Standards (4) Grades 3-5 CCSS.Math.Content.5.OA.A.1 See anchor statement Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols.All “CCSS.Math.Content.5.OA.A.1” resources See all (2) Grades 3-5 standards Grades 6-8 CCSS.Math.Content.6.EE.A.2.b See anchor statement Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity.All “CCSS.Math.Content.6.EE.A.2.b” resources See all (2) Grades 6-8 standards College and Career Readiness Standards for Adult Education (4) Grades 3-5 EE.A.2.c See anchor statement Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations).All “EE.A.2.c” resources See all (4) Grades 3-5 standards Grades 6-8 EE.A.2.c See anchor statement Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations).All “EE.A.2.c” resources See all (4) Grades 6-8 standards Next: Standards Support Materials for Teachers USING THIS RESOURCE Teaching Tips \| Order of Operations: PEMDAS You May Also Like ##### Make It Math Interactive 2:50 ##### Order of Operations Video 2:51 ##### Area of a Triangle: Musically Interpreted Video 2:45 ##### Solving Linear Equations with Negative Numbers Video Explore related topics MathematicsK-8 MathematicsExpressions & EquationsOperations & Algebraic ThinkingApplying Arithmetic to Algebraic ExpressionsWriting and Interpreting Numerical Expressions More from the Math at the Core: Middle School Collection Collection 344 2:09 ##### Volume of Shapes Composed of Right Rectangular Prisms Video 2:27 ##### Volume of Right Rectangular Prisms Video 3:30 ##### Modeling Fraction and Mixed Number Division Using Arrays Video ##### Locating Points on the Cartesian Graph Interactive ##### Horizontal and Vertical Distances on the Cartesian Graph Interactive See the Collection Made Possible Through producer, contributor funder Unlock the Power of PBS LearningMedia Create a free account to gain full access to the website Save & Organize Resources See State Standards Manage Classes & Assignments Sync with Google Classroom Create Lessons Customized Dashboard Get More Features Free Student site In partnership with Connect With Us Sign up for our weekly PreK-12 newsletter for the latest classroom resources, news, and more. Sign up! Learn More About Teachers’ Lounge Blog Educator Recognition Contact Us Newsletters Help © 2025 PBS & WGBH Educational Foundation. All rights reserved. Privacy Policy\|Terms of Use Top Print Teaching Tips \| Order of Operations: PEMDAS Learning OutcomesStudents will be able to evaluate arithmetic expressions with parentheses, brackets, and braces perform arithmetic operations in the conventional order without parentheses perform arithmetic operations involving whole-number exponents in the conventional order without parentheses (sixth grade only) define the mathematical terms below Common Core State Standards: 5.OA.A.1, 5.OA.A.2, 6.EE.A.2.b, 6.EE.A.2.c Vocabulary: Mnemonics, acronyms, order of operations, exponents, square root Materials: Per student: pencil and paper Procedure1. Introduction (5 minutes, whole group and pairs)Record on the board: 4 • 3 + 6 – 5 • 2 Ask students to take a moment to evaluate the expression with a partner. Then, gather solution strategies from the group. If no one notes that multiplication should be performed first, remind students yourself. Next, ask students to find a way to add parentheses to the expression so that it evaluates to 14. Circulate quickly to check that students are comfortable using parentheses and to review as needed. Challenge students who finish early to add parentheses so that the expression evaluates to as large a number as possible. (Possible solutions: 4 • (3 + 6 – 5) • 2 = 32; with parentheses and brackets: [4 • (3 + 6) –5] • 2 = 62) 2. Video (10 minutes, whole group and pairs)Ask students to share any mnemonics or acronyms that they use for remembering order of operations. Record their suggestions on the board. If no one has ideas, record and explain PEMDAS. Give students a few minutes to solve the following problem in pairs: 4 2 (17 – 15)/(3 + 1) – 5 Note: If students are less comfortable working with whole-number exponents, replace 4 2 with 16. When all or most have arrived at a solution, show students the video. If needed, pause at 1:20 to clarify what a square root is. If some students have had difficulty evaluating expressions in the lesson so far, you might also pause after each solution step beginning at 2:20 to confirm that students are able to follow along. Talk through the steps with the group. 3. Fewest 5s (10 minutes, pairs)Have students work in pairs to find arithmetical expressions (including exponents if they are comfortable using them) that evaluate to the numbers 1–10. Expressions may only use 5s—no other numbers. Students try to use the fewest 5s possible. They may use parentheses and, if needed, brackets and braces in their expressions. Their expressions must follow the order of operations. Go through an example with the group, for instance, (5 + 5)/(5 + 5) = 1. Ask students, Can anyone find an expression that evaluates to 1 using fewer 5s? Extension: As an optional variation for a greater challenge, present the following to students: 5 + 5 • 5 – 5 ÷ 5 Ask students to use parenthesis to modify this expression in as many ways as possible. They may use more than one pair of parentheses or brackets. They should evaluate each expression that they create and check that they only place parentheses where needed. For instance, two of the many ways to modify the expression are: (5 + 5) • (5 – 5) ÷ 5 = 0, and (5 + 5) • 5 – (5 ÷ 5) = 49 4. Conclusion (10 minutes, small groups and whole group)Have pairs join together in small groups to share their solutions and check that the order of operations is correct. Those who solved the “fewest 5s” problem should compare the number of 5s used in expressions for each number to see which pair used the fewest. Those who solved the challenge should check that the other pair used parentheses appropriately. To wrap up, ask students to reflect on the value of a using a standard order for performing operations: Why is a standard order of operations important? What would happen if we didn’t have a consistent order in which operations are performed? If needed, point out that as students have seen during the lesson, some arithmetic expressions can be interpreted in many different ways. PEMDAS provides a basis for common understanding. Activity Extension: Have students come up with and share with the class a personally meaningful mnemonic for order of operations. For example, they might relate this to a favorite pastime or use words from native languages. This activity is based on work developed at TERC.
34	https://stackoverflow.com/questions/15120659/largest-sum-of-k-elements-not-larger-than-m	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Largest sum of k elements not larger than m Ask Question Asked Modified 10 years, 8 months ago Viewed 2k times 3 This problem is from a programing competition, and I can't manage to solve it in acceptable time. You are given an array a of n integers. Find the largest sum s of exactly k elements (not necessarily continuous) that does not exceed a given integer m (s < m). Constraints: 0 < k <= n < 100 m < 3000 0 < a[i] < 100 Info: A solution is guaranteed to exist for the given input. Now, I guess my best bet is a DP approach, but couldn't come up with the correct formula. algorithm Share Improve this question edited Jan 21, 2015 at 18:32 user2555451 asked Feb 27, 2013 at 19:28 DenisDenis 1481010 bronze badges 5 1 You say programming competition. So there are probably additional constraints, e.g. on the language or on the runtime. If runtime constraints can be ignored just determine ALL k-element subsets, compute the sums and take the best of the bunch. Udo Klein – Udo Klein 2013-02-27 19:36:39 +00:00 Commented Feb 27, 2013 at 19:36 @UdoKlein As it is in programing competitions, there are constraints :) for C++ time limit is 3 seconds, for Java it's 5 seconds. Memory constrainst are of no worry (256 MB i think) Denis – Denis 2013-02-27 19:38:13 +00:00 Commented Feb 27, 2013 at 19:38 @UdoKlein - programming competition implies that an optimal solution is required, or at least something better than brute force - which would probably take years to finish on any PC with the given constraints. IVlad – IVlad 2013-02-27 19:39:27 +00:00 Commented Feb 27, 2013 at 19:39 @Debnis: so if the constraints are known, please put them into the question. Udo Klein – Udo Klein 2013-02-27 19:43:57 +00:00 Commented Feb 27, 2013 at 19:43 This is a contest from hackerrank.com recursive – recursive 2013-02-27 20:09:55 +00:00 Commented Feb 27, 2013 at 20:09 Add a comment \| 4 Answers 4 Reset to default 2 I would try two things. They are both based on the following idea: If we can solve the problem of deciding if there are k elements that sum exactly to p, then we can binary search for the answer in [1, m]. 1. Optimized bruteforce Simply sort your array and cut your search short when the current sum exceeds p. The idea is that you will generally only have to backtrack very little, since the sorted array should help eliminate bad solutions early. To be honest, I doubt this will be fast enough however. 2. A randomized algorithm Keep a used array of size k. Randomly assign elements to it. While their sum is not p, randomly replace an element with another and make sure to update their sum in constant time. Keep doing this a maximum of e times (experiment with its value for best results, the complexity will be O(e log m) in the end, so it can probably go quite high), if you couldn't get to sum p during this time, assume that it is not possible. Alternatively, forget the binary search. Run the randomized algorithm directly and return the largest valid sum it finds in e runs or until your allocated running time ends. I am not sure how DP would efficiently keep track of the number of elements used in the sum. I think the randomized algorithm is worth a shot since it is easy to implement. Share Improve this answer edited Feb 27, 2013 at 20:08 answered Feb 27, 2013 at 20:01 IVladIVlad 43.6k1313 gold badges115115 silver badges183183 bronze badges 1 Comment Denis Denis Dough, i completely forgot i can sort the array since the order doesn't matter. The 1. solution already did the trick (everything below 0.3 seconds) since the input array is quite small. But yes, on higher order inputs the brute force would not be acceptable at all. Thanks! 2 Both of the accepted methods are inferior. Also, this is not a problem type that can be solved by DP. The following is the correct method illustrated via example: imagine a = { 2, 3, 5, 9, 11, 14, 17, 23 } (hence n = 8), k = 3, and s = 30 Sort the array a. Define three pointers into the array, P1, P2, and P3 going from 1 to n. P1 < P2 < P3 Set P3 to a_max (here 23), P1 to 1, and P2 to 2. Calculate the sum s (here 23 + 2 + 3 = 28). If s > S, then decrease P3 by one and try again until you find a solution. If P3 < 3, then there is no solution. Store your first solution as best known solution so far (BKSSF). Next, increase P2 until s > S. If you find a better solution update BKSSF. Decrease P2 by one. Next increase P1 until s > S. Update if you find a better solution. Now go back to P2 and decrease it by one. Then increase P1 until s > S. etc. You can see this is a recursive algorithm, in which for every increase or decrease, there is one or more corresponding decreases, increases. This algorithm will be much, much faster than the attempts above. Share Improve this answer answered Mar 1, 2013 at 20:59 Tyler DurdenTyler Durden 11.6k1111 gold badges7575 silver badges135135 bronze badges 1 Comment Shivendra Shivendra Tyler Durden provide some code please ! it's not clear the algorithm! 0 for l <= k and r <= s: V[l][r] = true iff it is possible to choose exactly l elements that sum up to r. V = true for i in 1..n: V'[][] - initialize with false for l in 0..k-1: for r in 0..s: if V[l][r] and s + a[i] <= s: V'[l + 1][r + a[i]] = true V \|= V' That gives you all achievable sums in O(k n s). Share Improve this answer answered Mar 1, 2013 at 22:00 lopeklopek 52255 silver badges1212 bronze badges Comments 0 I think Tyler Durden had the right idea. But you don't have to sum -all- the elements, and you can basically do it greedily, so you can cut down the loop a lot. In C++: ``` include #include using namespace std; #define FI(n) for(int i=0;i<(n);i++) int m, n, k; int a[] = { 12, 43, 1, 4, 3, 5, 13, 34, 24, 22, 31 }, e; inline int max(int i) { return n-k+i+1; } void print(int e[], int ii, int sum) { cout << sum << '\t'; FI(ii+1) cout << e[i]<<','; cout<<'\n'; } bool desc(int a, int b) { return a>b; } int solve() { sort(a, a+n, desc); cout <<"a="; FI(n) cout << a[i]<<','; cout<<"\nsum\tindexes\n"; int i,sum; i = e = sum = 0; print (e,i,a); while(1) { while (e[i]=m) e[i]++; if (e[i]==max(i)) { if (!i) return -1; // FAIL cout<<""; print (e,i,sum); sum -= a[e[--i]++]; } else // sum+a[e[i]] ``` For m=36 it gives the output n,k,m=11 3 36 a=43,34,31,24,22,13,12,5,4,3,1, sum indexes 43 0, 34 1, 34 1,10, 31 2, 35 2,8, 35 2,8,11, 34 2,9, 35 2,9,10, 35 For m=37 it gives n,k,m=11 3 37 a=43,34,31,24,22,13,12,5,4,3,1, sum indexes 43 0, 34 1, 34 1,10, 31 2, 36 2,7, 36 2,7,11, 35 2,8, 36 2,8,10, 36 (One last try: for m=39 it also gives the right answer, 38) Output: the last number is the sum and the line above it has the indexes. Lines with an asterisk are before backtracking, so the last index of the line is one too high. Runtime should be O(kn). Sorry for the hard-to-understand code. I can clean it up and provide explanation upon request but I have another project due at the moment ;). Share Improve this answer answered May 8, 2013 at 9:30 SephraSephra 2322 bronze badges Comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm See similar questions with these tags. 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35	https://www.quora.com/For-a-0-b-0-how-do-you-prove-that-a-b-if-and-only-if-sqrt-a-sqrt-b	Something went wrong. Wait a moment and try again. Inequalities (general) Algebraic Proofs Square and Square Root Basic Algebra Proofs (mathematics) Mathematical Inequalities Square Roots (functions) 5 For a>0, b>0 how do you prove that a<b if and only if sqrt(a)<sqrt(b)? Garfy Johnson Studied Physics and Electrical Engineering at Linköping University (Graduated 2010) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 21.8K answers and 7.6M answer views · Updated Feb 4 Definition of square root : sqrt(a)sqrt(a) = a Now multiply both sides of the inequality first by sqrt(a) and we get: sqrt(a)sqrt(a) < sqrt(a)sqrt(b) We simplify using the definition of sqrt a < sqrt(a)sqrt(b) then instead doing the same with sqrt(b) we get sqrt(a)sqrt(b) < b If we put these together : a < sqrt(a)sqrt(b) < b as there is a number between a and b they too must also be of different size. If you are curious you may know that this number sqrt(a)sqrt(b) is called the geometric mean or geometric average of the numbers a and b. As a mean it will always lie between, never be bigger than t Definition of square root : sqrt(a)sqrt(a) = a Now multiply both sides of the inequality first by sqrt(a) and we get: sqrt(a)sqrt(a) < sqrt(a)sqrt(b) We simplify using the definition of sqrt a < sqrt(a)sqrt(b) then instead doing the same with sqrt(b) we get sqrt(a)sqrt(b) < b If we put these together : a < sqrt(a)sqrt(b) < b as there is a number between a and b they too must also be of different size. If you are curious you may know that this number sqrt(a)sqrt(b) is called the geometric mean or geometric average of the numbers a and b. As a mean it will always lie between, never be bigger than the biggest or smaller than the smallest. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. 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If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions How do I prove that for ? How do I prove that ab > 0 if and only if a>0 and b>0 or a<0 and B<0? How do I prove that if a<0 and b<0, then ab>0? How do I prove that a = 0 if and only if -a = 0? How can I find all positive integers such that ? Him Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.5K answers and 478.5K answer views · 2y If, ✓a < ✓b (✓a — ✓b) < 0 Multiplying both sides by (✓a + ✓b), which is fine since it is positive, (✓a — ✓b)(✓a + ✓b) < 0 a — b < 0 a < b Reverse these steps to get the reverse implication. Reuven Harmelin Studied Mathematics at טכניון (Graduated 1978) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2.3K answers and 1.9M answer views · Updated 2y The inequality a<b can be written as follows: 0<b-a= =(sqrt(b)-sqrt(a))(sqrt(b)+sqrt(a)) and since the sum of the two square roots is positive, it follows that also sqrt(b)-sqrt(a)>0 Hence sqrt(b)>sqrt(a). You can easily reverse that proof and show that if sqrt(b)-sqrt(a)>0 Then b-a>0 Sohel Zibara Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views · 2y For x > 0, f'(x) = 1 / (2 × sqrt(x)) > 0 where f(x) = sqrt(x). Thus f(x) = sqrt(x) i strictly increasing which means that for x and y > 0 such that x < y, we have f(x) < f(y). Related questions Is 0<a<b proof that 1/b<1/a? How do you prove "if $b>0$, then $a^2<b$ if and only if $-\sqrt {b} < a$ and $a < \sqrt {b} $." (calculus, algebra precalculus, math)? Is (a>b) equal to (a-b>0)? Let [math]a,b,c>0,[/math] and [math]a+b+c=3.[/math] How do I show that : [math]{\frac {a\sqrt {\sqrt {b}+\sqrt {c}}}{\sqrt {a+\sqrt {bc}}}}+{\frac {b \sqrt {\sqrt {c}+\sqrt {a}}}{\sqrt {b+\sqrt {ac}}}}+{\frac {c\sqrt { \sqrt {a}+\sqrt {b}}}{\sqrt {c+\sqrt {ab}}}}\leq 3[/math] ? How can you prove if a>0, b>0 and a<b, then a²<b²? Alex Moon BS in Pure Mathematics, Michigan State University · Author has 3.4K answers and 2.1M answer views · Apr 15 By the assumptions of [math] a,b [/math] positive and [math] \sqrt{a} < \sqrt{b} [/math] we have: [math] \displaystyle \sqrt{a}\sqrt{a} < \sqrt{a}\sqrt{b} \tag{} [/math] [math] \displaystyle \sqrt{a}\sqrt{b} < \sqrt{b}\sqrt{b} \tag{} [/math] [math] \displaystyle \text{aka:} \tag{} [/math] [math] \displaystyle a = \sqrt{a}\sqrt{a} < \sqrt{a}\sqrt{b} < \sqrt{b}\sqrt{b} = b \tag{} [/math] which shows our implication in one direction, as for the other: Consider [math] f(x) = \sqrt{x} > 0 [/math] we note that [math] f'(x) = 1/2f(x) > 0 [/math] which means [math] f [/math] is increasing over the positive real numbers, aka [math] 0 < a < b ⇒︎ f(a) < f(b) [/math] which is equivalent to what we wanted to show. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? 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Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · Feb 18 Write [math]\sqrt{a}=x[/math] and [math]\sqrt{b}=y,[/math] so you want to show that [math]x<y[/math] if and only if [math]x^2<y^2[/math] which is the same as [math]y-x>0[/math] if and only if [math]y^2-x^2>0[/math] Since [math]y^2-x^2=(y-x)(y+x)[/math] and [math]y+x>0,[/math] we see that [math]y-x>0[/math] implies math(y+x)>0[/math] and also that math(y+x)>0[/math] implies math(y+x)(y+x)^{-1}>0[/math] Done. Doug Dillon Ph.D. Mathematics · Upvoted by David Vanderschel , PhD Mathematics & Physics, Rice (1970) · Author has 12.4K answers and 11.4M answer views · 5y Related How do I prove that [math]\sqrt{a}+\sqrt{b}\le\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}[/math] for [math]a,b>0[/math] ? Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Roo Life is like Maths, if it is too easy, then something's going wrong · Author has 131 answers and 523.2K answer views · 8y Related If [math]a<0[/math] and [math]b>0[/math] then [math]\sqrt{a}\cdot\sqrt{b}[/math] is equal to? Hope it helps!! Hope it helps!! Robby Goetschalckx Computer scientist for 11+ years and passionate about math since childhood. · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 6.4K answers and 9.2M answer views · 5y Related How do you prove that if a,b>0, a a^2<b^2? A different approach from the answers so far: If [math]a, b > 0[/math] , then math > 0[/math] . If [math]a < b[/math] , then [math]b - a > 0[/math] . If math > 0[/math] and math > 0[/math] , then math(b - a) > 0[/math] . math(b - a) = b^2 - a^2[/math] . If [math]b^2 - a^2 > 0[/math] , then [math]a^2 < b^2[/math] . Sponsored by CDW Corporation What’s the best way to protect your growing infrastructure? Enable an AI-powered defense with converged networking and security solutions from Fortinet and CDW. Pedro Moore Gaissler Student at University of Toronto · Author has 311 answers and 326.7K answer views · 5y Related How do you prove that if a,b>0, a a^2<b^2? Simple. Let [math]a,b>0[/math] and [math]a<b[/math]. Since [math]a>0[/math] we can multiply both sides of the inequality by [math]a,[/math] obtaining [math]a^2 < ab[/math] and the order of the inequality remains intact because [math]a[/math] is positive. Repeating the process for [math]b[/math] we have that [math]ab<b^2[/math]. By putting the two inequalities together we get [math]a^2 < ab < b^2 [/math] [math]\implies a^2 < b^2[/math] since inequalities are transitive. Michael Carroll M. S. in Mathematics, New Mexico State University (Graduated 1968) · Author has 556 answers and 1.2M answer views · 6y Related Can you prove that a+b>sqrt ((a^2+b^2)) for a>0,b>0? I am using the rule that the positive square root of positive numbers preserves order. [math]a > 0,b > 0 \Rightarrow a + b > 0,ab > 0\[/math] [math]{(a + b)^2} = {a^2} + 2ab + {b^2}\[/math] [math]{a^2} + 2ab + {b^2} > {a^2} + {b^2}\[/math] [math]{(a + b)^2} > {a^2} + {b^2}\[/math] [math]a + b > \sqrt {{a^2} + {b^2}} [/math] Brian Powell B.S. in Mathematics & Education, East Carolina University (Graduated 1981) · Author has 1.6K answers and 2.1M answer views · 5y Related How do you prove that if a,b>0, a a^2<b^2? Given that a < b, there must exist some positive number, k, such that a + k = b. Consider b^2. By substitution, b^2 = (a + k)^2 or a^2 + 2ka + k^2 (by expanding the product). Multiplication is closed under the set of positive real numbers. Therefore, a^2 + 2ka + k^2 must be greater than a^2. Therefore, b^2 must be greater than a^2 (by substitution) Since b^2 > a^2, it follows that a^2 < b^2.. Ron Davis I earn my living with mathematics. · Author has 6.7K answers and 18.8M answer views · Updated 5y Related How do I prove that [math]\sqrt{a}+\sqrt{b}\le\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}[/math] for [math]a,b>0[/math] ? Prove [math]\sqrt{a} + \sqrt{b} \le \dfrac a {\sqrt{b}} + \dfrac b {\sqrt{a}} [/math] [math]\forall ~ a,b>0[/math] and [math]\sqrt{a},\sqrt{b}>0.[/math] Multiply both sides by [math]\sqrt{a} \, \sqrt{b}:[/math] [math]a\,\sqrt{b} + b\,\sqrt{a} \le a\,\sqrt{a} + b\,\sqrt{b}.[/math] Subtract the right-hand side from both sides: [math]a\,\sqrt{b} + b\,\sqrt{a} - a\,\sqrt{a} - b\,\sqrt{b} \le 0.[/math] Factor the left-hand side: [math]\left ( a - b \right )\left ( \sqrt{b} - \sqrt{a} \right ) \le 0.[/math] If [math]a = b[/math] we get the equality. For any other relationship, the two factors have opposite signs. However, if you use the negative roots instead of the positive roots, you get the opp Prove [math]\sqrt{a} + \sqrt{b} \le \dfrac a {\sqrt{b}} + \dfrac b {\sqrt{a}} [/math] [math]\forall ~ a,b>0[/math] and [math]\sqrt{a},\sqrt{b}>0.[/math] Proof: Multiply both sides by [math]\sqrt{a} \, \sqrt{b}:[/math] [math]a\,\sqrt{b} + b\,\sqrt{a} \le a\,\sqrt{a} + b\,\sqrt{b}.[/math] Subtract the right-hand side from both sides: [math]a\,\sqrt{b} + b\,\sqrt{a} - a\,\sqrt{a} - b\,\sqrt{b} \le 0.[/math] Factor the left-hand side: [math]\left ( a - b \right )\left ( \sqrt{b} - \sqrt{a} \right ) \le 0.[/math] If [math]a = b[/math] we get the equality. For any other relationship, the two factors have opposite signs. Q.E.D. However, if you use the negative roots instead of the positive roots, you get the opposite result, so the stipulation of positive roots is necessary in the statement of the problem. Related questions How do I prove that for ? How do I prove that ab > 0 if and only if a>0 and b>0 or a<0 and B<0? How do I prove that if a<0 and b<0, then ab>0? How do I prove that a = 0 if and only if -a = 0? How can I find all positive integers such that ? Is 0<a<b proof that 1/b<1/a? How do you prove "if $b>0$, then $a^2<b$ if and only if $-\sqrt {b} < a$ and $a < \sqrt {b} $." (calculus, algebra precalculus, math)? Is (a>b) equal to (a-b>0)? Let and How do I show that : ? How can you prove if a>0, b>0 and a<b, then a²<b²? If a=0 and b=0, then how would you prove that a/b=2? Let a, b, c ＞ 0. How do I prove that ? How do I prove this theorem: if a>0, then 1/a>0? Let a>0 and b>0. How do I prove that a<b‘b’? How can I prove that a>0 ,b>a and x>0 then a+x/b+x>a/b? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
36	https://reference.wolfram.com/language/ref/JacobiNS.html	JacobiNS: Jacobi reciprocal elliptic sine function ns—Wolfram Documentation Products Wolfram\|One The definitive Wolfram Language and notebook experience Mathematica The original technical computing environment Wolfram Notebook Assistant + LLM Kit All-in-one AI assistance for your Wolfram experience System Modeler Wolfram Player Finance Platform Wolfram Engine Enterprise Private Cloud Application Server Wolfram\|Alpha Notebook Edition Wolfram Cloud App Wolfram Player App More mobile apps Core Technologies of Wolfram Products Wolfram Language Computable Data Wolfram Notebooks AI & Linguistic Understanding Deployment Options Wolfram Cloud wolframscript Wolfram Engine Community Edition Wolfram LLM API WSTPServer Wolfram\|Alpha APIs From the Community Function Repository Community Paclet Repository Example Repository Neural Net Repository Prompt Repository Wolfram Demonstrations Data Repository Group & Organizational Licensing All Products Consulting & Solutions We deliver solutions for the AI era—combining symbolic computation, data-driven insights and deep technical expertise Data & Computational Intelligence Model-Based Design Algorithm Development Wolfram\|Alpha for Business Blockchain Technology Education Technology Quantum Computation WolframConsulting.com Wolfram Solutions Data Science Artificial Intelligence Biosciences Healthcare Intelligence Sustainable Energy Control Systems Enterprise Wolfram\|Alpha Blockchain Labs More Wolfram Solutions Wolfram Solutions For Education Research Universities Colleges & Teaching Universities Junior & Community Colleges High Schools Educational Technology Computer-Based Math More Solutions for Education Contact Us Learning& Support Get Started Wolfram Language Introduction Fast Intro for Programmers Fast Intro for Math Students Wolfram Language Documentation More Learning Highlighted Core Areas Demonstrations YouTube Daily Study Groups Wolfram Schools and Programs Books Grow Your Skills Wolfram U Courses in computing, science, life and more Community Learn, solve problems and share ideas. 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Blog News, views and insights from Wolfram Resources for Software Developers Tech Support Contact Us Support FAQs More Learning Highlighted Core Areas Demonstrations YouTube Daily Study Groups Wolfram Schools and Programs Books Support FAQs Contact Us Company About Wolfram Career Center All Sites & Resources Connect & Follow Contact Us Work with Us Student Ambassador Initiative Wolfram for Startups Student Opportunities Jobs Using Wolfram Language Educational Programs for Adults Summer School Winter School Educational Programs for Youth Middle School Camp High School Research Program Computational Adventures Read Stephen Wolfram's Writings Wolfram Blog Wolfram Tech \| Books Wolfram Media Complex Systems Educational Resources Wolfram MathWorld Wolfram in STEM/STEAM Wolfram Challenges Wolfram Problem Generator Wolfram Initiatives Wolfram Science Wolfram Foundation History of Mathematics Project Events Stephen Wolfram Livestreams Online & In-Person Events Contact Us Connect & Follow Wolfram\|Alpha Wolfram Cloud Your Account User Portal Wolfram Language& System Documentation Center JacobiNS See Also InverseJacobiNS JacobiSN JacobiNC JacobiND Related Guides Elliptic Functions Tech Notes Elliptic Integrals and Elliptic Functions See Also InverseJacobiNS JacobiSN JacobiNC JacobiND Related Guides Elliptic Functions Tech Notes Elliptic Integrals and Elliptic Functions JacobiNS[u,m] gives the Jacobi elliptic function . Details Examples Basic Examples Scope Numerical Evaluation Specific Values Visualization Function Properties Differentiation Integration Series Expansions Function Identities and Simplifications Function Representations Applications Properties & Relations Possible Issues See Also Tech Notes Related Guides Related Links History Cite this Page BUILT-IN SYMBOL See Also InverseJacobiNS JacobiSN JacobiNC JacobiND Related Guides Elliptic Functions Tech Notes Elliptic Integrals and Elliptic Functions See Also InverseJacobiNS JacobiSN JacobiNC JacobiND Related Guides Elliptic Functions Tech Notes Elliptic Integrals and Elliptic Functions JacobiNS Copy to clipboard. ✖ JacobiNS JacobiNS[u,m] Copy to clipboard. ✖ JacobiNS[u,m] gives the Jacobi elliptic function . Details Mathematical function, suitable for both symbolic and numerical manipulation. , where . is a doubly periodic function in u with periods and , where is the elliptic integral EllipticK. JacobiNS is a meromorphic function in both arguments. For certain special arguments, JacobiNS automatically evaluates to exact values. JacobiNS can be evaluated to arbitrary numerical precision. JacobiNS automatically threads over lists. Examples open all close all Basic Examples(4)Summary of the most common use cases Evaluate numerically: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot the function over a subset of the reals: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot over a subset of the complexes: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Series expansions about the origin: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Scope(34)Survey of the scope of standard use cases Numerical Evaluation(5) Evaluate numerically to high precision: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 The precision of the output tracks the precision of the input: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Evaluate for complex arguments: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Evaluate JacobiNS efficiently at high precision: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Compute average-case statistical intervals using Around: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Compute the elementwise values of an array: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Or compute the matrix JacobiNS function using MatrixFunction: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Specific Values(3) Simple exact values are generated automatically: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Some poles of JacobiNS: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Find a local minimum of JacobiNS as a root of : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Visualization(3) Plot the JacobiNS functions for various parameter values: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot JacobiNS as a function of its parameter : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot the real part of : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot the imaginary part of : Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Function Properties(8) JacobiNS is -periodic along the real axis: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 JacobiNS is -periodic along the imaginary axis: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 JacobiNS is an odd function in its first argument: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 JacobiNS is not an analytic function: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 It has both singularities and discontinuities: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Copy to clipboard. In:=3 ✖ Direct link to example Out=3 is neither nondecreasing nor nonincreasing: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 is not injective for any fixed : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Copy to clipboard. In:=3 ✖ Direct link to example Out=3 It is injective for : Copy to clipboard. In:=4 ✖ Direct link to example Out=4 Copy to clipboard. In:=5 ✖ Direct link to example Out=5 JacobiNS is not surjective for any fixed : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 JacobiNS is neither non-negative nor non-positive: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 JacobiNS is neither convex nor concave: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Differentiation(3) First derivative: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Higher derivatives: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot higher derivatives for : Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Derivative with respect to : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Integration(3) Indefinite integral of JacobiNS: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Definite integral of an odd function over the interval centered at the origin is 0: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 More integrals: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Series Expansions(3) Series expansion for : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot the first three approximations for around : Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Taylor expansion for : Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Plot the first three approximations for around : Copy to clipboard. In:=2 ✖ Direct link to example Out=2 JacobiNS can be applied to power series: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Function Identities and Simplifications(3) Parity transformation and periodicity relations are automatically applied: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Identity involving JacobiCS: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Argument simplifications: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Function Representations(3) Representation in terms of Csc of JacobiAmplitude: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Relation to other Jacobi elliptic functions: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 TraditionalForm formatting: Copy to clipboard. In:=1 ✖ Direct link to example Applications(5)Sample problems that can be solved with this function Map a rectangle conformally onto the lower half‐plane: Copy to clipboard. In:=1 ✖ Direct link to example Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Solution of the pendulum equation: Copy to clipboard. In:=1 ✖ Direct link to example Check the solution: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Plot solutions: Copy to clipboard. In:=3 ✖ Direct link to example Out=3 Closed form of iterates of the Katsura–Fukuda map: Copy to clipboard. In:=1 ✖ Direct link to example Compare the closed form with explicit iterations: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Copy to clipboard. In:=3 ✖ Direct link to example Out=3 Plot a few hundred iterates: Copy to clipboard. In:=4 ✖ Direct link to example Out=4 Solution of the sinh‐Gordon equation : Copy to clipboard. In:=1 ✖ Direct link to example Check the solution: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Plot the solution: Copy to clipboard. In:=3 ✖ Direct link to example Out=3 Hierarchy of solutions of the nonlinear diffusion equation : Copy to clipboard. In:=1 ✖ Direct link to example Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Verify these functions: Copy to clipboard. In:=3 ✖ Direct link to example Out=3 Properties & Relations(2)Properties of the function, and connections to other functions Compose with inverse functions: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Use PowerExpand to disregard multivaluedness of the inverse function: Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Solve a transcendental equation: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Possible Issues(2)Common pitfalls and unexpected behavior Machine-precision input may be insufficient to give the correct answer: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 Currently only simple simplification rules are built in for Jacobi functions: Copy to clipboard. In:=1 ✖ Direct link to example Out=1 Copy to clipboard. In:=2 ✖ Direct link to example Out=2 See Also InverseJacobiNSJacobiSNJacobiNCJacobiND Tech Notes ▪ Elliptic Integrals and Elliptic Functions Related Guides ▪ Elliptic Functions Related Links MathWorld History Introduced in 1988 (1.0) Cite this as:Wolfram Research (1988), JacobiNS, Wolfram Language function, Copy to clipboard. ✖ Wolfram Research (1988), JacobiNS, Wolfram Language function, Text Wolfram Research (1988), JacobiNS, Wolfram Language function, Copy to clipboard. ✖ Wolfram Research (1988), JacobiNS, Wolfram Language function, CMS Wolfram Language. 1988. "JacobiNS." Wolfram Language & System Documentation Center. Wolfram Research. Copy to clipboard. ✖ Wolfram Language. 1988. "JacobiNS." Wolfram Language & System Documentation Center. Wolfram Research. APA Wolfram Language. (1988). JacobiNS. Wolfram Language & System Documentation Center. Retrieved from Copy to clipboard. ✖ Wolfram Language. (1988). JacobiNS. Wolfram Language & System Documentation Center. Retrieved from BibTeX @misc{reference.wolfram_2025_jacobins, author="Wolfram Research", title="{JacobiNS}", year="1988", howpublished="\url{ note=[Accessed: 28-September-2025]} Copy to clipboard. ✖ @misc{reference.wolfram_2025_jacobins, author="Wolfram Research", title="{JacobiNS}", year="1988", howpublished="\url{ note=[Accessed: 28-September-2025]} BibLaTeX @online{reference.wolfram_2025_jacobins, organization={Wolfram Research}, title={JacobiNS}, year={1988}, url={ note=[Accessed: 28-September-2025]} Copy to clipboard. ✖ @online{reference.wolfram_2025_jacobins, organization={Wolfram Research}, title={JacobiNS}, year={1988}, url={ note=[Accessed: 28-September-2025]} Give FeedbackTop Introduction for Programmers Introductory Book Wolfram Function Repository\|Wolfram Data Repository\|Wolfram Data Drop\|Wolfram Language Products Top Products Wolfram\|One Mathematica Notebook Assistant + LLM Kit System Modeler Wolfram\|Alpha Notebook Edition Wolfram\|Alpha Pro Mobile Apps Wolfram Player Wolfram Engine Volume & Site Licensing Server Deployment Options Consulting Wolfram Consulting Repositories Data Repository Function Repository Community Paclet Repository Neural Net Repository Prompt Repository Wolfram Language Example Repository Notebook Archive Wolfram GitHub Learning Wolfram U Wolfram Language Documentation Webinars & Training Educational Programs Wolfram Language Introduction Fast Introduction for Programmers Fast Introduction for Math Students Books Wolfram Community Wolfram Blog Public Resources Wolfram\|Alpha Wolfram Problem Generator Wolfram Challenges Computer-Based Math Computational Thinking Computational Adventures Demonstrations Project Wolfram Data Drop MathWorld Wolfram Science Wolfram Media Publishing For Customers Online Store Product Downloads User Portal Your Account Support FAQ Contact Support Company About Wolfram Careers Contact Events Wolfram CommunityWolfram Blog Legal&Privacy Policy WolframAlpha.com\|WolframCloud.com © 2025Wolfram © 2025Wolfram\|Legal&Privacy Policy\| English English 日本語简体中文 Find out if you already have access to Wolfram tech through your organization ×
37	https://www.satra.com/test_methods/detail.php?id=161	SATRA TM205: Hardness of rubber, polyurethane and plastics - durometer method EN go ENZH Webinars and Online Resources Test Method: SATRA TM205 Hardness of rubber, polyurethane and plastics - durometer method August 2016 (2017) SCOPE This test method is intended to determine the hardness of a material using one of four durometer types, Shore A, Shore D, Asker C or Shore OO. The method is applicable to all types of compact or microcellular polymeric materials from soft foams through flexible polymers to hard plastics. The method is not applicable to the testing of coated fabrics. PRINCIPLE A specified indentor is forced into the material under specified conditions, and the hardness is indicated on a scale derived from the penetration of the indentor into the surface. The Shore A durometer is used for measuring most footwear outsole and midsole materials, the Shore D for very hard materials such as plastic top-pieces or football studs and the Asker C and Shore OO for very soft materials such as footbed foams and some midsole materials. Hardness measurements may be based on initial indentation, indentation after a specified period of time, or both. REFERENCES BS 5131-2.7: 1980 (1992) – Methods of test for footwear and footwear materials - The preparation of test pieces from soling materials for physical testing. HISTORY First issued February 1999. Revised August 2016, June 2017. AVAILABILITY This test method is available to members and non-members. PRICE (ex. VAT) This price is for a printed copy of the Test Method. Members: £80 Non-members: £230 TYPE General and physical properties (other than strength) USAGES Solings - polyurethane Solings - PVC, TR and other thermoplastic Solings - rubber, EVA and other crosslinked Add to basketGo back Resources SATRA corporate brochure Test Methods catalogue Home Products & Services back Test at SATRA back Start testing SATRA Test Methods Footwear PPE Personal Protective Equipment Furniture Automotive components Floor Coverings Materials and Clothing Leathergoods and Luggage Toys Chemical Analytical Testing (CAT) Sample Submission Also back SATRA Membership SATRA Quality Mark CE Mark (PPE) SATRA Bulletin articles SATRA Spotlight articles Test Equipment back Test Equipment Overview Test Machine Index Types of Testing Test Consumables Index Also back Building a bespoke laboratory SATRA Laboratory Accreditation SATRA Test Methods SATRA Bulletin articles Certification & Accreditation back Overview SATRA Laboratory Accreditation Accredited Laboratory List Leather Grading Accreditation Accredited Grader List Cutting Award Accredited Cutter List CE Mark back CE Mark (PPE) SAFT Footwear Technologist SATRA Quality Mark SATRA Sole Bond Award SATRA Diamond Certification Also back SATRA Bulletin articles SATRA Spotlight articles Training Test Equipment Research back Research Overview Slip resistance Footwear scanning, fitting and sizing Extreme outdoor performance wear Product development Consultancy Ecotextyle project Also back SATRA Quality Mark SATRA Bulletin articles Production efficiency back Production efficiency overview Time labour costing and balancing Cost effective use of materials Production Stitching Improvement Also back Leather Grading Accreditation Cutting Award SATRA Bulletin articles Training back Overview Webinars and Online Resources SAFT Footwear Technologist Basic shoemaking Shoe fit REACH chemicals regulation Furniture flammability Events and Seminars Introduction to Sustainability Also back SATRA Bulletin articles SATRA Spotlight articles Certification and Accreditation Events News Publications back SATRA Bulletin SATRA Bulletin Media Pack SATRA Technical Brief SATRA Test Methods Global Foot Dimensions Basic Shoemaking SATRA Spotlight Also back Use of SATRA articles About back About SATRA The SATRA Story SATRA Membership Furnishings Membership SATRA Management Team SATRA’s culture SATRA 2021 video Working at SATRA About our services back Footwear testing and research PPE certification and testing Test Equipment Chemical Analytical Testing (CAT) Contact SATRA Membership Member Access back SATRA member access Enter your email address please Continue If your email address is qualified we will email you a link to the Members content. For further guidance on gaining access click here. In addition to online content SATRA Members have access to a wide range of products and services. View Member Benefits CONNECT WITH US Sales and Services Privacy Policy\|Website Privacy Policy\|Terms of Use\|Terms of Business \| Site Map\|Careers\|Contact This website uses cookies. By using this site, you consent to their usage according to your browser’s current settings.
38	https://rexera.com/blog/what-is-annexation-in-real-estate/	Skip to content Navigating all real estate transactions, as title and escrow professionals know, can get complicated. Coordinating property closings with precision and expertise is a must, but every now and then, you are likely to stumble upon a type of property document that you are not comfortable deciphering yet, like real estate annexation. However, understanding what annexation in real estate is and its nuances brings you one step closer to ensuring seamless and legally sound transactions. In this article, we dive into the realm of real estate annexation and its profound impact on your work. By unraveling the complexities of annexation, you’ll gain invaluable insights into how it shapes property ownership, title examinations, and escrow transactions. Let’s jump right in. Table of Contents What Is Annexation? 3 Real Estate Annexation Terms Title and Escrow Should Get Familiar With 1. Fixture 2. Intent 3. Legal Implications Types of Real Estate Annexations Fixture Annexation Land Annexation Involuntary Annexation Voluntary Annexation Why Is Annexation Important? Benefits of Annexation Challenges of Annexation How Does Property Annexation Work? Implications of Real Estate Annexation for Title and Escrow How Do You Know If a Property Has Been Annexed? Summary What Is Annexation? Annexation in real estate refers to the process by which personal property is affixed or attached to real property, effectively becoming part of the real estate. This transformation turns the attached personal property into a fixture, impacting property ownership, title examinations, and escrow transactions. In simpler terms, when personal items are permanently attached to a property, they become part of that property. This has a big impact on how the property is described and how the closing process is handled. You need to pay close attention to these details and communicate them clearly to your customers to make sure the property transactions are accurate and legally sound. 3 Real Estate Annexation Terms Title and Escrow Should Get Familiar With As you process documents around the property annexations, you’ll inevitably happen upon legal terms that you’ll need to understand and explain to your clients if necessary. Here are three common annexation terms and their definitions. 1. Fixture When personal property is annexed to a property, it becomes a fixture. Fixtures are items that were once personal property but have become real property by virtue of being attached to the land or the improvements on the land. 2. Intent The key factor in determining whether an item has been annexed to real property is the intent of the annexing party. If the attachment was meant to be permanent and was made with the intention of improving the real estate, it is generally considered to be annexed. 3. Legal Implications Annexation can have legal implications related to property ownership, transfers, and responsibilities. For example, in a real estate sale, fixtures are typically included in the sale unless specifically excluded. Types of Real Estate Annexations In real estate, several types of annexations can occur, each with distinct implications for property ownership and transactions. The primary types of real estate annexation include: Fixture Annexation This type of annexation occurs when personal property becomes affixed to real property in a way that it becomes a permanent part of the real estate. Examples include built-in appliances, light fixtures, and heating systems. Land Annexation Land annexation involves the expansion of municipal boundaries to incorporate adjacent or nearby land. This type of annexation can impact property taxes, local regulations, and access to municipal services, thereby influencing the value and potential use of the affected properties. Involuntary Annexation Involuntary annexation occurs when a government entity imposes annexation on property owners within a specific area, often leading to changes in property rights, taxation, and regulatory oversight. Voluntary Annexation Voluntary annexation occurs when property owners petition to have their land annexed into a neighboring municipality. This type of annexation can be sought for various reasons, such as accessing municipal services, obtaining better zoning regulations, or seeking potential economic benefits. Why Is Annexation Important? Annexation plays a critical role in shaping property ownership, transactions, regulatory obligations, and the overall value and utility of real estate, making it essential for professionals in the industry to understand it thoroughly. Benefits of Annexation Access to Municipal Services: Annexation can provide access to improved municipal services such as water, sewer, and public safety, enhancing the quality of life for property owners and residents. Zoning and Development Opportunities: Annexation may open up new zoning and development opportunities, allowing for potential expansion and economic growth within the annexed area. Increased Property Value: Annexation can lead to an increase in property value due to access to better infrastructure, services, and potential for development. Regulatory Consistency: Annexation can bring properties under a consistent set of local regulations, ensuring uniformity in land use and development standards. Community Representation: Annexation may provide annexed residents with representation and a voice in local government decisions, fostering a sense of community and belonging. Challenges of Annexation Tax Implications: Annexation can lead to changes in property tax obligations, potentially resulting in increased tax burdens for property owners in the annexed area. Regulatory Changes: Annexation may subject properties to different regulatory frameworks, requiring property owners to adhere to new zoning, building codes, and land use regulations. Community Concerns: Annexation can be met with resistance from existing residents and property owners who may be apprehensive about changes in governance, taxation, and land use policies. Infrastructure Costs: Municipalities may face increased infrastructure and service provision costs as a result of annexation, impacting budgetary considerations and resource allocation. Land Use Conflicts: Annexation can lead to conflicts over land use and development priorities between the annexing municipality and existing property owners. How Does Property Annexation Work? The specific process of annexation can vary depending on the laws and regulations of the relevant jurisdiction. However, the general steps involved in the annexation process typically include the following: Petition or Proposal: Annexation can begin with a petition from property owners within the area to be annexed, or the governing body of the municipality may initiate it. Preliminary Studies and Planning: Before proceeding with annexation, the municipality may conduct studies and evaluations to assess the potential impact of the annexation on urban services (like access to sanitary sewers, water, fire protection, parks, open space, recreation and streets, roads and mass transit), infrastructure, and the community. Public Notice and Hearings: Public notices are often issued to inform the public about the proposed annexation. Public hearings may also be held to allow affected property owners and residents to express their views and concerns regarding the annexation. Legislative Approval: Depending on the jurisdiction, annexation may require approval by the governing body of the municipality, such as the city council or county board of commissioners. Boundary Adjustment and Legal Procedures: Once approved, the legal boundaries of the municipality are adjusted to include the annexed area. This involves updating legal descriptions, property records, and other relevant documentation. Service Extension and Integration: The annexing municipality may then begin the process of integrating the annexed area into its service provision systems, such as water, sewer, and public safety services. Impact on Property Owners: Property owners in the annexed area may experience changes in taxation, regulatory requirements, and access to municipal services as a result of the annexation. To stay on top of any complications arising from property annexations, title, and escrow agents should have a close understanding of how annexation works in general, even if there are differences from county to county. Implications of Real Estate Annexation for Title and Escrow Annexation can impact the title examination process conducted by title agents. When personal property is annexed to real property and becomes a fixture, it may affect the legal description of the property and the title report. Title agents need to ensure that any annexed fixtures are appropriately reflected in the title report and related documents. Annexations may also affect the ownership rights and interests in the property. Title agents must ascertain the status of annexed fixtures to accurately determine the current ownership rights associated with the real estate. This is crucial for establishing a clear title. In escrow transactions, the presence of annexed fixtures may require clear documentation and disclosure. Escrow agents need to ensure that any agreements related to annexed fixtures, such as inclusion or exclusion from the sale, are properly addressed in the escrow instructions and closing documents. Annexations play a role in the handling of escrow funds, as well. Especially if there are disputes or uncertainties regarding the status of annexed fixtures, escrow agents may need to account for any financial adjustments related to fixtures and ensure that the parties involved in the transaction are in agreement regarding the treatment of the annexed property. Last but not least, title and escrow agents should communicate effectively with all parties involved in the real estate transaction regarding the presence of annexed fixtures. Clear disclosure and documentation regarding the status of annexed property can help prevent misunderstandings and potential legal issues. How Do You Know If a Property Has Been Annexed? You can determine if a property has been annexed through various means, including Public Records: Local government offices, such as the county recorder’s office or assessor’s office, maintain public records related to property ownership and boundaries. These records often indicate whether a property has been annexed into a municipality. Online Resources: Many municipalities provide online resources, such as interactive maps or property search tools, that allow title professionals to view property boundaries and annexation status. Contacting the Municipality: Municipal planning or zoning departments can provide information regarding the annexation status of a property. They may have records, maps, or databases that can confirm whether a property has been annexed. Title Search: When conducting a title search for a property, the title report may contain information about any annexation-related changes to the property’s legal description or ownership status. Consulting Legal Professionals: Real estate attorneys can assist in determining the annexation status of property by reviewing relevant legal documents and public records. These resources should be able to help you gather all necessary information if the property you’re working to close has any annexations that might play a role in the transaction. Summary In this article, we’ve covered the essential processes underscoring how extending the boundaries of a property and adding fixtures affect real estate transactions. Additionally, we looked at how adding fixtures and extending the boundaries of real estate properties affect the work of title and escrow professionals. We offered steps you can take to ensure annexations won’t put a spoke in the wheel of your real estate transactions. We hope this comprehensive overview is the concise tool that helps you confidently navigate property closings. In the meantime, stay on top of the most detailed real estate transaction tips and guides by signing up for Rexera’s blog. We share actionable insights for real estate professionals from real estate professionals. Blog Posts Success Stories Resources Blog Posts Success Stories Our Products Title & Escrow Lenders SFR Investors
39	https://brainly.com/question/18994225	[FREE] Consider the end behavior of the function G(x) = 4\|x - 2\| - 3. As x approaches negative infinity, G(x) - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +27,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +44,2k Ace exams faster, with practice that adapts to you Practice Worksheets +8,8k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Consider the end behavior of the function G(x)=4∣x−2∣−3. As x approaches negative infinity, G(x) approaches infinity. As x approaches positive infinity, G(x) approaches infinity. 1 See answer Explain with Learning Companion NEW Asked by krauseh4267 • 11/06/2020 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 469231047 people 469M 5.0 32 Upload your school material for a more relevant answer Positive; Positive Explanation According to the question, the following information is given G(x) = 4 \|x - 2\| - 3 Based on the above information As it can be seen that there is a modulus that means if there is any negative value that it can become a positive value Therefore the answer is positive and positive hence, the same is to be considered Answered by andromache •23.2K answers•469.2M people helped Thanks 32 5.0 (25 votes) 1 Expert-Verified⬈(opens in a new tab) This answer helped 469231047 people 469M 5.0 30 Upload your school material for a more relevant answer The function G(x)=4∣x−2∣−3 approaches infinity as x approaches both negative and positive infinity. This is because the absolute value function causes the output to increase without bound in both directions. Thus, the end behavior is characterized by positive infinity in both cases. Explanation To analyze the end behavior of the function G(x)=4∣x−2∣−3, we will evaluate the function as x approaches both negative and positive infinity. Step 1: Understand the Absolute Value The absolute value function ∣x−2∣ behaves differently depending on whether x is less than or greater than 2. If x<2, then ∣x−2∣=2−x. If x≥2, then ∣x−2∣=x−2. Step 2: Evaluate as x→−∞ As x approaches negative infinity, we find that x<2. Thus, we can substitute in the expression for ∣x−2∣: G(x)=4(2−x)−3 G(x)=8−4 x−3 G(x)=−4 x+5 As x→−∞, the term −4 x approaches positive infinity. Therefore, G(x)→∞. Step 3: Evaluate as x→+∞ As x approaches positive infinity, we have x≥2, so we substitute the other expression for the absolute value: G(x)=4(x−2)−3 G(x)=4 x−8−3 G(x)=4 x−11 As x→+∞, the term 4 x also approaches positive infinity, meaning G(x)→∞. Conclusion: Hence, As x approaches negative infinity, G(x) approaches infinity. As x approaches positive infinity, G(x) also approaches infinity. Examples & Evidence For example, if you plug in values like x=−10 and x=10 into the function, you'll see that both result in large positive values. This illustrates the end behavior of approaching infinity as mentioned. The analysis is based on understanding the properties of absolute values and their impact on the function's growth as we analyze limits approaching infinity. Thanks 30 5.0 (22 votes) Advertisement krauseh4267 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.4 9 Which statement describes the end behavior of this function? g(x) = 1/2\|x - 3\| - 7 A. As x approaches positive infinity, g(x) approaches negative infinity. B. As x approaches negative infinity, g(x) approaches negative infinity. C. As x approaches positive infinity, g(x) approaches positive infinity. D. As x approaches negative infinity, g(x) is no longer continuous. Community Answer 5.0 1 Consider functions f and g below. [tex]g(x) = -x^{2} + 2x + 4[/tex] Which of the following statements is true? A. As x approaches infinity, the values of f(x) and g(x) both approach infinity. B. As x approaches infinity, the values of f(x) and g(x) both approach negative infinity. C. As x approaches infinity, the value of f(x) approaches infinity and the value of g(x) approaches negative infinity. D. As x approaches infinity, the value of f(x) approaches negative infinity and the value of g(x) approaches infinity. Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Below is a relation represented as a set of ordered pairs. {(5,2),(−3,−6),(1,4),(−6,8),(2,0)} a) Is the relation above a function? Why or why not? b) What is the inverse of the relation as a set of ordered pairs? c) Is the inverse a function? Why or why not? The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the function h(t)=300−16 t 2. Which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall? A. h(3)−h(0) B. h(3 3)−h(3 0) C. 3 h(3) D. 3 h(3)−h(0) Which system of equations can be used to find the roots of the equation 4 x 2=x 3+2 x ? $\begin{array}{l} y=-4 x^2 \ y=x^3+2 x \end{array}$ $\begin{array}{l} y=4 x^2 \ y=x^3+2 x \end{array}$ $\begin{array}{l} y=x^3-4 x^2+2 x \ y=0 \end{array}$ $\begin{array}{l} y=4 x^2 \ y=-x^3-2 x \end{array}$ The diagonal of a square is x units. What is the area of the square in terms of x? A. 2 1x 2 square units B. x 2 square units C. 2 x square units D. 2 1x square units If f(x)=2(18−9 x)+4 x, what is the value of f(7)? 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40	https://courses.lumenlearning.com/waymakercollegealgebra/chapter/describe-the-end-behavior-of-power-functions/	End Behavior of Power Functions \| College Algebra Skip to main content College Algebra Module 9: Power and Polynomial Functions Search for: End Behavior of Power Functions Learning Outcomes Identify a power function. Describe the end behavior of a power function given its equation or graph. Three birds on a cliff with the sun rising in the background. Functions discussed in this module can be used to model populations of various animals, including birds. (credit: Jason Bay, Flickr) Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown below. Year2009 2010 2011 2012 2013 Bird Population800 897 992 1,083 1,169 The population can be estimated using the function P(t)=−0.3 t 3+97 t+800 P(t)=−0.3 t 3+97 t+800, where P(t)P(t) represents the bird population on the island t years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. Identifying Power Functions In order to better understand the bird problem, we need to understand a specific type of function. A power functionis a function with a single term that is the product of a real number,coefficient, and variable raised to a fixed real number power. Keep in mind a number that multiplies a variable raised to an exponent is known as a coefficient. As an example, consider functions for area or volume. The function for the area of a circle with radius r r is: A(r)=π r 2 A(r)=π r 2 and the function for the volume of a sphere with radius r is: V(r)=4 3 π r 3 V(r)=4 3 π r 3 Both of these are examples of power functions because they consist of a coefficient, π π or 4 3 π 4 3 π, multiplied by a variable r raised to a power. A General Note: Power FunctionS A power function is a function that can be represented in the form f(x)=a x n f(x)=a x n where a and n are real numbers and a is known as the coefficient. Q & A Is f(x)=2 x f(x)=2 x a power function? No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function. Example: Identifying Power Functions Which of the following functions are power functions? f(x)=1 Constant function f(x)=x Identify function f(x)=x 2 Quadratic function f(x)=x 3 Cubic function f(x)=1 x Reciprocal function f(x)=1 x 2 Reciprocal squared function f(x)=√x Square root function f(x)=3√x Cube root function f(x)=1 Constant function f(x)=x Identify function f(x)=x 2 Quadratic function f(x)=x 3 Cubic function f(x)=1 x Reciprocal function f(x)=1 x 2 Reciprocal squared function f(x)=x Square root function f(x)=x 3 Cube root function Show Solution All of the listed functions are power functions. The constant and identity functions are power functions because they can be written as f(x)=x 0 f(x)=x 0 and f(x)=x 1 f(x)=x 1 respectively. The quadratic and cubic functions are power functions with whole number powers f(x)=x 2 f(x)=x 2 and f(x)=x 3 f(x)=x 3. The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f(x)=x−1 f(x)=x−1 and f(x)=x−2 f(x)=x−2. The square and cube root functions are power functions with fractional powers because they can be written as f(x)=x 1/2 f(x)=x 1/2 or f(x)=x 1/3 f(x)=x 1/3. Try It Which functions are power functions? f(x)=2 x 2⋅4 x 3 g(x)=−x 5+5 x 3−4 x h(x)=2 x 5−1 3 x 2+4 f(x)=2 x 2⋅4 x 3 g(x)=−x 5+5 x 3−4 x h(x)=2 x 5−1 3 x 2+4 Show Solution f(x)f(x)is a power function because it can be written as f(x)=8 x 5 f(x)=8 x 5.The other functions are not power functions. Identifying End Behavior of Power Functions The graph below shows the graphs of f(x)=x 2,g(x)=x 4 f(x)=x 2,g(x)=x 4, h(x)=x 6 h(x)=x 6, k(x)=x 8 k(x)=x 8, and p(x)=x 10 p(x)=x 10 which are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin. To describe the behavior as numbers become larger and larger, we use the idea of infinity. We use the symbol ∞∞ for positive infinity and −∞−∞ for negative infinity. When we say that “x approaches infinity,” which can be symbolically written as x→∞x→∞, we are describing a behavior; we are saying that x is increasing without bound. With even-powered power functions, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as x x approaches positive or negative infinity, the f(x)f(x) values increase without bound. In symbolic form, we could write as x→±∞,f(x)→∞as x→±∞,f(x)→∞ The graph below shows f(x)=x 3,g(x)=x 5,h(x)=x 7,k(x)=x 9,and p(x)=x 11 f(x)=x 3,g(x)=x 5,h(x)=x 7,k(x)=x 9,and p(x)=x 11, which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function. As the power increases, the graphs flatten near the origin and become steeper away from the origin. These examples illustrate that functions of the form f(x)=x n f(x)=x n reveal symmetry of one kind or another. First, in the even-powered power functions, we see that even functions of the form f(x)=x n,n even,f(x)=x n,n even, are symmetric about the y-axis. In the odd-powered power functions, we see that odd functions of the form f(x)=x n,n odd,f(x)=x n,n odd, are symmetric about the origin. For these odd power functions, as x approaches negative infinity, f(x)f(x)decreases without bound. As x approaches positive infinity, f(x)f(x)increases without bound. In symbolic form we write as x→−∞,f(x)→−∞as x→∞,f(x)→∞as x→−∞,f(x)→−∞as x→∞,f(x)→∞ The behavior of the graph of a function as the input values get very small ( x→−∞x→−∞ ) and get very large ( x→∞x→∞ ) is referred to as the end behavior of the function. We can use words or symbols to describe end behavior. The table below shows the end behavior of power functions of the form f(x)=a x n f(x)=a x n where n n is a non-negative integer depending on the power and the constant. \| \| Even power \| Odd power \| --- \| Positive constant a> 0 \| \| \| \| Negative constant a< 0 \| \| \| How To: Given a power function f(x)=a x n f(x)=a x n where n n is a non-negative integer, identify the end behavior. Determine whether the power is even or odd. Determine whether the constant is positive or negative. Use the above graphs to identify the end behavior. Example: Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f(x)=x 8 f(x)=x 8. Show Solution The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x(input)approaches infinity, f(x)f(x) (output) increases without bound. We write as x→∞,f(x)→∞x→∞,f(x)→∞. As x approaches negative infinity, the output increases without bound. In symbolic form, as x→−∞,f(x)→∞x→−∞,f(x)→∞. We can graphically represent the function. Example: Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f(x)=−x 9 f(x)=−x 9. Show Solution The exponent of the power function is 9 (an odd number). Because the coefficient is –1 (negative), the graph is the reflection about the x-axis of the graph of f(x)=x 9 f(x)=x 9. The graph shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write as x→−∞,f(x)→∞x→−∞,f(x)→∞ and as x→∞,f(x)→−∞x→∞,f(x)→−∞. Try It Describe in words and symbols the end behavior of f(x)=−5 x 4 f(x)=−5 x 4. Show Solution As x approaches positive or negative infinity, f(x)f(x) decreases without bound: as x→±∞,f(x)→−∞x→±∞,f(x)→−∞ because of the negative coefficient. Contribute! Did you have an idea for improving this content? We’d love your input. Improve this pageLearn More Candela Citations CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at Question ID 69337. Authored by: Shahbazian, Roy. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 15940. Authored by: Sousa, James. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Ex: End Behavior or Long Run Behavior of Functions. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at Question ID 69337. Authored by: Shahbazian, Roy. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 15940. Authored by: Sousa, James. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Ex: End Behavior or Long Run Behavior of Functions. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution PreviousNext
41	https://www.cuemath.com/numbers/comparing-decimals/	LearnPracticeDownload Comparing Decimals Comparing decimals means finding out the larger and the smaller decimal number in a given set of numbers. Decimal numbers can be compared in the same way as we compare other numbers. However, we need to remember that the digits that are placed after the decimal point also need to be considered. These digits have place values starting from tenths, followed by hundredths, then thousandths, and so on. Let us learn more about comparing decimals in this article. \| \| \| --- \| \| 1. \| What is Comparing Decimals? \| \| 2. \| Comparing Decimals and Fractions \| \| 3. \| Comparing Decimals on a Number Line \| \| 4. \| Comparing Decimals to Hundredths place \| \| 5. \| FAQs on Comparing Decimals \| What is Comparing Decimals? Comparing decimals is similar to comparing other whole numbers in which we start comparing the digits with the greatest place value. We place the given decimal numbers in a place value chart and start the comparison. If the digits on the greatest place value are the same, we move on to the digits in the next place to the right. We keep comparing digits until we reach the digits that are different. Let us understand this with the help of the following example. Example: Compare 0.64 and 0.362 Solution: Step 1: Compare the whole number part, which is the ones digit. If the numbers are the same, go to the next step. In this case, the ones digits has 0 in both the numbers. So, we move on to the next place to the right. Step 2: Compare the tenths place, which is the place to the right of the decimal point. When we compare the value in the tenths place, we see that 6 is greater than 3. At this step itself, we get to know that 0.64 is greater than 0.362. Therefore, we do not need to move on to the hundredths digit for further comparison. Step 3: Hence, we conclude that 0.64 > 0.362 Comparing Decimals and Fractions When we need to compare decimals and fractions, we first convert the given fraction to a decimal number and then compare the numbers using the same procedure. Example: Compare 3/4 and 0.728 Solution: First, let us convert 3/4 to a decimal number by dividing 3 by 4. So, 3 ÷ 4 = 0.75. Now, we have both the numbers in the decimal form. So, let us compare 0.75 and 0.728 using the method given above. Step 1: Compare the whole number part, which is the ones digit. In this case, the ones digits has 0 in both the numbers. So, we will move on to the next place value to the right. Step 2: Compare the tenths place, which is the first place to the right of the decimal point. When we compare the value on the tenths digit, we see that both the numbers have 7. So we move on to the hundredths place. Step 3: Compare the hundredths place.Now, when we compare the value at the hundredths place, we see that 5 is greater than 2. At this step, we get to know that 0.75 is greater than 0.728. Therefore, we do not need to move on to the thousandths digit for further comparison. Step 3: Hence, we conclude that 0.75 > 0.728 which means 3/4 > 0.728 Comparing Decimals on a Number Line When we compare decimals on a number line, we follow the basic rule of the number line which says that, as we move to the right the value of the numbers increases. For example, if we need to compare 6.5 and 6.7, we mark the decimal numbers on the number line in such a way that both the numbers are included. We need to focus between 6 and 7 because both the given numbers lie between 6 and 7. We mark 6 at the left end and 7 at the right end. Then, we mark all the numbers in between, to scale, writing 6.5 midway between 6 and 7. After marking the other tenths, we see that 6.7 comes to the right side of 6.5, therefore 6.5 < 6.7. Hence, it be concluded that 6.5 < 6.7 because 6.7 comes to the right of 6.5 on the number line. Comparing Decimals to Hundredths place We have already seen that we compare decimal numbers starting from the whole number part and then we move on to the digits given after the decimal point. Let us see how to compare decimals to the hundredths place. Example: Compare 8.362 and 8.391 to find the greater number. Solution: Let us follow the same steps as shown above. Step 1: Compare the whole number part. In this case, the whole number part has 8 in both the numbers. So, we will move on to the next place value. Step 2: Compare the tenths place, which is the first place to the right of the decimal point. When we compare the value on the tenths digit, we see that both the numbers have 3. At this point, we move on to the hundredths place. Step 3: Compare the hundredths place.Now, when we compare the value at the hundredths place, we see that 9 is greater than 6. So, we get to know that 8.391 is greater than 8.362. Therefore, we do not need to move on to the thousandths digit for further comparison. Step 3: Hence, we conclude that 8.362 < 8.391 Related Links Check out the following pages related to comparing decimals. Decimals Decimals and Fractions Adding Decimals Comparing Decimals Examples Example 1: Which decimal number is greater between the two: 5.612 or 5.071? Solution: Comparing decimals starts with the comparison of the whole number part. We see that the ones place in both the numbers has equal value (5). So, we move on to see the tenths digit, which is 6 and 0 in the respective numbers. At this step, we can say that 5.612 is greater than 5.071. Hence, 5.612 > 5.071 2. Example 2: Compare the decimals from least to greatest: 17.102, 17.243 and 17.05 Solution: Comparing decimals from least to greatest means arranging the numbers in ascending order. So, when we start the comparison with the whole number part, we see that it is equal (17 in all the numbers). We move on to the tenths digit and find that all the numbers are different, and among 1, 2, and 0, the smallest number is 0. So, 17.05 is the least number among these. After this, we compare 17.102 and 17.243. We see that 17.102 < 17.243. Therefore, the given numbers can be arranged as 17.05, 17.102, and 17.243 in the order of least to greatest. View Answer > go to slidego to slide Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions Check Answer > go to slidego to slide FAQs on Comparing Decimals How do you Compare Decimals? Decimals are compared in the same way as we compare other whole numbers. The only point to be remembered is that we also need to consider the place values given after the decimal point. These place values start with tenths, followed by hundredths and thousandths, and so on. First, we compare the digits before the decimal point. If these digits are equal, then we move on to compare the digits after the decimal. If they are unequal, the comparison is done at that step itself and we don't move further for any comparison. In other words, we keep comparing the digits to the right until we get a set of unequal digits to compare. What are the Rules for Comparing Decimals? There are certain rules for comparing decimals. Let us understand them with the help of the following example. For example, let us compare 5.274 and 5.237 Step 1: First, compare the whole number part. If the numbers are the same, go to the next step. In this case, the whole number part (the ones digit) has 5 in both the numbers. So, we move on to the next place value to the right. Step 2: Compare the tenths place, which is the digit to the right of the decimal point. When we compare the value on the tenths digit, we see that both the numbers have 2. So, we again move on to the next place value which is the hundredths place. Step 3: Compare the hundredths place.Now, when we compare the value at the hundredths place, we see that 7 is greater than 3. At this step itself, we get to know that 5.274 is greater than 5.237. Therefore, we do not need to move on to the thousandths digit for further comparison. Step 3: Hence, we conclude that 5.274 > 5.237 How to Compare Decimals and Fractions? When we need to compare decimals and fractions, we first convert the given fraction to a decimal number and then compare them using the same procedure. For example, let us compare 0.528 and 3/7. In this case, we will convert 3/7 to a decimal number by dividing 3 by 7. So, 3 ÷ 7 = 0.428. Now, we can compare 0.428 and 0.528 because both the numbers are in the decimal form. So, we start the comparison from the whole number part, which is 0 in both the numbers. So, we move on to the tenths digit which is different in both the numbers. At this step, we can say that 0.528 is greater than 0.428. We do not need to go to the next digit for any further comparison. Hence 3/7 < 0.528. How to Compare Decimals from Least to Greatest? Comparing decimals from least to greatest means arranging them in ascending order in which the smallest number comes first, followed by the greater numbers. For example, let us compare a given set of numbers from least to greatest. The given numbers are 1.002, 0.112, 1.102. We will start the comparison from the largest place value, which is ones in this case. So, we can see that 0.112 has the least value in ones place, which means this is the smallest number in all, therefore, we will place it first. Now, we will compare the remaining two numbers, 1.002 and 1.102, in which the ones place is equal. So, we compare the tenths place which is 0 and 1 in the respective numbers. This shows that 1.002 is the smaller number. So, we will place it as the next number in the list, followed by 1.102. Hence, the given numbers can be arranged from least to greatest as, 0.112, 1.002, 1.102. How to Compare Decimals on a Number Line? When we compare decimals on a number line, we follow the basic rule of the number line which says that, as we move to the right, the value of the numbers increases. For example, if we need to compare 4.3 and 4.7, we mark the decimal numbers on the number line in such a way that both the numbers are included. We see that 4.7 comes to the right of 4.3. Therefore, 4.7 is greater than 4.3 How to Compare Decimals to Hundredths Place? We compare decimals to the hundredths place when we have checked all the place values before it, that is, the whole number part, and the tenths place. If the numbers in these places are equal, we come to the hundredths place for comparing the numbers. Let us understand this with an example. Let us compare 7.14 and 7.16 Step 1: First, let us compare the whole number part. In this case, the whole number part has 7 in both the numbers. So, we will move on to the next place value. Step 2: Now, when we compare the tenths place, which is the first place to the right of the decimal point, we find that both the numbers have 1. At this point, we move on to the hundredths place. Step 3: We compare the hundredths place and see that 6 is greater than 4. So, we get to know that 7.16 is greater than 7.14. Step 3: Hence, we conclude that 7.16 > 7.14 How to Compare Decimals to Thousandths Place? Comparing decimals to thousandths place is similar to the comparison to hundredths place. In this case, when we have compared the numbers up to the place of the hundredths digit, and we find that both the numbers are equal, we come to the thousandths place for the comparison. On this place, if the digits are different, we can compare the numbers and get to know the larger number. 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42	https://math.stackexchange.com/questions/79738/proving-the-trig-identity-ei-alpha-ei-beta-2ei-alpha-beta-2	trigonometry - Proving the Trig Identity $e^{i \alpha} + e^{i \beta} = 2e^{i(\alpha + \beta)/2}\cos((\alpha-\beta)/2)$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Proving the Trig Identity e i α+e i β=2 e i(α+β)/2 cos((α−β)/2)e i α+e i β=2 e i(α+β)/2 cos⁡((α−β)/2) Ask Question Asked 13 years, 11 months ago Modified13 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. We're having trouble proving this trig identity: e i α+e i β=2 e i(α+β)2 cos(α−β 2)e i α+e i β=2 e i(α+β)2 cos⁡(α−β 2) We've tried various manipulations using basic trig identities (too lengthy to show here) and come up with nasty equations that don't lead anywhere. Can you point us in the right direction? trigonometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 7, 2011 at 4:04 Michael Hardy 1 asked Nov 7, 2011 at 3:32 smackcranesmackcrane 3,311 2 2 gold badges 26 26 silver badges 35 35 bronze badges 2 I assume α,β α,β are real.N. S. –N. S. 2011-11-07 03:34:35 +00:00 Commented Nov 7, 2011 at 3:34 @N.S. yes, they are smackcrane –smackcrane 2011-11-07 03:35:37 +00:00 Commented Nov 7, 2011 at 3:35 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. You need to show that cos(α)+cos(β)+i[sin(α)+sin(β)]=2[cos(α+β 2)+i sin(α+β 2)]cos(α−β 2).cos⁡(α)+cos⁡(β)+i[sin⁡(α)+sin⁡(β)]=2[cos⁡(α+β 2)+i sin⁡(α+β 2)]cos⁡(α−β 2). Identify the real and imaginary parts, and use the formulas for cos(x)cos(y)cos⁡(x)cos⁡(y) and sin(x)cos(y)sin⁡(x)cos⁡(y) respectively... Alternately, you can use that α=α+β 2+α−β 2 α=α+β 2+α−β 2 and β=α+β 2−α−β 2 β=α+β 2−α−β 2. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 7, 2011 at 3:38 N. S.N. S. 135k 12 12 gold badges 150 150 silver badges 271 271 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. If you use the fact that cos((α−β)/2)=exp(i(α−β)/2)+exp(−i(α−β)/2)2 cos⁡((α−β)/2)=exp⁡(i(α−β)/2)+exp⁡(−i(α−β)/2)2 lots of things cancel. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 7, 2011 at 3:42 Ross MillikanRoss Millikan 384k 28 28 gold badges 264 264 silver badges 472 472 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Just work on the right side, it is easy to show using definition of cosine over the complex numbers: c o s(z)=e i z+e−i z 2 c o s(z)=e i z+e−i z 2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 7, 2011 at 3:43 mboratkomboratko 4,603 28 28 silver badges 36 36 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry See similar questions with these tags. 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43	https://www.mdpi.com/2077-0383/13/14/3996	Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management Next Article in Journal Understanding Gastric GIST: From Pathophysiology to Personalized Treatment Next Article in Special Issue Adenomyosis: An Update Concerning Diagnosis, Treatment, and Fertility Previous Article in Journal Arthroscopic Assessment of Temporomandibular Joint Pathologies—Is It Possible for Non-Specialists in Arthroscopy? 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Villanacci, R. D’Alessandro, S. He, X. Grisafi, G. Ferrari, S. Maria Candiani, M. on Google Scholar Ottolina, J. Villanacci, R. D’Alessandro, S. He, X. Grisafi, G. Ferrari, S. Maria Candiani, M. on PubMed Ottolina, J. Villanacci, R. D’Alessandro, S. He, X. Grisafi, G. Ferrari, S. Maria Candiani, M. /ajax/scifeed/subscribe Article Views 6378 Citations 3 Table of Contents Abstract Introduction Materials and Methods Results Discussion Conclusions Author Contributions Funding Institutional Review Board Statement Informed Consent Statement Data Availability Statement Conflicts of Interest References Altmetricshare Shareannouncement Helpformat_quote Citequestion_answer Discuss in SciProfiles Need Help? Support Find support for a specific problem in the support section of our website. Get Support Feedback Please let us know what you think of our products and services. Give Feedback Information Visit our dedicated information section to learn more about MDPI. Get Information clear JSmol Viewer clear first_page Download PDF settings Order Article Reprints Font Type: Arial Georgia Verdana Font Size: Aa Aa Aa Line Spacing:    Column Width:    Background: Open Access Review Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management by Jessica Ottolina Jessica Ottolina SciProfilesScilitPreprints.orgGoogle Scholar 1,, Roberta Villanacci Roberta Villanacci SciProfilesScilitPreprints.orgGoogle Scholar 1, Sara D’Alessandro Sara D’Alessandro SciProfilesScilitPreprints.orgGoogle Scholar 1, Xuemin He Xuemin He SciProfilesScilitPreprints.orgGoogle Scholar 2, Giorgia Grisafi Giorgia Grisafi SciProfilesScilitPreprints.orgGoogle Scholar 1, Stefano Maria Ferrari Stefano Maria Ferrari SciProfilesScilitPreprints.orgGoogle Scholar 1 and Massimo Candiani Massimo Candiani SciProfilesScilitPreprints.orgGoogle Scholar 1 1 IRCCS San Raffaele Scientific Institute, Via Olgettina 60, 20132 Milan, Italy 2 Obstetrics and Gynecology Unit, Department of Biomedical Sciences and Human Oncology, University of “Aldo Moro”, 70124 Bari, Italy Author to whom correspondence should be addressed. J. Clin. Med.2024, 13(14), 3996; Submission received: 10 April 2024 / Revised: 17 June 2024 / Accepted: 19 June 2024 / Published: 9 July 2024 (This article belongs to the Special Issue Endometriosis and Adenomyosis: Modern Concepts on Clinical Outcomes, Treatment and Management) Download keyboard_arrow_down Download PDF Download PDF with Cover Download XML Download Epub Browse Figure Versions Notes Abstract Endometriosis and adenomyosis are complex gynecological conditions characterized by diverse clinical presentations, including superficial peritoneal endometriosis (SPE), ovarian endometrioma (OMA), and deep infiltrating endometriosis (DIE). The hallmark features of these pathologies involve the manifestation of pain symptoms and infertility, and approximately 30% of patients are asymptomatic. Despite ongoing research, definitive treatments for these conditions remain elusive, and clinical management primarily revolves around medical or surgical interventions. Recent advancements in our understanding of the efficacy of various treatment modalities, including medical therapy and surgical interventions, have provided clinicians with valuable insights into pain relief and fertility preservation. This review aims to provide an updated overview of the latest literature on clinical outcomes, treatment options, and management strategies for different types of endometriosis. By synthesizing the newest available data, this review seeks to inform clinicians and guide decision making based on factors such as patients’ symptom severity, childbearing desire, and overall health. Keywords: endometriosis; superficial peritoneal endometriosis; ovarian endometriosis; deep infiltrating endometriosis; adenomyosis 1. Introduction Endometriosis is a chronic inflammatory disease that significantly impacts a patient’s quality of life, social interactions, and work. It is primarily characterized by pain, often associated with the menstrual cycle, and infertility. The exact prevalence of endometriosis is unknown but estimates range from 2% to 10% of the general female population and up to 70% of infertile women [1,2]. Endometriosis is often categorized as peritoneal endometriosis, deep infiltrating endometriosis (DIE), or ovarian endometriosis based on the type and location of lesions. Adenomyosis is another related condition characterized by the infiltration of endometrial glands and stroma into the myometrium, leading to the hypertrophy of adjacent myometrial tissue. Definitive treatment for endometriosis and adenomyosis is lacking, and management typically involves medical or surgical interventions. In certain cases, international and national guidelines agree on first-line treatment options, but in other cases, there remains no consensus within the scientific community. New evidence from recent studies has helped improve our understanding and treatment of endometriosis-related pain and infertility . This narrative review aims to provide an overview of the latest data, analyzing recent guidelines, randomized controlled trials (RCTs), cohort studies, and meta-analyses on the clinical outcomes, treatment, and management of different types of endometriosis. 2. Materials and Methods A comprehensive literature search was conducted in PubMed, covering the period from January 2000 to December 2023. The research utilized combinations of the search terms “ovarian endometriosis”, “superficial peritoneal endometriosis”, “deep infiltrating endometriosis”, and “adenomyosis”, in conjunction with “infertility”, “pain”, and “quality of life”. The inclusion criteria were limited to publications in English. The selection criteria encompassed all published randomized controlled trials (RCTs) and non-randomized studies (NRSs), including observational studies, prospective and retrospective cohort studies, and case-control studies, specifically focusing on the treatment of pain and infertility in endometriosis. Both original research articles and review articles were considered. Additional relevant articles were identified by examining the references of the retrieved reviews and utilizing PubMed’s “similar articles” function for all of the selected records. After removing duplicates and articles that were not pertinent to the topic, a total of 59 papers were identified for inclusion in this narrative review (Figure 1). Figure 1. PRISMA flow diagram of the article selection process. 3. Results 3.1. Endometriosis 3.1.1. Ovarian Endometriosis Ovarian endometriomas or “chocolate cysts” are ovarian cysts filled with menstrual blood. The endometrioma wall usually comprises a flattened columnar epithelium with an endometrial-type stroma, surrounded by fibroreactive tissue with haemosiderin-laden macrophages. However, histologic analyses have shown that endometrial-like tissue often is minimal or absent (especially in larger cysts), and only fibrous tissue can be detected . Endometriomas affect approximately 17–44% of patients with endometriosis . These cysts are most often unilateral, with a mean cyst wall thickness of 1.2–1.5 mm. When bilateral, endometriomas are associated with more extensive disease and posterior cul-de-sac obliteration. The left ovary is more commonly affected, potentially due to the anatomical position of the sigmoid colon, which allows endometrial fragments from the uterine cavity to persist and implant on the left ovary. Ovarian endometriomas may be associated with infertility, dysmenorrhea, chronic pelvic pain, ovarian masses, or bowel upset (i.e., swelling or fullness) . Management, Treatment, and Clinical Outcomes Evidence-based guidelines recommend medical therapy as the first-line approach for patients with endometrioma-related pain who do not desire pregnancy. Medical treatment can improve patients’ quality of life and reduce cyst size and generally poses a low level of risk . Surgery is considered for patients with large endometriomas (e.g., size > 3 cm), those whose symptoms do not adequately improve or worsen, or those whose imaging results suggest malignancy. Medications that can reduce the size and symptoms of endometriomas include progestins, oral estrogen–progestin contraceptive pills, and gonadotropin-releasing hormone (GnRH) agonists or antagonists. The choice of medication is based on patient preferences, including considerations related to contraception use, side effects, drug availability, and cost. For example, in young women, oral contraceptives (OCs) are often preferred because they provide adequate estrogen support, which is essential for improving their quality of life . It is advisable to use preparations with the lowest possible estrogen content to prevent endometrial proliferation . Continuous-dose OC regimens may provide additional benefits over cyclic ones. A continuous OC schedule is in fact associated with lower dysmenorrhea and/or endometrioma recurrence rates compared with a cyclic schedule . Progestin monotherapy (e.g., dienogest or norethindrone acetate (NETA)) is historically favored in women who do not respond to combined hormone therapy or as a first-line therapy in smokers over 35 years old, women with a history of migraines, and those with predisposing risk factors for thromboembolic events . Patients with more severe symptoms or whose symptoms are not adequately controlled with the above treatments are offered a trial of GnRH antagonists . GnRH antagonists (e.g., elagolix or relugolix) suppress pituitary gonadotropin hormone production and create a dose-dependent hypoestrogenic state to inhibit endometriotic cell proliferation. Compared to GnRH analogues, antagonists do not provoke the initial flare-up phase, and the onset of their therapeutic effect is rapid . They also have the advantage of being administered orally. Symptom relief and adverse events, such as vasomotor phenomena, vaginal atrophy, and bone loss, are also dose-dependent. When used with add-back hormone therapy, adverse effects such as menopause-like symptoms and bone density decrease are reduced. According to Wang et al. , two recently FDA-approved doses of elagolix for the management of moderate-to-severe pain associated with endometriosis were both proven to be cost effective compared to leuprolide acetate (i.e., GnRh analogues) over a time frame of 1–2 years. However, studies comparing the efficacy of GnRH antagonists with OC and progestins are required. For patients with infertility, additional considerations are necessary. These women can be managed with either surgery or in vitro fertilization (IVF) . Currently, the decision between surgery and IVF is shared and tailored to the patient’s specific circumstances, including their history of previous surgeries, pain symptoms, age, ovarian reserve testing results, and semen analysis . The lack of randomized trials comparing these two strategies makes precise estimations challenging. Surgery might increase the chances of natural conception and relieve symptoms [17,18]. IVF may be more effective but involves higher costs and specific risks during the IVF procedure and pregnancy for women with endometriosis who do not undergo surgery. The presence of the endometrioma may in fact interfere with ovarian responsiveness to hyperstimulation and oocyte competence, the retrieval of the oocytes may be more difficult and riskier, the disease may progress during the procedure, one’s pregnancy outcome may be affected, and there is a risk of missing occult malignancies with cancer development later in life . An ongoing randomized controlled trial (RCT) by Ottolina et al. aims to provide clinically relevant findings in this context . When performing surgery on women with ovarian endometriosis, clinicians may consider both cystectomy and CO 2 laser vaporization, as both techniques have similar recurrence and reproductive rates [21,22]. In CO 2 laser vaporization, the “pseudo-capsule” of the cyst is not removed but ablated with the CO 2 laser. This laser is highly selective and precise, with a minimal depth of tissue penetration and little lateral thermal spread, reducing the risk of unintended thermal damage. Additionally, the CO 2 laser allows for the continuous visualization of the section plane between healthy and endometriosis-affected tissue. These properties are crucial for preserving the surrounding viable ovarian tissue. The CO 2 laser also simultaneously cauterizes bleeding tissue, providing effective hemostasis without the risks associated with cautery. The introduction of the CO 2 fiber laser has made the treatment of endometriotic cysts feasible even for non-expert surgeons . 3.1.2. Superficial Peritoneal Endometriosis Superficial peritoneal endometriosis is characterized by the presence of ectopic endometrium-like tissue outside the uterus, extending up to 5 mm under the peritoneal pelvic surface and/or the serosa of the pelvic viscera . The prevalence of this condition is difficult to assess due to the poor sensitivity of non-invasive diagnostic tools, such as transvaginal ultrasound . Despite this, it is the most common subtype of endometriosis, present in up to 80% of women affected by the disease and representing the only clinical finding in 30% of laparoscopically confirmed endometriosis cases [25,26]. Clinically, superficial endometriosis can be asymptomatic in 40% of cases and it is found when women undergo laparoscopies for other clinical reasons . However, it has been associated with primary infertility, moderate to severe dysmenorrhea, and deep dyspareunia . Given these associations, superficial endometriosis should be suspected in women of reproductive age presenting with infertility and/or painful symptoms, even in the absence of diagnostic clinical findings, and empiric treatment should be initiated. Management, Treatment, and Clinical Outcomes For symptomatic women with suspected superficial endometriosis who do not desire pregnancy, the first-line treatment recommended by international guidelines is medical hormonal therapy. Options include combined hormonal contraceptives, progestins, GnRH agonists, or GnRH antagonists, all of which aim to suppress estrogen stimulation. The choice of treatment is based on their effectiveness, side effects, long-term safety, cost, and availability . However, up to 30% of women do not achieve symptom relief with medical therapy, and some patients have contraindications to hormonal treatment [1,28]. In such cases, a laparoscopy should be considered. Minimally invasive surgical techniques, such as excision and ablation through coagulation or CO 2 laser vaporization, are available. Current guidelines recommend surgical excision over ablation to reduce endometriosis-associated pain and provide histological confirmation . Recent randomized controlled trials (RCTs) have compared the effectiveness of laparoscopic excision and ablation for superficial endometriosis-related pain. One trial found a significant improvement in dyspareunia after ablation at 6 months, but this improvement was not sustained at 12 months, with excision and ablation showing similar levels of effectiveness overall for pain management [29,30,31]. Superficial endometriosis is also associated with primary infertility, though not necessarily due to a diminished ovarian reserve, as serum antimüllerian levels and antral follicle counts are comparable to those of women without endometriosis . Their infertility may instead be due to factors such as distorted anatomy, inflammation, and epigenetic changes . The management of infertility in women with superficial endometriosis remains debated. The European Society of Human Reproduction and Embryology (ESHRE) recommends surgical treatment for endometriosis-associated infertility in stage I/II endometriosis due to an improvement in the rate of spontaneous ongoing pregnancy, though an increase in live birth rates has not been demonstrated . Assisted reproductive technologies (ART) can be considered before or after surgery if there are issues such as compromised tubal function, male factor infertility, or a poor prognosis . Thus, decisions on managing infertility related to superficial endometriosis should be individualized, considering the presence of painful symptoms, the patient’s age, preferences, and available fertility treatment options . 3.1.3. Deep Infiltrating Endometriosis Deep infiltrating endometriosis (DIE) is the most severe form of endometriosis, characterized by the involvement of either retroperitoneal structures or the peritoneum to a depth of at least 5 mm . It affects approximately 1% of women of reproductive age and up to 20% of women diagnosed with endometriosis . Common locations include the torus uteri, posterior fornix, uterosacral ligaments, vagina, bowel, and urinary tract. DIE significantly reduces the chances of natural conception, likely due to a combination of anatomical distortion, dyspareunia, and pelvic inflammation. Management, Treatment, and Clinical Outcomes When creating a treatment plan for DIE, clinicians must consider the patient’s symptoms, age, desire for future pregnancy, location and type of lesions, and the risks and benefits of both medical and surgical therapies. Evidence-based guidelines recommend medical therapy as the first approach for patients with deep endometriosis-related pain who do not desire pregnancy [38,39]. Exceptions include patients with significant bowel or ureteral obstruction, who require immediate surgical intervention . Medical treatment can substantially improve symptoms (approximately 70–80%) and avoid the risks associated with surgery, such as genital tract fistulae, bladder dysfunction, or bowel anastomotic leakage . Patients eligible for hormonal treatment should be counseled about the potential need for surgery in case of treatment failure (approximately 10%) and the possible development of a bowel or ureteral obstruction (1–2%). Both combined estrogen–progestin and progestin-only regimens are effective in reducing symptoms. Some authors suggest progestin-only regimens as a first-line treatment (e.g., norethindrone acetate [NETA] (2.5 mg) or dienogest (2 mg)) . Additional medical treatment options include the short-term use of gonadotropin-releasing hormone agonists . In cases of infertility-related endometriosis, clinicians must choose between surgical intervention and assisted reproductive technologies (ARTs). The impact of surgery for DIE on fertility is controversial, with the literature presenting heterogeneous findings. A Brazilian prospective study comparing IVF outcomes in women who underwent extensive endometriosis excision before IVF (64 patients) and those who only underwent IVF (105 patients) found significantly higher implantation (32.1% vs. 19%) and pregnancy rates (41% vs. 24%) in the surgical group, despite higher r-FSH dosages and fewer oocytes being retrieved . However, the study did not specify the types of endometriotic lesions in both groups. A retrospective observational study evaluated factors associated with pregnancy during the first two IVF attempts in infertile women with posterior deep endometriosis (lesions in the torus, uterosacral ligaments, recto-uterine space, rectovaginal septum, and/or bowels), finding a 48.7% pregnancy rate after two IVF attempts, with lower chances of pregnancy in the presence of a recto-uterine nodule [44,45]. Operative laparoscopy may be a treatment option for symptomatic patients wishing to conceive, though some patients may require surgery prior to ART to free the ovaries from adhesions, facilitating oocyte retrieval. Surgical decisions should be made by a multidisciplinary team with a high level of surgical expertise to minimize risks and maximize benefits. There is no consensus on the best surgical treatments for various forms of DIE. The critical point is to balance radicality and conservativity to improve patient quality of life while minimizing surgical complications . Surgeons should identify, isolate, and remove all sites of endometriosis in the pelvis, following nerve-sparing principles, especially in posterior compartment DIE . Sparing hypogastric nerves and inferior hypogastric plexuses is crucial to avoid autonomic nervous injury, which can cause urinary, anorectal, and sexual dysfunction . Surgical techniques for bowel endometriosis include shaving, discoid excision, or segmental resection . A recent meta-analysis found that shaving had the lowest postoperative complication rate but a higher incidence of recurrence compared to discoid excision and segmental resection [49,50,51]. An RCT (EndoRE trial) comparing conservative (shaving and disc excision) and radical (segmental bowel resection) approaches found high conception rates during follow-up (82.4% with radical and 85% with conservative approaches), with a 75.7% live birth rate and a 57.4% spontaneous pregnancy rate (SPR) [50,52]. Positive results were previously obtained by Bourdel et al., who reported a 49% SPR after shaving and 67% after resection, and by Donnez et al., who reported a 57% SPR after the shaving technique [53,54]. However, two prospective studies did not show significant differences [55,56]. An observational study by Stepniewska et al. found a cumulative pregnancy rate of 58% among women younger than 30 years and 45% in those aged 30–34 years following laparoscopic colorectal resection for severe intestinal symptoms in infertile women with endometriosis . More recently, a retrospective matched cohort study concluded that first-line surgery may be a good option for women with colorectal endometriosis-associated infertility compared with IVF without surgery . However, the groups were heterogeneous in terms of the surgery performed, including rectal shaving or segmental colorectal surgery based on infiltration extent, ovarian endometrioma cystectomy or ablation using plasma energy, salpingectomy, uterosacral ligament resection, partial colpectomy, partial bladder resection, ureterolysis, and ureteral reimplantation. Regarding urinary tract endometriosis, surgical techniques depend on the type and location of lesions . Ureterolysis is preferred for extrinsic ureteral endometriosis, while ureteral resection with end-to-end anastomosis is necessary for intrinsic ureteral endometriosis when conservative approaches fail. For lesions near the vesical–ureteral junction, ureteral reimplantation is considered. Surgery for bladder endometriosis involves a partial cystectomy, with or without the opening of the bladder cavity . Bladder DIE often coexists with other locations, complicating assessments of its impact on infertility and surgical outcomes . Retrospective studies have reported spontaneous pregnancy rates (SPR) ranging from 30% to 86%, with no significant differences between complete or partial thickness bladder DIE [61,62]. In summary, managing DIE involves careful consideration of various factors, with a multidisciplinary approach to optimize patient outcomes. Both medical and surgical treatments can be used and tailored to individual patient needs and circumstances. 3.2. Adenomyosis Uterine adenomyosis involves the infiltration of endometrial glands and stroma into the myometrium, resulting in the hypertrophy of adjacent myometrial tissue . This condition presents as a heterogeneous pathology, with diffuse, focal, and adenomyoma phenotypes within the myometrium. The exact cause of adenomyosis remains unknown, but recent theories suggest the mechanism of tissue injury and repair (TIAR) and stem cell theory. The dysfunction or absence of the endomyometrial junctional zone (JZ) may lead to internal or intrinsic adenomyosis through the invagination of the basal layer of the endometrium into the myometrium . However, an alternative theory suggests that adenomyotic lesions may develop from the metaplasia of misplaced embryonic pluripotent Mullerian remnants or indeed from the differentiation of adult stem cells . Currently, adenomyosis diagnosis primarily relies on non-invasive imaging techniques, particularly pelvic ultrasound and pelvic MRI, both of which demonstrate high levels of sensitivity and specificity. In a meta-analysis involving over 3300 patients with confirmed adenomyosis, transvaginal ultrasound (TVUS) exhibited a sensitivity and specificity of 81% and 87%, respectively, while MRI showed a 71% sensitivity and 91% specificity . A consensus-based practical classification of adenomyosis has been proposed by T. Van Den Bosch et al., based on ultrasound findings. This classification system includes the following: Identification of the presence of adenomyosis using the MUSA criteria ; Determination of the location of the adenomyosis; Differentiation between focal and diffuse disease; Discrimination between cystic and non-cystic lesions; Determination of myometrial layer involvement; Classification of disease extent as mild, moderate, or severe; Measurement of lesion size. However, international consensus on adenomyosis classification is still lacking. Moreover, studies have highlighted associations between sonographic features of adenomyosis and clinical presentation. Diffuse adenomyosis of the internal myometrium tends to occur more frequently in older, multiparous women, often associated with prior uterine surgery . Conversely, focal adenomyosis of the external myometrium is more prevalent in younger women and is often linked to deep endometriosis . The presence of abnormal uterine bleeding and infertility varies according to adenomyosis phenotype, with diffuse adenomyosis being more strongly associated with abnormal bleeding and external adenomyosis more prevalent in cases of infertility . Assessing adenomyosis’s impact on reproductive outcomes is challenging due to the limited differentiation between endometriosis and adenomyosis in many studies. However, evidence suggests that adenomyosis may have a greater impact on the likelihood of conception and the risk of miscarriage than endometriosis alone. Additionally, women with adenomyosis face an increased risk of developing preeclampsia and delivering small-for-gestational-age (SGA) infants, likely due to the mechanisms of poor placentation and endothelial inflammation . In summary, the diagnosis and classification of adenomyosis rely heavily on non-invasive imaging techniques, although international consensus on classification is lacking. Adenomyosis phenotype correlates with clinical presentation, and further research is needed to better understand its impact on reproductive outcomes. Management, Treatment, and Clinical Outcome The management of adenomyosis focuses on symptom alleviation, improving patients’ quality of life, and addressing associated complications (e.g., ureteral obstructions). Treatment choices depend on factors such as symptom severity, fertility desires, and overall health. Hormonal therapies are commonly used to alleviate adenomyosis symptoms by suppressing menstruation, reducing endometrial tissue growth, and diminishing inflammatory cytokines and prostaglandins. The most effective first-line therapy for symptomatic patients not desiring pregnancy is the 52 mg levonorgestrel-releasing intrauterine device (LNG 52). It acts directly on the uterus, maintains low systemic hormone levels, and offers long-acting, user-independent administration . Observational studies indicate its efficacy in improving heavy menstrual bleeding and dysmenorrhea associated with adenomyosis . Other hormonal strategies, such as progestins, oral contraceptive pills, and gonadotropin-releasing hormone (GnRH) agonists and antagonists, are effective alternatives for managing heavy menstrual bleeding and dysmenorrhea in patients unable to use LNG 52. However, data on their efficacy are limited. Adenomyomectomy may be considered for symptomatic patients with focal adenomyosis who desire fertility preservation. However, adenomyomas are challenging to excise due to difficulties in establishing a surgical plane and the woody consistency of the adenomyotic uterus. Additionally, suturing is difficult in this environment. Even if performed by expert surgeons, adenomyomectomy carries a risk of uterine rupture in future pregnancies, reported to be around 4% [73,74]. For patients who have completed childbearing or for whom medical therapies are ineffective or contraindicated, hysterectomies remain as a definitive treatment option . A hysterectomy may be considered a valid option in cases of adenomyosis, especially when other treatments have failed to alleviate symptoms such as severe pain and heavy menstrual bleeding. However, its effectiveness in resolving pain is not guaranteed. The persistence of pain can result from residual endometriotic lesions, central sensitization, or hormonal influences if the ovaries are retained. Therefore, a comprehensive evaluation and an individualized treatment plan, often involving a combination of surgical and non-surgical approaches, are essential for optimal management. Uterine artery embolization (UAE) is a non-operative alternative treatment for adenomyosis, aimed at reducing associated symptoms. Studies have shown a 25% decrease in uterine volume and improvement in abnormal bleeding and overall symptoms following UAE [76,77]. Acupuncture has been explored as a complementary therapy for managing adenomyosis symptoms. However, evidence regarding its effectiveness remains limited and inconclusive . The goal of adenomyosis management is to alleviate symptoms, improve patient quality of life, and address complications while considering patient preferences and fertility desires. Treatment outcomes vary depending on the chosen approach: hormonal therapies offer symptom relief, adenomyomectomies preserve fertility, and hysterectomies are used when other treatments have failed to alleviate severe pain and heavy menstrual bleeding. Non-operative options like UAE and complementary therapies like acupuncture offer alternative symptom management strategies. Further research is needed to assess the long-term efficacy and outcomes of these treatments. 3.3. Alternative Approaches for Managing Endometriosis-Related Symptoms Holistic, complementary, and alternative approaches for managing endometriosis often focus on relieving symptoms and improving overall well-being. Some approaches include the following: Dietary Changes: Certain diets, such as an anti-inflammatory diet rich in fruits, vegetables, whole grains, and healthy fats, may help alleviate symptoms. Some people find relief by avoiding trigger foods like dairy or gluten. Acupuncture: Acupuncture, a traditional Chinese medicine practice, involves inserting thin needles into specific points on the body to alleviate pain and promote relaxation. Some women with endometriosis find acupuncture helpful for managing pain and stress. Herbal Supplements: Certain herbs like turmeric, ginger, and chasteberry (vitex) are believed to have anti-inflammatory properties and may help reduce pain associated with endometriosis. However, it is essential to consult with a healthcare provider before trying herbal supplements, as they can interact with medications or have side effects. Mind–Body Practices: Techniques such as yoga, meditation, and mindfulness can help manage stress, improve sleep quality, and promote relaxation. These practices may help reduce pain and enhance overall well-being in individuals with endometriosis. Physical Therapy: Pelvic floor physical therapy can help address pelvic pain and dysfunction associated with endometriosis. Therapists use techniques such as manual therapy, stretching, and strengthening exercises to alleviate pain and improve pelvic function. Supplements: Some supplements, such as omega-3 fatty acids, magnesium, and vitamin D, may help reduce inflammation and alleviate symptoms of endometriosis. However, it is essential to consult with a healthcare provider before taking supplements to ensure they are safe and appropriate for one’s specific situation. Stress Management: Chronic stress can exacerbate symptoms of endometriosis. Practices like deep breathing exercises, progressive muscle relaxation, and biofeedback can help manage stress levels and reduce the impact of stress on symptoms. It is important to note that while these holistic approaches may provide symptom relief for some individuals, they should not replace conventional medical treatment. 4. Discussion Endometriosis and adenomyosis present complex challenges in terms of their diagnosis, management, and treatment, often requiring a multidisciplinary approach. Despite extensive research, certain aspects of their management remain controversial, necessitating ongoing studies to refine treatment strategies and improve patient outcomes. Need for Specialized Care: The complexity of endometriosis and adenomyosis necessitates referral to specialized facilities in which patients can benefit from specific high-quality management, from early diagnosis to the treatment of severe disease. Given the significant impact of endometriosis on women’s daily lives and its economic burden, gynecological societies and international organizations advocate for the establishment of expert centers throughout the country, formally accredited by health authorities, ideally as part of a National Health Plan . Challenges in Managing Infertility: Infertility associated with endometriosis poses a significant challenge. The decision between surgery and in vitro fertilization (IVF) depends on various factors, including previous surgical history, pain symptoms, age, and ovarian reserve. Robust prospective studies, such as multicenter randomized controlled trials (RCTs), are essential to compare the effectiveness of these approaches and guide clinical decision making. Ongoing Clinical Trials: The ongoing multicenter RCTs comparing IVF and surgery for ovarian endometriosis and deep infiltrating endometriosis (DIE) will provide valuable insights into their respective benefits and risks, particularly in terms of live birth rates . Additionally, this research will also include an experimental part aimed at assessing whether the systemic inflammatory environment of endometriosis may have a detrimental impact on the quality of folliculogenesis and embryological development. Future Directions in Adenomyosis Research: Adenomyosis poses unique challenges due to its heterogeneity and limited treatment options, especially in cases in which pregnancy is desired. The lack of RCTs comparing different treatment strategies highlights the need for further research to identify optimal management approaches. Studies investigating the role of gonadotropin-releasing hormone (GnRH) agonists in improving assisted reproductive technology (ART) outcomes in patients with severe adenomyosis offer promising insights. However, more research is needed to establish standardized protocols and evaluate their efficacy. Obstetric Complications: Women with severe endometriosis and adenomyosis face increased risks of obstetric complications, including preeclampsia, antepartum hemorrhage, preterm birth, and miscarriage. Understanding these associations is crucial for optimizing prenatal care and reducing maternal and fetal risks. 5. Conclusions Ongoing research efforts, including RCTs and multicenter cohort studies, are essential for advancing our understanding and management of endometriosis and adenomyosis. Collaborative initiatives involving multidisciplinary teams will facilitate the development of evidence-based guidelines and improve patient care outcomes. Author Contributions Conceptualization, J.O. and R.V.; methodology, J.O. and R.V.; validation, S.M.F. and M.C.; resources, R.V., G.G., S.D. and X.H.; writing—original draft preparation, J.O., R.V., G.G., S.D. and X.H.; writing—review and editing, J.O. and R.V.; visualization, S.M.F. and M.C.; supervision, S.M.F. and M.C. All authors have read and agreed to the published version of the manuscript. Funding This research received no external funding. Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. Data Availability Statement Not applicable. Conflicts of Interest The authors declare no conflicts of interest. 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Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management. Journal of Clinical Medicine. 2024; 13(14):3996. Chicago/Turabian Style Ottolina, Jessica, Roberta Villanacci, Sara D’Alessandro, Xuemin He, Giorgia Grisafi, Stefano Maria Ferrari, and Massimo Candiani. 2024. "Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management" Journal of Clinical Medicine 13, no. 14: 3996. APA Style Ottolina, J., Villanacci, R., D’Alessandro, S., He, X., Grisafi, G., Ferrari, S. M., & Candiani, M. (2024). Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management. Journal of Clinical Medicine, 13(14), 3996. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. Article Metrics No No Article Access Statistics For more information on the journal statistics, click here. Multiple requests from the same IP address are counted as one view. Zoom\|Orient\|As Lines\|As Sticks\|As Cartoon\|As Surface\|Previous Scene\|Next Scene Cite Export citation file: BibTeX) MDPI and ACS Style Ottolina, J.; Villanacci, R.; D’Alessandro, S.; He, X.; Grisafi, G.; Ferrari, S.M.; Candiani, M. Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management. J. Clin. Med.2024, 13, 3996. AMA Style Ottolina J, Villanacci R, D’Alessandro S, He X, Grisafi G, Ferrari SM, Candiani M. Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management. Journal of Clinical Medicine. 2024; 13(14):3996. Chicago/Turabian Style Ottolina, Jessica, Roberta Villanacci, Sara D’Alessandro, Xuemin He, Giorgia Grisafi, Stefano Maria Ferrari, and Massimo Candiani. 2024. "Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management" Journal of Clinical Medicine 13, no. 14: 3996. APA Style Ottolina, J., Villanacci, R., D’Alessandro, S., He, X., Grisafi, G., Ferrari, S. M., & Candiani, M. (2024). Endometriosis and Adenomyosis: Modern Concepts of Their Clinical Outcomes, Treatment, and Management. Journal of Clinical Medicine, 13(14), 3996. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. clear J. Clin. 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44	https://stackoverflow.com/questions/9859737/matrix-with-equal-sum-of-rows-and-columns	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Matrix with equal sum of rows and columns Ask Question Asked Modified 13 years, 5 months ago Viewed 10k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I have NxM matrix with integer elements, greater or equal than 0. From any cell I can transfer 1 to another one (-1 to the source cell, +1 to the destination). Using this operation, I have to make sums for all rows and columns equal. The question is how to find the minimal amount of such operations to achieve my task. During the processing cells may be negative. For example, for 1 1 2 2 1 0 1 1 0 0 1 1 1 1 1 2 The answer is 3. P.s.: I've tried to solve it on my own, but came only to brute-force solution. algorithm math matrix Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Mar 25, 2012 at 11:37 Mixo123 asked Mar 25, 2012 at 11:20 Mixo123Mixo123 8111 silver badge88 bronze badges 3 2 @Mixo123: this kind of question gets good answers if the questioner shows some evidence of having tackled the problem on his/her own. Right now it looks as if you just want us to do your work for you. Some SOers will, many won't. – High Performance Mark Commented Mar 25, 2012 at 11:33 I have tried, but no ideas. Only brute-force solution – Mixo123 Commented Mar 25, 2012 at 11:36 1 Are you sure that the problem always has solution (looks like sum of all elements un the matrix should divide over M and N)? What about Nx1 matrix? Or you're getting matrix which can be sorted in this way? – Eugeny89 Commented Mar 25, 2012 at 12:01 Add a comment \| 3 Answers 3 Reset to default This answer is useful 3 Save this answer. Show activity on this post. First, find the expected sum per row and per column 1. ``` rowSum = totalSum / numRows colSum = totalSum / numCols ``` Then, iterate through the rows and the columns and compute the following values: ``` rowDelta = 0 for each row r if sum(r) > rowSum rowDelta += sum(r) - rowSum colDelta = 0 for each col c if sum(c) > colSum colDelta += sum(c) - colSum ``` The number of the minimum moves to equilibrate all the rows and columns is: ``` minMoves = max(rowDelta, colDelta) ``` This works because you have to transfer from rows that exceed rowSum into rows that don't exceed it, and from columns that exceed colSum into columns that don't exceed it. If initially rowDelta was lower than colDelta, then you will attain a stage where you equilibrated all the rows, but the columns are still not equilibrated. At this case, you will continue transferring from cells to other cells in the same row. The same applies if initially colDelta was lower than rowDelta, and that's why we selected the maximum between them as the expected result. 1 If totalSum is not a multiple of numRows or numCols, then the problem has no solution. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 25, 2012 at 12:30 answered Mar 25, 2012 at 12:06 schsch 27.5k33 gold badges7272 silver badges8484 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Let us consider the one dimensional case: you have an array of numbers and you are allowed a single operation: take 1 from the value of one of the elements of the array and add it to other element. The goal is to make all elements equal with minimal operations. Here the solution is simple: you choose random "too big number" and add one to random "too small" number. Let me now describe how this relates to the problem at hand. You can easily calculate the sum that is needed for every column and every row. This is the total sum of all elements in the matrix divided by the number of columns or rows respectively. From then on you can calculate which rows and columns need to be reduced and which - increased. see here: ``` 1 1 2 2 -2 1 0 1 1 +1 0 0 1 1 +2 1 1 1 2 -1 +1+2-1-2 Expected sum of a row: 4 Expected sum of a column: 4 ``` So now we generate two arrays: the array of displacements in the rows: -2,+1,+2,-1 and the number of displacements in the columns: +1,+2,-1,-2. For this two arrays we solve the simpler task described above. It is obvious that we can not solve the initial problem in fewer steps than the ones required for the simpler task (otherwise the balance in the columns or rows will not be 0). However I will prove that the initial task can be solved in exactly as many steps as is the maximum of steps needed to solve the task for the columns and rows: Every step in the simpler task generates two indices i and j: the index from which to subtract and the index to which to add. Lets assume in a step in the column task we have indices ci and cj and in the row task we have indices ri and rj. Then we assign a correspondence of this in the initial task: take 1 from (ci, ri) and add one to (cj, rj). At certain point we will reach a situation in which there might be still more steps in, say, the columns task and no more in the rows task. So we get ci and cj, but what do we do for ri and rj? We just choose ri=rj so that we do not screw up the row calculations. In this solution I am making use of the fact I am allow to obtain negative numbers in the matrix. Now lets demonstrate: ``` Solution for columns: 4->1;3->2;4->2 Solution for rows: 1->3;1->3;2->4 Total solution: (4,1)->(1,3);(3,1)->(2,3);(4,2)->(2,4) ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 25, 2012 at 12:26 answered Mar 25, 2012 at 11:44 Boris StrandjevBoris Strandjev 47.1k1616 gold badges112112 silver badges136136 bronze badges 5 And what if all rows are '+' and all columns are neutral or '-'? – Mixo123 Commented Mar 25, 2012 at 11:48 @boris Maybe it will be better distinguish between row state and column state for each cell. – UmNyobe Commented Mar 25, 2012 at 11:52 @boris,Will it be polite to suggest a (possible) improvement as an answer? – Eugeny89 Commented Mar 25, 2012 at 12:08 Ok, fine. Now I have new, improved solution – Boris Strandjev Commented Mar 25, 2012 at 12:27 @Eugeny89, of course it will be. Actually you should have already, because now I posted my own improved version – Boris Strandjev Commented Mar 25, 2012 at 12:28 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Supose thar r1 is the index of a row with maximal sum, while r2 is the row with minimal sum. c1 column with maximal sum and c2 column with minimal. You need to repeat the following operation: if Matrix[r1][c1] == Matrix[r2][c2] we're done! Otherwise, Matrix[r1][c1] -= 1 and Matrix[r2][c2] += 1 Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Mar 25, 2012 at 13:09 Eugeny89Eugeny89 3,7631010 gold badges5555 silver badges105105 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm math matrix See similar questions with these tags. 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45	https://www.thinkinghard.com/blog/CubeRoots	How to Do Cube Roots of 9 Digit Numbers in Your Head a blog about things that I've been thinking hard about blog portfolio email about How to Do Cube Roots of 9 Digit Numbers in Your Head 19 September, 2005 cube root of a 9 digit number To get the first digit, calculate the cube root of the first 3 digits. To get the last digit, calculate the cube root mod 10. To get the middle digit, calculate the cube root mod 11. tags: mathematics Update (23 April 2006): try out the new Practice Page What is a Cube Root? (Just in case you don't already know.) A number cubed is that number times itself times itself. For example, 4 cubed, normally written 4 3, is 4 × 4 × 4 = 64. The cube root goes in the opposite direction, so the cube root of 64 is 4. To give a larger example, 657 cubed = 657 3 = 657 × 657 × 657 = 283593393, so the cube root of 283593393 is 657. Doing it in your head Now I'll have to admit that I used a calculator to calculate that 657 3 = 283593393 (actually I'm lying – I did it in Python, but that's beside the point). But I am claiming that you can learn to calculate the cube root of numbers like 283593393, by mental arithmetic alone. First Cheat When I say "the cube root of numbers like 283593393", there is of course something special about 283593393, which is that it is a cube of a whole number, so it's cube root is a whole number. Although there are 999999999 numbers up to 9 digits long, only 999 of these numbers have cube roots that are whole numbers. But if you are giving a display of your mental arithmetic powers to an audience, this is no problem. Give someone in your audience a calculator. Tell them to think of a 3 digit number, multiply it by itself twice and tell you the answer (which will be 7 to 9 digits long). You will then tell them what the original number was. Don't ask them to supply a random 9 digit number, because a random 9 digit number probably won't be a cube of a whole number, and you will be in trouble, and the following strategies won't work. Cube Roots up to 3 Digits If you had to calculate the cube root of a 3 digit number, it would be quite easy. Here is a table of cubes from 1 to 10: nn 3 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 9 729 10 1000 To do cube roots of 9 digit numbers, you are going to have to be familiar with these first 10 cubes. So you might as well memorise them. And if you have memorised them, then there isn't really any trick to calculating the cube roots. Cube Roots up to 6 Digits What is the cube root of 300763? The cube root of a 6 digit number has to be less than 100, so you could consider memorising the first 100 cubes. However that is a bit much, and it is also unnecessary, so I won't suggest that you memorise that many cubes. Let's look at the first 3 digits, which are "300". Now 300 is greater than 216 = 6 3 and less than 343 = 7 3. Which means that 300763 is greater than 216000 = 60 3 and less than 343000 = 70 3. So we know that our answer is somewhere between 60 and 70. To get the last digit easily, we need to use modular arithmetic, and to make it really easy, we will do arithmetic modulo 10. All this means is doing addition, subtraction and multiplication, keeping only the last digit of the answer. A modulo 10 version of the previous table is as follows: nn 3 mod 10 1 1 2 8 3 7 4 4 5 5 6 6 7 3 8 2 9 9 0 0 A curious pattern emerges: if we cube a number modulo 10 twice, we get back to where we started. So 8 3 = 512 = 2 (mod 10, where "mod" is an abbreviation for "modulo"), and 2 3 = 8 = 8 (mod 10). Applying this pattern to our problem cube 300763, we deduce that the last digit must be 3 3 = 27 = 7 (mod 10). So the answer must be a number in between 60 and 70 with last digit 7, which can only be 67. Cube Roots up to 9 Digits Using the same techniques, we can easily calculate the first and last digits of the cube root of a 9 digit number (assuming, as already stated, that it is known to be the cube of a whole number). So, for example, looking at 580093704, we compare 580 to 512 and 729, to deduce that the cube root is between 800 and 900. The last digit is 4, so the last digit of the cube root must be 4 3 = 64 = 4 (mod 10). All that is left to find is the middle digit. Unfortunately we have extracted all the information we can out of size comparisons and mod 10 arithmetic, and we need to think of something else. One "something else" is a different modulus. However this means doing long divisions in your head to get remainders, which is possibly a bit much work. We need a modulus for which there is an easy shortcut. And it has to be bigger than 10, if it is to uniquely choose the one correct answer from 10 possibilities. There are two major candidates, which are 9 and 11. Modulo 9 is easy to calculate, because you calculate the remainder of a number divided by 9 by just adding the digits. For example 580093704 = 5+8+0+0+9+3+7+0+4 = 36 mod 9 = 3+6 (mod 9) = 9 (mod 9) = 0. One problem with 9 is that it isn't bigger than 10. However, this might not be completely fatal. There is a short cut for modulo 11, but it is a little harder. Starting from the right end, you alternatively add and subtract digits. So 580093704 = 4 - 0 + 7 - 3 + 9 - 0 + 0 - 8 + 5 = 14 = 4 - 1 = 3 (modulo 11). (In this case we luckily ended up with a positive number less than 11, but you might have to adjust the answer to a number in this range by adding or subtracting 11.) To make use of these moduli, we need to calculate the tables of cubes mod 9 and 11: nn 3n 3 mod 9n 3 mod 11 0 0 0 0 1 1 1 1 2 8 8 8 3 27 0 5 4 64 1 9 5 125 8 4 6 216 0 7 7 343 1 2 8 512 8 6 9 729 0 3 10 1000 1 10 Unfortunately there is a problem with mod 9, which is that there are only 3 possible cubes mod 9, which means that going the other way there are still going to be 3 or 4 choices. Mod 11 works better, and we can see that each value from 0 to 10 occurs only once in the list of cubes modulo 11 (from 0 to 10), which means that cube roots can be calculated modulo 11. Returning back to the worked example, 580093704 = 3 modulo 11, so (looking at the table, which, by the way, you will have to memorise) its cube root = 9 modulo 11. We've already determined that the first digit is 8 and the last digit is 4. If we think of the cube root as 8 x 4, for some x, then we have the equation 4 - x + 8 = 9 modulo 11, i.e. 12 - x = 9 modulo 11, so x = 3 modulo 11. Which gives a final (and correct) answer of 834. Conclusion So there you have it, how to calculate cube roots of 9 digit numbers in your head. There is a certain amount of memorisation and fiddly addition and subtraction to do, and you probably need to be confident with these before you try it out as a party trick. I suggest that it will also be much easier if you make sure that you can see the number you are calculating the cube root of, so that you are not saddled with the additional mental load of having to memorise that number. If you liked this page ... If you liked this page, you might also like my PrimeShooter TM Space-Invaders-style Javascript game, where you have to kill the attacking numbers by firing prime factors at them until they are reduced to 1. Vote for or comment on this article on Reddit or Hacker News ... © 2005 Philip Dorrell
46	https://www.youtube.com/watch?v=3jgVo0B5ulw	Elapsed Time Word Problems - Complete SIS4Teachers 1380 subscribers 5 likes Description 504 views Posted: 2 Apr 2020 Complete Version.- 5 problems. These might be one of the trickiest types of word problems that show up in our students' work! But with the right visual model, they're a piece of cake! See how to create a number line and use it to correctly determine the amount of elapsed time! Transcript: Intro hi everyone this is Kristin Hilty from making that make sense today we're gonna focus on how to help your kids create a visual when working on elapsed time word problems elapsed time is something that every family can work on at home in graves three four and five this is a concept that our kids continue to struggle with and we see presented to them throughout the entire school year so one of the best ways to help them calculate how much time has gone from one point in time to another is to have them create a number line and using that to show the period of time that we're talking about so I have given you five different word problems i centered them all around this coronavirus pandemic that we've all been living through and Number Line we're gonna use these so that I can show you a couple different ways that this could look so the first problem we're gonna work on today during the corona virus at home schooling mrs. Kapil gets on her computer at seven o'clock in the morning she works until her kids get out of bed at 10:15 a.m. how long does she work in the morning so my question that I'm looking for is how long does mrs. Kapil work on her computer in the morning so I want kids to find the question and think about what is it that we have to answer we want to know the period of time mrs. Kapil is working on her computer in the morning what we want them to do here is draw a number line so I'm gonna draw my number line and I'm gonna put my arrows moving in both directions and I'm gonna look at the information I know I know what time she starts and I know what time she ends so over here on the left hand side I'm gonna go ahead and put her beginning time which we know is 7 o'clock a.m. over and what I'm going to do here is I'm actually going to check off that information because we went ahead and we used it so we showed where to place it we put a checkmark over it I also know that she continues to work until 10:15 in the morning so over here on the right hand side I'm going to go ahead and label that 10:15 a.m. and I'm going to check off that information because I've used that now what I want to do on my number line is I want to see how far is it from 7:00 in the morning until 10:15 in the morning so what we're gonna have our kids do is we're gonna have them make what we call jumps now some of your children are gonna be able to jump all the way from seven o'clock all the way over here to where they think it would be ten o'clock that's perfectly fine most of our kids especially when they're learning how to do a elapsed time can't make that big of a jump so what we're gonna do is I could go ahead and I could make a jump from 7 o'clock to 8 o'clock so I'm going to go ahead and label that 8 o'clock and I know that that here is a distance of 1 hour from 8 o'clock I can make another jump over to what I think would be 9 o'clock and that also is a distance of 1 hour from 9 o'clock I can jump another hour to get myself over here to 10 o'clock and I should be putting AMS underneath these and I know that that is also a distance of 1 hour now that I'm at 10 o'clock some of your children might need to do little jumps of 5 minutes and they could do 5 10 15 some of them can just do a smaller jump here and they know that that is a distance of 15 minutes now you can see how I label my hour and how I label my minutes now what we need to do is we need to calculate how much time pass I can put my hours together and I can see that that is a distance of 3 hours and then I have my 15 minutes so mrs. Kapil worked for three hours and 15 minutes in the morning putting that together now if you look at question number two I built off of that one here but now what I said is today she decided to take a break from 8:15 until 8:45 how long did she work today so on this specific day she worked for three hours and 15 minutes what I want to know here is if she puts a break in there now how long did she work some kids will be able to do go ahead and do that math and figure out that this is 30 minutes and go ahead and subtract it but what would this look like if we were to put this on a number line like we did here my question is how long does she work today and I have to use my information from back here in question 1 to begin to build this out so I'm also going to go ahead and draw my number line here and I know that her starting time was 7 o'clock in the morning so I'm going to go ahead and put that starting time back here again and I know that our ending time is still 10 15 in the morning so I know that she still ends at that time so now what I want to do is I know that I have to put this break in here from 8:15 to 8:45 well I know I can still jump from 7 o'clock to 8 o'clock so I'm still gonna go ahead and do that jump here and I know that that is still one hour but now I have to pay attention to this right here so what I'm gonna do here is I'm gonna do a small jump from 8:00 to 8:15 and I know she's still working during that time so I'm gonna go ahead and put that 15 minutes right here now I can check this information off because I put that on there and I know that this break lasts until 8:45 so I know that that's 8:45 when her break ends so I can go ahead and I can put that here now it really doesn't matter in this situation how long this time is just the fact that I put this break on here so I don't want to put that time above there because that's not part of the time that she worked from 8:45 what I'm gonna do at this point is I'm gonna make another 15-minute jump here to show that that gets me to nine o'clock in the morning so this I'm gonna label is my 15-minute jump to get me back over to 9:00 now from nine o'clock I continue to do my hour jump again and that's gonna get me over here to 10 o'clock in the morning and I know that that's one hour and then I can do my 15-minute jump again to get me over here to my 10:15 so now when I look at putting this together I see that I have an hour here and I have an hour here so I can go ahead and put that together for two hours and now I have my 15 minutes my 15 minutes in my 15 minutes now some kids might need to write this out and put 15 plus 15 plus 15 and some might combine this together to get us to 30 and then our 30 plus our 15 to get us over here to our 45 so I can see she worked 2 hours 45 minutes I could have them calculate I could add again an additional question what is the difference in the time she worked on this day to this day here and we can see that 30 minute difference which is what was represented right here on that break piece that we had all right we're gonna move into a Multiple Number Lines different type of a problem and now we're going to get to where we might need to use some multiple number lines to help us see what's going on here part of the problem with a word problem like this is all of the reading that is required to take place here and that's where kids get into trouble so how can we chunk this up a little bit to make it a little bit easier to start thinking about let's go ahead and read through the whole problem first Brett woke up at 9 o'clock a.m. to start his online learning he worked until 10:30 a.m. after he ate breakfast he went back to work at 10:50 a.m. his lunch was ready at 12 o'clock 40 minutes later he went outside to play for 30 minutes at that point he went back to working on his schoolwork for another 1 hour and 15 minutes what time did Brett stop working on his schoolwork for the day and how much time did he spend working so here we have two questions that we need to answer what time did Brett stop working on his schoolwork for the day and how much time did he spend working so we have a two different questions that were answering let's go and let's just start with the very first thing we know we know that Brett wakes up at 9 o'clock to start his online learning let's go ahead and let's just make a number line to represent that we know and I'm gonna actually make this one smaller cuz I'm gonna do some calculations over here because I'm gonna break this up a little bit he wakes up at 9 o'clock in the morning I'm gonna go ahead and label that 9 o'clock in the morning check off that information this is we're checking off what you've used is really going to come in handy so that kids can see with what information they still have left to work with he worked until 10:30 a.m. so we know that this is complete work time here he worked until 10:30 a.m. I might over here label this as work because he worked during that whole time after he ate breakfast he went back to work at 10:50 so we know he goes back to work so we're going to this number line start at 10:50 we know he goes back to work and his lunch was ready at 12 so we know that that one goes until 12 o'clock so again this is another work one that he does here and I can check off my 1050 and I can check off my 12 o'clock hmm 40 minutes later he went outside to play for 30 minutes so now we have to figure out how much time passes here 40 minutes later he went outside to play for 30 minutes well if his lunch was ready at noon 40 minutes later we know that that's gonna be 12:40 so this bar we're gonna put here at 12:40 check off my 40 minutes later he went outside to play for 30 minutes well at this point I need to figure out what time that's going to get me to so I'm gonna go ahead and I'm gonna do I know he played for 30 minutes so if I'm at 12:40 I'm gonna jump 10 minutes 10 minutes 10 minutes because I know that that's gonna be 30 minutes so if I do my jumps 12:40 and 10 minutes that gets me to 1250 1250 in 10 minutes is going to get me to 1 o'clock 1 o'clock in 10 minutes gets me to 110 all right so we know that that's at 110 at that point he goes back to working on his schoolwork I'm gonna go ahead and add another sheet of paper underneath here so I can continue to make these so at that point he goes back to working on his schoolwork I'm gonna draw another number line here we know this is 1:10 p.m. and I can check that off for another 1 hour and 15 minutes so we have to do a 1 hour in 15 minutes I'm gonna go ahead and draw that one hour jump and then I'm going to do I'm gonna do a 5 10 15 minutes so that's gonna get me my 15 folks these with HP 5 5 5 and 5 so that that gets me that 1 hour and 15 minutes so what I want to know is what time is it right here because that's what time you stopped working on the schoolwork and then I want to know how much time did he spend working well if we remember this here wasn't actually work so I'm not gonna label that one work but I am going to label this one work because that's when he went back to work again over here if I fill this in 110 in one hour that's gonna get me to 210 and now I have to do my 5-minute jumps this is going to get me to 215 220 this over here is going to get me to 225 p.m. so he stops working at 2:25 p.m. so we were able to answer that question now we have to figure out how much time did he spend working so you can see there's a couple different ways I had to work through this one I drew my number lines but some of them I went ahead and put my jumps on and some of these I haven't calculated the time for yet this one I know he worked for one hour 15 minutes I know that one I have to figure out this one here 9:00 to 10:30 I can go ahead and do my jump from 9:00 to 10:00 o'clock I know that's one hour and then this is going to be thirty minutes so this amount of time was one hour 30 minutes 10:52 12 o'clock well I'm gonna start at 10:50 and I'm gonna jump 10 minutes to get me to 11 o'clock and then I can do my our jump to get me from 11 o'clock to 12 o'clock so I know that that one is one hour 10 minutes so now when I look at all my work that I have over here I can go ahead I can put my hours together and I know my hours here are three hours and my 15 minutes in my 10 minutes I might start with that and put that together for 25 minutes so now if I have 30 minutes and 25 minutes I can put that together for 3 hours 55 minutes as I said that one was more of a complex one because we had a lot of different parts working within that one there it's a great way to have your kids calculate their schedule have them do their beginning time have them do their ending time where do you take breaks for lunch so this is a fun one that you can kind of play with at your own house and have your kids calculate how much time do they spend on e-learning a day or how much time do they get to play outside if you have them playing outside in multiple variables any place that you can think of to calculate that a last time it's really going to help that skill your teachers will really appreciate that all right this number four is again a little bit of a different type of a problem here Taylor went to babysit for a nurse she arrived at her house at 2:30 p.m. and left at 11 o'clock p.m. how much time did she spend babysitting and if she earned $8 per hour how much did she make so how big can we have another two-part question how much time did she spend babysitting and if she earned $8 per hour how much money did she make well first let's answer that first question how much time does she spend babysitting let's go ahead and draw our number line and we can put her beginning time of 2:30 and her ending time of 11 o'clock we can check off the information we used so 2:30 until 11 o'clock well 2:30 I'm going to go ahead and do my half-hour jump here some kids might want to jump on the just start jumping hourly and they might jump hourly from 2 33 34 35 36 38 39 30 10 30 and then do their 30-minute jump there's nothing wrong with that I just like to get to the even hour as soon as I can so that's why I do that jump first but either way that kids see to do it perfectly fine now three o'clock I'm trying to get all the way over to 11 o'clock well I know that from 3 o'clock I'm actually going to jump to 10 because I like working that number 10 so I'm gonna jump over here to ten o'clock could you do hourly jumps between here yes absolutely but I know from three o'clock to ten o'clock I know that that is seven hours so then from 10 o'clock to 11 o'clock that gives me a nice one hour jump here I can now put that time together and I can see that Taylor worked for 8 hours 30 minutes for that nurse our second part of the question if she earned eight hours $8.00 per hour how much did she make well what makes this tricky here is we do have this thirty minutes that she worked over here so I know that she worked eight hours at $8 an hour so I know I can calculate that if she works eight hours for $8.00 an hour I know that she made $64 for that part now we have this thirty minutes well she's not gonna make eight dollars for that thirty minutes so we now have to think she has this half hour here so for that half hour how much would she learn how much would she earn so we have to do that eight dollars divided in half so we know that for that part she makes four dollars now I can put that together and Taylor earned a total of sixty eight dollars while she babysat for that nurse now if I really knew Taylor well I would tell you Taylor actually did that for free to babysit for that nurse because they have been doing so much great work for us lately all right our last question is about everybody's favorite doctor right now doctor Falchi doctor Falchi was flying to visit hard-working doctors and nurses he boarded his plane at 11 a.m. the flight lasted 2 hours and 15 minutes when he landed it took him 20 minutes to get his luggage and 35 minutes to drive to the hospital what time was it when he arrived at the hospital and how much time did it take him to get to the hospital all right what time was it when he arrived at the Hospital how much time did it take him to get to the hospital I promise you your sheet has the he on it that my sheet does not all right so doctor foul she was flying so again we have another two-part question was flying to visit hard-working doctors and nurses there's no information in that sentence that I need to use he boarded his plane at 11:00 a.m. he boarded his plane at 11:00 a.m. that's my starting point for the day so right here 11 o'clock a.m. starting point the flight lasted 2 hours and 15 minutes okay let's show that one hour two hours 15 minutes so I can label that 1 hour 1 hour 15 minutes let's go ahead and figure out what time that is so 11 o'clock this is now going to get me to 12 o'clock p.m. this is why you want to have your kids labeling that AM PM so we can see when we jump from one to the other from here we have another hour that gets me to 1 o'clock p.m. 15 minutes that gets me to 1:15 so 2 hours 15 minutes we know it's now 1:15 when he landed it took him 20 minutes to get his luggage Oh 20 minutes to get his luggage well I'm gonna actually since I'm at 1:15 I'm gonna break that 20 up into 15 and 5 why am I gonna do that because I know if I do 15 minutes here that's gonna get me to 1:30 and then I just have to do a little 5:1 here and that's gonna get me to 1:35 p.m. so I can break that up in intervals that make sense to me took him 20 minutes to get his luggage and 35 minutes to drive to the hospital 35 minutes to drive to the hospital I'm at 1:35 here hmm I'm actually gonna do 10 minutes because that's gonna get me to 145 and then I'm gonna do another fifteen minutes because that gets me to two o'clock and I know this here is 25 minutes but I have to go 35 that's gonna leave me with another 10 minutes to do here which gets me to to 10 p.m. and I can check off that 35 minutes here what time was it when he arrived to the hospital look we just figured that out it's 2:10 when he arrived at the hospital how much time did it take him to get to the hospital well now that so we have to go back and count all of our increments here what I see is easy what jumps up to me right off the bat is I can put these together right here and I know that that gets me to two hours so what some kids will do is they'll take what they have here which is what I was starting to do and add all of it together that's that's perfectly fine another thing we can do is I could actually draw a second bar underneath here and I know that this is 11 o'clock didn't really have to draw it as far as I did but I went ahead and did that and this is 210 what I can do now is I can jump from 11 to 12 that gets me to 12 o'clock and now I can go from 12 o'clock to 2 o'clock and that's two hours and then I have this right here which is ten minutes and now do you see how that's a lot easier to put together to answer that second question so what I needed to do and I had to figure out what time this was going to be first before I could do any of this work absolutely I could go through and I could count up all of the ways that I broke it up in that one it's still going to get me the same answer three hours in ten minutes but for my mind visually it was a lot easier for me to redraw that and just calculate that time out and I can see that that is now three hours ten minutes again elapsed time it's a skill that can be worked on throughout you know at home it's not something that we just have to work on at school when you tell your kids to go outside and play for 25 minutes have them look at the clock you have to come inside in 25 minutes tell me what time do you need to come inside or if I want you to be back inside at 4:15 and it's 3:40 right now how long can you go outside and play ask your kids questions like that have them calculate how much time if you're running errands how much time did you spend at each store how much time did you spend all together running those errands so this is a life skill that we do use over time and that is why it is important that our kids do learn how to do that I hope you enjoy this any questions that you have please don't hesitate to put in the comment section and we will be sure to answer those for you have a great day you you
47	https://www.webelements.com/silicon/atoms.html	Silicon - 14Si ▸▸ 🇬🇧 Silicon 🇺🇦 Кремній 🇨🇳 矽 🇳🇱 Silicium 🇫🇷 Silicium 🇩🇪 Silicium 🇮🇱 צורן 🇮🇹 Silicio 🇯🇵 ケイ素 🇵🇹 Silício 🇪🇸 Silicio 🇸🇪 Kisel 🇷🇺 Кремний Silicon - 14Si: properties of free atoms ▸▸ Si Essentials Physical properties Electron shell data Atom sizes Electronegativity Isotopes and NMR Crystal structure Thermochemistry History Uses Geology Biology Binary compounds Compound properties Element reactions List all Si properties Silicon atoms have 14 electrons and the shell structure is 2.8.4. The ground state electron configuration of ground state gaseous neutral silicon is [Ne].3s2.3p2 and the term symbol is 3P0. Atomic spectrum A representation of the atomic spectrum of silicon. Ionisation Energies and electron affinity The electron affinity of silicon is 133.6 kJ mol‑1. The ionisation energies of silicon are given below. Ionisation energies of silicon \| Ionisation energy number \| Enthalpy / kJ mol‑1 \| \| 1st \| 786.52 \| \| 2nd \| 1577.13 \| \| 3rd \| 3231.58 \| \| 4th \| 4355.52 \| \| 5th \| 16091 \| \| 6th \| 19806 \| \| 7th \| 23790 \| \| 8th \| 29292 \| \| 9th \| 33893 \| \| 10th \| 38727 \| \| 11th \| 45953 \| \| 12th \| 50502 (calculated) \| \| 13th \| 235198 \| \| 14th \| 257922 \| Ionisation energies of silicon. Effective Nuclear Charges The following are "Clementi-Raimondi" effective nuclear charges, Zeff. Follow the hyperlinks for more details and for graphs in various formats. Effective nuclear charges for silicon \| 1s \| 13.5745 \| \| \| \| \| \| \| \| 2s \| 9.02 \| 2p \| 9.95 \| \| \| \| \| \| 3s \| 4.90 \| 3p \| 4.29 \| 3d \| (no data) \| \| \| \| 4s \| (no data) \| 4p \| (no data) \| 4d \| (no data) \| 4f \| (no data) \| \| 5s \| (no data) \| 5p \| (no data) \| 5d \| (no data) \| \| \| \| 6s \| (no data) \| 6p \| (no data) \| \| \| \| \| \| 7s \| \| \| \| \| \| \| \| References These effective nuclear charges, Zeff, are adapted from the following references: E. Clementi and D.L.Raimondi, J. Chem. Phys. 1963, 38, 2686. E. Clementi, D.L.Raimondi, and W.P. Reinhardt, J. Chem. Phys. 1967, 47, 1300. Electron binding energies Electron binding energies for silicon. All values of electron binding energies are given in eV. The binding energies are quoted relative to the vacuum level for rare gases and H2, N2, O2, F2, and Cl2 molecules; relative to the Fermi level for metals; and relative to the top of the valence band for semiconductors. \| Label \| Orbital \| eV [literature reference] \| \| K \| 1s \| 1839 \| \| L I \| 2s \| 149.7 [2, values derived from reference 1] \| \| L II \| 2p1/2 \| 99.8 \| \| L III \| 2p3/2 \| 99.2 \| Notes I am grateful to Gwyn Williams (Jefferson Laboratory, Virginia, USA) who provided the electron binding energy data. The data are adapted from references 1-3. They are tabulated elsewhere on the WWW (reference 4) and in paper form (reference 5). References J. A. Bearden and A. F. Burr, "Reevaluation of X-Ray Atomic Energy Levels," Rev. Mod. Phys., 1967, 39, 125. M. Cardona and L. Ley, Eds., Photoemission in Solids I: General Principles (Springer-Verlag, Berlin) with additional corrections, 1978. Gwyn Williams WWW table of values D.R. Lide, (Ed.) in Chemical Rubber Company handbook of chemistry and physics, CRC Press, Boca Raton, Florida, USA, 81st edition, 2000. J. C. Fuggle and N. Mårtensson, "Core-Level Binding Energies in Metals," J. Electron Spectrosc. Relat. Phenom., 1980, 21, 275.
48	https://philosophy.stackexchange.com/questions/46390/in-predicate-logic-do-we-necessarily-have-to-restrict-the-domain-of-discourse	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams In predicate logic, do we necessarily have to restrict the domain of discourse? Ask Question Asked Modified 7 years, 10 months ago Viewed 2k times 7 Can we combine formalized statements in predicate logic from widely different domains of discourse (say, one regarding integers, one regarding the fruits in one's garden) by using logical rules/theorems and still predicate logic stays truth-preserving? Sure, if we use identical free logical variables referring to different things (in the case of objects) or having a different sense (in the case of predicates), we may easily get to falsehoods, yet strictly speaking we didn't get them from truths. Can it be assumed that free variables carry along their semantic content faithfully, freed from their original context? And that then the correct combination by predicate logic never gets us from truth to falsehood? Of course, the resulting formalized statements, then back-translated to natural language may sound like written by EugÃ¨ne Ionesco. For example, This apple is prime or it is not prime yet isn't such a sentence, in the strict sense, true? So is the view that the domain of discourse for predicate logic can be extended to anything that (in a pragmatic way) exists, respectable? does it have a name? which philosophers of logic held it? Who rejected it? logic philosophy-of-logic Share Improve this question asked Oct 5, 2017 at 10:17 viuserviuser 5,49322 gold badges1919 silver badges5959 bronze badges 12 3 In general, variables are not restricted. We can do it explicitly using suitable predicate letters, like e.g. x (Natural(x) x¥0) (usually abbreviated as: xN (x¥0). Otherwise, the domain is part of the interpretation of a formula: with different interpretations, the meaning (and truth value) of a formula will change. Mauro ALLEGRANZA – Mauro ALLEGRANZA 2017-10-05 11:15:53 +00:00 Commented Oct 5, 2017 at 11:15 1 The view that the domain of discourse for predicate logic must encompass everything was formulated by Frege and Russell. Today, the "standard" view is due mainly to Tarski, with a precursor in the Algebra of Logic tradition : Boole and Schroder. Mauro ALLEGRANZA – Mauro ALLEGRANZA 2017-10-05 12:45:45 +00:00 Commented Oct 5, 2017 at 12:45 My answer philosophy.stackexchange.com/a/46369/935 may or may not be helpful user935 – user935 2017-10-05 14:36:47 +00:00 Commented Oct 5, 2017 at 14:36 No philosophers reject predicate logic afaik, but many would deny its ability to fully describe the world. As Bradley notes, in metaphysics predication seem necessary yet is illegitimate (since it forces on us a distinction between predicate and predicated that leads to false representations). user20253 – user20253 2017-10-05 14:53:29 +00:00 Commented Oct 5, 2017 at 14:53 1 @PeterJ I think the question is less about predicate logic generally and more about the question of restrictivism regarding quantification. Dennis – Dennis 2017-10-05 16:08:03 +00:00 Commented Oct 5, 2017 at 16:08 \| Show 7 more comments 2 Answers 2 Reset to default 7 There seem to be two parts to your question: one concerning whether all quantification is restricted, and another concerning whether free variables can carry contextual information that serves to restrict their admissible valuations. I'll address each in turn. Unrestricted Quantification It seems like your question is regarding the possibility of completely unrestricted quantification. The mainstream view is that such unrestricted quantification over everything is possible. In practice, when doing model theory we often construct models with determinate (restricted) domains to show model existence and consistency of a theory by way of its satisfiability. This model-theoretic convention, though, doesn't answer the question of whether all quantification must be restricted. The doubter of unrestricted quantification holds a view termed restrictivism. Typically, restrictivists derive motivation for their view from various set-theoretic and property-theoretic paradoxes. They claim that unrestricted quantification over, e.g., sets is impossible because the domain of quantification would be the set of all sets, which leads to paradox in standard set theories. Given the standard move of allowing proper classes to serve as a domain, or opting for a plural logic interpretation of the domain as the "plurality of all sets", this seems like a case of taking the semantics too seriously. That domains are sets is an artifact of model theory but not a necessary postulate. More sophisticated restrictivists realize that with suitable large cardinal axioms, what was a proper class without such large cardinals becomes a set. They typically complain, then, that the notion of set is indefinitely extensible (an objection popularized by Michael Dummett, who used the notion to push for intuitionistic logic/mathematics). The idea is that anytime you think you have "all" of the sets, you could always admit more by accepting some larger cardinal and extending the height of the cumulative hierarchy. They argue that since there's no non-arbitrary stopping point, and this process can continue on indefinitely, we are never in a position to quantify over a determinate domain of all sets. Such a view, however, seems to tie our abilities to quantify over a domain to our ability to describe that domain uniquely (at least up to isomorphism) -- something the Platonist will surely deny. The more common response to the paradoxes (though the restrictivist will complain such a response is objectionably ad hoc) is to say that they only show us that some predicates (like "set of all sets") do not have sets as extensions -- not that unrestricted quantification over sets is impossible. (This is more or less the idea behind Zermelo's approach in formulating the axiom of separation. Such predicates are rules out since you must always "separate" a set from a set -- ensuring the extensions of any predicates used in instances of the axioms are sets.) Additionally, restrictivism is often claimed to suffer from a paradox of self-defeat akin to the flaw afflicting the verificationist theory of meaning. There seems to be no way to state the view without rendering it false. Consider: (R) Every quantified statement is restricted. (Typically they'd claim this holds of necessity, but I've eliminated any modal notions for simplicity and because the additional complications don't help the restrictivist.) Now there is a dilemma: The use of "every" in (R) is restricted, and so (R) does not capture the restrictivist's view -- it's compatible with instances of unrestricted quantification that fall outside the scope of the restricted "every". The use of "every" in (R) is unrestricted, and so (R) is false since it provides its own counterexample. How to coherently state restrictivism is a subject of controversy, with many dismissing the view as incoherent. The primary source for this topic is Rayo and Uzquiano's (eds.) Absolute Generality. Contextually Restricted Free Variables The second question is largely separate from the first. There are a few uses of such contextually restricted free variables, off the top of my head these divide into (at least) two major categories: Dynamic Semantics and (some) presentations of Quantified Modal Logic. I'll start with dynamic semantics. Dynamic Semantics In dynamic semantics the central idea is that context serves to restrict the admissible variable assignments. It introduces quantification over variable assignment functions. Atomic formulae are of the form P(x) and are interpreted as the set of all assignments which assign to x an object that is P. Existential quantification acts to "reset" the set of admissible assignments to be those that differ from the original set of assignments by a renaming of the relevant variable/constant according to the (usually newly, otherwise the renaming is the trivial identity) introduced variable bound by the existential quantifier. This is the approach of Dynamic Predicate Logic, which takes these "actions" on assignments to provide the meaning of quantified statements (hence "dynamic"). A different formulation of dynamic semantics is given by Discourse Representation Theory, tracing back to work of Lauri Kartunnen from the 60's (1968?) but finding more complete development in the early 80's in the work of Irene Heim and the work of Hans Kamp (Kamp's work and its descendants are what is usually meant in modern references to "discourse representation theory"). The main innovation of DRT was to introduce the notion of "discourse referents" to treat phenomena like anaphoric reference. Anaphora occurs when an expression (typically a pronoun) depends for its reference on some previously introduced individual -- a referent derived from the preceding discourse. Consider: A man walked into the bar. He asked for a drink. The idea is that the phrase "a man" introduced some particular but underspecified individual into the discourse and that "he" refers to this individual. This is treated via "discourse representation structures" (DRSs). The first sentence is given the following DRS: [x: Man(x), Walked-into-bar(x)] Here, the variable "x" denotes a discourse referent introduced by "a man". Note that this is a particularly individual and so not simply what would be conveyed by an existential statement. The second sentence has the following DRS: [x: Ordered-a-drink(x)] (With the understanding that the gendered pronoun supplies additional information marking x as male. I've put x in bold to indicated that it's unresolved and must be identified with a previously introduced discourse referent.) These two DRSs are joined via the operation of "merge" to produce the DRS for the whole discourse (the two sentences): [x, x: Man(x), Walked-into-bar(x), Ordered-a-drink(x), x = x] The discourse referent "x" is often glossed as a free variable whose value is supplied by context (i.e., context restricts the admissible assignments). Quantified Modal Logic Finally, some presentation of quantified modal logic -- notably Fitting and Mendelsohn's First-Order Modal Logic -- use contextually restricted free variables, under the name of parameters, to deal with names in modal discourse. A parameter differs from an ordinary free variable in that it is never allowed to be bound by a quantifier in the course of a proof -- its value is always supplied via contextual restriction on admissible assignments. This is similar to the use of "dummy constants" in natural deduction systems for predicate logic. When you move from an existential claiming that "Something is P" to the claim that P(a), you are required to use a "fresh constant" -- a constant not previously introduced in your proof (without having been already discharged). The value of a here isn't really constant at all, it denotes some arbitrary thing that is P and is indifferent as to which P you choose. P then serves to restrict the assignments to a and in some systems free variables with such restrictions are used in place of dummy constants. Summary So, to conclude, it is possible to assume that free variables "carry contextual information", even across a change in context, but the details of how this works will depend on the particular framework. Certainly it is a common phenomenon, even if not usually explicitly treated in introductory discussions of predicate logic. As to unrestricted quantification, the majority view is that it is possible and coherent. The view does, however, have its detractors. It tends to be treated as almost a matter of dogma that unrestricted quantification is coherent and unproblematic, with the burden being placed on detractors to argue against it and formulate a coherent alternative. Share Improve this answer edited Jun 17, 2020 at 8:34 CommunityBot 1 answered Oct 5, 2017 at 17:37 DennisDennis 4,72611 gold badge2424 silver badges4040 bronze badges 21 2 I am not sure I understand, the bigger problem with unrestricted domains is that there are no predicates to speak of that would apply to everything. SEP article you linked says the same:"In typical contexts, the use of the quantifier everything is tacitly restricted to a domain of contextually salient objects... But the fact that unrestricted quantification is relatively uncommon is no reason to doubt it is attainable in certain contexts". Even "indefinitely extensible" classes in math would not contain candy as elements. The main problem seems to be not incoherence but pointlessness. Conifold – Conifold 2017-10-05 18:50:02 +00:00 Commented Oct 5, 2017 at 18:50 1 @Conifold I see, I haven't encountered that objection. Most defenders of unrestricted quantification as a possibility are happy to admit that most quantificational utterances contain (at least tacit) restrictions. The "usefulness" would typically be in formulating metaphysical theses. For instance, nominalism might be stated as the claim that "Everything is concrete"/"There are no abstract objects". For that to capture the nominalist's view the quantifiers would need to be unrestricted. I'm not sure if there are non-metaphysical claims that require unrestricted quantifiers.... Dennis – Dennis 2017-10-05 19:26:05 +00:00 Commented Oct 5, 2017 at 19:26 1 Maybe empiricism of a sort "All things are knowable through the senses" (roughly)? I took OP's question about whether unrestricted quantification was respectable to be more a matter of whether it's coherent or tenable, but I admit to not knowing what to make of the parenthetical suggesting a notion of "existing in a pragmatic way". Dennis – Dennis 2017-10-05 19:26:45 +00:00 Commented Oct 5, 2017 at 19:26 1 @wolf-revo-cats Oh, that's a bit different. You have in mind a sort of Carnapian set-up where different metaphysical claims might exist relative to different "frameworks". Neither the nominalist nor the platonist will accept such a set-up, since the nominalist holds that no matter how you loosen up restrictions on quantification, you'll never find abstract objects in your domain. The platonist holds that we can restrict to avoid talking about abstract objects, but then we're not talking about all there is.... Dennis – Dennis 2017-10-07 02:30:23 +00:00 Commented Oct 7, 2017 at 2:30 1 Some have argued for a non-ontological notion of quantification whereby you can quantify over things without committing yourself to their existence -- it's a sort of modern Meinongian position. The logic to capture that would be free logic which allows an "inner domain" and an "outer domain", where the nominalist could claim it's only the entities in the inner domain that really exist. But the key would be to keep the "existence loaded" quantifiers separate from the "free lunch" quantifiers. Dennis – Dennis 2017-10-07 02:35:00 +00:00 Commented Oct 7, 2017 at 2:35 \| Show 16 more comments 7 I am going to assume extra context provided by OPs earlier related question, Does fictional discourse pose special difficulties for logic? Classical devices Even the classical set theory has a simple device for restricting the range of quantifiers without changing the domain of discourse, xS P(x) restricts the domain to S. If the use of set theory is objectionable it can be simulated within the predicate logic itself, let x S be denoted S(x), then the above formula is equivalent to x (S(x) § P(x)), which blocks attempts to infer existence outside of S. Similarly, xS P(x) is simulated by x (S(x) P(x)). If x is not in S the antecedent of the implication is false, so it is trivially true, meaning that whatever happens ouside of S has no effect on the truth of the sentence. Thus, we can have a unified universe of discourse but de facto restrict it to subsets/subclasses when statements are made. Of course, in conversational language such restrictions are often left implicit, but this is a problem not so much for logic or semantics as for linguistic pragmatics, which studies speech acts and how speakers infer implicit restrictions from the context, like prior conversation, environment, shared background, etc. In programming languages the requisite background is made explicit by splitting variables into data types, and modularization, restricting the use of a variable to within the procedure where it is defined, and its subprocedures, as in object-oriented programming. Fictionalization Sometimes a more radical change of domain is called for, where S is not saliently described as either a set or a class, for example with fictional discourses. The role of x S is then taken over by modal operators of fictionalization introduced by Lewis in Truth in Fiction which apply to sentences rather than to variables, so the Platonist make believe that fictional objects populate some kinds of sets is not needed. But the mechanism of blocking an unwelcome mixing of discourses is the same in spirit: But if we prefix the operator "In the fiction f" to some of the original premisses and not to others, or if we take some but not all of the premisses as tacitly prefixed, then in general neither the original conclusion nor the prefixed conclusion "In the fiction f will follow. The premiss that Holmes lived at 221B Baker Street was true only if taken as prefixed. The premiss that the only building at 221B Baker Street was a bank, on the other hand, was true only if taken as unprefixed, for in the stories there was no bank there but rather a rooming house. Taking the premisses as we naturally would in the ways that make them true, nothing follows: neither the unprefixed conclusion that Holmes lived in a bank nor the prefixed conclusion that in the stories he lived in a bank. Taking both premisses as unprefixed, the unprefixed conclusion follows but the first premiss is false. Taking both premisses as prefixed, the prefixed conclusion follows but the second premiss is false. So this apple is not prime, but according to Ionesco it is prime, no contradiction. Rosen adapted Lewiss framework to modal semantics of possible worlds for those unwilling to accept the explosion in ontology it requires. All possible worlds and objects in them are reduced to fictions, see his Modal Fiction. Meinongian universe Finally, there is an option opposite to fictionalism, which adopts the most prolific ontology imaginable, that of Meinong, with golden mountains and round squares admitted on a par with desks and chairs. A coherent formal version of it was worked out by Parsons in A Prolegomenon to Meinongian Semantics. In a way, the mechanics is not that different from fictionalism or Russells paraphrase: Individuals are not sets of properties. However, corresponding to each individual is a unique set of properties-the set of properties that the individual has. I call this set the "correlate" of the individual¦ Likewise, Meinongian objects are not sets of properties either, but I'll talk as if they were. More literally, sets of properties (i.e., subsets of P) will "represent" objects in this theory, much as certain sets of sets represent numbers for logicists. [¦] There are incomplete objects. For example, the gold mountain is not determinate with respect to any properties except for its goldness and its mountainhood¦ There are supposed to be objects that don't exist. These will simply be sets of properties that aren't the correlates of any individuals¦ There are impossible objects, for example, the round square. This object may be represented by {roundness, squareness}. The round square happens to be incomplete, but there are complete impossible objects too. [¦] I intend 'there are' in the broadest sense, the sense of 'some'. In particular, my quantifiers are to range over all objects¦ It is clear from M4 that Meinong would not regard such a quantifier as existentially loaded. This is perhaps the closest option to the unrestricted uberdomain asked for in the OP. When restriction of context is needed one can simply use the classical restriction device described above, there are no qualms to be had in Meinongianism over what sets may contain. Pace Kant, there is also an existence predicate in this formalization, and blocking incoherence with impossible objects is simple: their properties may be inconsistent, but the existence predicate is false on them. "There are" and "there exist" come apart here. Considering that Meinongian objects are essentially identified with collections of properties, and to us those pretty much manifest as indefinite descriptions, i.e. conjunctions of descriptive sentences, we get something very much like fictionalism in disguise. So primality is not among this apple's properties, but Ionesco-primality is. There are also apples that combine primality and non-primality, they just happen not to exist. Share Improve this answer edited Nov 8, 2017 at 1:37 answered Oct 5, 2017 at 20:45 ConifoldConifold 44.8k44 gold badges102102 silver badges187187 bronze badges 1 2 +1 thanks for linking to OP's other question, I've been meaning to get to that one. Taught half of a class on fictional discourse, should have some good material for OP once I get my computer sorted and find those lecture notes. Dennis – Dennis 2017-10-05 21:00:40 +00:00 Commented Oct 5, 2017 at 21:00 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions logic philosophy-of-logic See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked Does fictional discourse pose special difficulties for logic? Related Is first-order logic the only fundamental logic? 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49	https://math.stackexchange.com/questions/2932756/understanding-conditional-probability-math-is-fun	Understanding Conditional Probability (Math is Fun) - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Understanding Conditional Probability (Math is Fun) Ask Question Asked 7 years ago Modified5 years, 9 months ago Viewed 5k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. I have trouble understanding a simple concept from Math is Fun. STATEMENT: 70% of your friends like Chocolate, and 35% like Chocolate AND like Strawberry. What percent of those who like Chocolate also like Strawberry? SOLUTION: P(Strawberry\|Chocolate) = P(Chocolate and Strawberry) / P(Chocolate) 0.35 / 0.7 = 50% . Hence 50% of your friends who like Chocolate also like Strawberry WHAT I CAN'T UNDERSTAND: What is the difference between the statements that "35% of friends like both Chocolate and Strawberry" and "friends who like Chocolate also like Strawberry". Just a play of words, practically they seem exact same to me. Am I missing something ? probability Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Sep 27, 2018 at 7:07 ArnuldOnDataArnuldOnData 357 1 1 gold badge 4 4 silver badges 26 26 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. The small difference in the words implies a difference in the denominator in the calculation, while the numerator stays the same. Suppose you in fact have 20 20 friends of whom 14 14 like Chocolate, and of these 7 7 like Chocolate and Strawberry. The proportion of all your friends who like both is 7 20=35%7 20=35% The proportion of your Chocolate-liking friends who also like Strawberry is 7 14=50%7 14=50% This second calculation is equivalent to 7/20 14/20=35%70%7/20 14/20=35%70%, and you would get the same percentage result if you in fact have 100 100 or 2000 2000 friends Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 27, 2018 at 7:28 HenryHenry 171k 10 10 gold badges 139 139 silver badges 297 297 bronze badges 1 So, 7 people who like both will always come from 14-group and rest of the 6 people like only Strawberry and not Chocolate. and 7/14 means 50% . They just removed the people who not like Chocolate. So conditional probability means removal of certain group/category of people ?ArnuldOnData –ArnuldOnData 2018-09-28 02:52:47 +00:00 Commented Sep 28, 2018 at 2:52 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The difference is in what comes before. For chocolate and strawberry the question is explicitly looking at everyone, and then within them, who likes both. The second bit; those who like chocolate also like strawberry, means that you are looking at a smaller group to begin with. You look at those who like chocolate, and within that smaller group find those who like both. Obviously those who like chocolate who also like strawberry are in the group of those who like chocolate and strawberry. So you're right in saying that they mean the same thing - to an extent. It just changes how you view the "population". Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 27, 2018 at 7:16 MRobinsonMRobinson 1,732 10 10 silver badges 19 19 bronze badges 3 Let me know if that doesn't make any sense and I'll try and reword MRobinson –MRobinson 2018-09-27 07:17:06 +00:00 Commented Sep 27, 2018 at 7:17 You explanation is simple enough. Understanding this in practical terms is bit difficult. Why you used the sentence " they mean the same thing - to an extent". If you look at 100% of the people then those who like both C & S will always come from C group because C group contains all the people who like both C and S. So, it is just that number of total people have been reduced from 100 to 70 because all the people who do not like C have been removed from the total. Am I correct ?ArnuldOnData –ArnuldOnData 2018-09-28 02:46:23 +00:00 Commented Sep 28, 2018 at 2:46 @Arnuld Yes, exactly. It can be important to look past the language in questions like this and think about what they actually want MRobinson –MRobinson 2018-09-28 07:15:58 +00:00 Commented Sep 28, 2018 at 7:15 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I think there is a confusion about m e a s u r e s m e a s u r e s: we do not discuss the s h a r e s h a r e or number of people in the population, we discuss another measure, called p r o b a b i l i t y p r o b a b i l i t y, which sometimes can be viewed as a share or proportion, but not necessarily. In your case 70%70% means 'pick a person from the population, what's the chance he likes chocolate' and 35%35% means 'pick a person from the population, what's the chance he likes strawberry'. 35%35% means 'pick a person from the group of people that like chocolate (subset of the population), what's the chance he likes the strawberry'. So you restrict the s e l e c t i o n s e l e c t i o n to a subset of the population. Therefore it is not an absolute number that matters, but a 'chance to observe an effect', of 'likelihood to observing the effect' not by examining the whole population, but a subset thereof. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 28, 2018 at 14:53 answered Sep 27, 2018 at 9:52 AlexAlex 19.4k 4 4 gold badges 30 30 silver badges 46 46 bronze badges 6 I am sorry but I find it very hard to relate conditional probability to practical life situation . People who like Strawberry and Chocolate are exactly the same ones who like Chocolate and Strawberry. Does the order of priority matter ? I like both but I find Chocolate tastier than Strawberry. Is this what P(S t r a w b e r r y\|C h o c o l a t e)P(S t r a w b e r r y\|C h o c o l a t e) means ? In this case it is no longer a game of numbers but emotions/taste came into the picture.ArnuldOnData –ArnuldOnData 2018-09-27 17:05:48 +00:00 Commented Sep 27, 2018 at 17:05 @Arnuld Preference between the two doesn't matter. P(S t r a w b e r r y\|C h o c o l a t e)P(S t r a w b e r r y\|C h o c o l a t e) means probability they like strawberry given they like chocolate. So you look at all the people who like chocolate, and see what proportion of those like strawberry.MRobinson –MRobinson 2018-09-28 07:17:11 +00:00 Commented Sep 28, 2018 at 7:17 @MRobinson I think the OP has already heard 'like strawberry given they like chocolate'. What I think he doesn't understand that we are not talking about ABSOLUTE numbers, but the l i k e l i h o o d l i k e l i h o o d to observe a strawberry lover whether we consider the whole population vs subset (choc lovers). Does this make sense?Alex –Alex 2018-09-28 11:49:24 +00:00 Commented Sep 28, 2018 at 11:49 @Alex Of course it makes sense, that's what I outlined in the second sentence MRobinson –MRobinson 2018-09-28 12:58:22 +00:00 Commented Sep 28, 2018 at 12:58 @MRobinson sorry the last question was more directed at the OP. I think it's better to look at probability as measure of l i k e l i h o o d l i k e l i h o o d or o p i n i o n o p i n i o n rather than share. proportion or absolute number. Our o p i n i o n o p i n i o n of the preference of strawberry changes when we consider the subset of the population rather than the whole population.Alex –Alex 2018-09-28 14:52:49 +00:00 Commented Sep 28, 2018 at 14:52 \|Show 1 more comment You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability See similar questions with these tags. 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50	https://pmc.ncbi.nlm.nih.gov/articles/PMC10458759/	Postmortem Analysis of Vitreous Urea Nitrogen, Creatinine, and Magnesium of Renal and Post-Renal Disease in Cats - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Toxics . 2023 Aug 10;11(8):685. doi: 10.3390/toxics11080685 Search in PMC Search in PubMed View in NLM Catalog Add to search Postmortem Analysis of Vitreous Urea Nitrogen, Creatinine, and Magnesium of Renal and Post-Renal Disease in Cats Adam W Stern Adam W Stern 1 College of Veterinary Medicine, University of Florida, Gainesville, FL 32608, USA; daliana.roig@ufl.edu (D.R.); cvalerio@ufl.edu (C.V.); tdenagamage@ufl.edu (T.D.) Find articles by Adam W Stern 1,, Daliana Roig Daliana Roig 1 College of Veterinary Medicine, University of Florida, Gainesville, FL 32608, USA; daliana.roig@ufl.edu (D.R.); cvalerio@ufl.edu (C.V.); tdenagamage@ufl.edu (T.D.) Find articles by Daliana Roig 1, Courtney Valerio Courtney Valerio 1 College of Veterinary Medicine, University of Florida, Gainesville, FL 32608, USA; daliana.roig@ufl.edu (D.R.); cvalerio@ufl.edu (C.V.); tdenagamage@ufl.edu (T.D.) Find articles by Courtney Valerio 1, Thomas Denagamage Thomas Denagamage 1 College of Veterinary Medicine, University of Florida, Gainesville, FL 32608, USA; daliana.roig@ufl.edu (D.R.); cvalerio@ufl.edu (C.V.); tdenagamage@ufl.edu (T.D.) Find articles by Thomas Denagamage 1 Editors: Davor Željěžić 1, Thomas Göen 1 Author information Article notes Copyright and License information 1 College of Veterinary Medicine, University of Florida, Gainesville, FL 32608, USA; daliana.roig@ufl.edu (D.R.); cvalerio@ufl.edu (C.V.); tdenagamage@ufl.edu (T.D.) Correspondence: adamstern@ufl.edu; Tel.: +1-352-294-4077 Roles Davor Željěžić: Academic Editor Thomas Göen: Academic Editor Received 2023 May 8; Revised 2023 Jul 11; Accepted 2023 Aug 9; Collection date 2023 Aug. © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC10458759 PMID: 37624190 Abstract Analysis of vitreous urea nitrogen (VUN) and creatinine (CREA) has been shown to be a useful indicator of azotemia in human postmortem examinations. Previous studies in cats, dogs, horses, and cattle have shown a good correlation between serum and postmortem ocular fluid for urea and creatinine. The aim of this study was to evaluate differences in VUN, CREA, and magnesium (MG) concentrations in vitreous humor collected postmortem from cats that presented with renal or post-renal disease and cats without renal or post-renal disease. Nine cats with renal or post-renal disease and twenty cats without renal or post-renal disease that underwent autopsy were used in this study. Collection of postmortem vitreous humor was performed, and vitreous samples were frozen at −80 °C for a minimum of 24 h prior to analysis using an in-clinic dry chemistry analyzer. Overall, there were statistically significant differences for VUN, CREA, and MG between cats with renal or post-renal disease and cats without renal or post-renal disease. Analysis of vitreous humor from cats suspected to have renal or post-renal disease can provide useful diagnostic information pertaining to renal function and issues in the urethra and urinary bladder. Future studies exploring postmortem vitreous chemistry in cats are warranted. Keywords: vitreous, postmortem, cat, creatinine, urea nitrogen, magnesium, biochemistry 1. Introduction Azotemia is defined as an increase in urea nitrogen and/or creatinine and can be due to pre-renal, renal, or post-renal causes. Potential causes of azotemia in cats include hypovolemia or dehydration (pre-renal azotemia), acute or chronic renal failure (renal), or post-renal obstruction (post-renal azotemia). In cats, one of the most common causes of acute renal failure is the ingestion of ethylene glycol, which is found in antifreeze and is of forensic importance given the high mortality rate in cats and ease of access to this substance. In the clinical setting, routine biochemical analysis of antemortem serum or plasma will allow for the identification of elevations of blood urea nitrogen (BUN) and/or creatinine (CREA). Unfortunately, there are instances in which routine biochemical analysis could not be performed prior to the death of a cat including financial constraints and the sudden death of the cat. It is not possible to assess for azotemia using postmortem biochemical analysis of serum or plasma and there are no studies in cats evaluating the usage of postmortem vitreous chemistry for the diagnosis of azotemia postmortem. Comparatively in humans, postmortem vitreous humor analysis has been shown to assist pathologists in determining the cause of death . For example, postmortem analysis of vitreous humor has been used for the analysis of electrolytes, creatinine, urea nitrogen, and glucose [2,3]. In cats, there is a strong correlation between serum and postmortem ocular fluid for the concentration of urea nitrogen and creatinine during the early postmortem period (at least 24 h after death) . In cats, a study using the histomorphology of feline chronic kidney disease identified correlations between antemortem serum markers of renal dysfunction and identified that interstitial fibrosis best correlated with the severity of azotemia . In contrast to chronic renal failure, in instances of acute renal failure, interstitial fibrosis is not a typical pathologic finding and no assumptions pertaining to azotemia from histologic examination can be made. In addition to azotemia, hypermagnesemia has been reported in cases of acute renal disease and obstructive uropathy in cats . The aim of this retrospective study was to assess cats with reported vitreous urea nitrogen (VUN), CREA, and magnesium (MG) levels of vitreous humor of cats with renal or post-renal disease and compare them to cats without renal or post-renal disease (normal). 2. Materials and Methods Vitreous humor was collected from cats that were presented for autopsy to the University of Florida’s Veterinary Forensic Pathology and Diagnostic Pathology services. A review of the submitted medical history and a postmortem examination was performed on each animal according to standard operating procedure to determine the cause of death. This study did not require approval from the Institutional Animal Care and Use Committee as all samples were collected postmortem and none of the animals were euthanized for purposes related to this study. Vitreous humor was collected from each cat within 48 h of death. Vitreous humor was collected from the vitreous body following insertion of a 16-gauge needle through the lateral aspect of the sclera and gentle aspiration into a 3 mL syringe. Samples were put into a sterile serum activator tube (Becton Dickinson, East Rutherford, NJ, USA). Samples were frozen at −80 °C for a minimum of 1 h prior to analysis. Only samples free of blood contamination were used. Samples were prepared through thawing and then spun in a microcentrifuge (Heska Corporation, Loveland, CO, USA) for 5 min, and the supernatant was drawn up and placed in a 0.5 mL red-top tube (Heska Corporation, Loveland, CO, USA) for analysis. Fluid analyses included VUN (mg/dL), CREA (mg/dL), and MG (mg/dL). Analytes were measured on a Heska Element DC Veterinary Chemistry Analyzer (Heska Corporation, Loveland, CO, USA) according to the manufacturer’s instructions for serum samples. If test results exceeded the measuring range (defined by the manufacturer), the vitreous sample was diluted according to the manufacturer’s instructions using a dilution factor (1:4) and re-measured. The sample dilution could not be carried out for one case as the entire sample was consumed during initial testing and the test result exceeding the measuring range was assigned the maximum quantitative value. The magnesium concentration for one cat was not measured. Data normality was assessed for VUN, CREA and MG in diseased and normal cats by using the Shapiro–Wilk normality test. Descriptive statistics were calculated to present the mean and standard deviation. Two-sample t-test was used to compare mean differences of VUN, CREA, and MG levels between cats with renal or post-renal disease and normal cats. Results are presented as mean difference, standard error of the mean difference, and p-value. Tests were considered significant if p< 0.05. Statistical analysis was performed using Statistix 10 (Analytical Software, Tallahassee, FL, USA). 3. Results Twenty-nine adult cats were used in this study. Nine cats had renal or post-renal disease including renal papillary necrosis, acute renal failure due to ethylene glycol intoxication, or obstructive uropathy. Twenty cats that did not have renal or post-renal disease and died from non-renal causes including vehicular trauma or predation were used for comparison. VUN and CREA were higher in the cats with renal or post-renal disease than in the cats without renal or post-renal disease. The MG levels in four out of eight cats with renal or post-renal disease were higher than those in the cats without renal or post-renal disease. The mean VUN, CREA, and MG in the cats with renal or post-renal disease were 245.56 mg/dL, 13.92 mg/dL, and 3.46 mg/dL, respectively (Table 1). The mean VUN, CREA, and MG in the cats without renal or post-renal disease were 23.65 mg/dL, 1.03 mg/dL, and 1.61 mg/dL, respectively (Table 1). Results of the two-sample t-test were statistically significant for the differences between the VUN (p = 0.0003), CREA (p = 0.0002), and MG (p = 0.04) in the cats with renal or post-renal disease compared to cats without renal or post-renal disease. A preliminary sample size calculation was not performed. However, post hoc power analysis for two-sample t-tests showed 99% and 100% power to detect VUN and CREA, respectively, in the samples selected in this study, while the power to detect MG was lower at 57%. Table 1. Descriptive statistics of VUN, creatinine, and magnesium levels in renal/post-renal disease and control cats, and mean difference, SE, and p-value for group differences. \| \| Disease Cats \| Control Cats \| \| \| \| :---: :---: :---: \| \| Variable \| N \| Mean \| SD \| N \| Mean \| SD \| Mean Difference \| SE \| p-Value \| \| VUN \| 9 \| 245.56 \| 112.14 \| 20 \| 23.65 \| 5.93 \| 221.9 \| 24.58 \| 0.0003 \| \| Creatinine \| 9 \| 13.92 \| 6.14 \| 20 \| 1.03 \| 0.67 \| 12.8 \| 1.36 \| 0.0002 \| \| Magnesium \| 8 \| 3.46 \| 2.13 \| 20 \| 1.61 \| 0.51 \| 1.8 \| 0.49 \| 0.04 \| Open in a new tab VUN = vitreous urea nitrogen, SD = standard deviation, SE = standard error; p-value for two-sample t-test is <0.05. 4. Discussion This study evaluated the differences between VUN, CREA, and MG measurements collected from the vitreous body in cats with renal and post-renal disease and compared the results to cats without renal and post-renal disease. A comparison of VUN and CREA from samples collected from cats with renal or post-renal disease revealed higher levels of both analytes could be detected in the vitreous humor, similar to the results of antemortem biochemical analysis of serum or plasma. Half of the cats with renal or post-renal disease had elevated levels of MG within the vitreous humor. The results of this study suggest that in deceased cats with renal or post-renal disease, VUN and CREA are likely to be elevated, and concurrently, MG could also be elevated. The ability to perform biochemical analysis of vitreous humor using samples collected at autopsy will be useful to assess renal function rather than relying solely on the subjective assessments of histopathology of the kidney. Postmortem vitreous chemistry in human medicine is considered a valuable tool to help support diagnoses made at autopsy as different analytes can be helpful in determining cause of death [2,3]. Multiple studies have concluded that urea nitrogen is stable and “closely approximate the antemortem serum levels” . It has been shown that VUN, CREA, and uric acid can all be used to diagnose chronic kidney disease and end-stage renal failure in humans and are considered relatively stable analytes postmortem . A comparison of antemortem serum and the postmortem vitreous humor in cats, dogs, and cattle showed a strong correlation between serum and postmortem eye fluid for urea nitrogen and CREA collected at necropsy at least 24 h after death . In cattle, it has been shown that vitreous humor can be used to detect magnesium imbalances for at least 48 h postmortem The postmortem analysis of vitreous humor from eight cows with uremia showed that postmortem analysis of vitreous humor could be successfully used to diagnose azotemia . In the current study, 100% of cats with elevated VUN and CREA died from renal or post-renal disease. Magnesium is considered to be the “forgotten” ion. A postmortem study in humans showed that vitreous humor MG did not correlate with age, postmortem interval, or other vitreous electrolytes and was stable during the early postmortem period (up to 75 h postmortem) . In a feline study, serum hypermagnesemia was observed in 16/42 (38.1%) and hypomagnesemia in 6/42 (14.3%) cats with chronic renal disease . In another study, hypermagnesemia was detected only in cats with renal disease, obstructive uropathy, or neoplastic disease . In the current study, 50% of cats developed hypermagnesemia within the vitreous humor, and these cats had renal papillary necrosis (one cat), ethylene glycol toxicity (two cats), and urethral obstruction (one cat). It is possible that diffusion of MG into the vitreous humor takes time, and since the time of onset for disease was not known for all cases, it is possible that if the cats had survived longer, vitreous MG levels would have become elevated. The Heska Element DC Veterinary Chemistry Analyzer has been shown to be able to analyze postmortem vitreous humor from cats . Although there are no standard reference intervals for VUN, CREA, or MG in cats, we found that normal cats, those without renal or post-renal disease, had a mean VUN level of 23.47 mg/dL (range: 14.6–32.6 mg/dL), a mean CREA level of 1.61 mg/dL (range: 0.2–2.3 mg/dL), and a mean MG level of 1.26 mg/dL (range: 0.3–2.3 mg/dL). The range of VUN in these normal cats was similar to the reported reference ranges for BUN (15–32 mg/dL) according to the Heska Element DC Veterinary Chemistry Analyzer. The range of CREA and MG in normal cats was at or below the reported reference range for CREA (0.8–1.8 mg/dL) and MG (1.8–2.7 mg/dL) according to the Heska Element DC Veterinary Chemistry Analyzer. Although the normal ranges of CREA and MG tend to overlap lower levels, this does not affect the possibility of utilizing these analytes to assess for azotemia or hypermagnesemia. 5. Conclusions This study indicated that cats with renal or post-renal disease, such as ethylene glycol toxicity or urethral obstruction, can have significant VUN, CREA, and occasionally MG elevations compared to cats that do not have renal or post-renal disease. The results of this study provide evidence that the pathologist can use feline vitreous humor to assess for antemortem azotemia using vitreous chemistry rather than hypothesize that changes within the kidneys could have resulted in azotemia. Although we were able to detect a statistically significant difference between VUN, CREA, and MG measurements in these cats, a larger sample size could allow further statistical analyses and the grouping of patients by various disease processes. Therefore, future larger studies analyzing postmortem vitreous humor in cats as well as other species, including dogs, cattle, horses, and marine mammals, are needed. A recent study using delphinid vitreous humor demonstrated that the Abaxis VetScan VS2 (VetScan) chemistry analyzer had acceptable precision for multiple analytes , and marine mammals should be considered. Another area of future study should include the effects of storage (freezing) and decomposition of animals as it is not uncommon for bodies to be frozen for storage or for animals to be discovered several days after their death. Additionally, the assessment of cats with clinically documented pre-renal azotemia such as dehydration would also be beneficial as dehydration is a difficult diagnosis to make solely based on autopsy findings. Another process to explore is fatal cases of heatstroke in cats and dogs since in humans, Zhu et al. showed that elevated vitreous CREA levels can reflect skeletal muscle damage in cases of heat stroke . Further vitreous chemistry research would benefit veterinary pathologists performing autopsies as vitreous chemistry can be utilized to assess for clinically relevant azotemia. Acknowledgments The authors would like to thank Elizabeth Stocker for her assistance with sample processing. Author Contributions Conceptualization, A.W.S.; methodology, A.W.S.; formal analysis, A.W.S. and T.D.; data curation, A.W.S., D.R. and C.V.; writing—original draft preparation, A.W.S.; writing—review and editing, D.R., C.V. and T.D. All authors have read and agreed to the published version of the manuscript. Institutional Review Board Statement Institutional Animal Care and Use Committee review and approval were waived for this study as all samples were collected postmortem and none of the animals were euthanized for purposes related to this study. Informed Consent Statement Not applicable. Data Availability Statement The data that support the findings of this study are available from the corresponding author, A.W.S., upon reasonable request. Conflicts of Interest The authors declare no conflict of interest. Funding Statement This research was partially funded by a grant from the Summerlee Foundation and the UF Foundation “A Cat Has No Name” fund (025309). 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Articles from Toxics are provided here courtesy of Multidisciplinary Digital Publishing Institute (MDPI) ACTIONS View on publisher site PDF (204.2 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Materials and Methods 3. Results 4. Discussion 5. 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51	https://dictionary.justia.com/capacity-to-contract	capacity to contract Definition, Meaning & Usage \| Justia Legal Dictionary Log InSign Up Find a Lawyer Ask a Lawyer Research the Law Law Schools Laws & Regs Newsletters Marketing Solutions Justia Connect Pro Membership Practice Membership Public Membership Justia Lawyer Directory Platinum Placements Gold Placements Justia Elevate SEO Websites Blogs Justia Amplify PPC Management Google Business Profile Social Media Justia Onward Blog JustiaDictionary capacity to contract Search rocket docketA court known for quickly resolving cases, often by strictly adhering to deadlines unconditional dischargeBeing freed from a duty or obligation with no conditions attached firewallA divider within a financial institution that prevents the sharing of confidential information between banking and brokerage operations Texas two-step bankruptcyA legal strategy involving Texas corporate law where a company splits into two entities so that one entity can declare bankruptcy to manage liabilities such as lawsuits while the other can be used to protect the company’s assets capacity to contract Definition of "capacity to contract" 1. The legal ability to form a binding agreement, generally excluding minors and those with significant mental impairments who cannot comprehend agreement terms How to use "capacity to contract" in a sentence 1. The law often challenges the capacity to contract of minors under 18. 2. Due to severe dementia, her capacity to contract was questioned in court. 3. His capacity to contract was unchallenged since he was financially and mentally capable. Provide Feedback Browse Our Legal Dictionary #ABCDEFGHIJKLMNOPQRSTUVWXYZ Justia Legal Resources Find a Lawyer Bankruptcy Lawyers Business Lawyers Criminal Lawyers Employment Lawyers Estate Planning Lawyers Family Lawyers Personal Injury Lawyers More... Individuals Bankruptcy Criminal Divorce DUI Estate Planning Family Law Personal Injury More... Business Business Formation Business Operations Employment Intellectual Property International Trade Real Estate Tax Law More... Law Schools Dictionary Admissions Financial Aid Course Outlines Law Journals Blogs Employment More... U.S. Federal Law U.S. Constitution U.S. Code Regulations Supreme Court Circuit Courts District Courts Dockets & Filings More... U.S. State Law State Constitutions State Codes State Case Law California Florida New York Texas More... Other Databases Legal Jobs Legal Blogs Business Forms Product Recalls Patents Trademarks Countries More... Marketing Solutions Justia Connect Membership Justia Lawyer Directory Justia Premium Placements Justia Elevate (SEO, Websites) Justia Amplify (PPC, GBP) Justia Onward Blog Testimonials More... This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. © 2025 JustiaJustia ConnectLegal PortalCompanyHelpTerms of ServicePrivacy PolicyMarketing Solutions
52	https://suresolv.com/brain-teaser/liar-and-truth-teller-riddle-step-step-easy-solution	Liar and truth-teller riddle step by step solution \| Suresolv Jump to Navigation Brain teaser Maze puzzles Riddles Inventive puzzles Reasoning puzzles Paradoxes Mathematical puzzles Logic puzzles Number lock puzzles Missing number puzzles River crossing puzzles Ball weighing puzzles Matchstick puzzles Problem solving Real life problem solving Problem solving techniques Problem solving skills Problem solving tools Health Cough Sleep Critical thinking HOW TOs Innovation Innovation concepts Systematic innovation Inventive principles Innovations around us Explorations Urbanization and Environment Global warming Trees and Plants Mahua AI Inquiries Our life Basics Basic maths Reasoning and logic analysis Exams School math Efficient Math problem solving Exam guidelines SSC CGL Tier II SSC CGL SSC CHSL SSC CPO Bank PO UGC/CSIR Net Maths WBCS Main Variety Sudoku Sudoku beginner level 1 and 2 Medium Sudoku level 3 Hard Sudoku level 4 NYTimes Sudoku hard Expert Sudoku Books Reflections Innovative creations Unusual thoughts You are here Home » Brain teaser » Liar and truth-teller riddle with step by step easy solution Submitted by Atanu Chaudhuri on Wed, 18/11/2020 - 23:32 Think methodically with common sense reasoning to solve the liar and truth-teller riddle In liar and truth-teller riddle, to know the safe path, a traveler can ask only one question to any of a habitual liar or a truth-teller. What to ask? Story of the liar and truth-teller Passing through many a land, a traveler came upon a deep jungle where the path split into two and went into the depths of the jungle. He had heard from a wise man that such a forest lay ahead. The wise man told him, “One path will lead to the warmth of a friendly village, but the other will lead you to the den of hungry tigers and sure death.” Now only he remembered the rest of what the wise old man told him, “When you are at the fork trying to figure out the safe path, two men will suddenly appear ready to help you. Beware, one of them will be a habitual liar—his answer to any question must always be a lie, but the other will just be the reverse—he always will tell you the truth.” The wise man finished, “Don’t forget, you can ask only one question to any of the two to find the correct safe path. You won’t get a second chance. They will understand your question, know the safe path, and know the nature of answering of each other but will answer only with YES or NO.” Recommended time for you to find the safe path: 15 minutes. Comments This is a classic logic puzzle from old times. Logicians who juggle with pure logic by choice will tell you the right answer in no time. But we are not logicians. We are common folks who use common sense logic and deductive reasoning in our own way. If you are not a logician, have a go. The experience will be enriching. You’ll get better results if you imagine yourself as the traveler. Systematic Solution to the Liar and truth-teller Riddle—First stage Key Pattern Discovery of the Nature of the Answer for Choosing the safe path Without going deep into the riddle, you will take up an initial trial in mind. You will visualize what could be the answers from the two if you ask the simple, straightforward question, Do you think the path to my left is the safe path to the village of friendly people? If indeed the left path IS the safe one, The liar will answer: NO, and the Truth-teller: YES. Without knowing which one of them is telling the truth, you won’t be able to decide. Thinking more you realize you have, Two unknown paths, and two men eager to help, but again you don’t know their nature of habitual answering. With such combinations of unknowns, you understand, it will be impossible to form the right question if you think in conventional way. You realize, You must form the question NOT in a simple way. Without wasting time on what type of question you have to ask, your attention shifts to the ANSWER itself. This is a natural way to solve problems—to analyze and understand all characteristics of the end result first, comparing it with given information. Note:The method is End State Analysis Approach, an often-used natural problem solving technique packed with power. Thinking more in this direction, you ask yourself the most important QUESTION at the moment, What must be the NATURE OF THE ANSWER from the two for me to know the safe path? You have already experienced that the answer to a simple question from one helper will be NO and the other, YES, just the reverse. You make a firm conclusion, If the answers from the two are REVERSE to each other, you won’t find the safe path. No going away from this. The conclusion is a fact and an inviolable truth. Ah ha, you couldn’t imagine earlier it would come to this pass. Yes, for you to be sure about the safe path, Whatever be the question, answers from both the helpers must be the SAME in every situation. This is a revelation to you and in problem solving terminology is discovery of the key pattern. You have now a precise requirement of their answers to the single question (or precise requirement specification). Naturally, the answer will be YES or NO. But if it is NO, both will answer NO. Same must also be true for YES for knowing safe path. This is the first breakthrough. You know half the battle you have won by knowing all about the answer that can be known. Nature of answer fully identified, it is time to shift focus of attention to the NATURE OF THE QUESTION. Systematic Solution to the Liar and truth-teller Riddle—Second stage Process of Knowing Nature of the Question to ask You realize you have to think in a NEW way to make the second breakthrough by understanding precisely the nature of the question—the requirement specification of the question. What can be a new form of a question! As you think more on this, the requirement of ‘NEW form’ strikes you and aided by the focus of the new form of question you reason, Well, the usual simple form of the question has been a single question for any combination of situation. All forms of such a question would have the same end result of failure. WHAT CAN BE CHANGED ABOUT THE QUESTION? Property change analysis technique The technique of exploring new ways to change a key property (in this case, the number of parts in the question) of the key entity (in this case, the question itself) often produce great results for quick and innovative solution to the problem. This is called, Property change analysis. Of course, you can very well change the number of parts in the single combined question, especially as you remember from your experience, Instead of a single question, two questions can easily be combined to form the single compound question. When you were younger, your neighbor uncle once asked when you opened the door, Is your father home? Is he well? Your father was home and well. So you answered, Yes. One answer to two questions. You further realize, if either your father were not home or not well or both, you might have simply said, No. That must be the new way that will help you know the safe path, Instead of one-part single question, you will ask a single question with two component questions joined together by AND, the easiest way to join two questions. And you know from elementary knowledge in language that such a compound question is easy to form—your neighbor uncle could have asked such a single question, Is your father home and well? You think you have more or less solved the problem and that too by using no difficult technique or concept. You have followed just simple common sense problem solving techniques coupled with concepts drawn out of everyday real life experiences and elementary knowledge of language. What have you achieved till now? You know now both the form of the question to ask and nature of the answer. The only task left is to create such a question with such an answer in reality. Systematic Solution to the Liar and truth-teller Riddle—Third stage Forming the Question by using the Answer and Question Specifications With clear idea of the nature of the answer and form of the question, a possible question to be asked could be, Are you the truth-teller, and do you think the path to my left is the safe path? Possible scenarios are, Situation 1: The helper asked is the truth-teller and the left path is the safe path: Final answer:YES. Situation 2:The helper asked is the liar and the left path is the safe path; his answer to first part of the question will be NO. So answer to the combined question will truly be NO. Habitual liar, being what he is, cannot but reverse this NO to YES. Hisfinal answer will be YES. So if the answer to your question is YES till now, you may assume that the left path is the safe path. Rest of the possible results when the left path is not the safe path are, Situation 3: The helper asked is the truth-teller and the left path is NOT the safe path: Final answer: NO. Situation 4:The helper asked is the liar and the left path is NOT the safe path. Correct answer to both the first and second parts of the question being NO, the habitual liar will reverse it to YES. Final answer: YES. The answer differs from the answer by the other helper. It violates the requirement specification of the answer. Just on the brink of success, you find that joining of two questions simply by ‘and’ won’t fully work. There is a challenge yet to overcome at the third stage of using the specifications of the answer and question toform the right way to combine the two question parts. You assumed an easy way to join the two questions without thinking much on how to join. But the effort is not all in vain—you will get important clues on how to combine by analyzing the results of the trial. Systematic Solution to the Liar and truth-teller Riddle—Final stage: Identifying Crucial Requirement of Joining two Questions You wonder, why and where did combining two questions by ‘and’ fail? As you concentrate on finding answer to this question by analyzing the result of your last try, you realize, Answers to the component question 1 and question 2 are independent of each other, and No need to think about the answer of the truth-teller, because primarily the goal is to force the liar to reverse the answer of the truth-teller twice to match his final answer. Liar is the most important person in the solution. In the Situation 2, using ‘and’ for joining, you could indeed force the liar to reverse the answer of the truth-teller twice. But in the Situation 4, where true answer to both part of questions were NO, the liar reversed each to YES and combined also to YES. HE was happy to think that he had indeed reversed the true combined answer and did justice to his habit! It leads you to discover the breakthrough idea. To force the liar to reverse true answer twice by two questions, One question must be DEPENDENT on the other. So, the successful method of joining the two questions must make sure that, The liar will reverse correct answer to the INDEPENDENT first question. Thinking that he has answered true to his habit, he will face the second DEPENDENT question. As a habitual liar, he will now have no other choice than toreverse answer to the first independent question once more by the second question. By two reversals, answers will be correct and same as the truth-teller. What is the other method of joining two questions that will achieve this result? Again, your common experience of using the language helps you for the final breakthrough. There must beonly one real question but asked twice in the commonly used form, What will be your answer if I ask you whether the QUESTION is true? The first of the two component questions the liar must answer by the rules of the language is, ‘whether the QUESTION is true’. The liar REVERSES the correct answer to this first component question and forms the INTERMEDIATE RESULT in line with his nature of answering. Now he faces the second component question, What will be your answer if INTERMEDIATE RESULT is true? He will have no other choice than to reverse the intermediate result true to his nature and form the final result. The final result becomes the REVERSE OF INTERMEDIATE RESULT. As INTERMEDIATE RESULT has itself been reversed once from correct result, the final result returns to the value of the correct result again by this second reversal. Following is the schematic of this mechanism, This is pure logic, but with reason, method and two trials, you have learned enough on how the single compound question must be formed and asked to any of the two helpers to find the safe path. With confidence, you finally form the two-part single question that will lead you to know the safe path, What will be your answer if I ask you whether the path to my left is the safe path? You ask this question to any of the two helpers. If the answer is YES, you take the left path, and if it is NO, you take the right path. There cannot be any other possibility—you have indeed forced the liar to reverse the true answer twice to match the final answer of the truth-teller in both the cases. Knowing that answer to the question from both helpers will be same, you asked the question to any of the two helpers. Comment Without going into the second trial, you might have discovered the most important PRIMARY REQUIREMENT OF ACTION—the liar must be forced to reverse the true result twice. This discovery is not difficult to make if you realize early that solving the riddle depends only on finding the right answer to the question, How to make the liar answer same as the truth-teller in any situation! The problem solving techniques, concepts and common knowledge used End state analysis: Analyzing the desired result or last action first. Goal is to gain more knowledge about the last action for achieving the desired result. Refining requirement specification in steps: Knowing precise requirements of the answer first and the question second simplified the steps to the solution greatly. Question, analysis and answer or QAA technique: The problem is simplified stage by stage by asking a series of relevant questions, analyzing problem details for getting the answer and actually forming the answer. Property change analysis technique: Exploring how many ways the key property of the key entity can be changed and assessing promise of each change often proves to be crucial in solving a complex problem. Great innovation can be created by this technique. Elementary knowledge in language: Basic domain concepts on combining two component questions to a single compound question, but in different ways. Well-formed trials or experiments: To learn more of the problem to identify patterns and create methods. Step by step deductive reasoning: By using all the above and discovering key patterns of information for solving the problem with complete confidence. End note Our focus all through the above solution process has been to think as the traveler in a simple way and find the safe path step by step using systematic reasoning and problem solving techniques drawn out of common knowledge and experience. Know how to solve difficult problems easily without wasting time on random attempts Our ebook on puzzle solutions by innovative methods will show you just that. Puzzles for Adults: 50 Brain Teasers with Step-by-Step Solutions: Boost Your Power of Problem Solving (BUY from Apple books, Barnes & Noble, Rokuten Kobo, Vivlio, Angus & Robertson, Tolino, PayHip and others) BUY the eBookAmazonKindle version here, from Google Playhere and Paperbackhere. Second book on Innovative Solutions to Matchstick Puzzles from Suresolv BUY Creative Matchstick Puzzles Innovative Solutions eBook Amazon Kindle version BUY the paperback here. Puzzles you may enjoy Easy to hard brain teasers with systematic solutions Challenging brain teasers with solutions: Long list. This will always be the most up-to-date full list with the brain teasers classified into categories that can be browsed separately. 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53	https://math24.files.wordpress.com/2013/02/ch1-folland.pdf	Real Analysis Chapter 1 Solutions Jonathan Conder 3. (a) Let M be an inﬁnite σ-algebra of subsets of some set X. There exists a countably inﬁnite subcollection C ⊆M, and we may choose C to be closed under taking complements (adding in missing complements if necessary). For each x ∈X, deﬁne Dx := ∩{C ∈C \| x ∈C}, so that Dx ∈M. Let x, y ∈X and suppose y ∈Dx. Then y ∈C for all C ∈C with x ∈C. Moreover, if y ∈C for some C ∈C, then y / ∈Cc ∈C, so x / ∈Cc and hence x ∈C. This implies that, if C ∈C, then x ∈C iﬀy ∈C. In particular Dx = Dy. If x, y ∈X and there exists z ∈Dx ∩Dy, then Dx = Dz = Dy, so the collection D := {Dx \| x ∈X} is pairwise disjoint. If C ∈C, then C = ∪{Dx \| x ∈C}, since x ∈Dx ⊆C for all x ∈C. Therefore, the image of the map ∪: 2D →2X contains C, so there exists a surjection from some subset of 2D onto C. If D were ﬁnite, then 2D would be as well, which is a contradiction because C is inﬁnite. Therefore D is an inﬁnite, pairwise disjoint subcollection of M. (b) Let M be an inﬁnite σ-algebra of subsets of some set X. By the previous exercise, there exists an inﬁnite, pairwise disjoint subcollection D of M. Since Q is countable, there exists an injection q : Q →D. Deﬁne r : 2Q →M by r(A) := ∪a∈Aq(a). Then r is well-deﬁned because each A ⊆Q is countable, and injective because D is pairwise disjoint. There exists an injection from R →2Q, for example the map x 7→(−∞, x] ∩Q (which is injective by the least upper bound property of R). Composing these injections gives an injective map from R →M, which shows that card(M) ≥c. 4. Let A be an algebra which is closed under countable increasing unions. To show that A is a σ-algebra, it suﬃces to show that it is closed under arbitrary countable unions. To this end, let {En}n∈N be a countable collection of members of A. For each n ∈N deﬁne Fn := ∪n k=1Ek. Since A is an algebra it is clear that (Fn)n∈N is an increasing sequence of members of A, so ∪n∈NEn = ∪n∈NFn ∈A as required. Conversely, it is plain that every σ-algebra is closed under countable increasing unions. 5. Let E be a collection of subsets of some set X, and let M be the σ-algebra generated by E. Deﬁne N := ∪{A \| A is the σ-algebra generated by some countable subcollection of E}. It is clear that ∅, X ∈N, E ⊆N and N is closed under taking complements. Let {En}n∈N be a countable collection of members of N. For each n ∈N there exists a countable subcollection En of E such that En is in the σ-algebra generated by En. Since union of countably many countable sets is countable, F := ∪n∈NEn is a countable subcollection of E. Let A be the σ-algebra generated by F, so that A ⊆N. For each n ∈N the σ-algebra generated by En is a subset of A, because A is a σ-algebra containing En. This implies that En ∈A for all n ∈N, and hence ∪n∈NEn ∈A ⊆N. Therefore N is closed under countable unions, so it is a σ-algebra containing E. This implies that M ⊆N. Conversely, let E ∈N. Then E belongs to the σ-algebra A generated by some countable subcollection of E. Since M is a σ-algebra containing this subcollection, A ⊆M and hence E ∈M. This shows that N ⊆M and hence N = M. 6. Note that µ(∅) = µ(∅∪∅) = µ(∅) = 0. Given a sequence (An)n∈N of disjoint sets in M, there exists a sequence (En)n∈N of sets in M and a sequence (Fn)n∈N of subsets of measure zero sets from M such that An = En ∪Fn for all n ∈N. Note that (En)n∈N is pairwise disjoint, and ∪n∈NFn is a subset of a measure zero set. Therefore µ(∪n∈NAn) = µ((∪n∈NEn) ∪(∪n∈NFn)) = µ(∪n∈NEn) = X n∈N µ(En) = X n∈N µ(An). This shows that µ is a measure. Let A ⊆X, and suppose there exists B ∈M such that A ⊆B and µ(B) = 0. Then B = E ∪F for some E ∈M and F a subset of a measure zero set N ∈M. It follows that A ⊆E ∪N, where E ∪N ∈M and µ(E ∪N) ≤µ(E) + µ(N) = µ(E) + 0 ≤µ(B) = 0. Therefore A = ∅∪A ∈M, which shows that µ is complete. 1 Real Analysis Chapter 1 Solutions Jonathan Conder Let λ : M →[0, ∞] be another measure which extends µ, and let A ∈M. Then A = E ∪F for some E ∈M and F a subset of a measure zero set N ∈M. It follows that µ(E) = λ(E) ≤λ(A) ≤λ(E) + λ(F) ≤λ(E) + λ(N) = µ(E) + µ(N) = µ(E). Therefore λ(A) = µ(E) = µ(A), so λ = µ. This shows that µ is the unique measure on M which extends µ. 7. Set µ := Pn j=1 ajµj, and note that µ(∅) = 0. If {Ei}i∈N is a pairwise disjoint collection of members of M, then µ(∪i∈NEi) = n X j=1 ajµj(∪i∈NEi) = n X j=1 aj X i∈N µj(Ei) = X i∈N n X j=1 ajµj(Ei) = X i∈N µ(Ei) (note that we are allowed to change the order of summation because all the terms are non-negative; alternatively you can consider the convergent/divergent cases separately). This shows that µ is a measure on (X, M). 8. Let (X, M, µ) be a measure space and {En}n∈N a countable collection of measurable sets. Then (∩∞ k=nEk)n∈N is an increasing sequence of measurable sets with µ(∩∞ k=nEk) ≤µ(En) for all n ∈N, so µ(lim inf n∈N En) = µ(∪n∈N(∩∞ k=nEk)) = lim n→∞µ(∩∞ k=nEk) = lim inf n→∞µ(∩∞ k=nEk) ≤lim inf n→∞µ(En). Moreover (∪∞ k=nEk)n∈N is a decreasing sequence of measurable sets with µ(∪∞ k=nEk) ≥µ(En), so µ(lim sup n∈N En) = µ(∩n∈N(∪∞ k=nEk)) = lim n→∞µ(∪∞ k=nEk) = lim sup n→∞µ(∪∞ k=nEk) ≥lim sup n→∞µ(En) provided that the ﬁrst term of the sequence has ﬁnite measure, i.e. µ(∪∞ k=1Ek) < ∞. 9. Let (X, M, µ) be a measure space and E, F ∈M. Then E = (E ∩F) ⊔(E \ F) and (E \ F) ⊔F = E ∪F, so µ(E) + µ(F) = µ(E ∩F) + µ(E \ F) + µ(F) = µ(E ∩F) + µ(E ∪F). 10. Clearly µE(∅) = µ(∅∩E) = µ(∅) = 0. If {En}n∈N is a pairwise disjoint collection of members of M, then µE(∪n∈NEn) = µ((∪n∈NEn) ∩E) = µ(∪n∈N(En ∩E)) = X n∈N µ(En ∩E) = X n∈N µE(En). Therefore µE is a measure. 11. Suppose that µ is continuous from below. Let {En}n∈N be a pairwise disjoint collection of measurable sets, and for each n ∈N deﬁne Fn := ∪n i=1En. Since µ is continuous from below and ﬁnitely additive, µ(∪i∈NEi) = µ(∪n∈NFn) = lim n→∞µ(Fn) = lim n→∞ n X i=1 µ(Ei) = ∞ X i=1 µ(Ei). This shows that µ is a measure. Conversely, if µ is a measure it is continuous from below by Theorem 1.8. Now suppose that µ(X) < ∞and µ is continuous from above. If (En)n∈N is an increasing sequence of measurable sets, then (Ec n)n∈N is a decreasing sequence of measurable sets, which implies that µ(∪n∈NEn) = µ(X) −µ((∪n∈NEn)c) = µ(X) −µ(∩n∈NEc n) = µ(X) −lim n→∞µ(Ec n) = lim n→∞µ(En). This shows that µ is continuous from below, so it is a measure by the previous paragraph. Conversely, if µ is a (ﬁnite) measure it is continuous from above by Theorem 1.8. 2 Real Analysis Chapter 1 Solutions Jonathan Conder 14. Suppose for a contradiction that there exists C ∈(0, ∞) such that every measurable subset F ⊆E satisﬁes µ(F) ≤C or µ(F) = ∞. Set M := sup{µ(F) \| F ⊆E is measurable and µ(F) < ∞}, and note that 0 ≤M ≤C. For each n ∈N there exists a measurable subset En ⊆E such that M −n−1 ≤µ(En) < ∞. Set Fn := ∪n i=1En for each n ∈N and deﬁne F := ∪n∈NFn. Note that M −n−1 ≤µ(En) ≤µ(F) and also µ(Fn) ≤Pn i=1 µ(En) < ∞for all n ∈N, so M ≤µ(F) = limn→∞µ(Fn) ≤M. This shows that µ(F) = M, so µ(E \ F) = ∞. Since µ is semiﬁnite, there exists a measurable subset A ⊆E \ F such that 0 < µ(A) < ∞. This contradicts the deﬁnition of M, because A ∪F ⊆E but µ(F) < µ(A) + µ(F) = µ(A ∪F) < ∞. Therefore, for any C ∈(0, ∞) there exists a measurable subset F ⊆E such that C < µ(F) < ∞. 17. Let A, B ⊆X be disjoint µ∗-measurable sets, and let E ⊆X. Then (A ∪B) ∩Ac = B and hence µ∗(E ∩(A ∪B)) = µ∗(E ∩(A ∪B) ∩A) + µ∗(E ∩(A ∪B) ∩Ac) = µ∗(E ∩A) + µ∗(E ∩B). It follows by induction that Pn k=1 µ∗(E ∩Ak) = µ∗(E ∩(∪n k=1Ak)) ≤µ∗(E ∩(∪∞ k=1Ak)) for all n ∈N. Therefore ∞ X k=1 µ∗(E ∩Ak) ≤µ∗(E ∩(∪∞ k=1Ak)), which implies that µ∗(E ∩(∪∞ k=1Ak)) = P∞ k=1 µ∗(E ∩Ak) because µ∗is subadditive. 18. (a) Let E ⊆X and ε ∈(0, ∞). If µ∗(E) = ∞then X ∈A ⊆Aσ, E ⊆X and µ∗(X) ≤µ∗(E) + ε. Otherwise µ∗(E) = inf (X n∈N µ0(An) An ∈A for all n ∈N and E ⊆∪n∈NAn ) , so there exists a sequence (An)n∈N in A such that E ⊆∪n∈NAn and P n∈N µ0(An) ≤µ∗(E) + ε. If A := ∪n∈NAn, then A ∈Aσ, E ⊆A and µ∗(A) ≤P n∈N µ0(An) ≤µ∗(E) + ε. (b) Let E ⊆X such that µ∗(E) < ∞. Suppose that E is µ∗-measurable. For each n ∈N there exists Bn ∈Aσ such that E ⊆Bn and µ∗(Bn) ≤µ∗(E) + n−1. Deﬁne B := ∩n∈NBn, so that B ∈Aσδ and E ⊆B. Since µ∗(E) < ∞, it follows that µ∗(B \ E) ≤µ∗(Bn ∩Ec) = µ∗(Bn) −µ∗(Bn ∩E) = µ∗(Bn) −µ∗(E) ≤n−1 for all n ∈N. This implies that µ∗(B \ E) = 0. Conversely, suppose there exists B ∈Aσδ such that E ⊆B and µ∗(B \ E) = 0. If F ⊆X, then µ∗(F ∩E) + µ∗(F ∩Ec) ≤µ∗(F ∩B) + µ∗(F ∩Bc) + µ∗(F ∩(B \ E)) ≤µ∗(F ∩B) + µ∗(F ∩Bc) = µ∗(F) because B ∈Aσδ is µ∗-measurable. Clearly µ∗(F) ≤µ∗(F ∩E) + µ∗(F ∩Ec), so E is µ∗-measurable. 19. If E ⊆X is µ∗-measurable, then µ∗(X) = µ∗(X ∩E)+µ∗(X ∩Ec) = µ∗(E)+µ∗(Ec) by deﬁnition, so µ∗(E) = µ∗(E). Conversely, let E ⊆X and suppose that µ∗(E) = µ∗(E). By the previous exercise, for each n ∈N there exist An, Bn ∈Aσ such that E ⊆An, Ec ⊆Bn, µ∗(An) ≤µ∗(E) + n−1 and µ∗(Bn) ≤µ∗(Ec) + n−1. If A := ∩n∈NAn, then µ∗(A \ E) ≤µ∗(An ∩Bn) = µ∗(An) −µ∗(An \ Bn) = µ∗(An) −µ∗(Bc n) = µ∗(An) −(µ∗(X) −µ∗(Bn)) ≤µ∗(E) + n−1 −µ∗(X) + µ∗(Ec) + n−1 3 Real Analysis Chapter 1 Solutions Jonathan Conder = µ∗(E) −µ∗(E) + 2n−1 = 2n−1 for all n ∈N, and hence µ∗(A\E) = 0. Moreover E ⊆A and A ∈Aσδ, so by the previous exercise E is µ∗-measurable. 20. (a) If (An)n∈N is a sequence in M∗such that E ⊆∪n∈NAn, then µ∗(E) ≤µ∗(∪n∈NAn) ≤ X n∈N µ∗(An) = X n∈N µ(An). Therefore µ∗(E) ≤µ+(E), by deﬁnition. If there exists A ∈M∗such that E ⊆A and µ∗(A) = µ∗(E), then µ+(E) ≤µ(A) + ∞ X n=2 µ(∅) = µ∗(A) = µ∗(E), so µ∗(E) = µ+(E). Conversely, suppose that µ∗(E) = µ+(E). By exercise 18 part (a), for each n ∈N there exists An ∈M∗ σ = M∗such that E ⊆An and µ+(An) ≤µ+(E) + n−1. Deﬁne A := ∩n∈NAn, so that A ∈M∗ and E ⊆A. Moreover µ+(A) ≤µ+(An) ≤µ+(E) + n−1 for all n ∈N, so µ+(A) ≤µ+(E). It follows that µ∗(A) = µ+(A) = µ+(E) = µ∗(E), because A ∈M∗and E ⊆A. (b) Let µ0 be a premeasure on an algebra A ⊆M∗such that µ∗is induced by µ0, and let E ⊆X. By exercise 18 part (a), for each n ∈N there exists An ∈Aσ ⊆M∗such that E ⊆An and µ∗(An) ≤µ∗(E) + n−1. Deﬁne A := ∩n∈NAn, so that A ∈M∗and E ⊆A. Moreover µ∗(A) ≤µ∗(An) ≤µ∗(E) + n−1 for all n ∈N, so µ∗(A) ≤µ∗(E). It follows that µ∗(A) = µ∗(E), because E ⊆A. By the previous exercise, this implies that µ∗(E) = µ+(E). Therefore µ∗= µ+. (c) Deﬁne µ∗: 2X →[0, ∞] by µ∗(∅) = 0, µ∗({0}) = 2, µ∗({1}) = 2 and µ∗(X) = 3. Clearly µ∗(A) ≤µ∗(B) for all A ⊆B ⊆X. Moreover µ∗is subadditive, because µ∗(X) < µ∗({0}) + µ∗({1}). Therefore µ∗is an outer measure. Note that {0} is not µ∗-measurable, because µ∗(X) = 3 < 2 + 2 = µ∗(X ∩{0}) + µ∗(X ∩{0}c). This implies that µ+({0}) = inf{P n∈N µ∗(An) \| An ⊆X for all n ∈N and An = X for some n ∈N} = µ∗(X) = 3 ̸= 2 = µ∗({0}). 21. Let E ⊆X be locally µ∗-measurable, and let A ⊆X. If µ∗(A) = ∞, then clearly µ∗(A ∩E) + µ∗(A ∩Ec) ≤µ∗(A). Otherwise, by exercise 18 part (a), for each n ∈N there exists a µ∗-measurable set An ⊆X such that A ⊆An and µ∗(An) ≤µ∗(A) + n−1. In particular, if n ∈N then µ∗(An) < ∞, and hence An ∩E and An ∩Ec = An ∩(An ∩E)c are µ∗-measurable (because E is locally µ∗-measurable). It follows that µ∗(A ∩E) + µ∗(A ∩Ec) ≤µ∗(An ∩E) + µ∗(An ∩Ec) = µ(An ∩E) + µ(An ∩Ec) = µ(An) = µ∗(An) ≤µ∗(A) + n−1 for all n ∈N, so µ∗(A ∩E) + µ∗(A ∩Ec) ≤µ∗(A). Clearly µ∗(A) ≤µ∗(A ∩E) + µ∗(A ∩Ec) in either case, so E is µ∗-measurable. This shows that µ is saturated. 22. (a) Clearly µ\|M = µ, so by exercise 6 from section 1.3 it suﬃces to show that M∗= M. To this end, let A ∈M. Then A = E ∪F for some E ∈M and F ⊆X such that F ⊆B for some B ∈M with µ(B) = 0. Clearly A ⊆E ∪B ∈M = Mσδ and µ∗((E ∪B) \ A) ≤µ∗(B) = 0. By exercise 18 part (c), it follows that A ∈M∗. Conversely, let A ∈M∗. By exercise 18 part (c) there exists B ∈Mσδ = M such that A ⊆B and µ∗(B \ A) = 0. Since B \ A ∈M∗, exercise 18 part (b) implies that B \ A ⊆E and µ∗(E \ (B \ A)) = 0 for some E ∈M. Note that B \ E ∈M and A ∩E ∈M∗. Moreover µ∗(A ∩E) ≤µ∗(E \ (B \ A)) = 0, so there exists F ∈M such that A ∩E ⊆F and µ∗(F \ (A ∩E)) = 0, by exercise 18 part (b). It follows that µ(F) = µ∗(F) ≤µ∗(F \ (A ∩E)) + µ∗(A ∩E) = 0. 4 Real Analysis Chapter 1 Solutions Jonathan Conder Since A = (B \ E) ∪(A ∩E), it follows that A ∈M. Therefore M∗= M. (b) Let e E ∈e M, and b µ be the completion of µ. Given A ∈M∗with µ(A) < ∞, the converse of the previous exercise implies that A ∈M (using exercise 18 part (b) instead of part (c)). Hence A = E ∪F for some E ∈M and F ⊆X such that F ⊆B for some B ∈M with µ(B) = 0. In particular b µ(A) = µ(E) = µ∗(E) ≤µ∗(A) < ∞. Therefore e E ∩A ∈M, as e E is locally b µ-measurable. Since µ∗( e E ∩A) < ∞the previous exercise implies that e E ∩A ∈M∗(again using exercise 18 part (b) instead of part (c)). This shows that e E is locally µ∗-measurable, so e E ∈M∗by exercise 21. Conversely, let e E ∈M∗and A ∈M be such that b µ(A) < ∞. Then A = E ∪F for some E ∈M and F ⊆X such that F ⊆B for some B ∈M with µ(B) = 0. It follows that µ∗( e E ∩A) ≤µ∗(A) ≤µ∗(E) + µ∗(F) = µ(E) + 0 = b µ(A) < ∞. This implies that A ∈M∗(by the previous exercise using part (b) instead of part (c)), so e E ∩A ∈M∗and hence e E ∩A ∈M (again by the previous exercise). This shows that e E is locally b µ-measurable, so e E ∈e M. Therefore e M = M∗. By exercise 6 from section 1.3, e b µ and µ agree on M. If E ∈e M \ M, then e b µ(E) = ∞by deﬁnition. Moreover, if E ∈e M = M∗and µ(E) < ∞, then E ∈M by the previous exercise (using part (b) instead of part (c)). This implies that e b µ and µ also agree on e M \ M. 23. (a) Let E := {(a, b] ∩Q \| a, b ∈R}. Clearly ∅= (0, 0] ∩Q ∈E. If (a1, b1] ∩Q ∈E and (a2, b2] ∩Q ∈E then their intersection is (max{a1, a2}, min{b1, b2}] ∩Q ∈E. Moreover, the complement of (a, b] ∩Q ∈E is ((−∞, a] ∩Q) ∪ ((b, ∞] ∩Q), which is a disjoint union of elements of E provided that a ≤b. If a > b then the complement of (a, b] ∩Q is just (−∞, ∞] ∩Q ∈E. This shows that E is an elementary family of subsets of Q, so the collection of ﬁnite disjoint unions of members of E is an algebra. If (a1, b1] ∩Q ∈E and (a2, b2] ∩Q ∈E are not disjoint, then their union is (min{a1, a2}, max{b1, b2}] ∩Q ∈E. Therefore A is the collection of ﬁnite disjoint unions of members of E, so A is an algebra. (b) Let M be the σ-algebra generated by A. Since Q is countable and M is closed under countable unions, it suﬃces to show that {x} ∈M for all x ∈Q. Given x ∈Q and n ∈N, it is clear that (x −1 n, x] ∩Q ∈M. Therefore {x} = ∩n∈N((x −1 n, x] ∩Q) ∈M, as required. (c) By deﬁnition µ0(∅) = 0. Let (En)n∈N be a sequence of disjoint members of A whose union lies in A. If En = ∅ for all n ∈N, then ∪n∈NEn = ∅and hence µ0(∪n∈NEn) = 0 = P n∈N µ0(En). Otherwise ∪n∈NEn ̸= ∅, and Em ̸= ∅for some m ∈N, so µ0(∪n∈N) = ∞and P n∈N µ0(En) ≥µ0(Em) = ∞. Therefore µ0 is a premeasure on A. Deﬁne µ1, µ2 : 2Q →[0, ∞] by µ1(E) =    ∞, E ̸= ∅ 0, E = ∅ and µ1(E) =    ∞, E contains 2−nm for some m ∈Z and n ∈N 0, otherwise. Clearly µ1\|A = µ0 = µ2\|A , because every non-empty interval contains a rational of the form 2−nm for some m ∈Z and n ∈N. For the same reason that µ0 is a premeasure, µ1 is a measure. A very similar argument implies that µ2 is a measure. But µ1 ̸= µ2 because µ1({3−1}) = ∞whereas µ2({3−1}) = 0. 24. (a) Since µ(A) + µ(Ac ∩B) = µ(A ∪B) = µ(B) + µ(Bc ∩A), by symmetry it suﬃces to show that µ(Ac ∩B) = 0. To this end, note that E ⊆A ∪Bc = (Ac ∩B)c (if a member of E is not in Bc, it is in B ∩E = A ∩E ⊆A). So µ(X) = µ∗(X) = µ∗(E) ≤µ∗((Ac ∩B)c) = µ((Ac ∩B)c) = µ(X) −µ(Ac ∩B), and hence µ(Ac ∩B) = 0 as required. 5 Real Analysis Chapter 1 Solutions Jonathan Conder (b) It is clear that ME is a σ-algebra on E and that ν(∅) = 0. Let {En}n∈N be a pairwise disjoint subset of ME. For each n ∈N there exists An ∈M such that En = An ∩E. Set A := ∪i∈N ∪∞ j=i+1 (Ai ∩Aj) and deﬁne Bn := An \ A for each n ∈N. It is easily checked that {Bn}n∈N is a pairwise disjoint subset of M and that En = Bn ∩E for all n ∈N. Therefore ν(∪n∈NEn) = ν(∪n∈N(Bn ∩E)) = ν((∪n∈NBn) ∩E) = µ(∪n∈NBn) = X n∈N µ(Bn) = X n∈N ν(En), which shows that ν is a measure on ME. 25. Let (Cn)n∈N be a sequence of compact intervals covering R, and ﬁx n ∈N. There exists a Gδ set Vn ∈Mµ and a null set Nn ∈Mµ such that E ∩Cn = Vn \ Nn. Let (Vnk)k∈N be a sequence of open sets such that Vn = ∩k∈NVnk. For each k ∈N deﬁne an open set Unk := Vnk ∪Cc n, and set U := ∩n,k∈NUnk. Then U is a Gδ set and U \ E ⊆∪n∈NNn. Indeed, if x ∈U \ E there exists n ∈N such that x ∈Cn, which implies that x ∈Vnk for all k ∈N and hence x ∈Vn but x / ∈E ∩Cn = Vn \ Nn. Moreover E ⊆U, because E ⊆Unk for all n, k ∈N. It follows that E = U \ (U \ E), where U \ E is a null set. If n ∈N, then E ∩Cn = Hn ∪Nn for some Fσ set Hn ∈Mµ and some null set Nn ∈Mµ. Clearly ∪n∈NHn and ∪n∈NNn are respectively Fσ and null sets. Moreover E = ∪n∈N(E ∩Cn) = (∪n∈NHn) ∪(∪n∈NNn). 26. Let E ∈Mµ and suppose that µ(E) < ∞. Given ε ∈(0, ∞), there exists an open set U ∈Mµ such that E ⊆U and µ(U) < µ(E) + ε 2. Let (Un)n∈N be a sequence of disjoint open intervals such that ∪n∈NUn = U. Then X n∈N µ(Un) = µ(U) < ∞, so there exists N ∈N such that P∞ n=N+1 µ(Un) < ε 2. Deﬁne A := ∪N n=1Un. It follows that µ(E△A) ≤µ(E \ A) + µ(A \ E) ≤µ(E \ U) + µ(U \ A) + µ(U \ E) = 0 + ∞ X n=N+1 µ(Un) + µ(U) −µ(E) < ε. 28. Let a, b ∈R. Since µF is continuous from above, µF ([a, b]) = µF (∩n∈N(a −n−1, b]) = lim n→∞µF ((a −n−1, b]) = lim n→∞(F(b) −F(a −n−1)) = F(b) −F(a−). It follows that µF ({a}) = µF ([a, a]) = F(a)−F(a−), in which case µF ([a, b)) = µF ([a, b])−µF ({b}) = F(b−)−F(a−) and µF ((a, b)) = µF ([a, b)) −µF ({a}) = F(b−) −F(a). 29. (a) Suppose that E ⊆N but m(E) > 0. Deﬁne R := Q ∩[0, 1), and for each r ∈R set Er := E + r. Clearly each Er is measurable with m(Er) = m(E), and ∪r∈REr ⊆[0, 2). Let r, s ∈R and suppose that Er intersects Es. Then there exists t ∈Er ∩Es, so that t −r, t −s ∈E ⊆N. Since t −r = (t −s) + (s −r) and s −r ∈Q, the deﬁnition of N implies that t −s = t −r. Therefore r = s, which shows that {Er}r∈R is pairwise disjoint. Hence ∞= X r∈R m(E) = X r∈R m(Er) = m(∪r∈REr) ≤m([0, 2)) = 2, which is a contradiction. Therefore m(E) = 0. 6 Real Analysis Chapter 1 Solutions Jonathan Conder (b) Suppose that m(E) > 0, but every subset of E is measurable. Since E = ∪n∈Z(E ∩[n, n + 1)), there exists n ∈Z such that m(E ∩[n, n + 1)) > 0. Deﬁne F := (E ∩[n, n + 1)) −n, so that F ⊆[0, 1), m(F) > 0 and every subset of F is measurable. Also deﬁne R := Q ∩[−1, 1], and for each r ∈R set Nr := N + r. It is clear that [0, 1) ⊆∪r∈RNr, and hence F = ∪r∈R(F ∩Nr). If r ∈R then F ∩Nr ⊆F and (F ∩Nr) −r ⊆N, which implies that both subsets are measurable (by containment in F and translational invariance) and have measure zero (by the previous exercise and translational invariance). Therefore m(F) ≤P r∈R m(F ∩Nr) = P r∈R 0 = 0, which is a contradiction so not every subset of E is measurable. 30. Let E ∈L with m(E) > 0, and suppose there exists α ∈(0, 1) such that m(E ∩I) ≤αm(I) for all open intervals I. Without loss of generality m(E) < ∞(since m is semiﬁnite, we may replace E by a subset of ﬁnite positive measure). Deﬁne ε := m(E)(1 −α), so that ε > 0. Since E ∈L and m is outer regular, there exists an open set U ⊆R such that E ⊆U and m(U) < m(E) + ε. As U is open, there exists a pairwise disjoint collection {Ii}i∈N of open intervals such that U = ∪i∈NIi, and these intervals are bounded because m(U) < ∞. If if i ∈N then m(Ii) = m(Ii \ E) + m(E ∩Ii) ≤m(Ii \ E) + αm(Ii) and hence (1 −α)m(Ii) ≤m(Ii \ E). Since ∪i∈N(Ii \ E) = U \ E has measure m(U) −m(E), it follows that (1 −α)m(U) = (1 −α) X i∈N m(Ii) = X i∈N (1 −α)m(Ii) ≤ X i∈N m(Ii \ E) = m(∪i∈N(Ii \ E)) < ε = (1 −α)m(E) and hence m(U) < m(E) ≤m(U), which is impossible. Thus, for each α ∈(0, 1) there exists an open interval I such that m(E ∩I) > αm(I). The same clearly holds for α ∈(−∞, 0]. 31. Let E ∈L with m(E) > 0, and set α := 3 4. By the previous exercise there exists a open interval I ⊆R with endpoints a, b ∈R such that m(E ∩I) > αm(I). Suppose there exists x ∈(−1 2m(I), 1 2m(I)) such that x / ∈E −E. Then −x / ∈E −E, and clearly 0 ∈E −E, so we may assume x > 0. Moreover E ⊆(Ec + x) ∩(Ec −x). Note that (a, a + 2x] ⊆I, because 2x < m(I) = b −a. Deﬁne n := max{k ∈N \| a + 2kx < b}, so that m(E ∩(a, a + 2nx]) = n X k=1 (m(E ∩(a + 2(k −1)x, a + (2k −1)x]) + m(E ∩(a + (2k −1)x, a + 2kx])) ≤ n X k=1 (m((Ec −x) ∩(a + 2(k −1)x, a + (2k −1)x]) + m((Ec + x) ∩(a + (2k −1)x, a + 2kx])) ≤ n X k=1 (m(Ec ∩(a + (2k −1)x, a + 2kx] −x) + m(Ec ∩(a + 2(k −1)x, a + (2k −1)x] + x)) ≤ n X k=1 (m(Ec ∩(a + (2k −1)x, a + 2kx]) + m(Ec ∩(a + 2(k −1)x, a + (2k −1)x])) ≤m(Ec ∩(a, a + 2nx]). This implies that m(E ∩(a, a + 2nx]) ≤1 2(m(E ∩(a, a + 2nx]) + 1 2m(Ec ∩(a, a + 2nx])) = 1 2m((a, a + 2nx]) = nx. Note that 4nx ≥m(I), since otherwise k := 2n satisﬁes a + 2kx < b (but k > n). It follows that m(E ∩I) ≤m(E ∩(a, a + 2nx]) + m((a + 2nx, b)) ≤m(I) −nx ≤m(I) −m(I) 4 = 3m(I) 4 = αm(I). This is a contradiction, so there does not exist x ∈(−1 2m(I), 1 2m(I)) such that x / ∈E −E. 7 Real Analysis Chapter 1 Solutions Jonathan Conder 33. Choose a surjection q : N →Q ∩[0, 1], and for each n ∈N deﬁne In := (q(n) −3−n, q(n) + 3−n). For each n ∈N, deﬁne Dn = In \ (∪∞ k=n+1Ik). If m, n ∈N and m < n, then Dm ∩Dn = ∅because Dn ⊆In ⊆∪∞ k=m+1Ik. Moreover 2 · 3−n = m(In) ≤m(Dn) + m(∪∞ k=n+1Ik) ≤m(Dn) + ∞ X k=n+1 m(Ik) = m(Dn) + ∞ X k=n+1 2 · 3−k = m(Dn) + 3−n, and hence m(Dn) ≥3−n, for all n ∈N. For each n ∈N there exists a Borel set An ⊆Dn such that 0 < m(An) < 3−n, which can be found by intersecting Dn with an interval of the form [3−n−1k, 3−n−1(k + 1)] for some k ∈Z (not all of the intersections can have zero measure). Deﬁne A := [0, 1] ∩(∪n∈NAn), and let I be a subinterval of [0, 1] with midpoint c. There exists n ∈N such that 4 · 3−n < m(I), and there are inﬁnitely many rationals in (c −3−n, c + 3−n), so there exists k ∈N with k ≥n and q(k) ∈(c −3−n, c + 3−n). It follows that Ik ⊆(c −2 · 3−n, c + 2 · 3n) ⊆I. Therefore Ak ⊆A ∩I (since Ak ⊆Ik ⊆[0, 1]), so m(A ∩I) ≥m(Ak) > 0. Moreover m(A ∩I) = m((∪n∈NAn) ∩I) = m(Ak) + m((∪n∈N{k}An) ∩I) < m(Dk) + m(I \ Dk) = m(I), because m(Ak) < m(Dk) and ∪n∈N{k}An ⊆∪n∈N{k}Dn ⊆Dc k. 8
54	https://www.merriam-webster.com/thesaurus/revealable	Synonyms of reveal as in to disclose as in to show reveal verb verb Synonyms & Similar Words Antonyms & Near Antonyms Synonyms & Similar Words Antonyms & Near Antonyms Synonym Chooser How does the verb reveal differ from other similar words? Some common synonyms of reveal are betray, disclose, divulge, and tell. While all these words mean "to make known what has been or should be concealed," reveal may apply to supernatural or inspired revelation of truths beyond the range of ordinary human vision or reason. When is betray a more appropriate choice than reveal? The words betray and reveal are synonyms, but do differ in nuance. Specifically, betray implies a divulging that represents a breach of faith or an involuntary or unconscious disclosure. When can disclose be used instead of reveal? While in some cases nearly identical to reveal, disclose may imply a discovering but more often an imparting of information previously kept secret. When could divulge be used to replace reveal? While the synonyms divulge and reveal are close in meaning, divulge implies a disclosure involving some impropriety or breach of confidence. In what contexts can tell take the place of reveal? The synonyms tell and reveal are sometimes interchangeable, but tell implies an imparting of necessary or useful information. Example Sentences Browse Nearby Words Cite this Entry “Reveal.” Merriam-Webster.com Thesaurus, Merriam-Webster, Accessed 28 Sep. 2025. Share More from Merriam-Webster on reveal Nglish: Translation of reveal for Spanish Speakers Subscribe to America's largest dictionary and get thousands more definitions and advanced search—ad free! More from Merriam-Webster Can you solve 4 words at once? Can you solve 4 words at once? Word of the Day kerfuffle See Definitions and Examples » Get Word of the Day daily email! Popular in Grammar & Usage Is it 'autumn' or 'fall'? Using Bullet Points ( • ) Merriam-Webster’s Great Big List of Words You Love to Hate How to Use Em Dashes (—), En Dashes (–) , and Hyphens (-) A Guide to Using Semicolons Popular in Wordplay Ye Olde Nincompoop: Old-Fashioned Words for 'Stupid' Great Big List of Beautiful and Useless Words, Vol. 3 'Za' and 9 Other Words to Help You Win at SCRABBLE 12 Words Whose History Will Surprise You More Words with Remarkable Origins Popular Is it 'autumn' or 'fall'? Ye Olde Nincompoop: Old-Fashioned Words for 'Stupid' Great Big List of Beautiful and Useless Words, Vol. 3 Games & Quizzes Learn a new word every day. Delivered to your inbox! © 2025 Merriam-Webster, Incorporated
55	https://www.cdc.gov/mmwr/preview/mmwrhtml/00056324.htm	\| \| \| \| \| \| \| \| --- --- --- \| Skip Navigation LinksSkip Navigation Links \| \| \| \| \| \| \| \| \| \| \| Centers for Disease Control and Prevention \| \| \| \| Safer Healthier People Blue White \| \| \| \| \| Blue White \| \| bottom curve \| \| CDC Home \| Search \| Health Topics A-Z \| spacer \| spacer \| \| \| \| \| --- \| \| spacer \| \| \| Blue curve \| MMWR spacer \| \| \| \| \| \| \| \| \| Centers for Disease Control and Prevention \| \| \| \| Safer Healthier People Blue White \| \| \| \| Blue White \| \| bottom curve \| \| \| \| --- \| \| spacer \| \| \| Blue curve \| MMWR spacer \| \| \| \| spacer \| \| \| \| spacer \| \| \| \| \| --- \| \| Persons using assistive technology might not be able to fully access information in this file. For assistance, please send e-mail to: mmwrq@cdc.gov. Type 508 Accommodation and the title of the report in the subject line of e-mail. Outbreak of Vibrio parahaemolyticus Infection Associated with Eating Raw Oysters and Clams Harvested from Long Island Sound -- Connecticut, New Jersey, and New York, 1998 During July-September 1998, an outbreak of Vibrio parahaemolyticus infections associated with consumption of oysters and clams harvested from Long Island Sound occurred among residents of Connecticut, New Jersey, and New York. This is the first reported outbreak of V. parahaemolyticus linked to consumption of shellfish harvested from New York waters. This report summarizes the investigation of this outbreak. On August 10, 1998, a New York City resident with toxigenic V. cholerae O1 infection who had not traveled recently was reported to the New York City Department of Health (NYCDOH). NYCDOH initiated an investigation to determine the most likely source of the infection. Using a broadcast facsimile, NYCDOH contacted all Queens County laboratories on August 12 and, on August 26, asked selected infectious diseases physicians and all New York City hospitals and laboratories to consider V. cholerae as a potential cause of diarrhea and to report any confirmed or suspected Vibrio infections to the NYCDOH. Although no additional V. cholerae infections were reported, 23 culture-confirmed cases of V. parahaemolyticus were reported among residents of Connecticut, New Jersey, and New York. Dates of illness onset ranged from July 21 through September 17 (Figure_1). An investigation coordinated by the New York State Department of Health determined that 22 of 23 ill persons had eaten or handled oysters, clams, or crustaceans: 16 ate raw oysters or clams, two ate steamed crabs, one ate crab cakes, one ate boiled crabs and lobsters, one ate lobster roll, and one handled live crabs. The median onset of illness following consumption of shellfish was 19 hours (range: 12-52 hours). Clinical histories were available for 19 of the 23 ill persons; 17 (89%) had gastroenteritis and two (11%) had bloodstream infections with lower extremity edema and bullae. Among patients with gastroenteritis, reported clinical symptoms included diarrhea (100%), abdominal cramps (94%), nausea (94%), vomiting (82%), fever (47%), bloody stools (29%), headache (24%), and myalgia (24%). Median duration of gastrointestinal illness was 5 days. Traceback investigations by local and state health departments identified the site of harvest for oysters or clams eaten by 11 of the 16 patients. Oysters or clams eaten by eight patients were harvested from Oyster Bay, off New York's Long Island Sound, during August 4-27. Shellfish tags from oysters and clams eaten by the other three persons indicated harvest areas elsewhere off Long Island or, in one case, Washington state (1) . During the outbreak period, mean surface water temperature measurements from 15 Oyster Bay stations was 77.2 F (25.1 C), compared with cooler 1997 and 1996 measurements (74.1 F {23.4 C} and 69.4 F {20.7 C}, respectively). On September 10, the New York State Department of Environmental Conservation (NYSDEC) closed Oyster Bay to harvesting of shellfish and recalled shellfish harvested from that area after August 10. Laboratory testing of 12 V. parahaemolyticus clinical isolates, including the eight traced to Oyster Bay, identified O3:K6 serotype. Pulsed-field gel electrophoresis (PFGE) performed on four clinical isolates at the New York City Bureau of Labs indicated that three isolates epidemiologically linked to Oyster Bay had indistinguishable PFGE patterns, and the other isolate not linked to Oyster Bay had a distinctly different pattern. Oysters harvested on five occasions from Oyster Bay during September 11-October 14 contained V. parahaemolyticus at less than or equal to 120 colony forming units {cfu} per gram of oyster meat. None of these environmental isolates matched the outbreak strain or other clinical isolates by PFGE. On the basis of these results and a decline in water temperature to 63.5 F (17.5 C), NYSDEC reopened Oyster Bay to commercial shellfish harvesting on October 22. No additional culture-confirmed cases of V. parahaemolyticus infection have been reported. Reported by: E Wechsler, C D'Aleo, VA Hill, J Hopper, D Myers-Wiley, E O'Keeffe, J Jacobs, F Guido, A Huang, MD, Westchester County Health Dept, New Rochelle; SN Dodt, B Rowan, M Sherman, A Greenberg, MD, Div of Disease Control, Nassau County Dept of Health, Mineola; D Schneider, B Noone, L Fanella, BR Williamson, E Dinda, M Mayer, MD, Suffolk County Dept of Health Svcs, Hauppauge; M Backer, A Agasan, MD, Enteric Pathogens Laboratory, L Kornstein, PhD, Environmental Microbiology Laboratory, New York City Bur of Laboratories; F Stavinsky, Bur of Environmental Investigations; B Neal, D Edwards, M Haroon, D Hurley, L Colbert, J Miller, MD, B Mojica, MD, New York City Dept of Health; E Carloni, B Devine, M Cambridge, Bur of Community Sanitation and Food Protection; T Root, D Schoonmaker, M Shayegani, Wadsworth Laboratories, Albany; W Hastback, New York State Dept of Environmental Conservation; B Wallace, MD, S Kondracki, Bur of Communicable Disease Control; P Smith, MD, State Epidemiologist, New York State Dept of Health. S Matiuck, K Pilot, M Acharya, Bur of Labs; G Wolf, W Manley, C Genese, J Brooks, MD, Acting State Epidemiologist, New Jersey Dept of Health. Z Dembek, PhD, J Hadler, MD, State Epidemiologist, Connecticut Dept of Public Health. Center for Food Safety and Applied Nutrition, Food and Drug Administration. Fish and Wildlife Svc, US Dept of Agriculture. Foodborne and Diarrheal Diseases Br, Div of Bacterial and Mycotic Diseases, National Center for Infectious Diseases; State Br, Div of Applied Public Health Training, Epidemiology Program Office; and EIS officers, CDC. Editorial Note Editorial Note: This is the fourth multistate outbreak of V. parahaemolyticus infections in the United States since 1997, and the first associated with shellfish harvested from the northeast Atlantic Ocean. Before 1997, foodborne outbreaks caused by V. parahaemolyticus had been infrequently reported in the United States (1). During 1997-1998, multistate outbreaks of V. parahaemolyticus were associated with consumption of raw or undercooked oysters harvested from the Pacific Northwest and Texas (2; CDC, unpublished data, 1998). V. parahaemolyticus is a halophilic, gram-negative bacterium that naturally inhabits marine and estuarine waters. V. parahaemolyticus infections are usually acquired by persons who eat raw or undercooked shellfish, particularly oysters, or whose skin wounds are exposed to warm seawater. The most common clinical manifestation of infection is self-limited gastroenteritis, but infections may result in septicemia that can be life threatening (3,4). The concentration of V. parahaemolyticus in seawater increases with increasing water temperature and corresponds with a seasonal increase in sporadically occurring cases in warmer months (4). This outbreak and the recent outbreaks of V. parahaemolyticus infections in the Pacific Northwest and Texas occurred during summer months. To reduce the risk for V. parahaemolyticus and other shellfish-associated infections, persons should avoid eating raw or undercooked shellfish, particularly during warmer months. Monitoring of environmental conditions, such as water temperature and salinity, may help determine when shellfish harvesting areas should be closed and re-opened to harvesting. Guidelines regulating the harvesting of oysters and clams rely on quantitative measurement of V. parahaemolyticus levels in oyster or clam meat. However, data from recent outbreaks may require revision of these guidelines. The recommended action level of V. parahaemolyticus per gram of oyster meat that must be detected in the absence of human illness before closing oyster beds is greater than 10,000 cfu/g. Oyster samples that were harvested from implicated beds in the Pacific Northwest in 1997 and Oyster Bay in 1998 yielded less than 200 V. parahaemolyticus cfu/g of oyster meat, indicating that human illness can occur at levels much lower than the current action level. Infection with V. parahaemolyticus is not a notifiable condition in most states, including New York. This outbreak was detected only coincidentally because of enhanced surveillance during an investigation of a case of V. cholerae O1. Health-care providers treating patients with gastroenteritis who have a history of recent ingestion of raw or undercooked shellfish should consider Vibrio infection and request a stool culture specifically for Vibrio. Clinical laboratories should use thiosulfate-citrate-bile salts-sucrose agar (TCBS), a selective medium for culturing for Vibrio spp., when culturing stool specimens for Vibrio and should consider using TCBS for routine screening of all stools specimens, at least during summer months. CDC coordinates a passive Gulf Coast Vibrio surveillance system and the Foodborne Diseases Active Surveillance Network (FoodNet) to monitor the incidence of Vibrio infections. Because of these multistate outbreaks, all states should consider making infections with V. parahaemolyticus and other vibrioses reportable, with referral of clinical isolates to public health laboratories for confirmation and strain subtyping. References 1. Bean NH, Goulding JS, Lao C, Angulo FJ. Surveillance for foodborne-disease outbreaks -- United States, 1988-1992. MMWR 1996;45(no. SS-5). 2. CDC. Outbreak of Vibrio parahaemolyticus infections associated with eating raw oysters -- Pacific Northwest, 1997. MMWR 1998;47;457-62. 3. Levine WC, Griffin PM, and the Gulf Coast Vibrio Working Group. Vibrio infections on the Gulf Coast: results of first year of regional surveillance. J Infect Dis 1993;167:479-83. 4. Hlady WG, Klontz KC. The epidemiology of Vibrio infections in Florida, 1981-1993. J Infect Dis 1996;173:1176-83. The shipper that provided the oysters harvested elsewhere in Long Island also had received oysters from Oyster Bay at approximately the same time. Although comingling of shellfish is against state regulations, it is known to occur. Figure_1 Figure_1 Return to top. Disclaimer All MMWR HTML versions of articles are electronic conversions from ASCII text into HTML. This conversion may have resulted in character translation or format errors in the HTML version. Users should not rely on this HTML document, but are referred to the electronic PDF version and/or the original MMWR paper copy for the official text, figures, and tables. An original paper copy of this issue can be obtained from the Superintendent of Documents, U.S. Government Printing Office (GPO), Washington, DC 20402-9371; telephone: (202) 512-1800. Contact GPO for current prices. Questions or messages regarding errors in formatting should be addressed to mmwrq@cdc.gov. Page converted: 01/28/99 \| \| \| \| \| \| \| \| HOME \| ABOUT MMWR \| MMWR SEARCH \| DOWNLOADS \| RSS \| CONTACT POLICY \| DISCLAIMER \| ACCESSIBILITY \| \| \| \| \| \| \| \| \| \| --- --- \| \| Safer, Healthier People \| \| USA.GovDHHS \| Department of Health and Human Services \| \| This page last reviewed 5/2/01 \| Persons using assistive technology might not be able to fully access information in this file. For assistance, please send e-mail to: mmwrq@cdc.gov. Type 508 Accommodation and the title of the report in the subject line of e-mail. Outbreak of Vibrio parahaemolyticus Infection Associated with Eating Raw Oysters and Clams Harvested from Long Island Sound -- Connecticut, New Jersey, and New York, 1998 During July-September 1998, an outbreak of Vibrio parahaemolyticus infections associated with consumption of oysters and clams harvested from Long Island Sound occurred among residents of Connecticut, New Jersey, and New York. This is the first reported outbreak of V. parahaemolyticus linked to consumption of shellfish harvested from New York waters. This report summarizes the investigation of this outbreak. On August 10, 1998, a New York City resident with toxigenic V. cholerae O1 infection who had not traveled recently was reported to the New York City Department of Health (NYCDOH). NYCDOH initiated an investigation to determine the most likely source of the infection. Using a broadcast facsimile, NYCDOH contacted all Queens County laboratories on August 12 and, on August 26, asked selected infectious diseases physicians and all New York City hospitals and laboratories to consider V. cholerae as a potential cause of diarrhea and to report any confirmed or suspected Vibrio infections to the NYCDOH. Although no additional V. cholerae infections were reported, 23 culture-confirmed cases of V. parahaemolyticus were reported among residents of Connecticut, New Jersey, and New York. Dates of illness onset ranged from July 21 through September 17 (Figure_1). An investigation coordinated by the New York State Department of Health determined that 22 of 23 ill persons had eaten or handled oysters, clams, or crustaceans: 16 ate raw oysters or clams, two ate steamed crabs, one ate crab cakes, one ate boiled crabs and lobsters, one ate lobster roll, and one handled live crabs. The median onset of illness following consumption of shellfish was 19 hours (range: 12-52 hours). Clinical histories were available for 19 of the 23 ill persons; 17 (89%) had gastroenteritis and two (11%) had bloodstream infections with lower extremity edema and bullae. Among patients with gastroenteritis, reported clinical symptoms included diarrhea (100%), abdominal cramps (94%), nausea (94%), vomiting (82%), fever (47%), bloody stools (29%), headache (24%), and myalgia (24%). Median duration of gastrointestinal illness was 5 days. Traceback investigations by local and state health departments identified the site of harvest for oysters or clams eaten by 11 of the 16 patients. Oysters or clams eaten by eight patients were harvested from Oyster Bay, off New York's Long Island Sound, during August 4-27. Shellfish tags from oysters and clams eaten by the other three persons indicated harvest areas elsewhere off Long Island or, in one case, Washington state (1) . During the outbreak period, mean surface water temperature measurements from 15 Oyster Bay stations was 77.2 F (25.1 C), compared with cooler 1997 and 1996 measurements (74.1 F {23.4 C} and 69.4 F {20.7 C}, respectively). On September 10, the New York State Department of Environmental Conservation (NYSDEC) closed Oyster Bay to harvesting of shellfish and recalled shellfish harvested from that area after August 10. Laboratory testing of 12 V. parahaemolyticus clinical isolates, including the eight traced to Oyster Bay, identified O3:K6 serotype. Pulsed-field gel electrophoresis (PFGE) performed on four clinical isolates at the New York City Bureau of Labs indicated that three isolates epidemiologically linked to Oyster Bay had indistinguishable PFGE patterns, and the other isolate not linked to Oyster Bay had a distinctly different pattern. Oysters harvested on five occasions from Oyster Bay during September 11-October 14 contained V. parahaemolyticus at less than or equal to 120 colony forming units {cfu} per gram of oyster meat. None of these environmental isolates matched the outbreak strain or other clinical isolates by PFGE. On the basis of these results and a decline in water temperature to 63.5 F (17.5 C), NYSDEC reopened Oyster Bay to commercial shellfish harvesting on October 22. No additional culture-confirmed cases of V. parahaemolyticus infection have been reported. Reported by: E Wechsler, C D'Aleo, VA Hill, J Hopper, D Myers-Wiley, E O'Keeffe, J Jacobs, F Guido, A Huang, MD, Westchester County Health Dept, New Rochelle; SN Dodt, B Rowan, M Sherman, A Greenberg, MD, Div of Disease Control, Nassau County Dept of Health, Mineola; D Schneider, B Noone, L Fanella, BR Williamson, E Dinda, M Mayer, MD, Suffolk County Dept of Health Svcs, Hauppauge; M Backer, A Agasan, MD, Enteric Pathogens Laboratory, L Kornstein, PhD, Environmental Microbiology Laboratory, New York City Bur of Laboratories; F Stavinsky, Bur of Environmental Investigations; B Neal, D Edwards, M Haroon, D Hurley, L Colbert, J Miller, MD, B Mojica, MD, New York City Dept of Health; E Carloni, B Devine, M Cambridge, Bur of Community Sanitation and Food Protection; T Root, D Schoonmaker, M Shayegani, Wadsworth Laboratories, Albany; W Hastback, New York State Dept of Environmental Conservation; B Wallace, MD, S Kondracki, Bur of Communicable Disease Control; P Smith, MD, State Epidemiologist, New York State Dept of Health. S Matiuck, K Pilot, M Acharya, Bur of Labs; G Wolf, W Manley, C Genese, J Brooks, MD, Acting State Epidemiologist, New Jersey Dept of Health. Z Dembek, PhD, J Hadler, MD, State Epidemiologist, Connecticut Dept of Public Health. Center for Food Safety and Applied Nutrition, Food and Drug Administration. Fish and Wildlife Svc, US Dept of Agriculture. Foodborne and Diarrheal Diseases Br, Div of Bacterial and Mycotic Diseases, National Center for Infectious Diseases; State Br, Div of Applied Public Health Training, Epidemiology Program Office; and EIS officers, CDC. Editorial Note Editorial Note: This is the fourth multistate outbreak of V. parahaemolyticus infections in the United States since 1997, and the first associated with shellfish harvested from the northeast Atlantic Ocean. Before 1997, foodborne outbreaks caused by V. parahaemolyticus had been infrequently reported in the United States (1). During 1997-1998, multistate outbreaks of V. parahaemolyticus were associated with consumption of raw or undercooked oysters harvested from the Pacific Northwest and Texas (2; CDC, unpublished data, 1998). V. parahaemolyticus is a halophilic, gram-negative bacterium that naturally inhabits marine and estuarine waters. V. parahaemolyticus infections are usually acquired by persons who eat raw or undercooked shellfish, particularly oysters, or whose skin wounds are exposed to warm seawater. The most common clinical manifestation of infection is self-limited gastroenteritis, but infections may result in septicemia that can be life threatening (3,4). The concentration of V. parahaemolyticus in seawater increases with increasing water temperature and corresponds with a seasonal increase in sporadically occurring cases in warmer months (4). This outbreak and the recent outbreaks of V. parahaemolyticus infections in the Pacific Northwest and Texas occurred during summer months. To reduce the risk for V. parahaemolyticus and other shellfish-associated infections, persons should avoid eating raw or undercooked shellfish, particularly during warmer months. Monitoring of environmental conditions, such as water temperature and salinity, may help determine when shellfish harvesting areas should be closed and re-opened to harvesting. Guidelines regulating the harvesting of oysters and clams rely on quantitative measurement of V. parahaemolyticus levels in oyster or clam meat. However, data from recent outbreaks may require revision of these guidelines. The recommended action level of V. parahaemolyticus per gram of oyster meat that must be detected in the absence of human illness before closing oyster beds is greater than 10,000 cfu/g. Oyster samples that were harvested from implicated beds in the Pacific Northwest in 1997 and Oyster Bay in 1998 yielded less than 200 V. parahaemolyticus cfu/g of oyster meat, indicating that human illness can occur at levels much lower than the current action level. Infection with V. parahaemolyticus is not a notifiable condition in most states, including New York. This outbreak was detected only coincidentally because of enhanced surveillance during an investigation of a case of V. cholerae O1. Health-care providers treating patients with gastroenteritis who have a history of recent ingestion of raw or undercooked shellfish should consider Vibrio infection and request a stool culture specifically for Vibrio. Clinical laboratories should use thiosulfate-citrate-bile salts-sucrose agar (TCBS), a selective medium for culturing for Vibrio spp., when culturing stool specimens for Vibrio and should consider using TCBS for routine screening of all stools specimens, at least during summer months. CDC coordinates a passive Gulf Coast Vibrio surveillance system and the Foodborne Diseases Active Surveillance Network (FoodNet) to monitor the incidence of Vibrio infections. Because of these multistate outbreaks, all states should consider making infections with V. parahaemolyticus and other vibrioses reportable, with referral of clinical isolates to public health laboratories for confirmation and strain subtyping. References The shipper that provided the oysters harvested elsewhere in Long Island also had received oysters from Oyster Bay at approximately the same time. Although comingling of shellfish is against state regulations, it is known to occur. Figure_1 Return to top. Disclaimer All MMWR HTML versions of articles are electronic conversions from ASCII text into HTML. This conversion may have resulted in character translation or format errors in the HTML version. Users should not rely on this HTML document, but are referred to the electronic PDF version and/or the original MMWR paper copy for the official text, figures, and tables. An original paper copy of this issue can be obtained from the Superintendent of Documents, U.S. Government Printing Office (GPO), Washington, DC 20402-9371; telephone: (202) 512-1800. Contact GPO for current prices. Page converted: 01/28/99 \| \| \| \| \| HOME \| ABOUT MMWR \| MMWR SEARCH \| DOWNLOADS \| RSS \| CONTACT POLICY \| DISCLAIMER \| ACCESSIBILITY \| \| \| \| \| \| \| \| \| \| --- --- \| \| Safer, Healthier People \| \| USA.GovDHHS \| Department of Health and Human Services \| \| HOME \| ABOUT MMWR \| MMWR SEARCH \| DOWNLOADS \| RSS \| CONTACT POLICY \| DISCLAIMER \| ACCESSIBILITY \| \| \| \| \| \| --- --- \| \| Safer, Healthier People \| \| USA.GovDHHS \| Department of Health and Human Services \| Department of Health and Human Services This page last reviewed 5/2/01
56	https://www.showbiz411.com/2017/07/28/liz-smith-is-forever-the-greatest-gossip-columnist-of-all-time-is-still-watching-us-so-be-careful?share=email&nb=1	Browse: Company: Support Showbiz411 Support independent journalism, free from the trades and other publications that are part of the tinsel town machine. For 12 years, Showbiz411.com has been covering Hollywood, Broadway, the music business and the business of celebrity. Ads are our main source of funding, but contributions (not tax exempt) from readers who enjoy the scoops, exclusives, and fact based reports are always welcome and very appreciated. To inquire about ads, email us at showbiz411@gmail.com. Liz Smith Is Forever: The Greatest Gossip Columnist of All Time Is Still Watching Us, So Be Careful Share Today the New York Times ran an interview with Liz by John Leland– weird title— The Rise and Fall of Liz Smith. The Greatest Gossip columnist of all time is 94. She gives one of these interviews once a year. Last summer, I did it for the New York Observer. Who is 94 anyway? Carl Reiner, Betty White, and a handful of celebrities. Liz taught me everything I know about writing about celebrities. She couldn’t be more dear to me. I spoke to her this morning, and she’s too busy to make appointments. Rise and fall? Never! It’s Rise and Rise! Here’s what I wrote last June, 2016. I took the accompanying photo of Liz, Cate Blanchett and the late Philip Seymour Hoffman at a lunch for “Blue Jasmine.” Liz Smith is calling. It’s mid-July, she’s supposed to be on vacation, she’s 93 years old and it’s 93 degrees outside. Liz: “Honey, I can’t find Denis, and I’m trying to find out who the publicist is for The Front Page. It’s opening soon.” Me: “Liz, I don’t think it’s opening until the fall. [The revival begins performances in September.]” Liz: “That’s all right. Do you have the number?” And that’s the way it is for Liz Smith, the Grande Dame of Gossip, the great powerhouse behind Literacy Partners and Living Landmarks, a fixture in New York society and the entertainment world since the ’70s when her daily column ran at different times in the Daily News, The New York Post and New York Newsday. She is planning her fall schedule. The Denis she speaks of is Denis Ferrara, her trusty aide and specialist on topics like Madonna and Liza and Cher. Ferrara now—after a couple of decades—shares her byline on their daily column for NY Social Diary and the 20 or more newspapers across the country where they are still syndicated. In early July, Liz, Denis and Mary Jo McDonough took a break from the daily grind so Liz could move for the first time in around 40 years—from her famous perch at a large apartment she called home East 38th Street and Third Avenue to a new, smaller one on Park Avenue and 63rd Street This forced them also to leave Liz’s beloved watering hole and meeting place, El Rio Grande restaurant, which features a Tex side and a Mex side. Forever the longhorn, Liz’s table is on the Tex side, where she and often old pal Iris Love like the margaritas strong and tangy. I have regularly been drunk under the table by these ladies, only to see them stand without wobbling and scoot away in a matter of minutes. I was summoned to El Rio Grande a few days before the move for one last rodeo. Liz swears she’ll return, but let’s face it—63rd Street is pretty far away even if you’re 63, forget about 93. At 93, Liz, I am happy to report, is completely “all there.” People ask me, “How is she?” as if I will respond that she’s gone gaga. Far from it (although she gets a kick out of Lady Gaga). Being 93 and a grande dame is still hard work. Liz suffered a minor stroke last winter and recuperated at a friend’s apartment. She’s had falls, too, and uses an aluminum walker, reluctantly and only if admonished. She is truly unsinkable, though, a real combination of a sweet-as-pie, tough-talking Texan and an indefatigable New Yorker. Her honey blonde hair has finally gone to a soft gray, and her blue eyes are more incandescent than ever. This summer has energized her as one of her favorite all time subjects has returned full force to the front burner: Donald Trump. Twenty-six years ago, Trump’s divorce from wife Ivana took Liz from Guilty Pleasure byline to Front Page Newsmaker. A photo of Liz and Ivana Trump at lunch in the middle of the storm over Marla Maples (mother of the lovely Wharton student Tiffany who spoke at the RNC two weeks ago) propelled them all into New York legend. It was bigger even than Liz’s long-time public feud with Frank Sinatra. Looking around El Rio Grande, Liz lays out exactly where we are: “I think politics has become the new show business,” she says. She remembers when she called Donald in 1990 to inquire discreetly about rumors his marriage was falling apart. She wrote in her memoir Natural Blonde: “I liked the Trumps. They had three little kids, and I didn’t want to be the one to notify Ivana that her husband was playing around. It just wasn’t my style. I figured my warning shot would bring Donald to his senses. (Such fools we scriveners be.)” But Trump’s indecision about fessing up led to front-page mayhem very quickly. Months later Ivana invited Liz to lunch at Le Grenouille. The rest, as they say, is misery. Leaving the East 52nd eatery, Liz and Ivana were mobbed by paparazzi. They landed on the front page of the News. Stars were born. Refilling her margarita, she reminisces: “Parker Ladd and Arnold Scaasi said, ‘You should go meet the Trumps.’ I said, ‘What are the Trumps?’ They said, ‘They’re very rich and aspiring to rise, and they will love you, and you will get a lot of money for charity.’ So he introduces me. Then I hang around with them for about a year. I liked her, but I couldn’t understand a word she ever said. I liked him because he reminded me of my brothers. I was amused by him. He would take me under the arm and introduce me to famous people. He’d say, ‘Isn’t she the greatest?’ He took me to prizefights and all this crap. I gave them their money’s worth, and I flew on their plane. I realized he thought he owned me. He didn’t own me. But everything for journalists is access.” I ask, “Is this the Donald Trump we knew back then? What about when he calls Elizabeth Warren Pocahontas?” Liz doesn’t miss a beat. “I would say yes. He [Trump] will do anything [to win]…He’s one of the great public actors of all time…He said he would buy the New York Daily News in order to fire me. It was the greatest thing. He made me world famous.” She is resolute that this is the man she knew all those years ago. “I think we’re at fault in our innocence. We never saw anything like him. But he is exactly is like what he was. And his family was really nice. His mother, his father, his brother who vanished—Robert.” (Trump’s brother Robert, once married to Blaine and on the society pages every day, is AWOL from the campaign as is his sister, Judge Maryanne Trump Barry.) Denis: “You had no idea the day after Ivana’s lunch at Le Grenouille you’d be on the front page of the Daily News.” Liz: “It was like The Day of the Locust.” The Trumps are far behind in the window of Liz Smith’s long ride through celebrity. Over the years she “made” a number of people including 60 Minutes correspondent Mike Wallace (who she worked for in the 1950s), actresses Elaine Stritch and Holland Taylor, director Joel Schumacher and Barbara Walters. For a quarter century, everyone wanted “to be in Liz.” A mere mention traveled faster than anything on Twitter or Snapchat today. By the time you arrived for lunch at the Russian Tea Room, the Four Seasons or Trattoria dell’Arte, the buzz was loud. Liz Smith created celebrities. For decades, Barbara Walters walked on water in Liz Smith’s column. Then Liz lost her print outlet in The New York Post when Editor-in-Chief Col Allan (who inexplicably hated her) cut her from the paper in February 2009. Liz felt Barbara dropped her because her usefulness was over—even though Liz’s column was still online and still syndicated. Last spring Liz made some scathing remarks about Walters to a Hollywood trade writer. “Barbara called me up after that, and she came over for dinner. That’s the last I’ve heard from her.” With Liz out of the paper, it was lost on no one that Walters transferred her affections to the Post’s remaining gossip diva Cindy Adams, her new best friend. It may have been just as well. Adams is seen frequently assisting Walters at theater and other events since the 86-year-old former newscaster seems to have trouble walking. So does Liz. But she says, “I feel great, I really do. I just can’t walk unassisted.” Liz’s antipathy from Allan didn’t extend to other NewsCorp executives. Roger Ailes (now a pariah) was a steadfast friend who put Liz on his payroll and used her on TV as often as possible. Rupert Murdoch, though Allan’s boss, didn’t demand to keep her in the Post. But she ran into him last winter in Mustique when his romance with her old friend Jerry Hall was blossoming. They got along famously. As for Cindy Adams, her forever rival, they are more frenemies than anything else. When Cindy’s late husband comic Joey Adams took a spill in public, Liz testified on his behalf. “I loved Joey,” she says, and she has deep admiration for Cindy’s perseverance. But a lot of the Liz Smith world has changed. Stritch is gone; so is Arnold Scaasi who pushed her into Literacy Partners. “I raised $30 million for them over the years,” she says, which is true, and she poured herself into hosting and publicizing their events. Her inner circle consists of Ferrara, as well as the irrepressible world-renowned archeologist Iris Love. To my observation, Smith—married twice to men, including one who just disappeared and had to be declared dead—and Love’s relationship transcends time and definition. They don’t live together. They squabble, purr and laugh. They are family. Liz’s extended family includes her friend of 28 years, NBC correspondent Cynthia McFadden. Liz met Cynthia pre-Court TV when she was a lawyer and working for TV producer Fred Friendly. Cynthia put Liz on a media panel with heavyweights like ABC News’s Peter Jennings and former Washington Post owner Katherine Graham. “They asked me, ‘Where do you get your best tips?’ I said from other journalists, many from The New York Times. I said everyone who works there knows everything. They can’t print anything. Most of them are my friends. And Peter Jennings said, ‘We’re your friends!’ ” The friendship with McFadden stuck and grew. How would she describe Cynthia? Liz says, “The most ethical person I know.” Earlier in July, when the moving trucks were coming and the boxes were being assembled, the promise of a return to the daily grind seemed like it go either way. But on July 18, Liz and Denis roared back to life. The drum beats on with Ferrara working electronically from Hoboken and Smith operating her computer from uptown. Their sense of humor keeps them going. When the break was coming, Ferrara devised a snappy salutation in Liz’s name about leaving El Rio Grande: “No more of those delectable fresh tostadas or those gasoline-fueled margaritas. (If I were younger I might say that the latter deprivation is a good thing, but…I am NOT younger, and one only lives once, or so I am told. If this is not true, I’ll be sure to come back as—a margarita!)” Donate to Showbiz411.com Showbiz411 is now in its 13th year of providing breaking and exclusive entertainment news. This is an independent site, unlike the many Hollywood trades that are owned by one company. To continue providing news that takes a fresh look at what's going on in movies, music, theater, etc, advertising is our basis. Reader donations would be greatly appreciated, too. They are just another facet of keeping fact based journalism alive. Thank you Read more NFL Makes Bad Choice with Bad Bunny for Super Bowl Half Time Game Pop Poop Out: First Maroon 5, Now Ed Sheeran Sales Disappear for New Album as Generational End Comes in Music Cycle Global Citizen Concert Yields Huge Drop of Views on YouTube From Last Year as The Weeknd Bows Out of Show, Cardi B Headlines Box Office: Leo, Sean Penn Win the “Battle” with $22.4 Mil Opening Weekend, Sony Resorts to Old Spider Man Trio for $2.1 Mil Take Brian Cox (aka Logan Roy), June Squibb, Lucy Liu, More Hit Star Studded BAFTA NY Tea Party at Swanky Mandarin Oriental In Other News NFL Makes Bad Choice with Bad Bunny for Super Bowl Half Time Game Pop Poop Out: First Maroon 5, Now Ed Sheeran Sales Disappear for New Album as Generational End Comes in Music Cycle Global Citizen Concert Yields Huge Drop of Views on YouTube From Last Year as The Weeknd Bows Out of Show, Cardi B Headlines Box Office: Leo, Sean Penn Win the “Battle” with $22.4 Mil Opening Weekend, Sony Resorts to Old Spider Man Trio for $2.1 Mil Take Hollywood to the Hudson Company Trending NFL Makes Bad Choice with Bad Bunny for Super Bowl Half Time Game Pop Poop Out: First Maroon 5, Now Ed Sheeran Sales Disappear for New Album as Generational End Comes in Music Cycle Global Citizen Concert Yields Huge Drop of Views on YouTube From Last Year as The Weeknd Bows Out of Show, Cardi B Headlines Categories © Showbiz411.com
57	https://en.wikipedia.org/wiki/Bacteriophage	Jump to content Bacteriophage العربية Azərbaycanca বাংলা 閩南語 / Bn-lm-gí Беларуская Български Bosanski Català Čeština Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Íslenska Italiano עברית ಕನ್ನಡ ქართული Қазақша Kiswahili Kurdî Кыргызча Latina Latviešu Lietuvių Magyar Македонски მარგალური Nederlands 日本語 Norsk bokmål Norsk nynorsk Occitan Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ Polski Português Romnă Русиньскый Русский Simple English Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska தமிழ் ไทย Тоҷикӣ Türkçe Українська Tiếng Việt 吴语粵語中文 Edit links From Wikipedia, the free encyclopedia Virus that infects bacteria "Phage" redirects here. For other uses, see Phage (disambiguation). A bacteriophage (/bækˈtɪərioʊfeɪdʒ/), also known informally as a phage (/ˈfeɪdʒ/), is a virus that infects and replicates within bacteria. The term is derived from Ancient Greek φαγεῖν (phagein) 'to devour' and bacteria. Bacteriophages are composed of proteins that encapsulate a DNA or RNA genome, and may have structures that are either simple or elaborate. Their genomes may encode as few as four genes (e.g. MS2) and as many as hundreds of genes. Phages replicate within the bacterium following the injection of their genome into its cytoplasm. Bacteriophages are among the most common and diverse entities in the biosphere. Bacteriophages are ubiquitous viruses, found wherever bacteria exist. It is estimated there are more than 1031 bacteriophages on the planet, more than every other organism on Earth, including bacteria, combined. Viruses are the most abundant biological entity in the water column of the world's oceans, and the second largest component of biomass after prokaryotes, where up to 9x108 virions per millilitre have been found in microbial mats at the surface, and up to 70% of marine bacteria may be infected by bacteriophages. Bacteriophages were used from the 1920s as an alternative to antibiotics in the former Soviet Union and Central Europe, as well as in France and Brazil. They are seen as a possible therapy against multi-drug-resistant strains of many bacteria. Bacteriophages are known to interact with the immune system both indirectly via bacterial expression of phage-encoded proteins and directly by influencing innate immunity and bacterial clearance. Phage–host interactions are becoming increasingly important areas of research. Classification [edit] Bacterial viruses lack common ancestry and, for that reason, are classified in many unrelated taxa, listed hereafter: In the realm Duplodnaviria, the class Caudoviricetes contains bacterial viruses. Unlike the other taxa listed here, Caudoviricetes does not exclusively contain bacterial viruses; archaeal viruses are also included in the class. Caudoviruses are also called tailed viruses or head-tail viruses, and they are often sorted into three types based on tail morphology: podoviruses (short tail), myoviruses (long, contractile tail), and siphoviruses (long, non-contractile tail). In the realm Monodnaviria, the kingdoms Loebvirae and Sangervirae contain bacterial viruses.[note 1] In the realm Riboviria, the phylum Artimaviricota,[note 2] the class Vidaverviricetes,[note 3] the class Leviviricetes, and possibly the family Picobirnaviridae[note 4] contain bacterial viruses. In the realm Singelaviria, the family Matsushitaviridae contains bacterial viruses. In the realm Varidnaviria, the class Ainoaviricetes,[note 5] the order Vinavirales, and the subphylum Prepoliviricotina[note 6] contain bacterial viruses. Lastly, the families Obscuriviridae and Plasmaviridae, which are unassigned to higher taxa, are bacterial virus families. The taxonomy of the aforementioned taxa can be visualized as follows, with bacterial virus taxa in bold: Realm: Duplodnaviria Kingdom: Heunggongvirae Phylum: Uroviricota Class: Caudoviricetes Realm: Monodnaviria Kingdom: Loebvirae Kingdom: Sangervirae Realm: Riboviria Kingdom: Orthornavirae Phylum: Artimaviricota Phylum: Duplornaviricota Class: Vidaverviricetes Phylum: Lenarviricota Class: Leviviricetes Phylum: Pisuviricota Class: Duplopiviricetes Order: Durnavirales Family: Picobirnaviridae Realm: Singelaviria Kingdom: Helvetiavirae Phylum: Dividoviricota Class: Laserviricetes Order: Halopanivirales Family: Matsushitaviridae Realm: Varidnaviria Kingdom: Abadenavirae Phylum: Produgelaviricota Class: Ainoaviricetes Class: Belvinaviricetes Order: Vinavirales Kingdom: Bamfordvirae Phylum: Preplasmiviricota Subphylum: Prepoliviricotina Unassigned taxa: Obscuriviridae and Plasmaviridae History [edit] In 1896, Ernest Hanbury Hankin reported that something in the waters of the Ganges and Yamuna rivers in India had a marked antibacterial action against cholera and it could pass through a very fine porcelain Chamberland filter. In 1915, British bacteriologist Frederick Twort, superintendent of the Brown Institution of London, discovered a small agent that infected and killed bacteria. He believed the agent must be one of the following: a stage in the life cycle of the bacteria an enzyme produced by the bacteria themselves, or a virus that grew on and destroyed the bacteria Twort's research was interrupted by the onset of World War I, as well as a shortage of funding and the discoveries of antibiotics. Independently, French-Canadian microbiologist Félix d'Hérelle, working at the Pasteur Institute in Paris, announced on 3 September 1917 that he had discovered "an invisible, antagonistic microbe of the dysentery bacillus". For d'Hérelle, there was no question as to the nature of his discovery: "In a flash I had understood: what caused my clear spots was in fact an invisible microbe... a virus parasitic on bacteria." D'Hérelle called the virus a bacteriophage, a bacterium-eater (from the Greek phagein, meaning "to devour"). He also recorded a dramatic account of a man suffering from dysentery who was restored to good health by the bacteriophages. It was d'Hérelle who conducted much research into bacteriophages and introduced the concept of phage therapy. In 1919, in Paris, France, d'Hérelle conducted the first clinical application of a bacteriophage, with the first reported use in the United States being in 1922. Nobel prizes awarded for phage research [edit] In 1969, Max Delbrück, Alfred Hershey, and Salvador Luria were awarded the Nobel Prize in Physiology or Medicine for their discoveries of the replication of viruses and their genetic structure. Specifically the work of Hershey, as contributor to the Hershey–Chase experiment in 1952, provided convincing evidence that DNA, not protein, was the genetic material of life. Delbrück and Luria carried out the Luria–Delbrück experiment which demonstrated statistically that mutations in bacteria occur randomly and thus follow Darwinian rather than Lamarckian principles. In 2018, George Smith and Gregory Winter were awarded Nobel Prize in Chemistry for the phage display of peptides and antibodies. Uses [edit] Phage therapy [edit] Main article: Phage therapy Phages were discovered to be antibacterial agents and were used in the former Soviet Republic of Georgia (pioneered there by Giorgi Eliava with help from the co-discoverer of bacteriophages, Félix d'Hérelle) during the 1920s and 1930s for treating bacterial infections. D'Herelle "quickly learned that bacteriophages are found wherever bacteria thrive: in sewers, in rivers that catch waste runoff from pipes, and in the stools of convalescent patients." They had widespread use, including treatment of soldiers in the Red Army. However, they were abandoned for general use in the West for several reasons: Antibiotics were discovered and marketed widely. They were easier to make, store, and prescribe. Medical trials of phages were carried out, but a basic lack of understanding of phages raised questions about the validity of these trials. Publication of research in the Soviet Union was mainly in the Russian or Georgian languages and for many years was not followed internationally. The Soviet technology was widely discouraged and in some cases illegal due to the red scare. The use of phages has continued since the end of the Cold War in Russia, Georgia, and elsewhere in Central and Eastern Europe. The first regulated, randomized, double-blind clinical trial was reported in the Journal of Wound Care in June 2009, which evaluated the safety and efficacy of a bacteriophage cocktail to treat infected venous ulcers of the leg in human patients. The FDA approved the study as a Phase I clinical trial. The study's results demonstrated the safety of therapeutic application of bacteriophages, but did not show efficacy. The authors explained that the use of certain chemicals that are part of standard wound care (e.g. lactoferrin or silver) may have interfered with bacteriophage viability. Shortly after that, another controlled clinical trial in Western Europe (treatment of ear infections caused by Pseudomonas aeruginosa) was reported in the journal Clinical Otolaryngology in August 2009. The study concludes that bacteriophage preparations were safe and effective for treatment of chronic ear infections in humans. Additionally, there have been numerous animal and other experimental clinical trials evaluating the efficacy of bacteriophages for various diseases, such as infected burns and wounds, and cystic fibrosis-associated lung infections, among others. On the other hand, phages of Inoviridae have been shown to complicate biofilms involved in pneumonia and cystic fibrosis and to shelter the bacteria from drugs meant to eradicate disease, thus promoting persistent infection. Meanwhile, bacteriophage researchers have been developing engineered viruses to overcome antibiotic resistance, and engineering the phage genes responsible for coding enzymes that degrade the biofilm matrix, phage structural proteins, and the enzymes responsible for lysis of the bacterial cell wall. There have been results showing that T4 phages that are small in size and short-tailed can be helpful in detecting E. coli in the human body. Therapeutic efficacy of a phage cocktail was evaluated in a mouse model with nasal infection of multi-drug-resistant (MDR) A. baumannii. Mice treated with the phage cocktail showed a 2.3-fold higher survival rate compared to those untreated at seven days post-infection. In 2017, a 68-year-old diabetic patient with necrotizing pancreatitis complicated by a pseudocyst infected with MDR A. baumannii strains was being treated with a cocktail of Azithromycin, Rifampicin, and Colistin for 4 months without results and overall rapidly declining health. Because discussion had begun of the clinical futility of further treatment, an Emergency Investigational New Drug (eIND) was filed as a last effort to at the very least gain valuable medical data from the situation, and approved, so he was subjected to phage therapy using a percutaneously (PC) injected cocktail containing nine different phages that had been identified as effective against the primary infection strain by rapid isolation and testing techniques (a process which took under a day). This proved effective for a very brief period, although the patient remained unresponsive and his health continued to worsen; soon isolates of a strain of A. baumannii were being collected from drainage of the cyst that showed resistance to this cocktail, and a second cocktail which was tested to be effective against this new strain was added, this time by intravenous (IV) injection as it had become clear that the infection was more pervasive than originally thought. Once on the combination of the IV and PC therapy the patient's downward clinical trajectory reversed, and within two days he had awoken from his coma and become responsive. As his immune system began to function he had to be temporarily removed from the cocktail because his fever was spiking to over 104 °F (40 °C), but after two days the phage cocktails were re-introduced at levels he was able to tolerate. The original three-antibiotic cocktail was replaced by minocycline after the bacterial strain was found not to be resistant to this and he rapidly regained full lucidity, although he was not discharged from the hospital until roughly 145 days after phage therapy began. Towards the end of the therapy it was discovered that the bacteria had become resistant to both of the original phage cocktails, but they were continued because they seemed to be preventing minocycline resistance from developing in the bacterial samples collected so were having a useful synergistic effect. Other [edit] Food industry [edit] Phages have increasingly been used to safen food products and to forestall spoilage bacteria. Since 2006, the United States Food and Drug Administration (FDA) and United States Department of Agriculture (USDA) have approved several bacteriophage products. LMP-102 (Intralytix) was approved for treating ready-to-eat (RTE) poultry and meat products. In that same year, the FDA approved LISTEX (developed and produced by Micreos) using bacteriophages on cheese to kill Listeria monocytogenes bacteria, in order to give them generally recognized as safe (GRAS) status. In July 2007, the same bacteriophage were approved for use on all food products. In 2011 USDA confirmed that LISTEX is a clean label processing aid and is included in USDA. Research in the field of food safety is continuing to see if lytic phages are a viable option to control other food-borne pathogens in various food products. Water indicators [edit] Bacteriophages, including those specific to Escherichia coli, have been employed as indicators of fecal contamination in water sources. Due to their shared structural and biological characteristics, coliphages can serve as proxies for viral fecal contamination and the presence of pathogenic viruses such as rotavirus, norovirus, and HAV. Research conducted on wastewater treatment systems has revealed significant disparities in the behavior of coliphages compared to fecal coliforms, demonstrating a distinct correlation with the recovery of pathogenic viruses at the treatment's conclusion. Establishing a secure discharge threshold, studies have determined that discharges below 3000 PFU/100 mL are considered safe in terms of limiting the release of pathogenic viruses. Diagnostics [edit] In 2011, the FDA cleared the first bacteriophage-based product for in vitro diagnostic use. The KeyPath MRSA/MSSA Blood Culture Test uses a cocktail of bacteriophage to detect Staphylococcus aureus in positive blood cultures and determine methicillin resistance or susceptibility. The test returns results in about five hours, compared to two to three days for standard microbial identification and susceptibility test methods. It was the first accelerated antibiotic-susceptibility test approved by the FDA. Counteracting bioweapons and toxins [edit] Government agencies in the West have for several years been looking to Georgia and the former Soviet Union for help with exploiting phages for counteracting bioweapons and toxins, such as anthrax and botulism. Developments are continuing among research groups in the U.S. Other uses include spray application in horticulture for protecting plants and vegetable produce from decay and the spread of bacterial disease. Other applications for bacteriophages are as biocides for environmental surfaces, e.g., in hospitals, and as preventative treatments for catheters and medical devices before use in clinical settings. The technology for phages to be applied to dry surfaces, e.g., uniforms, curtains, or even sutures for surgery now exists. Clinical trials reported in Clinical Otolaryngology show success in veterinary treatment of pet dogs with otitis. Bacterium sensing and identification [edit] The sensing of phage-triggered ion cascades (SEPTIC) bacterium sensing and identification method uses the ion emission and its dynamics during phage infection and offers high specificity and speed for detection. Phage display [edit] Phage display is a different use of phages involving a library of phages with a variable peptide linked to a surface protein. Each phage genome encodes the variant of the protein displayed on its surface (hence the name), providing a link between the peptide variant and its encoding gene. Variant phages from the library may be selected through their binding affinity to an immobilized molecule (e.g., botulism toxin) to neutralize it. The bound, selected phages can be multiplied by reinfecting a susceptible bacterial strain, thus allowing them to retrieve the peptides encoded in them for further study. Antimicrobial drug discovery [edit] Phage proteins often have antimicrobial activity and may serve as leads for peptidomimetics, i.e. drugs that mimic peptides. Phage-ligand technology makes use of phage proteins for various applications, such as binding of bacteria and bacterial components (e.g. endotoxin) and lysis of bacteria. Basic research [edit] Bacteriophages are important model organisms for studying principles of evolution and ecology. Detriments [edit] Dairy industry [edit] Bacteriophages present in the environment can cause cheese to not ferment. In order to avoid this, mixed-strain starter cultures and culture rotation regimes can be used. Genetic engineering of culture microbes – especially Lactococcus lactis and Streptococcus thermophilus – have been studied for genetic analysis and modification to improve phage resistance. This has especially focused on plasmid and recombinant chromosomal modifications. Some research has focused on the potential of bacteriophages as antimicrobial against foodborne pathogens and biofilm formation within the dairy industry. As the spread of antibiotic resistance is a main concern within the dairy industry, phages can serve as a promising alternative. Replication [edit] The life cycle of bacteriophages tends to be either a lytic cycle or a lysogenic cycle. In addition, some phages display pseudolysogenic behaviors. With lytic phages such as the T4 phage, bacterial cells are broken open (lysed) and destroyed after immediate replication of the virion. As soon as the cell is destroyed, the phage progeny can find new hosts to infect. Lytic phages are more suitable for phage therapy. Some lytic phages undergo a phenomenon known as lysis inhibition, where completed phage progeny will not immediately lyse out of the cell if extracellular phage concentrations are high. This mechanism is not identical to that of the temperate phage going dormant and usually is temporary. In contrast, the lysogenic cycle does not result in immediate lysing of the host cell. Those phages able to undergo lysogeny are known as temperate phages. Their viral genome will integrate with host DNA and replicate along with it, relatively harmlessly, or may even become established as a plasmid. The virus remains dormant until host conditions deteriorate, perhaps due to depletion of nutrients, then, the endogenous phages (known as prophages) become active. At this point they initiate the reproductive cycle, resulting in lysis of the host cell. As the lysogenic cycle allows the host cell to continue to survive and reproduce, the virus is replicated in all offspring of the cell. An example of a bacteriophage known to follow the lysogenic cycle and the lytic cycle is the phage lambda of E. coli. Sometimes prophages may provide benefits to the host bacterium while they are dormant by adding new functions to the bacterial genome, in a phenomenon called lysogenic conversion. Examples are the conversion of harmless strains of Corynebacterium diphtheriae or Vibrio cholerae by bacteriophages to highly virulent ones that cause diphtheria or cholera, respectively. Strategies to combat certain bacterial infections by targeting these toxin-encoding prophages have been proposed. Attachment and penetration [edit] Bacterial cells are protected by a cell wall of polysaccharides, which are important virulence factors protecting bacterial cells against both immune host defenses and antibiotics. To enter a host cell, bacteriophages bind to specific receptors on the surface of bacteria, including lipopolysaccharides, teichoic acids, proteins, or even flagella. This specificity means a bacteriophage can infect only certain bacteria bearing receptors to which they can bind, which in turn, determines the phage's host range. Polysaccharide-degrading enzymes are virion-associated proteins that enzymatically degrade the capsular outer layer of their hosts at the initial step of a tightly programmed phage infection process.[citation needed] Host growth conditions also influence the ability of the phage to attach and invade them. As phage virions do not move independently, they must rely on random encounters with the correct receptors when in solution, such as blood, lymphatic circulation, irrigation, soil water, etc.[citation needed] Myovirus bacteriophages use a hypodermic syringe-like motion to inject their genetic material into the cell. After contacting the appropriate receptor, the tail fibers flex to bring the base plate closer to the surface of the cell. This is known as reversible binding. Once attached completely, irreversible binding is initiated and the tail contracts, possibly with the help of ATP present in the tail, injecting genetic material through the bacterial membrane. The injection is accomplished through a sort of bending motion in the shaft by going to the side, contracting closer to the cell and pushing back up. Podoviruses lack an elongated tail sheath like that of a myovirus, so instead, they use their small, tooth-like tail fibers enzymatically to degrade a portion of the cell membrane before inserting their genetic material. Synthesis of proteins and nucleic acid [edit] Within minutes, bacterial ribosomes start translating viral mRNA into protein. For RNA-based phages, RNA replicase is synthesized early in the process. Proteins modify the bacterial RNA polymerase so it preferentially transcribes viral mRNA. The host's normal synthesis of proteins and nucleic acids is disrupted, and it is forced to manufacture viral products instead. These products go on to become part of new virions within the cell, helper proteins that contribute to the assemblage of new virions, or proteins involved in cell lysis. In 1972, Walter Fiers (University of Ghent, Belgium) was the first to establish the complete nucleotide sequence of a gene and in 1976, of the viral genome of bacteriophage MS2. Some dsDNA bacteriophages encode ribosomal proteins, which are thought to modulate protein translation during phage infection. Virion assembly [edit] In the case of the T4 phage, the construction of new virus particles involves the assistance of helper proteins that act catalytically during phage morphogenesis. The base plates are assembled first, with the tails being built upon them afterward. The head capsids, constructed separately, will spontaneously assemble with the tails. During assembly of the phage T4 virion, the morphogenetic proteins encoded by the phage genes interact with each other in a characteristic sequence. Maintaining an appropriate balance in the amounts of each of these proteins produced during viral infection appears to be critical for normal phage T4 morphogenesis. The DNA is packed efficiently within the heads. The whole process takes about 15 minutes. Early studies of bactioriophage T4 (1962–1964) provided an opportunity to gain understanding of virtually all of the genes that are essential for growth of the bacteriophage under laboratory conditions. These studies were made possible by the availability of two classes of conditional lethal mutants. One class of such mutants was referred to as amber mutants. The other class of conditional lethal mutants was referred to as temperature-sensitive mutants Studies of these two classes of mutants led to considerable insight into the functions and interactions of the proteins employed in the machinery of DNA replication, repair and recombination, and on how viruses are assembled from protein and nucleic acid components (molecular morphogenesis). Release of virions [edit] Phages may be released via cell lysis, by extrusion, or, in a few cases, by budding. Lysis, by tailed phages, is achieved by an enzyme called endolysin, which attacks and breaks down the cell wall peptidoglycan. An altogether different phage type, the filamentous phage, makes the host cell continually secrete new virus particles. Released virions are described as free, and, unless defective, are capable of infecting a new bacterium. Budding is associated with certain Mycoplasma phages. In contrast to virion release, phages displaying a lysogenic cycle do not kill the host and instead become long-term residents as prophages. Communication [edit] Research in 2017 revealed that the bacteriophage Φ3T makes a short viral protein that signals other bacteriophages to lie dormant instead of killing the host bacterium. Arbitrium is the name given to this protein by the researchers who discovered it. Genome structure [edit] Given the millions of different phages in the environment, phage genomes come in a variety of forms and sizes. RNA phages such as MS2 have the smallest genomes, with only a few kilobases. However, some DNA phages such as T4 may have large genomes with hundreds of genes; the size and shape of the capsid varies along with the size of the genome. The largest bacteriophage genomes reach a size of 735 kb. Bacteriophage genomes can be highly mosaic, i.e. the genome of many phage species appear to be composed of numerous individual modules. These modules may be found in other phage species in different arrangements. Mycobacteriophages, bacteriophages with mycobacterial hosts, have provided excellent examples of this mosaicism. In these mycobacteriophages, genetic assortment may be the result of repeated instances of site-specific recombination and illegitimate recombination (the result of phage genome acquisition of bacterial host genetic sequences). Evolutionary mechanisms shaping the genomes of bacterial viruses vary between different families and depend upon the type of the nucleic acid, characteristics of the virion structure, as well as the mode of the viral life cycle. Some marine roseobacter phages, also known as roseophages, contain deoxyuridine (dU) instead of deoxythymidine (dT) in their genomic DNA. There is some evidence that this unusual component is a mechanism to evade bacterial defense mechanisms such as restriction endonucleases and CRISPR/Cas systems which evolved to recognize and cleave sequences within invading phages, thereby inactivating them. Other phages have long been known to use unusual nucleotides. In 1963, Takahashi and Marmur identified a Bacillus phage that has dU substituting dT in its genome, and in 1977, Kirnos et al. identified a cyanophage containing 2-aminoadenine (Z) instead of adenine (A). Systems biology [edit] The field of systems biology investigates the complex networks of interactions within an organism, usually using computational tools and modeling. For example, a phage genome that enters into a bacterial host cell may express hundreds of phage proteins which will affect the expression of numerous host genes or the host's metabolism. All of these complex interactions can be described and simulated in computer models. For instance, infection of Pseudomonas aeruginosa by the temperate phage PaP3 changed the expression of 38% (2160/5633) of its host's genes. Many of these effects are probably indirect, hence the challenge becomes to identify the direct interactions among bacteria and phage. Several attempts have been made to map protein–protein interactions among phage and their host. For instance, bacteriophage lambda was found to interact with its host, E. coli, by dozens of interactions. Again, the significance of many of these interactions remains unclear, but these studies suggest that there most likely are several key interactions and many indirect interactions whose role remains uncharacterized. Host resistance and anti-phage defense [edit] Bacteriophages are a major threat to bacteria and prokaryotes have evolved numerous mechanisms to block infection (host resistance) or to block the replication of bacteriophages within host cells (anti-phage defense). Some examples include The CRISPR system Retrons and the anti-toxin system encoded by them. The Thoeris defense system is known to deploy a unique strategy for bacterial antiphage resistance via NAD+ degradation. The Hailong anti-phage defense system consists of a two-gene operon encoding a transmembrane ion channel effector (HalA) and a nucleotidyltransferase (NTase, HalB). An infecting phage can activate HalA, triggering membrane depolarization and thus cell death. Although the infected cell is likely going to die, this protects the bacterial population from further spread. Bacteriophage–host symbiosis [edit] Temperate phages are bacteriophages that integrate their genetic material into the host as extrachromosomal episomes or as a prophage during a lysogenic cycle. Some temperate phages can confer fitness advantages to their host in numerous ways, including giving antibiotic resistance through the transfer or introduction of antibiotic resistance genes (ARGs), protecting hosts from phagocytosis, protecting hosts from secondary infection through superinfection exclusion, enhancing host pathogenicity, or enhancing bacterial metabolism or growth. Bacteriophage–host symbiosis may benefit bacteria by providing selective advantages while passively replicating the phage genome. In the environment [edit] Main article: Marine bacteriophage Metagenomics has allowed the in-water detection of bacteriophages that was not possible previously. Also, bacteriophages have been used in hydrological tracing and modelling in river systems, especially where surface water and groundwater interactions occur. The use of phages is preferred to the more conventional dye marker because they are significantly less absorbed when passing through ground waters and they are readily detected at very low concentrations. Non-polluted water may contain approximately 2×108 bacteriophages per ml. Bacteriophages are thought to contribute extensively to horizontal gene transfer in natural environments, principally via transduction, but also via transformation. Metagenomics-based studies also have revealed that viromes from a variety of environments harbor antibiotic-resistance genes, including those that could confer multidrug resistance. Recent findings have mapped the complex and intertwined arsenal of anti-phage defense tools in environmental bacteria. In humans [edit] Although phages do not infect humans, there are countless phage particles in the human body, given the extensive human microbiome. One's phage population has been called the human phageome, including the "healthy gut phageome" (HGP) and the "diseased human phageome" (DHP). The active phageome of a healthy human (i.e., actively replicating as opposed to nonreplicating, integrated prophage) has been estimated to comprise dozens to thousands of different viruses. There is evidence that bacteriophages and bacteria interact in the human gut microbiome both antagonistically and beneficially. Preliminary studies have indicated that common bacteriophages are found in 62% of healthy individuals on average, while their prevalence was reduced by 42% and 54% on average in patients with ulcerative colitis (UC) and Crohn's disease (CD). Abundance of phages may also decline in the elderly. The most common phages in the human intestine, found worldwide, are crAssphages. CrAssphages are transmitted from mother to child soon after birth, and there is some evidence suggesting that they may be transmitted locally. Each person develops their own unique crAssphage clusters. CrAss-like phages also may be present in primates besides humans. Commonly studied bacteriophages [edit] Among the countless phages, only a few have been studied in detail, including some historically important phage that were discovered in the early days of microbial genetics. These, especially the T-phage, helped to discover important principles of gene structure and function. 186 phage λ phage Φ6 phage Φ29 phage ΦX174 Bacteriophage φCb5 G4 phage M13 phage MS2 phage (23–28 nm in size) N4 phage P1 phage P2 phage P4 phage R17 phage T2 phage T4 phage (169 kbp genome, 200 nm long) T7 phage T12 phage Bacteriophage databases and resources [edit] Phagesdb Phagescope See also [edit] Viruses portal Antibiotic Bacterivore CrAssphage CRISPR DNA viruses Macrophage Phage ecology Phage monographs (a comprehensive listing of phage and phage-associated monographs, 1921–present) Phagemid Polyphage RNA viruses Transduction Viriome Virophage, viruses that infect other viruses Notes [edit] ^ Loebvirae is monotypic down to its sole order, Tubulavirales, and Sangervirae is monotypic down to its sole family, Microviridae. ^ Artimaviricota contains just one species, Atsuirnavirus caloris. ^ Vidaverviricetes is monotypic down to its sole family, Cystoviridae. ^ Picobirnaviruses are thought to be animal viruses, but there is evidence that they may be bacterial viruses. ^ Ainoaviricetes contains just one species, Finnlakevirus FLiP. ^ Prepoliviricotina is monotypic down to its sole family, Tectiviridae. 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PMC 10767790. PMID 37904614. Bibliography [edit] Hauser AR, Mecsas J, Moir DT (July 2016). "Beyond Antibiotics: New Therapeutic Approaches for Bacterial Infections". Clinical Infectious Diseases. 63 (1): 89–95. doi:10.1093/cid/ciw200. PMC 4901866. PMID 27025826. Strathdee S, Patterson T (2019). The Perfect Predator. Hachette Books. ISBN 978-0-316-41808-9. Häusler T (2006). Viruses vs. superbugs: a solution to the antibiotics crisis?. London: Macmillan. ISBN 978-1-4039-8764-8. External links [edit] Wikimedia Commons has media related to Bacteriophages. Wikiquote has quotations related to Bacteriophage. Abedon ST. "The Bacteriophage Ecology Group". The Ohio State University. Archived from the original on 3 June 2013. Tourterel C, Blouin Y. "Bacteriophages illustrations and genomics". Orsay phage web site. Archived from the original on 29 October 2013. Retrieved 24 October 2013. "QuipStories: Bacteriophages get a foothold on their prey" (PDF). PDBe. Flatow I (April 2008). "Using 'Phage' Viruses to Help Fight Infection". Science Friday podcast. NPR. Archived from the original on 17 April 2008. "Animation of a scientifically correct T4 bacteriophage targeting E. coli bacteria". YouTube. 21 May 2019. "T4 Bacteriophage targeting E. coli bacteria". Animation by Hybrid Animation Medical. 21 December 2009. Bacteriophages: What are they. 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58	https://www.sciencedirect.com/science/article/abs/pii/S0016236120318421	Effect of flame speed on knocking characteristics for SI engine under critical knocking conditions - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (28) Cited by (37) Fuel Volume 282, 15 December 2020, 118846 Full Length Article Effect of flame speed on knocking characteristics for SI engine under critical knocking conditions Author links open overlay panel Lin Chen, Ren Zhang, Haiqiao Wei, Jiaying Pan Show more Add to Mendeley Share Cite rights and content Highlights •Effect of flame propagation speed on engine knock was optically studied. •Closing the SCV can greatly increase in-cylinder turbulence intensity thus flame speed. •Only auto-ignition with high HRR can induce engine knock. •Fast flame is more likely to induce engine knock at critical knock conditions. Abstract Engine knock essentially involves the interplay between end-gas auto-ignition and primary flame propagation. However, literature shows inconsistent correlations between turbulent flame speed and knocking intensity. In this study, different levels of turbulent intensity were achieved by modifying in-cylinder swirl ratios, and the effect of turbulent flame speed on knocking characteristics was firstly investigated by using an optical engine under critical knocking conditions. In-cylinder swirl ratios were controlled by a swirl control valve (SCV) and particle image velocimetry (PIV) experiments were employed to quantify the turbulent flow velocity under different swirl ratio conditions. High-speed photography and instantaneous pressure acquisition were synchronously employed for auto-ignition and knocking measurement. The experimental results show that end-gas auto-ignition is a sufficient condition for engine knock and engine knock is strongly associated with the peak heat release rate when auto-ignition occurs. Under critical knocking conditions, fast turbulent flame propagation always results in advanced auto-ignition timing and concentrated heat release and thereby severe auto-ignition. Consequently, the higher flame speed promotes heavier knocking combustion under enhanced turbulent intensity conditions. The current study shall provide useful insights into the nature of engine knock and the regulation of engine combustion. Introduction Downsizing combining supercharging is one of the most effective technologies in energy saving. However, under high load conditions, the severe thermodynamic condition might promote the possibility of auto-ignition in the end-gas region . For traditional high-compression-ratio spark-ignition (SI) engines, the auto-ignition ahead of the flame front always results in engine knock. The strong local over-pressure caused by extremely rapid heat release in the auto-ignition region will lead to engine body damage . With the development of advanced compression combustion strategies, the auto-ignition is controllable without knock and favorite for high efficiency . Because the rapid combustion rate of auto-ignition can gain more concentrated heat release , through which power output and thermal efficiency can be improved . Therefore, it is necessary to study the inner mechanism between auto-ignition and knocking characteristics for further improving engine thermal efficiency. Theoretically, engine knock is induced by end-gas auto-ignition, which involves complex competition between the ignition delay period of the mixture in end-gas region and the time of SI flame propagation . If the SI flame propagates fast enough, auto-ignition will vanish and thus knock can be suppressed . More or less based on the idea, it is proposed to increase the flame speed to suppress engine knock , . Based on the idea of increasing in-cylinder turbulence, Taiga et al. investigated the effect of different compression flow fields on knocking characteristics. The results show that end-gas auto-ignition is suppressed under fast flame propagation conditions. Using a gasoline engine, Heywood et al. also contributed the knock suppression to the fast flame speed by the hydrogen addition. Our recent simulation results also show that the formation of auto-ignition can be inhibited due to the enhanced mass and heat transfer under high turbulence intensity conditions . However, there are other studies showing that fast flame makes the end-gas pressure and temperature rise more rapidly and thereby more advanced auto-ignition and heavy knock . For example, Chen et al. compared the mass fraction burned (MFB) duration between normal and knocking cycles and found that the combustion rate for knocking cycles is much higher than that of normal cycles. Additionally, based on a turbocharged SI engine under different flame speed conditions, it is found that higher flame speed induces more severe end-gas auto-ignition and knocking events. Recently, Yu et al. numerically found that knock intensity behaves a non-monotonic correlation with flame speed and knock intensity increases firstly and then decreases with the increase of flame speed. To sum up, there are still many ambiguities and conflicts between flame speed and engine knock despite numerous studies, which need to be further studied especially in real optical engines. In the past decades, more and more advanced combustion modes were proposed and visualization technology was also been used in the exploration of auto-ignition with various knocking characteristics. For example, it is found that the end-gas auto-ignition burns smoothly under critical knocking conditions , . Since no pressure oscillation occurs, the auto-ignition is favorite for high efficiency. Besides, the location of auto-ignition spots was found to be located near the flame front, which is also different from traditional knocking auto-ignition . These phenomena are totally different from the thought that large UBMF (the mass fraction of in-cylinder unburned mixture when end-gas auto-ignition occurs) will induce heavy knock . More recently, our optical results show that auto-ignition doesn’t necessarily induce knock if auto-ignition flame is controllable and the speed is comparable to the SI flame . The above studies have revealed that different auto-ignition results in different levels of knock intensity (pressure oscillations). However, due to the lack of actual engine experiments, most previous work has neglected the effect of spark-ignited flame propagation. In the current work, the objective is to further experimentally clarify the effect of different in-cylinder flow fields on knocking characteristics with considering spark-ignited flame speed and the inner mechanism between end-gas auto-ignition and knock intensity is also revealed under critical knocking conditions. To achieve this, the high compression ratio optical engine was equipped with a swirl control valve (SCV). Therefore, different in-cylinder turbulence intensities are achieved to change the turbulent flame speed. Furthermore, high-speed direct photography was used to evaluate flame propagation speed and knocking pressures were obtained by a pressure sensor. The explanation of synchronization measurement will be introduced later. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Experimental apparatus and procedure For the current experiments, a specially designed optical engine was used. The engine is single-cylinder, four-valve and combined with non-intrusive optical equipment. In this section, the general specification of the engine and the operating conditions were provided firstly with addressing the swirl control valve. Then, the optical experiments including PIV and direct photography were introduced, which are the main parts of the experiment in this study. Finally, the data analysis of the study Overall knocking characteristics In the current study, a MAPO value of 0.1 MPa was selected as the knocking threshold and knocking combustion can be defined as the condition if the knocking cycle exceeds 10% . First, based on the low turbulence intensity case (Swirl 1), a spark sweep was conducted to obtain the knock limited spark timing (after which the knocking cycle is above 10% ). Then the critical knocking condition was defined as the knock limited spark timing (−10.5 CAD) under low turbulence intensity conditions. Conclusions To clarify the role of flame propagation speed in end-gas auto-ignition and knock intensity under different turbulence intensity conditions, a high compression ratio optical engine with a swirl control valve was adopted under critical knocking conditions. High-speed direct photography was applied to quantify SI flame propagation and auto-ignition propagation, and meanwhile, the simultaneous pressure acquisition was applied to quantify knocking characteristics. The experimental results are as CRediT authorship contribution statement Lin Chen: Conceptualization, Data curation, Formal analysis, Writing - original draft. Ren Zhang: Data curation, Formal analysis. Haiqiao Wei: Funding acquisition, Project administration, Validation. Jiaying Pan: Writing - review & editing. Declaration of Competing Interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Acknowledgement This study was supported by the National Key R&D Program of China (2017YFE0102800) and National Natural Science Foundation of China (51706152, 51825603). Recommended articles References (28) Z. Wang et al. Knocking combustion in spark-ignition engines Prog Energy Combust Sci (2017) H. Xu et al. The relief of energy convergence of shock waves by using the concave combustion chamber under severe knock Energy Convers Manage (2018) L. Chen et al. Effects of partitioned fuel distribution on auto-ignition and knocking under spark assisted compression ignition conditions Appl Energy (2020) L. Chen et al. Numerical investigations on the effects of turbulence intensity on knocking combustion in a downsized gasoline engine Energy (2019) P.G. Hill et al. The effects of swirl and tumble on combustion in spark ignition engines Prog Energy Combust Sci (1994) F. Zhao et al. Automotive spark-ignited direct-injection gasoline engines Prog Energy Combust Sci (1999) H. Wei et al. Effect of turbulent mixing on the end gas auto-ignition of n -heptane/air mixtures under IC engine-relevant conditions Combust Flame (2016) L. Chen et al. Numerical analysis of knocking characteristics and heat release under different turbulence intensities in a gasoline engine Appl Therm Eng (2019) Y. Chen et al. A study on the influence of burning rate on engine knock from empirical data and simulation Combust Flame (2015) H. Yu et al. Effects of flame propagation speed and chamber size on end-gas autoignition Proc Combust Inst (2017) J.V. Pastor et al. A spectroscopy study of gasoline partially premixed compression ignition spark assisted combustion Appl Energy (2013) J. Martz et al. Combustion regime of a reacting front propagating into an auto-igniting mixture Proc Combust Inst (2011) A. Robert et al. LES study of deflagration to detonation mechanisms in a downsized spark ignition engine Combust Flame (2015) L. Chen et al. Understanding the correlation between auto-ignition, heat release and knocking characteristics through optical engines with high compression ratio Fuel (2020) View more references Cited by (37) Investigation of rapid flame front controlled knock combustion and its suppression in natural gas dual-fuel marine engine 2023, Energy Citation Excerpt : Some researchers believe that faster flame spread will facilitate thorough combustion of the fuel mixture prior to the end-gas auto-ignition, thereby suppressing knock . Others, however, believe that a rapid flame spread can significantly increase the temperature and pressure at the end-gas, thereby promoting auto-ignition . However, the research on the influence of swirl on the knock in large-scale combustion chambers, particularly the new-type knock, is lacking. Show abstract In this study, a new-type knock mechanism for large-bore marine engines, which differs from the existing end-gas auto-ignition knock theory and detonation wave theory, is fully revealed, using a high time resolution dynamic pressure difference method. The results indicate that the disparate enhancement effects of the flame front on the high-pressure region and the low-pressure region result in an increasing pressure difference, eventually forming the knock. The distinctions between this knock and the conventional end-gas auto-ignition knock were examined. Subsequently, the different functions of swirl in the two knock mechanisms were investigated. The results illustrate that, increasing swirl in end-gas auto-ignition knock can accelerate the termination of flame surface propagation and reduce the end-gas temperature by shortening the end-gas heat accumulation, thereby decreasing the knock intensity. However, reducing swirl in the new-type knock can diminish the combustion intensity of the flame front, thereby reducing its enhancement of the pressure wave. Consequently, a reduction to one-fourth of the initial swirl level can reduce knock intensity by 91%. Particularly, a novel knock suppression method of positioning the jet flame in the opposite direction of the swirl is proposed, which can avoid the local rapid combustion to effectively eliminate knock while maintaining power. ### Understanding the difference in combustion and flame propagation characteristics between ammonia and methane using an optical SI engine 2022, Fuel Show abstract As a carbon-free hydrogen-carrier fuel with a high content of hydrogen, ammonia (NH 3) exhibits unfavorable combustion properties such as low flammability and large cyclic variations, requiring further investigation in actual engines. In this work, for the first time, using a single-cylinder optical spark ignition (SI) engine with a high compression ratio, engine performance as well as combustion and flame propagation characteristics were investigated. Synchronization measurement of in-cylinder pressure and high-speed photography was performed, and the combustion characteristics were compared between ammonia and methane under the same operating conditions. The results show that the combustion stability and power capability of ammonia are significantly lower than that of methane. Further analysis combining combustion phasing and flame propagation characteristics indicates that the burning time loss mainly comes from the initial flame development process rather than the main combustion duration which results in a deficit in engine performance. The average probability maps of the flame location indicate that the flame front propagation of the two fuels emerges with obviously different characteristics. Quantitative flame solving shows that NH 3 flame propagation features a higher flame stretch sensitivity which is macroscopically reflected in the cyclic variations. Furthermore, the quantification of flame response to turbulence was solved and observed much higher effects for NH 3 flame than methane, which indicates NH 3 flame propagation is mainly driven by the flame response to turbulence while the flame propagation of methane exhibits higher sensitivity to temperature than ammonia. The current study shall provide insights into the application of ammonia in the internal combustion engine. ### Numerical investigation on knock characteristics and mechanism of large-bore natural gas dual-fuel marine engine 2022, Fuel Show abstract This work revealed the unique knock characteristics and mechanism of large-bore natural gas dual-fuel Marine engine with flame jet and strong swirl by a novel three-dimensional pressure difference method to further reduce the knock tendency. The knock characteristics were investigated by pressure monitoring points distributed in different spaces. The results indicate that, owing to the large cylinder bore, almost no pressure oscillations were found in the center position, whereas, severe pressure oscillations appeared at the edge. The pressure difference results illustrate that the movement of the high-pressure area and the low-pressure area tends to the low-frequency resonance mode (1,0), which manifests as an axial rotation following the swirl motion direction. Furthermore, the pressure difference method combined with the maximum amplitude of pressure oscillation (MAPO) method and energy distribution slices was adopted to explore the knock mechanism. The results show that the large-bore natural gas engine has a special knock mechanism which can neither be attributed to the end-gas self-ignition theory nor completely flame acceleration theory. The knock mechanism is that the flame jets from the prechambers cause the uneven initial pressure distribution, and the uneven pressure distribution is strengthened by the uneven heat release on the flame surface. This knock mechanism was confirmed by changing the pilot fuel mass and equivalence ratio subsequently. By the way, a knock suppression method by reducing the pilot fuel was proposed based on the knock mechanism. ### Optical experiments on the effect of turbulent jet ignition on lean burning and engine knocking 2022, Fuel Citation Excerpt : Fig. 12 shows a more intuitive display about the flame area and flame centroid paths for the two chosen cycles. The obtained flame is limited once the flame reaches the optical window boundary since there is an asymmetry of the flame propagation . It can be observed that before −8 CAD, SI flame has a bigger flame area, which is due to that the flame of TJI initiates inside the pre-chamber. Show abstract Recently the turbulent jet ignition (TJI) configured by pre-chamber has been paid more attention as it can effectively achieve rapid combustion thus high thermal efficiency. In this work, using a single-cylinder optical engine with high compression ratios, the effects of turbulent jet ignition on lean burning and engine knocking were comprehensively investigated, with the conventional spark-ignition scenarios for comparisons. The results show that the pre-chamber with larger volume results in stronger flame jets and a higher promoting effect on the combustion. However, due to the higher heat dissipation and throttling loss, the pre-chamber with larger volume results in lower IMEP. For lean combustion, the optimized TJI (small volume) can reduce the combustion instability and improve the thermal efficiency when compared to spark ignition. The main reason is that TJI can increase the in-cylinder flame speed and achieve a concentrated HRR. For knocking combustion, TJI increases the knock tendency and knock intensity. The main reason is that the accelerated flame advances the auto-ignition and induces a high energy density. In summary, the effect of TJI on engine performance differs under different load conditions and the TJI strategy should be optimized especially under heavy load conditions. ### Experimental investigation on reactivity-controlled compression ignition (RCCI) combustion characteristic of n-heptane/ammonia based on an optical engine 2023, International Journal of Engine Research ### Improving Thermal Efficiency of Internal Combustion Engines: Recent Progress and Remaining Challenges 2022, Energies View all citing articles on Scopus View full text © 2020 Elsevier Ltd. All rights reserved. Recommended articles Split diesel injection effect on knocking of natural gas/diesel dual-fuel engine at high load conditions Applied Energy, Volume 279, 2020, Article 115828 Amin Yousefi, …, Madjid Birouk ### An experimental and kinetic modeling study on the low-temperature oxidation of oxymethylene ether-2 (OME-2) by means of stabilized cool flames Combustion and Flame, Volume 253, 2023, Article 112792 Kevin De Ras, …, Kevin M.Van Geem ### Experimental investigation of the stochastic nature of end-gas autoignition with detonation development in confined combustion chamber Combustion and Flame, Volume 210, 2019, pp. 324-338 Zhao Jianfu, …, Wei Haiqiao ### Effects of direct-injected hydrogen addition on methane combustion performance in an optical SI engine with high compression-ratio International Journal of Hydrogen Energy, Volume 45, Issue 4, 2020, pp. 3284-3293 Ren Zhang, …, Changwen Liu ### Experimental analysis of super-knock occurrence based on a spark ignition engine with high compression ratio Energy, Volume 165, Part B, 2018, pp. 68-75 Lei Zhou, …, Rui Chen ### Effects of partitioned fuel distribution on auto-ignition and knocking under spark assisted compression ignition conditions Applied Energy, Volume 260, 2020, Article 114269 Lin Chen, …, Haiqiao Wei Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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59	https://www.scribd.com/document/847713276/CSE421DigitalControlL08	CSE421 Digital Control L08 \| PDF \| Algorithms \| Algebra Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 28 views 30 pages CSE421 Digital Control L08 The document discusses the stability of discrete-time systems in digital control, focusing on methods such as factorization, the Jury Test, and the Routh-Hurwitz criterion. It provides examp… Full description Uploaded by aamir ahmed AI-enhanced title and description Go to previous items Go to next items Download Save Save CSE421DigitalControlL08 For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save CSE421DigitalControlL08 For Later You are on page 1/ 30 Search Fullscreen Digital Control CSE421 Asst. Prof. Dr.Ing. Mohammed Nour A. Ahmed mnahmed@eng.zu.edu.eg Lecture 8: Stability of Dis crete Systems Copyright ©2016 Dr.Ing. Mohammed Nour Abdelgwad Ahmed as part of the course work and learning material. All Rights Reserved.Where otherwise noted, this work is licensed undera Creative CommonsAttribution-NonCommercial-ShareAlike 4.0 International License. Zagazig University \| Faculty of Engineering \| Computer and Systems Engineering Department \| Zagazig, Egypt adDownload to read ad-free Lecture: 8 Stability of Discrete Systems Factorization Jury T est Routh–Hurwitz Criterion Mohammed Ahmed (Asst. Prof. Dr.Ing.)Digital Control2 / 29 adDownload to read ad-free Stability of Discrete Systems Suppose that we have the following transfer function of a closed-loop discrete-time system: Y ( z ) R ( z )= G ( z ) 1 + GH ( z )= N ( z ) D ( z ) The system is stable if all poles lie inside the unit circle in z-plane. roots of the characteristic equation D ( z ) = 0 Mohammed Ahmed (Asst. Prof. Dr.Ing.)Digital Control3 / 29 adDownload to read ad-free Stability of Discrete Systems There are several methods to check the stability of a discrete-time system such as:Factorizing D ( z ) and ﬁnding its roots.Jury T est.Routh–Hurwitz criterion . Mohammed Ahmed (Asst. Prof. Dr.Ing.)Digital Control4 / 29 adDownload to read ad-free Factorizing the Characteristic Equation The direct method to check system stability is to factorize the characteristic equation, ◮ determine its roots, and check if their magnitudes are all less than 1. it is not usually easy to factorize the characteristic equation by hand we can use MATLAB command roots . Mohammed Ahmed (Asst. Prof. Dr.Ing.)Digital Control5 / 29 adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Jury Stability Test 50% (2) Jury Stability Test 7 pages DERTS Lec11 No ratings yet DERTS Lec11 19 pages lJNIT-4: (Stability Analysis) No ratings yet lJNIT-4: (Stability Analysis) 8 pages 10 EMI 09 System Stability No ratings yet 10 EMI 09 System Stability 25 pages Jury Stability Test No ratings yet Jury Stability Test 12 pages Chap 7 Exercise No ratings yet Chap 7 Exercise 5 pages Jury's Test No ratings yet Jury's Test 21 pages Digital Control System Stability No ratings yet Digital Control System Stability 66 pages Module 4 Stability in Discrete Time Systems No ratings yet Module 4 Stability in Discrete Time Systems 14 pages Lecture 2 No ratings yet Lecture 2 10 pages Stability: Page - 1 No ratings yet Stability: Page - 1 4 pages Lecture 3 - Dc-Peeed No ratings yet Lecture 3 - Dc-Peeed 39 pages 6 - Lecture 6 - Jury Stability - (2nd Term 2021-2022) No ratings yet 6 - Lecture 6 - Jury Stability - (2nd Term 2021-2022) 12 pages Alina Shah Dcs Lab 4 No ratings yet Alina Shah Dcs Lab 4 5 pages Chapter - 04ICE No ratings yet Chapter - 04ICE 59 pages Control Systems - Stability Analysis No ratings yet Control Systems - Stability Analysis 10 pages Transient and Steady State Response Analysis 2 No ratings yet Transient and Steady State Response Analysis 2 37 pages Chapter 4 No ratings yet Chapter 4 38 pages LCS Module-3 No ratings yet LCS Module-3 24 pages 5 - Digital Control Systems - Stability No ratings yet 5 - Digital Control Systems - Stability 26 pages Stability Analysis Methods Guide No ratings yet Stability Analysis Methods Guide 10 pages Digital Control System Stability No ratings yet Digital Control System Stability 11 pages System Stability No ratings yet System Stability 24 pages 6 Stability No ratings yet 6 Stability 20 pages 12 Stability No ratings yet 12 Stability 41 pages Routh-Hurwitz Stability Criterion Guide No ratings yet Routh-Hurwitz Stability Criterion Guide 19 pages 6 Stability of Discrete-Time Systems - Complete No ratings yet 6 Stability of Discrete-Time Systems - Complete 40 pages Advanced Digital Control Syst EE554 No ratings yet Advanced Digital Control Syst EE554 21 pages Control Por Computador: December 10, 2013 No ratings yet Control Por Computador: December 10, 2013 41 pages 4.0 Stability Analysis Techniques No ratings yet 4.0 Stability Analysis Techniques 21 pages Chapter+6+-+Stability 03272024 No ratings yet Chapter+6+-+Stability 03272024 34 pages Chapter 3 Stability of DCS No ratings yet Chapter 3 Stability of DCS 34 pages I. 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60	https://math.stackexchange.com/questions/908753/how-to-generate-log-function-that-intersects-at-0-1-and-1-0	calculus - How to generate $\log$ function that intersects at $(0,1)$ and $(1,0)$? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to generate log log function that intersects at (0,1)(0,1) and (1,0)(1,0)? Ask Question Asked 11 years, 1 month ago Modified11 years, 1 month ago Viewed 1k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I apologize for any incorrect or missing formatting, first time posting in the math stack exchange. It's been a few years since I've done any kind of calculus, so I remember nothing at all, which is probably the reason why I find my self stumped so early in my calculations. I'm looking to generate a logarithmic algorithm that will follow a plot that will pass the following points: (0,1)(0,1) and (1,0)(1,0). My end goal is to generate a programming function that will return a very simple repulsion force ceofficient value based on a distance between two points. I would like, however, that the force drop (logarithmic?) the further these two points are, and grow (exponentially?) the closer they get. This is in no way an accurate calculation I'm looking for, but merely a non-linear function to return me a scalar coefficient between 0 and 1 that I can manipulate later on. I want the function to cross at (1,0)(1,0) and (0,1)(0,1) so I get a coefficient between 0 0 and 1 1. In my function, any Y value lower than 0 0 will be floored at 0 0 and likewise for Y Y values over 1 1 (where x<0 x<0, which is possible since my X X value in my algorithm will likely be an offsetted value of the distance between two points). That being said, with the following few assumptions: y=0 y=0 when x=1 x=1 y=1 y=1 when x=0 x=0 y<0 y<0 when x>1 x>1 y>1 y>1 when x<0 x<0 y=A l o g(x+B)+C y=A l o g(x+B)+C assuming log is log based 10 I tried to deduce the A,B,C A,B,C constants in order to generate a formula. Here is what I've done so far: From assumption #1, I can generate 1=A l o g(B)+C 1=A l o g(B)+C From assumption #2, I can generate 0=A l o g(1+B)+C 0=A l o g(1+B)+C I make both equations equals by transforming equation from 1. to 0=A l o g(B)+C−1 0=A l o g(B)+C−1 Making the following equality: A l o g(B)+C−1=A l o g(1+B)+C A l o g(B)+C−1=A l o g(1+B)+C I can scratch both C C constants out (is it safe to assume C=0 C=0, or C C can be equal to any arbitrary value, without changing the equation's plot?): A l o g(B)−1=A l o g(1+B)A l o g(B)−1=A l o g(1+B) Dividing each part by A gets me: A l o g(B)A−1 A=A l o g(1+B)A A l o g(B)A−1 A=A l o g(1+B)A Allowing me to reduce both logs: log(B)−1 A=l o g(1+B)log⁡(B)−1 A=l o g(1+B) If I put both logs on the same side: l o g(B)−l o g(1+B)=1 A l o g(B)−l o g(1+B)=1 A I can combine them into: log(B 1+B)=1 A log⁡(B 1+B)=1 A And I'd say this is where I'm stuck; I have two unknowns (A A and B B) with one equation to solve. I don't know where to go from here. What's the next step? calculus algorithms logarithms Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 25, 2014 at 14:56 PrusprusPrusprus asked Aug 25, 2014 at 14:26 PrusprusPrusprus 143 5 5 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. If you have the functional form f(x)=A log(x+B)+C f(x)=A log⁡(x+B)+C (with three parameters), and impose only the two constraints f(0)=1 f(0)=1 and f(1)=0 f(1)=0, then your problem is underdetermined and you should expect to have a free parameter remaining. What you've done so far is correct. f(0)=A log B+C=1 f(1)=A log(1+B)+C=0.f(0)=A log⁡B+C=1 f(1)=A log⁡(1+B)+C=0. Subtracting one equation from the other gives A log B 1+B=1 A log⁡B 1+B=1 or A=1/log(B/(1+B)),A=1/log⁡(B/(1+B)), and C=−A log(1+B)=−log(1+B)/log(B/(1+B)).C=−A log⁡(1+B)=−log⁡(1+B)/log⁡(B/(1+B)). You are free to choose B B as you like and then solve for A A and C C. For instance, let B=1 B=1. Then A=1/log(1/2)=−1/log 2 A=1/log⁡(1/2)=−1/log⁡2 and C=−log 2/log(1/2)=1 C=−log⁡2/log⁡(1/2)=1, so your function is f(x)=−log(x+1)log 2+1=1−log 2(x+1)=log 2(2 x+1)=1−log 2(x+1),f(x)=−log⁡(x+1)log⁡2+1=1−log 2⁡(x+1)=log 2⁡(2 x+1)=1−log 2⁡(x+1), which you can readily see meets your constraints and is logarithmic. Or, let B=1/9 B=1/9. Then A=−1 A=−1 and C=log 10(10/9)=1−log 10 9 C=log 10⁡(10/9)=1−log 10⁡9, so f(x)=−log 10(x+1 9)+1−log 10 9=1−log 10(9 x+1).f(x)=−log 10⁡(x+1 9)+1−log 10⁡9=1−log 10⁡(9 x+1). As you can see, a nice parameterization of the entire family of solutions is f β(x)=1−log β+1(β x+1),f β(x)=1−log β+1⁡(β x+1), valid for any positive β β. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 25, 2014 at 17:34 answered Aug 25, 2014 at 14:50 mjqxxxxmjqxxxx 43.4k 3 3 gold badges 64 64 silver badges 115 115 bronze badges 2 Is this the correct graph? wolframalpha.com/input/?i=log+base+2+of+%282%2F%28x%2B1%29%29Prusprus –Prusprus 2014-08-25 15:07:19 +00:00 Commented Aug 25, 2014 at 15:07 1 If you enter "log base 2 of (2/(x+1)) from x=0 to x=1", you will see just the relevant range of x x.mjqxxxx –mjqxxxx 2014-08-25 17:26:38 +00:00 Commented Aug 25, 2014 at 17:26 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The answer is that each choice of A,B,C A,B,C that satisfy the conditions you've produced will give you a transformed log function that satisfies your conditions, and there are many such choices of A,B,C A,B,C that do this. That doesn't mean that we can set any particular variable to any value, however; for example, since log log is increasing, 1 A=log(B 1+B)<log 1=0 1 A=log⁡(B 1+B)<log⁡1=0 (but it turns out we can take A A to be any negative number, and this parametrizes all of the functions that satisfy your constraint). As an aside, let me point out that attraction-repulsion laws in nature generally don't take such a form, but that's a separate issue from the mathematics itself. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 25, 2014 at 14:50 Travis WillseTravis Willse 108k 13 13 gold badges 145 145 silver badges 283 283 bronze badges 1 I've edited the question to explain that a this is no way a representative calculation of natural attraction-repulsion, but a simple calculation that can produce for me a coefficient that I can use to assure that two objects will try to repulse at small distances.Prusprus –Prusprus 2014-08-25 14:58:04 +00:00 Commented Aug 25, 2014 at 14:58 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. I suggest that you instead use y=1(x+1)13 y=1(x+1)13 You can change the exponent 13 13 so as to modify the shape. This gives the correct sort of behaviour, it is significantly faster to evaluate computationally than logarithms, and it is physically more realistic (when you push real atoms together they repel with a force ∼r−13∼r−13). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 25, 2014 at 15:10 answered Aug 25, 2014 at 14:40 lemonlemon 3,558 17 17 silver badges 20 20 bronze badges 7 Sorry, for the confusion. It's the first time I write a 'physics' algorithm, so I base myself on the cause/effect I see on screen to understand what I'm doing - surely my terminology isn't correct. Y will be the force coefficient, which I will force between 0 and 1 by placing the floor at 0 and the ceiling at 1 on its value. I'm not so much looking for a vectorial value; rather, simply just a scalar value that I can later implement in the directions I wish. I am therefore running this logarithmic function twice, one for the dX and another for the dY, to get both X and Y force coefficients.Prusprus –Prusprus 2014-08-25 14:45:24 +00:00 Commented Aug 25, 2014 at 14:45 As for the logarithm function, there may be a different function that may give me a similar result, I just don't know what these are. What I like about the log function is that it would give me a slope that is more aggressive the closer I approach of x = 0, x being the distance between two given points.Prusprus –Prusprus 2014-08-25 14:48:00 +00:00 Commented Aug 25, 2014 at 14:48 @Prusprus If y(0)=1 y(0)=1 then you're not going to see any 'aggressive' increase in the slope.lemon –lemon 2014-08-25 14:49:43 +00:00 Commented Aug 25, 2014 at 14:49 @Prusprus I strongly recommend that you try the Lennard-Jones functional form.lemon –lemon 2014-08-25 14:50:46 +00:00 Commented Aug 25, 2014 at 14:50 1 wolframalpha.com/input/…lemon –lemon 2014-08-25 14:55:27 +00:00 Commented Aug 25, 2014 at 14:55 \|Show 2 more comments You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus algorithms logarithms See similar questions with these tags. 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61	https://www.ck12.org/flexi/cbse-math/hyperbola/how-to-find-the-foci-of-a-hyperbola/	Flexi answers - How to find the foci of a hyperbola? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Hyperbola Question How to find the foci of a hyperbola? Flexi Says: To find the foci of a hyperbola, follow these steps: Identify the standard form of the hyperbola equation: For a horizontal hyperbola: (x−h)2 a 2−(y−k)2 b 2=1 For a vertical hyperbola: (y−k)2 a 2−(x−h)2 b 2=1 Determine the center (h,k) of the hyperbola from the equation. Calculate c, the distance from the center to each focus, using the formula: c=a 2+b 2 4. Find the foci: For a horizontal hyperbola, the foci are at (h±c,k). For a vertical hyperbola, the foci are at (h,k±c). Example: For the hyperbola (x−3)2 16−(y+2)2 9=1: Center: (h,k)=(3,−2) a 2=16, b 2=9 Calculate c: c=16+9=25=5 Foci: (3±5,−2) or (8,−2) and (−2,−2) Final Answer: The foci are (8,−2) and (−2,−2). Analogy / Example Try Asking: Find the standard equation of the hyperbola which satisfies the given conditions: Vertices (-2, 8) and (8, 8), a focus (12, 8)How is the parametric form of the tangent equation of a hyperbola is formulated?What are the foci of the hyperbola – x^2 + y^2 = 2? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
62	https://www.youtube.com/watch?v=yD48aFuG0fU	Equilateral and Isosceles \| Triangle and its properties \| NCERT Math Class 7 \| Khan Academy Khan Academy India - English 548000 subscribers 7 likes Description 820 views Posted: 2 Apr 2024 In this video we discuss the properties of the equilateral and isosceles triangles. Khan Academy is a free learning platform for Class 1-12 students with videos, exercises, and tests for maths, science, and more subjects. Our content is aligned to CBSE syllabus and available in Hindi, English, and many more regional languages. Experience the joy of easy, seamless, accessible learning anywhere, anytime with Khan Academy. Subscribe to our YouTube channel - As a 501(c)(3) nonprofit organization, we would love your help! Donate here: Created by Devashish Phadnis 1 comments Transcript: let us recall the properties of equilateral triangle and isoceles triangle an equilateral triangle is shown on the left and the triangle is triangle ABC and when we say it's an equilateral triangle that means all the sides of the triangle are equal they could be of any length but all of them are equal so AB is equal to BC is equal to AC one more interesting property of an equilateral triangle is that all the angles of lateral triangle are equal as well and they are all of the same measure and they are all 60° so angle AB C is equal to angle BCA is equal to angle C A and all are equal to 60° on the right hand side we have triangle pqr and it's an isoceles triangle as we can see it's an isoc triangle because by definition two of the sides of of triangle pqr are equal so whenever two sides of any triangle are equal in length it's called as an isoceles triangle and in this case PQ is equal to p r they could be of any length and now another interesting property of an isoceles triangle is that the angles opposite to the equal sides are also equal so angle P RQ is opposite side PQ and angle pqr is opposite to the side PR and therefore these are equal so angle pqr is equal to angle P RQ in equilateral triangle we just saw that all the angles are 60° that is not the case in isoceles triangle so the measures of these angles can be anything but they are acute angles there is an interesting note that I would like to share with you every equilateral triangle is an isos triangle because by definition if all three sides are equal at least two sides are equal as well and therefore every equilateral triangle is by default an isoceles triangle but VI Versa is not true that means every isoceles triangle is not an equilateral triangle
63	https://projecteuclid.org/journals/kodai-mathematical-journal/volume-16/issue-2/How-to-define-singular-solutions/10.2996/kmj/1138039786.pdf	S. IZUMIYA AND J. YU KODAI MATH. J. 16 (1993), 227-234 HOW TO DEFINE SINGULAR SOLUTIONS BY SHYUICHI IZUMIYA AND JIANMING YU Abstract We shall give a rigorous definition of singular solutions of ordinary differential equations of the form F(x,y,dy/dx)—0. Our main result clarifies the geometric meaning of such a definition. All arguments are elementary. 0. Introduction. How to define singular solution? We shall consider this philosophical and stimulative question. However, we stick to ordinary differential equations of the form F(x, y, dy/dx)=0. Even for such elementary differential equations, the above question is still significant. In classical treatises of equations (Caratheodory , Courant-Hilbert , Forsyth , , Ince , Petrovski ) the discussions of equations with singular solutions are informal. In these, a "general solution" of the differential equation F(x, y, dy/dx)—0 is defined to be an one-parameter family of solutions and a "singular solution" is a solution which is not contained in the "general solution". However, this definition of singular solutions is very confused as the following example shows: Example 0.1. Consider the equation y=2p-x — p2, where p—dyldx. In the "general solution" is given by y=2p-x-p\ where c is a parameter. It is clear that y=0 is also a solution, but it is not contained in the "general solution". Then y=0 must be the "singular solution". On the other hand, we have a two parameter family of solutions: If we fix />(0)=d^0 and put c=c2c2 u then we have x=(2/3)p+(c/p) and Received November 2, 1992. 227 228 SHYUICHI IZUMIYA AND JIANMING YU y—2p'X — p 2. But, if we fix />(O)=Ci=O, then we have y=0. Moreover, if we consider this family of solutions around a point (#„, 0, 0), xQφ0, then we have the relation x(0)=(2/3)Cι+cz=x0f so that, we have an one-parameter family of "solutions" around (x0, 0, 0): Of course, ;y=0 is contained in this family of solutions. In this note we intend to give a rigorous definition of singular solutions of first order ordinary differential equations of the form F(x, y, dy/dx)=Q. In M. and T. Fukuda tried to give a rigorous definition of singular solutions of higher order ordinary differential equations. Their definition is nearly correct, however the definition of singular solutions is strongly depend on the definition of general solutions. The vagueness of the definition of singular solutions is caused by that of general solutions, so that we should begin to define "general solutions" in the correct way. The main results of this note can be generalized to the case of partial differential equations (, ). In these articles, we need some techniques of contact geometry. Here, we shall only use a purely elementary method; most of the arguments are contained in the course of ad-vanced calculus in the university. All functions and mappings considered here are differentiable of class C°°, unless stated otherwise. 1. Basic notions. We consider a first order ordinary differential equation of the form F(x, y, dy/dx)=0. If we put p=dy/dx, we may consider F as a function of ( χ> y> P) and assume that F is a smooth function defined on an open subset U in R 3 such that grad FΦ0 at any point (x, y, P)<ΞU. Then S=F" 1(0) is a smooth surface in U. We now define the notion of solutions. A smooth solution of F=0 is a smooth function y=f(x) defined on an interval (α b)czR such that F(x, f(x), f'(x))=0. This is the classical notion of solutions of the equation F=0. The following is the geometric generalization of the notion of solution due to Lie. A geometric solution of F— 0 is a smooth regular curve γ: (α b)-^F~\0) such that y'{t)—p(t)x'{t) and ϊ(t)=(x(t), y(f), p(f))ϊn the canonical coordinates system of R\ Here, we say that γ is regular if 7"(0=('(0, 3>'(0, P'(t))Φ(0, 0, 0) at any t^(a b). In the terminology of contact geometry, the above curve is called a Legendrian curve (see ). Then we can prove the following simple lemma (just an exercise for students). LEMMA 1.1. Let γ: (a b)->F~\0) be a geometric solution. Suppose that x'(t) φQ at any t<Ξ(a b). Then there exist real numbers c, d, diffeomorphism φ:(c d) HOW TO DEFINE SINGULAR SOLUTIONS 229 >(α b) and a smooth function f defined on (c d) such that ϊ°φ(x)=(x, f{x), /'( According to the above property, we may define the notion of singular point of solutions. We say that tQ is a geometric singular point of the solution y if x /(ίo)=θ. Thus γ is multivalued around the geometric singular point. It is clear that £0 is a geometric singular point of y if and only if (x'(f0), y'(fo))= (0, 0). On the other hand, there exists a notion of the Legendrian transformation by which a dual relationship can be set up between one equation and another. We adopt another coordinate system (X,Y,P) of R by X—p, Y—x-p—y, P—x. We refer to the diffeomorphism L:R 3-+R defined by L(x, y, p)= (P, x-p—y, x) as a Legendre transformation. By the definition, we have L-\Xf Y, P)=(P, X P-Y, X). If we apply the Legendre transformation to our equation, we obtain a new equation F(X, Y, P)=Fo^L)-ί(X} Y, P)=F(Pt X-P-Y, X)=0 in the new coordinate system (X, Y, P). If we calculate partial derivatives at the point (Xo, Yo> Po) corresponding to (o> yo> Po)> we can show the following: r 0, Po)=(Fx+p-Fy)(xo, y0, p0) FKXo, Yo, Po)=-Fy(xQ, y0, p0) FKXo, Yo, Po)=(Fp + x-Fy)(xQ, y0, p0). The following lemma is quite simple but important in the later section. LEMMA 1.2. (1) Let γ:(a b)-F-\0) be a geometric solution of F=0. Then L°γ: (a b)-F'K0) is a geometric solution of F=0. (2) // t0 is a geometric singular point of y, then tQ is a geometric non-singular point of L°y. If the equation F=0 satisfies FpΦθ at (x0, y0, p0), then we can locally re-write this equation in the form p—f{xf y), where / is a smooth function by the implicit function theorem. This form is far more convenient than the original one, because there exists the classical existence theorem of solutions. By the above argument, if F=0 satisfies Fx+p-Fyψ§ at (#„, yQ, pQ), then the Legendre transformed equation F=0 of F=0 satisfies F$Φθ so that we get a solution of F—0. Since F=F°(L), then we have a geometric solution of F=0 by Lemma 1.2. Then the point (x0, y0, pQ) at which F=Fp=0 or F=FP= Fx + p'Fy=0 are satisfied has special meanings. We call (x0, yo, po) a π-smgular point of F=0 if F=Fp=0 at (x0, y0, p0) and a contact singular point of F=0 if F=Fp-Fx^rp'Fy=0. We denote Σπ{F) as the set of τr-singular points, ΣC(F) as the set of contact singular points and DF — π(Σπ(F)) as the discriminant set 230 SHYUICHI IZUMIYA AND JIANMING YU of F=Q, where π(x, y, p)—(xy y). The following lemma shows the importance of the contact singular set Σe(F). LEMMA 1.3. Let γ:(a b)-F-\Q) be a smooth regular curve. If I m a g e d ΣC(F), then γ is a geometric solution. Proof. Denote γ(t)=(x(t), y(f), p(t)), then we have () F(x(f), y(t), p(t))=Fp(x(t), y(t), p(t))=(Fx+p FvXx(t), y(t), p(t))=O. If Fy=0 at γ(t), then Fp=Fx=0 at γ(f) by the above equality. This contradicts to the assumption that gradF^O, so that we have Fv(x(t), y(f), p(f))Φθ. Cal-culating the derivative of F(x(t), y(t), />())=0 with respect to t, we have ), y(t)y p(t)) x'(t)+Fy(x(jt), y(t), p(t))>y'(t)=O. It follows that ). y(f)> ^ w " Fv(x(t), y(t), p(t)) Λ W By the relation (), we have M)=z Fx(x(t), y(f), Pit)) A ] Fyixit), yit), Pit))' so that we have the relation y'(t)=p(t) x'(t). 2. Results. In order to avoid the confusion as in Example 0.1, we now introduce the following notion. Let Γ:ia b)Xia β)-»F~ 1(0) be an one-parameter family of geometric solutions of F=0. We say that Γ is a complete solution if (xt yt Pt\ rank =2 \Xc yc pj at any point (f, c)e(α b)x(a β), where Γ(t, c)=(x(t, c), yit, c), pit, c)) and c is a parameter. If the image of Γ contains a point (x0, y, poϊ^F-^O), we call it a complete solution around (xo> 3Ό, Po)> In some classical textbooks (cf. ), the above term is used in a different sense. However, we adopt the above defini-tion according to the terminology in the theory of first-order partial differential equations (, ). We say that an equation F=0 is completely integrable around (jt0, yQ) pϋ) if there exists a complete solution of F=Q around (xo> y0, p0). The equation in Example 0.1 is not completely mtegrable around the origin. We now state a basic property of complete mtegrable equations. PROPOSITION 2.1. Suppose that F=0 is completely integrable around ix0, y0, Po)> HOW TO DEFINE SINGNLAR SOLUTIONS 231 then (x0, y0, po)^Σc(F) or ΣC(F) is an l-dimensιonal submanifold around (x0, 3Ό, Po)> Proof. Let Γ: (a b)X(a β^F-'φ) be a complete solution of F=0 around (o, yQ, Po). Suppose that (x0, y<>, PO)ΪΞΣC(F), then FyΦθ at (0, y0, p0) by the same reason as that in the proof of Lemma 1.3. We now differentiate the both sides of the equality F(x(f, c), y(t, c), p(t, c)) =0 with respect to c, where Γ(t, c)=(x(t, c), y(t, c), p(t, c)) and c is a para-meter. Then we have Fx xc+Fy yc+Fp pe=0. It follows from the definition that Γ(t, c)^Σc(F) if and only if Fy'(yc-p'Xc)=0. Since FυΦθ at Γ(t, c), the above equality is equivalent to yc(t, c)—p{t, c) xc{t, c)=0. Thus we have Σe(F)={(x(t, c\ y(f, c), p(t, c))\yc(t, c)-p(t, c)-xc(t, c)=0}. Since Γ is a complete solution, we have yt(f, c)=p(t, c) xt(t, c), so that ytc(t, c)=pc(t, c) xt(t, c)+p(t, c)-xtc(t, c). On the other hand, Γ'\Σe(F)) is defined by the equation f(f, c)=yc(t, c) — p(t, c). Xc(t, c)=0. If we have ft(t, c)=yct(t, c)-pt(t, c)-xc(t, c)+p(t, c)-xet(t, c) =0, then pt(t, c) xc(t, c)—pc(t, c) xt(t, c)=Q. This contradicts to the fact that (xt yt Pt\ rank 1=2. \^β yc pel Hence, ft(t, c)Φθ at (f, c)^Γ~\Σc(F)). This completes the proof. If FpΦθ at (x0, y0, p0), it is already mentioned in §1 and that there exists a unique solution of F=0 by the classical existence theorem and this gives a complete solution around (x0, yOf p0). We can also have a complete solution of F=0 around (x0, y0, po)£Σc(F) by the argument about the Legendre transfor-mation in §1. In fact, we can prove local uniqueness of complete solutions around any points, however we do not need this fact. Then we omit the proof. We now give a rigorous definition of singular solutions. Let γ: (a b)-+ F~\0) be a geometric solution such that Γ(o)=(o, yo, ίo) We say that γ is a singular solution of F=0 around (xOf yQ, pQ), if it is never contained in any complete solution around (x0, yo, Po)> We also say that γ is a singular solution of F=0 in the strict sense around (x0, y0, p0) for any open subinterval (c d)a (a b), y(c d) is never contained in any complete solutions of F— 0. Returning to the equation y=2p-x — p 2 in Example 0.1, we can easily show that ΣC(F)= {(0, 0, 0)}. By Proposition 2.1, this equation is not completely inte-grable around (0, 0, 0). It follows that the solution y—0 is the singular solution around (0, 0, 0). However, it is not a singular solution in the strict sense. The following theorem describes the relation between singular solutions and singular solutions in the strict sense. THEOREM 2.2. For an equation F=0 and a geometric solution γ: (α b)—> 232 SHYUICHI IZUMIYA AND JIANMING YU F~ ι(0) such that γ(to)=(xo, yQ, p0), the following are equivalent. (0) F=0 is completely integrable around (x0, yQ, p0) and γ is a singular solu-tion of F=0 around (xQ, y0, p0). (1) γ is a singular solution of F—0 in the strict sense. (2) There exists a complete solution around any point of γ((a b)) such that each member is transverse to y. (3) Image γ^Σc(F). Proof. (3)=K2). By the same reason as that of the proof of Lemma 1.3, we have Fyφ0 at γ(0). By the implicit function theorem, the equation F=0 can be rewritten as y = h(x, p) around γ(fo)f so that, we have Σc(y-h(x, p))={(x, h(x, p), p)\hp{x, p)=hx{x, p)-p=0. Then we may distinguish two cases: (a) hxp(x, p)-l=0, (b) hxp(x, p)-lΦθ. Case (a). Since hxp=hpx=lφ0 at (x(t0), P(t0)), the set Σπ(y-h(x, p))={(x, h(x, p), p)\hp=0} is a smooth curve, γ is also a smooth curve, then we have γ((a b))=Σc(y — h(x, p)) —Σπ(y — h{xy p)). It follows that there exists a smooth function μ around (x(t0), p(to)) such that μ does not vanish on such a neighbourhood and hx — p ^μ-hp. We now consider a vector field on the (x, jf))-plane defined by dx μ' dp' By the same reason as in Example 0.1, the flows of the vector field V gives a complete solution of y — h(x, />)=0 around γ(t0). On the other hand, by the previous arguments, the curve (x(t), p(t)) is given by the equation hp—0 near (x(t0), p(t0)). If we calculate the canonical inner product of grad hp and V, then we have =l-μ-hpp. Differentiate the equation hx — p—μ-hp with respect to p, then we have μ hpp =0 at ((„), p(t0)). It follows that so that V does not tangent to 7. This means that each member of the complete solution is transverse to γ around p(f0). Case (b). In this case the set {(x, h(x, p), p)\hx{x, ρ)—ρ=zQ} is a smooth curve near ((f0), y(U), p(t0)). By the same reason as that of the case (a), there exists a smooth function λ around (x(t0), p(tQ)) which never vanish and satisfies hpz=χ.(hx-p). We adopt a vector field V=λ-d/dx—d/dp on (x, />)-plane, so that the flow of V gives a required complete solution. HOW TO DEFINE SINGULAR SOLUTIONS 233 If y is not a singular solution in the strict sense, then there exist an intervals (α' b')cz(a b), (a' β')^{a β) and a complete solution Γ:(a' b')X (a' β')-+F-ι(0) such that ΓCo=γ for some co^(a' β'). By Proposition 2.1 and the previous proof, there exists ί oe(α' b') such that Γ(U, cQ)£Σc(F). By the local uniqueness of the complete solution outside of ΣC(F), the complete solution Γ is equal to the original complete solution on some open neighbourhood of Γ(to, Co) which is given by the condition (2). However, each member of this complete solution must be transverse to γ=ΓCo. This gives a contradiction. (1H(3). If Image γφΣe(F), then there exists to^(a b) such that y(to)£Σc(F). It follows that FpφQ or Fx+p-FyΦθ at γ(t0). In both cases, there exists a unique complete solution of F=0 which contains γ around γ(t0). This contradicts to the definition of singular solutions in the strict sense. Here, we only proved that Image y tzΣc(F) instead of the condition (3), how-ever, by the previous proof and Proposition 2.1, we can assert that Image y— ΣC(F). (0)=H3). Let Γ: (a b)X(a i8)->F" 1(0) be a complete solution of F=0 around (o, 3 > o , po)^Σc(F). If ImageyφΣc(F)y then there exists to^(a b) such that y(tQ)(£Σc(F). It follows that FpΦθ or Fx+p-FyΦθ at γ(f0). In both cases, there exists a unique complete solution of F=0 which contains y around y(t0). This complete solution is equal to Γ around y(t0). Hence, Image γ=-Image ΓCQ around t0, where ΓC(β)—Γ(t, c0). This solution eventually reaches (xOf y0, p0), so that it is contained in Γ around (xQ, y0, p0). This contradicts to the definition of singular solutions. It is clear that (1) and (2)=H0). This completes the proof. The following is a classical example of an equation with singular solution. Example 2.3. The Clairaut equation: y = x p+f(p). The singular solution is Σc(F)=Σπ(F) and the discriminant set DF is the envelope of the complete solution y = x-c+f(c). In classical treatises, it has been considered that Σπ(F) was a strong can-didate for the singular solution as the Clairaut equation shows. However, Theorem 2.2 asserts that ΣC(F) is the singular solution in the strict sense. Example 2.4. y = x-We can calculate that Σπ(F)= {(x, y, p)\y = x-p 2+p s and p (2x+3p)=0} and Σc(F)={(x, y, p)\y :=p=0]. By Theorem 2.2, this equation is completely inte-grable around (0, 0, 0) and the singular solution is given by y = p=0. Acknowledgment. A part of this work was done during the first author's stay at the University of Liverpool. The first author would like to thank all members of the Department of Pure Mathematics for giving him a chance to talk about these subjects at the internal colloquium. This work was completed 234 SHYUICHI IZUMIYA AND JIANMING YU during the authors' stay at Guizhou University in China. The authors are also grateful to Professor Guobin Zhang for his kind hospitality. REFERENCES [ 1 ] V.I. ARNOL'D, Contact geometry and wave propagation, Monographic de LΈn-seignement Mathematique 34 (1989). [ 2 ] C. CARATHEODORY, Calculus of Variations and Partial Differential Equations of First Order, Part I, Partial Differential Equations of the First Order, Holden-Day, 1965. [ 3 ] R. COURANT AND D. HILBERT, Methods of mathematical physics I, II, Wiley, New York, 1962. [ 4 ] A. R. FORSYTH, A Treatise on differential equations, Macmillan and Co, 1885. [ 5 ] A. R. FORSYTH, Theory of differential equations, Part III partial differential equations. Cambridge Univ. Press, London, 1906. [ 6 ] M. FUKUDA AND T. FUKUDA, Singular solutions of ordinary differential equa-tions, The Yokohama Math. Jour. 15 (1977), 41-58. [ 7 ] E. L. INCE, Ordinary differential equations, Dover, 1926. [ 8 ] S. IZUMIYA, Singular solutions of first order differential equations, to appear in Bull. London Math. Soc. [ 9 ] I.G. PETROVSKI, Ordinary differential equations, Prentice-Hall, 1996. J. Yu, On singular solutions of completely integrable partial differential equa-tions of first order, in preparation. DEPARTMENT OF MATHEMATICS, FACULTY OF SCIENCE, HOKKAIDO UNIVERSITY, SAPPORO 060, JAPAN INSTITUTE OF MATHEMATICS, ACADEMIA SINICA, BEIJING 100080, P.R. CHINA
64	https://lantsandlaminins.com/sensitivity-specificity-and-predictive-values-mcq/	Sensitivity, Specificity and Predictive Values MCQ – LaNts and Laminins Skip to content LaNts and Laminins The Hamill lab blog Menu Blog Science Roomexpand child menu A very friendly introduction to Basement Membranes Laminins – a friendly intro! LaNts – a friendly intro Taking control of a cancer-associated “splicing switch” LaNt alpha31 in the ocular surface; a new player in corneal wound repair? LaNt alpha31 distribution in human tissues Are you really as good at applying sunscreen as you think? Can you replace your sunscreen with SPF containing moisturisers? UV photography; how to guide LaNt a31 – a breast cancer marker and potential drug target Join the team Core Cell and Molecular Biology + Genetics Quizzesexpand child menu Guide to identifying mode of inheritance from genetic pedigrees Cell Biology and Genetics Foundation MCQ – 1 Core Genetics Multiple Choice Quiz Q1 Genetic Pedigree Multiple Choice Quiz – 1 Genetic Pedigree Multiple Choice Quiz – 2 Genetic Pedigree Multiple Choice Quiz – 3 Genetic Pedigree – Multiple Choice Quiz – 4 Multiple Choice Quiz – Genetic Pedigrees – 5 – ABO and Rh blood group Genetics Pedigrees MCQ 6 – Real Example – LOCS Mitosis and Meiosis MCQ (genetics/cell biology) Transcription, translation, protein and gene regulation MCQ Sensitivity, Specificity and Predictive Values MCQ Mean, Median, Mode MCQ Hallmarks of Cancer MCQ Lab Skills Quizzesexpand child menu MCQ working in the lab MCQ – Lab calculations MCQ Microscopy MCQ protein work MCQ cell culture MCQ RNA/DNA techniques MCQ Flow cytometry and FACS Science grammar quiz! Lab Skill Quiz – Data interpretation I Lab skills quiz – data interpretation MCQ #2 Lab skills MCQ – Data interpretation #3 Worked examples videos: basic lab calculations Writing Guidesexpand child menu Downloadable Manuscript Template Doc Titles Preparing Scientific Posters Abstracts Introductions Materials and Methods Results Figure Preparation Choosing a graph type Figure legends Results Discussions Quick tips-methods Quick tips-results “Life in the Lab” book Experimental Design Guideexpand child menu Statistics test – flow chart Experimental Design – Part 1 – Hypotheses Experimental Design – Part 2 – Measurements Experimental Design – Part 3 – Controls Experimental Design – part 4 – Assigning groups Experimental Design – Part 5 – Sample size determination MRes BLS linksexpand child menu MRes: Introduction to Lab Skills Practical Class MRes – RT-PCR (video) MRes – Western Blotting Worked examples videos: basic lab calculations The LaNts and Laminins team LaNt wars: the LaNt mouse saga… Sensitivity, Specificity and Predictive Values MCQ This test will get into the terms and applications surrounding the predictive value of medical tests. We’ll start with some quick definitions then move onto some harder calculations/applied knowledge questions. Which term defines the proportion of individuals who test negative for a condition who actually do not have the condition? Sensitivity Specificity Positive Predictive Value Negative Predictive Value Share this: Click to share on Facebook (Opens in new window)Facebook Click to share on X (Opens in new window)X More Click to share on WhatsApp (Opens in new window)WhatsApp Click to share on LinkedIn (Opens in new window)LinkedIn Click to share on Pinterest (Opens in new window)Pinterest Click to share on Pocket (Opens in new window)Pocket Click to share on Telegram (Opens in new window)Telegram Click to share on Reddit (Opens in new window)Reddit Like Loading... Recent Posts They grow up so fast – Dzhalil + Louie at ILCAMS day Life in the Lab: The Ultimate Guidebook for Aspiring Researchers – SALE NOW ON They Grow Up so Fast – Congratulations to Dr Fawziah Asiri Showing off microscopy images at Pint of Science Team bowling Blog Stats 667,288 hits Lab Life Filming for lab skills teaching Drs Hamill, Iorio and Sheridan Search for: Translate Powered by Translate Blog Stats 667,288 hits Subscribe to Blog via Email Enter your email address to subscribe to this blog and receive notifications of new posts by email. Email Address: Subscribe Join 2,691 other subscribers [privacy-do-not-sell-link] Blog at WordPress.com. SubscribeSubscribed LaNts and Laminins Join 160 other subscribers Sign me up Already have a WordPress.com account? Log in now. LaNts and Laminins SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments... Write a Comment... Email (Required) Name (Required) Website %d Original text Rate this translation Your feedback will be used to help improve Google Translate
65	http://www.jci.org/articles/view/117430/files/pdf	C5a-induced Expression of P-selectin in Endothelial Cells Kimberly E. Foreman," Ara A. Vaporciyan, Brian K. Bonish, Michael L. Jones, Kent J. Johnson," M. Michael Glovsky,§ Susan M. Eddy, and Peter A. Ward Parke-Davis Research Division of Warner-Lambert Company, Ann Arbor, Michigan 48105; 'Department of Pathology, University of Michigan Medical School, Ann Arbor, Michigan 48109-0602; and §Asthma and Allergy Center, Huntington Memorial Hospital, Pasadena, California 91105-3022 Abstract Human umbilical vein endothelial cells have recently been shown to respond to C5a with increases in intracellular Ca2", production of D-myo-inositol 1,4,5-triphosphate and superoxide anion generation. In the current studies, C5a has been found to cause in a time- and dose-dependent man- ner rapid expression of endothelial P-selectin, secretion of von Willebrand factor, and adhesiveness for human neutro- phils. The effects of C5a in P-selectin expression and adhe- siveness of neutrophils were similar to the effects of hista-mine and thrombin on endothelial cells. The adhesiveness of C5a-stimulated endothelium for neutrophils was blocked by anti-P-selectin, but not by antibodies to intercellular ad-hesion molecule 1, E-selectin, or CD18. A cell-based ELISA technique has confirmed upregulation of P-selectin in endo- thelial cells exposed to C5a. Binding of C5a to endothelial cells has been demonstrated, with molecules bound being - 10% of those binding to neutrophils. By a reverse tran- scriptase-PCR technique, endothelial cells have been shown to contain mRNA for the C5a receptor. These data suggest that C5a may be an important inflammatory mediator for the early adhesive interactions between neutrophils and en- dothelial cells in the acute inflammatory response. (J. Clin. Invest. 1994. 94:1147-1155.) Key words: adhesion - comple- ment endothelial cells neutrophils receptors Introduction During acute inflammation, neutrophils first adhere to vascular endothelium and then migrate into the surrounding tissues. It has been proposed that at least three distinct steps are involved in neutrophil recruitment (1, 2). First, there must be "activa- tion" of the endothelium, defined as appearance of adhesion- promoting factors. A transient and reversible intermittent adhe- sion results in the "rolling" of unactivated neutrophils along the surface of activated endothelium. This is followed by the firm adhesion of neutrophils to endothelial cells. At some point, either before or after this adhesive interaction with the endothe- lium, neutrophils appear to become activated. Finally, neutro- Address correspondence to Peter A. Ward, M.D., The University of Michigan Medical School, Department of Pathology, 1301 Catherine Street, Box 0602, Ann Arbor, MI 48109-0602. Received for publication 14 February 1994 and in revised form 6 May 1994. phils migrate through endothelial cell junctions into the sur- rounding tissue. The initial rolling of the neutrophils along the endothelial cell surface appears to be mediated by members of the selectin family of adhesion molecules (L-, E-, and P-selec- tin) (3). The /2 integrins (CDlla/CD18, CDllb/CD18, and CDllc/CD18) expressed on neutrophils are believed to medi- ate firm adhesion through interaction with intercellular adhesion molecules (ICAM) 1 and 2' expressed on the activated endothe- lium (3). Recently it has been suggested that the next step involves interaction of neutrophils with platelet-endothelial cell adhesion factor present in basolateral junctions of endothelial cells (4, 5). The final step, migration of the neutrophils into the area of inflammation, may be mediated by chemotactic fac- tors such as C5a, LTB4, and platelet-activating factor (6). P-selectin was originally identified as an a-granule mem- brane protein in platelets capable of being translocated to the cell membrane after platelet activation and degranulation (7, 8). Also, P-selectin is produced constitutively and stored intra- cellularly in Weibel-Palade bodies of human umbilical vein endothelial cells (HUVECs) (9, 10). Exposure of endothelium to mediators such as thrombin, histamine, H202, or the comple- ment membrane attack complex (C5b-C9) results in rapid upregulation of P-selectin expression on the cell surface (9, 1 1-13). Except for H202, the induced expression of P-selectin is short lived, peaking at 5-10 min and returning to basal levels within 30-60 min due, apparently, to internalization of P-selec- tin (11). In the case of H202 stimulation of endothelial cells, P-selectin expression appears to be stable (12). Unactivated human blood neutrophils adhere in a CD 18-independent manner to endothelial cells stimulated under conditions known to pro- mote P-selectin expression ( 12, 14-16). Recent studies suggest that glycoproteins with sialylated 0-linked oligosaccharides may be the neutrophil ligands involved in P-selectin-dependent adhesion (17, 18), although sulfated glycoproteins and glyco- lipids have also been identified provisionally as neutrophil "counter-receptors" for P-selectin (17). In general, ligands for P-selectin interact in a Ca2' dependent manner with the NH2-terminal lectin binding domain of P-selectin. It has been shown that the binding of Ca2+ to the lectin binding domain results in a conformational change in P-selectin, promoting ligand recog- nition (19, 20). Previous studies in our laboratory have demonstrated that anti-P-selectin antibodies are protective in acute lung injury due to systemic complement activation after infusion of cobra venom factor into rats (21). Injury in this model reaches a peak within 30 min, is dependent both on complement activation as 1. Abbreviations used in this paper: BCECF-AM, 2 ',7 '-bis- (2-carboxy- ethyl)-5(and-6)-carboxyfluorescein, acetoxymethyl; C5aR, C5a recep- tor; HUVECs, human umbilical vein endothelial cells; ICAM-l and -2, intercellular adhesion molecules 1 and 2; ORF, open reading frame. C5a Receptor-induced Expression of P-selectin in Endothelial Cells 1147 J. Clin. Invest. © The American Society for Clinical Investigation, Inc. 0021-9738/94/09/1147/09 $2.00 Volume 94, September 1994, 1147-1155 well as on neutrophil-generated toxic oxygen products, and is associated with lung vascular expression of P-selectin. Anti- body-induced blockade of P-selectin greatly reduces neutrophil accumulation in the lung and diminishes the intensity of injury (21). The following in vitro studies were designed to explore further the role of the anaphylatoxin peptide C5a on P-selectin expression and function in HUVECs. C5a has been found to cause in HUVECs a rapid increase in adhesion of unactivated neutrophils to the endothelial cells. This increased adhesiveness is mediated by CSa-induced P-selectin expression on the endo-thelium. Evidence is also provided for binding of C5a to endo-thelial cells and for the presence of mRNA for CSa receptor (C5aR) in these endothelial cells. Methods Chemicals and reagents. Recombinant human C5a was a gift of Dr. Henry Showell (Pfizer Central Research, Groton, CT). The ED50 for the chemotactic activity of this CSa preparation was 1-5 nM. This was based on four separate determinations, using polycarbonate membranes (polyvinyl chloride-free) with 3-psm pores. Human blood neutrophil migration occurred over a 30-min period. Neutrophils were quantitated by fluoremetric analysis as recently described (22). cDNA for human neutrophil C5aR was the gift of Drs. Craig and Norma Gerard (Chil-dren's Hospital, Boston, MA). Murine monoclonal antibodies against human P-selectin (PNB1.6 and PB1.3) were the gift of Dr. James C. Paulson (Cytel Corporation, San Diego, CA). Monoclonal antibody to human CD18 was provided by Dr. C. Wayne Smith (Baylor University, Houston, TX). The human anti-E-selectin monoclonal antibody (CL3) has been described previously (23). Anti-ICAM-1 antibody was ob-tained from Research and Diagnostic Systems (Minneapolis, MN). An- tibodies to human FcyRII (IV.3) and FcyRIll (3G8) were purchased from Medarex, Inc. (West Lebanon, NH), while TNF-a was obtained from Peprotech (Rocky Hill, NJ). The 2',7'-bis-(2-carboxyethyl)- 5(and-6)-carboxyfluorescein, acetoxymethyl (BCECF-AM) used for fluorescent labeling of neutrophils was purchased from Molecular Probes (Eugene, OR). Silicone oil (SF1250; General Electric) was the gift of Scot Westerbeek (Matteson-Ridolfi Incorporated, Riverview, MI). Radioiodinated C5a was purchased from NEN-Dupont (Wilming- ton, DE). Other reagents were obtained from Sigma Chemical Co. (St. Louis, MO). Endothelial cell preparation. HUVECs were isolated by collagenase treatment of freshly obtained human umbilical cords and plated on gelatin-coated tissue culture dishes (Falcon Co., Lincoln Park, NJ) (24). Dulbecco's modified Eagle's media with Ham's F-12 and 20% heat- inactivated fetal bovine serum (FBS) supplemented with 2 mM L-gluta-mine, 100 U/ml penicillin, 100 ,g/ml streptomycin, 0.25 sg/ml Fun- gizone, 25 jig/ml endothelial cell growth supplement (Collaborative Research, Bedford, MA), and 15 U/ml bovine heparin was used to maintain the cells. HUVECs were used between the first and third passage and were characterized by a cobblestone appearance and specific staining for von Willebrand factor (vWf). Preparation of neutrophils. Whole blood was obtained from healthy human volunteers and anticoagulated with citrate dextrose solution. Neutrophils were isolated by gradient centrifugation over Ficoll-Hy- paque (Pharmadia, Uppsala, Sweden) followed by 1% dextran sedimen-tation for 1 h to separate neutrophils from erythrocytes. Remaining erythrocytes wqre removed by hypotonic lysis and the neutrophils were washed twice i Hanks' balanced salt solution containing 0.1% bovine serum albumin (HBSS-BSA). Cells were resuspended to 5 x 106 cells/ ml in HBSS- SA, and preparations contained > 95% neutrophils. Fluorescent labeling of neutrophils. 50-pig aliquots of BCECF-AM were freshly di solved in 50 ji dimethylsulfoxide. This was then added to the neutroph l suspension at a final concentration of 1 mM BCECF- AM. The mixture was incubated for 30 min at 37°C, and then the cells were washed twice with 40C HBSS-BSA. The cells were then resuspended in HUVEC media at 1 x 106 cells/ml for use in the adhe- sion assay. Assay of neutrophil adherence to HUVEC monolayers. An assay for measuring static adhesion of fluorescent-labeled neutrophils to HUVEC monolayers has been recently described (25). Briefly, HUVECs were plated at 5 X 104 cells/well in 96-well flat-bottom fibronectin-coated plates and grown to confluence at 370C with 5% CO2. The cells were washed twice with fresh HUVEC media. Recombinant C5a or TNF-a was added to the appropriate wells (as indicated) and incubated at 370C with 5% CO2. The cells were then washed twice with warm media and 1 x 105 fluorescent-labeled neutrophils were added to each well. Astandard curve relating cell number to fluorescence was prepared on each plate by performing twofold serial dilutions of the neutrophils from 1 x 105 to 6.25 x 103 cells/well. Microtiter plates were incubated for 30 min at 370C with 5% CO2. The nonadherent neutrophils were then removed with gentle washing. Remaining fluorescence was measured with a fluorescence measurement system (Cytofluor model 2300; Milli- pore, Bedford, MA) using an excitation filter at 485 nm and emission filter at 530 nm. In experiments using monoclonal antibodies to block neutrophil binding, antibodies (10-20 /Ag/ml) were added to the wells containing the neutrophils. All antibodies were used as intact murine IgG, monoclonal antibodies. Fluorescent-labeled neutrophils were treated for 15 min at room temperature with 10 jsg/ml antibodies to block Fc receptor interactions immediately before their addition to mi- crotiter wells. ELISA for adhesion molecule expression. An ELISA to determine adhesion molecule expression on monolayers of endothelial cells was developed in this laboratory (26). Briefly, HEUVECs were plated at 5 x 104 cells/well in 96-well flat-bottom fibronectin-coated plates and grown to confluence at 37°C with 5% CO2. The cells were washed with warm HUVEC media and stimulated as described in the text at 37°C. After stimulation, the endothelial cell monolayers were washed and fixed with 1% paraformaldehyde. 5% nonfat dry milk was used to reduce nonspecific binding and the cells incubated with primary antibody (1 Ag/ml) for 45 min. The cells were washed and a peroxidase-conjugated rabbit anti-mouse IgG (Dako Corporation, Carpinteria, CA) was added for 40 min. After washing, the substrate (o-phenylenediamine dihydro- chloride) was added for 30 min and the reaction was quenched with 3 M sulfuric acid. Optical density was determined at 490 nm using an automated microplate reader (EL340; Bio-Tek Instruments, Winooski, VT). Preparation of CSa des arg. 100 1lA of a 10 /.M stock solution of CSa was incubated for 30 min at 37°C with 5 U carboxypeptidase B. The carboxypeptidase B was then heat inactivated by incubating the mixture for 30 min at 56°C. ELISA for vWf HUVECs were grown to confluence in a 150-mm gelatin-coated tissue culture dish. The cells were washed twice with Dulbecco's modified Eagle's media and incubated for 15 min with indicated inflammatory mediators. Supernatants were collected, filter sterilized to remove any contaminating HUVECs, and concentrated to 1 ml. Twofold serial dilutions of the supernatants were incubated for 2.5 h in wells of a 96-well plate coated with anti-human vWf and blocked with BSA. A standard curve of vWf concentration was prepared on each plate by performing twofold serial dilutions of human reference plasma from 2,500 to 2.4 pg vWf. The plates were washed and a peroxi- dase-conjugated rabbit anti-human vWf was added for 2 h. After wash- ing, the substrate (o-phenylenediamine dihydrochloride) was added for 10 min and the reaction was quenched with 3 M sulfuric acid. Optical density was determined at 490 nm using an automated microplate reader. CSa binding studies. Binding of C5a to neutrophils and HUVECs was performed using a silicone oil technique similar to that described previously (27). Briefly, HUVECs were removed from tissue culture dishes using 0.02% EDTA and washed twice in binding buffer (50 mM Hepes, pH 7.4, containing 1 mM CaCl2, 5 mM MgCl2, 0.5% BSA, 10 mM glucose, 0.1% bacitracin, 1 mg/ml leupeptin, and 2 sg/ml aproti- nin) (28). 2 x 106 cells were incubated for 60 min on ice with 10 nM C5a supplemented with 0.1 nM radioiodinated C5a (- 60,000 cpm) in a final volume of 300 Al. 100-yl aliquots of the cell suspension were 1148 Foreman, Vaporciyan, Bonish, Jones, Johnson, Glovsky, Eddy, and Ward layered on 200 ul of silicone oil in a 400-M1 polyethylene microcentri- fuge tube and centrifuged for 2 min in a microfuge (Beckman Instru- ments, Inc., Fullerton, CA). Near the pellets, the tubes were transected with a razor blade separating the pellet containing cell-associated radio- iodinated C5a from unbound C5a in the supernatant fluid. Cell-associ- ated and free radioactivity were measured in a gamma counter and molecules of C5a bound per cell were calculated. C5a binding to neutro- phils was measured in a manner similar to that used for HUVECs, with the exception that 5 x 106 neutrophils were used in the assay. Nonspe- cific binding to both cell types was measured in the presence of 1,000- fold excess unlabeled C5a (29). Detection of mRNA for the C5a receptor. Total RNA was isolated using Trizol Reagent (Gibco, Gaithersburg, MD) using 1 X 107 HUVECs (30). mRNA from the HUVECs was subject to reverse tran- scription using a cDNA cycle kit (Invitrogen, San Diego, CA) with oligo dTs as a primer. Single-stranded cDNA was then amplified over 35 cycles using PCR with primers directed against the full open reading frame (1,053 bases) or a fragment of the open reading frame (636 bases) of the oligonucleotide coding for the neutrophil C5a receptor. cDNA from the neutrophil C5a receptor was used as a positive control for the C5a receptor primers. PCR products were analyzed on a 0.9% agarose gel and the bands were visualized with ethidium bromide. Statistical analysis. Each ELISA and cell adhesion assay experiment contained four to six replicates and was repeated three times with similar results. Data from separate experiments were not merged because of intrinsic differences in the optical density values from experiment to experiment, reflective of baseline immunoperoxidase products. In con- trast, cell adhesion studies between experiments were highly reproduc- ible relative to values of nonstimulated and stimulated HUVECs. For brevity and clarity, we have routinely presented data from individual experiments. For each data point, values (mean±SEM) represent results from replicates (4-6). Where the error bars are not visible in the figures, this is due to the very small value of the error bar. Where not displayed, SEM values are described in the figure legends. The C5a binding experi- ments were performed in triplicate and in nine independent experiments. Each data point represents the mean from a representative experi- ment±SEM. Statistical differences between groups were measured by analysis of variance. Results Histamine-induced P-selectin expression on HUVECs and in- creased neutrophil adhesion. HUVEC monolayers were stimu-lated with increasing concentrations of histamine for 5 mi, followed by immediate washing, fixation, and determination of P-selectin expression with a cell-based ELISA. As shown by the data in Fig. 1 A, a dose-dependent increase in P-selectin expression occurred after stimulation with histamine. To deter- mine neutrophil adhesion, HUVEC monolayers were incubated for 5 min with increasing concentrations of histamine, followed by washing and addition of neutrophils. Concentrations of hista- mine of - 0.1 mM did not promote neutrophil adhesion (Fig. 1 B). Incubation of the endothelial cells with concentrations of histamine of 2 500 yM resulted in significant increases in neutrophil adherence, with a plateau in cell adhesion appearing between 1 and 5 mM concentrations of histamine. Upregulation of P-selectin expression after exposure of endothelial cells to 5 mM histamine was rapid, reaching a peak within 5 min, with a return toward basal levels by 30 min (Fig. 2). Similar results have been previously reported using a direct immunochemical detection ("5I-antibody) of P-selectin expression on histamine- stimulated endothelial cells (11). Thrombin-induced P-selectin expression on HUVECs. As expected, incubation of HUVECs with increasing concentra- tions of thrombin for S min resulted in a dose-dependent in- 0.48- EC 00) 0 a -a ._ 0 0.46- 0.44- 0.42- A. P-selectin Expression p<0.05 0l .01 0. .... ...1 0.001 0.01 0.1 I - 0 C. 0 2 ED d0 10 100 "l 10100 B. Neutrophil Adherence p<0.001 0.001 0.01 0.1 10 100 Histamine (mM) Figure 1. Dose-responses for histamine stimulation of HUVEC mono- layers. Cells were incubated with increasing concentrations of histamine for 5 min followed by washing. (A) Endothelial cell P-selectin expres- sion was measured by an ELISA. (B) Human blood neutrophils were added and adhesion evaluated 30 min later. All comparisons were to non-histamine-exposed HUVECs. crease in P-selectin expression (Fig. 3). Concentrations of thrombin of 2 0.1 U/ml resulted in significant upregulation of P-selectin on HUVECs, with a plateau between 1 and 10 U thrombin/ml. Again, these results are similar to those in which a different immunochemical analysis has been employed ( 11). CSa-induced P-selectin expression on HUVECs and in- creased neutrophil adhesion. HUVEC monolayers were incu-bated with increasing concentrations of C5a for 10 min, fol- lowed by washing and evaluation of P-selectin expression. C5a promoted a dose-dependent increase in P-selectin expression on HUVECs (Fig. 4 A). C5a concentrations of < 25 nM did not result in increased expression, while C5a concentrations of 2 50 nM resulted in significant upregulation of P-selectin (P < 0.05). When 250 nM C5a des arg was used, the increase in P-selectin was abolished (Fig. 4 A) as was the increased cell adhesion (Fig. 4 B). The dose response profile with neutrophil adhesion showed a similar profile (Fig. 4 B). The relatively high concentrations of C5a required to induce endothelial cell responsiveness was not due to poor activity of the C5a prepara- C5a Receptor-induced Expression of P-selectin in Endothelial Cells 1149 0.48-0 0.46 - 0.44- 0.42 - 0.40 - E 00) C,) 0 am -c 0 . 0o 10 20 0.575 - 0.550- 0.525- p<0.05 30 Duration (min) of Exposure to Histamine Figure 2. Upregulation of endothelial cell P-selectin expression after incubation with 5 mM histamine for 1-30 min. P-selectin expression was measured by an ELISA method. All comparisons were to non- histamine-exposed HUVECs. tion, because the same preparation of C5a induced neutrophil chemotactic responsiveness in the range of 1-5 nM (see above). This activity is consistent with the biological activity of CSa purified from activated human serum (31). In companion studies in which endothelial cell monolayers were exposed to a range of concentrations of C5a for 10 min and then washed, neutrophil adhesion to CSa-stimulated HUVECs was increased by as much as 5.8-fold, in comparison to unstimulated cells (18.6±1.0 vs. 3.2±0.3%). Studies were performed to determine the time course for P- selectin expression as well as neutrophil adhesion to HUVECs. Endothelial cells were incubated with 250 nM C5a for 2, 5, 10, 15, or 30 min, followed by washing. As shown in Fig. 5 A, P-selectin expression was upregulated rapidly, demonstrating 0.47- 0.46- 0.45- 0.44- 0.43-0.42 - 0.41 - p<0.02 . 0.001 0.01 0.1 110 100 Thrombin (units/ml) Figure 3. Dose-responses for thrombin stimulation of HUVEC mono- layers. Cells were incubated with increasing concentration of thrombin for 5 min followed by washing. P-selectin expression was measured by an ELISA method. Comparisons were to non-thrombin-exposed HU- VECs. 20-15 - 0) 0 10 - 75 ~0 5- A. P-selectin Expression p<0.05 .p C5a 10 B. Neutrophil Adherence p<0.02 10 T0 C5a des arg .... I 100 1000 T I °C5a des arg 100 1000 C5a (nM) Figure 4. Dose-responses for C5a stimulation of HUVEC monolayers. Cells were incubated with increasing concentrations of C5a for 10 min followed by washing. (A) Endothelial cell P-selectin expression was measured by an ELISA method. (B) Neutrophils were added and adhe- sion evaluated 30 min later. Neutrophil adhesion to unstimulated HU- VECs was 3.2±0.3%. For both P-selectin and neutrophil adhesion mea- surements, CSa des arg showed no activity. All comparisons were to non-C5a-treated HUVECs. increased expression (P < 0.001) over basal levels after 5-, 10-, and 15-min exposure to CSa, with a return to baseline by 30 min. These data demonstrate that CSa, like histamine, can rapidly upregulate P-selectin expression on endothelial cells. A similar time course was observed for neutrophil adhesion to HUVECs incubated with 250 nM CSa for various periods of time. Rapid and significant increases (P < 0.01) in neutrophil adhesion were demonstrated with HUVECs incubated with CSa for 5, 10, and 15 min (Fig. S B). HUVECs exposed for 30 min to CSa also bound significantly more neutrophils than unstimulated HUVECs, although at this time point, neutrophil adhesion was returning toward baseline levels. Stabilization of P-selectin expression on HUVECs by re- moval of CSa. After stimulation of endothelial cell monolayers, the neutrophil adhesion assay is associated with a 30-min incu- bation during which the neutrophils settle onto and interact with the endothelial cells. Experiments were designed to determine if continuous presence of CSa would affect P-selectin expres- 1150 Foreman, Vaporciyan, Bonish, Jones, Johnson, Glovsky, Eddy, and Ward 0 c 0) co cx U) a) 0 0. 0 Ec 0 0) Z.ra C 5) e 0 0 0.600- 0.40. C 0.4 C C. 0.35--0 p<0 001 0.30 , 010 20 30 15-B. Neutrophil Adhesion T 0 0 Ne i 5- vp<0.0l 0o 010 20 Duration (min) of Exposure to C5a 30 Figure 5. Time course for C5a-mediated P-selectin expression and neu- trophil adhesion to HUVEC monolayers. Endothelial cells were incu- bated with 250 nM C5a for 0-30 min followed by washing. (A) P- selectin expression on HUVECs was measured with an ELISA method. (B) Neutrophils were added and adhesion measured 30 min later. Error bars in Afor the first two points were 0.20 and 0.10, respectively. All comparisons were to the time 0 values. sion on endothelial cells. It should be noted that in the adhesion assay endothelial cells were incubated with CSa for a given period of time, following which CSa was removed by washing the monolayers before addition of neutrophils. Accordingly, for these experiments endothelial cell monolayers were treated with 250 nM CSa and then washed after 10 or 30 min. Monolayers were fixed, as indicated, by addition of paraformaldehyde (as described above) and P-selectin expression was evaluated by the ELISA assay. The results are shown in Fig. 6, where it is evident that increased P-selectin expression was similar, pro- vided the monolayers were washed 10 min after addition of CSa and regardless of time of fixation. In sharp contrast, the continuing presence of CSa for 30 min resulted in a significant (P < 0.03) loss in increased expression of P-selectin. Reasons for the differences in stability of P-selectin expression are not known. These data suggest that continued presence of C5a may result in loss of P-selectin. In view of these data, in all subse- quent experiments, unless otherwise indicated, endothelial cell activation with 250 nM C5a occurred for a 10-min period fol- lowed by washing. CSa-induced vWf secretion from HUVECs. Confluent HU- VECs were incubated for 15 min with a range of concentrations of C5a, histamine, thrombin, or media (as a negative control), and the supernatant fluids were then evaluated by ELISA for secreted vWf. Incubation of HUVECs with C5a (10, 100, and 250 nM) induced vWf secretion that increased significantly compared with that released from unstimulated HUVECs (Fig. 7). Stimulation of HUVECs with histamine (1 and 5 mM) or thrombin (10 U/ml) also resulted in increased vWf release, a finding that has been described previously ( 11). P-selectin dependency of neutrophil adherence to CSa-stim- ulated HUVECs. HUVEC monolayers were incubated with 250 nM C5a for 10 min, followed by washing and addition of neutro- phils. Antibodies were added to the HUVEC monolayers, at a final concentration of 20 /ug/ml, followed by addition of neutro- phils (which had been pretreated with blocking antibodies to FcyRH and FcyRLII). Under these conditions, anti-P-selectin antibody reduced neutrophil adhesion by 59% (P < 0.001), while antibodies to E-selectin, ICAM-1, and CD18, as well as a combination of these antibodies (at 20 ig/ml each) did not affect neutrophil binding to the HUVECs (Fig. 8 A). To demon- strate the blocking capabilities of the antibodies to E-selectin, ICAM-1, and CD18, neutrophil adhesion was evaluated in HU- VECs stimulated with 25 ng/ml TNF-a for 2 h (in the presence of 10 1ug/ml of these antibodies). Anti-E-selectin, anti-ICAM- 1, and anti-CD18 antibodies all decreased significantly neutro- phil binding (P < 0.001) to TNF-a-stimulated HUVECs, with a combination of these antibodies reducing neutrophil adhesion by 32% (P < 0.001) (Fig. 8 B). These results are similar to those reported previously (25). In contrast, anti-P-selectin did not affect neutrophil adhesion to TNF-a-stimulated HUVECs (Fig. 8 B). Lack of expression of E-selectin in CSa-stimulated HU- VECs. HUVECs were stimulated with 25 ng/ml TNF-a, 250 nM C5a, or 5 mM histamine for 2, 5, 10, 15, 30, or 120 min, washed, and immediately evaluated for E-selectin expression. As expected, incubation of endothelial cells with TNF-a for 2 h resulted in a significant increase in E-selectin expression compared with baseline values (optical density of 1.731±0.02 vs. 0.669±0.012, P < 0.001) (data not shown). In contrast, incubation of HUVECs with CSa or histamine did not promote E-selectin upregulation at 4 h (data not shown). Thus, CSa stimulation of HUVECs results in upregulation of P-selectin, but not E-selectin. Binding of CSa to HUVECs. 2 X 106 HUVECs were incu-bated for 60 min on ice with 10 nM CSa supplemented with 0.1 nM radioiodinated C5a (- 60,000 cpm). Aliquots of the cell suspension were layered on silicone oil and centrifuged to separate bound and free radioligand. The results revealed an average of 4,940±360 molecules of CSa bound per cell. The cells bound - 3% of the total radioligand added; 40-50% of this amount of C5a failed to bind in the presence of 1,000- fold excess unlabeled CSa. Binding of the same radioiodinated CSa to neutrophils revealed - 70,000 CSa molecules per cell (data not shown). These findings were characteristic of results obtained in nine separate experiments. Because of technical problems it has not been possible to obtain data sufficient for Scatchard analysis and computation of binding affinities. These data suggest that CSa binds to endothelial cells, but the data are insufficient to prove receptor-mediated binding. Detection in endothelial cells of mRNA for CSa receptor. cDNA was prepared from HTVEC mRNA using reverse tran- C5a Receptor-induced Expression of P-selectin in Endothelial Cells 1151 p=n.s. p<0.05 I p=n.s. p<0.03 I~~~~ L1T No C5a for 10 min, C5a for 10 min, Stimulation wash, then fix wash, media for 30 min, then fix Incubation Conditions scriptase. The cDNA was then amplified using PCR with pri-mers directed against either the full open reading frame (full ORF, 1,053 bases) or a fragment of the open reading frame (partial ORF, 636 bases) for the neutrophil C5aR. A positive control of cDNA for the neutrophil C5aR was used for PCR, and a single band at - 1,000 bases was visualized using the full ORF primers, while a band at - 600 bases was seen with the partial ORF primers (Fig. 9, A and B, lane 2). PCR products from unstimulated HUVECs as well as TNF-a- or IFN--y- stimulated cells each showed a band similar to neutrophil CSa receptor control at - 1,000 bases using the full ORF primers and - 600 bases using the partial ORF primers (Fig. 9, A and B, lanes 3-5). Thus, TNF-a and IL-1 had no obvious effects on HUVEC content of mRNA for C5aR. Identical results were found using either passage one or passage three HUVECs (data not shown). p<0.02 K K K K El No C5a El C5a, 25OnM I\\ C5a for 30 min, wash, then fix. Figure 6. Stabilization of P-selectin expression on HUVECs. Endothelial cells were incubated with 250 nM C5a for 10 or 30 min, followed by washing (and fixation with 1% paraformalde- hyde, as indicated) and an additional 30 min incubation in media. P-selectin expression was assayed with the ELISA method. Ccussion ,ently published data have indicated that CSa will cause lothelial cells (human umbilical vein or rat pulmonary artery lothelial cells) to activate protein kinase C and generate D- o-inositol 1,4,5-triphosphate, to produce O2 and to convert kthine dehydrogenase to xanthine oxidase (32, 33). The data the current report indicate that C5a also will cause rapid ression of P-selectin and adherence of neutrophils. It is also dent from this report that C5a causes another feature of lothelial cell activation: secretion of vWf (occurring within min). Based on the use of blocking antibodies, the adhesion- rmoting process induced by endothelial cell activation with a has been shown to be P-selectin dependent. The failure of ibodies to ICAM-1 and CD18 to block this adhesion process it to block TNF-a-induced adhesiveness for neutrophils) sug- Figure 7. Release of vWf from HUVEC monolayers. Cells were stimulated for 15 min with various inflammatory me- diators and released vWf factor in the supernatant was mea- sured with an ELISA method. Statistically significant dif-ference (P < 0.05) with respect to the media control. Where not displayed, error bars were < 0.003. 1152 Foreman, Vaporciyan, Bonish, Jones, Johnson, Glovsky, Eddy, and Ward 0.32 - 0.31- C 0 0) C,i C] 0. 0 0.3 - 0.29- 0.28- 0.27- 1.00'¶ 0.75- 0.50- 0.25. E 00) co C 0 0 S%F'o I I ~ . I~ -WI .J6 IzI O No Cta O Cta, 250nM for 10 min. A. C5a-Sbmulated HUVECs T r S KXX Negative Poaitive ani- ant- anti- anti- MAb Control Control P-sealctin E-sealctin ICAM1 CD18 Combination 80 B. TNF-,iumukdtd HUVECs -Po0.00m 60-50- Negve Posltive ant-anti-anti- Control Contrl P-seadctin E-sealctin ICAM-1 (34-36). The products in these preparations would likely have been C5a des arg, which, in our hands, has diminished activity for endothelial cells but is chemotactically active for neutrophils via the C5a des arg-vitamin D-binding protein complex (37, 38). It is known that C5a des arg has 1/10 to 1/20 the chemo- tactic activity for human blood neutrophils compared with C5a (39). It is also known that monocytes respond chemotactically to C5a and C5a des arg in a quantitatively similar manner (40). On balance, it would appear that the loss of activity of C5a des arg for upregulation of endothelial P-selectin and for neutrophil adhesion of treated endothelial cell monolayers is consistent with the interpretation that C5a des arg is a poor agonist for endothelial cells when compared with C5a. It appears that as- sembly of C5b-C9 on the surface of the endothelial cell can result in P-selectin upregulation, although this also ultimately causes some cytotoxicity as measured by release of lactic dehy- drogenase (13). The ability of complement activation products to cause rapid upregulation of endothelial P-selectin may ex- plain the ability of neutrophils to adhere rapidly to endothelium anG MAb CD1I Combnation Figure 8. Monoclonal antibody blockade of neutrophil adherence to C5a-stimulated HUVEC monolayers (A) or TNF-a stimulated mono- layers (B). Endothelial cells were stimulated with 250 nM C5a for 10 min followed by washing and fixation. Neutrophils suspended in media containing 20 tg/ml of blocking antibodies to P- or E-selectin, ICAM- 1, CD18, or a combination of antibodies (to E-selectin, ICAM-1, and CD18) was added to the monolayers and adhesion measured 30 min later. Error bars in A for the first and third bars were 0.08 and 0.02, respectively. Other endothelial monolayers were also stimulated with TNF-a (25 ng/ml) for 2 h followed by washing. Neutrophils, suspended in media containing 20 ,g/ml of blocking antibodies to P- or E-selectin, ICAM-1, CD18, or a combination of antibodies (to E-selectin, ICAM- 1, and CD18) was added to the monolayers and adhesion measured 30 min later. Adherence of neutrophils to unstimulated HUVECs (negative control) in this experiment was 4.1% (B). P values were determined by comparison to the positive control groups. gests that the adhesion-promoting process can not be attributed to nonspecific binding of CSa to endothelial cells, followed by activation of neutrophils, with resultant upregulation of CD Ilb/CDl8 and cell adherence via the ICAM-l -dependent pathway. Binding studies with `2I-C5a reveal specific binding, al- though the small number of binding sites (10% of the number found on human neutrophils) and problems with insufficient levels of specific radioactivity in the CSa preparations makes it difficult to obtain data sufficient for Scatchard analysis and calculation of binding affinities. Nevertheless, specific binding of CSa to HUVECs has been consistently found in our studies. The ability to demonstrate by reverse transcriptase-PCR tech- nology the presence of mRNA for the C5aR, using probes to the C5aR of human neutrophils, strengthens the case for the presence in HUVECs of C5aR. To date, we have determined the C5aR is not upregulated in HUVECs stimulated with TNF- a, IL-1/, or IFN-y (data not shown). In earlier studies, two complement-related products have been described as being able to facilitate adhesion of neutrophils to endothelial cells. "C5a fragments" (zymosan-activated se-rum or a fraction thereof) were described as having the ability to modify neutrophils to enhance adhesion with endothelial cells A -- 800 -~ 400 B 800 400 Figure 9. Detection of mRNA for C5a receptor in HUVECs. cDNA was prepared from total cellular RNA isolated from HUVECs. The cDNA was amplified using PCR with partial ORF primers (A) or full ORF primers (B), and the PCR product was evaluated on a 0.9% agarose gel. A and B: Lanes 1 and 6, base pair ladder markers (100 and multiples thereof); lane 2, cDNA for C5aR obtained from HL-60 cells; lane 3, unstimulated HUVECs; lane 4, TNF-a-stimulated HUVECs; lane 5, IFN-y stimulated HUVECs. C5a Receptor-induced Expression of P-selectin in Endothelial Cells 1153 15 - ;- 10 cs c LD D 8after injury, as opposed to the much slower onset of adhesion requiring upregulation of ICAM-1 and E-selectin. In prelimi- nary experiments we have demonstrated that activation of whole human serum by cobra venom factor yields an activity that will cause upregulation of P-selectin on HUVEC (Foreman, K. E., and P. A. Ward, unpublished data). Whether this might be due to C5a des arg with its cofactor (vitamin D-binding protein [see above]), the membrane attack complex, or some other activation product remains to be determined. Another pathway of rapid adhesion has been shown in vitro when iC3b is deposited on endothelial cells after the addition of fresh human serum to HUVECs that have been previously exposed to cobra venom factor conjugated to Ulex europaeus L In this case, neutrophil adhesion to HUVECs peaks 10-15 min after addition of the complement source, and the adhesion is blocked by the presence of antibodies to CD lb or CD18 (41). The adhesion process has been attributed to CR3 (Mac- 1 or CDl lb/CDl 8) on the neutrophil. With respect to neutrophils, C5a binds to C5aR, inducing polarization, pseudopod formation, upregulation of CDlla/ CD18, granule enzyme release, and generation of H202 and superoxide anion. Each of these effects may enhance binding, transmigration, and effector function of neutrophils recruited into an inflammatory site. In addition, C5a can bind to peripheral blood basophils, causing release of histamine from cytoplasmic granules (42). In turn, this could also lead to upregulation of endothelial P-selectin. It can be surmised from these observa- tions that C5a has potent agonist activities for both endothelial cells as well as for phagocytic cells and that the combination of these effector cell responses results in substantial proin- flammatory outcomes. Obviously, determining which of these pathways may be involved in vivo is difficult. Nevertheless, the data suggest that C5a may play several roles in the acute inflammatory response. It may, as described in this report, cause rapid upregulation of P-selectin, facilitating the earliest steps in the adhesion process; C5a can directly activate neutrophils to cause upregulation of CDlla/CD18, facilitating adhesion of neutrophils to endothelial ICAM-1; finally C5a may be a key chemoattractant for the transmigration of neutrophils in comple- ment-dependent inflammation. The other dimension in neutrophil adhesion to the activated endothelium (resulting in the ultimate transmigration of neutro- phils) relates to evidence that, in the adhesion-promoting pro-cess, neutrophil activation (as assessed by increased intracellu- lar Ca2' and production of oxygen metabolites) occurs because of endothelial expression of platelet-activating factor and/or IL-8 (43, 44). This activation process may prime neutrophils for more effective transmigrational response to chemotactic factors. It remains to be demonstrated that C5a interaction with HUVEC is through a receptor-mediated mechanism. The evi- dence for specific binding and the detection in HUVEC of mRNA for C5aR support this possibility but do not prove it. The earlier lack of evidence for binding of CSa to HUVEC involved an entirely different binding assay for CSa (45). Our own experience has suggested that the technical approaches may affect CSa binding to HUVEC. For instance, neither sus- pensions of HUVEC nor monolayers have ever, in our hands, yielded evidence for specific binding of C5a (Foreman, K. E., and P. A. Ward, unpublished observations). It may be that technical problems have not been overcome entirely. In any case, definitive demonstration of C5aR-dependent binding of CSa to HUVEC remains to be demonstrated. Acknowledgments We thank Dr. David Ginsburg (University of Michigan Medical School) for performing the vWf ELISA. We thank Dr. Henry Showell (Pfizer Central Research, Groton, CT) for supplying recombinant human C5a, Drs. Craig and Norma Gerard (Children's Hospital, Boston, MA) for supplying cDNA for the human neutrophil C5aR, Dr. James C. Paulson (Cytel Corporation) for providing antibodies to human P-selectin, Dr. C. Wayne Smith (Baylor University) for providing antibody to human CD18, and Scot Westerbeek (Matteson-Ridolfi Incorporated) for the gift of General Electric silicone oil SF1250. K. E. 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66	https://en.wikipedia.org/wiki/Transcendental_number	Jump to content Search Contents (Top) 1 History 2 Properties 3 Numbers proven to be transcendental 4 Conjectured transcendental numbers 5 Proofs for specific numbers 5.1 A proof that e is transcendental 5.1.1 Lemma 1 5.1.2 Lemma 2 5.1.3 Conclusion 5.2 The transcendence of π 6 See also 7 Notes 8 References 9 Sources 10 External links Transcendental number العربية Azərbaycanca বাংলা Беларуская Български Català Чӑвашла Čeština Cymraeg Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Hrvatski Ido Bahasa Indonesia Íslenska Italiano עברית ქართული Қазақша Latina Latviešu Lietuvių Lombard Magyar Malagasy മലയാളം Bahasa Melayu Mirandés Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Português Romnă Русский Scots Sicilianu Simple English Slovenčina Slovenščina Српски / srpski Suomi Svenska தமிழ் ไทย Тоҷикӣ Türkçe Українська اردو Tiếng Việt West-Vlams 吴语粵語中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia In mathematics, a non-algebraic number "U-number" redirects here. For thermal conductivity, see R-value (insulation) § U-value. In mathematics, a transcendental number is a real or complex number that is not algebraic: that is, not the root of a non-zero polynomial with integer (or, equivalently, rational) coefficients. The best-known transcendental numbers are π and e. The quality of a number being transcendental is called transcendence. Though only a few classes of transcendental numbers are known, partly because it can be extremely difficult to show that a given number is transcendental, transcendental numbers are not rare: indeed, almost all real and complex numbers are transcendental, since the algebraic numbers form a countable set, while the set of real numbers ⁠⁠ and the set of complex numbers ⁠⁠ are both uncountable sets, and therefore larger than any countable set. All transcendental real numbers (also known as real transcendental numbers or transcendental irrational numbers) are irrational numbers, since all rational numbers are algebraic. The converse is not true: Not all irrational numbers are transcendental. Hence, the set of real numbers consists of non-overlapping sets of rational, algebraic irrational, and transcendental real numbers. For example, the square root of 2 is an irrational number, but it is not a transcendental number as it is a root of the polynomial equation x2 − 2 = 0. The golden ratio (denoted or ) is another irrational number that is not transcendental, as it is a root of the polynomial equation x2 − x − 1 = 0. History [edit] The name "transcendental" comes from Latin trānscendere 'to climb over or beyond, surmount', and was first used for the mathematical concept in Leibniz's 1682 paper in which he proved that sin x is not an algebraic function of x. Euler, in the eighteenth century, was probably the first person to define transcendental numbers in the modern sense. Johann Heinrich Lambert conjectured that e and π were both transcendental numbers in his 1768 paper proving the number π is irrational, and proposed a tentative sketch proof that π is transcendental. Joseph Liouville first proved the existence of transcendental numbers in 1844, and in 1851 gave the first decimal examples such as the Liouville constant in which the nth digit after the decimal point is 1 if n = k! (k factorial) for some k and 0 otherwise. In other words, the nth digit of this number is 1 only if n is one of 1! = 1, 2! = 2, 3! = 6, 4! = 24, etc. Liouville showed that this number belongs to a class of transcendental numbers that can be more closely approximated by rational numbers than can any irrational algebraic number, and this class of numbers is called the Liouville numbers. Liouville showed that all Liouville numbers are transcendental. The first number to be proven transcendental without having been specifically constructed for the purpose of proving transcendental numbers' existence was e, by Charles Hermite in 1873. In 1874 Georg Cantor proved that the algebraic numbers are countable and the real numbers are uncountable. He also gave a new method for constructing transcendental numbers. Although this was already implied by his proof of the countability of the algebraic numbers, Cantor also published a construction that proves there are as many transcendental numbers as there are real numbers.[a] Cantor's work established the ubiquity of transcendental numbers. In 1882 Ferdinand von Lindemann published the first complete proof that π is transcendental. He first proved that ea is transcendental if a is a non-zero algebraic number. Then, since eiπ = −1 is algebraic (see Euler's identity), iπ must be transcendental. But since i is algebraic, π must therefore be transcendental. This approach was generalized by Karl Weierstrass to what is now known as the Lindemann–Weierstrass theorem. The transcendence of π implies that geometric constructions involving compass and straightedge only cannot produce certain results, for example squaring the circle. In 1900 David Hilbert posed a question about transcendental numbers, Hilbert's seventh problem: If a is an algebraic number that is not 0 or 1, and b is an irrational algebraic number, is ab necessarily transcendental? The affirmative answer was provided in 1934 by the Gelfond–Schneider theorem. This work was extended by Alan Baker in the 1960s in his work on lower bounds for linear forms in any number of logarithms (of algebraic numbers). Properties [edit] A transcendental number is a (possibly complex) number that is not the root of any integer polynomial. Every real transcendental number must also be irrational, since every rational number is the root of some integer polynomial of degree one. The set of transcendental numbers is uncountably infinite. Since the polynomials with rational coefficients are countable, and since each such polynomial has a finite number of zeroes, the algebraic numbers must also be countable. However, Cantor's diagonal argument proves that the real numbers (and therefore also the complex numbers) are uncountable. Since the real numbers are the union of algebraic and transcendental numbers, it is impossible for both subsets to be countable. This makes the transcendental numbers uncountable. No rational number is transcendental and all real transcendental numbers are irrational. The irrational numbers contain all the real transcendental numbers and a subset of the algebraic numbers, including the quadratic irrationals and other forms of algebraic irrationals. Applying any non-constant single-variable algebraic function to a transcendental argument yields a transcendental value. For example, from knowing that π is transcendental, it can be immediately deduced that numbers such as , , , and are transcendental as well. However, an algebraic function of several variables may yield an algebraic number when applied to transcendental numbers if these numbers are not algebraically independent. For example, π and (1 − π) are both transcendental, but π + (1 − π) = 1 is obviously not. It is unknown whether e + π, for example, is transcendental, though at least one of e + π and eπ must be transcendental. More generally, for any two transcendental numbers a and b, at least one of a + b and ab must be transcendental. To see this, consider the polynomial (x − a)(x − b) = x2 − (a + b) x + a b . If (a + b) and a b were both algebraic, then this would be a polynomial with algebraic coefficients. Because algebraic numbers form an algebraically closed field, this would imply that the roots of the polynomial, a and b, must be algebraic. But this is a contradiction, and thus it must be the case that at least one of the coefficients is transcendental. The non-computable numbers are a strict subset of the transcendental numbers. All Liouville numbers are transcendental, but not vice versa. Any Liouville number must have unbounded partial quotients in its simple continued fraction expansion. Using a counting argument one can show that there exist transcendental numbers which have bounded partial quotients and hence are not Liouville numbers. Using the explicit continued fraction expansion of e, one can show that e is not a Liouville number (although the partial quotients in its continued fraction expansion are unbounded). Kurt Mahler showed in 1953 that π is also not a Liouville number. It is conjectured that all infinite continued fractions with bounded terms, that have a "simple" structure, and that are not eventually periodic are transcendental (in other words, algebraic irrational roots of at least third degree polynomials do not have apparent pattern in their continued fraction expansions, since eventually periodic continued fractions correspond to quadratic irrationals, see Hermite's problem). Numbers proven to be transcendental [edit] Numbers proven to be transcendental: π (by the Lindemann–Weierstrass theorem). if a is algebraic and nonzero (by the Lindemann–Weierstrass theorem), in particular Euler's number e. where n is a positive integer; in particular Gelfond's constant (by the Gelfond–Schneider theorem). Algebraic combinations of π and such as and (following from their algebraic independence). where a is algebraic but not 0 or 1, and b is irrational algebraic, in particular the Gelfond–Schneider constant (by the Gelfond–Schneider theorem). The natural logarithm ln(a) if a is algebraic and not equal to 0 or 1, for any branch of the logarithm function (by the Lindemann–Weierstrass theorem). if a and b are positive integers not both powers of the same integer, and a is not equal to 1 (by the Gelfond–Schneider theorem). All numbers of the form are transcendental, where are algebraic for all and are non-zero algebraic for all (by Baker's theorem). The trigonometric functions sin(x), cos(x) and their hyperbolic counterparts, for any nonzero algebraic number x, expressed in radians (by the Lindemann–Weierstrass theorem). Non-zero results of the inverse trigonometric functions arcsin(x), arccos(x) and their hyperbolic counterparts, for any algebraic number x (by the Lindemann–Weierstrass theorem). , for rational x such that . The fixed point of the cosine function (also referred to as the Dottie number d) – the unique real solution to the equation , where x is in radians (by the Lindemann–Weierstrass theorem). if a is algebraic and nonzero, for any branch of the Lambert W function (by the Lindemann–Weierstrass theorem), in particular the omega constant Ω. if both a and the order r are algebraic such that , for any branch of the generalized Lambert W function. , the square super-root of any natural number is either an integer or transcendental (by the Gelfond–Schneider theorem). Values of the gamma function of rational numbers that are of the form or . Algebraic combinations of π and or of π and such as the lemniscate constant (following from their respective algebraic independences). The values of Beta function if and are non-integer rational numbers. The Bessel function of the first kind , its first derivative, and the quotient are transcendental when ν is rational and x is algebraic and nonzero, and all nonzero roots of and are transcendental when ν is rational. The number , where and are Bessel functions and γ is the Euler–Mascheroni constant. Values of the Fibonacci zeta function at positive even arguments. Any Liouville number, in particular: Liouville's constant . Numbers with irrationality measure larger than 2, such as the Champernowne constant and Cahen's constant (by Roth's theorem). Numbers artificially constructed not to be algebraic periods. Any non-computable number, in particular: Chaitin's constant. Constructed irrational numbers which are not simply normal in any base. Any number for which the digits with respect to some fixed base form a Sturmian word. The Prouhet–Thue–Morse constant and the related rabbit constant. The Komornik–Loreti constant. The paperfolding constant (also named as "Gaussian Liouville number"). The values of the infinite series with fast convergence rate as defined by Y. Gao and J. Gao, such as . Any number of the form (where , are polynomials in variables and , is algebraic and , is any integer greater than 1). Numbers of the form and For b > 1 where is the floor function. The numbers and with only two different decimal digits whose nonzero digit positions are given by the Moser–de Bruijn sequence and its double. The values of the Rogers-Ramanujan continued fraction where is algebraic and . The lemniscatic values of theta function (under the same conditions for ) are also transcendental. j(q) where is algebraic but not imaginary quadratic (i.e, the exceptional set of this function is the number field whose degree of extension over is 2). The constants and in the formula for first index of occurrence of Gijswijt's sequence, where k is any integer greater than 1. Conjectured transcendental numbers [edit] Numbers which have yet to be proven to be either transcendental or algebraic: Most nontrivial combinations of two or more transcendental numbers are themselves not known to be transcendental or even irrational: eπ, e + π, ππ, ee, πe, π√2, eπ2. It has been shown that both e + π and π/e do not satisfy any polynomial equation of degree ≤ 8 and integer coefficients of average size 109. At least one of the numbers ee and ee2 is transcendental. Since the field of algebraic numbers is algebraically closed and e and π are roots of the polynomial x2 - (e + π)x + eπ, at least one of the numbers eπ and e + π is transcendental. Schanuel's conjecture would imply that all of the above numbers are transcendental and algebraically independent. The Euler–Mascheroni constant γ: In 2010 it has been shown that an infinite list of Euler-Lehmer constants (which includes γ/4) contains at most one algebraic number. In 2012 it was shown that at least one of γ and the Gompertz constant δ is transcendental. The values of the Riemann zeta function ζ(n) at odd positive integers ; in particular Apéry's constant ζ(3), which is known to be irrational. For the other numbers ζ(5), ζ(7), ζ(9), ... even this is not known. The values of the Dirichlet beta function β(n) at even positive integers ; in particular Catalan's Constant β(2). (None of them are known to be irrational.) Values of the Gamma Function Γ(1/n) for positive integers and are not known to be irrational, let alone transcendental. For at least one the numbers Γ(1/n) and Γ(2/n) is transcendental. Any number given by some kind of limit that is not obviously algebraic. Proofs for specific numbers [edit] A proof that e is transcendental [edit] The first proof that the base of the natural logarithms, e, is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following: Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients c0, c1, ..., cn satisfying the equation: It is difficult to make use of the integer status of these coefficients when multiplied by a power of the irrational e, but we can absorb those powers into an integral which “mostly” will assume integer values. For a positive integer k, define the polynomial and multiply both sides of the above equation by to arrive at the equation: By splitting respective domains of integration, this equation can be written in the form where Here P will turn out to be an integer, but more importantly it grows quickly with k. Lemma 1 [edit] There are arbitrarily large k such that is a non-zero integer. Proof. Recall the standard integral (case of the Gamma function) valid for any natural number . More generally, : if then . This would allow us to compute exactly, because any term of can be rewritten as through a change of variables. Hence That latter sum is a polynomial in with integer coefficients, i.e., it is a linear combination of powers with integer coefficients. Hence the number is a linear combination (with those same integer coefficients) of factorials ; in particular is an integer. Smaller factorials divide larger factorials, so the smallest occurring in that linear combination will also divide the whole of . We get that from the lowest power term appearing with a nonzero coefficient in , but this smallest exponent is also the multiplicity of as a root of this polynomial. is chosen to have multiplicity of the root and multiplicity of the roots for , so that smallest exponent is for and for with . Therefore divides . To establish the last claim in the lemma, that is nonzero, it is sufficient to prove that does not divide . To that end, let be any prime larger than and . We know from the above that divides each of for , so in particular all of those are divisible by . It comes down to the first term . We have (see falling and rising factorials) and those higher degree terms all give rise to factorials or larger. Hence That right hand side is a product of nonzero integer factors less than the prime , therefore that product is not divisible by , and the same holds for ; in particular cannot be zero. Lemma 2 [edit] For sufficiently large k, . Proof. Note that where u(x), v(x) are continuous functions of x for all x, so are bounded on the interval [0, n]. That is, there are constants G, H > 0 such that So each of those integrals composing Q is bounded, the worst case being It is now possible to bound the sum Q as well: where M is a constant not depending on k. It follows that finishing the proof of this lemma. Conclusion [edit] Choosing a value of k that satisfies both lemmas leads to a non-zero integer added to a vanishingly small quantity being equal to zero: an impossibility. It follows that the original assumption, that e can satisfy a polynomial equation with integer coefficients, is also impossible; that is, e is transcendental. The transcendence of π [edit] A similar strategy, different from Lindemann's original approach, can be used to show that the number π is transcendental. Besides the gamma-function and some estimates as in the proof for e, facts about symmetric polynomials play a vital role in the proof. For detailed information concerning the proofs of the transcendence of π and e, see the references and external links. See also [edit] Mathematics portal Transcendental number theory, the study of questions related to transcendental numbers Transcendental element, generalization of transcendental numbers in abstract algebra Gelfond–Schneider theorem Diophantine approximation Periods, a countable set of numbers (including all algebraic and some transcendental numbers) which may be defined by integral equations. Hartmanis–Stearns conjecture, a conjectural transcendence criterion based on complexity of computing the number's expansion \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Number systems \| Complex \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Real \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| \| Rational \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Integer \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| --- --- --- \| \| Natural \| \| \| \| Zero: 0 \| \| One: 1 \| \| Prime numbers \| \| Composite numbers \| \| \| \| Negative integers \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| Fraction \| \| \| \| Finite decimal \| \| Dyadic (finite binary) \| \| Repeating decimal \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- \| Irrational \| \| \| \| Algebraic irrational \| \| Irrational period \| \| Transcendental \| \| \| \| \| \| Imaginary \| \| \| \| Notes [edit] ^ Cantor's construction builds a one-to-one correspondence between the set of transcendental numbers and the set of real numbers. In this article, Cantor only applies his construction to the set of irrational numbers. References [edit] ^ Pickover, Cliff. "The 15 most famous transcendental numbers". sprott.physics.wisc.edu. Retrieved 2020-01-23. ^ Shidlovskii, Andrei B. (June 2011). Transcendental Numbers. Walter de Gruyter. p. 1. ISBN 9783110889055. ^ a b Bunday, B. D.; Mulholland, H. (20 May 2014). Pure Mathematics for Advanced Level. Butterworth-Heinemann. ISBN 978-1-4831-0613-7. Retrieved 21 March 2021. ^ Baker, A. (1964). "On Mahler's classification of transcendental numbers". Acta Mathematica. 111: 97–120. doi:10.1007/bf02391010. S2CID 122023355. ^ Heuer, Nicolaus; Loeh, Clara (1 November 2019). "Transcendental simplicial volumes". arXiv:1911.06386 [math.GT]. ^ "Real number". Encyclopædia Britannica. mathematics. Retrieved 2020-08-11. ^ "transcendental". 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Lectures on Transcendental Numbers. Lecture Notes in Mathematics. Vol. 546. Springer. ISBN 978-3-540-07986-6. Zbl 0332.10019. Natarajan, Saradha [in French]; Thangadurai, Ravindranathan (2020). Pillars of Transcendental Number Theory. Springer Verlag. ISBN 978-981-15-4154-4. Pytheas Fogg, N. (2002). Berthé, V.; Ferenczi, Sébastien; Mauduit, Christian; Siegel, A. (eds.). Substitutions in dynamics, arithmetics and combinatorics. Lecture Notes in Mathematics. Vol. 1794. Springer. ISBN 978-3-540-44141-0. Zbl 1014.11015. Shallit, J. (15–26 July 1996). "Number theory and formal languages". In Hejhal, D.A.; Friedman, Joel; Gutzwiller, M.C.; Odlyzko, A.M. (eds.). Emerging Applications of Number Theory. IMA Summer Program. The IMA Volumes in Mathematics and its Applications. Vol. 109. Minneapolis, MN: Springer (published 1999). pp. 547–570. ISBN 978-0-387-98824-5. External links [edit] English Wikisource has original text related to this article: Über die Transzendenz der Zahlen e und π. (in German) Weisstein, Eric W. "Transcendental Number". MathWorld. Weisstein, Eric W. "Liouville Number". MathWorld. Weisstein, Eric W. "Liouville's Constant". MathWorld. "Proof that e is transcendental". planetmath.org. "Proof that the Liouville constant is transcendental". deanlmoore.com. Retrieved 2018-11-12. Fritsch, R. (29 March 1988). Transzendenz von e im Leistungskurs? [Transcendence of e in advanced courses?] (PDF). Rahmen der 79. Hauptversammlung des Deutschen Vereins zur Förderung des mathematischen und naturwissenschaftlichen Unterrichts [79th Annual, General Meeting of the German Association for the Promotion of Mathematics and Science Education]. Der mathematische und naturwissenschaftliche Unterricht (in German). Vol. 42. Kiel, DE (published 1989). pp. 75–80 (presentation), 375–376 (responses). Archived from the original (PDF) on 2011-07-16 – via University of Munich (mathematik.uni-muenchen.de ). — Proof that e is transcendental, in German. Fritsch, R. (2003). "Hilberts Beweis der Transzendenz der Ludolphschen Zahl π" (PDF). Дифференциальная геометрия многообразий фигур (in German). 34: 144–148. Archived from the original (PDF) on 2011-07-16 – via University of Munich (mathematik.uni-muenchen.de/~fritsch). \| v t e Irrational numbers \| \| Chaitin's (Ω) Liouville Prime (ρ) Omega Cahen Logarithm of 2 Dottie Lemniscate (ϖ) Twelfth root of 2 Apéry's (ζ(3)) Cube root of 2 Plastic ratio (ρ) Square root of 2 Supergolden ratio (ψ) Erdős–Borwein (E) Golden ratio (φ) Square root of 3 Supersilver ratio (ς) Square root of 5 Silver ratio (σ) Square root of 6 Square root of 7 Euler's (e) Pi (π) Square root of 10 \| \| Schizophrenic Transcendental Trigonometric \| \| v t e Number systems \| \| Sets of definable numbers \| Natural numbers () Integers () Rational numbers () Constructible numbers Algebraic numbers () Closed-form numbers Periods () Computable numbers Arithmetical numbers Set-theoretically definable numbers Gaussian integers + Gaussian rationals Eisenstein integers \| \| Composition algebras \| Division algebras: Real numbers () Complex numbers () Quaternions () Octonions () \| \| Splittypes \| Over : Split-complex numbers Split-quaternions Split-octonions Over : Bicomplex numbers Biquaternions Bioctonions \| \| Other hypercomplex \| Dual numbers Dual quaternions Dual-complex numbers Hyperbolic quaternions Sedenions () Trigintaduonions () Split-biquaternions Multicomplex numbers Geometric algebra/Clifford algebra + Algebra of physical space + Spacetime algebra + Plane-based geometric algebra \| \| Infinities and infinitesimals \| Cardinal numbers Extended natural numbers Extended real numbers + Projective Extended complex numbers Hyperreal numbers Levi-Civita field Ordinal numbers Supernatural numbers Surreal numbers Superreal numbers \| \| Other types \| Irrational numbers Fuzzy numbers Transcendental numbers p-adic numbers (p-adic solenoids) Profinite integers Normal numbers \| \| Classification List \| \| v t e Number theory \| \| Fields \| Algebraic number theory (class field theory, non-abelian class field theory, Iwasawa theory, Iwasawa–Tate theory, Kummer theory) Analytic number theory (analytic theory of L-functions, probabilistic number theory, sieve theory) Geometric number theory Computational number theory Transcendental number theory Diophantine geometry (Arakelov theory, Hodge–Arakelov theory) Arithmetic combinatorics (additive number theory) Arithmetic geometry (anabelian geometry, p-adic Hodge theory) Arithmetic topology Arithmetic dynamics \| \| Key concepts \| Numbers 0 Natural numbers Unity Prime numbers Composite numbers Rational numbers Irrational numbers Algebraic numbers Transcendental numbers p-adic numbers (p-adic analysis) Arithmetic Modular arithmetic Chinese remainder theorem Arithmetic functions \| \| Advanced concepts \| Quadratic forms Modular forms L-functions Diophantine equations Diophantine approximation Irrationality measure Simple continued fractions \| \| Category List of topics List of recreational topics Wikibook Wikiversity \| \| Authority control databases \| \| National \| United States France BnF data Japan Czech Republic Israel \| \| Other \| Yale LUX \| Retrieved from " Category: Transcendental numbers Hidden categories: CS1 Russian-language sources (ru) CS1 German-language sources (de) Articles with short description Short description is different from Wikidata CS1 interwiki-linked names CS1: long volume value Wikisource templates with missing id Articles containing proofs Transcendental number Add topic
67	https://takeuforward.org/data-structure/find-second-smallest-and-second-largest-element-in-an-array/	Published Time: Sun Mar 24 2024 16:39:35 GMT+0000 (Coordinated Universal Time) Find Second Smallest and Second Largest Element in an array - Tutorial Close Welcome back to takeUforward Sign in with GoogleSign in with Google Or login with your email Forgot Password? Login Don’t you have an account? Create account Open in New TabClose Close Report a Bug Please provide details of the bug you encountered. Name Email Phone Number Bug Description Submit Open in New TabClose Find Second Smallest and Second Largest Element in an array 71616 Problem Statement: Given an array, find the second smallest and second largest element in the array. Print ‘-1’ in the event that either of them doesn’t exist. Examples Example 1: Input: [1,2,4,7,7,5] Output: Second Smallest : 2 Second Largest : 5 Explanation: The elements are as follows 1,2,3,5,7,7 and hence second largest of these is 5 and second smallest is 2 Example 2: Input: Output: Second Smallest : -1 Second Largest : -1 Explanation: Since there is only one element in the array, it is the largest and smallest element present in the array. There is no second largest or second smallest element present. Practice: Solve Problem Disclaimer: Don’t jump directly to the solution, try it out yourself first. Brute-Force Approach Better Approach Optimal Approach Expand any one approach by clicking the given options in the bar. Clicking one approach on bar, closes all other expands. You can manually expand more than one approach at a time Brute Force Approach Algorithm / Intuition Solution 1: (Brute Force) [this approach only works if there are no duplicates] Intuition: What do we do to find the largest or the smallest element present in an array? We ideally sort them and the first element would be the smallest of all while the last element would be the largest. Can we find the second-smallest and second-largest using a similar approach? Approach: Sort the array in ascending order The element present at the second index is the second smallest element The element present at the second index from the end is the second largest element Code C++ Java Python JavaScript Copy ```cpp include using namespace std; void getElements(int arr[],int n) { if(n==0 \|\| n==1) cout<<-1<<" "<<-1<<endl; // edge case when only one element is present in array sort(arr,arr+n); int small=arr; int large=arr[n-2]; cout<<"Second smallest is "<<small<<endl; cout<<"Second largest is "<<large<<endl; } int main() { int arr[]={1,2,4,6,7,5}; int n=sizeof(arr)/sizeof(arr); getElements(arr,n); return 0; } ``` Output: Second smallest is 2 Second largest is 6 ```java import java.io.; import java.util.Arrays; class Test { static private void getElements(int[] arr, int n) { if (n == 0 \|\| n==1) { System.out.print(-1); System.out.print(" "); System.out.print(-1); System.out.print("\n"); } Arrays.sort(arr); int small = arr; int large = arr[n - 2]; System.out.println("Second smallest is "+small); System.out.println("Second largest is "+large); } public static void main(String[] args) { int[] arr = {1, 2, 4, 6, 7, 5}; int n = arr.length; getElements(arr, n); } } ``` Output: Second smallest is 2 Second largest is 6 ```python def getElements(arr, n): if n == 0 or n == 1: print(-1, -1) # edge case when only one element is present in array arr.sort() small = arr large = arr[n-2] print("Second smallest is", small) print("Second largest is", large) if name == 'main': arr = [1, 2, 4, 6, 7, 5] n = len(arr) getElements(arr, n) ``` Output: Second smallest is 2 Second largest is 6 ```javascript function getElements(arr) { if (arr.length === 0 \|\| arr.length === 1) { console.log(-1 + " " + -1); // Edge case when only one element is present in the array return; } arr.sort((a, b) => a - b); let small = arr; let large = arr[arr.length - 2]; console.log("Second smallest is " + small); console.log("Second largest is " + large); } const arr = [1, 2, 4, 6, 7, 5]; getElements(arr); ``` Output: Second smallest is 2 Second largest is 6 Complexity Analysis Time Complexity: O(NlogN), For sorting the array Space Complexity: O(1) Better Approach Algorithm / Intuition Solution 2(Better Solution) Intuition: Even though we want to have just the second smallest and largest elements, we are still sorting the entire array for that and thus increasing the time complexity. Can we somehow try to not sort the array and still get our answer? Approach: Find the smallest and largest element in the array in a single traversal After this, we once again traverse the array and find an element that is just greater than the smallest element we just found. Similarly, we would find the largest element which is just smaller than the largest element we just found Indeed, this is our second smallest and second largest element. Code C++ Java Python JavaScript ```cpp include using namespace std; void getElements(int arr[],int n) { if(n==0 \|\| n==1) cout<<-1<<" "<<-1<<endl; // edge case when only one element is present in array int small=INT_MAX,second_small=INT_MAX; int large=INT_MIN,second_large=INT_MIN; int i; for(i=0;i<n;i++) { small=min(small,arr[i]); large=max(large,arr[i]); } for(i=0;i<n;i++) { if(arr[i]second_large && arr[i]!=large) second_large=arr[i]; } cout<<"Second smallest is "<<second_small<<endl; cout<<"Second largest is "<<second_large<<endl; } int main() { int arr[]={1,2,4,6,7,5}; int n=sizeof(arr)/sizeof(arr); getElements(arr,n); return 0; } ``` Output: Second smallest is 2 Second largest is 6 ```java import java.io.; import java.util.Arrays; class Test { static private void getElements(int[] arr, int n) { if (n == 0 \|\| n==1) { System.out.print(-1); System.out.print(" "); System.out.print(-1); System.out.print("\n"); } int small = Integer.MAX_VALUE; int second_small = Integer.MAX_VALUE; int large = Integer.MIN_VALUE; int second_large = Integer.MIN_VALUE; int i; for (i = 0;i < n;i++) { small = Math.min(small,arr[i]); large = Math.max(large,arr[i]); } for (i = 0;i < n;i++) { if (arr[i] < second_small && arr[i] != small) { second_small = arr[i]; } if (arr[i] > second_large && arr[i] != large) { second_large = arr[i]; } } System.out.println("Second smallest is "+second_small); System.out.println("Second largest is "+second_large); } public static void main(String[] args) { int[] arr = {1, 2, 4, 6, 7, 5}; int n = arr.length; getElements(arr, n); } } ``` Output: Second smallest is 2 Second largest is 6 ```python def getElements(arr, n): if n == 0 or n == 1: print(-1, -1) # edge case when only one element is present in array small = float('inf') second_small = float('inf') large = float('-inf') second_large = float('-inf') for i in range(n): small = min(small, arr[i]) large = max(large, arr[i]) for i in range(n): if arr[i] < second_small and arr[i] != small: second_small = arr[i] if arr[i] > second_large and arr[i] != large: second_large = arr[i] print("Second smallest is", second_small) print("Second largest is", second_large) if name == 'main': arr = [1, 2, 4, 6, 7, 5] n = len(arr) getElements(arr, n) ``` Output: Second smallest is 2 Second largest is 6 ```javascript function getElements(arr) { if (arr.length === 0 \|\| arr.length === 1) { console.log(-1 + " " + -1); // Edge case when only one element is present in the array return; } let small = Infinity; let second_small = Infinity; let large = -Infinity; let second_large = -Infinity; for (let i = 0; i < arr.length; i++) { small = Math.min(small, arr[i]); large = Math.max(large, arr[i]); } for (let i = 0; i < arr.length; i++) { if (arr[i] < second_small && arr[i] !== small) second_small = arr[i]; if (arr[i] > second_large && arr[i] !== large) second_large = arr[i]; } console.log("Second smallest is " + second_small); console.log("Second largest is " + second_large); } const arr = [1, 2, 4, 6, 7, 5]; getElements(arr); ``` Output: Second smallest is 2 Second largest is 6 Complexity Analysis Time Complexity: O(N), We do two linear traversals in our array Space Complexity: O(1) Optimal Approach Algorithm / Intuition Solution 3(Best Solution) Intuition: In the previous solution, even though we were able to bring down the time complexity to O(N), we still needed to do two traversals to find our answer. Can we do this in a single traversal by using smart comparisons on the go? Approach: We would require four variables: small,second_small, large, and second_large. Variable small and second_small are initialized to INT_MAX while large and second_large are initialized to INT_MIN. Second Smallest Algo: If the current element is smaller than ‘small’, then we update second_small and small variables Else if the current element is smaller than ‘second_small’ then we update the variable ‘second_small’ Once we traverse the entire array, we would find the second smallest element in the variable second_small. Here’s a quick demonstration of the same. Second Largest Algo: If the current element is larger than ‘large’ then update second_large and large variables Else if the current element is larger than ‘second_large’ then we update the variable second_large. Once we traverse the entire array, we would find the second largest element in the variable second_large. Here’s a quick demonstration of the same. Code C++ Java Python JavaScript ```cpp include using namespace std; int secondSmallest(int arr[],int n) { if(n<2) return -1; int small = INT_MAX; int second_small = INT_MAX; int i; for(i = 0; i < n; i++) { if(arr[i] < small) { second_small = small; small = arr[i]; } else if(arr[i] < second_small && arr[i] != small) { second_small = arr[i]; } } return second_small; } int secondLargest(int arr[],int n) { if(n<2) return -1; int large=INT_MIN,second_large=INT_MIN; int i; for (i = 0; i < n; i++) { if (arr[i] > large) { second_large = large; large = arr[i]; } else if (arr[i] > second_large && arr[i] != large) { second_large = arr[i]; } } return second_large; } int main() { int arr[]={1,2,4,7,7,5}; int n=sizeof(arr)/sizeof(arr); int sS=secondSmallest(arr,n); int sL=secondLargest(arr,n); cout<<"Second smallest is "<<sS<<endl; cout<<"Second largest is "<<sL<<endl; return 0; } ``` Output: Second smallest is 2 Second largest is 5 ```java import java.io.; class Test { static private int secondSmallest(int[] arr, int n) { if (n < 2) { return -1; } int small = Integer.MAX_VALUE; int second_small = Integer.MAX_VALUE; int i; for (i = 0; i < n; i++) { if (arr[i] < small) { second_small = small; small = arr[i]; } else if (arr[i] < second_small && arr[i] != small) { second_small = arr[i]; } } return second_small; } static private int secondLargest(int[] arr, int n) { if(n<2) return -1; int large = Integer.MIN_VALUE; int second_large = Integer.MIN_VALUE; int i; for (i = 0; i < n; i++) { if (arr[i] > large) { second_large = large; large = arr[i]; } else if (arr[i] > second_large && arr[i] != large) { second_large = arr[i]; } } return second_large; } public static void main(String[] args) { int[] arr = {1, 2, 4, 7, 7, 5}; int n = arr.length; int sS = secondSmallest(arr, n); int sL = secondLargest(arr, n); System.out.println("Second smallest is "+sS); System.out.println("Second largest is "+sL); } } ``` Output: Second smallest is 2 Second largest is 5 ```python def secondSmallest(arr, n): if (n < 2): return -1 small = float('inf') second_small = float('inf') for i in range(n): if (arr[i] < small): second_small = small small = arr[i] elif (arr[i] < second_small and arr[i] != small): second_small = arr[i] return second_small def secondLargest(arr, n): if (n < 2): return -1 large = float('-inf') second_large = float('-inf') for i in range(n): if (arr[i] > large): second_large = large large = arr[i] elif (arr[i] > second_large and arr[i] != large): second_large = arr[i] return second_large if name == "main": arr = [1, 2, 4, 7, 7, 5] n = len(arr) sS = secondSmallest(arr, n) sL = secondLargest(arr, n) print("Second smallest is", sS) print("Second largest is", sL) ``` Output: Second smallest is 2 Second largest is 5 ```javascript function secondSmallest(arr) { if (arr.length < 2) return -1; let small = Infinity; let second_small = Infinity; for (let i = 0; i < arr.length; i++) { if (arr[i] < small) { second_small = small; small = arr[i]; } else if (arr[i] < second_small && arr[i] !== small) { second_small = arr[i]; } } return second_small; } function secondLargest(arr) { if (arr.length < 2) return -1; let large = -Infinity; let second_large = -Infinity; for (let i = 0; i < arr.length; i++) { if (arr[i] > large) { second_large = large; large = arr[i]; } else if (arr[i] > second_large && arr[i] !== large) { second_large = arr[i]; } } return second_large; } const arr = [1, 2, 4, 7, 7, 5]; const sS = secondSmallest(arr); const sL = secondLargest(arr); console.log("Second smallest is " + sS); console.log("Second largest is " + sL); ``` Output: Second smallest is 2 Second largest is 5 Complexity Analysis Time Complexity: O(N), Single-pass solution Space Complexity: O(1) Video Explanation Special thanks to Naman Daga __and Sudip Ghosh___ _for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article Profile Blogs DSA Sheets Striver's A2Z Sheet Striver's SDE Sheet Striver's 79 Sheet Blind 75 Sheet System Design Striver's DSA Playlist Arrays Series DP Series Graph Series LinkedList Series SDE Core Sheets DBMS Operating System Computer Networks Striver's CP Sheet Saved Notes CW Fullstack Final Expand / collapse sidebar Open sidebar Login Find Second Smallest and Second Largest Element in an array 71616 Problem Statement: Given an array, find the second smallest and second largest element in the array. Print ‘-1’ in the event that either of them doesn’t exist. Examples Example 1: Input: [1,2,4,7,7,5] Output: Second Smallest : 2 Second Largest : 5 Explanation: The elements are as follows 1,2,3,5,7,7 and hence second largest of these is 5 and second smallest is 2 Example 2: Input: Output: Second Smallest : -1 Second Largest : -1 Explanation: Since there is only one element in the array, it is the largest and smallest element present in the array. There is no second largest or second smallest element present. Practice: Solve Problem Disclaimer: Don’t jump directly to the solution, try it out yourself first. Brute-Force Approach Better Approach Optimal Approach Expand any one approach by clicking the given options in the bar. Clicking one approach on bar, closes all other expands. You can manually expand more than one approach at a time Brute Force Approach Algorithm / Intuition Solution 1: (Brute Force) [this approach only works if there are no duplicates] Intuition: What do we do to find the largest or the smallest element present in an array? We ideally sort them and the first element would be the smallest of all while the last element would be the largest. Can we find the second-smallest and second-largest using a similar approach? Approach: Sort the array in ascending order The element present at the second index is the second smallest element The element present at the second index from the end is the second largest element Code C++ Java Python JavaScript ```cpp include using namespace std; void getElements(int arr[],int n) { if(n==0 \|\| n==1) cout<<-1<<" "<<-1<<endl; // edge case when only one element is present in array sort(arr,arr+n); int small=arr; int large=arr[n-2]; cout<<"Second smallest is "<<small<<endl; cout<<"Second largest is "<<large<<endl; } int main() { int arr[]={1,2,4,6,7,5}; int n=sizeof(arr)/sizeof(arr); getElements(arr,n); return 0; } ``` Output: Second smallest is 2 Second largest is 6 ```java import java.io.; import java.util.Arrays; class Test { static private void getElements(int[] arr, int n) { if (n == 0 \|\| n==1) { System.out.print(-1); System.out.print(" "); System.out.print(-1); System.out.print("\n"); } Arrays.sort(arr); int small = arr; int large = arr[n - 2]; System.out.println("Second smallest is "+small); System.out.println("Second largest is "+large); } public static void main(String[] args) { int[] arr = {1, 2, 4, 6, 7, 5}; int n = arr.length; getElements(arr, n); } } ``` Output: Second smallest is 2 Second largest is 6 ```python def getElements(arr, n): if n == 0 or n == 1: print(-1, -1) # edge case when only one element is present in array arr.sort() small = arr large = arr[n-2] print("Second smallest is", small) print("Second largest is", large) if name == 'main': arr = [1, 2, 4, 6, 7, 5] n = len(arr) getElements(arr, n) ``` Output: Second smallest is 2 Second largest is 6 ```javascript function getElements(arr) { if (arr.length === 0 \|\| arr.length === 1) { console.log(-1 + " " + -1); // Edge case when only one element is present in the array return; } arr.sort((a, b) => a - b); let small = arr; let large = arr[arr.length - 2]; console.log("Second smallest is " + small); console.log("Second largest is " + large); } const arr = [1, 2, 4, 6, 7, 5]; getElements(arr); ``` Output: Second smallest is 2 Second largest is 6 Complexity Analysis Time Complexity: O(NlogN), For sorting the array Space Complexity: O(1) Better Approach Algorithm / Intuition Solution 2(Better Solution) Intuition: Even though we want to have just the second smallest and largest elements, we are still sorting the entire array for that and thus increasing the time complexity. Can we somehow try to not sort the array and still get our answer? Approach: Find the smallest and largest element in the array in a single traversal After this, we once again traverse the array and find an element that is just greater than the smallest element we just found. Similarly, we would find the largest element which is just smaller than the largest element we just found Indeed, this is our second smallest and second largest element. Code C++ Java Python JavaScript ```cpp include using namespace std; void getElements(int arr[],int n) { if(n==0 \|\| n==1) cout<<-1<<" "<<-1<<endl; // edge case when only one element is present in array int small=INT_MAX,second_small=INT_MAX; int large=INT_MIN,second_large=INT_MIN; int i; for(i=0;i<n;i++) { small=min(small,arr[i]); large=max(large,arr[i]); } for(i=0;i<n;i++) { if(arr[i]second_large && arr[i]!=large) second_large=arr[i]; } cout<<"Second smallest is "<<second_small<<endl; cout<<"Second largest is "<<second_large<<endl; } int main() { int arr[]={1,2,4,6,7,5}; int n=sizeof(arr)/sizeof(arr); getElements(arr,n); return 0; } ``` Output: Second smallest is 2 Second largest is 6 ```java import java.io.; import java.util.Arrays; class Test { static private void getElements(int[] arr, int n) { if (n == 0 \|\| n==1) { System.out.print(-1); System.out.print(" "); System.out.print(-1); System.out.print("\n"); } int small = Integer.MAX_VALUE; int second_small = Integer.MAX_VALUE; int large = Integer.MIN_VALUE; int second_large = Integer.MIN_VALUE; int i; for (i = 0;i < n;i++) { small = Math.min(small,arr[i]); large = Math.max(large,arr[i]); } for (i = 0;i < n;i++) { if (arr[i] < second_small && arr[i] != small) { second_small = arr[i]; } if (arr[i] > second_large && arr[i] != large) { second_large = arr[i]; } } System.out.println("Second smallest is "+second_small); System.out.println("Second largest is "+second_large); } public static void main(String[] args) { int[] arr = {1, 2, 4, 6, 7, 5}; int n = arr.length; getElements(arr, n); } } ``` Output: Second smallest is 2 Second largest is 6 ```python def getElements(arr, n): if n == 0 or n == 1: print(-1, -1) # edge case when only one element is present in array small = float('inf') second_small = float('inf') large = float('-inf') second_large = float('-inf') for i in range(n): small = min(small, arr[i]) large = max(large, arr[i]) for i in range(n): if arr[i] < second_small and arr[i] != small: second_small = arr[i] if arr[i] > second_large and arr[i] != large: second_large = arr[i] print("Second smallest is", second_small) print("Second largest is", second_large) if name == 'main': arr = [1, 2, 4, 6, 7, 5] n = len(arr) getElements(arr, n) ``` Output: Second smallest is 2 Second largest is 6 ```javascript function getElements(arr) { if (arr.length === 0 \|\| arr.length === 1) { console.log(-1 + " " + -1); // Edge case when only one element is present in the array return; } let small = Infinity; let second_small = Infinity; let large = -Infinity; let second_large = -Infinity; for (let i = 0; i < arr.length; i++) { small = Math.min(small, arr[i]); large = Math.max(large, arr[i]); } for (let i = 0; i < arr.length; i++) { if (arr[i] < second_small && arr[i] !== small) second_small = arr[i]; if (arr[i] > second_large && arr[i] !== large) second_large = arr[i]; } console.log("Second smallest is " + second_small); console.log("Second largest is " + second_large); } const arr = [1, 2, 4, 6, 7, 5]; getElements(arr); ``` Output: Second smallest is 2 Second largest is 6 Complexity Analysis Time Complexity: O(N), We do two linear traversals in our array Space Complexity: O(1) Optimal Approach Algorithm / Intuition Solution 3(Best Solution) Intuition: In the previous solution, even though we were able to bring down the time complexity to O(N), we still needed to do two traversals to find our answer. Can we do this in a single traversal by using smart comparisons on the go? Approach: We would require four variables: small,second_small, large, and second_large. Variable small and second_small are initialized to INT_MAX while large and second_large are initialized to INT_MIN. Second Smallest Algo: If the current element is smaller than ‘small’, then we update second_small and small variables Else if the current element is smaller than ‘second_small’ then we update the variable ‘second_small’ Once we traverse the entire array, we would find the second smallest element in the variable second_small. Here’s a quick demonstration of the same. Second Largest Algo: If the current element is larger than ‘large’ then update second_large and large variables Else if the current element is larger than ‘second_large’ then we update the variable second_large. Once we traverse the entire array, we would find the second largest element in the variable second_large. Here’s a quick demonstration of the same. Code C++ Java Python JavaScript ```cpp include using namespace std; int secondSmallest(int arr[],int n) { if(n<2) return -1; int small = INT_MAX; int second_small = INT_MAX; int i; for(i = 0; i < n; i++) { if(arr[i] < small) { second_small = small; small = arr[i]; } else if(arr[i] < second_small && arr[i] != small) { second_small = arr[i]; } } return second_small; } int secondLargest(int arr[],int n) { if(n<2) return -1; int large=INT_MIN,second_large=INT_MIN; int i; for (i = 0; i < n; i++) { if (arr[i] > large) { second_large = large; large = arr[i]; } else if (arr[i] > second_large && arr[i] != large) { second_large = arr[i]; } } return second_large; } int main() { int arr[]={1,2,4,7,7,5}; int n=sizeof(arr)/sizeof(arr); int sS=secondSmallest(arr,n); int sL=secondLargest(arr,n); cout<<"Second smallest is "<<sS<<endl; cout<<"Second largest is "<<sL<<endl; return 0; } ``` Output: Second smallest is 2 Second largest is 5 ```java import java.io.; class Test { static private int secondSmallest(int[] arr, int n) { if (n < 2) { return -1; } int small = Integer.MAX_VALUE; int second_small = Integer.MAX_VALUE; int i; for (i = 0; i < n; i++) { if (arr[i] < small) { second_small = small; small = arr[i]; } else if (arr[i] < second_small && arr[i] != small) { second_small = arr[i]; } } return second_small; } static private int secondLargest(int[] arr, int n) { if(n<2) return -1; int large = Integer.MIN_VALUE; int second_large = Integer.MIN_VALUE; int i; for (i = 0; i < n; i++) { if (arr[i] > large) { second_large = large; large = arr[i]; } else if (arr[i] > second_large && arr[i] != large) { second_large = arr[i]; } } return second_large; } public static void main(String[] args) { int[] arr = {1, 2, 4, 7, 7, 5}; int n = arr.length; int sS = secondSmallest(arr, n); int sL = secondLargest(arr, n); System.out.println("Second smallest is "+sS); System.out.println("Second largest is "+sL); } } ``` Output: Second smallest is 2 Second largest is 5 ```python def secondSmallest(arr, n): if (n < 2): return -1 small = float('inf') second_small = float('inf') for i in range(n): if (arr[i] < small): second_small = small small = arr[i] elif (arr[i] < second_small and arr[i] != small): second_small = arr[i] return second_small def secondLargest(arr, n): if (n < 2): return -1 large = float('-inf') second_large = float('-inf') for i in range(n): if (arr[i] > large): second_large = large large = arr[i] elif (arr[i] > second_large and arr[i] != large): second_large = arr[i] return second_large if name == "main": arr = [1, 2, 4, 7, 7, 5] n = len(arr) sS = secondSmallest(arr, n) sL = secondLargest(arr, n) print("Second smallest is", sS) print("Second largest is", sL) ``` Output: Second smallest is 2 Second largest is 5 ```javascript function secondSmallest(arr) { if (arr.length < 2) return -1; let small = Infinity; let second_small = Infinity; for (let i = 0; i < arr.length; i++) { if (arr[i] < small) { second_small = small; small = arr[i]; } else if (arr[i] < second_small && arr[i] !== small) { second_small = arr[i]; } } return second_small; } function secondLargest(arr) { if (arr.length < 2) return -1; let large = -Infinity; let second_large = -Infinity; for (let i = 0; i < arr.length; i++) { if (arr[i] > large) { second_large = large; large = arr[i]; } else if (arr[i] > second_large && arr[i] !== large) { second_large = arr[i]; } } return second_large; } const arr = [1, 2, 4, 7, 7, 5]; const sS = secondSmallest(arr); const sL = secondLargest(arr); console.log("Second smallest is " + sS); console.log("Second largest is " + sL); ``` Output: Second smallest is 2 Second largest is 5 Complexity Analysis Time Complexity: O(N), Single-pass solution Space Complexity: O(1) Video Explanation Special thanks to Naman Daga __and Sudip Ghosh___ _for contributing to this article on takeUforward. 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68	https://dsp.stackexchange.com/questions/75901/butterworth-filter-higher-order	Butterworth filter higher order - Signal Processing Stack Exchange Join Signal Processing By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Signal Processing helpchat Signal Processing Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Companies Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Butterworth filter higher order Ask Question Asked 4 years, 3 months ago Modified4 years, 3 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I am new (very beginner) to signal processing and I am trying to understand how the Butterworth filter operates. According to my current understanding of filters, there seems to be a tradeoff between the the passband ripples and the transition band. Like if the former decreases, the latter increases and vice-versa. Now this makes sense from the definition of the Butterworth filter - a maximally flat frequency response at the expense of a wider transition band. What I don't understand is that how is it possible that by increasing the order of the Butterworth filter, the transition band will decrease but the passband remains flat?? Also what does an unstable filter mean? filters butterworth Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Jun 19, 2021 at 21:12 Peter K.♦ 26.2k 9 9 gold badges 49 49 silver badges 94 94 bronze badges asked Jun 19, 2021 at 20:45 DumbDumb 1 1 1 bronze badge 3 Welcome to SE.SP! Butterworth filters do not have "ripples" in the passband. That's the point of maximal flatness: the first and higher order derivatives are all zero at a given frequency. Can you explain what you mean by "ripples" ?Peter K. –Peter K.♦ 2021-06-19 21:15:01 +00:00 Commented Jun 19, 2021 at 21:15 by ripples i mean fluctuations in the passband. So even if you keep increasing the order, the passband will remain flat? but then why don't we just keep increasing the order of the BW until we reach the ideal 'brickwall' filter in practice?Dumb –Dumb 2021-06-19 21:22:56 +00:00 Commented Jun 19, 2021 at 21:22 The the delay, computational cost, and sensitivity to numerical noise (quantization and rounding causing the calculation to "blow up") become greater with a higher order filter. So the ideal brickwall would take forever, and require infinite precision arithmetic thus requiring more petabytes than exist in any computer.hotpaw2 –hotpaw2 2021-06-20 00:14:11 +00:00 Commented Jun 20, 2021 at 0:14 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. According to my current understanding of filters, there seems to be a tradeoff between the the passband ripples and the transition band. No. Butterworth filters have neither passband nor stopband ripples. What I don't understand is that how is it possible that by increasing the order of the Butterworth filter, the transition band will decrease but the passband remains flat?? The magnitude squared of the transfer function of an analog Butterworth lowpass filter of order N N with a cutoff frequency of f 0 f 0 is \|H(f)\|2=1 1+(f/f 0)2 N\|H(f)\|2=1 1+(f/f 0)2 N The passband is determined f/f 0≪1 f/f 0≪1 in which case we simply get \|H(f)\|≈1\|H(f)\|≈1, so there is no ripple . For f/f 0≫1 f/f 0≫1 we get \|H(f)\|≈(f 0/f)N\|H(f)\|≈(f 0/f)N so it falls with the frequency raised to the order of the filter. The higher the order, the quicker it falls. Please note that is \|H(f)\|2\|H(f)\|2 falls monotonic over all frequencies. There are no ripples anywhere. Also what does an unstable filter mean? Certain types of filters (like a Butterworth) are recursive, i.e. they feed part of their output back into the input. This can create a "run-away" condition. Let's look a simple difference equation y[n]=x[n]+0.5⋅y[n−1]y[n]=x[n]+0.5⋅y[n−1] If we feed a unit impulse into this system the output will be 1, 0.5, 0.25, 0.125 ... It keeps going for a while but will die down over time. Now let's look at an unstable example. y[n]=x[n]+2⋅y[n−1]y[n]=x[n]+2⋅y[n−1] For the same unit impulse input the output will be 1, 2, 4, 8, 16, 32, ... so there is unbounded growth. That's unstable. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jun 20, 2021 at 6:38 robert bristow-johnson 22.5k 4 4 gold badges 41 41 silver badges 82 82 bronze badges answered Jun 20, 2021 at 1:06 HilmarHilmar 50.3k 1 1 gold badge 33 33 silver badges 68 68 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Answering part #2 of your question: In a well conditioned numerical computation, the quantization and rounding errors stay small with respect to the size of the desire result. An unstable filter can do the opposite, where tiny numerical noise incurred during doing the DSP arithmetic can rapidly increase until it becomes vastly larger than any desired result. Since real computers have finite memory, an unstable filter might make a garbage result impossible to avoid. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Jun 20, 2021 at 0:23 hotpaw2hotpaw2 35.9k 9 9 gold badges 49 49 silver badges 93 93 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Signal Processing Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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69	https://artofproblemsolving.com/downloads/printable_post_collections/5171?srsltid=AfmBOoqnt_Z53Bl7pmYvJW_q5JR0gd25XTnt5wfLWasXtCx5_jPg_d8p	AoPS Community 2011 All-Russian Olympiad All-Russian Olympiad 2011 www.artofproblemsolving.com/community/c5171 by 3333, sartt, Bugi – Grade 9 Day 1 1 A quadratic trinomial P (x) with the x2 coefficient of one is such, that P (x) and P (P (P (x))) share a root. Prove that P (0) ∗ P (1) = 0 . 2 Given is an acute angled triangle ABC . A circle going through B and the triangle’s circumcenter, O, intersects BC and BA at points P and Q respectively. Prove that the intersection of the heights of the triangle P OQ lies on line AC . 3 A convex 2011-gon is drawn on the board. Peter keeps drawing its diagonals in such a way, that each newly drawn diagonal intersected no more than one of the already drawn diagonals. What is the greatest number of diagonals that Peter can draw? 4 Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers? Day 2 1 For some 2011 natural numbers, all the 2010 ·2011 2 possible sums were written out on a board. Could it have happened that exactly one third of the written numbers were divisible by three and also exactly one third of them give a remainder of one when divided by three? 2 In the notebooks of Peter and Nick, two numbers are written. Initially, these two numbers are 1 and 2 for Peter and 3 and 4 for Nick. Once a minute, Peter writes a quadratic trinomial f (x),the roots of which are the two numbers in his notebook, while Nick writes a quadratic trinomial g(x) the roots of which are the numbers in his notebook. If the equation f (x) = g(x) has two distinct roots, one of the two boys replaces the numbers in his notebook by those two roots. Otherwise, nothing happens. If Peter once made one of his numbers 5, what did the other one of his numbers become? 3 Let ABC be an equilateral triangle. A point T is chosen on AC and on arcs AB and BC of the circumcircle of ABC , M and N are chosen respectively, so that M T is parallel to BC and N T is parallel to AB . Segments AN and M T intersect at point X, while CM and N T intersect in point Y . Prove that the perimeters of the polygons AXY C and XM BN Y are the same. ©2024 AoPS Incorporated 1AoPS Community 2011 All-Russian Olympiad 4 There are some counters in some cells of 100 × 100 board. Call a cell nice if there are an even number of counters in adjacent cells. Can exactly one cell be nice ? K. Knop – Grade 10 Day 1 1 In every cell of a table with n rows and ten columns, a digit is written. It is known that for every row A and any two columns, you can always find a row that has different digits from A only when it intersects with two columns. Prove that n ≥ 512 . 2 Nine quadratics, x2 + a1x + b1, x 2 + a2x + b2, ..., x 2 + a9x + b9 are written on the board. The sequences a1, a 2, ..., a 9 and b1, b 2, ..., b 9 are arithmetic. The sum of all nine quadratics has at least one real root. What is the the greatest possible number of original quadratics that can have no real roots? 3 The graph G is not 3-coloured. Prove that G can be divided into two graphs M and N such that M is not 2-coloured and N is not 1-coloured. V. Dolnikov 4 Perimeter of triangle ABC is 4. Point X is marked at ray AB and point Y is marked at ray AC such that AX = AY = 1 . Line segments BC and XY intersectat point M . Prove that perimeter of one of triangles ABM or ACM is 2.(V. Shmarov). Day 2 1 Given are 10 distinct real numbers. Kyle wrote down the square of the difference for each pair of those numbers in his notebook, while Peter wrote in his notebook the absolute value of the differences of the squares of these numbers. Is it possible for the two boys to have the same set of 45 numbers in their notebooks? 2 Given is an acute triangle ABC . Its heights BB 1 and CC 1 are extended past points B1 and C1.On these extensions, points P and Q are chosen, such that angle P AQ is right. Let AF be a height of triangle AP Q . Prove that angle BF C is a right angle. 3 For positive integers a > b > 1, define xn = an − 1 bn − 1 Find the least d such that for any a, b , the sequence xn does not contain d consecutive prime numbers. ©2024 AoPS Incorporated 2AoPS Community 2011 All-Russian Olympiad V. Senderov 4 A 2010 × 2010 board is divided into corner-shaped figures of three cells. Prove that it is possible to mark one cell in each figure such that each row and each column will have the same number of marked cells. I. Bogdanov & O. Podlipsky – Grade 11 Day 1 1 Two natural numbers d and d′, where d′ > d , are both divisors of n. Prove that d′ > d + d2 n . 2 On side BC of parallelogram ABCD (A is acute) lies point T so that triangle AT D is an acute triangle. Let O1, O2, and O3 be the circumcenters of triangles ABT , DAT , and CDT respectively. Prove that the orthocenter of triangle O1O2O3 lies on line AD . 3 There are 999 scientists. Every 2 scientists are both interested in exactly 1 topic and for each topic there are exactly 3 scientists that are interested in that topic. Prove that it is possible to choose 250 topics such that every scientist is interested in at most 1 theme. A. Magazinov 4 Ten cars are moving at the road. There are some cities at the road. Each car is moving with some constant speed through cities and with some different constant speed outside the cities (different cars may move with different speed). There are 2011 points at the road. Cars don’t overtake at the points. Prove that there are 2 points such that cars pass through these points in the same order. S. Berlov Day 2 1 Given are two distinct monic cubics F (x) and G(x). All roots of the equations F (x) = 0 , G(x) = 0 and F (x) = G(x) are written down. There are eight numbers written. Prove that the greatest of them and the least of them cannot be both roots of the polynomial F (x). 2 There are more than n2 stones on the table. Peter and Vasya play a game, Peter starts. Each turn, a player can take any prime number less than n stones, or any multiple of n stones, or 1 stone. Prove that Peter always can take the last stone (regardless of Vasya’s strategy). S Berlov ©2024 AoPS Incorporated 3AoPS Community 2011 All-Russian Olympiad 3 Let P (a) be the largest prime positive divisor of a2 + 1 . Prove that exist infinitely many positive integers a, b, c such that P (a) = P (b) = P (c). A. Golovanov 4 Let N be the midpoint of arc ABC of the circumcircle of triangle ABC , let M be the midpoint of AC and let I1, I 2 be the incentres of triangles ABM and CBM . Prove that points I1, I 2, B, N lie on a circle. M. Kungojin ©2024 AoPS Incorporated 4
70	https://mathspace.co/textbooks/syllabuses/Syllabus-303/topics/Topic-5579/subtopics/Subtopic-74100/	Comparisons using equivalent fractions (2, 3, 4, 5, 6, 8, 10, 12, 100) \| Stage 2 Maths \| HK Primary Stage 2 \| Mathspace Book a Demo Topics F r a c t i o n s F r a c t i o n b a r s N u m b e r l i n e s A r e a m o d e l s F r a c t i o n s w i t h o b j e c t s C o m p a r i s o n s u s i n g m o d e l s C o m p a r i s o n s u s i n g n u m b e r l i n e s E q u i v a l e n t f r a c t i o n s(2,3,4,6,8) C o m p a r i s o n s u s i n g e q u i v a l e n t f r a c t i o n s(2,3,4,5,6,8,1 0,1 2,1 0 0) LessonPractice O r d e r i n g a n d c o u n t i n g w i t h f r a c t i o n s M i x e d n u m b e r s(n a m e,i d e n t i f y a n d c o m p a r e) A d d i n g t e n t h s a n d h u n d r e d t h s A d d i n g t e n t h s a n d h u n d r e d t h s(f r a c t i o n s a n d d e c i m a l s) S h a d e d F r a c t i o n s a n d N u m b e r L i n e s(U n i t F r a c t i o n s) C o m p a r i n g f r a c t i o n s F r a c t i o n s o n a n u m b e r l i n e E q u i v a l e n t F r a c t i o n s U s i n g b e n c h m a r k s E q u i v a l e n t f r a c t i o n s C o m p a r e a n d O r d e r F r a c t i o n s O r d e r i n g U n i t F r a c t i o n s O r d e r i n g F r a c t i o n s A d d i n g a n d s u b t r a c t i n g F r a c t i o n s E g y p t i a n F r a c t i o n s(I n v e s t i g a t i o n) T e n t h s a n d H u n d r e d t h s(e q u i v a l e n t f r a c t i o n s) A d d/s u b u s i n g n u m b e r l i n e s(i n c l m i x e d n u m b e r s)(2,3,4,5,6,8,1 0,1 2) T e n t h s,1 0 0 t h s a n d D e c i m a l s A d d/s u b u s i n g m o d e l s(i n c l m i x e d n u m b e r s)(2,3,4,5,6,8,1 0,1 2) N u m b e r s e n t e n c e s w i t h f r a c t i o n s A d d/s u b o f u n i t f r a c t i o n s A d d/s u b u s i n g s y m b o l s(i n c l m i x e d n u m b e r s)(2,3,4,5,6,8,1 0,1 2) A d d i n g a n d S u b t r a c t i n g M i x e d N u m b e r s A d d i t i o n a n d S u b t r a c t i o n o f L i k e F r a c t i o n s F r a c t i o n F l o w e r s(I n v e s t i g a t i o n) M u l t i p l e s o f u n i t f r a c t i o n s u s i n g n u m b e r l i n e s a n d s y m b o l s(A x 1/b)(2,3,4,5,6,8,1 0,1 2) C h e c k i n g R e a s o n a b l e n e s s o f F r a c t i o n S u m s C o m p a r e a d d/s u b s t a t e m e n t s w i t h f r a c t i o n s M u l t i p l i c a t i o n o f f r a c t i o n s u s i n g n u m b e r l i n e s a n d s y m b o l s(C x a/b)(2,3,4,5,6,8,1 0,1 2) A r e a s a s F r a c t i o n s S h a d e d F i g u r e s a n d N u m b e r L i n e s M u l t i p l i c a t i o n o f f r a c t i o n s(2,3,4,6,8,1 0,1 2,1 0 0) C o m p a r i s o n s u s i n g m o d e l s C o m p a r i n g f r a c t i o n s M u l t i p l y i n g F r a c t i o n s D i v i d i n g w h o l e n u m b e r s b y u n i t f r a c t i o n s F r a c t i o n s o f Q u a n t i t i e s(u n i t f r a c t i o n s) D i v i d i n g u n i t f r a c t i o n s b y w h o l e n u m b e r s F r a c t i o n s o f Q u a n t i t i e s(s i m p l e n o n-u n i t) E q u i v a l e n t F r a c t i o n s D i v i d i n g F r a c t i o n s O r d e r i n g f r a c t i o n s w i t h b e n c h m a r k s o n l y O r d e r i n g F r a c t i o n s N u m b e r s e n t e n c e s w i t h f r a c t i o n s A d d i t i o n a n d S u b t r a c t i o n s u s i n g n u m b e r l i n e s(2,3,4,5,6,8,1 0,1 2) A d d i t i o n a n d S u b t r a c t i o n u s i n g m o d e l s(2,3,4,5,6,8,1 0,1 2) A d d i n g a n d S u b t r a c t i n g f r a c t i o n s(e a s i l y r e l a t e d d e n o m i n a t o r s) C o m p a r e a d d/s u b s t a t e m e n t s w i t h f r a c t i o n s T y p e s o f F r a c t i o n s S i m p l i f y i n g F r a c t i o n s M i x e d q u e s t i o n s o n f r a c t i o n s(c o m p a r i n g,e q u i v a l e n c e,s i m p l i f y i n g a n d o r d e r i n g) M i x e d Q u e s t i o n s o n F r a c t i o n s Log inSign upBook a Demo Hong Kong Stage 2 Comparisons using equivalent fractions (2, 3, 4, 5, 6, 8, 10, 12, 100) LessonPractice Lesson Share You may have already learned about comparing fractions using area models. Now we will look at using equivalent fractions to compare the value of fractions. Equivalent fractions are fractions that look different, but have the same value. Such as, $$1 2 and $$2 4 As you can see in the diagrams, they have the same value, but look different. $$1 2 and $$2 4 Equivalent fractions are useful to be sure of the value compared to other fractions, so we can work out which fraction is bigger or smaller. Watch this video to look at comparing fractions using equivalent fractions. Video Player is loading. Play Video Play Skip Backward Mute Current Time 0:00 / Duration 5:50 Loaded: 1.13% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-5:50 1x Playback Rate 2x 1.5x 1x, selected 0.5x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected English Captions Español Captions Audio Track default, selected Picture-in-Picture 480p, selected Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Opacity Text Background Color Opacity Caption Area Background Color Opacity Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. To compare fractions we need to think of their number family to help change them into an equal number of pieces. A number family is three numbers that fit together to make facts using multiplication, for example, $$5, $$2 and $$1 0 because $$5×2=1 0. Using this fact, we can make: fifths into tenths by splitting each fifth into two equal parts or halves into tenths by splitting each half into five equal parts. Remember! When changing to an equivalent fraction we do not change the value of the fraction. To change into an equivalent fraction we use number families to relate the denominators. Worked Examples Question 1 We are going to compare the two fractions $$1 8 and $$1 4. Turn $$1 4 into a fraction in eighth s. We can now compare the two fractions $$1 8 and $$1 4. Which fraction is larger? The fractions are the same size. A $$1 8 B $$1 4 C Reveal Solution Watch video Question 2 We are going to compare the two fractions $$4 6 and $$1 3. Turn $$1 3 into a fraction in sixth s. Now plot $$4 6 and $$1 3 on the numberline: Which fraction is larger? $$1 3 A $$4 6 B Reveal Solution Watch video Question 3 We are going to compare the two fractions $$4 0 1 0 0 and $$2 5. Turn $$2 5 into a fraction in hundredths. Now compare $$4 0 1 0 0 and $$2 5 $$4 0 1 0 0=2 5 A $$4 0 1 0 0>2 5 B $$4 0 1 0 0<2 5 C Reveal Solution Watch video What is Mathspace About Mathspace
71	https://bluemountainmath.com/absolute-best-ways-to-teach-absolute-value/	Blue Mountain Math Making Math Fun with Engaging Activities Absolute Best Ways to Teach Absolute Value Like most people, I begin my lesson on absolute value with a story about distance. I love telling stories in class. Stories are a way to lessen stress in class and provide a way to have some fun. In my story, I drew a number line. I placed the school at zero and told the students that I live 10 blocks from the school. I then asked where if they could determine where I live. Of, course the first answer was at positive 10. Most students were happy with that answer. When I pressed for another answer, they caught on quick (they know me well). Absolute Value Stories This is important for students to understand. Absolute value represents the distance. Distance is always positive. The other story I tell is about my car. My odometer reads 10,000. I travel 100 miles to San Diego (I live in the Inland Empire in Southern California). What does my odometer read when I get to San Diego? Of course, it now reads 10,100 miles. So when I come home, I am going the other way. Do I now subtract 100 miles? (Students answered only if I am driving backward, lol). Students need concrete models. Telling stories about travel and location that are familiar will help. Different Teaching Methods In a recent meeting, the subject of teaching absolute value equations was raised. It was interesting to hear the methods that teachers preferred. Some were convinced that their method was the only one. This is the way I normally teach this. I explain that the value inside the absolute value bars can either be positive or negative, so we are going to have 2 answers. And I set up both equations as shown. Connecting Models to Concepts Another teacher took exception to this. She said it makes no sense to students. Her students need to know why it works. My way does not have an explanation. For her students the why is valuable. She demonstrated that she teaches it this way: I know what you are thinking. This is a distinction without a difference. But to her eyes, the second method makes sense to her. Because the expression inside the absolute value bars can be positive or negative, then the obvious procedure is to write the expression as positive or negative. This, she insists, can be explained to students. Using the Number Line Another teacher did not like either. He is visual. Teachers that love Geometry like pictures and diagrams. He teaches that absolute value is represented on the number line as the distance from zero. Which works fine when there is no number with x between the bars. When others began asking questions, it got a little confusing. I got confused. Another teacher to the rescue. He shared the following: This helps students see the equation. It provides some scaffolding for students that need models. This was the way he learned the concept and how he teaches it to students. Providing Practice I had never seen this version but I tried it, asking students to describe the answer in words and also provide a solution. Students found it easier. This worksheet formed a great introduction for students. This is the first introduction to absolute value in class and more practice is needed. But it was exciting to see teachers so enthusiastic about their strategies. As students are different, it is important to provide a variety of ways to learn. Resources you may enjoy: « Effective Student Goal Setting College Research Project » Leave a Reply Cancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed. Latest on Pinterest Latest on Instagram Bow Ties send a signal that you are smarter and mo My cat is upset that dogs have a day until I expla And for my southern relatives, have some fried chi AT last, a day for those of us who are over a cert I still love radio, mostly music. I used to listed Did you know that the Wright Brothers who are cred Why can't we have a fajita day each week like Taco We have already used our colored markers and penci Did you know that left handed people have brains t Latest on Facebook This message is only visible to admins. Problem displaying Facebook posts. Backup cache in use. Click to show error Error: Error validating access token: The session has been invalidated because the user changed their password or Facebook has changed the session for security reasons. Type: OAuthException
72	https://www.bennadel.com/blog/2240-creating-repeated-sequences-with-the-modulus-mod-operator.htm	Creating Repeated Sequences With The Modulus (MOD) Operator Skip to main content Ben Nadel . com Navigation Home Activity Podcast My Book About Me People Contact Light UI Dark UI Ben Nadel at the New York ColdFusion User Group (Mar. 2009) with: Aaron FossSimon Free ← Previous PhotoNext Photo → Creating Repeated Sequences With The Modulus (MOD) Operator By Ben Nadel Published 2011-08-10 in ColdFusion — Comments (13) This almost doesn't warrant its own post; but, it seems to be one of those small factoids that I can't keep in my head without writing it down. And so, a quick post to demonstrate creating repeated sequences with the modulus operator (MOD or %). To be fair, the modulus operator already creates repeated sequences; but, those sequences end with a zero. This is perfect for when you only care about one number in a particular sequence; but, when you want to deal with a sequence as a set of numbers, typically, you want your sequence to start at 1 and proceed to N. In order to end your sequence on "N", you have to add some math to your modulus calculation. But, before we look at that, let's take a look at a simple modulus loop: - Loop to 15 using a modulus of 3 to get a result at each index. - i# % 3 = #(i % 3)# view rawcode-1.cfm hosted with ❤ by GitHub Here, we are looping from 1 to 15 and outputting the mod-3 for each index. When we run this code, we get the following output: 1 % 3 = 1 2 % 3 = 2 3 % 3 = 0 4 % 3 = 1 5 % 3 = 2 6 % 3 = 0 7 % 3 = 1 8 % 3 = 2 9 % 3 = 0 10 % 3 = 1 11 % 3 = 2 12 % 3 = 0 13 % 3 = 1 14 % 3 = 2 15 % 3 = 0 As you can see, this creates the repeated sequence, {1,2,0}. Like I said, standard modulus sequences end with zero. This is great if you want to determine which index you are at (1 == first, 0 == last); but, if you wanted to get the index within the repeated sequence, your logic gets a bit more tricky. To create a repeated sequence that ends in N rather than zero, we have to add some more math to the mix: - Now, loop to 15 using a modulus of 3. However, this time, rather than repeating 1,2,0, we want to repeat 1,2,3. To do this, we have to use a slightly different formula: ((n - 1) % 3) + 1 - ((#i# - 1) % 3) + 1 = #(((i - 1) % 3) + 1)# view rawcode-2.cfm hosted with ❤ by GitHub This time, we are subtracting one from the index and then adding one to the result. When we run this code, we get the following output: ((1 - 1) % 3) + 1 = 1 ((2 - 1) % 3) + 1 = 2 ((3 - 1) % 3) + 1 = 3 ((4 - 1) % 3) + 1 = 1 ((5 - 1) % 3) + 1 = 2 ((6 - 1) % 3) + 1 = 3 ((7 - 1) % 3) + 1 = 1 ((8 - 1) % 3) + 1 = 2 ((9 - 1) % 3) + 1 = 3 ((10 - 1) % 3) + 1 = 1 ((11 - 1) % 3) + 1 = 2 ((12 - 1) % 3) + 1 = 3 ((13 - 1) % 3) + 1 = 1 ((14 - 1) % 3) + 1 = 2 ((15 - 1) % 3) + 1 = 3 As you can see, this time, we created the repeated sequence, {1,2,3}. With the extra math, we were able to end on N (3 in our case) rather than zero. And, of course, we could abstract this concept out into its own function: - Define arguments. - - loop to 15, repeating 1,2,3. - modSequence( #i#, 3 ) = #modSequence( i, 3 )# view rawcode-3.cfm hosted with ❤ by GitHub Here, we're simply encapsulating the details behind a function invocation. When we run the above code we get the following page output: modSequence( 1, 3 ) = 1 modSequence( 2, 3 ) = 2 modSequence( 3, 3 ) = 3 modSequence( 4, 3 ) = 1 modSequence( 5, 3 ) = 2 modSequence( 6, 3 ) = 3 modSequence( 7, 3 ) = 1 modSequence( 8, 3 ) = 2 modSequence( 9, 3 ) = 3 modSequence( 10, 3 ) = 1 modSequence( 11, 3 ) = 2 modSequence( 12, 3 ) = 3 modSequence( 13, 3 ) = 1 modSequence( 14, 3 ) = 2 modSequence( 15, 3 ) = 3 The modulus operator is pretty awesome, no doubt. Hopefully, I'll now be able to remember how to use it to create a more "standard" repeated sequence. And if I can't, at least I know where to look. Want to use code from this post? Check out the license. Short link: You Might Also Enjoy Some of My Other Posts The Integer Division Operator In ColdFusion And On The Seventh Row, MOD Created 1, And It Was Good Reader Comments AnnaAug 10, 2011 at 11:10 AM 383 Comments When I first started programming, I thought the MOD operator was the coolest, most awesome and interesting thing. I still love it to this day. Thanks for writing about it. You get extra brownie points, as well, for adding extra math. :-) Ben NadelAug 10, 2011 at 1:35 PM 16,080 Comments @Anna, Yeah, the mod operator is totally awesome :) It's one of my favorite operators (which may very well be the geekiest thing I say all day). AnnaAug 10, 2011 at 3:13 PM 383 Comments @Ben, haha...you can be geeky, that's ok. I am geeky all day, every day, probably, but it's gotten so bad with me, that I don't even really notice it anymore, and don't recognize it when I am. WebManWalkingAug 10, 2011 at 7:43 PM 290 Comments @Ben, If you cfloop from -6 to 6, instead of from 1 to 15, you'll see that % is actually a remainder operator, not a true modulo. In a true modulo, -5 % 3 would 1, not -2. It can make a difference if you're trying to generate a random number within a range. Say you want to generate a random number in the range of 0 to 59, to represent the number of seconds to wait before doing something. Ordinarily, % 60 does that quite nicely. But if the first operand accidentally goes negative somehow, you instead get a number in the range of -59 to 0. Just something to watch out for. WebManWalkingAug 10, 2011 at 8:02 PM 290 Comments This might be a bit more visual. Switching to code for its monofont. Remainder versus true modulo: i i \ 3 i % 3 true i mod 3 -6 -2 0 0 -5 -1 -2 1 -4 -1 -1 2 -3 -1 0 0 -2 0 -2 1 -1 0 -1 2 0 0 0 0 1 0 1 1 2 0 2 2 3 1 0 0 4 1 1 1 5 1 2 2 6 2 0 0 You can take integer quotient times divisor and add in the remainder to get the dividend, on either side of 0. You can't do that with true modulo. WebManWalkingAug 10, 2011 at 8:03 PM 290 Comments Aw, shucks. I wish you would allow pre in addition to code. AnnaAug 15, 2011 at 11:41 AM 383 Comments @Ben, I forgot to mention that I like the other operators, too. It's just that by the time I got to the mod operator, I had used the others so much, I had just grown kind of tired of them. They were kind of like old news at that point. But I still like them. Just think the mod is super-cool. (and can be helpful for a lot of solutions to different problems you can't really solve easily in other ways). RandallAug 18, 2011 at 9:41 AM 167 Comments We use MOD to place things in row, three columns wide. Thus, if there are eight items, the output ends up looking like: 1 2 3 4 5 6 7 8 is that what prompted this post, Ben? Ben NadelOct 29, 2011 at 10:02 PM 16,080 Comments @WebManWalking, It's funny you mention that - I almost never think about numbers in the negative. I know that sounds odd; but, I can't even remember the last time that I really had to work with negative numbers. ... I am sure that is a ridiculous thought - I am sure I use them all the time :) @Randall, That's the kind of situation where I really started to use MOD for the first time on a regular basis. When I was using Table tags to lay things out, especially, I remember I used to use the MOD operator to determine where I should add the closing / opening TR tags. Paul RoweFeb 13, 2013 at 3:22 PM 29 Comments Another particularly useful way to use the MOD operator is to break items up into an arbitrary number of columns. Say you have an unknown number of elements and you want to display them in three reasonably even columns, you would divide the number of elements by the number of columns, round up or down (so you're working with an integer), and start a new column when your MODULUS is zero. Ben NadelFeb 14, 2013 at 9:51 AM 16,080 Comments @Paul, I believe that splitting columns and rows was actually one of the first things I ever used it for in production. I vaguely remember having code like: ``` .. Start new row. ``` ... something like that. Paul RoweFeb 14, 2013 at 10:38 AM 29 Comments @Ben That's about what the code would look like. I was thinking about, say, three left-floating DIVs with unordered lists containing the items you want to display. In that situation, your modulus divisor is not constant, but calculated based on the number of records. I suppose you could also do this sort of thing with news articles or classified ads based on the number of "inches" for each article or some other arbitrary unit of measuring length, thus simulating the columnar layout of a newspaper. Babak ShokouhiOct 13, 2014 at 4:59 PM 1 Comments How about a sequence like this? How would you go about it? 45678123 Or 891234567 Oh my chickens, this post is old! Hit me up on Twitter if you want to discuss it further. I believe in love. I believe in compassion. I believe in human rights. I believe that we can afford to give more of these gifts to the world around us because it _costs us nothing_to be decent and kind and understanding. And, I want you to know that when you land on this site, you are accepted for who you are, no matter how you identify, what truths you live, or whatever kind of goofy shit makes you feel alive! Rock on with your bad self! — Ben Nadel Ben Nadel © 2025. All content is the property of Ben Nadel. For use of code see license. Back to Top About Ben Nadel I am the co-founder and a principal engineer at InVision App, Inc — the world's leading online whiteboard and productivity platform powering the future of work. I also rock out in JavaScript and ColdFusion 24x7 and I dream about chained Promises resolving asynchronously. Social Links GitHub Twitter LinkedIn Facebook Site Links Home Activity Podcast Projects About Me Contact Me People Utilities InVision RSS Feed My Other Projects Big Sexy Poems Dig Deep Fitness Incident Commander Feature Flags (Book)
73	https://stackoverflow.com/questions/18790284/how-to-calculate-the-number-of-coprime-subsets-of-the-set-1-2-3-n	algorithm - How to calculate the number of coprime subsets of the set {1,2,3,..,n} - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to calculate the number of coprime subsets of the set {1,2,3,..,n} Ask Question Asked 12 years ago Modified3 years, 11 months ago Viewed 22k times This question shows research effort; it is useful and clear 28 Save this question. Show activity on this post. I am solving this task (problem I). The statement is: How many subsets of the set {1, 2, 3, ..., n} are coprime? A set of integers is called coprime if every two of its elements are coprime. Two integers are coprime if their greatest common divisor equals 1. Input First line of input contains two integers n and m (1 <= n <= 3000, 1 <= m <= 10^9 + 9) Output Output the number of coprime subsets of {1, 2, 3, ..., n} modulo m. Example input: 4 7 output: 5 There are 12 coprime subsets of {1,2,3,4}: {}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {3,4}, {1,2,3}, {1,3,4}. I think it can be solved by using prime numbers. (keeping track of if we used each prime numbers) ..but I'm not sure. Can I get some hints to solve this task? You can found this sequence at here : algorithm integer primes Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Sep 21, 2013 at 8:50 Love PaperLove Paper asked Sep 13, 2013 at 15:42 Love PaperLove Paper 481 1 1 gold badge 4 4 silver badges 16 16 bronze badges 36 1 If this problem is too hard, try solving a simpler problem. How many pairs of numbers less than or equal to n are coprime? Or simpler still: how many numbers less than n are coprime to n?Colonel Panic –Colonel Panic 2013-09-17 15:47:16 +00:00 Commented Sep 17, 2013 at 15:47 2 The coprime requirement immediately made me think of the Euler totient 2013-09-20 12:19:49 +00:00 Commented Sep 20, 2013 at 12:19 1 Here is the relevant paper. math.clemson.edu/~calkin/Papers/calkin_granville.pdf I believe theorem 2 is what you are looking for, good luck.Percy Fawcett –Percy Fawcett 2013-09-21 02:20:12 +00:00 Commented Sep 21, 2013 at 2:20 2 @octref, I just found this from the OEIS(A084422, and this is the table of the sequence. For example, you can see that there are 374855124868136960 coprime subsets when n = 200.Love Paper –Love Paper 2013-09-21 08:49:50 +00:00 Commented Sep 21, 2013 at 8:49 3 @BrettHale Well, if you think of a set being coprime if there does not exist two distinct elements with gcd > 1, singleton and and empty sets would work.Dennis Meng –Dennis Meng 2013-09-22 16:15:45 +00:00 Commented Sep 22, 2013 at 16:15 \|Show 31 more comments 9 Answers 9 Sorted by: Reset to default This answer is useful 17 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by Love Paper Show activity on this post. Okay, here’s the goods. The C program that follows gets n=3000 in less than 5 seconds for me. My hat’s off to the team(s) that solved this problem in a competitive setting. The algorithm is based on the idea of treating the small and large primes differently. A prime is small if its square is at most n. Otherwise, it’s large. Observe that each number less than or equal to n has at most one large prime factor. We make a table indexed by pairs. The first component of each pair specifies the number of large primes in use. The second component of each pair specifies the set of small primes in use. The value at a particular index is the number of solutions with that usage pattern not containing 1 or a large prime (we count those later by multiplying by the appropriate power of 2). We iterate downward over numbers j with no large prime factors. At the beginning of each iteration, the table contains the counts for subsets of j..n. There are two additions in the inner loop. The first accounts for extending subsets by j itself, which does not increase the number of large primes in use. The second accounts for extending subsets by j times a large prime, which does. The number of suitable large primes is the number of large primes not greater than n/j, minus the number of large primes already in use, since the downward iteration implies that each large prime already in use is not greater than n/j. At the end, we sum the table entries. Each subset counted in the table gives rise to 2 k subsets where k is one plus the number of unused large primes, as 1 and each unused large prime can be included or excluded independently. ```cpp / assumes int, long are 32, 64 bits respectively / include include enum { NMAX = 3000 }; static int n; static long m; static unsigned smallfactors[NMAX + 1]; static int prime[NMAX - 1]; static int primecount; static int smallprimecount; static int largeprimefactor[NMAX + 1]; static int largeprimecount[NMAX + 1]; static long table; static void eratosthenes(void) { int i; for (i = 2; i i <= n; i++) { int j; if (smallfactors[i]) continue; for (j = i; j <= n; j += i) smallfactors[j] \|= 1U << primecount; prime[primecount++] = i; } smallprimecount = primecount; for (; i <= n; i++) { if (!smallfactors[i]) prime[primecount++] = i; } if (0) { int k; for (k = 0; k < primecount; k++) printf("%d\n", prime[k]); } } static void makelargeprimefactor(void) { int i; for (i = smallprimecount; i < primecount; i++) { int p = prime[i]; int j; for (j = p; j <= n; j += p) largeprimefactor[j] = p; } } static void makelargeprimecount(void) { int i = 1; int j; for (j = primecount; j > smallprimecount; j--) { for (; i <= n / prime[j - 1]; i++) { largeprimecount[i] = j - smallprimecount; } } if (0) { for (i = 1; i <= n; i++) printf("%d %d\n", i, largeprimecount[i]); } } static void maketable(void) { int i; int j; table = calloc(smallprimecount + 1, sizeof table); for (i = 0; i <= smallprimecount; i++) { table[i] = calloc(1U << smallprimecount, sizeof table[i]); } table[0U] = 1L % m; for (j = n; j >= 2; j--) { int lpc = largeprimecount[j]; unsigned sf = smallfactors[j]; if (largeprimefactor[j]) continue; for (i = 0; i < smallprimecount; i++) { long cur = table[i]; long next = table[i + 1]; unsigned f; for (f = sf; f < (1U << smallprimecount); f = (f + 1U) \| sf) { cur[f] = (cur[f] + cur[f & ~sf]) % m; } if (lpc - i <= 0) continue; for (f = sf; f < (1U << smallprimecount); f = (f + 1U) \| sf) { next[f] = (next[f] + cur[f & ~sf] (lpc - i)) % m; } } } } static long timesexp2mod(long x, int y) { long z = 2L % m; for (; y > 0; y >>= 1) { if (y & 1) x = (x z) % m; z = (z z) % m; } return x; } static long computetotal(void) { long total = 0L; int i; for (i = 0; i <= smallprimecount; i++) { unsigned f; for (f = 0U; f < (1U << smallprimecount); f++) { total = (total + timesexp2mod(table[i][f], largeprimecount - i + 1)) % m; } } return total; } int main(void) { scanf("%d%ld", &n, &m); eratosthenes(); makelargeprimefactor(); makelargeprimecount(); maketable(); if (0) { int i; for (i = 0; i < 100; i++) { printf("%d %ld\n", i, timesexp2mod(1L, i)); } } printf("%ld\n", computetotal()); return EXIT_SUCCESS; } ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Oct 13, 2021 at 11:51 Athari 33.9k 17 17 gold badges 111 111 silver badges 152 152 bronze badges answered Sep 25, 2013 at 3:45 David EisenstatDavid Eisenstat 65.7k 7 7 gold badges 66 66 silver badges 126 126 bronze badges 5 Comments Add a comment justhalf justhalfOver a year ago I don't get your step 4. Why {7, 9, 10} becomes {9,2}? 2013-09-25T03:51:17.723Z+00:00 0 Reply Copy link David Eisenstat David EisenstatOver a year ago @justhalf It becomes {1, 9, 2} after dividing out the large primes 7 and 5, then {9, 2}. 2013-09-25T13:51:18.497Z+00:00 1 Reply Copy link TooTone TooToneOver a year ago good going-- I'm more impressed that you got this working on your own than in a team. 2013-09-25T20:24:02.91Z+00:00 0 Reply Copy link Love Paper Love PaperOver a year ago Great! I could understand your explanation. Thanks for your reply:) 2013-09-25T21:07:27.39Z+00:00 0 Reply Copy link גלעד ברקן גלעד ברקןOver a year ago It seems like a bit of old news - although this problem has held my curiosity for the better part of a month - but could you help me understand how your program (efficiently) counts the "number of solutions with a particular usage pattern?" I haven't been able to get my head around it. In any case, reading it is teaching me new things. Very smart. 2013-10-15T13:56:41.527Z+00:00 1 Reply Copy link Add a comment This answer is useful 8 Save this answer. Show activity on this post. Here's an answer which goes through the first 200 elements in the sequence in less than a second, giving the right answer 200 → 374855124868136960. With optimizations (see edit 1), it can calculate the first 500 entries in under 90s, which is quick -- although @David Eisenstat's answer is likely to be better if it can be developed. I think takes a different approach to the algorithms given so far, including my own original answer, so I'm posting it separately. After optimizing, I realized that I was really coding up a graph problem so I rewrote the solution as a graph implementation (see edit 2). The graph implementation allows some more optimizations, is a lot more elegant, more than an order of magnitude faster, and scales better: it calculates f(600) in 1.5s, compared to 27s. The main idea here is to use a recursion relationship. For any set, the number of subsets meeting the criterion are the sum of: the number of subsets with one element removed; and the number of subsets with that element definitely included. In the second case, when the element is definitely included, any other elements which aren't coprime with it must be removed. Efficiency issues: I've chosen to remove the largest element to maximize the chance of that element already being coprime to all the others, in which case only one rather than two recursive calls need to be made. Caching / memoization helps. Code below. ```cpp include include include include include include include const int PRIMES[] = // { 2, 3, 5, ... ..., 2969, 2971, 2999 }; const int NPRIMES = sizeof(PRIMES) / sizeof(int); typedef std::set intset; typedef std::vector intsetvec; const int MAXCALC = 200; // answer at intsetvec primeFactors(MAXCALC +1); typedef std::vector intvec; // Caching / memoization typedef std::map intvec2dbl; intvec2dbl set2NumCoPrimeSets; double NumCoPrimeSets(const intvec& set) { if (set.empty()) return 1; // Caching / memoization const intvec2dbl::const_iterator i = set2NumCoPrimeSets.find(set); if (i != set2NumCoPrimeSets.end()) return i->second; // Result is the number of coprime sets in: // setA, the set that definitely has the first element of the input present // + setB, the set the doesn't have the first element of the input present // Because setA definitely has the first element, we remove elements it isn't coprime with // We also remove the first element: as this is definitely present it doesn't make any // difference to the number of sets intvec setA(set); const int firstNum = setA.begin(); const intset& factors = primeFactors[firstNum]; for(int factor : factors) { setA.erase(std::remove_if(setA.begin(), setA.end(), [factor] (int i) { return i % factor == 0; } ), setA.end()); } // If the first element was already coprime with the rest, then we have setA = setB // and we can do a single call (m=2). Otherwise we have two recursive calls. double m = 1; double c = 0; assert(set.size() - setA.size() > 0); if (set.size() - setA.size() > 1) { intvec setB(set); setB.erase(setB.begin()); c = NumCoPrimeSets(setB); } else { // first elt coprime with rest m = 2; } const double numCoPrimeSets = m NumCoPrimeSets(setA) + c; // Caching / memoization set2NumCoPrimeSets.insert(intvec2dbl::value_type(set, numCoPrimeSets)); return numCoPrimeSets; } int main(int argc, char argv[]) { // Calculate prime numbers that factor into each number upto MAXCALC primeFactors.insert(1); // convenient for(int i=2; i<=MAXCALC; ++i) { for(int j=0; j<NPRIMES; ++j) { if (i % PRIMES[j] == 0) { primeFactors[i].insert(PRIMES[j]); } } } const clock_t start = clock(); for(int n=1; n<=MAXCALC; ++n) { intvec v; for(int i=n; i>0; --i) { // reverse order to reduce recursion v.push_back(i); } const clock_t now = clock(); const clock_t ms = now - start; const double numCoPrimeSubsets = NumCoPrimeSets(v); std::cout << n << ", " << std::fixed << numCoPrimeSubsets << ", " << ms << "\n"; } return 0; } ``` Time characteristics look a lot better than my first answer. But still won't go upto 3000 in 5s! Edit 1 There are some interesting optimizations that can be made to this method. Overall this gives a 4x improvement for larger n. All numbers in the set that are already coprime can be removed in one single preprocessing step: if m number are removed, then the original set has 2 m factor times more combinations than the reduced one (because for each coprime, you can either have it in or out of the set independently of other elements). Most importantly, it is possible to choose an element to remove that is anywhere in the set. It turns out that removing the most connected element works best. The recursive relationship that was previously used can be generalized to remove more than one element where all elements removed have the same prime factors. E.g. for the set {2, 3, 15, 19, 45}, the numbers 15 and 45 have the same prime factors of 3 and 5. There are 2 numbers removed at once, and so the number of subsets for {2, 3, 15, 19, 45} = twice the number of combinations for either 15 or 45 present (for the set {2, 19} because 3 has to be absent if either 15 or 45 are present) + the number of subsets for 15 and 45 absent (for the set {2, 3, 19}) Using short for the number type improved performance by about 10%. Finally, it is also possible to transform sets into sets with equivalent prime factors, in the hope of getting better cache hits by standardizing the sets. E.g., { 3, 9, 15} is equivalent (isomorphic) to 2, 4, 6. This was the most radical idea but probably had the least effect on performance. It's probably a lot easier to understand it with a concrete example. I've chosen the set {1..12} which is large enough to get a feel for how it works but small enough so that it's comprehensible. ```cpp NumCoPrimeSets({ 1 2 3 4 5 6 7 8 9 10 11 12 }) Removed 3 coprimes, giving set { 2 3 4 5 6 8 9 10 12 } multiplication factor now 8 Removing the most connected number 12 with 8 connections To get setA, remove all numbers which have any of the prime factors { 2 3 } setA = { 5 } To get setB, remove 2 numbers which have exactly the prime factors { 2 3 } setB = { 2 3 4 5 8 9 10 } Recursing on 2 NumCoPrimeSets(setA) + NumCoPrimeSets(setB) NumCoPrimeSets({ 5 }) Base case return the multiplier, which is 2 NumCoPrimeSets({ 2 3 4 5 8 9 10 }) Removing the most connected number 10 with 4 connections To get setA, remove all numbers which have any of the prime factors { 2 5 } setA = { 3 9 } To get setB, remove 1 numbers which have exactly the prime factors { 2 5 } setB = { 2 3 4 5 8 9 } Recursing on 1 NumCoPrimeSets(setA) + NumCoPrimeSets(setB) NumCoPrimeSets({ 3 9 }) Transformed 2 primes, giving new set { 2 4 } Removing the most connected number 4 with 1 connections To get setA, remove all numbers which have any of the prime factors { 2 } setA = { } To get setB, remove 2 numbers which have exactly the prime factors { 2 } setB = { } Recursing on 2 NumCoPrimeSets(setA) + NumCoPrimeSets(setB) NumCoPrimeSets({ }) Base case return the multiplier, which is 1 NumCoPrimeSets({ }) Base case return the multiplier, which is 1 Returned from recursing on 2 NumCoPrimeSets({ }) + NumCoPrimeSets({ }) Caching for{ 2 4 }: 3 = 2 1 + 1 Returning for{ 3 9 }: 3 = 1 3 NumCoPrimeSets({ 2 3 4 5 8 9 }) Removed 1 coprimes, giving set { 2 3 4 8 9 } multiplication factor now 2 Removing the most connected number 8 with 2 connections To get setA, remove all numbers which have any of the prime factors { 2 } setA = { 3 9 } To get setB, remove 3 numbers which have exactly the prime factors { 2 } setB = { 3 9 } Recursing on 3 NumCoPrimeSets(setA) + NumCoPrimeSets(setB) NumCoPrimeSets({ 3 9 }) Transformed 2 primes, giving new set { 2 4 } Cache hit, returning 3 = 1 3 NumCoPrimeSets({ 3 9 }) Transformed 2 primes, giving new set { 2 4 } Cache hit, returning 3 = 1 3 Returned from recursing on 3 NumCoPrimeSets({ 3 9 }) + NumCoPrimeSets({ 3 9 }) Caching for{ 2 3 4 8 9 }: 12 = 3 3 + 3 Returning for{ 2 3 4 5 8 9 }: 24 = 2 12 Returned from recursing on 1 NumCoPrimeSets({ 3 9 }) + NumCoPrimeSets({ 2 3 4 5 8 9 }) Caching for{ 2 3 4 5 8 9 10 }: 27 = 1 3 + 24 Returning for{ 2 3 4 5 8 9 10 }: 27 = 1 27 Returned from recursing on 2 NumCoPrimeSets({ 5 }) + NumCoPrimeSets({ 2 3 4 5 8 9 10 }) Caching for{ 2 3 4 5 6 8 9 10 12 }: 31 = 2 2 + 27 Returning for{ 1 2 3 4 5 6 7 8 9 10 11 12 }: 248 = 8 31 ``` Code below ```cpp include include include include include include include include include include typedef short numtype; const numtype PRIMES[] = // ... const numtype NPRIMES = sizeof(PRIMES) / sizeof(numtype); typedef std::set numset; typedef std::vector numsetvec; const numtype MAXCALC = 200; // answer at numsetvec primeFactors(MAXCALC +1); typedef std::vector numvec; // Caching / memoization typedef std::map numvec2dbl; numvec2dbl set2NumCoPrimeSets; double NumCoPrimeSets(const numvec& initialSet) { // Preprocessing step: remove numbers which are already coprime typedef std::unordered_map num2numvec; num2numvec prime2Elts; for(numtype num : initialSet) { const numset& factors = primeFactors[num]; for(numtype factor : factors) { prime2Elts[factor].push_back(num); } } numset eltsToRemove(initialSet.begin(), initialSet.end()); typedef std::vector<std::pair<numtype,int>> numintvec; numvec primesRemaining; for(const num2numvec::value_type& primeElts : prime2Elts) { if (primeElts.second.size() > 1) { for (numtype num : primeElts.second) { eltsToRemove.erase(num); } primesRemaining.push_back(primeElts.first); } } double mult = pow(2.0, eltsToRemove.size()); if (eltsToRemove.size() == initialSet.size()) return mult; // Do the removal by creating a new set numvec set; for(numtype num : initialSet) { if (eltsToRemove.find(num) == eltsToRemove.end()) { set.push_back(num); } } // Transform to use a smaller set of primes before checking the cache // (beta code but it seems to work, mostly!) std::sort(primesRemaining.begin(), primesRemaining.end()); numvec::const_iterator p = primesRemaining.begin(); for(int j=0; p!= primesRemaining.end() && j<NPRIMES; ++p, ++j) { const numtype primeRemaining = p; if (primeRemaining != PRIMES[j]) { for(numtype& num : set) { while (num % primeRemaining == 0) { num = num / primeRemaining PRIMES[j]; } } } } // Caching / memoization const numvec2dbl::const_iterator i = set2NumCoPrimeSets.find(set); if (i != set2NumCoPrimeSets.end()) return mult i->second; // Remove the most connected number typedef std::unordered_map<numtype, int> num2int; num2int num2ConnectionCount; for(numvec::const_iterator srcIt=set.begin(); srcIt!=set.end(); ++srcIt) { const numtype src = srcIt; const numset& srcFactors = primeFactors[src]; for(numvec::const_iterator tgtIt=srcIt +1; tgtIt!=set.end(); ++tgtIt) { for(numtype factor : srcFactors) { const numtype tgt = tgtIt; if (tgt % factor == 0) { num2ConnectionCount[src]++; num2ConnectionCount[tgt]++; } } } } num2int::const_iterator connCountIt = num2ConnectionCount.begin(); numtype numToErase = connCountIt->first; int maxConnCount = connCountIt->second; for (; connCountIt!=num2ConnectionCount.end(); ++connCountIt) { if (connCountIt->second > maxConnCount \|\| connCountIt->second == maxConnCount && connCountIt->first > numToErase) { numToErase = connCountIt->first; maxConnCount = connCountIt->second; } } // Result is the number of coprime sets in: // setA, the set that definitely has a chosen element of the input present // + setB, the set the doesn't have the chosen element(s) of the input present // Because setA definitely has a chosen element, we remove elements it isn't coprime with // We also remove the chosen element(s): as they are definitely present it doesn't make any // difference to the number of sets numvec setA(set); const numset& factors = primeFactors[numToErase]; for(numtype factor : factors) { setA.erase(std::remove_if(setA.begin(), setA.end(), [factor] (numtype i) { return i % factor == 0; } ), setA.end()); } // setB: remove all elements which have the same prime factors numvec setB(set); setB.erase(std::remove_if(setB.begin(), setB.end(), [&factors] (numtype i) { return primeFactors[i] == factors; }), setB.end()); const size_t numEltsWithSamePrimeFactors = (set.size() - setB.size()); const double numCoPrimeSets = numEltsWithSamePrimeFactors NumCoPrimeSets(setA) + NumCoPrimeSets(setB); // Caching / memoization set2NumCoPrimeSets.insert(numvec2dbl::value_type(set, numCoPrimeSets)); return mult numCoPrimeSets; } int main(int argc, char argv[]) { // Calculate prime numbers that factor into each number upto MAXCALC for(numtype i=2; i<=MAXCALC; ++i) { for(numtype j=0; j<NPRIMES; ++j) { if (i % PRIMES[j] == 0) { primeFactors[i].insert(PRIMES[j]); } } } const clock_t start = clock(); std::ofstream fout("out.txt"); for(numtype n=0; n<=MAXCALC; ++n) { numvec v; for(numtype i=1; i<=n; ++i) { v.push_back(i); } const clock_t now = clock(); const clock_t ms = now - start; const double numCoPrimeSubsets = NumCoPrimeSets(v); fout << n << ", " << std::fixed << numCoPrimeSubsets << ", " << ms << "\n"; std::cout << n << ", " << std::fixed << numCoPrimeSubsets << ", " << ms << "\n"; } return 0; } ``` It's possible to process upto n=600 in about 5 minutes. Time still looks exponential however, doubling every 50 to 60 n or so. The graph for calculating just one n is shown below. Edit 2 This solution is much more naturally implemented in terms of a graph. Two more optimizations arose: Most importantly, if the graph G can be partitioned into two sets A and B such that there are no connections between A and B, then coprimes(G) = coprimes(A) coprimes(B). Secondly, it's possible to collapse all numbers for a set of prime factors into a single node, so the value for the node is the count of numbers for its prime factors. In the code below the Graph class holds the original adjacency matrix and the node values, and the FilteredGraph class holds the current list of remaining nodes as a bitset so that as nodes are removed, the new adjacency matrix can be calculated by bit masking (and there is relatively little data to pass in the recursion). ```cpp include "Primes.h" include include include include include include include include include // Graph declaration const int MAXGROUPS = 1462; // empirically determined class Graph { typedef std::bitset bitset; typedef std::vector adjmatrix; typedef std::vector intvec; public: Graph(int numNodes) : m_nodeValues(numNodes), m_adjMatrix(numNodes) {} void SetNodeValue(int i, int v) { m_nodeValues[i] = v; } void SetConnection(int i, int j) { m_adjMatrix[i][j] = true; m_adjMatrix[j][i] = true; } int size() const { return m_nodeValues.size(); } private: adjmatrix m_adjMatrix; intvec m_nodeValues; friend class FilteredGraph; }; class FilteredGraph { typedef Graph::bitset bitset; public: FilteredGraph(const Graph unfiltered); int FirstNode() const; int RemoveNode(int node); void RemoveNodesConnectedTo(int node); double RemoveDisconnectedNodes(); bool AttemptPartition(FilteredGraph FilteredGraph); size_t Hash() const { return std::hash<bitset>()(m_includedNodes); } bool operator==(const FilteredGraph& x) const { return x.m_includedNodes == m_includedNodes && x.m_unfiltered == m_unfiltered; } private: bitset RawAdjRow(int i) const { return m_unfiltered->m_adjMatrix[i]; } bitset AdjRow(int i) const { return RawAdjRow(i) & m_includedNodes; } int NodeValue(int i) const { return m_unfiltered->m_nodeValues[i]; } const Graph m_unfiltered; bitset m_includedNodes; }; // Cache namespace std { template<> class hash { public: size_t operator()(const FilteredGraph & x) const { return x.Hash(); } }; } typedef std::unordered_map graph2double; graph2double cache; // MAIN FUNCTION double NumCoPrimesSubSets(const FilteredGraph& graph) { graph2double::const_iterator cacheIt = cache.find(graph); if (cacheIt != cache.end()) return cacheIt->second; double rc = 1; FilteredGraph A(graph); FilteredGraph B(graph); if (A.AttemptPartition(&B)) { rc = NumCoPrimesSubSets(A); A = B; } const int nodeToRemove = A.FirstNode(); if (nodeToRemove < 0) // empty graph return 1; // Graph B is the graph with a node removed B.RemoveNode(nodeToRemove); // Graph A is the graph with the node present -- and hence connected nodes removed A.RemoveNodesConnectedTo(nodeToRemove); // The number of numbers in the node is the number of times it can be reused const double removedNodeValue = A.RemoveNode(nodeToRemove); const double A_disconnectedNodesMult = A.RemoveDisconnectedNodes(); const double B_disconnectedNodesMult = B.RemoveDisconnectedNodes(); const double A_coprimes = NumCoPrimesSubSets(A); const double B_coprimes = NumCoPrimesSubSets(B); rc = removedNodeValue A_disconnectedNodesMult A_coprimes + B_disconnectedNodesMult B_coprimes; cache.insert(graph2double::value_type(graph, rc)); return rc; } // Program entry point int Sequence2Graph(Graph ppGraph, int n); int main(int argc, char argv[]) { const clock_t start = clock(); int n=800; // runs in approx 6s on my machine Graph pGraph = nullptr; const int coPrimesRemoved = Sequence2Graph(&pGraph, n); const double coPrimesMultiplier = pow(2,coPrimesRemoved); const FilteredGraph filteredGraph(pGraph); const double numCoPrimeSubsets = coPrimesMultiplier NumCoPrimesSubSets(filteredGraph); delete pGraph; cache.clear(); // as it stands the cache can't cope with other Graph objects, e.g. for other n const clock_t now = clock(); const clock_t ms = now - start; std::cout << n << ", " << std::fixed << numCoPrimeSubsets << ", " << ms << "\n"; return 0; } // Graph implementation FilteredGraph::FilteredGraph(const Graph unfiltered) : m_unfiltered(unfiltered) { for(int i=0; isize(); ++i) { m_includedNodes.set(i); } } int FilteredGraph::FirstNode() const { int firstNode=0; for(; firstNodesize() && !m_includedNodes.test(firstNode); ++firstNode) { } if (firstNode == m_unfiltered->size()) return -1; return firstNode; } int FilteredGraph::RemoveNode(int node) { m_includedNodes.set(node, false); return NodeValue(node); } void FilteredGraph::RemoveNodesConnectedTo(const int node) { const bitset notConnected = ~RawAdjRow(node); m_includedNodes &= notConnected; } double FilteredGraph::RemoveDisconnectedNodes() { double mult = 1.0; for(int i=0; isize(); ++i) { if (m_includedNodes.test(i)) { const int conn = AdjRow(i).count(); if (conn == 0) { m_includedNodes.set(i, false);; mult = (NodeValue(i) +1); } } } return mult; } bool FilteredGraph::AttemptPartition(FilteredGraph pOther) { typedef std::vector intvec; intvec includedNodesCache; includedNodesCache.reserve(m_unfiltered->size()); for(int i=0; isize(); ++i) { if (m_includedNodes.test(i)) { includedNodesCache.push_back(i); } } if (includedNodesCache.empty()) return false; const int startNode= includedNodesCache; bitset currFoundNodes; currFoundNodes.set(startNode); bitset foundNodes; do { foundNodes \|= currFoundNodes; bitset newFoundNodes; for(int i : includedNodesCache) { if (currFoundNodes.test(i)) { newFoundNodes \|= AdjRow(i); } } newFoundNodes &= ~ foundNodes; currFoundNodes = newFoundNodes; } while(currFoundNodes.count() > 0); const size_t foundCount = foundNodes.count(); const size_t thisCount = m_includedNodes.count(); const bool isConnected = foundCount == thisCount; if (!isConnected) { if (foundCount < thisCount) { pOther->m_includedNodes = foundNodes; m_includedNodes &= ~foundNodes; } else { pOther->m_includedNodes = m_includedNodes; pOther->m_includedNodes &= ~foundNodes; m_includedNodes = foundNodes; } } return !isConnected; } // Initialization code to convert sequence from 1 to n into graph typedef short numtype; typedef std::set numset; bool setIntersect(const numset& setA, const numset& setB) { for(int a : setA) { if (setB.find(a) != setB.end()) return true; } return false; } int Sequence2Graph(Graph ppGraph, int n) { typedef std::map numset2int; numset2int factors2count; int coPrimesRemoved = n>0; // for {1} // Calculate all sets of prime factors, and how many numbers belong to each set for(numtype i=2; i<=n; ++i) { if ((i > n/2) && (std::find(PRIMES, PRIMES+NPRIMES, i) !=PRIMES+NPRIMES)) { ++coPrimesRemoved; } else { numset factors; for(numtype j=0; j<NPRIMES && PRIMES[j]<n; ++j) { if (i % PRIMES[j] == 0) { factors.insert(PRIMES[j]); } } factors2count[factors]++; } } // Create graph Graph& pGraph = ppGraph; pGraph = new Graph(factors2count.size()); int srcNodeNum = 0; for(numset2int::const_iterator i = factors2count.begin(); i!=factors2count.end(); ++i) { pGraph->SetNodeValue(srcNodeNum, i->second); numset2int::const_iterator j = i; int tgtNodeNum = srcNodeNum+1; for(++j; j!=factors2count.end(); ++j) { if (setIntersect(i->first, j->first)) { pGraph->SetConnection(srcNodeNum, tgtNodeNum); } ++tgtNodeNum; } ++srcNodeNum; } return coPrimesRemoved; } ``` The graph for calculating coprimes(n) is shown below in red (with the old approach in black). Based on the (exponential) rate of increase observed, the prediction for n=3000 is 30 hours, assuming that the program doesn't blow up. This is starting to look computationally feasible, especially with more optimizations, but is nowhere near the 5s that is required! No doubt the required solution is short and sweet, but this has been fun... Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited May 23, 2017 at 12:33 CommunityBot 1 1 1 silver badge answered Sep 23, 2013 at 4:10 TooToneTooTone 8,247 5 5 gold badges 40 40 silver badges 71 71 bronze badges Comments Add a comment This answer is useful 2 Save this answer. Show activity on this post. Here's something rather straightforward in Haskell, which takes about 2 seconds for n=200 and slows exponentially. ``` {-# OPTIONS_GHC -O2 #-} f n = 2^(length second + 1) (g [] first 0) where second = filter (\x -> isPrime x && x > div n 2) [2..n] first = filter (flip notElem second) [2..n] isPrime k = null [ x \| x <- [2..floor . sqrt . fromIntegral $ k], k modx == 0] g s rrs depth \| null rrs = 2^(length s - depth) \| not $ and (map ((==1) . gcd r) s) = g s rs depth + g s' rs' (depth + 1) \| otherwise = g (r:s) rs depth where r:rs = rrs s' = r : filter ((==1) . gcd r) s rs' = filter ((==1) . gcd r) rs ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Oct 14, 2013 at 3:06 community wiki 23 revsgroovy 9 Comments Add a comment David Eisenstat David EisenstatOver a year ago This reminds me a bit of my first attempt. I suspect that there will be far too many partitions of the primes for large n. 2013-09-28T14:41:20.547Z+00:00 0 Reply Copy link גלעד ברקן גלעד ברקןOver a year ago @DavidEisenstat thanks for checking it out. haskell: length $ factorsets 3000 => 1823 (different powers of one prime are counted as one factorset) Wouldn't that mean summing less than 1823 overlapping sets of at most length 431? 2013-09-28T14:51:03.287Z+00:00 0 Reply Copy link David Eisenstat David EisenstatOver a year ago I guess I've failed to infer your algorithm from examples. What do the overlapping sets look like for n = 20? 2013-09-28T14:59:13.617Z+00:00 0 Reply Copy link David Eisenstat David EisenstatOver a year ago Wait, what was that other number? Is the number of terms larger than the factor sets? 2013-09-28T15:44:40.587Z+00:00 0 Reply Copy link גלעד ברקן גלעד ברקןOver a year ago @DavidEisenstat I think I got what you mean...a factorset of 2 3 would be the same as 2 2 3 in terms of whether it can be grouped together with, say 5 7. That's what I meant by 1823 factorsets. 2013-09-28T16:13:48.487Z+00:00 0 Reply Copy link Add a comment\|Show 4 more comments This answer is useful 1 Save this answer. Show activity on this post. Here's an approach that gets the given sequence upto n=62 in under 5s (with optimizations it runs n=75 in 5s, however note my second attempt at this problem does better). I'm assuming the modulo part of the problem is just to do with avoiding numerical errors as the function gets large, so I'm ignoring it for now. The approach is based on the fact that we can choose at most one number in a subset for each prime. We start with the first prime, 2. If we don't include 2, then we have 1 combination for this prime. If we do include 2, then we have as many combinations as they are numbers divisible by 2. Then we go onto the second prime, 3, and decide whether or not to include that. If we don't include it, we have 1 combination for this prime. If we do include 2, then we have as many combinations as they are numbers divisible by 3. ... and so on. Taking the example {1,2,3,4}, we map the numbers in the set onto primes as follows. I've included 1 as a prime as it makes the exposition easier at this stage. cpp 1 → {1} 2 → {2,4} 3 → {3} We have 2 combinations for the "prime" 1 (don't include it or 1), 3 combinations for the prime 2 (don't include it or 2 or 4), and 2 combinations for 3 (don't include it or 3). So the number of subsets is 2 3 2 = 12. Similarly for {1,2,3,4,5} we have cpp 1 → {1} 2 → {2,4} 3 → {3} 5 → {5} giving 2 3 2 2= 24. But for {1,2,3,4,5,6}, things aren't so straightforward. We have cpp 1 → {1} 2 → {2,4,6} 3 → {3} 5 → {5} but if we choose the number 6 for the prime 2, we can't choose a number for the prime 3 (as a footnote, in my first approach, which I may come back to, I treated this as though the choices for 3 were cut in half when we chose 6, so I used 3.5 rather than 4 for the number of combinations for the prime 2 and 2 3.5 2 2 = 28 gave the right answer. I couldn't get this approach to work beyond n=17, however.) The way I dealt with this is to split up the processing for each set of prime factors at each level. So {2,4} have prime factors {2}, whereas {6} has prime factors {2,3}. Omitting the spurious entry for 1 (which isn't a prime), we now have cpp 2 → {{2}→{2,4}, {2,3}→6} 3 → {{3}→{3}} 5 → {{5}→{5}} Now there are three paths to calculate the number of combinations, one path where we don't select the prime 2, and two paths where we do: through {2}→{2,4} and through {2,3}→6. The first path has 1 2 2 = 4 combinations because we can either select 3 or not, and we can either select 5 or not. Similarly the second has 2 2 2 = 8 combinations noting that we can choose either 2 or 4. The third has 1 1 2 = 2 combinations, because we only have one choice for the prime 3 -- not to use it. In total this gives us 4 + 8 + 2 = 14 combinations (as an optimization note that the first and second paths could have been collapsed together to get 3 2 2 = 12). We also have the choice of selecting 1 or not, so the total number of combinations is 2 14 = 28. C++ code to recursively run through the paths is below. (It's C++11, written on Visual Studio 2012, but ought to work on other gcc as I haven't included anything platform specific). ```cpp include include include include include include include const int PRIMES[] = // { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199 }; const int NPRIMES = sizeof(PRIMES) / sizeof(int); typedef std::vector intvec; typedef std::set intset; typedef std::vector> intsetvec; struct FactorSetNumbers { intset factorSet; intvec numbers; // we only need to store numbers.size(), but nice to see the vec itself FactorSetNumbers() {} FactorSetNumbers(const intset& factorSet_, int n) : factorSet(factorSet_) { numbers.push_back(n); } }; typedef std::vector factorset2numbers; typedef std::vector factorset2numbersArray; double NumCoPrimeSubsets( const factorset2numbersArray& factorSet2Numbers4FirstPrime, int primeIndex, const intset& excludedPrimes) { const factorset2numbers& factorSet2Numbers = factorSet2Numbers4FirstPrime[primeIndex]; if (factorSet2Numbers.empty()) return 1; // Firstly, we may choose not to use this prime number at all double numCoPrimeSubSets = NumCoPrimeSubsets(factorSet2Numbers4FirstPrime, primeIndex + 1, excludedPrimes); // Optimization: if we're not excluding anything, then we can collapse // the above call and the first call in the loop below together factorset2numbers::const_iterator i = factorSet2Numbers.begin(); if (excludedPrimes.empty()) { const FactorSetNumbers& factorSetNumbers = i; assert(factorSetNumbers.factorSet.size() == 1); numCoPrimeSubSets = (1 + factorSetNumbers.numbers.size()); ++i; } // We are using this prime number. The number of subsets for this prime number is the sum of // the number of subsets for each set of integers whose factors don't include an excluded factor for(; i!=factorSet2Numbers.end(); ++i) { const FactorSetNumbers& factorSetNumbers = i; intset intersect; std::set_intersection(excludedPrimes.begin(),excludedPrimes.end(), factorSetNumbers.factorSet.begin(),factorSetNumbers.factorSet.end(), std::inserter(intersect,intersect.begin())); if (intersect.empty()) { intset unionExcludedPrimes; std::set_union(excludedPrimes.begin(),excludedPrimes.end(), factorSetNumbers.factorSet.begin(),factorSetNumbers.factorSet.end(), std::inserter(unionExcludedPrimes,unionExcludedPrimes.begin())); // Optimization: don't exclude on current first prime, // because can't possibly occur later on unionExcludedPrimes.erase(unionExcludedPrimes.begin()); numCoPrimeSubSets += factorSetNumbers.numbers.size() NumCoPrimeSubsets(factorSet2Numbers4FirstPrime, primeIndex + 1, unionExcludedPrimes); } } return numCoPrimeSubSets; } int main(int argc, char argv[]) { const int MAXCALC = 80; intsetvec primeFactors(MAXCALC +1); // Calculate prime numbers that factor into each number upto MAXCALC for(int i=2; i<=MAXCALC; ++i) { for(int j=0; j<NPRIMES; ++j) { if (i % PRIMES[j] == 0) { primeFactors[i].insert(PRIMES[j]); } } } const clock_t start = clock(); factorset2numbersArray factorSet2Numbers4FirstPrime(NPRIMES); for(int n=2; n<=MAXCALC; ++n) { { // For each prime, store all the numbers whose first prime factor is that prime // E.g. for the prime 2, for n<=20, we store // {2}, { 2, 4, 8, 16 } // {2, 3}, { 6, 12, 18 } // {2, 5}, { 5, 10, 20 } // {2, 7}, { 14 } const int firstPrime = primeFactors[n].begin(); const int firstPrimeIndex = std::find(PRIMES, PRIMES + NPRIMES, firstPrime) - PRIMES; factorset2numbers& factorSet2Numbers = factorSet2Numbers4FirstPrime[firstPrimeIndex]; const factorset2numbers::iterator findFactorSet = std::find_if(factorSet2Numbers.begin(), factorSet2Numbers.end(), & { return x.factorSet == primeFactors[n]; }); if (findFactorSet == factorSet2Numbers.end()) { factorSet2Numbers.push_back(FactorSetNumbers(primeFactors[n], n)); } else { findFactorSet->numbers.push_back(n); } // The number of coprime subsets is the number of coprime subsets for the first prime number, // starting with an empty exclusion list const double numCoPrimeSubSetsForNEquals1 = 2; const double numCoPrimeSubsets = numCoPrimeSubSetsForNEquals1 NumCoPrimeSubsets(factorSet2Numbers4FirstPrime, 0, // primeIndex intset()); // excludedPrimes const clock_t now = clock(); const clock_t ms = now - start; std::cout << n << ", " << std::fixed << numCoPrimeSubsets << ", " << ms << "\n"; } } return 0; } ``` Timings: computes the sequence upto 40 in <0.1s, the sequence upto 50 in 0.5s, to 60 in 2.5s, to 70 in 20s, and to 80 in 157s. Although this certainly seems to output the right numbers it is, as might be expected, too costly. In particular it takes at least exponential time (and quite possibly combinatorial time). Clearly this approach doesn't scale as required. However there may be something here which gives other people ideas (or rules out this approach as a failure). It seems like there are two possibilities: This approach can be made to work, because of some combination of the following. There are some clever mathematical optimizations I haven't spotted which eliminate calculations altogether. There are some efficiency optimizations, e.g. use bitset rather than set. Caching. This seems most promising, in that it might be possible to change the recursive call structure into a tree structure, which could be incrementally updated (where a parent-child relationship indicates multiply, and sibling relationship indicates add). This approach can't be made to work. There is some approach that is largely unrelated to this one. It's possible that the first approach I used might be made to work. This was much more of a "closed form" solution that worked very efficiently for upto n=17 and failed at n=18 and above, being out by a small number. I spent a long time writing up the patterns and trying to figure out why it suddenly failed for n=18, but couldn't see it. I may come back to this, or I will include it as an alternative answer if anyone is interested. Edit: I've made some optimizations using a few tricks try to avoid redoing existing calculations where possible and the code is about 10x faster. Sounds good, but it's only a constant improvement. What's really needed is some insight into this problem -- e.g. can we base #subsets(n+1) on #subsets(n)? Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited May 23, 2017 at 11:53 CommunityBot 1 1 1 silver badge answered Sep 22, 2013 at 15:33 TooToneTooTone 8,247 5 5 gold badges 40 40 silver badges 71 71 bronze badges 4 Comments Add a comment justhalf justhalfOver a year ago My idea was something like this, but instead I calculated the difference between subset(n) and subset(n+1). My idea was to calculate: how many subsets can n+1 be included from the previous subset(n) subsets? If n+1 is prime, the answer is clearly 2subset(n), otherwise, we need to calculate as you have done here. I'm surprised that you can actually implement this in quite a short code in C++. I implemented this in Python with code about the same length as yours. (btw, my machine is fast, it can calculate up to n=160 in under 5s) 2013-09-23T01:26:43.207Z+00:00 0 Reply Copy link TooTone TooToneOver a year ago @justhalf That sounds good. I did implement the 2subset(n) optimization (which saved about 30%). I suspect you might have done a better job than me, especially as Python is interpreted, and it might be worth posting / explaining what you have done. I think "how much can you do in 5s" is a reasonable judge of an algorithm especially if we can't get a version without exponential / combinatorial explosion. 2013-09-23T02:09:36.7Z+00:00 0 Reply Copy link justhalf justhalfOver a year ago I'm more interested in your attempt to give a closed form solution. Can you tell us more about it? 2013-09-23T03:03:28.487Z+00:00 0 Reply Copy link TooTone TooToneOver a year ago @justhalf It might be superceded by the formula just posted... If you'r e still interested let me know tomorrow but it's a bit tenuous and very hacky. I wrote another answer just now which uses a recursive formulation based on removing an element from the set. I think your approach is probably more similar to my first answer, however. 2013-09-23T04:12:37.597Z+00:00 0 Reply Copy link Add a comment This answer is useful 0 Save this answer. Show activity on this post. This is how I would do it: Find the prime factors mod m of the numbers up to n Create a queue q of sets, and add to it the empty set, and set counter to 1 While the queue is not empty, pop an element X from the queue For all the numbers k from max(X) to n, check if the factors of the set intersect with the factors of the numer. If not, add to the queue X U k and increment counter by 1. Otherwise, go to the next k. Return counter Two important things must be pointed out: You don't need the factorization of the numbers up to n, but just their prime factors, that means, for 12 you just need 2 and 3. This way checking if 2 numbers are coprime becomes checking if the intersection of two sets is empty. You can keep track of the "set of factors" of each new set you create, this way you won't have to test every new number against every other number in the set, but just intersct his prime factors set against the one of the whole set. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 13, 2013 at 18:09 answered Sep 13, 2013 at 16:09 SaveSave 12k 1 1 gold badge 20 20 silver badges 23 23 bronze badges 4 Comments Add a comment James Waldby - jwpat7 James Waldby - jwpat7Over a year ago You haven't defined what set k is from, its initial value, how to get next k, etc. Also specify algorithm complexity; looks like it might be O(s·w) if there are a total of s coprime subsets and it takes work w to check an intersect. Since s is like O(2ⁿ), your method might be slow. Probably an O(2ⁿ mod m) method exists. 2013-09-13T17:46:32.02Z+00:00 0 Reply Copy link Save SaveOver a year ago @jwpat7 k is not a set, is just a number between max(X) and n. and according to my calculation, the cost should be O(sn^3), where s is the cost of intersecting 2 subsets : in fact, if you consider the worst case scenario, you'll have to check n numbers against all the substes of size 1, n-1 against those of size to and so on. 2013-09-13T18:16:15.51Z+00:00 0 Reply Copy link Dennis Meng Dennis MengOver a year ago How can you make sure that you will always have enough room to maintain that queue? 2013-09-20T01:08:23.163Z+00:00 0 Reply Copy link user1952500 user1952500Over a year ago The 'mod m' mentioned in the problem is for the final answer i.e., number of sets mod m. The prime factors mod m will result in something else. 2013-09-25T20:48:01.117Z+00:00 0 Reply Copy link Add a comment This answer is useful 0 Save this answer. Show activity on this post. Here is a way in O(n2^p), where p is the number of primes under n. Not making use of the modulus. ``` class FailureCoprimeSubsetCounter{ int[] primes;//list of primes under n PrimeSet[] primeSets;//all 2^primes.length //A set of primes under n. And a count which goes with it. class PrimeSet{ BitSet id;//flag x is 1 iff prime[x] is a member of this PrimeSet long tally;//number of coprime sets that do not have a factor among these primes and do among all the other primes //that is, we count the number of coprime sets whose maximal coprime subset of primes[] is described by this object PrimeSet(int np){...} } int coprimeSubsets(int n){ //... initialization ... for(int k=1; k<=n; k++){ PrimeSet p = listToPrimeSet(PrimeFactorizer.factorize(k)); for(int i=0; i<Math.pow(2,primes.length); i++){ //if p AND primes[i] is empty //add primes[i].tally to PrimeSet[ p OR primes[i] ] } } //return sum of all the tallies } } ``` But since this is a competition problem, there must be a quicker and dirtier solution. And being that this method needs exponential time and space and there are 430 primes under 3000, something more elegant too. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 20, 2013 at 20:06 Michael SMichael S 1,885 1 1 gold badge 18 18 silver badges 19 19 bronze badges 3 Comments Add a comment Dennis Meng Dennis MengOver a year ago Your code seems to imply that you want to make primeSets contain 2^430 elements, which is already more than the number of atoms in the known universe. 2013-09-20T21:01:58.047Z+00:00 0 Reply Copy link Michael S Michael SOver a year ago Didn't I mention that? :p 2013-09-20T21:40:14.907Z+00:00 0 Reply Copy link Dennis Meng Dennis MengOver a year ago Sure, but all things considered; I'd much rather have a solution that takes a ton of time but would otherwise be able to run on a normal machine. A solution isn't really a solution if it can't run without crashing due to lack of memory. 2013-09-20T21:49:56.197Z+00:00 0 Reply Copy link Add a comment This answer is useful 0 Save this answer. Show activity on this post. EDIT: A recursive approach added. Solves in 5 seconds upto n = 50. ``` include include using namespace std; int coPrime = {0}; int n, m; // function that checks whether a new integer is coprime with all //elements in the set S. bool areCoprime ( int p, vector& v ) { for ( int i = 0; i < v.size(); i++ ) { if ( !coPrime[v[i]][p] ) return false; } return true; } // implementation of Euclid's GCD between a and b bool isCoprimeNumbers( int a, int b ) { for ( ; ; ) { if (!(a %= b)) return b == 1 ; if (!(b %= a)) return a == 1 ; } } int subsets( vector& coprimeList, int index ) { int count = 0; for ( int i = index+1; i <= n; i++ ) { if ( areCoprime( i, coprimeList ) ) { count = ( count + 1 ) % m; vector<int> newVec( coprimeList ); newVec.push_back( i ); count = ( count + subsets( newVec, i ) ) % m; } } return count; } int main() { cin >> n >> m; int count = 1; // empty set count += n; // sets with 1 element each. // build coPrime matrix for ( int i = 1; i <= 3000; i++ ) for ( int j = i+1; j <= 3000; j++ ) if ( isCoprimeNumbers( i, j ) ) coPrime[i][j] = 1; // find sets beginning with i for ( int i = 1; i <= n; i++ ) { vector<int> empty; empty.push_back( i ); count = ( count + subsets( empty, i ) ) % m; } cout << count << endl; return 0; } ``` A naive approach can be (for N = 3000): Step1: Build Boolean matrix Build a list of prime numbers from 2 to 1500. For each number 1 to 3000, build a set of its prime factors. Compare each pair of sets and get a boolean matrix that states whether element i and j are mutually coprime (1) or not (0). Step 2: Calculate the number of coprime sets of length k (k = 0 to 3000) Initialize count = 1 (empty set). Now k = 1. Maintain a list of sets of length k. Build 3000 sets containing only that particular element. (increment the count) Scan each element from k to 3000 and see if a new set can be formed by adding it to any of the existing sets of length k. Note: some newly formed sets may or may not be identical. If you use set of sets, then only unique sets shall be stored. Delete all sets that still have a length k. Increment count by the current number of unique sets. k = k + 1 and goto step 3. Alternatively, you can maintain a list of products of each of the elements in a set and check whether the new element is coprime with the product. In that case, you do not need to store the Boolean matrix. The complexity for the above algorithm seems somewhere between O(n^2) and O(n^3). Full Code in C++: (improvement: condition added that element should be checked in a set only if it is > than the largest value in the set). ``` include include include using namespace std; int coPrime = {0}; // function that checks whether a new integer is coprime with all //elements in the set S. bool areCoprime ( int p, set S ) { set<int>::iterator it_set; for ( it_set = S.begin(); it_set != S.end(); it_set++ ) { if ( !coPrime[p][it_set] ) return false; } return true; } // implementation of Euclid's GCD between a and b bool isCoprimeNumbers( int a, int b ) { for ( ; ; ) { if (!(a %= b)) return b == 1 ; if (!(b %= a)) return a == 1 ; } } int main() { int n, m; cin >> n >> m; int count = 1; // empty set set< set<int> > setOfSets; set< set<int> >::iterator it_setOfSets; // build coPrime matrix for ( int i = 1; i <= 3000; i++ ) for ( int j = 1; j <= 3000; j++ ) if ( i != j && isCoprimeNumbers( i, j ) ) coPrime[i][j] = 1; // build set of sets containing 1 element. for ( int i = 1; i <= n; i++ ) { set<int> newSet; newSet.insert( i ); setOfSets.insert( newSet ); count = (count + 1) % m; } // Make sets of length k for ( int k = 2; k <= n; k++ ) { // Scane each element from k to n set< set<int> > newSetOfSets; for ( int i = k; i <= n; i++ ) { //Scan each existing set. it_setOfSets = setOfSets.begin(); for ( ; it_setOfSets != setOfSets.end(); it_setOfSets++ ) { if ( i > (( it_setOfSets ).rbegin()) && areCoprime( i, it_setOfSets ) ) { set<int> newSet( it_setOfSets ); newSet.insert( i ); newSetOfSets.insert( newSet ); } } } count = ( count + newSetOfSets.size() ) % m; setOfSets = newSetOfSets; } cout << count << endl; return 0; } ``` The above code seems to give correct result but consumes a lot of time: Say M is large enough: For N = 4, count = 12. (almost instantaneous) For N = 20, count = 3232. (2-3 seconds) For N = 25, count = 11168. (2-3 seconds) For N = 30, count = 31232 (4 seconds) For N = 40, count = 214272 (30 seconds) Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 24, 2013 at 6:03 answered Sep 21, 2013 at 4:33 Abhishek BansalAbhishek Bansal 12.7k 4 4 gold badges 34 34 silver badges 46 46 bronze badges 1 Comment Add a comment TooTone TooToneOver a year ago Precomputing a matrix to check whether numbers are corrected is a very nice idea; I wish I'd thought of that. I'm not sure about enumerating each set individually though -- I think for an efficient solution you need to count them in groups somehow. 2013-09-24T01:13:59.11Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. Here's the different approach I mentioned earlier. It is indeed much faster than the one I used before. It can calculate up to coprime_subsets(117) in less than 5 seconds, using an online interpreter(ideone). The code builds the possible sets starting from the empty set, and inserting every number into all possible subsets. ```python primes_to_3000 = set([2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999]) primes up to sqrt(3000), used for factoring numbers primes = set([2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53]) factors = [set() for _ in xrange(3001)] for p in primes: for n in xrange(p, 3001, p): factors[n].add(p) def coprime_subsets(highest): count = 1 used = {frozenset(): 1} for n in xrange(1, highest+1): if n in primes_to_3000: # insert the primes into all sets count <<= 1 if n < 54: used.update({k.union({n}): v for k, v in used.iteritems()}) else: for k in used: used[k] = 2 else: for k in used: # only insert into subsets that don't share any prime factors if not factors[n].intersection(k): count += used[k] used[k.union(factors[n])] += used[k] return count ``` Here's my idea and an implementation in python. It seems to be slow, but I'm not sure if it's just the way I was testing(using an online interpreter)... It could be that running it on a "real" computer might make a difference, but I can't test that at the moment. There are two parts to this approach: Pre-generate a list of prime factors Create a cached recursive function for determining the number of possible subsets: For each number, check its factors to see if it can be added to the subset If it can be added, get the count for the recursive case, and add to total After that, I guess you just take the modulo... Here's my python implementation(improved version): ```python primes up to 1500 primes = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499 factors = [set() for _ in xrange(3001)] for p in primes: for n in xrange(p, 3001, p): factors[n].add(p) def coprime_subsets(highest, current=1, factors_used=frozenset(), cache={}): """ Determine the number of possible coprime subsets of numbers, using numbers starting at index current. factor_product is used for determining if a number can be added to the current subset. """ if (current, factors_used) in cache: return cache[current, factors_used] count = 1 for n in xrange(current, highest+1): if factors_used.intersection(factors[n]): continue count += coprime_subsets(highest, n+1, factors_used.union(factors[n])) cache[current, factors_used] = count return count ``` I also have another idea, which I will try implementing if I get the time. I believe a different approach might be quite a bit faster. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 25, 2013 at 21:05 answered Sep 22, 2013 at 16:33 stranacstranac 28.4k 5 5 gold badges 27 27 silver badges 32 32 bronze badges 1 Comment Add a comment stranac stranacOver a year ago Added a third version, using a different approach. 2013-09-25T21:06:49.563Z+00:00 0 Reply Copy link This answer is useful -1 Save this answer. Show activity on this post. It looks like the proposed answers, as well as the preamble to the question, address a question different from the one asked. The question was: Output the number of coprime subsets of {1, 2, 3, ..., n} modulo m. The proposed answers attempt to address another one: Output the number of coprime subsets of {1, 2, 3, ..., n}. These questions are NOT equivalent. The 1st one deals with the finite ring Z_m, and the 2nd one deals with the ring of integers Z which has completely different arithmetic. Moreover, the definition "Two integers are coprime if their greatest common divisor equals 1" in the preamble to the question is not applicable to Z_m: finite rings don't have arithmetically stable comparison, so there's no such thing as a "greatest" common divisor. The same objection applies to the example in the question: 3 and 4 are NOT relatively prime modulo 7 because both are divisible by 2 modulo 7: 4=(22)%7 and 3=(25)%7. In fact, Z_m arithmetic is so weird that one can give the answer in O(1) time, at least for prime m: for any n and prime m there are NO coprime pairs modulo m. Here's why: Non-zero elements of Z_m form a cyclic group of order m-1, which implies that for any non-zero elements a and b from Z_m one can write a=bc for some c in Z_m. This proves that there are no coprime pairs in Z_m for prime m. From the example in the question: let's take a look at {2, 3} mod 7 and {3, 4} mod 7: 2=(33)%7 and 3=(46)%7. Therefore {2,3} are not coprime in Z_7 (both are divisible by 3) and {3,4} are not coprime in Z_7 (both are divisible by 4). Now let's consider the case of non-prime m. Write ma as a product of prime powers m=p_1^i_1...p_k^i_k. If a and b have a common prime factor than they are clearly not coprime. If at least one of them, say b, does not divide any of the primes p_1,...,p_k then a and b have a common factor for roughly the same reason as in the case of prime m: b would be a multiplicative unit of Z_m, and therefore a would be divisible by b in Z_m. So it remains to consider the case when m is composite and a and b are divisible by distinct prime factors of m, let's say a is divisible by p and b is divisible by q. In this case they indeed can be coprimes. For example, 2 and 3 modulo 6 are coprimes. So the question for coprime pairs boils down to these steps: Finding prime factors of m that are less than n. If none then there are no coprime pairs. Enumerating products of powers of these prime factors that remain the factors of m that are less than n. Computing the number of Z-comprime pairs among thoses. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 25, 2013 at 18:22 answered Sep 25, 2013 at 17:33 MichaelMichael 6,011 3 3 gold badges 39 39 silver badges 63 63 bronze badges 6 Comments Add a comment David Eisenstat David EisenstatOver a year ago The explanation for the sample test case in the linked description contradicts your interpretation. 2013-09-25T19:37:27.09Z+00:00 1 Reply Copy link linski linskiOver a year ago This is really neat observation, I'm upvoting solely because of the notion of the ring arithmetic. However, if you take a look in the original linked pdf you will see that the exact notation is : "Output the number of coprime subsets of {1, 2, 3, ..., n}, modulo m" - with a comma before modulo operator. 2013-09-25T19:39:26.243Z+00:00 0 Reply Copy link Michael MichaelOver a year ago I didn't quite get the significance of the comma. As for arithmetic in Z or modulo m, both make sense for practical applications. For example, if the problem originated in cryptography it would be perfectly reasonable to ask about coprimes modulo the product of the keys. Given the contradiction between modulo m and the example only the person who asked the question can tell which interpretation needs to be answered... 2013-09-25T20:02:14.647Z+00:00 0 Reply Copy link TooTone TooToneOver a year ago I think the modulo is there to avoid rounding errors so that a precise answer can be given and assessed in the competitive setting. 2013-09-25T20:26:10.297Z+00:00 0 Reply Copy link Brett Hale Brett HaleOver a year ago The link to the original question is quite clear. 2013-09-25T23:23:00.31Z+00:00 0 Reply Copy link Add a comment\|Show 1 more comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! 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74	https://www.sciencedirect.com/topics/medicine-and-dentistry/tinea-cruris	Skip to Main content My account Sign in Tinea cruris In subject area:Medicine and Dentistry Tinea cruris is defined as an erythematous, scaly eruption that typically occurs in the inguinal creases and medial thighs, usually presenting symmetrically and potentially with papular margins. AI generated definition based on: Nelson Pediatric Symptom-Based Diagnosis, 2018 How useful is this definition? Add to Mendeley Also in subject areas: Agricultural and Biological Sciences Pharmacology, Toxicology and Pharmaceutical Science Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Vulvar Dermatoses: Papulosquamous Diseases 2008, Obstetric and Gynecologic Dermatology (Third Edition)Libby Edwards Tinea cruris Tinea cruris, also called “jock itch,” is a cutaneous fungal infection of hair-bearing skin. Epidemiology and clinical manifestations Occurring primarily in men, particularly in the setting of tinea pedis and fungal nail infections, tinea cruris is a disease of postpubertal individuals, and it occurs more often in humid and warm climates. The most common symptom associated with tinea cruris is pruritus, especially on the proximal, medial thighs. In addition, the rash is unsightly to some. Pain is generally not associated. Classic tinea cruris presents as well-demarcated, red, lightly scaling plaques covering the proximal, medial thighs, with accentuation of the scale peripherally that produces an annular configuration (Figure 17.10). Generally, tinea cruris does not prominently affect the labia majora or mons, except in women using topical corticosteroids or who are otherwise immunosuppressed. Tinea cruris affecting skin with terminal hair sometimes involves the follicular epithelium (fungal folliculitis), resulting in red nodules, pustules, and/or crusted papules within the more typical plaques of tinea cruris. Tinea cruris does not affect the modified mucous membranes of the vulva. Individuals using topical corticosteroids on their tinea cruris may exhibit tinea cruris that is morphologically atypical (tinea incognito), showing poorly demarcated borders, coalescing arcuate plaques, and fungal folliculitis. At these times, the peripheral accentuation of scale may be less obvious (Figure 17.11). Differential diagnosis All causes of red vulvar plaques should be considered in the differential diagnosis (Table 17.3). These include lichen simplex chronicus, psoriasis, candidiasis, contact dermatitis, and seborrheic dermatitis. A microscopic examination of a fungal preparation of skin confirms the diagnosis by showing branching hyphae without budding yeast. Uncertain or recalcitrant cases can be confirmed by dermatophyte culture. Laboratory abnormalities and histology The only laboratory abnormality seen in patients with tinea cruris is a positive fungal culture, or a fungal preparation of scale revealing dermatophytosis microscopically. A biopsy of tinea cruris is generally not required for diagnosis. However, histology shows organisms within the stratum corneum that are most easily visualized using special stains such as the periodic acid–Schiff reaction. There are often also changes of dermatitis, to include spongiosis. Pathogenesis The cause of tinea cruris is infection with a dermatophyte, with Tricophyton rubrum, T. mentagrophytes, or Epidermophyton floccosum being the most common. These are indistinguishable microscopically. Therapy and prognosis The treatment of tinea cruris consists of the administration of any of several antifungal oral or topical medications. All azoles are effective, whereas nystatin exhibits no activity against tinea. Terbinafine (Lamisil) and ciclopirox (Loprox) are active agents as well. These are applied once or twice a day, and are appropriate therapy when tinea cruris is mild, limited in extent, and shows no signs of fungal folliculitis. When abundant coarse hair or folliculitis is present, or when tinea is extensive, patients respond best to oral medication. Options include the azoles, fluconazole (Diflucan) 100–200 mg weekly for 2–4 weeks, or itraconazole (Sporonox), 100 mg/day or ketoconazole, 200 mg/day until clear. Griseofulvin 500 mg twice daily is effective as well, but nausea and headaches are often limiting side-effects. Oral terbinafine 250 mg daily until clear is effective and well tolerated. All of these medications, except for griseofulvin, also treat Candida in case of a misdiagnosis. Patients should be advised that recurrence can occur, particularly in the setting of onychomycosis.These patients should be instructed to restart a topical azole cream or terbinafine cream immediately with the first sign of recurrence. View chapterExplore book Read full chapter URL: Book2008, Obstetric and Gynecologic Dermatology (Third Edition)Libby Edwards Chapter Skin Disorders due to Fungi 2011, Hurwitz Clinical Pediatric Dermatology (Fourth Edition)Amy S. Paller MD, Anthony J. Mancini MD Tinea cruris Tinea cruris (‘jock itch’) is an extremely common superficial fungal infection of the groin and upper thighs. It is seen primarily in male adolescents and adults, and occurs less commonly in females. Tinea cruris is most symptomatic in hot, humid weather and is most frequently noted in obese individuals or those subject to vigorous physical activity and chafing. Tight-fitting clothing such as athletic supporters, jockey shorts, wet bathing suits, and panty hose may contribute to this condition as well. The three most common dermatophytes to result in tinea cruris are E. floccosum, T. rubrum, and T. mentagrophytes. Tinea pedis is a common co-existing condition, possibly related to autoinoculation of the dermatophyte with clothing that comes into contact with the feet. Tinea cruris presents as sharply marginated, erythematous plaques with an elevated border of scaling, pustules, or vesicles. It is usually but not always bilaterally symmetric, and involves the intertriginous folds near the scrotum, the upper inner thighs (Fig. 17.20), and occasionally the perianal regions, buttocks and abdomen. The scrotum and labia majora are usually spared, and if they are involved or satellite papulopustules are present, the diagnosis of candidiasis (Fig. 17.21) should be considered. The lesions of tinea cruris may vary in color from red to brown, and central clearing may be present. In chronic infection the redness and scaling may be slight, the active margin may be subtle or ill defined, and lichenification may be present. Tinea cruris must be differentiated from intertrigo, seborrheic dermatitis, psoriasis, irritant contact dermatitis, allergic contact dermatitis (generally due to therapy), or erythrasma (a superficial dermatosis caused by the diphtheroid Corynebacterium minutissimum). A characteristic coral-red fluorescence under Wood's light examination is helpful in distinguishing erythrasma (see Ch. 14). The diagnosis of tinea cruris can be confirmed by a potassium hydroxide wet-mount microscopic examination of cutaneous scrapings or by fungal culture. Topical therapy (as discussed for tinea corporis) usually suffices for tinea cruris, and is applied for 3–4 weeks. Other useful measures include reducing excessive chafing and irritation by the use of loose-fitting cotton underclothing, drying thoroughly following bathing or perspiration, and weight loss. The use of an absorbent antifungal powder (i.e., Micatin, Tinactin, or ZeaSorb-AF) is sometimes helpful, and oral antifungal therapy is occasionally indicated for severe or recalcitrant disease. Tinea pedis, if present, should also be adequately treated as a preventative measure. View chapterExplore book Read full chapter URL: Book2011, Hurwitz Clinical Pediatric Dermatology (Fourth Edition)Amy S. Paller MD, Anthony J. Mancini MD Chapter Skin Infections 2010, Diagnostic Pathology of Infectious DiseaseCarlos Nicolas Prieto-Granada, ... Martin C. MihmJr. Tinea Cruris Tinea cruris is an infection of the groin most commonly seen in adult men.461 T. rubrum and Epidermophyton floccosum are the most frequent etiologic agents.452 Lesions are usually bilateral, erythematous to reddish-brown, and sharply demarcated. The periphery of these lesions shows fine scaling, with papules and pustules being seen in more inflammatory cases.461,469 View chapterExplore book Read full chapter URL: Book2010, Diagnostic Pathology of Infectious DiseaseCarlos Nicolas Prieto-Granada, ... Martin C. MihmJr. Chapter Dermatophytosis 2006, Tropical Infectious Diseases (Second Edition)MICHAEL B. SMITH, MICHAEL R. MCGINNIS Tinea Cruris Tinea cruris is an infection involving the groin, perineum, and perianal region that is prevalent in people living in tropical regions, where high humidity contributes to skin maceration. Both males and females may develop tinea cruris, but it is more common in males. Tinea cruris often occurs in conjunction with tinea pedis. The major pathogens are T. rubrum (the most common etiologic agent worldwide), E. floccosum, and T. mentagrophytes. The initial lesion is circinate and becomes serpiginous. Lesions associated with E. floccosum develop distinct margins and raised borders with randomly dispersed vesicles or vesiculopustules containing serous exudate. The center of the lesion is brownish to red, with branny furfuraceous scales. In males, the infection typically begins on the thigh where it contacts the scrotum. A pruritic, erythematous rash develops with the infection extending downward on the inner thighs, often on the left side. The gluteal and pubic areas are often involved. If T. rubrum is the etiologic agent, it may extend to the waist, buttocks, and thighs. T. mentagrophytes infection may extend to the chest, back, legs, and feet. Acute infections are associated with intense itching. Predisposing factors include diabetes, neurodermatitis, and friction between skin folds in obese people, perspiration, humidity, and irritants that contribute to skin maceration. The etiologic agent may originate from skin scales containing viable fungi, animal sources, and fomites. Epidemics usually involve people who are close to each other such as in barracks, dormitories, and locker rooms. View chapterExplore book Read full chapter URL: Book2006, Tropical Infectious Diseases (Second Edition)MICHAEL B. SMITH, MICHAEL R. MCGINNIS Chapter Etiologic Agents of Infectious Diseases 2008, Principles and Practice of Pediatric Infectious Disease (Third Edition)Moise Levy, John Browning TINEA CRURIS (JOCK ITCH) Tinea cruris is an acute or subacute superficial fungal infection of the perineum and inner aspect of the thighs that occurs primarily in adolescent and young adult males. An increased incidence of infection in postpubertal females who wear close-fitting exercise garments or panty hose has been noted in recent years. Although the infection occurs worldwide, it is more common in individuals who reside in warm, humid climates. T. rubrum, T. mentagrophytes, and Epidermophyton floccosum are the most common agents causing tinea cruris. E. floccosum is most common worldwide; however, in recent years T. rubrum commonly causes disease in the United States.3 The appearance of tinea cruris is variable. The acute form of the disease is characterized by red to reddish brown, usually symmetrical macules or patches with well-dermarcated borders; fine, oozing vesicles are present. Although lesions are not painful, the involved skin is tender and local pruritus is common. In the subacute and chronic forms of infection, the involved skin has a brownish discoloration with well-defined margins, plaques, and desquamation. Although the clinical appearance and symptoms of tinea cruris are distinctive, the condition must be differentiated from primary irritant or allergic contact dermatitis, erythrasma (Corynebacterium minutissimum infection), seborrheic dermatitis, psoriasis, and candidal intertrigo. Often, secondary infection with Candida albicans develops in obese individuals with chronic tinea cruris. When the diagnosis is uncertain, a skin scraping from the advancing edge of the lesion should be obtained and examined microscopically using a KOH preparation for the presence of hyphae. Erythrasma can be readily distinguished from tinea cruris by its coral-colored fluorescence when exposed to ultraviolet light. Treatment of tinea cruris consists of drying weeping areas of skin with Burow compresses and application of topical antifungal agents. When tinea cruris occurs with a concurrent dermatophyte infection of the fingernails or toenails (onychomycosis), the infection may not completely clear (due to reinfection) until the onychomycosis has been adequately treated. View chapterExplore book Read full chapter URL: Book2008, Principles and Practice of Pediatric Infectious Disease (Third Edition)Moise Levy, John Browning Chapter Acquired Rashes in the Older Child 2018, Nelson Pediatric Symptom-Based DiagnosisKristen E. Holland, Paula J. Soung Tinea Cruris Tinea cruris, uncommon before adolescence, is an erythematous, scaly eruption involving the inguinal creases and medial thighs. The eruption is usually symmetric, and sometimes the margins are papular. This infection may resemble candidiasis, in which there is also scrotal erythema. Erythrasma, an uncommon superficial bacterial infection caused by Corynebacterium minutissimum, may also mimic tinea cruris. The coral red fluorescence seen on Wood lamp examination is diagnostic of erythrasma, which can be further differentiated from tinea cruris by a negative KOH preparation and fungal culture. Dermatophyte infections of the skin can usually be successfully managed with topical antifungal agents such as clotrimazole, econazole, ciclopirox, tolnaftate, or terbinafine creams or lotions. These medications are applied twice daily for approximately 2-4 weeks. They should be continued for several days after clinical resolution is apparent. Widespread eruptions or treatment failures may necessitate systemic antifungal therapy, such as griseofulvin, fluconazole, itraconazole, or terbinafine. View chapterExplore book Read full chapter URL: Book2018, Nelson Pediatric Symptom-Based DiagnosisKristen E. Holland, Paula J. Soung Chapter Etiologic Agents of Infectious Diseases 2012, Principles and Practice of Pediatric Infectious Diseases (Fourth Edition)Caroline Diane Sarah Piggott, Sheila Fallon Friedlander Tinea Cruris (Jock Itch) Tinea cruris is an acute or subacute superficial fungal infection of the perineum and inner aspect of the thighs that occurs primarily in males. There is an increased incidence in postpubertal females who wear close-fitting exercise garments or panty hose. Increasing incidence of tinea cruris in adolescents may be related to increases in diabetes mellitus and obesity.40 Infection is more common in warm, humid climates. T. rubrum, T. mentagrophytes, and Epidermophyton floccosum are the most common pathogens. The incidence of T. rubrum is increasing in the U.S.6 Red to reddish brown, usually symmetrical scaly lesions with well-demarcated borders and/or fine, oozing vesicles are typical. Local pruritus is common. Chronically involved skin can manifest as brownish discoloration with well-defined borders. Differential diagnosis includes primary irritant or allergic contact dermatitis, erythrasma (Corynebacterium minutissimum infection), seborrheic dermatitis, psoriasis, and candidal intertrigo. Often, secondary infection with Candida albicans develops in obese individuals with chronic tinea cruris. KOH examination of specimen from the advancing edge of the lesion can show hyphae. Erythrasma can be identified by its coral red-colored fluorescence when exposed to ultraviolet light. Treatment consists of drying the weeping areas with Burow solution compresses and application of topical antifungal agents. For those with Candida superinfection, the addition of nystatin powder is helpful due to its anti-infective and drying properties. Nystatin powder should not be mixed with the antifungal cream as this can form a gritty irritating mixture. When tinea cruris occurs with a concurrent dermatophyte infection of the fingernails or toenails (onychomycosis), treatment of both disorders may be required for cure. View chapterExplore book Read full chapter URL: Book2012, Principles and Practice of Pediatric Infectious Diseases (Fourth Edition)Caroline Diane Sarah Piggott, Sheila Fallon Friedlander Review article Geriatric Dermatology Update 2024, Clinics in Geriatric MedicineSaniya Shaikh DO, Aditya Nellore MD Tinea Cruris Tinea cruris, or jock itch, presents as a red, scaly, annular plaque on the inguinal folds, perineum, and/or buttocks. It is often spread through clothes contaminated by other infected body parts while dressing. Sources of autoinoculation include tinea pedis, tinea manuum, and tinea unguium. Therefore, the hands, nails, and feet should be examined. Patients with these conditions should be encouraged to put on gloves or socks before dressing to prevent spread.17 Notably, tinea cruris is more common in men than women because the external genital male anatomy predisposes to warmth and moisture. The differential diagnosis includes candida intertrigo, erythrasma, and inverse psoriasis. Tinea cruris can be distinguished from candidal infection because it spares the penis and scrotum. Treatment is similar to that of tinea corporis.18 View article Read full article URL: Journal2024, Clinics in Geriatric MedicineSaniya Shaikh DO, Aditya Nellore MD Review article Geriatric Dermatology 2013, Clinics in Geriatric MedicineReena S. Varade BA, Nicole M. Burkemper MD Tinea Cruris Tinea cruris, commonly known as jock itch, involves the inguinal folds, perineum, and buttocks. It presents as a red to brown plaque with scale at the borders that may or may not have central clearing. It is often spread through coexisting fungal infections on other parts of the body. Onychomycosis and tinea pedis can be spread to the groin through contamination of clothing when the patient puts on undergarments. Tinea manuum and tinea unguium can also be the source of autoinoculation of tinea cruris therefore the hands and nails should also be examined.19 Tinea cruris is found more commonly in men than women because onychomycosis and tinea pedis are more common in men and the external genital male anatomy predisposes to warmth and moisture. Patients with onychomycosis should be encouraged to put on socks before dressing to prevent spread. The differential diagnosis includes candida intertrigo, erythrasma, and inverse psoriasis. Infections with Candida usually involve the penis and scrotum and tinea cruris spares those areas, which helps to distinguish between these 2 entities. The treatment of tinea cruris is similar to that of tinea corporis, with terbinafine or econazole cream used twice daily for 2 to 4 weeks.20 View article Read full article URL: Journal2013, Clinics in Geriatric MedicineReena S. Varade BA, Nicole M. Burkemper MD Chapter Vulvar Dermatoses: Papulosquamous Diseases 2008, Obstetric and Gynecologic Dermatology (Third Edition)Libby Edwards Epidemiology and clinical manifestations Occurring primarily in men, particularly in the setting of tinea pedis and fungal nail infections, tinea cruris is a disease of postpubertal individuals, and it occurs more often in humid and warm climates. The most common symptom associated with tinea cruris is pruritus, especially on the proximal, medial thighs. In addition, the rash is unsightly to some. Pain is generally not associated. Classic tinea cruris presents as well-demarcated, red, lightly scaling plaques covering the proximal, medial thighs, with accentuation of the scale peripherally that produces an annular configuration (Figure 17.10). Generally, tinea cruris does not prominently affect the labia majora or mons, except in women using topical corticosteroids or who are otherwise immunosuppressed. Tinea cruris affecting skin with terminal hair sometimes involves the follicular epithelium (fungal folliculitis), resulting in red nodules, pustules, and/or crusted papules within the more typical plaques of tinea cruris. Tinea cruris does not affect the modified mucous membranes of the vulva. Individuals using topical corticosteroids on their tinea cruris may exhibit tinea cruris that is morphologically atypical (tinea incognito), showing poorly demarcated borders, coalescing arcuate plaques, and fungal folliculitis. At these times, the peripheral accentuation of scale may be less obvious (Figure 17.11). View chapterExplore book Read full chapter URL: Book2008, Obstetric and Gynecologic Dermatology (Third Edition)Libby Edwards Related terms: Candidiasis Athlete's Foot Tinea corporis Tinea Capitis Dermatophytosis Onychomycosis Terbinafine Ketoconazole Tinea Dermatophyte View all Topics
75	https://fastercapital.com/es/contenido/Fijacion-de-precios-por-costo-marginal--una-estrategia-de-fijacion-de-precios-en-la-que-una-empresa-fija-el-precio-igual-al-costo-marginal-de-produccion.html	Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion - FasterCapital Página principal portafolio About Programas- [x] Programas Raise Capital Mega Financing Cofundador tecnológico Franquicia de Negocios Financiamiento Inmobiliario Servicios de propiedad intelectual Iniciar negocios en los Emiratos Árabes Unidos Verificar financiación Haga crecer su startup Idea al producto Visa de inicio Servicios- [x] Servicios Fuentes de financiamiento Venture Capital Ángel Capital Préstamos comerciales Subvenciones iniciales Servicios de financiación Valoración de inicio Business Plan Plataforma de presentación Modelo financiero Servicios tecnológicos Diseño de software Web Design Diseño de aplicaciones móviles CTO Services Servicios de crecimiento Ventas como servicio Marketing de contenidos Publicidad digital SEO Services Centro de aprendizaje- [x] Centro de aprendizaje Acerca de LearnHub Contenido Palabras clave Temas Preguntas Galería de infografías Pareja Contacto Página PrincipalContenidoFijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion Actualizado: 28 abril 202519 minutos Tabla de contenidos 1. Introducción a la fijación de precios por costo marginal 2. Comprensión del costo marginal de producción 3. Ventajas de la fijación de precios por costo marginal 4. Desafíos y limitaciones de la fijación de precios por costo marginal 5. Ejemplos de fijación de precios por costo marginal en la práctica 6. Comparación con otras estrategias de precios 7. Implementación de precios de costo marginal en las empresas 8. Impacto en la rentabilidad y la cuota de mercado 9. Conclusión y conclusiones clave Ayuda gratuita y descuentos deFaster Capital! Convertirse en un compañero Necesito ayuda en: Consiga emparejarse con más de 155.000 ángeles y 50.000 VC en todo el mundo. ¡Utilizamos nuestro sistema de inteligencia artificial y le presentamos a los inversores a través de cálidas presentaciones! 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Presupuesto disponible para mejorar tus ventas: Trabajamos con usted en marketing de contenidos, presencia en las redes sociales y lo ayudamos a encontrar consultores de marketing expertos y cubrir el 50% de los costos. ¿Que servicios necesitas? Presupuesto disponible para sus actividades de marketing: Nombre Completo Nombre de empresa Email País Whatsapp Comentario Los envíos de correo electrónico comercial serán respondidos dentro de 1 o 2 días hábiles. Los envíos de correos electrónicos personales tardarán más Entregar Algunas imágenes de este artículo aún no han sido traducidas. ¡Podrás volver a visitar esta página pronto para ver las imágenes en tu idioma! Fijación de precios por costo marginal: una estrategia de fijación de precios en la que una empresa fija el precio igual al costo marginal de producción 1. Introducción a la fijación de precios por costo marginal Comprensión del costo marginal (MC): Definición: El costo marginal representa el costo adicional incurrido al producir una unidad más de un producto o servicio. Considera el cambio incremental en el costo total debido a un aumento en la producción. Fórmula: MC = ΔTC / ΔQ, donde ΔTC es el cambio en el coste total y ΔQ es el cambio en la cantidad producida. Ejemplo: Imagina una panadería que produce pastelitos. Si el costo de los ingredientes y la mano de obra para el pastelito número 100 es de $5 y el costo del pastelito número 101 es de $5,50, el costo marginal de ese pastelito adicional es de $0,50. La estrategia de fijación de precios por costo marginal: Objetivo: La fijación de precios por costo marginal tiene como objetivo lograr eficiencia asignativa alineando los precios con el costo real de producción. Ventajas: Asignación de recursos: cuando los precios reflejan costos marginales, los consumidores toman decisiones de consumo que benefician a la sociedad en su conjunto. Sin pérdida de eficiencia: a diferencia de los precios de monopolio (donde los precios exceden el costo marginal), la fijación de precios de costo marginal minimiza la pérdida de eficiencia. Desafíos: beneficio a corto plazo: Fijar precios al costo marginal puede no maximizar los beneficios a corto plazo, especialmente si los costos fijos son significativos. Variabilidad: los costos marginales pueden fluctuar debido a cambios en los precios de los insumos, la tecnología o los niveles de producción. Escenarios de fijación de precios por costo marginal: Competencia perfecta: en mercados perfectamente competitivos, las empresas naturalmente fijan precios iguales a los costos marginales. La intensa competencia garantiza que ninguna empresa pueda cobrar más sin perder clientes. Servicios públicos: los servicios regulados (por ejemplo, agua, electricidad) a menudo utilizan precios de costo marginal para evitar comportamientos monopolísticos y garantizar un acceso justo. Consideraciones ambientales: la fijación de precios de costo marginal puede fomentar prácticas respetuosas con el medio ambiente. Por ejemplo, los impuestos al carbono alinean los precios con el costo ambiental de producción. Ejemplos: Tarifas de electricidad: las empresas de servicios públicos ajustan las tarifas de electricidad según la hora del día (hora pico o valle) para igualar los costos marginales. Transporte: las aerolíneas suelen ofrecer billetes de última hora a precios más bajos (más cercanos al coste marginal) para llenar los asientos vacíos. Licencias de software: algunas empresas de software cobran por usuario o por función, alineando los precios con el coste de proporcionar servicios adicionales. Críticas y controversias: Sostenibilidad a largo plazo: Los críticos argumentan que centrarse únicamente en el costo marginal descuida las inversiones y la innovación a largo plazo. Preocupaciones por la equidad: la fijación de precios de costo marginal puede no considerar las disparidades de ingresos o las implicaciones de bienestar social. Imperfecciones del mercado: en mercados imperfectamente competitivos, las empresas pueden tener poder de mercado, lo que hace que la fijación de precios por costo marginal sea menos factible. En resumen, la fijación de precios por costo marginal ofrece un enfoque pragmático para la fijación de precios, equilibrando la eficiencia, la equidad y la rentabilidad. Si bien tiene sus limitaciones, comprender la interacción entre costos, precios y bienestar social es crucial tanto para las empresas como para los formuladores de políticas. Introducción a la fijación de precios por costo marginal - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 2. Comprensión del costo marginal de producción ¿Qué es el costo marginal? En esencia, el costo marginal representa el costo adicional incurrido al producir una unidad más de un bien o servicio. Es el costo incremental asociado con la expansión de la producción. Imagine una panadería que produce deliciosos croissants. A medida que hornean más croissants, aumenta el costo de la harina, la mantequilla y la mano de obra. El costo adicional de hornear un croissant más es el costo marginal. La curva de costo marginal: Visualice un gráfico donde el eje x representa la cantidad producida y el eje y representa el costo. La curva de coste marginal normalmente tiene pendiente ascendente. Inicialmente, a medida que aumenta la producción, el costo marginal aumenta lentamente debido a las economías de escala. Sin embargo, más allá de cierto punto, aparecen rendimientos decrecientes y la curva se vuelve más pronunciada. Piense en ello como el punto en el que la cocina de la panadería se llena y la eficiencia del chef disminuye. Costos marginales a corto plazo versus costos marginales a largo plazo: El costo marginal a corto plazo considera costos variables (como materias primas y mano de obra) que cambian con los niveles de producción. Por ejemplo, si la panadería necesita contratar personal adicional durante las horas pico, ese es un costo a corto plazo. El costo marginal a largo plazo incluye tanto los costos variables como los fijos (como alquiler, maquinaria y gastos administrativos). Si la panadería decide ampliar su cocina, el costo marginal a largo plazo representa el nuevo horno y el espacio adicional. estrategias de precios y costo marginal: precio del costo marginal: algunas empresas adoptan una estrategia de fijación de precios en la que fijan el precio igual al costo marginal. Este enfoque garantiza que cada venta contribuya positivamente a cubrir los costos. Sin embargo, es posible que no maximice las ganancias. Maximización de beneficios: las empresas suelen fijar precios por encima del coste marginal para maximizar los beneficios. Si la panadería vende croissants a un precio superior a su coste marginal, el excedente contribuye a cubrir los costes fijos y generar beneficios. Ejemplos: servicios de viajes compartidos: Uber y Lyft consideran los costos marginales al fijar el precio de los viajes. El costo de un viaje adicional (combustible, tiempo del conductor) guía sus decisiones de precios. Licencias de software: las empresas que venden licencias de software suelen fijar precios en función de los costes marginales (espacio en el servidor, atención al cliente) para atraer a más usuarios. Fabricación: un fabricante de automóviles calcula el coste de producir un coche adicional (materiales, mano de obra en la línea de montaje) para determinar el precio. Críticas y limitaciones: Ignora los costos fijos: la fijación de precios de costo marginal ignora los costos fijos, que pueden ser sustanciales para algunas empresas. Demanda del mercado: fijar precios basándose únicamente en el costo marginal podría no alinearse con la demanda del mercado o la disposición a pagar del consumidor. Entorno dinámico: en los mercados dinámicos, los costos marginales pueden fluctuar rápidamente debido a los cambios en los precios de los insumos o los avances tecnológicos. En resumen, comprender el costo marginal permite a las empresas tomar decisiones informadas sobre los niveles de producción, los precios y la rentabilidad. Ya sea que opten por fijar el precio al costo marginal u optimizar para obtener ganancias, este concepto sigue siendo una piedra angular de la teoría económica. ¿Necesitas apoyo para solicitar préstamos? ------------------------------------------ FasterCapital te ayuda a solicitar préstamos empresariales a escala global, preparando tus documentos y poniéndote en contacto con los prestamistas Únete a nosotros! 3. Ventajas de la fijación de precios por costo marginal Eficiencia de costos: al fijar el precio igual al costo marginal, las empresas pueden asegurarse de cubrir sus costos variables y maximizar la eficiencia de costos. Esto ayuda a optimizar la asignación de recursos y minimizar el desperdicio. competencia en el mercado: La fijación de precios de costo marginal promueve una competencia sana en el mercado. Dado que el precio se basa en el costo real de producción, evita que las empresas cobren precios excesivos y las alienta a ofrecer precios competitivos para atraer clientes. Bienestar del consumidor: esta estrategia de precios beneficia a los consumidores ofreciéndoles productos o servicios a un precio que refleja fielmente el costo real de producción. Garantiza que no se cobre de más a los consumidores y promueve la asequibilidad. Penetración de mercado: la fijación de precios de costo marginal puede ser particularmente ventajosa para los nuevos participantes o las empresas que buscan penetrar en un mercado. Al fijar el precio igual al costo marginal, pueden atraer clientes con precios más bajos, ganar participación de mercado y establecer su presencia. Flexibilidad: esta estrategia de precios permite a las empresas ajustar los precios en función de los cambios en los costos de producción. Si el costo marginal disminuye, la empresa puede reducir el precio en consecuencia, generando beneficios tanto para la empresa como para los consumidores. Transparencia: La fijación de precios por costo marginal promueve la transparencia en la fijación de precios. Dado que el precio está directamente relacionado con el costo de producción, elimina cualquier ambigüedad o cargos ocultos, mejorando la confianza entre la empresa y los clientes. Para ilustrar el concepto, consideremos un ejemplo. Supongamos que una empresa fabrica teléfonos inteligentes y el costo marginal de producir cada unidad es de 200 dólares. Al adoptar el precio de costo marginal, la empresa fija el precio de venta de cada teléfono inteligente en 200 dólares. Este enfoque garantiza que el precio refleje con precisión el coste de producción, beneficiando tanto a la empresa como a los clientes. Ventajas de la fijación de precios por costo marginal - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 4. Desafíos y limitaciones de la fijación de precios por costo marginal Sensibilidad a las fluctuaciones de costes: Información: La fijación de precios por costo marginal responde en gran medida a los cambios en los costos de producción. Cualquier aumento en los costos de los insumos impacta directamente el precio de venta. Ejemplo: Considere una pequeña panadería que produce pan fresco diariamente. Si el coste de la harina o de la energía aumenta repentinamente, la panadería debe ajustar sus precios en consecuencia. Esta sensibilidad puede dar lugar a cambios frecuentes de precios, lo que puede confundir a los clientes. enfoque a corto plazo: Perspectiva: La fijación de precios por costo marginal tiende a priorizar la rentabilidad a corto plazo sobre la sostenibilidad a largo plazo. Las empresas pueden ignorar los costos fijos (como el alquiler, los salarios y el equipo) al fijar los precios. Ejemplo: una nueva empresa de software podría centrarse únicamente en cubrir costes variables (como el uso del servidor) sin considerar la necesidad de futuras inversiones en investigación y desarrollo. Ignora la demanda del mercado y la disposición a pagar del consumidor: Información: La fijación de precios por costo marginal supone que los consumidores están dispuestos a pagar el costo exacto de producción. Sin embargo, la demanda del mercado y las preferencias de los consumidoresjuegan un papel crucial en la determinación de los precios. Ejemplo: las marcas de lujo como Rolex o Gucci no fijan el precio de sus productos basándose únicamente en los costos de producción. Consideran el valor de la marca, la exclusividad y la percepción del consumidor. Socava la innovación y la calidad: Perspectiva: cuando las empresas se centran únicamente en cubrir costos variables, pueden dudar a la hora de invertir en mejoras de calidad o innovación. Ejemplo: un fabricante de automóviles podría evitar gastar en características de seguridad o tecnologías ecológicas si eso significa aumentar los costos de producción y afectar los precios marginales. Incapacidad para realizar subvenciones cruzadas: Información: La fijación de precios de costo marginal evita los subsidios cruzados, donde las ganancias de un producto/servicio subsidian las pérdidas de otro. Ejemplo: los sistemas de transporte público suelen utilizar subvenciones cruzadas. Las tarifas más bajas de los autobuses pueden compensarse con tarifas más altas de los trenes, lo que garantiza la viabilidad financiera general. Desafíos en las empresas multiproducto: Perspectiva: Las empresas que producen múltiples productos enfrentan complejidades. La fijación de precios de costo marginal para cada producto puede no conducir a una rentabilidad general óptima. Ejemplo: una empresa de electrónica que fabrica teléfonos inteligentes, tabletas y computadoras portátiles debe considerar la interacción de costos y demanda en todos los productos. Supuestos sobre las estructuras de costos: Información: La fijación de precios por costo marginal supone estructuras de costos lineales, donde cada unidad adicional cuesta lo mismo. En realidad, los costos pueden variar debido a economías de escala o deseconomías. Ejemplo: una fábrica que produce aparatos podría experimentar rendimientos decrecientes más allá de un cierto nivel de producción, lo que afectaría a los costes marginales. Poder de mercado y competencia: Perspectiva: En los mercados competitivos, las empresas no siempre pueden fijar precios basándose únicamente en los costos marginales. El poder de mercado, la lealtad a la marca y las acciones de los competidores influyen en las decisiones de fijación de precios. Ejemplo: si un jugador dominante reduce los precios significativamente, los competidores más pequeños pueden tener dificultades para igualar esos precios, lo que afecta su viabilidad. En resumen, si bien la fijación de precios por costo marginal simplifica las decisiones de fijación de precios, pasa por alto factores críticos como el comportamiento del consumidor, los objetivos a largo plazo y la dinámica del mercado. Las empresas deben lograr un equilibrio entre precios basados en costos y consideraciones estratégicas para prosperar en el mercado. Desafíos y limitaciones de la fijación de precios por costo marginal - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 5. Ejemplos de fijación de precios por costo marginal en la práctica La fijación de precios de costo marginal es una estrategia de fijación de precios empleada por las empresas donde el precio de un producto o servicio se fija igual al costo marginal de producción. Este enfoque tiene como objetivo maximizar la eficiencia y garantizar que los precios reflejen con precisión los costos incurridos en la producción de unidades adicionales. Exploremos algunos ejemplos de cómo se aplica en la práctica el precio del costo marginal: Servicios públicos: muchas empresas de servicios públicos, como los proveedores de agua y electricidad, adoptan precios de costo marginal. Fijan sus precios basándose en el costo incremental de producir y distribuir cada unidad adicional del servicio público. Esto garantiza que los consumidores paguen un precio justo que se alinee con el costo real de brindar el servicio. Aerolíneas: En la industria de las aerolíneas, el precio de costo marginal se utiliza comúnmente para fijar el precio de los asientos. Las aerolíneas consideran factores como los costos de combustible, los gastos de tripulación y los costos de mantenimiento para determinar el costo marginal de operar cada asiento adicional. Al fijar precios cercanos al costo marginal, las aerolíneas pueden optimizar los ingresos y al mismo tiempo llenar sus vuelos de manera eficiente. Servicios digitales: las empresas que ofrecen servicios digitales, como plataformas de streaming o proveedores de almacenamiento en la nube, suelen emplear precios de coste marginal. Estos servicios tienen costos marginales bajos ya que pueden atender a clientes adicionales sin incurrir en gastos adicionales significativos. Al fijar precios cercanos al costo marginal, estas empresas pueden atraer más clientes manteniendo la rentabilidad. Cadenas de comida rápida: algunas cadenas de comida rápida adoptan precios de costo marginal para ciertos elementos del menú. Por ejemplo, ofrecer precios con descuento durante las horas de menor actividad ayuda a utilizar el exceso de capacidad y minimizar el desperdicio. Al ajustar los precios en función del costo marginal de producir porciones adicionales, las cadenas de comida rápida pueden optimizar los ingresos y reducir el desperdicio de alimentos. Servicios de viajes compartidos: Las plataformas de viajes compartidos como Uber y Lyft utilizan precios de costo marginal para determinar las tarifas. El algoritmo de fijación de precios tiene en cuenta factores como la distancia, el tiempo y la demanda para calcular el coste marginal de ofrecer cada viaje. Esto permite fijar precios dinámicos que reflejan el costo real de brindar el servicio en un momento dado. Estos ejemplos resaltan cómo empresas de diversas industrias implementan precios de costo marginal para alinear los precios con los costos de producción. Al adoptar esta estrategia, las empresas pueden lograr eficiencia, optimizar los ingresos y ofrecer precios justos a los consumidores. Ejemplos de fijación de precios por costo marginal en la práctica - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 6. Comparación con otras estrategias de precios Precio de costo más: Descripción general: el precio de costo plus implica agregar un margen al costo de producción para determinar el precio de venta. Es sencillo y de uso común. Ventajas: Márgenes de beneficio predecibles: al establecer un margen de beneficio fijo, las empresas pueden garantizar márgenes de beneficio consistentes. Fácil Implementación: Calcular el precio es relativamente sencillo. Desventajas: Ignora la demanda: la fijación de precios de costo plus no considera la demanda del mercado ni la disposición a pagar del consumidor. Inflexible: Los precios permanecen estáticos incluso si las condiciones del mercado cambian. Ejemplo: una panadería calcula el coste de los ingredientes y la mano de obra de un pastel y añade un margen de beneficio del 30 % para llegar al precio de venta. Precios basados en el valor: Descripción general: los precios basados en el valor se centran en el valor percibido por el cliente. Alinea el precio con los beneficios recibidos. Ventajas: Centrado en el cliente: refleja lo que los clientes están dispuestos a pagar en función del valor percibido. Responsive: los precios se pueden ajustar en función de los cambios en las preferencias del cliente. Desventajas: Subjetivo: determinar el valor puede ser complicado; Varía de persona a persona. Intensivo en datos: requiere comprender en profundidad las preferencias del cliente. Ejemplo: una marca de relojes de lujo valora más sus productos porque los clientes los asocian con prestigio y artesanía. Precio de penetración: Descripción general: el precio de penetración implica establecer un precio inicial bajo para ganar participación de mercado rápidamente. Ventajas: Entrada al mercado: Ayuda a que nuevos productos penetren en mercados competitivos. Atrae clientes: los precios bajos atraen a consumidores sensibles al precio. Desventajas: Sacrificio de beneficios a corto plazo: los beneficios iniciales pueden ser bajos. Riesgo de percepción de marca: Percibida como de baja calidad por algunos consumidores. Ejemplo: un servicio de streaming ofrece un período de prueba gratuito para atraer a los usuarios a suscribirse. Precio de costo marginal: Descripción general: el precio de costo marginal establece el precio igual al costo marginal de producción. Ventajas: Eficiencia: Fomenta la asignación óptima de recursos. Equidad: Garantiza que los consumidores paguen lo que realmente cuesta producir una unidad adicional. Desventajas: Complejidad: Calcular los costos marginales con precisión puede resultar un desafío. Ignora los costos fijos: no tiene en cuenta los gastos generales fijos. Ejemplo: una empresa de viajes compartidos cobra a los pasajeros según el coste del combustible y el desgaste de cada viaje. Precios dinámicos: Descripción general: los precios dinámicos ajustan los precios en función de las condiciones del mercado, la demanda y la oferta en tiempo real. Ventajas: Ingresos Maximizados: Optimiza los precios para los períodos de máxima demanda. Adaptabilidad: Responde a las fluctuaciones de la demanda. Desventajas: Injusticia percibida: los clientes pueden sentirse manipulados. Algoritmos complejos: Requiere modelos de precios sofisticados. Ejemplo: las aerolíneas ajustan los precios de los billetes en función de factores como el tiempo de reserva, la disponibilidad de asientos y las fechas de viaje. Recuerde, cada estrategia de precios tiene su lugar en el panorama empresarial. La elección depende de factores como la industria, el tipo de producto y la dinámica competitiva. Mientras navega por el laberinto de precios, considere las compensaciones y seleccione la estrategia que mejor se alinee con sus objetivos comerciales. ¡No dudes en preguntar si quieres más ejemplos o más ideas! Comparación con otras estrategias de precios - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 7. Implementación de precios de costo marginal en las empresas La esencia de la fijación de precios por costo marginal En esencia, la fijación de precios por costo marginal consiste en alinear el precio de un producto o servicio con el costo incremental de producir una unidad adicional. A diferencia de los métodos tradicionales de fijación de precios que consideran los costos fijos y los costos promedio, los precios de costo marginal se centran únicamente en los costos variables incurridos al aumentar la producción. A continuación se presentan algunas ideas clave desde diferentes perspectivas: Eficiencia Económica: Defensores: Los economistas a menudo defienden la fijación de precios por costo marginal como un mecanismo eficiente. Al fijar el precio al costo marginal, las empresas asignan recursos de manera óptima, asegurando que los consumidores paguen un precio justo y que se minimice el exceso de capacidad. Críticos: Algunos argumentan que la fijación de precios por costo marginal ignora los costos fijos, que son esenciales para la sostenibilidad a largo plazo. Además, supone una competencia perfecta, que rara vez existe en los mercados del mundo real. Consideraciones a corto plazo frente a largo plazo: Beneficios a corto plazo: la implementación de precios de costo marginal puede generar ganancias a corto plazo. Por ejemplo, durante un período promocional, una empresa podría fijar precios iguales a los costos marginales para atraer más clientes. Riesgos a largo plazo: ignorar los costos fijos puede ser perjudicial a largo plazo. Si una empresa constantemente fija precios por debajo del costo total promedio, puede tener dificultades para cubrir los gastos fijos y poner en peligro su estabilidad financiera. Industrias y escenarios: Productos básicos: la fijación de precios por costo marginal prevalece en los mercados de productos básicos (por ejemplo, electricidad, agua) donde dominan los costos variables. Estas industrias a menudo enfrentan presiones regulatorias para mantener los precios cerca de los costos marginales. Productos de alta tecnología: en sectores en rápida evolución (por ejemplo, teléfonos inteligentes, software), las empresas pueden adoptar precios de costo marginal para ganar participación de mercado rápidamente. Bienes personalizados: para productos personalizados (por ejemplo, automóviles de lujo, trajes hechos a medida), las empresas rara vez utilizan el precio de costo marginal debido al énfasis en la singularidad y el valor de la marca. Desafíos y compensaciones: Asignación de costos: determinar el costo marginal preciso puede resultar complejo. Se deben considerar cuidadosamente los costos generales, los recursos compartidos y las economías de escala. Percepción del mercado: los consumidores suelen asociar precios más altos con mejor calidad. Fijar precios demasiado cerca de los costos marginales puede indicar baja calidad o desesperación. Margen de beneficio: si bien la fijación de precios de costo marginal garantiza la eficiencia, deja poco espacio para los márgenes de beneficio. Las empresas deben lograr un equilibrio entre recuperación de costos y competitividad. Ejemplos en la práctica Servicios de viajes compartidos (por ejemplo, Uber, Lyft): Estas plataformas ajustan dinámicamente los precios en función de la demanda (surge pricing). Si bien no se trata de una fijación de precios de costo estrictamente marginal, el concepto es similar. Una mayor demanda conduce a precios más altos, lo que refleja mayores costos variables (se necesitan más impulsores). Industria aérea: Las aerolíneas suelen utilizar el precio de coste marginal para la venta de billetes de última hora. Los asientos vacíos representan una pérdida de ingresos, por lo que los precios cercanos al coste marginal fomentan las reservas de última hora. Restaurantes de comida rápida: Los menús económicos con artículos de bajo precio (por ejemplo, hamburguesas a 1 dólar) demuestran precios de costo marginal. Estos artículos cubren costos variables y atraen a clientes preocupados por su presupuesto. En resumen, implementar la fijación de precios de costo marginal requiere un enfoque matizado. Las empresas deben sopesar las ganancias a corto plazo con los riesgos a largo plazo, considerar la dinámica de la industria y lograr un equilibrio entre eficiencia y rentabilidad. Al hacerlo, pueden navegar por el complejo panorama de precios y tomar decisiones informadas que beneficien tanto a sus resultados como a sus clientes. Recuerde, esta sección es solo el comienzo de nuestra exploración. ¡Hay mucho más por descubrir a medida que profundizamos en las complejidades de las estrategias de precios! Implementación de precios de costo marginal en las empresas - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 8. Impacto en la rentabilidad y la cuota de mercado Comprender el nexo: rentabilidad y cuota de mercado La rentabilidad y la participación de mercado están entrelazadas como dos bailarines de tango. Analicemos su relación: Estrategia de rentabilidad y precios: El Precio del Costo Marginal (MCP), como su nombre indica, fija precios iguales al coste marginal de producción. Este enfoque tiene como objetivo maximizar la eficiencia minimizando el desperdicio y garantizando que cada unidad adicional producida contribuya positivamente a las ganancias. Información 1: Cuando una empresa adopta MCP, puede lograr rentabilidad a corto plazo evitando sobreprecios y atrayendo clientes sensibles a los costos. Información 2: Sin embargo, depender únicamente de MCP puede generar desafíos a largo plazo, como una inversión insuficiente en innovación o mejora de la calidad. cuota de mercado y dinámica competitiva: La participación de mercado refleja la porción de las ventas totales de una empresa dentro de una industria o segmento de mercado específico. Es una métrica crucial para evaluar la competitividad. Información 3: Los precios agresivos (incluido el MCP) pueden aumentar la cuota de mercado al atraer a consumidores sensibles a los precios. Una mayor participación de mercado a menudo se traduce en economías de escala, poder de negociación y mayor visibilidad. Insight 4: Sin embargo, perseguir ciegamente la participación de mercado sin considerar la rentabilidad puede ser peligroso. Sacrificar márgenes por volumen podría conducir a inestabilidad financiera. Afrontando las compensaciones: equilibrando la rentabilidad y la cuota de mercado Ahora, profundicemos en los matices: metas a corto plazo frente a metas a largo plazo: Insight 5: MCP puede ser una victoria a corto plazo, especialmente para nuevas empresas o empresas que ingresan a un mercado competitivo. Ayuda a ganar tracción y construir una base de clientes. Ejemplo: un nuevo servicio de viajes compartidos que ofrece tarifas más bajas que los tradicionales atrae pasajeros rápidamente, pero enfrenta desafíos de rentabilidad en el futuro. elasticidad del precio y sensibilidad de la demanda: Insight 6: La demanda elástica (donde los consumidores son muy sensibles a los cambios de precios) favorece las ganancias de participación de mercado. La demanda inelástica (donde los cambios de precios tienen un impacto mínimo) permite márgenes más altos. Ejemplo: una aerolínea económica que utiliza MCP para llenar asientos vacíos durante las horas de menor actividad aumenta su cuota de mercado, pero debe gestionar los costes meticulosamente. Posicionamiento estratégico: Insight 7: Las empresas deben elegir sabiamente sus batallas. Algunas optan por precios superiores (mayores márgenes, menor participación de mercado) para posicionarse como marcas de lujo. Ejemplo: la estrategia de precios premium de Apple para los iPhone mantiene la rentabilidad al tiempo que preserva su imagen de marca. Reacciones de la competencia: Insight 8: Los competidores responden a los movimientos de precios. Si todo el mundo adopta el MCP, se convierte en una carrera hacia el fondo. Ejemplo: Los supermercados que participan en guerras de precios para ganar participación de mercado pueden perjudicar la rentabilidad de todos los actores. Conclusión En la compleja danza de la rentabilidad y la participación de mercado, las empresas deben lograr un equilibrio. MCP puede ser una herramienta poderosa, pero es esencial considerar las implicaciones a largo plazo, la dinámica competitiva y los objetivos estratégicos. Recuerde, el tango no se trata sólo de velocidad; se trata de gracia y precisión. Impacto en la rentabilidad y la cuota de mercado - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion 9. Conclusión y conclusiones clave En el ámbito de las estrategias de fijación de precios, la fijación de precios por coste marginal (MCP) destaca como un enfoque fascinante. Al fijar el precio igual al costo marginal de producción, las empresas pueden navegar por el complejo panorama de la oferta, la demanda y la rentabilidad. A medida que profundizamos en la conclusión de nuestra exploración, resumamos las conclusiones clave de esta estrategia. equilibrio entre eficiencia y rentabilidad: MCP tiene como objetivo lograr eficiencia económica asegurando que el precio cobrado por un producto o servicio se alinee estrechamente con el costo incremental de producir una unidad más. Este enfoque minimiza la pérdida de eficiencia y promueve la eficiencia asignativa. Sin embargo, las empresas deben lograr un delicado equilibrio. Si bien fijar precios al costo marginal garantiza la equidad, no siempre maximiza las ganancias. Entran en juego consideraciones estratégicas, poder de mercado y sostenibilidad a largo plazo. El contexto ideal para MCP: MCP funciona mejor en mercados competitivos donde las empresas carecen de un poder de mercado significativo. En tales entornos, los precios están impulsados por las fuerzas de la oferta y la demanda, y el MCP se alinea con la asignación eficiente de recursos. Cuando las empresas enfrentan competencia, fijar precios por encima del costo marginal puede llevar a una pérdida de participación de mercado. Por el contrario, fijar precios por debajo del costo marginal puede resultar en pérdidas a corto plazo. Desafíos y limitaciones: Costos fijos: MCP ignora los costos fijos (como alquiler, salarios y equipos) que no varían con los niveles de producción. Estos costos deben cubrirse para mantener el negocio en funcionamiento. Largo plazo versus corto plazo: MCP supone que las empresas pueden ajustar los niveles de producción instantáneamente. En realidad, las empresas enfrentan limitaciones debido a limitaciones de capacidad, retrasos en la producción y otros factores. Consideraciones de calidad: Centrarse únicamente en el costo marginal puede descuidar las mejoras de calidad. Invertir en mejores materiales, I+D o servicio al cliente puede mejorar la competitividad a largo plazo. Ejemplos en la práctica: Servicios públicos: los proveedores de agua, electricidad y gas natural suelen utilizar MCP. Cobran a los consumidores en función del costo de suministrar una unidad adicional. Sin embargo, los organismos reguladores garantizan que estas empresas cubran los costos fijos y mantengan la calidad del servicio. Industria aérea: algunas aerolíneas adoptan MCP para la venta de billetes de última hora. Cuando los asientos siguen sin venderse cerca de la salida, las aerolíneas pueden ofrecerlos a un costo marginal para maximizar los ingresos. Productos farmacéuticos: los fabricantes de medicamentos genéricos suelen poner precios a sus productos cercanos al coste marginal. Este enfoque beneficia a los consumidores al brindarles alternativas asequibles a los medicamentos de marca. Consideraciones éticas: MCP plantea cuestiones éticas. ¿Deberían las empresas priorizar el bienestar social sobre las ganancias? ¿Es justo cobrar precios más altos a segmentos con demanda inelástica (por ejemplo, medicamentos que salvan vidas)? Los defensores argumentan que el MCP promueve el bienestar social, especialmente para los bienes esenciales. Los críticos destacan el riesgo de una inversión insuficiente en innovación y calidad. Variaciones estratégicas: Tarifas en dos partes: las empresas cobran una tarifa fija (por ejemplo, suscripción de membresía) junto con un costo marginal por unidad. Este enfoque combina precios fijos y variables. precios de carga máxima: las empresas cobran precios más altos durante los períodos de demanda máxima (por ejemplo, electricidad durante las olas de calor del verano). Si bien no es un MCP puro, refleja fluctuaciones de costos. En resumen, la fijación de precios por costo marginal es una herramienta poderosa, pero su aplicación requiere matices. Las empresas deben sopesar la eficiencia económica, las consideraciones éticas y la viabilidad a largo plazo. Mientras nos despedimos de esta exploración, apreciemos la intrincada danza entre costo, valor e impacto social que da forma a las decisiones de fijación de precios. Conclusión y conclusiones clave - Fijacion de precios por costo marginal una estrategia de fijacion de precios en la que una empresa fija el precio igual al costo marginal de produccion Este blog se traduce automáticamente con la ayuda de nuestro servicio de inteligencia artificial. 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76	https://proofwiki.org/wiki/Continuous_Injection_of_Interval_is_Strictly_Monotone	Continuous Injection of Interval is Strictly Monotone From ProofWiki Jump to navigation Jump to search Theorem Let $I$ be a real interval. Let $f: I \to \R$ be an injective continuous real function. Then $f$ is strictly monotone. Proof Aiming for a contradiction, suppose $f$ is not strictly monotone. That is, there exist $x, y, z \in I$ with $x < y < z$ such that either: : $\map f x \le \map f y$ and $\map f y \ge \map f z$ or: : $\map f x \ge \map f y$ and $\map f y \le \map f z$ Suppose $\map f x \le \map f y$ and $\map f y \ge \map f z$. If $\map f x = \map f y$, or $\map f y = \map f z$, or $\map f x = \map f z$, $f$ is not injective, which is a contradiction. Thus, $\map f x < \map f y$ and $\map f y > \map f z$. Suppose $\map f x < \map f z$. That is: : $\map f x < \map f z < \map f y$ As $f$ is continuous on $I$, the Intermediate Value Theorem can be applied. Hence there exists $c \in \openint x y$ such that $\map f c = \map f z$. As $z \notin \openint x y$, we have $c \ne z$. So $f$ is not injective, which is a contradiction. Suppose instead $\map f x > \map f z$. That is: : $\map f z < \map f x < \map f y$ Again, as $f$ is continuous on $I$, the Intermediate Value Theorem can be applied. Then, there exists $c \in \openint y z$ such that $\map f c = \map f x$. So $f$ is not injective, which is a contradiction. If we suppose $\map f x \ge \map f y$ and $\map f y \le \map f z$, we reach a similar contradiction. By Proof by Contradiction, $f$ is strictly monotone. $\blacksquare$ Retrieved from " Categories: Proofs by Contradiction Proven Results Continuous Functions Navigation menu
77	https://edu.rsc.org/experiments/combustion-of-ethanol/1773.article	Combustion of ethanol \| Demonstration \| RSC Education Skip to main content Skip to navigation Resources Practical Experiments Mast navigation Register Sign In Search our site All All Resources Articles Search our site Search Menu Close menu Home I am a … Back to parent navigation item I am a … Primary teacher Secondary/FE teacher Early career or student teacher Technician HE teacher Student Resources Back to parent navigation item Resources Primary Secondary Higher education Curriculum support Practical Analysis Literacy in science teaching Periodic table Back to parent navigation item Periodic table Interactive periodic table Climate change and sustainability Careers Resources shop Collections Back to parent navigation item Collections Remote teaching support Starters for ten Screen experiments Assessment for learning Microscale chemistry Faces of chemistry Classic chemistry experiments Nuffield practical collection Anecdotes for chemistry teachers Literacy in science teaching More … Climate change and sustainability Alchemy On this day in chemistry Global experiments PhET interactive simulations Chemistry vignettes Context and problem based learning Journal of the month Chemistry and art Back to parent navigation item Chemistry and art Techniques Art analysis Pigments and colours Ancient art: today's technology Psychology and art theory Art and archaeology Artists as chemists The physics of restoration and conservation Cave art Ancient Egyptian art Ancient Greek art Ancient Roman art Classic chemistry demonstrations In search of solutions In search of more solutions Creative problem-solving in chemistry Solar spark Chemistry for non-specialists Health and safety in higher education Analytical chemistry introductions Exhibition chemistry Introductory maths for higher education Commercial skills for chemists Kitchen chemistry Journals how to guides Chemistry in health Chemistry in sport Chemistry in your cupboard Chocolate chemistry Adnoddau addysgu cemeg Cymraeg The chemistry of fireworks Festive chemistry Education in Chemistry Teach Chemistry Events Teacher PD Back to parent navigation item Teacher PD Courses Back to parent navigation item Courses Self-led Live online Selected PD articles PD for primary teachers PD for secondary teachers What we offer Chartered Science Teacher (CSciTeach) Teacher mentoring Enrichment Back to parent navigation item Enrichment UK Chemistry Olympiad Back to parent navigation item UK Chemistry Olympiad Who can enter? 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How does it work? Resources and past papers FAQs Top of the Bench Schools' Analyst How to prepare for the Chemistry Olympiad Chemistry Olympiad worked answers Chemistry Olympiad past papers Our work Regional support Education coordinators RSC Yusuf Hamied Inspirational Science Programme Policy Science Education Policy Alliance RSC Education News Supporting teacher training Interest groups More navigation items Combustion of alcohols In association with Nuffield Foundation No comments Illustrate the large energy changes that take place during the combustion of alcohols with this spectacular demonstration In this ‘alcohol gun’ experiment, students observe what happens when a mixture of ethanol vapour and air is ignited using an electric spark inside a corked plastic bottle. The resulting minor explosion fires the cork across the room. The demonstration can provide a useful illustration of the principle behind the internal combustion engine. It needs careful pre-lesson preparation, but only takes five minutes to perform. The class will need to be seated at a safe distance from the demonstration, while maintaining a good view of what is happening. Equipment Apparatus Eye protection for the teacher and students Safety screens Plastic fizzy drinks bottle, 500 cm 3 (see note 4 below) Cork, to fit the bottle Rubber stopper, size 17 Metal paper clips, 2 (see note 5) Leads, with a crocodile clip on one end, x2 EHT power pack (5 kV) or modified piezoelectric lighter (see notes 6 and 7) Retort stand, with boss and clamp G-clamp Chemicals Ethanol (IDA, Industrial Denatured Alcohol) (HIGHLY FLAMMABLE, HARMFUL) a few cm 3 Health, safety and technical notes Read our standard health and safety guidance. Wear eye protection throughout.The bang is not loud enough for students or the teacher to need to protect their ears. Ethanol (IDA, industrial denatured alcohol), CH 3 CH 2 OH(l),(HIGHLY FLAMMABLE, HARMFUL) – see CLEAPSS HazcardHC040A. Do NOT use any other alcohol other than ethanol (IDA). A fizzy drinks bottle is used as these are designed to withstand pressures up to 20 atmospheres. Do NOT use a bottle larger than 500 cm 3. Use a heated cork borer of suitable diameter to cut a hole in the base of the plastic bottle so that the size 17 rubber bung fits very tightly in the hole. Straighten both paper clips. Heat one clip until red-hot, and push through the rubber bung so that it protrudes about 5 mm from the narrow end of the bung. Repeat with the second clip so that it protrudes about 5 mm apart from the first to form the electrodes for a spark gap which will ignite the mixture. The ‘tails’ of the paper clips should be connected to a terminal block, as shown in the diagram below. It may be necessary to provide an insulating barrier between the exposed part of the paper clip ‘tails’ to avoid sparks occurring on the wrong side of the bung. Show Fullscreen Source: Royal Society of Chemistry How to build the spark generator using a rubber bung To complete the spark generator connect an EHT power pack to the other side of the terminal block shown in the diagram above. Adjust the spark gap so that a large spark occurs when the voltage is turned up to about 4.5 kV. This adjustment is important if the ‘gun’ is to fire reliably. Press the spark gap assembly firmly into the hole in the base of the bottle and check that sparking is still reliable. An alternative spark generator can be made from a piezoelectric gas lighter (see diagram below).To avoid the spark occurring between the lighter electrodes, bend one of electrodes well away from the other.Solder a pair of insulated leads onto each terminal of the lighter. Show Fullscreen Source: Royal Society of Chemistry How to build the spark generator using a piezoelectric gas lighter Alternatively, the spark gap of a piezoelectric lighter could be inserted directly into the hole in the bottle and glued and sealed in place. This does away with the need for leads, but leaves the teacher closer to the explosion. The demonstration should be practised by the teacher (with or without the cork in place) to establish the best electrode separation. If there are problems getting reliable sparking, ensure the spark assembly is dry – any moisture can cause the current to leak around the surface of the bung. Procedure Clamp the plastic bottle so that the mouth is pointing safely away from the class, the doorway and any vulnerable equipment or light fittings. Attach the retort stand to the bench with a G-clamp, and surround the bottle with safety screens on three sides. (A screen behind the bottle is important in case the ignition mechanism is ejected.) Show Fullscreen Source: Royal Society of Chemistry How to set up the ‘alcohol gun’ Squirt 1 cm 3 of ethanol into the bottle. Shake to ensure vaporisation of the alcohol and mixing of the vapour with air. Do NOT enrich the atmosphere with oxygen. The flash of the explosion can be seen if the room is darkened and is more easily seen if the bottle is transparent Insert the cork gently in the mouth of the bottle, ensure the bottle is securely clamped, and fire the ‘gun’ by turning up the EHT voltage or squeezing the piezoelectric lighter – the latter may have to be pressed several times before the mixture ignites. There will be a bang, and the cork will be fired a distance up to 5 m. After use, examine the bottle for damage. If there is any deformation or cracking, do not use again. To fire again the air in the bottle must be replenished by removing the sparking assembly and blowing through the bottle. It may be necessary to dry the sparking assembly with a tissue. If using EHT to fire the gun, after launch, switch off the mains. Do NOT touch any wiring or make adjustment if the voltmeter is still registering a voltage. Teaching notes A car spark plug could be available to show its similarity to the sparking assembly. The combustion energy for ethanol is –1367 kJ mol–1 corresponding to the equation: C 2 H 5 OH(l) + 3O 2(g) → 2CO 2(g) + 3H 2 O(l) The demonstration could be used in a variety of contexts: Learning about exothermic reactions, in terms of bonds - broken and bonds made Properties of alcohols Illustrating the internal combustion engine Ethanol as a fuel for cars Further information For a wide background discussion on the use of ethanol as a vehicle fuel, see the Wikipedia article on ethanol fuel. For those interested in the implications of this experiment for the safety of whisky distilleries, there is an appropriate article from the Health and Safety Executive:Potential explosion hazards due to evaporating ethanol in whisky distilleries, HSL/2003/08, Project Leader: H. S. Ledin, Author(s): H. S. Ledin MSc PhD DIC, Science Group: Fire and Explosion Group.The article does contain some interesting insights for teachers into the application of physical chemistry concepts to an industrial context. Additional information This is a resource from thePractical Chemistry project, developed by the Nuffield Foundation and the Royal Society of Chemistry. This collection of over 200 practical activities demonstrates a wide range of chemical concepts and processes. Each activity contains comprehensive information for teachers and technicians, including full technical notes and step-by-step procedures. Practical Chemistry activities accompanyPractical Physics andPractical Biology. © Nuffield Foundation and the Royal Society of Chemistry No comments Level 14-16 years 16-18 years Use Demonstrations Category Energy Fuels Reactions and synthesis Specification England A/AS level Edexcel Chemistry Topic 6: Organic Chemistry I Topic 6E: Alcohols 38 i. understand the reactions of alcohols with: oxygen in air (combustion) OCR Chemistry A Module 4: Core organic chemistry 4.2 Alcohols, haloalkanes and analysis 4.2.1 Alcohols b) combustion of alcohols Scotland Higher SQA Chemistry 3. Chemistry in society (c) Chemical energy National 5 SQA Chemistry Nature's chemistry Everyday consumer products Alcohols Alcohols are used as fuels as they are highly flammable and burn with very clean flames. Wales GCSE WJEC Chemistry Unit 2: CHEMICAL BONDING, APPLICATION OF CHEMICAL REACTIONS and ORGANIC CHEMISTRY 2.5 CRUDE OIL, FUELS AND ORGANIC CHEMISTRY (f) the combustion reactions of hydrocarbons and other fuels WJEC Combined science Unit 5: Chemistry 2 5.5 CRUDE OIL, FUELS AND CARBON COMPOUNDS (f) the combustion reactions of hydrocarbons and other fuels Northern Ireland A/AS level CCEA Chemistry Unit AS 2: Further Physical and inorganic Chemistry and an Introdution to Organic Chemistry 2.8 Energetics 2.8.1 define the terms exothermic and endothermic and understand that chemical reactions are usually accompanied by heat changes; 2.6 Alcohols 2.6.4 describe the complete and incomplete combustion of alcohols and their use as an alternative fuel; Advertisement Related articles Ideas How to build an equilibrium simulation 2025-07-30T06:00:00Z By Harry Martin Deepen your post-16 learners’ understanding of reversible chemical reactions while improving their IT skills News Reversing our reliance on fractional distillation of crude oil 2025-07-10T05:40:00Z By Nina Notman Scientists use semi-permeable polymer membrane and reverse osmosis to separate mixtures of hydrocarbons and save energy Job profile Tender and contract manager Adam uses his chemistry knowledge to help win business for his company. 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Sign inRegister More Experiments Class experiment Microscale titration \| 11–16 years Use this simple microscale titration experiment with integrated instructions to introduce your learners to titrations Class experiment Universal indicator microscale \| 11–16 years Introduce learners aged 11–16 to the concept of pH and address common misconceptions regarding the use of indicators Class experiment Electrolysis on a microscale \| 14–16 years Use this microscale version of the electrolysis of copper(II) chloride to safely explore a key practical experiment Load more articles Contact us Topics Issues Contributors Email alerts FAQs Safety Home About us Membership & professional community Campaigning & outreach Journals, books & databases News& events Locations & contacts Careers Teaching & learning Awards & funding Advertise Help & legal Privacy policy Terms & conditions FacebookBlueskyTwitterLinkedInYoutube © Royal Society of Chemistry 2025 Registered charity number: 207890 This website collects cookies to deliver a better user experience. 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78	https://www.pittsburghzoo.org/animal/elephant-african/	Skip to content Pittsburgh Zoo & Aquarium Connecting people to wildlife close African Elephant chevron_leftback to all animals Animal Quick Facts Size The African Savanna elephant is the largest living land-dwelling animal. Adults can range between 19 to 24 feet long, 8 to 11 feet tall and weigh 7000 to 12,000 pounds. #### Diet Grasses, tree limbs, leaves, tubers, and fruit. #### Color Gray. #### Lifespan Up to 60 years. #### Range African Savanna elephants can be found scattered across sub-Saharan Africa. They are found throughout central, eastern, and southern Africa. The African forest elephant is a separate species and can be found in the denser forests of the Congo basin and West Africa. #### Habitat Forests, partial deserts, and grasslands. Loxodonta africana Location in the Zoo: African Savanna Elephants also reside at the Zoo’s International Conservation Center African elephants are herbivorous animals, meaning they eat shrubbery, vines, herbs, tree leaves and bark. Elephants live in matriarchal societies with their offspring. Male elephants, called bulls, leave the family herd as they mature and join bachelor groups. An elephant pregnancy lasts about 22 months and elephant babies, called calves, are born weighing 200 to 250 pounds. Elephants can communicate via infrasound, below the range of human hearing. View Zoo Map
79	https://openstax.org/books/introductory-business-statistics-2e/pages/7-4-finite-population-correction-factor	Loading [MathJax]/jax/element/mml/optable/BasicLatin.js Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Introductory Business Statistics 2e 7.4 Finite Population Correction Factor Introductory Business Statistics 2e 7.4 Finite Population Correction Factor Search for key terms or text. We saw that the sample size has an important effect on the variance and thus the standard deviation of the sampling distribution. Also of interest is the proportion of the total population that has been sampled. We have assumed that the population is extremely large and that we have sampled a small part of the population. As the population becomes smaller and we sample a larger number of observations the sample observations are not independent of each other. To correct for the impact of this, the Finite Correction Factor can be used to adjust the variance of the sampling distribution. It is appropriate when more than 5% of the population is being sampled and the population has a known population size. There are cases when the population is known, and therefore the correction factor must be applied. The issue arises for both the sampling distribution of the means and the sampling distribution of proportions. The Finite Population Correction Factor for the variance of the means shown in the standardizing formula is: Z=x¯−µσn√⋅N−nN−1−−−−√ and for the variance of proportions is: σp'=p(1−p)n−−−−−−−√×N−nN−1−−−−−−√ The following examples show how to apply the factor. Sampling variances get adjusted using the above formula. Example 7.1 It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. Problem If the sample size is 225 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds. Solution N=4000,n=225,σ=10.37,µ=75.45,(x¯−µ)=±2 Z=x¯−µσn√⋅N−nN−1−−−−√=±210.37225√⋅4000−2254000−1−−−−−−√≈2.978 f(Z)≈2.978⋅2=0.997 Note that "differs by less" references the area on both sides of the mean within 2 pounds right or left. Try It 7.1 For the 5,000 students in University A, the average height of the students is 170 cm. The population standard deviation is 5 cm. If the sample size is 100 students, find the probability that the sample mean is within 1 cm of the true mean. Example 7.2 When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded their credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06. Problem Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit? Solution N=3000,n=360,p=0.06 σp'=p(1−p)n−−−−−−−√×N−nN−1−−−−−−√=0.06(1−0.06)360−−−−−−−−−−−−√×3000−3603000−1−−−−−−−−−√=0.0117 p1=10360=0.0278,p2=20360=0.0556 Z=p'−pp(1−p)n−−−−−√⋅N−nN−1−−−−√=0.0278−0.060.011744=−2.74 Z=p'−pp(1−p)n−−−−−√⋅N−nN−1−−−−√=0.0556−0.060.011744=−0.38 p(0.0278−0.060.011744<z<0.0556−0.060.011744)=p(−2.74<z<−0.38)=0.4969−0.1480=0.3489 Try It 7.2 There is a 0.01 probability that a student gets food poisoning eating food in a cafeteria. Suppose there are 2,000 orders in a day in the cafeteria. If we select 200 orders randomly, what is the probability that between 5 and 10 students will get food poisoning? Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Alexander Holmes, Barbara Illowsky, Susan Dean Publisher/website: OpenStax Book title: Introductory Business Statistics 2e Publication date: Dec 13, 2023 Location: Houston, Texas Book URL: Section URL: © Jun 25, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
80	http://hyperphysics.phy-astr.gsu.edu/hbase/pertab/cs.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Cesium \| \| \| \| \| --- --- \| \| Symbol \| Cs \| Atomic number \| 55 \| \| \| \| \| \| \| \| --- --- --- \| \| Atomic mass \| \| \| \| 132.905 \| \| Thermal conductivity \| \| \| \| 0.359 \| \| \| Atomic radius \| \| \| \| 3.34 \| \| Covalent radius \| \| \| \| 2.35 \| \| \| Density \| \| \| \| 1.87 \| \| Atomic volume \| \| \| \| 71.07 \| \| \| Melting point \| \| \| \| 301.55 \| \| Oxidation states \| \| \| \| 1 \| \| \| Boiling point \| \| \| \| 952 \| \| Stable isotopes \| \| \| \| 1 \| \| \| Heat of vapor. \| \| \| \| 67.74 \| \| Electronegativity \| \| \| \| 0.79 \| \| \| Heat of fusion \| \| \| \| 2.092 \| \| First ionization \| \| \| \| 3.894 \| \| \| Specific heat \| \| \| \| 0.24 \| \| Electrical conduct. \| \| \| \| ... \| \| \| \| \| Electron configuration \| \| \| \| --- \| \| General description \| Nuclear data \| \| IndexPeriodic TableChemistry concepts \| \| \| \| \| --- \| \| HyperPhysics \| R Nave \| \| Go Back \| \| \| \| \| \| --- --- \| \| Cesium The compounds of cesium are similar to those of potassium. This presents a particular hazard when radioactive cesium-137 contaminates the soil, as it did near Chernobyl. Growing plants require large amounts of potassium, and concentrate it from the soil. Since cesium mimics potassium, the radioactive cesium is also taken up and reconcentrated. The result is that plants grown on contaminated soil may contain a much higher concentration of radioactive cesium than the soil in which they are grown. The electron configuration of cesium consists of one lone 6s electron outside a perfectly symmetrical core of 54 other electrons characteristic of the noble element xenon. This fact has made it the foundation for mankind's best clock, the cesium atomic clock. \| \| \| --- \| \| Atomic data \| Nuclear data \| \| IndexPeriodic TableChemistry conceptsReferencePaulingCh. 26 \| \| \| \| \| --- \| \| HyperPhysics \| R Nave \| \| Go Back \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Cesium Nuclear Data \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Z \| A \| AtomicMass (u) \| NuclearMass(GeV/c2) \| Binding Energy(MeV) \| Spin \| NaturalAbund. \| Half-life \| Decay \| QMeV \| \| 55 \| 133 \| 132.905429 \| 123.7735 \| 1118.6 \| 7/2 \| 1.00 \| stable \| ... \| ... \| \| 55 \| 135 \| 134.905885 \| 125.6369 \| 1134.3 \| 7/2 \| ... \| 2.3My \| b- \| 0.21 \| \| 55 \| 137 \| 136.907074 \| 127.5011 \| 1149.3 \| 7/2 \| ... \| 30.17y \| b- \| 1.17 \| Cesium-137 is one of the most dangerous of the fission fragments from uranium-235 fission. \| \| \| Atomic data \| \| IndexPeriodic TableChemistry conceptsReferenceRohlf \| \| \| \| \| --- \| \| HyperPhysics \| R Nave \| \| Go Back \|
81	http://freeit.free.fr/The%20Civil%20Engineering%20Handbook,2003/0958%20ch20.pdf	© 2003 by CRC Press LLC 20 Stress Distribution 20.1 Elastic Theory (Continuum) Three-Dimensional Systems • Two-Dimensional Systems 20.2 Particulate Medium Two-Dimensional Systems • Three-Dimensional System • Distributed Vertical Loads at Surface From a microscopic point of view, all soil bodies are composed of discrete particles that are connected to each other by forces of mutual attraction and repulsion. Given an initial state of equilibrium, if an additional force system is applied, deformations may occur; particle arrangements may be altered; and changes in the distribution of the resultant forces may take place. Although the intensity of the generated forces may be high at points of contact, their range of inﬂuence is very short. Generally, effects extend only over a distance of molecular size or in the very near vicinity of the particles. The internal forces generated at these points by the induced loadings are called stresses. Because stresses reﬂect distributional changes induced by boundary loadings, they can be thought of as providing a measure of the transmission of induced energy throughout the body. The transformation from real soil bodies, composed of discrete particles, into a form such that useful deductions can be made through the exact processes of mathematics is accomplished by introducing the abstraction of a continuum, or continuous medium, and pleading statistical macroscopic equivalents through the introduc-tion of material properties. In concept, the soil is assumed to be continuously distributed in the region of space under consideration. This supposition then brings the continuous space required by mathemat-ical formulations and the material points of real bodies into conformity. This chapter will present some basic solutions for increases in vertical stresses due to some commonly encountered boundary loadings. Solutions will be presented both from the linear theory of elasticity and from particulate mechanics. Special efforts have been made to present the results in easy-to-use form. Many solutions can also be found in computer software packages. 20.1 Elastic Theory (Continuum) Three-Dimensional Systems Soil in the neighborhood of a point is called isotropic if its deﬁning parameters are the same in all directions emanating from that point. Isotropy reduces the number of elastic constants at a point to two: E, the modulus of elasticity; and m, Poisson’s ratio [Harr, 1966]. If the elastic constants are the same at all points within a region of a soil body, that region is said to be homogeneous. Invoking linear constitutive equations produces Milton E. Harr Purdue University 20-2 The Civil Engineering Handbook, Second Edition (20.1) where ei and si are the normal strains and stresses, respectively, and gi and ti are the shearing strains and stresses in the i = x, y, z directions, respectively. Force Normal to Surface (Boussinesq Problem) Assuming that the z direction coincides with the direction of gravity, the vertical stress under a concen-trated load P as shown in Fig. 20.1(a), where R2 = x2 + y2 + z2, is [Boussinesq, 1885] (20.2) It should be noted that the vertical normal stress (sz) is independent of the so-called elastic parameters. Equation (20.2) can also be written as (20.3) FIGURE 20.1 Concentrated force acting at and normal to surface; sz = IB P/z2. (b) 1.0 0.10 0.8 IB 0.08 0.6 0.06 0.4 0.04 0.2 0 0.1 0.2 0.1 2.0 0.3 0.4 0.5 3.0 0.6 0.7 4.0 0.8 0.9 5.0 1.0 0.02 0 1 r z 1 r z 1 r z r z r z 3 2π = IB 1 + 2 −5 . 2 0 1 r z z x y P R sz r z (a) ex 1 E --- sx m sy sz – ( ) – [ ] = ey 1 E --- sy m sx sz – ( ) – [ ] = ez 1 E --- sz m sx sy – ( ) – [ ] = g yz 2 1 m + ( ) E --------------------tyz = g xz 2 1 m + ( ) E --------------------txz = g xy 2 1 m + ( ) E --------------------txy = sz 3Pz 3 2pR 5 ------------= sz IB P z 2 ----= © 2003 by CRC Press LLC Stress Distribution 20-3 where A plot of the inﬂuence factor IB is given in Fig. 20.1(b). Because superposition is valid, the effects of a number of normal forces acting on the surface of a body can be accounted for by adding their relative inﬂuence values. Example 20.1 Find the vertical stress at point M in Fig. 20.2 due to the three loads shown acting in a line at the surface. Solution. For the 200 lb force r/z = 0 and, from Fig. 20.1(b), IB = 0.478. For the 100 lb forces, r/z = 20/10 = 2, IB = 0.009. Hence, the corresponding vertical stress at point M is By applying the principle of superposition, the increased stress due to distributed loadings over ﬂexible areas at the surface can be obtained by dividing the loading into increments (see Fig. 20.3) and treating each increment as a concentrated force. For a concentrated force parallel to the boundary surface (Fig. 20.4), the vertical stress [Cerruti, 1882] is (20.4) By combining Eqs. (20.2) and (20.4), the increase in vertical stress, consistent with the assumptions of linear elasticity, can be determined for a concentrated force at the surface with any inclination. FIGURE 20.2 Example 20.1. FIGURE 20.3 Distributed loads. 100 Ib 200 Ib 100 Ib 20' 20' 10' M y Plan x y Profile z IB 3 2p ------ 1 r z --Ë ¯ Ê ˆ 2 + 5/2 – = sz 200 0.478 ( ) 100 --------------------------2 100 ( ) 0.009 ( ) 100 ----------------------------------+ 0.974 lb/ft 2 = = sz 3Qxz 2 2pR 5 ---------------= © 2003 by CRC Press LLC 20-4 The Civil Engineering Handbook, Second Edition Uniform Flexible Load over Rectangular Area The vertical stress under the corner of a ﬂexible, uniformly loaded, rectangular area with sides a by b, as in Fig. 20.5(a), is (20.5) FIGURE 20.4 Concentrated force parallel to surface. FIGURE 20.5 Normal uniform load over rectangular area. Source: After Steinbrenner, W. 1936 A rational method for the determination of the vertical normal stresses under foundations. Proc. 1st Int. Conf. Soil Mech. Found. Eng., Vol. 2. x y z 0 Q z b a (a) m = m = 10 n = , q sz sz = qIR a b z b 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.10 IR 0.08 0.06 0.04 0.02 0 1 2 3 4 5 (b) n 6 7 8 9 4 3 2 1 1.4 6 10 sz q 2p ------mn 1 m 2 n 2 + + ------------------------------- 1 m 2 2n 2 + + 1 n 2 + ( ) m 2 n 2 + ( ) ------------------------------------------sin 1 – m m 2 mn 2 + 1 n 2 + ----------------------------------------------+ = © 2003 by CRC Press LLC Stress Distribution 20-5 where m = a/b and n = z/b. This can also be written as where IR is a dimensionless inﬂuence factor. Figure 20.5(b) gives a plot of IR as a function of the dimensionless ratios m and n. This form of the solution was given by Steinbrenner . By superposition, the distribution of vertical stress can be obtained anywhere under uniformly loaded, ﬂexible, rectangular loadings. Two cases will be examined. Vertical Stress for a Point Interior to a Loaded Area For this case, Fig. 20.6(a), the inﬂuence factor IR is determined from Fig. 20.5(b) for each of the rectan-gular areas (Roman numerals) with their corresponding m and n values and add them to obtain Vertical Stress for a Point Exterior to a Loaded Area For a point such as B in Fig. 20.6(b), the stress is computed as Inﬂuence Chart Normal Load Although the above procedure can also be used to approximate irregularly shaped loadings, an inﬂuence chart, developed by Newmark greatly reduces the work required. Such a chart is shown in Fig. 20.7. To use the chart, the shape of the loading is drawn (generally on tracing paper) to scale so that the length AB on the chart represents the depth z at which the vertical stress is desired. The scaled drawing is then oriented so that the point under which the stress is sought is directly over the center of the circles on the chart. The number of blocks covered by the area of loading multiplied by the inﬂuence value (each square is 0.001q for this chart) yields, for the vertical stress, that part of the uniformly distributed load. By repeating this procedure and varying the size of the drawings, the complete distribution of vertical stresses can be found with depth. A separate drawing of the area is required for each depth. Although the chart was developed for uniform loadings over the whole area, the effects of varying loads can be treated as a series of uniform loadings. Two-Dimensional Systems In soil mechanics and foundation engineering, problems such as the analysis of retaining walls or of continuous footings and slopes generally offer one dimension very large in comparison with the other two. Hence, if boundary forces are perpendicular to and independent of this dimension, all cross sections will be the same. In Fig. 20.8 the y dimension is taken to be large, and it is assumed that the state of affairs existing in the xz plane holds for all planes parallel to it. These conditions are said to deﬁne the state of plane strain. FIGURE 20.6 Vertical stress. (a) Interior. (b) Exterior. I I II II III III IV (a) (b) IV A B sz qIR = szA q IRI IRII – IRIII IRIV + + ( ) = szB q IRI III + IRII IV + IRIII – IRIV – + ( ) = © 2003 by CRC Press LLC 20-6 The Civil Engineering Handbook, Second Edition Inﬁnite Line Load Normal to Surface (Flamant Problem) Figure 20.9(a) shows a semi-inﬁnite plane with a concentrated load (line load) of intensity P (per unit run) normal to the surface. The solution, given by Flamant , is (20.6) In Fig. 20.9(b) is given a plot of this equation taken as sz = Isz(P/z). Also plotted are the inﬂuence factors for Isx and the tangential stress, It. Example 20.2 Find the vertical normal stress at a depth corresponding to point A in Fig. 20.10, due to the three normal line forces N1, N2 and N3. Solution. The inﬂuence curve is oriented so that the origin (point O) is directly above point A (as shown). The magnitude of the force multiplied by the ordinate of the sz curve immediately under it for P = 1 (such as under N1) gives that part of the stress at point A due to the particular force. By superposition the total vertical stress at point A is obtained as the algebraic sum of the contributions of each of the FIGURE 20.7 Inﬂuence chart for uniform vertical normal stress. Source: After Newmark, N. M. 1942. Inﬂuence charts for computation of stresses in elastic foundations. Univ. of Illinois Bull. 338. Scale distance = depth z Influence value = 0.001 B A O sz 2Pz 3 p x 2 z 2 + ( ) 2 --------------------------= B1C1 © 2003 by CRC Press LLC Stress Distribution 20-7 FIGURE 20.8 Examples of plane strain models. FIGURE 20.9 Inﬁnite line load, normal to surface. O O y x (b) (c) Retaining wall Continuous footing z y x (a) O Slope y x z z Iσz Iσz σz σx Iσx Iτ Iσx Iτ 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.5 1.0 1.5 (b) 2.0 2.5 x /z (a) x z z x M P , line load Iσz σz = P z P z P z Iσx σx = Iτ τxz = © 2003 by CRC Press LLC 20-8 The Civil Engineering Handbook, Second Edition forces. Thus, for the three forces N1, N2, and N3, the increase in vertical stress at point A is szA = . Superposition also permits the determination of the stress under any distribution of ﬂexible surface loading. For example, to obtain the vertical stress at point A in Fig. 20.10 under the distributed line load q(x), the load is ﬁrst divided into a number of increments, and each increment is then treated, as just described, as a concentrated force. Inﬁnite Strip of Width b An inﬂuence chart [Giroud, 1973] provides the stress sz at point P(x, z) in Fig. 20.11, due to the distributed vertical load q over a ﬂexible strip of width b in the form (20.7) Plots of this equation as well as the inﬂuence factor Isx are also shown. 20.2 Particulate Medium Two-Dimensional Systems Inﬁnite Line Load Normal to Surface On the basis of a probabilistic model, Harr gave the expression for the vertical stress due to a concentrated load (line load) of intensity P (per unit run) normal to the surface (see Fig. 20.9) as (20.8) The symbol – sz will be used to designate the vertical normal stress for the probabilistic model. The overlaid bar implies that this is the expected value of the stress. The customary symbol sz will be reserved for the linear theory of elasticity. Harr called the parameter n (Greek nu) the coefﬁcient of lateral stress and showed that it can be approximated by the conventional coefﬁcient of lateral earth pressure, K. For comparisons with the linear theory of elasticity, n ª 1/3. Example 20.3 Compare the distribution of the vertical normal stress of the particulate theory with that given by the theory of elasticity, Eq.(20.6), using n = Kactive = 1/3. Solution. Some numerical values of these two functions are given in Table 20.1. The correspondence is seen to be very good. FIGURE 20.10 Example 20.2. 1.0 0.8 0.6 0.4 Stress / (P/z) 0.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 A txz sx sz O P q(x) N2 N3 x z C1 N1 C3 C2 B3 B2 B1 N1 B1C1 ( ) N2 B2C2 ( ) N3 B3C3 ( ) + + ( ) sz Isz q ◊ = sz P z 2np ----------------- exp x 2 2nz 2 -----------– = © 2003 by CRC Press LLC Stress Distribution 20-9 Inﬁnite Strip of Width b The counterpart of Eq. (20.7), Fig. 20.11, for the particulate model is (20.9a) and under the center line (x = 0), (20.9b) FIGURE 20.11 Stresses under inﬁnite strip loading. Source: Giroud, J. P. 1973. Tables pour le Calcul des Fondations. 2 Vols. Dunod, Paris. σz σx ‘ = ∞ P z x z b (a) z x q b 2 x / Iσz, Iσx σz = q . Iσz σx = q . Iσx b 2z m = 1.0 m=5 m=2 m=1 m=5 m=2.5 m=2 m=0.33 m=0.25 m=0.2 m=1 m=0.1 m=0.5 m=0.33 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 1 2 3 4 (b) 5 sz q y x b/2 + z n -----------------y x b/2 – z n ----------------– Ó ˛ Ì ˝ Ï ¸ = sz 0 ( ) 2qy b 2z n -------------= © 2003 by CRC Press LLC 20-10 The Civil Engineering Handbook, Second Edition Values of the function y [ ] are given in Table 16.3 of Chapter 16. For example, y[0.92] = 0.321. In Fig. 20.12 a plot of – sz/q is given for a range of values (compare with Fig. 20.11 for n = 1/3). Example 20.4 Find the expected value for the vertical normal stress at point x = 2 ft, z = 4 ft for a uniformly distributed load q = 100 lb/ft2 acting over a strip 8 ft wide. Take n = p/8. Solution. Equation (20.9a) becomes for this case TABLE 20.1 Comparison of Particulate and Elastic Solutions (Inﬁnite Normal Line Load) 0.0 0.69 0.64 0.1 0.68 0.62 0.2 0.65 0.59 0.3 0.60 0.54 0.4 0.54 0.47 0.5 0.47 0.41 0.6 0.40 0.34 0.8 0.26 0.24 1.0 0.15 0.16 1.2 0.08 0.11 1.5 0.02 0.06 1.8 0.01 0.04 2.0 0.004 0.03 FIGURE 20.12 Uniform normal load over strip. x z --zsz P § 1 2np ----------Ë ¯ Ê ˆ 1/2 x 2 2nz 2 -----------– exp = zsz P § 2 p ---Ë ¯ Ê ˆ 1 x 2 z 2 ----+ 2 – = (a) b q x z x P(x, z) O sz /q 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 (b) b/2 b 1.5b 2b x 0 z 2v = 2.5b ÷ z 2v = 1.5b ÷ z 2v = b ÷ z 2v = b/2 ÷ z 2v = b/4 ÷ sz 100 y 2 4 + 4 0.63 ( ) ------------------y 2 4 – ( ) 4 0.63 ( ) ------------------– Ó ˛ Ì ˝ Ï ¸ = © 2003 by CRC Press LLC Stress Distribution 20-11 From Table 16.3 of Chapter 16, The theory of elasticity gives for the vertical stress in this case, from Fig. 20.11, sz = 73 lb/ft2. Multilayer System Given a system with N layers in which the i th layer has thickness hi and coefﬁcient of lateral stress ni, Fig. 20.13, Kandaurov showed that the equivalent thickness for the upper N – 1 layers can be found as (20.10) Hence, the expected vertical normal stress in the N th layer (zN is the vertical distance into the N th layer as measured from its upper boundary), for a line load P normal to the layers, will be (20.11) Example 20.5 A three-layer system is subjected to a line load of 9000 lb/ft with the following information: h1 = 1 ft, n1 = 0.4; h2 = 2 ft, n2 = 0.3; n3 = 0.2, h3 is unbounded. Find the expected value for the vertical normal stress 3 ft into the third layer immediately under the line load. Solution. From Eq. (20.10), the equivalent thickness is Thus, from Eq. (20.11) the expected vertical normal stress at a depth of 3 ft in the third layer, immediately under the line load (x = 0), is FIGURE 20.13 Multilayer system. 1 2 3 h1 h2 z2 z3 h3 O P sz (x, h1) x sz 100 0.4916 0.2881 + { } 78.0 lb/ft 2 = = hN 1 – h1 n1 nN ------h2 n2 nN ------L hN 1 – nN 1 – nN -----------+ + + = sz x z , ( ) P hN 1 – zN + ----------------------- 1 2pnN ------------- exp x 2 2nN hN 1 – zN + ( ) 2 ---------------------------------------– = h 1 0.4 0.2 -------2 0.3 0.2 -------+ 3.86 ft = = © 2003 by CRC Press LLC 20-12 The Civil Engineering Handbook, Second Edition The theory of elasticity would give for this case, assuming a single homogeneous layer, Eq. (20.6), sz = 954.9 lb/ft2. Three-Dimensional System Force Normal to Surface The probabilistic counterpart of the Boussinesq solution, Eq. (20.2), for the expected vertical normal stress is (20.12) where r2 = x2 + y2. Example 20.6 Find the coefﬁcient n in Eq. (20.12) that would yield the same value for the maximum expected value of normal vertical stress as that given by the theory of elasticity. Solution. For the theory of elasticity Eq. (20.2) gives as the maximum vertical stress Equation (20.12) gives Equating the two produces v = 1/3. Example 20.7 Compare the lateral attenuation of vertical normal stress for the probabilistic theory with that given by the theory of elasticity (for three dimensions) using v = 1/3. Solution. Substituting v = 1/3 into Eq. (20.12) obtains The theory of elasticity, Eq. (20.2), yields sz = 3P/2pz2(1 + r2/z2)–5/2. Some values of the two functions are given in Table 20.2. In Fig. 20.14 a nomograph of the expected vertical normal stress is given for a range of n values. Example 20.8 Find the expected value of the vertical normal stress at the point r = 6 ft, z = 10 ft if v = 1/5 under a concentrated vertical force of 100 lb. Solution. The arrow in Fig. 20.14 indicates that for the given conditions vz2–sz/P = 0.06. Hence, the expected value of the vertical normal stress is – sz = 0.065(100)(5)(1/100) = 0.33 lb/ft2. The theory of elasticity gives for this case, Fig. 20.1, sz = 0.22 lb/ft2. sz 9000 3.86 3 + ------------------- 1 2p 0.2 ( ) ------------------1169.7 lb/ft 2 = = sz P 2pnz 2 -------------- exp r 2 2nz 2 -----------– = szmax 3P 2pz 2 -----------= szmax P 2pnz 2 --------------= sz 3P/2pz 2 exp 3r 2/z 2 – [ ] = © 2003 by CRC Press LLC Stress Distribution 20-13 Distributed Vertical Loads at Surface Normal Uniform Load over a Rectangular Area The probabilistic counterpart of Eq. (20.5) for the expected vertical normal stress under the corner of a rectangular area with sides a by b, as in Fig. 20.5(a), is (20.13) Values of the function y [ ] are given in Table 16.3 of Chapter 16. FIGURE 20.14 Expected value of vertical stress, particulate model. TABLE 20.2 Comparison of Particulate and Elastic Solutions (Normal Point Load) r/z z2– sz/P z2sz/P 0.0 0.48 0.48 0.1 0.47 0.47 0.2 0.45 0.43 0.4 0.38 0.33 0.6 0.28 0.22 0.8 0.18 0.14 1.0 0.11 0.08 1.2 0.06 0.05 1.5 0.02 0.03 2.0 0.001 0.01 sz (r, z) r/zv = 1/3 r/zv = 1/4 r/zv = 1/5 r/zv = 1/6 r/zv = 1/7 r/zv = 1/8 x P r z y 0.18 1 5 v = ,r / z = 0.6 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0 1/3 1/4 1/5 v 1/6 1/7 1/8 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 0.16 vz2sz P sz q § y a z n ---------- y b z n ----------= © 2003 by CRC Press LLC 20-14 The Civil Engineering Handbook, Second Edition Example 20.9 A uniformly distributed vertical load of magnitude 25 lb/ft2 acts over an area 4 ft by 8 ft at the surface of a particulate medium. Find the expected vertical normal stress 6 ft below the center of the area. Take v = 1/3. Solution. From Eq. (20.13), taking the 4 ¥ 8 ft area as four areas, each with sides 2 ¥ 4 ft, we have and – sz = 8.2 lb/ft2. The theory of elasticity, as shown in Fig. 5(b), yields for this case sz = 7.3 lb/ft2. Uniform Normal Load over Circular Area The expected vertical normal stress under the center of a circular area, of radius a, subject to a uniform normal load q was obtained by Kandaurov as (20.14) The theory of elasticity gives for this case [Harr, 1966] (20.15) The inﬂuence chart shown in Fig. 20.15 was prepared using Eq. (20.14). The chart is used in the same way as was previously explained in reference to Fig. 20.7. For layered systems, the equivalent depth hN–1+ zN, Eq. (20.10) can be substituted for z. Example 20.10 Compare Eqs. (20.14) and (20.15) for v = 1/3. Solution. Some values are given in Table 20.3. Again we see that for v = 1/3 the correspondence between the two is very good for the special value of the coefﬁcient of lateral stress. FIGURE 20.15 Inﬂuence chart. 0 A B Influence value = 0.005 AB = z 2n ÷ sz 4 25 ( )y 2 6 1/3 --------------- y 4 6 1/3 ---------------100y 0.58 [ ]y 1.15 [ ] 100 0.219 ( ) 0.375 ( ) = = = sz z ( ) q 1 exp a 2 2nz 2 -----------– Ë ¯ Ê ˆ – = sz z ( ) q 1 1 a z § ( ) 2 1 + [ ] 3/2 -----------------------------------– = © 2003 by CRC Press LLC Stress Distribution 20-15 Parabolic Loading over Circular Area The parabolic form of the loading on the surface is deﬁned as (20.16a) For the special, but important, case under the center of the loading (r = 0), the expected vertical stress reduces to the expression (20.16b) Tangential (Horizontal Force at Surface) For a concentrated force of intensity Q parallel to the boundary surface (the equivalent of Cerruti’s solution, Fig. 20.4), Muller obtained the expression (20.17) Example 20.11 Compare the expression for the vertical stress in Eq. (20.17) with that given by Cerruti (Eq. 20.4) for the line y = 0, z = 1. Solution. Cerruti obtained sz = 3Qxz2/2 pR5 where R2 = x2 + y2 + z2. From Eq. (20.17) for y = 0, z = 1, It can be shown that limx Æ 0 – sz = 0. Some numerical values for two values of n are given in Table 20.4. It is seen that the vertical stresses for the developed theory are smaller than those given by the theory of elasticity in the near vicinity of the applied load. However, the general patterns of the distributions are quite similar away from the point of loading. TABLE 20.3 Comparison of Particulate and Elastic Solutions (Normal Vertical Distributed Load) a/z – sz/q sz/q 0.2 0.06 0.06 0.4 0.21 0.20 0.6 0.42 0.37 0.8 0.62 0.52 1.0 0.78 0.65 1.2 0.88 0.74 1.4 0.95 0.80 2.0 1.00 0.91 sz r 0 , ( ) q 1 r 2 a 2 ----– Ë ¯ Ê ˆ 0 r a £ £ 0 r a > Ó Ô Ì Ô Ï = sz 0 z , ( ) q 1 2nz 2 a 2 ----------- 1 exp a 2 2nz 2 -----------– Ë ¯ Ê ˆ – – Ó ˛ Ì ˝ Ï ¸ = sx Q 2pnx 2 --------------- exp y 2 z 2 + 2nx 2 ---------------– = 2psz Q ----------1 nx 2 -------- exp 1 2nx 2 -----------– = © 2003 by CRC Press LLC 20-16 The Civil Engineering Handbook, Second Edition Deﬁning Terms Homogeneous — Having elastic parameters the same at all points in a region of a body. Inﬂuence factor — A dimensionless cluster of factors generally obtained from a graphical presentation. Isotropic — Having deﬁning parameters the same in all directions emanating from a point. Plane strain — A two-dimensional simpliﬁcation used when there is little or no variation in strain in the direction normal to the plane. Strain — Changes in displacements due to changes in stress. Stress — Intensity of internal forces within a soil body induced by loadings. In the classical theory of elasticity stress is single-valued. The particulate theory deals with a distribution, the best mea-sure of which is the expected (mean) value. References Ahlvin, R. G., and Ulery, H. H. 1962. Tabulated values for determining the complete pattern of stresses, strains, and deﬂections beneath a uniform circular load on a homogeneous half space. Highway Res. Board Bull. 342. Barksdale, R., and Harr, M. E. 1966. An inﬂuence chart for vertical stress increase due to horizontal shear loadings. Highway Res. Board Rec. 108. Boussinesq, J. 1885. Applications des Potentiels à l’Étude de l’Équilibre et Mouvement des Solides Élastiques. Gauthier-Villard, Paris. Carothers, S. D. 1920. Direct determination of stresses. Proc. R. Soc., Series A, 97. Cerruti, V. 1882. Ricerche intorno al’equilibrio de’ corpi elastici isotropi, Reale Accademia dei Lincei, serie 3a Memoria della Classe di scienze ﬁsiche …, Vol. XIII, Rome. Egorov, K. E. 1940. Distribution of stresses in base under rigid strip footing. Mauchn. Issled, Stantsiya Fundamentstroya, no. 9. Fadum, R. E. 1948. Inﬂuence values for estimating stresses in elastic foundations. Proc. 2nd Int. Conf. Soil Mech. Found. Eng., Vol. 3. The Netherlands. Flamant, A. 1892. Comptes Rendus, Vol. 114, p. 1465. Florin, V. A. 1959. Fundamentals of Soil Mechanics. 2 Vols. Gosstroiizdat, Moscow. Giroud, J. P. 1973. Tables pour le Calcul des Fondations. 2 Vols. Dunod, Paris. Harr, M. E. 1966. Foundations of Theoretical Soil Mechanics. McGraw-Hill, New York. Harr, M. E. 1977. Mechanics of Particulate Media: A Probabilistic Approach. McGraw-Hill, New York. Harr, M. E. 1987. Reliability-Based Design in Civil Engineering. McGraw-Hill, New York. Harr, M. E., and Lovell, C. W., Jr, 1963. Vertical stresses under certain axisymmetrical loading. Highway Res. Board Rec. 39. Highway Research Board. 1962. Stress distribution in earth masses. Highway Res. Board Bull. 342. TABLE 20.4 Comparison of Particulate and Elastic Solutions (Concentrated Force Parallel to Boundary Surface) 2p – sz/Q x 2psz/Q v = 1/3 v = 1/2 0 0 0.00 0.00 0.25 0.64 0+ 0+ 0.5 0.86 0.03 0.15 0.75 0.74 0.37 0.60 1.0 0.53 0.67 0.74 1.5 0.24 0.68 0.57 2.0 0.11 0.52 0.39 2.5 0.05 0.38 0.27 © 2003 by CRC Press LLC Stress Distribution 20-17 Kandaurov, I. I. 1959. Theory of Discrete Distribution of Stress and Deformation … Izd. VATT (in Russian). Kandaurov, I. I. 1966. Mechanics of Discrete Media and Its Application to Construction. Izd. Liter. po Stroitel’stvu (in Russian). Translated into English by Zeidler, R. B. A. A. Balkema, Rotterdam, 1991. Kandaurov, I. I. 1988. Mekhanika Zernistykh Sred i Yeye Primeneniye v Stroitel’stve. Stroyizdat, Leningrad. Love, A. E. H. 1944. A Treatise on the Mathematical Theory of Elasticity. Dover Publications, Inc., New York. Michell, J. H. 1900. The stress distribution in an aeotropic solid with inﬁnite boundary plane. Proc. London Math. Soc., 32. Mindlin, R. D. 1936. Force at a point in the interior of a semi-inﬁnite solid. Physics 7. Muller, R. A. 1962. Statistical theory of transformation of stress in granular soil bases. Bases, Foundations, and Soil Mechanics, no. 4. Muller, R. A. 1963. Deformation of granular soil. Proc. All-Union Sci.-Res. Mining Inst., Leningrad. Newmark, N. M. 1940. Stress distributions in soils. Proc. Purdue Conf. on Soil Mech. and Its Appl., Lafayette, IN, July. Newmark, N. M. 1942. Inﬂuence charts for computation of stresses in elastic foundations. Univ. of Illinois Bull. 338. Newmark, N. M. 1947. Inﬂuence charts for computation of vertical displacements in elastic foundations. Univ. of Illinois, Eng. Exp. Stn. Bull. 45. Perloff, W. H., Galadi, G. Y., and Harr, M. E. 1967. Stresses and displacements within and under long elastic embankments. Highway Res. Rec. 181. Poulos, H. G., and Davis, E. H. 1974. Elastic Solutions for Soil and Rock, John Wiley & Sons, New York. Reprinted in 1991 by Centre for Geotechnical Research, University of Sydney, Sydney, NSW 2006, Australia. Reimbert, M. L., and Reimbert, A. M. 1974. Retaining Walls. Transactions Technological Publishing, Bay Village, OH. Steinbrenner, W. 1936. A rational method for the determination of the vertical normal stresses under foundations. Proc. Int. Conf. Soil Mech. Found. Eng. Vol. 2. Timoshenko, S. P. 1953. History of Strength of Materials. McGraw-Hill Book Company, New York. Todhunter, I., and Pearson, K. 1960. A History of the Theory of Elasticity and of the Strength of Materials. Dover Publications, Inc., New York (ﬁrst published by Cambridge University Press, London, 1886–1893). Turnbull, W. J., Maxwell, A. A., and Ahlvin R. G. 1961. Stresses and deﬂections in homogeneous soil masses. Proc. 5th Intl. Conf. Soil Mech. Found. Eng., Paris. Westergaard, H. M. 1938. A problem of elasticity suggested by a problem in soil mechanics: soft material reinforced by numerous strong horizontal sheets. In Contributions to the Mechanics of Solids, Dedicated to S. Timoshenko by His Friends on the Occasion of His Sixtieth Birthday Anniversary. The MacMillan Company, New York. Further Information Many compilations of elastic solutions are available in the literature; primary among these is that of Poulos and Davis , which has recently been reprinted (see references). Historical background can be found in Timoshenko , Todhunter and Pearson , and Love . Many particulate solutions can be found in Kandaurov and Harr . © 2003 by CRC Press LLC
82	http://ui.adsabs.harvard.edu/abs/2011APS..NWS.C1011G/abstract	Neutron Capture by Cadmium: Thermal Cross Sections and Resonance Integrals of ^106,108,110,112,114,116Cd - Astrophysics Data System Skip to main content Now on article abstract page Toggle navigation ads === Feedback Submit Updates General Feedback ORCID Sign in to ORCID to claim papers in the ADS. About About ADS What's New ADS Blog ADS Help Pages ADS Legacy Services Careers@ADS Sign Up Log In Search Bar to Enter New Query quick field:Author First Author Abstract Year Fulltext Select a field or operator All Search Terms Your search returned 0 results Your search returned 0 results Show Menu Full Text Sources view AbstractCitationsReferencesCo-ReadsSimilar PapersVolume ContentGraphicsMetricsExport Citation Neutron Capture by Cadmium: Thermal Cross Sections and Resonance Integrals of ^106,108,110,112,114,116Cd Show affiliations Loading affiliations affiliations loading Gicking, Allison M. (0); Krane, Kenneth S. (1) Abstract The neutron capture cross sections of the stable, even-mass Cd isotopes (A = 106, 108, 110, 112, 114, and 116) have been previously measured in sources of natural abundance or low enrichment, often making the results uncertain owing to the large absorption cross section of naturally occurring ^113Cd. Ambiguities in values of the isomeric branching ratios have also contributed to uncertainties in previous results. We have remeasured the Cd neutron capture cross sections using samples of greater than 90% isotopic enrichment irradiated in the OSU TRIGA reactor. Gamma-ray emission spectra were analyzed to determine the effective resonance integrals and thermal cross sections leading to eight radioactive ground and isomeric states in the Cd isotopes. Publication: American Physical Society, 13th Annual Meeting of the Northwest Section of the APS, October 20-22, 2011, abstract id. C1.011 Pub Date:October 2011 Bibcode: 2011APS..NWS.C1011G Copied! Feedback/Corrections? full text sources Publisher Electronic on-line publisher article (HTML) © The SAO Astrophysics Data System adshelp[at]cfa.harvard.edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement 80NSSC25M7105 The material contained in this document is based upon work supported by a National Aeronautics and Space Administration (NASA) grant or cooperative agreement. Any opinions, findings, conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of NASA. Resources About ADS ADS Help System Status What's New Careers@ADS Web Accessibility Policy Social @adsabs ADS Blog Project Switch to basic HTML Privacy Policy Terms of Use Smithsonian Astrophysical Observatory Smithsonian Institution NASA 🌓 × Back How may we help you? #### Missing/Incorrect Record Submit a missing record or correct an existing record.#### Missing References Submit missing references to an existing ADS record.#### Associated Articles Submit associated articles to an existing record (e.g. arXiv / published paper).#### General Feedback Send your comments and suggestions for improvements. Name Email Feedback Submit You can also reach us at adshelp [at] cfa.harvard.edu This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. Close without submitting
83	https://study.com/academy/lesson/tetrahedron-definition-formula.html	Tetrahedron \| Definition, Faces & Formula - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses NY Regents - Geometry Study Guide and Exam Prep Tetrahedron \| Definition, Faces & Formula Contributors: Lisa Gilbert, Shaun Ault Author Author: Lisa Gilbert Show more Instructor Instructor: Shaun Ault Show more Learn to define what a tetrahedron is. Discover the regular tetrahedron and the number of faces, edges, and vertices it has. Find the volume of a tetrahedron. Updated: 11/21/2023 Table of Contents Definition of a Tetrahedron How many Faces does a Tetrahedron have? Volume of Tetrahedron Applications of Tetrahedron Lesson Summary Show FAQ What are some examples of tetrahedra? The outer surface of the iconic Spaceship Earth at EPCOT in Walt Disney World is made up of thousands of tetrahedra. Additionally, Rubik's Cube created a tetrahedron puzzle similar to the cube but triangular in shape. How do you find the volume of a tetrahedron? The volume of a tetrahedron can be calculated by using this formula: V = 1/3(B)(h), where B is the area of the base and h is the height of the apex. The volume of a regular tetrahedron can be found with this formula: V=a 3/6 2 where a is the length of an edge. How would you describe a tetrahedron? A tetrahedron is a 3D figure with four triangular faces, four vertices, and six edges. 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Your next lesson will play in 10 seconds 0:01 Defining the Tetrahedron 0:38 Volume Formula in General 2:26 Volume of a Regular… 3:25 Surface Area 4:14 Lesson Summary QuizCourseView Video OnlySaveTimeline 58K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Triangular Pyramid \| Definition, Vertices & Formula 4:56 ##### Rectangular Pyramids \| Definition, Properties & Examples 4:02 ##### Triangular Pyramid \| Volume, Surface Area & Formula 5:54 ##### Rectangular Pyramid \| Volume, Slant Height & Surface Area 7:09 ##### Pyramids: Definition, Area & Volume 7:43 ##### Square Pyramid \| Overview, Properties & Formulas 3:59 ##### Surface Area & Volume of an Octahedron 3:44 ##### 3D Shapes \| Types, Properties & Examples 3:28 ##### Polyhedron \| Definition, Types & Examples 3:32 ##### Three-Dimensional Shapes \| Definition, Types & Characteristics 3:28 ##### 3-D Shapes: Lesson for Kids 2:59 ##### Octahedron \| Definition, Bases & Properties 5:08 ##### Centroid & Center of Mass of a Semicircle \| Overview & Examples 3:06 ##### Surface Area of a Hemisphere \| Overview & Formula 3:14 ##### Arc Length & Sector Area \| Definition, Formula & Examples 6:39 ##### Finding the Radius of a Circle \| Formula & Examples 3:45 ##### Area of a Sector \| Formula & Examples 6:33 ##### Volume & Surface Area of a Sphere \| Formula & Examples 4:19 ##### Hemisphere in Math \| Definition, Shape & Formula 2:49 ##### Indirect Measurement Meaning & Examples 4:37 ##### AP Calculus AB & BC: Exam Prep ##### ELM: CSU Math Study Guide ##### Geometry: High School ##### SAT Subject Test Mathematics Level 1: Practice and Study Guide ##### SAT Subject Test Mathematics Level 2: Practice and Study Guide ##### NY Regents Exam - Integrated Algebra: Test Prep & Practice ##### NY Regents - Geometry Study Guide and Exam Prep ##### Trigonometry: High School ##### McDougal Littell Geometry: Online Textbook Help ##### Prentice Hall Geometry: Online Textbook Help ##### SAT Mathematics Level 2: Help and Review ##### SAT Mathematics Level 1: Help and Review ##### Glencoe Algebra 1: Online Textbook Help ##### Holt McDougal Algebra I: Online Textbook Help ##### Algebra Connections: Online Textbook Help ##### Glencoe Pre-Algebra: Online Textbook Help ##### Basic Geometry: Help & Review ##### ORELA Middle Grades Mathematics Study Guide and Test Prep ##### WEST Middle Grades Mathematics (203) Study Guide and Test Prep ##### NMTA Middle Grades Mathematics (203): Practice & Study Guide Definition of a Tetrahedron --------------------------- A tetrahedron is a three-dimensional figure with three triangular sides and a triangular base. Sometimes referred to as a triangular pyramid or three-sided pyramid, a tetrahedron has six edges and four vertices. Tetrahedra are the simplest type of polyhedra. Other common types of polyhedra include pyramids, cubes, prisms, and octahedra. A tetrahedron with four triangular faces The outer surface of the iconic Spaceship Earth at EPCOT in Walt Disney World is covered with over 11,000 tetrahedra. Spaceship Earth at EPCOT is covered with over 11,000 tetrahedra. Properties of a Tetrahedron A tetrahedron has specific properties. These properties include: Faces: 4 triangular faces Edges: 6 edges Vertices: 4 vertices, where three faces and three edges meet at each Any shape that meets these criteria is a tetrahedron. Regular Tetrahedron A regular tetrahedron is one in which all four faces are equilateral triangles. The interior angles of an equilateral triangle all measure 60 degrees, and all three sides are of equal length. In a regular tetrahedron, all four faces are congruent, all four vertices are equidistant from each other, and all six edges are of equal length. Additionally, opposite edges of a regular tetrahedron are perpendicular to each other. All other tetrahedra are classified as irregular tetrahedra. Rubik's Cube created a puzzle similar to its original Rubik's cube but in the shape of a regular tetrahedron. Regular tetrahedron with 4 equal faces How many Faces does a Tetrahedron have? --------------------------------------- Unlike a typical pyramid, which has four triangular sides and one square or rectangular base, a tetrahedron has only three triangular sides and a triangular base. In other words, a tetrahedron has four faces, all of which are triangular in shape but are not necessarily congruent. Each face can be any type of triangle: isosceles, right, equilateral, scalene, obtuse, or acute. How many Vertices does a Triangular Pyramid have? A tetrahedron, or triangular pyramid, has a total of four vertices. Each vertex is the point at which three of the four faces meet. The top vertex, or the vertex opposite the base, is commonly referred to as the apex. Any four points not on the same plane in space, when connected, will form the vertices of a tetrahedron. How many Edges does a Triangular Pyramid have? A triangular pyramid has six edges. Each edge of a triangular pyramid is shared by two faces. For a regular tetrahedron, all six edges have the same length because each face is an equilateral triangle that is congruent to the other three faces. Volume of Tetrahedron --------------------- The volume of an object is measured as the amount of space within the boundaries of that object. The volume of a tetrahedron formula is the same as the formula for the volume of a pyramid. The volume of a tetrahedron is calculated by first finding the area of the base triangle and the height, or the perpendicular distance of the apex from the plane on which the base lies. Then, this formula is used to determine the volume: V=1/3(B)(h) where B is the area of the base and h is the height of the apex. Example: Find the volume of the tetrahedron in the image: B = 1/2(3)(5), h = 8 Solution: The area of the base triangle is A=1/2(3)(5) which is 7.5. The volume of this tetrahedron can be calculated as V=1/3(7.5)(8). So, the volume is 20 cubic units. The formula for finding the volume of a regular tetrahedron is V=a 3/6 2 where a is the length of an edge. Example: Find the volume of a regular tetrahedron whose edges are all equal to 4. Solution: Substitute the 4 in the formula for the a and calculate. V=4 3/6 2 The volume of a regular tetrahedron with an edge length of 4 is 7.54 cubic units. The volume of a tetrahedron can also be calculated when given a set of points for the vertices. Example: What is the volume of a tetrahedron with coordinates (0,0,0), (0,4,0), (0,0,3), and (3,0,0)? Solution: Because the base triangle vertices are at (0,0,0), (0,4,0), and (3,0,0), the area of the base is 6. The height of the apex is 3. So, the volume is (1/3)(6)(3), which is 6 cubic units. Surface Area of a Tetrahedron The surface area of a tetrahedron is the combined area of all four faces. The surface area of an irregular tetrahedron is the sum of the areas of all faces. The formula for calculating the surface area of a regular tetrahedron is S=3⋅a 2 where a is the length of the edge. Example: Using the same example as above, calculate the surface area of the regular tetrahedron whose edges are all equal to 4. Solution: Substitute 4 for a in the formula. S=3⋅4 2 The surface area of a regular tetrahedron with edge lengths of 4 is 27.71 square units. Applications of Tetrahedron --------------------------- The tetrahedral shape is found in many different applications. The calculations used in mathematics can be applied to architecture and chemistry. A tetrahedral structure is often incorporated into architecture. It forms a very rigid structure, allowing for strong builds that can bear a lot of weight. The tetrahedral shape is seen in floors, beams, and bridges. The tetrahedral shape can also be found naturally in the chemical shapes that molecules form. One of the common molecular shapes is the tetrahedral. The angle of a tetrahedron is 60 degrees. When a central vertex is added, the angle between the central vertices and the angles of a tetrahedron equals 109.5 degrees. This is what defines the tetrahedral molecular shape, where the molecules are 109.5 degrees apart from each other. Lesson Summary -------------- A tetrahedron is a type of polyhedron with four faces, six edges, and four vertices in a three-dimensional space. Tetrahedra are also known as three-sided pyramids or triangular pyramids. The faces of a tetrahedron can be any type of triangle. The vertex that is opposite the tetrahedron's base is called the apex. A regular tetrahedron is a specific type of tetrahedron whose faces are all congruent equilateral triangles with all interior angles measuring 60 degrees, whose edges are all of equal length, whose vertices are all equidistant from each other, and whose opposite edges are perpendicular to each other. The volume of a tetrahedron, or the amount of space contained within the shape, can be calculated using the formula V=1/3(B)(h) where B is the area of the base and h is the height of the apex. The formula used to determine the volume of a regular tetrahedron is V=a 3/6 2 where a is the length of an edge. Volume is measured in cubic units. The surface area of a tetrahedron is the sum of the areas of all four triangular faces. The formula for the surface area of a regular tetrahedron is S=3⋅a 2 where a is the length of the edges. Surface area is measured in square units. Video Transcript Defining the Tetrahedron A tetrahedron, or the plural tetrahedra, is simply a pyramid with a triangular base. So no, not like the ones in Egypt. It is a solid object with four triangular faces, three on the sides or lateral faces, and one on the bottom or the base, and four vertices or corners. If the faces are all congruent equilateral triangles, then the tetrahedron is called regular. In fact, you may have seen and held in your hands a regular tetrahedron. Some games use 4-sided dice in the shape of regular tetrahedra. Volume Formula in General Because the tetrahedron is a type of pyramid, its volume formula is the same as for all pyramids: In this formula, B is the area of the base, and h is the height. For example, a tetrahedron with a height of 10 inches and base triangle that has an area of 12 square inches, would have a volume of one third times 12 times 10. Volume equals 40 cubic inches. The next example is a little more challenging. Suppose there is a tetrahedron with vertices located at the points A = (0, 0, 0), B = (5, 0, 0), C = (0, 6, 0), and D = (0, 0, 7) . What's the volume of tetrahedron ABCD? It helps to remember a little bit of coordinate geometry in order to identify the lengths of segments AB, AC, and AD. Since these segments are parallel to the x, y, and z axes respectively, the lengths are easy to find: AB = 5, AC = 6, and AD = 7. Now move on to the volume computation. First find the base area. The base in this case is a right triangle, base AC = 6, and height AB = 5 - not to be confused with the height of the tetrahedron itself! So, the base area is (1/2)(6)(5) = 15. This is B in the volume formula above. Next, the height of the solid is h = AD = 7, so the volume of the tetrahedron is: V = (1/3)(15)(7) = 35 Volume of a Regular Tetrahedron The volume formula discussed above is really handy for situations in which the base area and height are given or easy to find. You might have to use a little trigonometry to find a missing height (either for the base triangle or for the tetrahedron itself) or Heron's formula for the area of the base triangle. All faces of the regular tetrahedron are equilateral triangles of side length a. Using techniques from geometry, one can find the base area and height of the tetrahedron and plug them into the volume formula to obtain: For example, the volume of a regular tetrahedron whose edge is a length of 3 is exactly: Surface Area Finding the surface area of a tetrahedron is just as easy as finding areas of triangles! All tetrahedra have four triangular faces, and the surface area of any solid is defined as the sum of the areas of all the faces. This works for any type of tetrahedron (regular or not), but the problems can be tricky. Using trigonometry and geometry to find any missing triangle heights may be necessary. On the other hand, there is an easy formula for the surface area of a regular tetrahedron of side length a: For example, the surface area of a tetrahedron of edge length 3 is exactly: Lesson Summary A tetrahedron is a triangular pyramid with 4 faces; each is a triangle. A regular tetrahedron has all edge lengths equal. There are formulas for the volume and surface area of a tetrahedron: To find the surface area of a tetrahedron, find the area for each of the four triangles. Then add together the areas of the four triangles. a = the length of one side Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now NY Regents - Geometry Study Guide and Exam Prep 14 chapters 128 lessons Chapter 1 High School Geometry: Foundations of Geometry Geometry Definition, History & Branches 4:36 min Inductive vs. Deductive Reasoning in Geometry \| Definition & Uses 4:59 min Thales & Pythagoras: Early Contributions to Geometry 5:14 min The Axiomatic System: Definition & Properties 5:17 min Euclid's Axiomatic Geometry: Developments & Postulates 5:58 min Undefined Terms in Geometry \| Definition & Examples 5:23 min Properties and Postulates of Geometric Figures 4:53 min Algebraic Laws and Geometric Postulates 5:37 min Chapter 2 High School Geometry: Logic in Mathematics Critical Thinking and Logic in Mathematics 4:27 min Circular Reasoning, Hasty Generalizations, False Causes & Limited Choice Fallacies 4:47 min Appeal to Ignorance Fallacy \| Definition & Examples 8:53 min Truth Value \| Definition, Propositions & Tables 9:49 min Conjunction vs. Disjunction in Math \| Overview & Characteristics 3:39 min Conditional Statement \| Definition & Examples 4:54 min Conditional Statements \| Converse, Inverse & Contrapositive 7:09 min Direct Proof Definition, Induction & Examples 7:11 min Geometric Proof Types & Formats \| What is a Proof in Geometry? 8:35 min Chapter 3 Introduction to Geometric Figures Line Segments & Rays \| Differences & Measurement 3:59 min Points, Lines & Angles in Geometry \| Definition & Examples 5:13 min Angle Relationships \| Overview & Types 10:28 min Geometric Constructions Using Lines and Angles 4:32 min Line Segment Bisection & Midpoint Theorem: Geometric Construction 4:39 min Division of Line Segment \| Parts & Measurements 5:22 min Parallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 min Constructing Perpendicular Lines in Geometry 3:39 min Constructing an Angle Bisector in Geometry 3:36 min Methods & Tools for Making Geometric Constructions 3:17 min Practice Making Geometric Constructions with Tools 4:17 min Constructing Equilateral Triangles, Squares, and Regular Hexagons Inscribed in Circles 5:00 min Chapter 4 High School Geometry: Similar Polygons Ratio & Proportion \| Meaning, Differences & Examples 5:17 min Geometric Mean \| Definition, Formula & Calculation 5:15 min Angle Bisector Theorem \| Definition & Examples 4:58 min Proportion \| Definition, Formula & Examples 5:22 min Similar Polygons Definition & Examples 8:00 min The Transitive Property of Similar Triangles 4:50 min Triangle Proportionality Theorem \| Overview, Proofs & Examples 4:53 min Constructing Similar Polygons 4:59 min Right Triangle \| Properties, Proof & Theorems 5:58 min Pythagorean Theorem \| Overview, Formula & Examples 7:33 min Converse of the Pythagorean Theorem \| Overview & Examples 5:02 min Similar Triangles & the AA Criterion 5:07 min What is a Polygon? \| Definition, Types & Examples 6:08 min Chapter 5 High School Geometry: Quadrilaterals Parallelograms \| Definition, Properties & Theorems 5:20 min Area of a Parallelogram \| Measurement, Formula & Examples 4:02 min Rhombus \| Definition, Properties & Measurements 4:24 min Area of a Rhombus \| Formula, Methods & Examples 6:30 min Kite in Geometry \| Definition, Shape & Properties 4:21 min Rectangles: Definition, Properties & Construction 4:08 min Area of a Rectangle \| Formula & Examples 4:40 min What is a Square? \| Definition, Properties & Examples 6:52 min Trapezoid \| Definition, Types & Attributes 4:24 min Area of a Trapezoid Formula & Examples 4:38 min Heron's Formula in Geometry \| Overview & Examples 5:54 min Chapter 6 High School Geometry: Circular Arcs and Circles Finding the Area & Circumference of a Circle 7:24 min Arc of a Circle \| Overview, Length & Examples 4:36 min Central and Inscribed Angles: Definitions and Examples 6:32 min Measure of an Arc: Process & Practice 4:51 min How to Find the Measure of an Inscribed Angle 5:09 min Tangent of a Circle \| Definition, Formula & Examples 3:52 min Measurements of Angles Involving Tangents, Chords & Secants 6:59 min Measurements of Lengths Involving Tangents, Chords and Secants 5:44 min Inscribed and Circumscribed Figures: Definition & Construction 6:32 min Arc Length & Sector Area \| Definition, Formula & Examples 6:39 min Circumscribed Circle of a Triangle \| Overview & Examples 3:53 min Chapter 7 High School Geometry: Analytical Geometry Parts of a Graph \| Labels & Examples 6:21 min The Cartesian Coordinate System \| Cartesian Graph & Examples 9:51 min Midpoint \| Formula & Examples 3:33 min Distance Formula \| Overview & Examples 5:27 min Slope of a Line \| Meaning, Calculation & Examples 6:19 min Geometric Proofs for Polygons 6:34 min Chapter 8 Triangles and Congruency Triangles in Geometry \| Definition, Types & Formulas 4:30 min Area of Triangles & Rectangles \| Formula, Calculation & Examples 5:43 min How to Classify Triangles \| Overview & Examples 5:44 min Perimeter of Triangles and Rectangles 8:54 min Interior & Exterior Angles of a Triangle \| Overview & Examples 5:25 min Similar Triangles \| Theorems, Formula & Examples 7:23 min ASA, SSS & SAS Triangle Postulates \| Properties & Examples 6:15 min Similar Triangles \| Definition, Application Problems & Examples 6:23 min Congruent Triangles \| Definition, Parts & Examples 5:19 min Perpendicular Bisector Theorem \| Converse & Examples 6:41 min Angle Bisector Theorem \| Proof & Examples 6:12 min Proving Congruent Isosceles Triangles 4:51 min Converse Statement \| What is the Converse of a Statement? 5:09 min Altitude, Median & Angle Bisector of a Triangle 4:50 min Properties of Concurrent Lines in a Triangle 6:17 min Chapter 9 Angles, Proofs, and Theorems for Triangles Angles and Triangles: Practice Problems 7:43 min Right Triangle Congruence Theorems \| Definition & Examples 7:00 min Constructing Triangles: Types of Geometric Construction 5:59 min Constructing the Median of a Triangle 4:47 min Angle-Angle-Side \| Theorem, Proof & Example 6:31 min Hypotenuse Angle Theorem \| Definition, Proof & Examples 5:50 min Hypotenuse Leg Theorem \| Definition, Proof & Examples 6:19 min Similarity Transformations in Corresponding Figures 7:28 min How to Prove Relationships in Figures using Congruence & Similarity 5:14 min Practice Proving Relationships using Congruence & Similarity 6:16 min Chapter 10 Parallel Lines and Polygons Euclid's Parallel Postulate \| Definition & Examples 4:25 min Parallel Lines and Transversals \| Definition & Diagrams 7:40 min Parallel Line Rules, Proofs & Examples 6:55 min Using Converse Statements to Prove Lines Are Parallel 6:46 min Characteristics of a Polygon \| Overview & Examples 4:25 min Regular Polygon \| Definition, Sides & Types 6:01 min Area of a Regular Polygon \| Formula & Examples 4:15 min Degrees in a Triangle \| Measurement & Examples 5:14 min Angles in a Polygon \| Measurement, Formula & Examples 6:00 min Chapter 11 Conic Sections Directrix & Focus of a Parabola \| Equation & Examples 4:47 min Equation of a Parabola \| Focus & Directrix Formula 6:16 min Foci of Ellipses & Hyperbolas \| Definition & Examples 6:11 min Ellipse Foci Formula & Calculations 5:08 min Hyperbola Equation \| Foci Formula, Parts & Example 7:07 min Practice with the Conic Sections 5:38 min Chapter 12 Geometric Solids Planes and the Polyhedron: Definition and Example 3:52 min Platonic Solids \| Definition, Properties & Types 4:39 min Prisms: Definition, Area & Volume 6:12 min Pyramids: Definition, Area & Volume 7:43 min Volume & Surface Area of a Cylinder \| Formula & Examples 5:09 min Volume of a Cone \| Definition, Formula & Examples 8:59 min Volume & Surface Area of a Sphere \| Formula & Examples 4:19 min Chapter 13 About the NY Regents Examinations What are the NYS Regents Exams? What are the NYS Regents Exams Requirements? NYS Regents Exam Schedule What are the NYS Regents Exams Locations? Advanced Regents Diploma Requirements What is the NYS Regents Scholarship? 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84	https://chem.libretexts.org/Courses/Monterey_Peninsula_College/CHEM_30A%3A_Introduction_to_Chemistry_for_Health_Sciences/12%3A_Gases/12.03%3A_The_Combined_Gas_Law	Published Time: 2020-12-14T01:25:36Z 12.3: The Combined Gas Law - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 12: Gases CHEM 30A: Introduction to Chemistry for Health Sciences { } { "12.01:Gases_and_Pressure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12.02:_Ideal_Gas_Law" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12.03:_The_Combined_Gas_Law" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12.04:_Dalton\'s_Law(Law_of_Partial_Pressures)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Introduction_to_Chemistry_the_Periodic_Table_and_Ionic_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Nuclear_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Measurements_Unit_Conversions_and_Density" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Structure_of_Molecular_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Intermolecular_Forces_and_Solutions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Organic_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Biochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Writing_Chemical_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Moles_and_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Types_of_Chemical_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Acids_Bases_and_Buffers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Gases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Appendix-_Algebra_Review" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Wed, 01 Jun 2022 20:28:49 GMT 12.3: The Combined Gas Law 288515 288515 Jessica Baum { } Anonymous Anonymous User 2 false false [ "article:topic", "showtoc:no", "license:ccbyncsa", "source-chem-16103", "program:hidden", "licenseversion:40" ] [ "article:topic", "showtoc:no", "license:ccbyncsa", "source-chem-16103", "program:hidden", "licenseversion:40" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Monterey Peninsula College 4. CHEM 30A: Introduction to Chemistry for Health Sciences 5. 12: Gases 6. 12.3: The Combined Gas Law Expand/collapse global location CHEM 30A: Introduction to Chemistry for Health Sciences Front Matter 1: Introduction to Chemistry, the Periodic Table, and Ionic Compounds 2: Nuclear Chemistry 3: Measurements, Unit Conversions, and Density 4: Structure of Molecular Compounds 5: Intermolecular Forces and Solutions 6: Organic Chemistry 7: Biochemistry 8: Writing Chemical Equations 9: Moles and Stoichiometry 10: Types of Chemical Reactions 11: Acids, Bases, and Buffers 12: Gases 13: Appendix- Algebra Review Back Matter 12.3: The Combined Gas Law Last updated Jun 1, 2022 Save as PDF 12.2: Ideal Gas Law 12.4: Dalton's Law (Law of Partial Pressures) Page ID 288515 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Combined Gas Law 1. Example 12.3.6 12.3.6 2. Example 12.3.7 12.3.7 Direct vs. Inverse Relationships Direct Relationships Example 12.3.4 12.3.4: Increasing Temperature Exercise 12.3.4 12.3.4 Example 12.3.5 12.3.5: Checking Your Work Exercise 12.3.5 12.3.5: Making a Prediction Inverse Relationships Example 12.3.1 12.3.1: Increasing Pressure in a Gas Exercise 12.3.1 12.3.1: Increasing Volume in a Gas Example 12.3.2 12.3.2: Making a Prediction Exercise 12.3.2 12.3.2 Example 12.3.3 12.3.3: Checking Your Work Exercise 12.3.3 12.3.3 To Your Health: Breathing Determining Relationships Dalton's Law of Partial Pressures Example 12.3.10 12.3.10 Application of Dalton's Law: Collecting Gases over Water Example 12.3.11 12.3.11 Henry’s Law Applicability of Henry's Law Application of Henry's Law: Scuba diving Career Focus: Respiratory Therapist Concept Review Exercises Answers Key Takeaway Exercises Answers Learning Objectives To predict the properties of gases using the gas laws. In the previous section you read about the ideal gas law which can be used to determine a missing property of a gas when the other three properties are known. This is helpful when one variable cannot be easily measured, such as the pressure inside a container that does not have a pressure gauge. In this section we will turn our focus to making predictions about gases that are changing. For instance, what is the new pressure of a gas when the temperature is increased? What is the new volume of a gas when the pressure is decreased? A systematic way to learn relationships between properties of gases is to control as many variables as possible and to study how one variable is affected when another changes. For instance, study how the pressure of a gas changes when the temperature is increased or decreased (and other variables such as the amount of gas and volume are kept constant). With four different possible variables (pressure, volume, temperature, and amount of gas) there are several possible relationships to study. Each can be represented by a mathematical equation. A more realistic scenario in which multiple properties are changing simultaneously can be represented by a mathematical equation that includes all of the variables that describe a gas. Such an equation is called the combined gas law because it is constructed from a combination of the individual gas laws (mathematical equations) that describe pairs of properties. In this section you will learn to use the combined gas law equation, including how to use it as a replacement for all of the individual gas laws that describe how different pairs of properties change individually. If the units of similar quantities are not the same, one of them must be converted to the other quantity’s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process. Combined Gas Law Other gas laws can be constructed, but we will focus on one in this section, the combined gas law. P 1 V 1 n 1 T 1=P 2 V 2 n 2 T 2(12.3.1)(12.3.1)P 1 V 1 n 1 T 1=P 2 V 2 n 2 T 2 Notice that this equation contains copies of each variable (i.e two pressures, two volumes, etc.). The subscript 1 indicates an initial value (before a change); a subscript 2 indicates a final value (after a change).As with the ideal gas law, the temperature must be expressed in Kelvin. For the other variables any relevant unit may be used, but the units must be paired. In other words, both pressures must have the same unit, both volumes must have the same unit, etc. Example 12.3.6 12.3.6 A sample of gas has P 1 = 1.50 atm, V 1 = 10.5 L, and T 1 = 300 K. What is the final volume if P 2 = 0.750 atm and T 2 = 350 K? Solution Using the combined gas law, substitute for five of the quantities. Since the amount of gas is not changing, n 1 and n 2 are the same and cancel out. (1.50 a t m)(10.5 L)300 K=(0.750 a t m)(V 2)350 K(1.50 a t m)(10.5 L)300 K=(0.750 a t m)(V 2)350 K We algebraically rearrange this expression to isolate V 2 on one side of the equation: V 2=(1.50 a t m)(10.5 L)(350 K)(300 K)(0.750 a t m)=24.5 L V 2=(1.50 a t m)(10.5 L)(350 K)(300 K)(0.750 a t m)=24.5 L Note how all the units cancel except the unit for volume. Exercise 12.3.6 12.3.6 A sample of gas has P 1 = 0.768 atm, V 1 = 10.5 L, and T 1 = 300 K. What is the final pressure if V 2 = 7.85 L and T 2 = 250 K? Answer 0.856 atm Example 12.3.7 12.3.7 A balloon containing a sample of gas has a temperature of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11°C and the pressure is 655 torr. What is the new volume of the balloon? Solution The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr: 22°C + 273 = 295 K = T 1 11°C + 273 = 284 K = T 2 1.09 a t m×760 t o r r 1 a t m=828 t o r r=P 1 1.09 a t m×760 t o r r 1 a t m=828 t o r r=P 1 Now we can substitute the quantities into the combined has law: (828 t o r r)(1,070 m L)295 K=(655 t o r r)×V f 284 K(828 t o r r)(1,070 m L)295 K=(655 t o r r)×V f 284 K To solve for V 2, we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left: (828 t o r r)(1,070 m L)(284 K)(295 K)(655 t o r r)=V 2(828 t o r r)(1,070 m L)(284 K)(295 K)(655 t o r r)=V 2 Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So V 2 = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL. Exercise 12.3.7 12.3.7 A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this gas has a pressure of 6.88 torr and a temperature of −15°C. What is the new volume of the gas? Answer 110,038 L Direct vs. Inverse Relationships If only two properties of a gas are changing, it is usually possible to make a qualitative prediction about the size of the unknown variable. This can be very helpful when checking your work! Direct Relationships Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. Because volume increases as temperature increases, these two properties are directly proportional.If you are solving a problem in which only temperature and volume are changing, then you can check your work to make sure that volume increases when temperature increases and vice versa. Example 12.3.4 12.3.4: Increasing Temperature What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant. Solution If the temperature of a gas sample is decreased, the volume decreases as well. Exercise 12.3.4 12.3.4 What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant. Answer The temperature increases. Example 12.3.5 12.3.5: Checking Your Work A gas sample at 20°C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. Solution Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles’s law. Thus, 20°C + 273 = 293 K = T 1 60°C + 273 = 333 K = T 2 Now we can substitute these values into Charles’s law, along with the initial volume of 20.0 L: 20.0 L 293 K=V 2 333 K 20.0 L 293 K=V 2 333 K Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel: (333 K)(20.0 L)293 K=V 2(333 K)(20.0 L)293 K=V 2 Solving for the final volume, V 2 = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant). Exercise 12.3.5 12.3.5: Making a Prediction A gas sample at 35°C has an initial volume of 5.06 L. What is its volume if the temperature is changed to −35°C? First make a prediction: will the new volume be higher or lower than 5.06 L? Then calculate the answer.Assume that the pressure and the amount of the gas remain constant. Answer The final volume should be lower than 5.06 L because when temperature decreases volume also decreases. The final volume is 3.91 L. Inverse Relationships If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Because volume decreases as pressure increases, these two properties are inversely proportional. If you are solving a problem in which only pressure and volume are changing, then you can check your work to make sure that volume decreases when pressure increases and vice versa. Example 12.3.1 12.3.1: Increasing Pressure in a Gas What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same. Solution If the pressure of a gas is increased, the volume decreases in response. Exercise 12.3.1 12.3.1: Increasing Volume in a Gas What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same. Answer If the volume of a gas is increased, the pressure decreases. Example 12.3.2 12.3.2: Making a Prediction If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant. First predict whether the answer will be larger or smaller than 7.02 L, then perform the calculation to see if you were correct. Solution The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is P 1, 7.02 L is V 1, and 0.987 atm is P 2. What we are looking for is the final volume—V 2. Therefore, substituting these values into P 1 V 1 = P 2 V 2: (Because the amount of gas and temperature are not changing n 1 and n 2 cancel out and T 1 and T 2 cancel out) (1.56 atm)(7.02 L) = (0.987 atm) × V 2 The expression has atmospheres on both sides of the equation, so they cancel algebraically: (1.56)(7.02 L) = (0.987) × V 2 Now we divide both sides of the expression by 0.987 to isolate V 2, the quantity we are seeking: (1.56)(7.02 L)0.987=V 2(1.56)(7.02 L)0.987=V 2 Performing the multiplication and division, we get the value of V 2, which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related. Exercise 12.3.2 12.3.2 If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant. Answer 8.48 atm Example 12.3.3 12.3.3: Checking Your Work If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 mm Hg? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant. Solution This example is similar to Example 12.3.2 12.3.2, except now the final pressure is expressed in mm Hg. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to mm Hg: 1.56 a t m×760 m m H g 1 a t m=1,190 m m H g 1.56 a t m×760 m m H g 1 a t m=1,190 m m H g Now we can use Boyle’s law: (1,190 mm Hg)(7.02 L) = (1,775 mm Hg) × V 2 Torr cancels algebraically from both sides of the equation, leaving (1,190)(7.02 L) = (1,775) × V 2 Now we divide both sides of the equation by 1,775 to isolate V 2 on one side. Solving for the final volume, V 2=(1,190)(7.02 L)1,775=4.71 L V 2=(1,190)(7.02 L)1,775=4.71 L Because the pressure increases, it makes sense that the volume decreases. The answer for the final volume is essentially the same if we converted the 1,775 mm Hg to atmospheres: 1,775 m m×1 a t m 760 m m H g=2.336 a t m 1,775 m m×1 a t m 760 m m H g=2.336 a t m. Using Boyle’s law: (1.56 atm)(7.02 L) = (2.335 atm) × V 2; V f=(1.56 a t m)(7.02 L)2.336 a t m=4.69 L V f=(1.56 a t m)(7.02 L)2.336 a t m=4.69 L. Exercise 12.3.3 12.3.3 If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant. Answer 575 torr. Since the volume decreased, it makes sense that the pressure increased because pressure and volume are inversely related. To Your Health: Breathing Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of the inverse relationship between volume and pressure (called Boyle’s law). The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume—just as predicted by Boyle’s law. The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight—only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5–1.0 L of air per normal breath. Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1–2 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle’s law. Determining Relationships Think about the two relationships that were just discussed: Volume and temperature are directly related: when one increases, the other also increases Pressure and volume are inversely related: when one increases, the other decreases Now take a look at the combined gas law and see if you can figure out whether relationships between other variables (e.g. pressure and temperature) will be direct or inverse. P 1 V 1 n 1 T 1=P 2 V 2 n 2 T 2(12.3.2)(12.3.2)P 1 V 1 n 1 T 1=P 2 V 2 n 2 T 2 You can determine whether a pair of properties are directly or inversely related by looking at their relative positions in the mathematical equation. If two variables must be multiplied to obtain a constant, like P 1 and V 1, then they are inversely proportional. If two variables must be divided to obtain a constant, like V 1 and T 1, then they are directly proportional. Dalton's Law of Partial Pressures The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure, a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently. The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas. The definition of Dalton's Law of Partial Pressures that address this is: The total pressure of a gas is the sum of the partial pressures of its components which is expressed algebraically as P t o t a l=P 1+P 2+P 3...=∑i P i(12.3.3)(12.3.3)P t o t a l=P 1+P 2+P 3...=∑i P i or, equivalently P t o t a l=R T V∑i n i(12.3.4)(12.3.4)P t o t a l=R T V∑i n i There is also a similar relationship based on volume fractions, known as Amagat's law of partial volumes. It is exactly analogous to Dalton's law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat's law holds only for ideal gases which must all be at the same temperature and pressure. Dalton's law has neither of these restrictions. Although Amagat's law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course. Example 12.3.10 12.3.10 Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened, What will be the pressure in the system? Which gas will be most abundant in the mixture? Assume that the temperature is uniform and that the volume of the connecting tubes is negligible. Solution The trick here is to note that the total number of moles n T and the temperature remain unchanged, so we can make use of Boyle's law PV = constant. We will work out the details for CO 2 only, denoted by subscripts a. For CO 2, P a V a=(2.13 a t m)(1.50 L)=3.19 L⋅a t m(12.3.5)(12.3.5)P a V a=(2.13 a t m)(1.50 L)=3.19 L⋅a t m Adding the PV products for each separate container, we obtain ∑i P i V i=6.36 L⋅a t m=n T R T(12.3.6)(12.3.6)∑i P i V i=6.36 L⋅a t m=n T R T We will call this sum P 1 V 1. After the stopcocks have been opened and the gases mix, the new conditions are denoted by P 2 V 2. From Boyle's law (??????, P 1 V 1=P 2 V 2=6.36 L⋅a t m(12.3.7)(12.3.7)P 1 V 1=P 2 V 2=6.36 L⋅a t m V 2=∑i V i=4.50 L(12.3.8)(12.3.8)V 2=∑i V i=4.50 L Solving for the final pressure P 2 we obtain (6.36 L-atm)/(4.50 L) = 1.41 atm. For part (b), note that the number of moles of each gas is n = PV/RT. The mole fraction of any one gas is X i = n i /n T. For CO 2, this works out to (3.19/RT) / (6.36/RT) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture. Dalton’s law states that in a gas mixture (P t o t a l P t o t a l) each gas will exert a pressure independent of the other gases (P n P n) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure (P t o t a l P t o t a l) by the gas percentage (%). P T o t a l=P 1+P 2+P 3+P 4+...+P n(12.3.9)(12.3.9)P T o t a l=P 1+P 2+P 3+P 4+...+P n or P n=% of individual gas n P T o t a l(12.3.10)(12.3.10)P n=% of individual gas n P T o t a l Table 12.3.1 12.3.1: Partial Pressures for the gases in air on a typical day\| Gas \| Partial Pressure (mm Hg) \| Percentage (%) \| --- \| Nitrogen, (N_2) \| P N 2 P N 2 = 594 \| 78 \| \| Oxygen, O 2 O 2 \| P O 2 P O 2= 160 \| 21 \| \| Carbon Dioxide, C O 2 C O 2 \| P C O 2 P C O 2 = 0.25 \| 0.033 \| \| Water Vapor, H 2 O H 2 O \| P H 2 O P H 2 O = 5.7 \| 0.75 \| \| Other trace gases \| P O t h e r P O t h e r = 0.05 \| 0.22 \| \| Total air \| P T o t a l P T o t a l = 760 \| 1 \| Application of Dalton's Law: Collecting Gases over Water A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part. Figure 12.3.1 12.3.1:An Apparatus for Collecting Gases by the Displacement of Water The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H 2 O. The partial pressure of H 2 O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas. Example 12.3.11 12.3.11 Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of O 2 O 2 produced. Solution From Dalton's law, P O 2=P t o t a l–P H 2 O=754–19.8=734 t o r r=0.966 a t m P O 2=P t o t a l–P H 2 O=754–19.8=734 t o r r=0.966 a t m Now use the Ideal Gas Law to convert to moles n=P V R T=(0.966 a t m)(0.155 L)(0.082 L a t m m o l−1 K−1)(295 K)=0.00619 m o l n=P V R T=(0.966 a t m)(0.155 L)(0.082 L a t m m o l−1 K−1)(295 K)=0.00619 m o l Henry’s Law Henry's law is one of the gas laws formulated by William Henry in 1803. It states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. To explain this law, Henry derived the equation: C=k P g a s(12.3.11)(12.3.11)C=k P g a s where Henry’s Law tells us that the greater the pressure of gas above the surface of a liquid, the higher the concentration of the gas in the liquid. Also, Henry’s law tells us that gases diffuse from areas of high gas concentration to areas of low gas concentration. Applicability of Henry's Law Henry's law only works if the molecules are at equilibrium. Henry's law does not work for gases at high pressures (e.g., N 2(g)N 2(g) at high pressure becomes very soluble and harmful when in the blood supply). Henry's law does not work if there is a chemical reaction between the solute and solvent (e.g., H C l(g)H C l(g) reacts with water by a dissociation reaction to generate H 3 O+H 3 O+ and C l−C l− ions). Application of Henry's Law: Scuba diving Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O 2 is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure. Figure 12.3.2 12.3.2: Scuba Dviging actively takes into account both Henry's and Dalton's Laws But at a total pressure of 2 atm, the partial pressure of O 2 O 2 in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the O 2 O 2 pressure of 0.8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of O 2 O 2. This can be achieved most simply by raising the nitrogen content, but high partial pressures of N 2 can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures. Career Focus: Respiratory Therapist Certain diseases—such as emphysema, lung cancer, and severe asthma—primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised. Most respiratory therapists must complete at least two years of college and earn an associate’s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor’s offices, nursing homes, or patient’s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy. Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body. Concept Review Exercises What properties do the gas laws help us predict? What makes the ideal gas law different from the other gas laws? Answers Gas laws relate four properties: pressure, volume, temperature, and number of moles. The ideal gas law does not require that the properties of a gas change. Key Takeaway The physical properties of gases are predictable using mathematical formulas known as gas laws. C C is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L) k k is Henry's law constant (often in units of M/atm) P g a s P g a s is the partial pressure of the gas (often in units of Atm) Exercises What conditions of a gas sample should remain constant for Boyle’s law to be used? What conditions of a gas sample should remain constant for Charles’s law to be used? Does the identity of a gas matter when using Boyle’s law? Why or why not? Does the identity of a gas matter when using Charles’s law? Why or why not? A sample of nitrogen gas is confined to a balloon that has a volume of 1.88 L and a pressure of 1.334 atm. What will be the volume of the balloon if the pressure is changed to 0.662 atm? Assume that the temperature and the amount of the gas remain constant. A sample of helium gas in a piston has a volume of 86.4 mL under a pressure of 447 torr. What will be the volume of the helium if the pressure on the piston is increased to 1,240 torr? Assume that the temperature and the amount of the gas remain constant. If a gas has an initial pressure of 24,650 Pa and an initial volume of 376 mL, what is the final volume if the pressure of the gas is changed to 775 torr? Assume that the amount and the temperature of the gas remain constant. A gas sample has an initial volume of 0.9550 L and an initial pressure of 564.5 torr. What would the final pressure of the gas be if the volume is changed to 587.0 mL? Assume that the amount and the temperature of the gas remain constant. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is 18°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and amount of the gas remain constant. A person draws a normal breath of about 1.00 L. If the initial temperature of the air is −10°C and the air warms to 37°C, what is the new volume of the air? Assume that the pressure and the amount of the gas remain constant. An air/gas vapor mix in an automobile cylinder has an initial temperature of 450 K and a volume of 12.7 cm 3. The gas mix is heated to 565°C. If pressure and amount are held constant, what is the final volume of the gas in cubic centimeters? Given the following conditions for a gas: V i = 0.665 L, T i = 23.6°C, V f = 1.034 L. What is T f in degrees Celsius and kelvins? Assuming the amount remains the same, what must be the final volume of a gas that has an initial volume of 387 mL, an initial pressure of 456 torr, an initial temperature of 65.0°C, a final pressure of 1.00 atm, and a final temperature of 300 K? When the nozzle of a spray can is depressed, 0.15 mL of gas expands to 0.44 mL, and its pressure drops from 788 torr to 1.00 atm. If the initial temperature of the gas is 22.0°C, what is the final temperature of the gas? Use the ideal gas law to show that 1 mol of a gas at STP has a volume of about 22.4 L. Use a standard conversion factor to determine a value of the ideal gas law constant R that has units of L•torr/mol•K. How many moles of gas are there in a 27.6 L sample at 298 K and a pressure of 1.44 atm? How many moles of gas are there in a 0.066 L sample at 298 K and a pressure of 0.154 atm? A 0.334 mol sample of carbon dioxide gas is confined to a volume of 20.0 L and has a pressure of 0.555 atm. What is the temperature of the carbon dioxide in kelvins and degrees Celsius? What must V be for a gas sample if n = 4.55 mol, P = 7.32 atm, and T = 285 K? What is the pressure of 0.0456 mol of Ne gas contained in a 7.50 L volume at 29°C? What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of −86.0°C? A mixture of the gases N 2 N 2, O 2 O 2, and A r A r has a total pressure of 760 mm Hg. If the partial pressure of N 2 N 2 is 220 mm Hg and of O 2 O 2 is 470 mm Hg, What is the partial pressure of A r A r? What percent of the gas above is Ar? Apply Henry’s Law to the diagram below to explain: why oxygen diffuses from the alveoli of the lungs into the blood and from the blood into the tissues of the body. why carbon dioxide diffuses from the tissues into the blood and from the blood into the alveoli and then finally out into the atmosphere. Answers temperature and amount of the gas pressure and amount of the gas The identity does not matter because the variables of Boyle’s law do not identify the gas. The identity does not matter because the variables of Charles law do not identify the gas. 3.79 L 31.1 mL 92.1 mL 918.4 torr 1.07 L 1.18 L 23.7 cm 3 461 K; 188 0 C 206 mL 835 K; 562 0 C The ideal gas law confirms that 22.4 L equals 1 mol. 760 t o r r 1 a t m 760 t o r r 1 a t m 1.63 mol 4.2 x 10-4 mol 405 K; 132°C 14.5 L 0.151 atm 18.2 atm 70 mm Hg 9.2% Gases diffuse from high concentration to low concentration (Henry's Law). The partial pressure of oxygen is high in the alveoliand low in the blood of the pulmonary capillaries. As a result, oxygen diffuses across the respiratory membrane from the alveoli into the blood. It's also higher partial pressure in the blood than in the tissues, hence it transfers to the tissues. On the other hand, carbon dioxide diffuses from the tissues (highest CO 2 partial pressure) and across the respiratory membrane from the blood into the alveoliand outto the atmosphere. 12.3: The Combined Gas Law is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. Back to top 12.2: Ideal Gas Law 12.4: Dalton's Law (Law of Partial Pressures) Was this article helpful? Yes No Recommended articles 12: Gases Article typeSection or PageLicenseCC BY-NC-SALicense Version4.0OER program or PublisherThe Publisher Who Must Not Be NamedShow Page TOCno on page Tags source-chem-16103 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 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85	https://see.stanford.edu/materials/lsocoee364a/hw3sol.pdf	EE364a, Winter 2007-08 Prof. S. Boyd EE364a Homework 3 solutions 3.42 Approximation width. Let f0, . . . , fn : R →R be given continuous functions. We consider the problem of approximating f0 as a linear combination of f1, . . . , fn. For x ∈Rn, we say that f = x1f1 + · · · + xnfn approximates f0 with tolerance ǫ > 0 over the interval [0, T] if \|f(t) −f0(t)\| ≤ǫ for 0 ≤t ≤T. Now we choose a ﬁxed tolerance ǫ > 0 and deﬁne the approximation width as the largest T such that f approximates f0 over the interval [0, T]: W(x) = sup{T \| \|x1f1(t) + · · · + xnfn(t) −f0(t)\| ≤ǫ for 0 ≤t ≤T}. Show that W is quasiconcave. Solution. To show that W is quasiconcave we show that the sets {x \| W(x) ≥α} are convex for all α. We have W(x) ≥α if and only if −ǫ ≤x1f1(t) + · · · + xnfn(t) −f0(t) ≤ǫ for all t ∈[0, α). Therefore the set {x \| W(x) ≥α} is an intersection of inﬁnitely many halfspaces (two for each t), hence a convex set. 3.54 Log-concavity of Gaussian cumulative distribution function. The cumulative distribu-tion function of a Gaussian random variable, f(x) = 1 √ 2π Z x −∞e−t2/2 dt, is log-concave. This follows from the general result that the convolution of two log-concave functions is log-concave. In this problem we guide you through a simple self-contained proof that f is log-concave. Recall that f is log-concave if and only if f ′′(x)f(x) ≤f ′(x)2 for all x. (a) Verify that f ′′(x)f(x) ≤f ′(x)2 for x ≥0. That leaves us the hard part, which is to show the inequality for x < 0. (b) Verify that for any t and x we have t2/2 ≥−x2/2 + xt. (c) Using part (b) show that e−t2/2 ≤ex2/2−xt. Conclude that Z x −∞e−t2/2 dt ≤ex2/2 Z x −∞e−xt dt. (d) Use part (c) to verify that f ′′(x)f(x) ≤f ′(x)2 for x ≤0. Solution. The derivatives of f are f ′(x) = e−x2/2/ √ 2π, f ′′(x) = −xe−x2/2/ √ 2π. 1 (a) f ′′(x) ≤0 for x ≥0. (b) Since t2/2 is convex we have t2/2 ≥x2/2 + x(t −x) = xt −x2/2. This is the general inequality g(t) ≥g(x) + g′(x)(t −x), which holds for any diﬀerentiable convex function, applied to g(t) = t2/2. Another (easier?) way to establish t2/2 ≤−x2/2 + xt is to note that t2/2 + x2/2 −xt = (1/2)(x −t)2 ≥0. Now just move x2/2 −xt to the other side. (c) Take exponentials and integrate. (d) This basic inequality reduces to −xe−x2/2 Z x −∞e−t2/2 dt ≤e−x2 i.e., Z x −∞e−t2/2 dt ≤e−x2/2 −x . This follows from part (c) because Z x −∞e−xt dt = e−x2 −x . 3.57 Show that the function f(X) = X−1 is matrix convex on Sn ++. Solution. We must show that for arbitrary v ∈Rn, the function g(X) = vTX−1v. is convex in X on Sn ++. This follows from example 3.4. 4.1 Consider the optimization problem minimize f0(x1, x2) subject to 2x1 + x2 ≥1 x1 + 3x2 ≥1 x1 ≥0, x2 ≥0. Make a sketch of the feasible set. For each of the following objective functions, give the optimal set and the optimal value. 2 (a) f0(x1, x2) = x1 + x2. (b) f0(x1, x2) = −x1 −x2. (c) f0(x1, x2) = x1. (d) f0(x1, x2) = max{x1, x2}. (e) f0(x1, x2) = x2 1 + 9x2 2. Solution. The feasible set is shown in the ﬁgure. x1 x2 (1, 0) (2/5, 1/5) (0, 1) (a) x⋆= (2/5, 1/5). (b) Unbounded below. (c) Xopt = {(0, x2) \| x2 ≥1}. (d) x⋆= (1/3, 1/3). (e) x⋆= (1/2, 1/6). This is optimal because it satisﬁes 2x1+x2 = 7/6 > 1, x1+3x2 = 1, and ∇f0(x⋆) = (1, 3) is perpendicular to the line x1 + 3x2 = 1. 4.4 [P. Parrilo] Symmetries and convex optimization. Suppose G = {Q1, . . . , Qk} ⊆Rn×n is a group, i.e., closed under products and inverse. We say that the function f : Rn →R is G-invariant, or symmetric with respect to G, if f(Qix) = f(x) holds for all x and i = 1, . . . , k. We deﬁne x = (1/k) Pk i=1 Qix, which is the average of x over its G-orbit. We deﬁne the ﬁxed subspace of G as F = {x \| Qix = x, i = 1, . . . , k}. (a) Show that for any x ∈Rn, we have x ∈F. (b) Show that if f : Rn →R is convex and G-invariant, then f(x) ≤f(x). 3 (c) We say the optimization problem minimize f0(x) subject to fi(x) ≤0, i = 1, . . . , m is G-invariant if the objective f0 is G-invariant, and the feasible set is G-invariant, which means f1(x) ≤0, . . . , fm(x) ≤0 = ⇒f1(Qix) ≤0, . . . , fm(Qix) ≤0, for i = 1, . . . , k. Show that if the problem is convex and G-invariant, and there exists an optimal point, then there exists an optimal point in F. In other words, we can adjoin the equality constraints x ∈F to the problem, without loss of generality. (d) As an example, suppose f is convex and symmetric, i.e., f(Px) = f(x) for every permutation P. Show that if f has a minimizer, then it has a minimizer of the form α1. (This means to minimize f over x ∈Rn, we can just as well minimize f(t1) over t ∈R.) Solution. (a) We ﬁrst observe that when you multiply each Qi by some ﬁxed Qj, you get a permutation of the Qi’s: QjQi = Qσ(i), i = 1, . . . , k, where σ is a permutation. This is a basic result in group theory, but it’s easy enough for us to show it. First we note that by closedness, each QjQi is equal to some Qs. Now suppose that QjQi = QkQi = Qs. Multiplying by Q−1 i on the right, we see that Qj = Qk. Thus the mapping from the index i to the index s is one-to-one, i.e., a permutation. Now we have Qjx = (1/k) k X i=1 QjQix = (1/k) k X i=1 Qσ(i)x = (1/k) k X i=1 Qix = x. This holds for j, so we have x ∈F. (b) Using convexity and invariance of f, f(x) ≤(1/k) k X i=1 f(Qix) = (1/k) k X i=1 f(x) = f(x). 4 (c) Suppose x⋆is an optimal solution. Then x⋆is feasible, with f0(x⋆) = f0((1/k) k X i=1 Qix) ≤ (1/k) k X i=1 f0(Qix) = f0(x⋆). Therefore x⋆is also optimal. (d) Suppose x⋆is a minimizer of f. Let x = (1/n!) P P Px⋆, where the sum is over all permutations. Since x is invariant under any permutation, we conclude that x = α1 for some α ∈R. By Jensen’s inequality we have f(x) ≤(1/n!) X P f(Px⋆) = f(x⋆), which shows that x is also a minimizer. 4.8 Some simple LPs. Give an explicit solution of each of the following LPs. (a) Minimizing a linear function over an aﬃne set. minimize cTx subject to Ax = b. Solution. We distinguish three possibilities. • The problem is infeasible (b ̸∈R(A)). The optimal value is ∞. • The problem is feasible, and c is orthogonal to the nullspace of A. We can decompose c as c = ATλ + ˆ c, Aˆ c = 0. (ˆ c is the component in the nullspace of A; ATλ is orthogonal to the nullspace.) If ˆ c = 0, then on the feasible set the objective function reduces to a constant: cTx = λTAx + ˆ cTx = λTb. The optimal value is λTb. All feasible solutions are optimal. • The problem is feasible, and c is not in the range of AT (ˆ c ̸= 0). The problem is unbounded (p⋆= −∞). To verify this, note that x = x0 −tˆ c is feasible for all t; as t goes to inﬁnity, the objective value decreases unboundedly. In summary, p⋆=      +∞ b ̸∈R(A) λTb c = ATλ for some λ −∞ otherwise. 5 (b) Minimizing a linear function over a halfspace. minimize cTx subject to aTx ≤b, where a ̸= 0. Solution. This problem is always feasible. The vector c can be decomposed into a component parallel to a and a component orthogonal to a: c = aλ + ˆ c, with aT ˆ c = 0. • If λ > 0, the problem is unbounded below. Choose x = −ta, and let t go to inﬁnity: cTx = −tcTa = −tλaTa →−∞ and aTx −b = −taTa −b ≤0 for large t, so x is feasible for large t. Intuitively, by going very far in the direction −a, we ﬁnd feasible points with arbitrarily negative objective values. • If ˆ c ̸= 0, the problem is unbounded below. Choose x = ba −tˆ c and let t go to inﬁnity. • If c = aλ for some λ ≤0, the optimal value is cTab = λb. In summary, the optimal value is p⋆= ( λb c = aλ for some λ ≤0 −∞ otherwise. (c) Minimizing a linear function over a rectangle. minimize cTx subject to l ⪯x ⪯u, where l and u satisfy l ⪯u. Solution. The objective and the constraints are separable: The objective is a sum of terms cixi, each dependent on one variable only; each constraint depends on only one variable. We can therefore solve the problem by minimizing over each component of x independently. The optimal x⋆ i minimizes cixi subject to the constraint li ≤xi ≤ui. If ci > 0, then x⋆ i = li; if ci < 0, then x⋆ i = ui; if ci = 0, then any xi in the interval [li, ui] is optimal. Therefore, the optimal value of the problem is p⋆= lTc+ + uTc−, where c+ i = max{ci, 0} and c− i = max{−ci, 0}. 6 (d) Minimizing a linear function over the probability simplex. minimize cTx subject to 1Tx = 1, x ⪰0. What happens if the equality constraint is replaced by an inequality 1Tx ≤1? We can interpret this LP as a simple portfolio optimization problem. The vector x represents the allocation of our total budget over diﬀerent assets, with xi the fraction invested in asset i. The return of each investment is ﬁxed and given by −ci, so our total return (which we want to maximize) is −cTx. If we replace the budget constraint 1Tx = 1 with an inequality 1Tx ≤1, we have the option of not investing a portion of the total budget. Solution. Suppose the components of c are sorted in increasing order with c1 = c2 = · · · = ck < ck+1 ≤· · · ≤cn. We have cTx ≥c1(1Tx) = cmin for all feasible x, with equality if and only if x1 + · · · + xk = 1, x1 ≥0, . . . , xk ≥0, xk+1 = · · · = xn = 0. We conclude that the optimal value is p⋆= c1 = cmin. In the investment interpre-tation this choice is quite obvious. If the returns are ﬁxed and known, we invest our total budget in the investment with the highest return. If we replace the equality with an inequality, the optimal value is equal to p⋆= min{0, cmin}. (If cmin ≤0, we make the same choice for x as above. Otherwise, we choose x = 0.) (e) Minimizing a linear function over a unit box with a total budget constraint. minimize cTx subject to 1Tx = α, 0 ⪯x ⪯1, where α is an integer between 0 and n. What happens if α is not an integer (but satisﬁes 0 ≤α ≤n)? What if we change the equality to an inequality 1Tx ≤α? Solution. We ﬁrst consider the case of integer α. Suppose c1 ≤· · · ≤ci−1 < ci = · · · = cα = · · · = ck < ck+1 ≤· · · ≤cn. The optimal value is c1 + c2 + · · · + cα 7 i.e., the sum of the smallest α elements of c. x is optimal if and only if x1 = · · · = xi−1 = 1, xi + · · · + xk = α −i + 1, xk+1 = · · · = xn = 0. If α is not an integer, the optimal value is p⋆= c1 + c2 + · · · + c⌊α⌋+ c1+⌊α⌋(α −⌊α⌋). In the case of an inequality constraint 1Tx ≤α, with α an integer between 0 and n, the optimal value is the sum of the α smallest nonpositive coeﬃcients of c. 4.17 Optimal activity levels. We consider the selection of n nonnegative activity levels, denoted x1, . . . , xn. These activities consume m resources, which are limited. Activity j consumes Aijxj of resource i, where Aij are given. The total resource consumption is additive, so the total of resource i consumed is ci = Pn j=1 Aijxj. (Ordinarily we have Aij ≥0, i.e., activity j consumes resource i. But we allow the possibility that Aij < 0, which means that activity j actually generates resource i as a by-product.) Each resource consumption is limited: we must have ci ≤cmax i , where cmax i are given. Each activity generates revenue, which is a piecewise-linear concave function of the activity level: rj(xj) = ( pjxj 0 ≤xj ≤qj pjqj + pdisc j (xj −qj) xj ≥qj. Here pj > 0 is the basic price, qj > 0 is the quantity discount level, and pdisc j is the quantity discount price, for (the product of) activity j. (We have 0 < pdisc j < pj.) The total revenue is the sum of the revenues associated with each activity, i.e., Pn j=1 rj(xj). The goal is to choose activity levels that maximize the total revenue while respecting the resource limits. Show how to formulate this problem as an LP. Solution. The basic problem can be expressed as maximize Pn j=1 rj(xj) subject to x ⪰0 Ax ⪯cmax. This is a convex optimization problem since the objective is concave and the constraints are a set of linear inequalities. To transform it to an equivalent LP, we ﬁrst express the revenue functions as rj(xj) = min{pjxj, pjqj + pdisc j (xj −qj)}, which holds since rj is concave. It follows that rj(xj) ≥uj if and only if pjxj ≥uj, pjqj + pdisc j (xj −qj) ≥uj. 8 We can form an LP as maximize 1Tu subject to x ⪰0 Ax ⪯cmax pjxj ≥uj, pjqj + pdisc j (xj −qj) ≥uj, j = 1, . . . , n, with variables x and u. To show that this LP is equivalent to the original problem, let us ﬁx x. The last set of constraints in the LP ensure that ui ≤ri(x), so we conclude that for every feasible x, u in the LP, the LP objective is less than or equal to the total revenue. On the other hand, we can always take ui = ri(x), in which case the two objectives are equal. 9 Solutions to additional exercises 1. Optimal activity levels. Solve the optimal activity level problem described in exercise 4.17 in Convex Optimization, for the instance with problem data A =         1 2 0 1 0 0 3 1 0 3 1 1 2 1 2 5 1 0 3 2         , cmax =         100 100 100 100 100         , p =      3 2 7 6     , pdisc =      2 1 4 2     , q =      4 10 5 10     . You can do this by forming the LP you found in your solution of exercise 4.17, or more directly, using cvx. Give the optimal activity levels, the revenue generated by each one, and the total revenue generated by the optimal solution. Also, give the average price per unit for each activity level, i.e., the ratio of the revenue associated with an activity, to the activity level. (These numbers should be between the basic and discounted prices for each activity.) Give a very brief story explaining, or at least commenting on, the solution you ﬁnd. Solution. The following Matlab/CVX code solves the problem. (Here we write the problem in a form close to its original statement, and let CVX do the work of refor-mulating it as an LP!) A=[ 1 2 0 1; 0 0 3 1; 0 3 1 1; 2 1 2 5; 1 0 3 2]; cmax=[100;100;100;100;100]; p=[3;2;7;6]; pdisc=[2;1;4;2]; q=[4; 10 ;5; 10]; cvx_begin variable x(4) maximize( sum(min(p.x,p.q+pdisc.(x-q))) ) subject to x >= 0; Ax <= cmax cvx_end x r=min(p.x,p.q+pdisc.(x-q)) 10 totr=sum(r) avgPrice=r./x The result of the code is x = 4.0000 22.5000 31.0000 1.5000 r = 12.0000 32.5000 139.0000 9.0000 totr = 192.5000 avgPrice = 3.0000 1.4444 4.4839 6.0000 We notice that the 3rd activity level is the highest and is also the one with the highest basic price. Since it also has a high discounted price its activity level is higher than the discount quantity level and it produces the highest contribution to the total revenue. The 4th activity has a discounted price which is substantially lower then the basic price and its activity is therefore lower that the discount quantity level. Moreover it require the use of a lot of resources and therefore its activity level is low. 2. Reformulating constraints in cvx. Each of the following cvx code fragments describes a convex constraint on the scalar variables x, y, and z, but violates the cvx rule set, 11 and so is invalid. Brieﬂy explain why each fragment is invalid. Then, rewrite each one in an equivalent form that conforms to the cvx rule set. In your reformulations, you can use linear equality and inequality constraints, and inequalities constructed using cvx functions. You can also introduce additional variables, or use LMIs. Be sure to explain (brieﬂy) why your reformulation is equivalent to the original constraint, if it is not obvious. Check your reformulations by creating a small problem that includes these constraints, and solving it using cvx. Your test problem doesn’t have to be feasible; it’s enough to verify that cvx processes your constraints without error. Remark. This looks like a problem about ‘how to use cvx software’, or ‘tricks for using cvx’. But it really checks whether you understand the various composition rules, convex analysis, and constraint reformulation rules. (a) norm( [ x + 2y , x - y ] ) == 0 (b) square( square( x + y ) ) <= x - y (c) 1/x + 1/y <= 1; x >= 0; y >= 0 (d) norm([ max( x , 1 ) , max( y , 2 ) ]) <= 3x + y (e) xy >= 1; x >= 0; y >= 0 (f) ( x + y )^2 / sqrt( y ) <= x - y + 5 (g) x^3 + y^3 <= 1; x>=0; y>=0 (h) x+z <= 1+sqrt(xy-z^2); x>=0; y>=0 Solution. (a) The lefthand side is correctly identiﬁed as convex, but equality constraints are only valid with aﬃne left and right hand sides. Since the norm of a vector is zero if and only if the vector is zero, we can express the constraint as x+2y==0; x-y==0, or simply x==0; y==0. (b) The problem is that square() can only accept aﬃne arguments, because it is convex, but not increasing. To correct this use square_pos() instead: square_pos( square( x + y ) ) <= x - y We can also reformulate this constraint by introducing an additional variable. variable t square( x+y ) <= t square( t ) <= x - y Note that, in general, decomposing the objective by introducing new variables doesn’t need to work. It works in this case because the outer square function is convex and monotonic over R+. Alternatively, we can rewrite the constraint as 12 ( x + y )^4 <= x - y (c) 1/x isn’t convex, unless you restrict the domain to R++. We can write this one as inv_pos(x)+inv_pos(y)<=1. The inv_pos function has domain R++ so the constraints x > 0, y > 0 are (implicitly) included. (d) The problem is that norm() can only accept aﬃne argument since it is convex but not increasing. One way to correct this is to introduce new variables u and v: norm( [ u , v ] ) <= 3x + y max( x , 1 ) <= u max( y , 2 ) <= v Decomposing the objective by introducing new variables work here because norm is convex and monotonic over R2 +, and in particular over (1, ∞) × (2, ∞). (e) xy isn’t concave, so this isn’t going to work as stated. But we can express the constraint as x>=inv_pos(y). (You can switch around x and y here.) Another solution is to write the constraint as geomean([x,y])>=1. We can also give an LMI representation: [ x 1; 1 y ] == semidefinite(2) (f) This fails when we attempt to divide a convex function by a concave one. We can write this as quad_over_lin(x+y,sqrt(y)) <= x-y+5 This works because quad_over_lin is monotone descreasing in the second argu-ment, so it can accept a concave function here, and sqrt is concave. (g) The function x3 + y3 is convex for x ≥0, y ≥0. But x3 isn’t convex for x < 0, so cvx is going to reject this statement. One way to rewrite this constraint is quad_pos_over_lin(square(x),x) + quad_pos_over_lin(square(y),y) <= 1 This works because quad_pos_over_lin is convex and increasing in its ﬁrst ar-gument, hence accepts a convex function in its ﬁrst argument. (The function quad_over_lin, however, is not increasing in its ﬁrst argument, and so won’t work.) Alternatively, and more simply, we can rewrite the constraint as pow_pos(x,3) + pow_pos(y,3) <= 1 (h) The problem here is that xy isn’t concave, which causes cvx to reject the state-ment. To correct this, notice that q xy −z2 = q y(x −z2/y), so we can reformulate the constraint as 13 x+z <= 1+geomean([x-quad_over_lin(z,y),y]) This works, since geomean is concave and nondecreasing in each argument. It therefore accepts a concave function in its ﬁrst argument. We can check our reformulations by writing the following feasibility problem in cvx (which is obviously infeasible) cvx_begin variables x y u v z x == 0; y == 0; ( x + y )^4 <= x - y; inv_pos(x) + inv_pos(y) <= 1; norm( [ u ; v ] ) <= 3x + y; max( x , 1 ) <= u; max( y , 2 ) <= v; x >= inv_pos(y); x >= 0; y >= 0; quad_over_lin(x + y , sqrt(y)) <= x - y + 5; pow_pos(x,3) + pow_pos(y,3) <= 1; x+z <= 1+geomean([x-quad_over_lin(z,y),y]) cvx_end 3. The illumination problem. This exercise concerns the illumination problem described in lecture 1 (pages 9–11). We’ll take Ides = 1 and pmax = 1, so the problem is minimize f0(p) = maxk=1,...,n \| log(aT k p)\| subject to 0 ≤pj ≤1, j = 1, . . . , m, (1) with variable p ∈Rn. You will compute several approximate solutions, and compare the results to the exact solution, for a speciﬁc problem instance. As mentioned in the lecture, the problem is equivalent to minimize maxk=1,...,n h(aT k p) subject to 0 ≤pj ≤1, j = 1, . . . , m, (2) where h(u) = max{u, 1/u} for u > 0. The function h, shown in the ﬁgure below, is nonlinear, nondiﬀerentiable, and convex. To see the equivalence between (1) and (2), we note that f0(p) = max k=1,...,n \| log(aT k p)\| = max k=1,...,n max{log(aT k p), log(1/aT k p)} 14 = log max k=1,...,n max{aT k p, 1/aT k p} = log max k=1,...,n h(aT k p), and since the logarithm is a monotonically increasing function, minimizing f0 is equiv-alent to minimizing maxk=1,...,n h(aT k p). 0 1 2 3 4 0 1 2 3 4 5 u h(u) The problem instance. The speciﬁc problem data are for the geometry shown below, using the formula akj = r−2 kj max{cos θkj, 0} from the lecture. There are 10 lamps (m = 10) and 20 patches (n = 20). We take Ides = 1 and pmax = 1. The problem data are given in the ﬁle illum_data.m on the course website. Running this script will construct the matrix A (which has rows aT k ), and plot the lamp/patch geometry as shown below. 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 15 Equal lamp powers. Take pj = γ for j = 1, . . . , m. Plot f0(p) versus γ over the interval [0, 1]. Graphically determine the optimal value of γ, and the associated objective value. You can evaluate the objective function f0(p) in Matlab as max(abs(log(Ap))). Least-squares with saturation. Solve the least-squares problem minimize Pn k=1(aT k p −1)2 = ∥Ap −1∥2 2. If the solution has negative values for some pi, set them to zero; if some values are greater than 1, set them to 1. Give the resulting value of f0(p). Least-squares solutions can be computed using the Matlab backslash operator: A\b returns the solution of the least-squares problem minimize ∥Ax −b∥2 2. Regularized least-squares. Solve the regularized least-squares problem minimize Pn k=1(aT k p −1)2 + ρ Pm j=1(pj −0.5)2 = ∥Ap −1∥2 2 + ρ∥p −(1/2)1∥2 2, where ρ > 0 is a parameter. Increase ρ until all coeﬃcients of p are in the interval [0, 1]. Give the resulting value of f0(p). You can use the backslash operator in Matlab to solve the regularized least-squares problem. Chebyshev approximation. Solve the problem minimize maxk=1,...,n \|aT k p −1\| = ∥Ap −1∥∞ subject to 0 ≤pj ≤1, j = 1, . . . , m. We can think of this problem as obtained by approximating the nonlinear function h(u) by a piecewise-linear function \|u −1\| + 1. As shown in the ﬁgure below, this is a good approximation around u = 1. 16 0 0.5 1 1.5 2 0 0.5 1 1.5 2 2.5 3 3.5 4 u \|u −1\| + 1 You can solve the Chebyshev approximation problem using cvx. The (convex) function ∥Ap −1∥∞can be expressed in cvx as norm(Ap-ones(n,1),inf). Give the resulting value of f0(p). Exact solution. Finally, use cvx to solve minimize maxk=1,...,n max(aT k p, 1/aT k p) subject to 0 ≤pj ≤1, j = 1, . . . , m exactly. You may ﬁnd the inv_pos() function useful. Give the resulting (optimal) value of f0(p). Solution: The following Matlab script ﬁnds the approximate solutions using the heuristic methods proposed, as well as the exact solution. % illum_sol: finds approximate and exact solutions of % the illumination problem clear all; % load input data illum_data; % heuristic method 1: equal lamp powers % --------------------------------------nopts=1000; p = logspace(-3,0,nopts); f = zeros(size(p)); for k=1:nopts f(k) = max(abs(log(Ap(k)ones(m,1)))); 17 end; [val_equal,imin] = min(f); p_equal = p(imin)ones(m,1); % heuristic method 2: least-squares with saturation % -------------------------------------------------p_ls_sat = A\ones(n,1); p_ls_sat = max(p_ls_sat,0); % rounding negative p_i to 0 p_ls_sat = min(p_ls_sat,1); % rounding p_i > 1 to 1 val_ls_sat = max(abs(log(Ap_ls_sat))); % heuristic method 3: regularized least-squares % ---------------------------------------------rhos = linspace(1e-3,1,nopts); crit = []; for j=1:nopts p = [A; sqrt(rhos(j))eye(m)][ones(n,1); sqrt(rhos(j))0.5ones(m,1)]; crit = [ crit norm(p-0.5,inf) ]; end idx = find(crit <= 0.5); rho = rhos(idx(1)); % smallest rho s.t. p is in [0,1] p_ls_reg = [A; sqrt(rho)eye(m)][ones(n,1); sqrt(rho)0.5ones(m,1)]; val_ls_reg = max(abs(log(Ap_ls_reg))); % heuristic method 4: chebyshev approximation % -------------------------------------------cvx_begin variable p_cheb(m) minimize(norm(Ap_cheb-1, inf)) subject to p_cheb >= 0 p_cheb <= 1 cvx_end val_cheb = max(abs(log(Ap_cheb))); % exact solution: % ---------------cvx_begin variable p_exact(m) minimize(max([Ap_exact; inv_pos(Ap_exact)])) subject to p_exact >= 0 18 p_exact <= 1 cvx_end val_exact = max(abs(log(Ap_exact))); % Results % -------[p_equal p_ls_sat p_ls_reg p_cheb p_exact] [val_equal val_ls_sat val_ls_reg val_cheb val_exact] The results are summarized in the following table. method 1 method 2 method 3 method 4 exact f0(p) 0.4693 0.8628 0.4439 0.4198 0.3575 p1 0.3448 1 0.5004 1 1 p2 0.3448 0 0.4777 0.1165 0.2023 p3 0.3448 1 0.0833 0 0 p4 0.3448 0 0.0002 0 0 p5 0.3448 0 0.4561 1 1 p6 0.3448 1 0.4354 0 0 p7 0.3448 0 0.4597 1 1 p8 0.3448 1 0.4307 0.0249 0.1882 p9 0.3448 0 0.4034 0 0 p10 0.3448 1 0.4526 1 1 19
86	https://busynessgirl.com/logarithm-graphs-in-wolfram-alpha/	Home About Contact Info Who am I? Career Education Teaching Experience Teaching Philosophy Presentations Publications Endorsements Resources Contemporary Algebra Collection Games Games for Trigonometry Games for Algebra Games for Elementary Games for Calculus Games for Fun Lame Math Games Useful Information Errata Tutorials SoTL and Math Ed Math Instructional Practices Useful Tools Handouts Mindmaps Videos Useful Tech Mobile Apps for Education Chrome Extensions Teaching with Tech Columns Business Ethics Resources Favorites Posts from Edge of Learning Posts from TaLDA Posts from Digital Life Posts from Play & Learn Posts from Mathematics India Travelogue Tweeting my Dissertation Speaking Sample Talks Talk Menu Workshop Menu Bio Consulting Select Page Logarithm Graphs in Wolfram Alpha Posted by busynessgirl \| Mar 16, 2010 \| Algebra, CAS, College Algebra, Digital Graphs and Diagrams, Teaching Math At the Wolfram Alpha Workshop at ICTCM, there was universal disappointment about the fact that you cannot get a graph of a logarithm that is only over the real numbers. We tried everything we could think of to remove the complex part of the graph. Personally, I have tried and tried and tried and tried to explain the problem with this in the feedback window for Wolfram Alpha, but been universally unsuccessful. Every time I suggest a change, I am told that the “After review, our internal development group believes the plots for input “log(x)” are correct.” … yes, I know that … that doesn’t mean it’s the answer that most people will be looking for. I find it ironic that “inverse of e^x” produces the graph we’d like to see, and even gives log(x) as an equivalent. But then ask for a graph of log(x) or ln(x) and the graph will always include the solution over the complex numbers. What’s worse is that W\|A inconsistently decides when to use reals only and when to use both complex and real numbers. For example, the output for y=ln(x), y=x includes the complex numbered plot, while the output for y=ln(x), y=2x-3 includes only the Reals. What!?! Actually, I have some idea why this is … it seems that in some cases, if the extra graph intersects the real part of the log graph, then you get reals only. If the graphs do not intersect, then you get real+complex. For example compare the output for y=ln(x), y=2x-3 to the output for y=ln(x), y=2x+5. On the other hand, when I tried to show a graph transformation, like y=ln(x) with y=ln(x)+4 (including the extra graph y=4x-3), I was back to getting the graphs with complex numbers again. Maddening. We spend a LOT of time in the algebra and precalculus levels working with transformations of graphs, understanding inverse functions, and specifics like the domain of a graph. We can’t use Wolfram Alpha for any of these topics with regards to logarithms because of the way the graphs look. I can live with the fact that W\|A uses log(x) instead of ln(x) … it’s not great, and is confusing to students, but I can explain it and live with it. But as long as the Wolfram Alpha graph includes the complex number system with no way to see the graphs on only the reals, we’ll have to pull out that old-fashioned graphing calculator to teach this section, and that’s a shame. I’ve also heard the argument that we should just include the domain we want to see in the W\|A input. For example, y=ln(x), x>0. But how is a student, learning logs for the first time, supposed to recognize that this is required? After all, the graph they see when they first try W\|A with y=ln(x) leads them to believe that y=ln(x) has a domain that includes all real numbers but zero. This argument also means that to show graph transformations, we need to use much more complicated graphing commands, restricting each domain separately (to tell you the truth, I have not yet figured out a way to do it … although I suspect it’s possible). It seems to me that there are two obvious solutions to this math teaching nightmare, and I can’t imagine why either one wouldn’t serve all parties using Wolfram Alpha (both high-level mathematicians, and the rest of us): Solution #1: Use a toggle-able option to see the graph with only reals or both complex and reals (I would prefer a default to the Real numbers graph, since my guess would be that the majority of the world’s population would be looking for that one). Solution #2: Display TWO graphs. Show a graph of the logarithm that is only on the real number system. Then, below it, show a graph that includes both the complex and real number systems. That’s all – end of rant. This is the one thing I absolutely hate about Wolfram Alpha. And I’m guessing that I’m not alone here. Please can’t we just find a solution without hearing “After review, our internal development group believes the plots for input “log(x)” are correct.” again? 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87	https://fiveable.me/critical-thinking/unit-9/truth-tables-logical-connectives/study-guide/2QvrZ06tWUNmUXnK	printables 💡Critical Thinking Unit 9 Review 9.2 Truth Tables and Logical Connectives 💡Critical Thinking Unit 9 Review 9.2 Truth Tables and Logical Connectives Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 APA 💡Critical Thinking Unit & Topic Study Guides 9.1 Symbolic Logic and Propositional Calculus 9.2 Truth Tables and Logical Connectives 9.3 Testing Validity Using Truth Tables Truth tables are essential tools in propositional logic, helping us analyze complex logical statements. They list all possible combinations of truth values for propositions, allowing us to evaluate the overall truth of compound statements. Logical connectives like conjunction (AND), disjunction (OR), and negation (NOT) form the building blocks of more complex logical expressions. Understanding these connectives is crucial for constructing and interpreting truth tables, as well as for logical reasoning in general. Truth Tables and Logical Connectives Understanding Truth Tables and Values Truth tables systematically list all possible combinations of truth values for logical statements Truth values represent the logical status of a proposition as either true (T) or false (F) Truth tables consist of columns for each proposition and rows for all possible combinations Construct truth tables by listing all possible combinations of truth values for the given propositions Use truth tables to evaluate complex logical statements and determine their overall truth value Analyzing Logical Statements Tautology refers to a compound statement that is always true regardless of the truth values of its components Identify tautologies in truth tables by checking if all rows in the final column are true Contradiction describes a compound statement that is always false regardless of the truth values of its components Recognize contradictions in truth tables when all rows in the final column are false Contingency applies to statements that can be either true or false depending on the truth values of their components Determine contingencies by examining truth tables for a mix of true and false values in the final column Basic Logical Connectives Conjunction and Disjunction Conjunction (AND) combines two propositions, resulting in a true statement only when both components are true Represent conjunction using the symbol ∧ or the word "and" Create truth tables for conjunctions by marking true only when both input columns are true Disjunction (OR) combines two propositions, resulting in a true statement when at least one component is true Denote disjunction using the symbol ∨ or the word "or" Construct truth tables for disjunctions by marking true when either or both input columns are true Inclusive OR allows for both components to be true, while exclusive OR (XOR) requires exactly one true component Negation Negation reverses the truth value of a proposition Represent negation using the symbol ¬ or the word "not" Create truth tables for negations by flipping the truth value of the input column Apply negation to compound statements by using De Morgan's Laws to distribute negation across conjunctions and disjunctions Use double negation to cancel out the effect of two consecutive negations (¬¬p is equivalent to p) Advanced Logical Connectives Conditional Statements Conditional (IF-THEN) statements express a relationship where one proposition implies another Represent conditionals using the symbol → or the phrase "if...then" Construct truth tables for conditionals by marking false only when the antecedent is true and the consequent is false Understand the difference between necessary and sufficient conditions in conditional statements Recognize variations of conditional statements (converse, inverse, contrapositive) Biconditional Statements Biconditional (IF AND ONLY IF) statements express logical equivalence between two propositions Denote biconditionals using the symbol ↔ or the phrase "if and only if" Create truth tables for biconditionals by marking true when both components have the same truth value Understand biconditionals as a combination of two conditional statements (p → q) and (q → p) Use biconditionals to express definitions and equivalences in mathematical and logical reasoning
88	https://www.youtube.com/watch?v=ZzKga4AKOFo	Find f(g(x)) and g(f(x)) From the Functions f(x) = 1/(x - 6) and g(x) = 7/x + 6 The Glaser Tutoring Company 81900 subscribers 115 likes Description 12886 views Posted: 28 Feb 2021 For the following exercises, use each pair of functions to find f(g(x)) and g(f(x)). Simplify your answers. f(x) = 1/(x - 6), g(x) = 7/x + 6 f(x) = 1/(x - 4), g(x) = 2/x + 4 Here are all of our Math Playlists: Functions: 📕Functions and Function Notation: 📗Domain and Range: 📘Rates of Change and Behavior of Graphs: 📙Composition of Functions: 📕Transformation of Functions: 📗Absolute Value Functions: Linear Functions: 📕Linear Functions: 📗Graphs of Linear Functions: This question(s) was provided by OpenStax™ (www.openstax.org) which is licensed under the Creative Commons Attribution 4.0 International License. ( OpenStax™ is a registered trademark, which was not involved in the production of, and does not endorse, this product. 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CompositionOfFunctions #FunctionsMath #Math 3 comments Transcript: for the following exercises use each pair of functions to find f of g of x and g of f of x and then we need to simplify our answers okay so composite functions i know that these are composite functions because of this type of notation where you have one function and then in parentheses you have another function composite functions is just a fancy way of saying you're taking two functions that are independent of each other and you're you know meshing them together you're mixing them together to get one single output one single answer okay now the trick to composite functions is if we look down at our tips and tricks composite functions we will always be working from inner to outer when i say inner to outer i mean parentheses so you work from the innermost parentheses to the outermost parentheses the innermost function to the outermost okay so let's categorize this by doing a and b for a on the left-hand side let's do f of g of x okay so as i look at this work from inner to outer so the innermost parentheses is the g of x function i'm not going to say specifically x because you have to include the whole function so the inner function is the g of x function and then once we solve that puzzle we move to the outermost function which is the f function here so for step number one all we do is we plug in an input it's usually a number to the inner function and then use our algebra and just solve so the inner function here and if i say number one just to keep with my ones and twos the g of x function because that's what it said here the g of x is oh well the g of x was x sorry seven over x plus six okay can't really simplify that um because they did give me a number so i have to move on to the next part the second part says i'm going to take that new input that i just solved for and plug it into the outer function so i'm going to basically just take this answer and input it into the outer function which is f so that's number two f of this seven over x plus six if you hear my dog in the background i apologize uh yeah he likes to bark something must be outside um so we're gonna plug this in and anytime that i see an x value in my outer function in this case it's the f function i'm going to replace the x with what i got over here so this would just be one over seven over x plus six minus six do we see that i plugged in this whole thing for the x value and now i just got to simplify so let's see a six minus a six bye-bye they go bye-bye and now i have f of seven over x plus six um this is the same thing as saying the final answer f of g of x which is 1 over 7 over x but remember when we are dividing by a fraction we always like to multiply by their reciprocal so i'm just going to take this and swap it so it would literally be 1 times x over 7 which would just be x over 7 and that is your final answer in the most simplified form so that is your first answer for the a part f of g of x now let's see i'm going to take the two functions and swap the composite function instead now it's going to be g of f of x let's see if we get the same answer the innermost function now is the f of x the outermost is the g one so i gotta work from inner to outer so for one i'm gonna take the f of x function what was f of x oh it was this one over x minus six can't simplify they didn't give me a number so i'm now going to take this and plug it into so if i say number two plug it into the outer function which was the g function and instead of g of x it's now going to be one over x minus six and whenever you see an x you just replace that here so this would be seven divided by one actually let me put the one over here and i will put it into a different color seven divided by one over x minus six and then you gotta keep with that plus six now we're gonna do the same thing as we did before we're dividing by a fraction so let's inverse it so g of 1 over x minus 6 equals 7 times by you're reverting this so this would just be x minus six over one which is just x minus six and then finish it out it's plus six so now we can simplify what we're doing is distributing so i'm gonna pull it all together i can say that the g 1 over x minus 6 which is the same thing as g of f of x is 7 times x is 7 x 7 times a negative 6 is minus 42 plus 6 sum it all up uh negative 42 plus 6 what is that that is 36 negative 36 so this would equal 7x minus 36 and that is your final answer so this is the simplified version if you want to just take that okay we are almost there first thing i just want to i just want to point out is these are not the same answer so the order of the composite function even if you're taking the same function the ordering matters all right so yeah we got this one last one this one's going to be quicker because we kind of get the hang of it so let's do f of g of x first inner function is the f of x then for part two we take that and put it into the outer function which is the f function so for one g of x is this right 2x well 2 divided by x plus 4. 2 over x plus 4. can't simplify that so i'm going to take this and plug it into my outer function for part two f of this whole thing two over four or two over x plus four and my f of x function was this but every time i see an x now i'm just going to write this so let's see we got 1 over 2 over x plus 4 minus 4 because i got to keep with this right so what can we simplify okay well plus 4 minus 4 that goes bye-bye they cancel each other out so now let's sum it up we have f of 2 over x that was a little ugly plus 4 equals it's the same thing as saying f of g of x which equals 1 over 2 over x but whenever we're dividing by a fraction we just like to take the inverse right we multiply by the inverse so this would be 1 times x over 2. one times x over two is x over two and there you go almost quick and easy now we just gotta do b um g of f of x inner to outer inner is now the f function plugging it into the g function so for one f of x equals this right 1 over x minus 4. can i simplify this no so what i'm going to do is i'm going to plug this in into my outer function of g so that's part two g of this now one over x minus four and for the g function whenever you see a x value you're plugging in that so it would be 2 over insert this because that was the x 1 over x minus 4 and then plus 4 so i'll put this in blue just to show you and now same thing as before i'm dividing by a fraction so i'm just going to do the inverse multiply by the inverse so g of 1 over x minus 4. is two times inverse it so it would be um x minus four and then plus four now we distribute i got a number here i got to distribute between these parentheses and we're going to get our final answer here so g of 1 over x minus 4 is the same thing as saying g of f of x which is two times x is two x two times a negative four is a negative eight plus four so this would equal two x minus four and if you wanted to you know simplify this you could even pull out the 2 here you know 2 x minus 2. it doesn't really matter to me so i'm going to highlight these they all mean the same thing awesome job guys and this is the answer and once again look at your final answers are they the same no they're not so yeah composite functions really matter when you're taking into consideration who's the inner function who's the outer function if you reach the end and you can understand this and i hope you can guys um let me know in the comments but you guys are amazing all right you're getting so good at math um keep going you guys got this um thank you so much for you know coming to this lesson i hope this helped uh like this if you you know liked it obviously if you want to follow us along you can subscribe and we'll have tons more math videos we got physics videos if you guys are taking physics you name it we got it for you guys all right well as of right now but we'll see what's what's happening in the future all right so i got guys i hope you have an awesome day um yeah thanks for tuning in see you later bye
89	https://www.mathsisfun.com/calculus/l-hopitals-rule.html	L'Hôpital's Rule L'Hôpital's Rule can help us calculate a limit that may otherwise be hard or impossible. L'Hôpital is pronounced "lopital". He was a French mathematician from the 1600s. It says that the limit when we divide one function by another is the same after we take the derivative of each function (with some special conditions shown later). In symbols we can write: limx→cf(x)g(x) = limx→cf’(x)g’(x) The limit as x approaches c of "f-of−x over g-of−x" equals the the limit as x approaches c of "f-dash-of−x over g-dash-of−x" All we did is add that little dash mark ’ on each function, which means to take the derivative. Example: limx→2x2+x−6x2−4 At x=2 we would normally get: 22+2−622−4 = 00 Which is indeterminate, so we are stuck. Or are we? Let's try L'Hôpital! Differentiate both top and bottom (see Derivative Rules): limx→2x2+x−6x2−4 = limx→22x+1−02x−0 Now we just substitute x=2 to get our answer: limx→22x+1−02x−0 = 54 Here is the graph, notice the "hole" at x=2: Note: we can also get this answer by factoring, see Evaluating Limits. Example: limx→∞exx2 Normally this is the result: limx→∞exx2 = ∞∞ Both head to infinity. Which is indeterminate. But let's differentiate both top and bottom (note that the derivative of ex is ex): limx→∞exx2 = limx→∞ex2x Hmmm, still not solved, both tending towards infinity. But we can use it again: limx→∞exx2 = limx→∞ex2x = limx→∞ex2 Now we have: limx→∞ex2 = ∞ It has shown us that ex grows much faster than x2. Cases We have already seen a 00 and ∞∞ example. Here are all the indeterminate forms that L'Hopital's Rule may be able to help with: 00 ∞∞ 0×∞ 1∞ 00 ∞0 ∞−∞ Conditions Differentiable For a limit approaching c, the original functions must be differentiable either side of c, but not necessarily at c. Likewise g’(x) is not equal to zero either side of c. The Limit Must Exist This limit must exist: limx→cf’(x)g’(x) Why? Well a good example is functions that never settle to a value. Example: limx→∞x+cos(x)x Which is a ∞∞ case. Let's differentiate top and bottom: limx→∞1−sin(x)1 And because it just wiggles up and down it never approaches any value. So that new limit does not exist! And so L'Hôpital's Rule is not usable in this case. BUT we can do this: limx→∞x+cos(x)x = limx→∞(1 + cos(x)x) As x goes to infinity then cos(x)x tends to between −1∞ and +1∞, and both tend to zero. And we are left with just the "1", so: limx→∞x+cos(x)x = limx→∞(1 + cos(x)x) = 1 Limits (An Introduction) Calculus Index Copyright © 2024 Rod Pierce
90	https://studfile.net/preview/4231936/page:16/	1.8. Растворы. Способы выражения концентрации растворов Файловый архив студентов. 1307 вузов, 5193 предметов. Войти / Регистрация ВойтиРегистрация Логин: Пароль: Забыли пароль? Email: Email повторно: Логин: Пароль: - [x] Принимаю пользовательское соглашение FAQОбратная связьВопросы и предложения Вузы Предметы Пользователи Добавить файлы Реферат AI Заказать работу Добавил: UploadОпубликованный материал нарушает ваши авторские права? Сообщите нам. Вуз: Российский государственный профессионально-педагогический университет Предмет: [НЕСОРТИРОВАННОЕ] Файл: ЗКТэССО / 5920.doc Скачиваний: 81 Добавлен: 29.05.2015 Размер: 1.18 Mб Скачать ☆ ►Содержание► X •Методические указания и контрольные задания по дисциплине «химия» •Введение •1. Методические указания по основным разделам курса химии •1.1. Электронное строение атома •Примеры решения задач •1.2. Периодический закон и периодическая система д.И. Менделеева •Относительная электроотрицательность элементов •Примеры решения задач •1.3. Химическая связь •Примеры решения задач •2S22p2 1s2 (типа He) •1.4. Классы неорганических соединений •Примеры решения задач •1.5. Элементы химической термодинамики и термохимии •Примеры решения задач •1.6. Химическая кинетика и химическое равновесие •Примеры решения задач •1.7. Электролитическая диссоциация. Реакции ионного обмена •Примеры решения задач •1.8. Растворы. Способы выражения концентрации растворов •Примеры решения задач •Р е ш е н и е. 1) Записываем выражение для молярной концентрации раствора NaOh : •Р е ш е н и е. 1) Вспомним, что молярная доля растворенного вещества равна: •Р е ш е н и е. 1) Найдем массу 100 мл 15% раствора h2so4: •1.9. Коллоидные растворы •Примеры решения задач •1.10. Растворы неэлектролитов •Примеры решения задач. •1.11. Окислительно-восстановительные реакции •Примеры решения задач •1.12. Электрохимические процессы в гетерогенных системах. Гальванические элементы •Примеры решения задач •1.13. Коррозия металлов •Примеры решения задач •1.14. Электролиз •Примеры решения задач •2. Контрольные задания •3. Варианты контрольных заданий •Литература <<<< Предыдущая345678910111213141516 / 27161718192021222324252627>Следующая >>> 1.8. Растворы. Способы выражения концентрации растворов Раствор – гомогенная (однородная) система, состоящая как минимум из двух компонентов, один из которых растворитель, другой – растворенное вещество. То есть состав раствора = растворитель + растворенное вещество. Например, водный раствор хлорида натрия состоит из двух компонентов: воды (растворителя) и хлорида натрия (растворенного вещества). Существует несколько способов выражения концентрации растворов. Молярная концентрация, или молярность(С M)– количество вещества (ν) растворенного компонента, содержащееся в 1 литре (дм 3) раствора: [моль/л], (14) Молярная концентрация эквивалентов (С экв), (нормальная концентрация или нормальность N) – количество вещества эквивалентов (ν экв) растворенного компонента, содержащееся в 1 литре (дм 3) раствора: С экв (N) = ν экв / V [моль-экв/л] (15) Моляльная концентрация или моляльность (b) – количество вещества (ν) растворенного компонента в 1 кг раствора: [моль/кг], (16) Массовая доля (ω)– отношение массы растворенного вещества (m р.в.) к массе раствора (m р-ра ) Ее рассчитывают, выражая в долях единицы или в процентах. Массовая доля, выраженная в процентах, называется процентной концентрацией: (17) Процентная концентрация показывает массу растворенного вещества, содержащегося в 100 г раствора. Например, ω( KOH ) = 3% означает, что в 100 г этого раствора содержится 3 г KOH и 97 г H 2 O . Молярная доля(N i) – отношение количества вещества растворенного компонента ( в-ва ) (или растворителя,  р-ля) к суммарному количеству вещества всех компонентов раствора. Например, в системе, состоящей из растворителя и одного растворенного вещества, молярная доля растворенного вещества равна: , (18) Молярная доля растворителя: (19) Примеры решения задач Пример 1. Определить молярную концентрацию раствора NaOH с массовой долей 10% и плотностью  =1,1 г/см 3. Р е ш е н и е. 1) Записываем выражение для молярной концентрации раствора NaOh : 2) 10%-ный раствор – это 10г NaOH в 100г раствора. Находим количество вещества NaOH, содержащееся в 10 г: , m NaOH = 10 г, M NaOH = 23 + 16 +1 = 40 г/моль , т.е. 0,25 моль NaOH содержится в 100 г раствора. 3) Находим объем раствора массой 100 г: m = V  ,= 0,091 л 4) Рассчитываем молярную концентрацию: Ответ: молярная концентрация раствора гидроксида натрия с массовой долей 10% составляет 2,74 моль/л Пример 2. Найти молярную долю растворенного вещества в растворе сахарозы с массовой долей 67%. Р е ш е н и е. 1) Вспомним, что молярная доля растворенного вещества равна: 67%-ный (по массе) раствор означает, что в 100 г раствора содержится 67 г сахарозы и 33 г воды. 2) Определяем количество вещества сахарозы и количество вещества воды: в-ва = 67/342 = 0,196 моль, р-ля = 33/18 = 1,83 моль Следовательно, молярная доля сахарозы равна: Ответ: молярная доля сахарозы в растворе сахарозы с массовой долей 67% составляет 0,097. Пример 3 . Какой объем серной кислоты с массовой долей 96% (плотностью 1,84 г/см 3) и какую массу воды нужно взять для приготовления 100 мл 15%-ного (по массе) раствора H 2 SO 4 ( = 1,10 г/см 3). Р е ш е н и е. 1) Найдем массу 100 мл 15% раствора h2so4: m H 2 SO 4-р-ра = V  = 100  1,10 =110 г. 2) Из формулы массовой доли находим массу серной кислоты, содержащейся в этом растворе: ; m H 2 SO 4 = 3) Найдем массу 96% раствора, содержащего 16,5 г H 2 SO 4: 4) Находим объем 96 %-ного раствора серной кислоты: Ответ: для приготовления 100мл 15%-ного раствора H 2 SO 4 потребуется 9,3 мл 96%-ного раствора серной кислоты и 110 – 16,5 = 93,5 г воды. <<<< Предыдущая345678910111213141516 / 27161718192021222324252627>Следующая >>> Соседние файлы в папке ЗКТэССО #29.05.2015 347.14 Кб582181.doc #29.05.2015 319.49 Кб712448.doc #29.05.2015 44.83 Кб553003(Возрастная физиология и психофизиология).docx #29.05.2015 260.1 Кб1133827(русский язык).doc #29.05.2015 154.62 Кб615541(История).doc #29.05.2015 1.18 Mб815920.doc #29.05.2015 4.51 Mб75Гулин сб. задач по физике.doc #29.05.2015 1 Mб54Инструкции по доступу к каталогу НБ РГППУ_ ИОС_Лань.docx #29.05.2015 140.29 Кб55Памятка читатею.doc ПомощьОбратная связьВопросы и предложенияПользовательское соглашениеПолитика конфиденциальности Ограничение Для продолжения скачивания необходимо пройти капчу:
91	https://www.teacherspayteachers.com/Product/Literal-Equations-Task-Cards-711014	Literal Equations Task Cards Description Literal Equations: This Literal Equations pack contains a total of forty (40) task cards divided into two (2) sets. These are great for individual practice, as a pair/group activity, or to use in math centers. What’s included? SET A - 20 Task Cards with literal equations that can be solved in 1 to 3 steps Note to Teachers: Problems #15, #16 and #18 require students to perform basic factoring to solve the literal equations. SET B – 20 Task Cards with literal equations that can be solved in 3 or more steps Note to Teachers: Problems #2, #12, #13, #14, #15, #16, #19 and #20 require students to perform basic factoring to solve the literal equations. More Literal Equations resources: Literal Equations Picture Puzzle Literal Equations Mazes Literal Equations Scavenger Hunt (Can be solved in 2 to 3 steps) We hope that you and your students enjoy these task cards. Our goal is to create quality math resources that will be helpful in your class instruction. If you found any errors/typos or have any questions about this product, please don’t hesitate to contact us at chilimath@yahoo.com. Thank you! Mike and Kathy © 2016 ChiliMath. The purchase of this product entitles the buyer a limited individual license and is non-transferable. This product is intended to be used by one teacher only. If you would like to share this math resource with your colleagues, you may revisit our TpT store and purchase additional licenses at a discounted price. This product may be printed or photocopied but may not be sold, transmitted, or put on the internet/intranet without a written permission from the authors. Thank you for respecting our work! Literal Equations Task Cards Reviews Questions & Answers
92	https://helpingwithmath.com/decomposing-fractions-study/	Decomposing Fractions \| Definition, Examples, Purpose & Test Skip to content Helping with Math Menu Home Membership All Worksheets Log In Menu By Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade End Of Year Test Booklets Common Core Worksheet Mapping By Age Age 4-6 Age 5-7 Age 6-8 Age 7-9 Age 8-10 Age 9-11 Age 10-12 Age 11-13 Age 12-14 By Topic Addition Algebra Algebraic Expressions Angles Area Basic Facts Decimals Division Equations Factors Fractions Functions Geometry Graph & Charts Integers Measurement Multiplication Number Sense Percentages Perimeter Place Value Polynomials Radicals Ratio Rounding Shapes Statistics Subtraction Time Volume Word Problems Math Skills Themes Animals & Living Things Celebrations Career/Work Education Fun Health & Fitness Seasonal Space Worksheets Calendar Common Core Mapping Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Common Core Exam Booklets Math Common Core Standards Additional Resources Math Calculators Math Questions Math Quizzes Math Flash Cards Math Table Charts Seasonal Worksheets Calendar Math Theory Algebra Calculus Decimals Factors Fractions Geometry Graphs & Charts Measurement & Time Numbers Number Sense Operations of Numbers Rounding Statistics & Probability Trigonometry Home » Math Theory » Fractions » Decomposing Fractions Decomposing Fractions Table of Contents Introduction Grade Appropriateness Math Domain Applicable Common Core Standards Definition of the Topic Key Concepts Discussion with Illustrative Examples Decomposing Fractions into Unit Fractions Decomposing Fractions into Different Denominators Examples with Solution Real-life Application with Solution Practice Test Frequently Asked Questions (FAQs) What is the purpose of decomposing fractions? Can all fractions be decomposed into unit fractions? Can you decompose fractions with unlike denominators? Why is decomposing fractions important in math? How can decomposing fractions help with the addition and subtraction of fractions? Recommended Worksheets Introduction Fractions are fundamental concepts in mathematics that students encounter early in their education. Decomposing fractions is an essential skill that helps students break down and simplify complex problems, making them more manageable. This article will comprehensively overview decomposing fractions, including grade appropriateness, applicable math domains and common core standards, key concepts, and examples with solutions. We will also discuss real-life applications and provide practice tests to solidify understanding. Grade Appropriateness Decomposing fractions is typically introduced in 4th grade and remains relevant throughout middle and high school as students study more advanced math concepts. Math Domain Decomposing fractions fall under the Number and Operations-Fractions domain in mathematics. Applicable Common Core Standards Decomposing fractions is aligned with the following Common Core Standards: 4.NF.B.3: Understand a fraction a/b with a > 1 as a sum of fractions 1/b. 5.NF.A.1: Add and subtract fractions with unlike denominators by replacing given fractions with equivalent fractions. Definition of the Topic Decomposing fractions refers to breaking down a fraction into a sum of simpler fractions. The goal is to express the original fraction in an equivalent form that makes computation or conceptual understanding easier. Key Concepts Equivalent fractions: Fractions with different numerators and denominators but represent the same value. Simplifying fractions: The process of reducing a fraction is by taking its numerator and denominator and dividing them by their greatest common factor. Least Common Denominator (LCD): The smallest multiple that two or more denominators share, used to find equivalent fractions with common denominators. Discussion with Illustrative Examples Decompose means ‘breaking apart’, ‘splitting up’, or ‘dividing into smaller parts’. Decomposing a fraction into smaller fractions adds all the smaller parts together, resulting in the initial fraction. Decomposing Fractions into Unit Fractions A unit fraction is a fraction with one as a numerator. For example, $\frac{1}{3}$,$\frac{1}{5}$,$\frac{1}{7}$,$\frac{1}{9}$, etc. When we break a fraction into unit fractions, we split it into equal parts. Let us say, for example, $\frac{3}{5}$=$\frac{1}{5}$+$\frac{1}{5}$+$\frac{1}{5}$, which is the same as three times the unit fraction $\frac{1}{5}$. Let us consider another fraction, say $\frac{7}{5}$; we can decompose it into a sum of unit fractions, as shown below. Decomposing Fractions into Different Denominators This method decomposes a fraction into smaller fractions with different denominators. We express the fraction as the sum of fractions which are not all unit fractions. Let us consider the fraction $\frac{7}{7}$ and decompose it into the sum of three fractions with different denominators. As shown below, by getting the sum of unit fractions, we can simplify fractions and obtain $\frac{7}{8}$=$\frac{1}{2}$+$\frac{1}{4}$+$\frac{1}{8}$. Examples with Solution Example 1 Decompose $\frac{5}{6}$ into a sum of unit fractions (fractions with a numerator of 1). Solution To decompose $\frac{5}{6}$ into a sum of unit fractions, we must have the sum of five one-sixths. $\frac{5}{6}$ =$\frac{1}{6}$ +$\frac{1}{6}$ +$\frac{1}{6}$ +$\frac{1}{6}$ +$\frac{1}{6}$ Example 2 Decompose $\frac{9}{11}$ as the sum of three equal fractions. Solution If we split the number into three equal parts, we have 9=3+3+3. Therefore, $\frac{9}{11}$=$\frac{3}{11}$+$\frac{3}{11}$+$\frac{3}{11}$ Example 3 Using two smaller fractions with different denominators, decompose $\frac{8}{15}$. Solution Since we must obtain two smaller fractions with different denominators, we have, $\frac{1}{15}$+$\frac{1}{15}$+$\frac{1}{15}$=$\frac{3}{15}$=$\frac{1}{15}$ and $\frac{1}{15}$+$\frac{1}{15}$+$\frac{1}{15}$+$\frac{1}{15}$+$\frac{1}{15}$=$\frac{1}{15}$=$\frac{1}{3}$. Therefore, $\frac{8}{15}$=$\frac{1}{5}$+$\frac{1}{3}$. Real-life Application with Solution Problem 1 Imagine that you and two friends want to share $\frac{6}{7}$ of a pizza equally. You can decompose the fraction into three equal parts to determine how much pizza each person will receive. Solution $\frac{6}{7}$÷3=($\frac{1}{7}$+$\frac{1}{7}$+$\frac{1}{7}$+$\frac{1}{7}$+$\frac{1}{7}$)÷3 =($\frac{1}{7}$+$\frac{1}{7}$)+($\frac{1}{7}$+$\frac{1}{7}$)+($\frac{1}{7}$+$\frac{1}{7}$) =$\frac{2}{7}$+$\frac{2}{7}$+$\frac{2}{7}$ Each person will receive $\frac{2}{7}$ of the pizza. Problem 2 Maria, Martin, and Mario ate $\frac{3}{8}$ of the pie. Write an equation to show how much of the pie each ate. Solution Dividing the numerator into three equal parts, we have 3=1+1+1. So, $\frac{3}{8}$=$\frac{1}{8}$+$\frac{1}{8}$+$\frac{1}{8}$ Hence, each of them ate $\frac{1}{8}$ of the pie. Practice Test Decompose $\frac{3}{4}$ into a sum of unit fractions. Decompose $\frac{2}{7}$ into a sum of smaller fractions with the same denominators. Decompose $\frac{11}{12}$ into a sum of unit fractions. Decompose $\frac{9}{24}$ into a sum of three smaller fractions with different denominators. Decompose $\frac{6}{8}$ into a sum of unit fractions. Decompose $\frac{13}{15}$ into a sum of unit fractions. Answers: $\frac{3}{4}$ =$\frac{1}{4}$ +$\frac{1}{4}$ +$\frac{1}{4}$ $\frac{6}{7}$=$\frac{2}{7}$ +$\frac{2}{7}$ +$\frac{2}{7}$ $\frac{11}{12}$ =$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ +$\frac{1}{12}$ $\frac{9}{24}$ =$\frac{1}{12}$ +$\frac{1}{8}$ +$\frac{1}{6}$ $\frac{6}{8}$ =$\frac{1}{8}$ +$\frac{1}{8}$ +$\frac{1}{8}$ +$\frac{1}{8}$ +$\frac{1}{8}$ +$\frac{1}{8}$ $\frac{13}{15}$ =$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ +$\frac{1}{15}$ Frequently Asked Questions (FAQs) What is the purpose of decomposing fractions? Decomposing fractions helps students understand the concept of fractions, simplifies complex problems and makes computation easier. Can all fractions be decomposed into unit fractions? Yes, all fractions can be decomposed into a sum of unit fractions. Can you decompose fractions with unlike denominators? Yes, you can decompose fractions with unlike denominators by finding a common denominator or by using a combination of different denominators. Why is decomposing fractions important in math? Decomposing fractions is important because it helps students develop a strong foundation in understanding fractions, which is crucial for more advanced math concepts. It also makes solving problems involving fractions easier by breaking them down into simpler components. How can decomposing fractions help with the addition and subtraction of fractions? Decomposing fractions can help students find equivalent fractions with common denominators, which is necessary for adding and subtracting fractions with unlike denominators. 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93	https://pdg.lbl.gov/2025/reviews/rpp2024-rev-accel-phys-colliders.pdf	1 31. Accelerator Physics of Colliders 31. Accelerator Physics of Colliders Revised July 2023 by V. Shiltsev (FNAL) and F. Zimmermann (CERN). This article provides background for the High-Energy Collider Parameter Tables that follow and some additional information; see in-depth review and a comprehensive list of references in ; citations below are limited to widely used textbooks and open access seminal papers and reviews. 31.1 Energy and Luminosity Collisions of two beams of particles accelerated to high energies E1,2 provide access to center-of-mass energies (c.m.e.) Ecme ≈2√E1E2, assuming a typically small or zero crossing angle. Most of the 31 colliders that have ever reached the operational stage (seven are operational now) used equal masses and energies of colliding particles, with c.m.e. equal to twice the beam energy Ecme = 2Eb. Other machines collide beams of unequal energies, such as electron-proton or electron-ion colliders, or asymmetric B-factories, that produce new short-lived particles, whose decays are more easily detected and analyzed with a Lorentz boost. In an accelerator, charged particles gain energy from an electric ﬁeld, which usually varies in time at a high frequency ranging from 100s of kHz to 10s of GHz. With proper phasing to the RF ﬁeld over distance l, the energy gain of a particle with charge Ze is proportional to the average accelerating gradient G, i.e. ∆Eb = ZeGl. In principle, the highest beam accelerating gradients achieved to date in operational machines or beam test facilities (G ≈100 MV/m in 12 GHz normal-conducting RF cavities and 31.5 MV/m in 1.3 GHz superconducting ones) allow accessing high energies over reasonably long linear accelerators (linacs), but cost considerations often call for minimization of RF acceleration via repeated use of the same RF system which, in that case, would boost the energy in small portions ∆Eb = ZeVRF per turn every time a particle passes through the total cavity voltage VRF. Such an arrangement can be realized either in the form of storage-ring circular colliders or also through novel schemes based on, e.g., recirculating linear accelerators (RLAs) with or without energy recovery. Circular colliders are by far the most common; here, the momentum and energy of ultra-relativistic particles are determined by the bending radius inside the dipole magnets, ρ, and by the average magnetic ﬁeld B of these magnets: p = ZeBρ or Eb [GeV] = 0.3Z(Bρ) [Tm] . (31.1) Such synchrotron condition assures approximately constant radius of the beam orbit during accel-eration. Transverse focusing by quadrupole magnets is needed to keep particles inside the rather limited space provided by the accelerator beam pipe passing through the magnet apertures. The maximum ﬁeld of normal-conducting (NC) magnets is about 2 T, due to the saturation of ferromag-netic materials, and, while this is suﬃcient for lower energy colliders, such as most e+e−storage rings, it is not adequate for frontier-energy hadron (or muon) beams, because of the implied need for excessively long accelerator tunnels and prohibitively high total magnet power consumption. The development of superconducting (SC) magnets that employ high electric current carrying Nb-Ti wires cooled by liquid helium below 5 K, opened up the way towards higher ﬁelds and to hadron colliders at record energies . For example, the 14 TeV c.m.e. LHC at CERN, uses double-bore SC magnets with a maximum ﬁeld of 8.3 T at a temperature of 1.9 K, in a tunnel of C = 26.7 km circumference (dipole-magnet bending radius ρ = 2800 m). The double-bore design allows acceler-ation of the same particle type in opposite directions and also the operation with diﬀerent particle species (e.g., protons and heavy ions) in the two apertures, while a single bore magnet implies the use of particles and antiparticles for the collider application. As the production of anti-particles is energy consuming and therefore limited, this concept opens the door to high-performance hadron S. Navas et al. (Particle Data Group), Phys. Rev. D 110, 030001 (2024) 31st May, 2024 10:11am 2 31. Accelerator Physics of Colliders colliders. The exploration of rare nuclear and high energy particle physics phenomena requires not only an appropriately high energy, but also a suﬃciently large number of detectable reactions. The number of events of interest Nexp is given by the product of the cross section of the reaction under study, σexp, and the time integral over the instantaneous luminosity, L: Nexp = σexp · Z L(t)dt. (31.2) In the Tables, luminosity is stated in the units of cm−2s−1. The integral on the right is referred to as integrated luminosity Lint, and, reﬂecting the smallness of typical particle-interaction cross-sections is often reported in units of inverse femto- or attobarn, e.g., 1 ab−1=1042 cm−2. Colliders usually employ bunched beams of particles with approximately Gaussian distributions, and for two bunches containing N1 and N2 particles colliding head-on with frequency fcoll, a basic expression for the luminosity is L = fcoll N1N2 4πσ∗ xσ∗ y F (31.3) where σ∗ x and σ∗ y characterize the rms transverse beam sizes in the horizontal and vertical directions at the interaction point, and F is a factor of order 1, that takes into account ineﬃcient geometric overlapping of the beams due to a crossing angle and ﬁnite bunch length, and dynamic eﬀects, such as the mutual focusing of the two beam during the collision (see below). Having nb bunches per beam increases the frequency of collisions fcoll = nbf0 where f0 is either the revolution frequency of a circular collider or the repetition rate of a linear one. To achieve a high luminosity, one, therefore, has to maximize the population and number of bunches, either producing these narrowly or focusing them tightly, and colliding them at high frequencies at dedicated locations, where products of their reactions can be registered by particle detectors. Subsequent sections in this report brieﬂy expand on the beam dynamics behind collider design, comment on the realization of collider performance in a selection of today’s facilities, and end with some remarks on future possibilities. 31.2 Beam Dynamics Given the enormous and highly concentrated power carried by modern high energy particle beams, the main concern of beam dynamics in colliders is stability of motion of i) individual particles in accelerators, ii) single high-intensity beams of many particles moving together, and iii) colliding beams [3–5]. 31.2.1 Single Particle Dynamics While a reference particle at the nominal energy proceeds along the design trajectory (reference orbit) mostly determined by transverse magnetic dipole ﬁelds, other particles in the bunch are kept close by through the focusing eﬀect of quadrupole ﬁelds. Assume that the reference particle carries a right-handed Cartesian coordinate system, with the co-moving z-coordinate pointed in the direction of motion along the reference trajectory, z = s −vt (with v the reference particle velocity, and t time). The independent variable is the distance s of the reference particle along this trajectory, rather than time t, and for simplicity this reference path is taken to be planar. The transverse coordinates are x (horizontal) and y (vertical), where {x, z} deﬁnes the plane of the reference trajectory. Several time scales are involved, and this is reﬂected in the approximations used in formulating the equations of motion. All of today’s high-energy colliders are alternating gradient synchrotrons or, respectively, storage rings and the shortest time scale is set by so-called betatron oscillations. The linearized equations of motion of a particle displaced from the reference trajectory are: 31st May, 2024 3 31. Accelerator Physics of Colliders x′′ + Kx(s)x = 0 , y′′ + Ky(s)y = 0 , z′ = −x/ρ(s) , with Kx ≡Ze p ∂By ∂x + 1 ρ2 and Ky ≡−Ze p ∂By ∂x (31.4) where ρ = p/ZeBy is the radius of curvature due to the ﬁeld on the reference orbit. The prime denotes d/ds and the Maxwell equation in vacuum ∇× B = 0 helps to eliminate Bx(s) using the relation ∂Bx/∂y = ∂By/∂x. In this linear approximation, the vertical magnetic ﬁeld By(s) in the (x, z)-plane contains only dipole and quadrupole terms, which are treated as static in time, but s-dependent. The solutions of the Hill’s equations (31.4) for x and y with a restoring force periodic in s are those of quasi-harmonic oscillators: x(s) = p 2Jxβx cos ψx , x′(s) = − s 2Jx βx [αx cos ψx + sin ψx] , (31.5) where the action Jx is a constant of integration, αx = αx(s) ≡−(1/2)dβx(s)/ds, and the envelope of oscillations is modulated by the beta-function βx(s). A solution of the same form describes the motion in y. The betatron oscillation phase advances according to dψx/ds = 1/βx; that is, 2πβx also plays the role of a local wavelength of oscillations along the orbit. An extremely important parameter for circular machines is the tune, Qx, which is the number of such oscillations per turn about the closed path: Qx = 1 2π I dψx = 1 2π I ds βx(s). (31.6) While the integer part of the tune [Qx,y] generally characterizes the extent of the focusing lattice, it is the fractional part of the tune {Qx,y} that needs to be well deﬁned and controlled by the machine operators in order to stay away from potentially detrimental resonances, which may occur under conditions of kQx +lQy = m, where k, l, and m are integers. For example, for the LHC a combina-tion of horizontal and vertical tunes — also called the working point — equal to (Qx, Qy)=(64.31, 59.32) has been selected, such that resonances up to the order of \|k\| + \|l\| = 10 or 12 are avoided. These resonances are driven by high order multipole components of the ﬁelds in the magnets, or by self-ﬁelds of the beam, or by the electromagnetic ﬁelds of the opposite bunch. Normally, the nonlinear components are very weak compared to linear ones, nevertheless, when the nonlinear res-onance condition is encountered, the amplitudes of particle oscillations could grow over the beam lifetime, resulting in the escape of the particles to the machine aperture, in the increase of the average beam size, or in both; either of these is highly undesirable phenomena. Careful analysis of nonlinear beam dynamics is instrumental in determining and optimizing the dynamic aperture, which is deﬁned as the maximum amplitude of a bounded particle motion. Neglecting for now all nonlinear eﬀects and usually small x −y coupling, and considering only the linear dynamics, the beta-function is well deﬁned and satisﬁes the following equation: 2βxβ′′ x −β′2 x + 4β2 xKx = 4 . (31.7) In a region free of magnetic ﬁelds, such as in the neighborhood of a collider interaction point (IP), usually occupied by particle detectors, a symmetric solution of Eq. (31.7) is a parabola: βx(s) = β∗ x + s2 β∗ x , (31.8) 31st May, 2024 4 31. Accelerator Physics of Colliders where, in this case, s denotes the longitudinal distance from the IP. The location of the beam waist usually coincides with the IP and corresponds to the minimum value of the beta-function β∗ x; the asterisk is used to indicate IP parameters. Note that individual quadrupole magnet focuses particles in one plane and defocuses in another, see Eq.(31.4), and a standard way to provide focusing in both planes is to employ an alternating gradient periodic focusing lattice, consisting of a sequence of equally-spaced quadrupoles with a magnetic ﬁeld gradient equal in magnitude, but alternating in sign (“focusing quadrupole - drift space - defocusing quadrupole - drift space” – known as a FODO cell). Eq. (31.7) has stable periodic solutions βx(s), βy(s) in both planes provided that the focal length of the quadrupoles is longer than half the focusing-lens spacing L, i.e., f = p/(eB2l) > L/2 (where l is the length of a quadrupole magnet, here assumed to be short l ≪L, and B2 ≡\|∂By/∂x\| the quadrupoles’ ﬁeld gradient). In that case, the beta-functions have maxima at the focusing quadrupoles and minima at the defocusing ones, equal to, for example, βmax,min = (2± √ 2)L in the case of f = L/ √ 2, which corresponds to a betatron phase advance ∆ψx,y = 90◦per FODO cell. Expressing the invariant Jx in terms of x, x′ yields Jx = 1 2 γxx2 + 2αxxx′ + βxx′2 = x2 + (αxx + βxx′)2 2βx (31.9) with γx = γx(s) ≡(1 + α2 x(s))/βx(s). In a periodic system, these Courant-Snyder parameters (frequently referred to as Twiss parameters) α(s), β(s), γ(s) are usually deﬁned by the focusing lattice; in a single pass system such as a linac, the parameters may be selected to match the x-x′ distribution of the input beam. For a given position s in the ring, the transverse particle motion in {x, x′ ≡dx/ds} phase space describes an ellipse, the area of which is 2πJx, where the horizontal action Jx is a constant of motion and independent of s. If the interior of that ellipse is populated by an ensemble of non-interacting and non-radiating particles, that area, given the name emittance, is constant over the trajectory as well and would only change with energy. In a typical case of the particle’s energy change rate being much slower than betatron motion, and considering a Hamiltonian system (i.e., a hadron collider or a linear collider, either without signiﬁcant synchrotron radiation), the adiabatic invariant R pxdx is conserved, and given that for small angles px = x′ · βγmc2, it is common practice to consider an energy-independent normalized emittance that is equal to the product of the emittance and relativistic factor βγ/π and denoted by εn. For a beam with a Gaussian distribution in {x, x′}, average action value ⟨Jx⟩and standard deviations σx, and σx′, the deﬁnition of the normalized rms emittance is εnx ≡βγ⟨Jx⟩= βγ σ2 x(s) βx(s) = βγ σ2 x′(s) γx(s) , (31.10) with a corresponding expression for the other transverse direction, y. The angular brackets denote an average over the beam distribution. For 1D Gaussian beam, 95% of the particles are contained within {x, x′} phase space area of 6πεn/(βγ). Normalized beam emittances are conserved over the acceleration cycle in linear, static focusing lattices Kx,y(s), and consequently, one would expect the same εn at the hadron (or linear) collider top energy as the one coming from the very initial low energy particle source. Unfortunately, that is rarely the case as many time-varying or nonlinear phenomena come into play. In an e−/e+ storage ring, the normalized emittance is not preserved during acceleration, but at each energy the beam’s equilibrium emittance is determined by the eﬀect of synchrotron radiation as a balance between radiation damping and quantum excitation . In such a ring, for a given accelerator optics, the normalized equilibrium emittance increases with the third power of the beam energy . 31st May, 2024 5 31. Accelerator Physics of Colliders As for the description of a particle’s longitudinal motion, one takes the fractional momentum deviation ∆p/p from that of the reference particle as the variable conjugate to z. The factors Kx,y and ρ in Hill’s equations (31.4) are dependent on momentum p, leading to a number of eﬀects: ﬁrst, the trajectory of oﬀ-momentum particles deviates by ∆x(s) = Dx(s)(∆p/p), where the dispersion function Dx(s) is determined by the magnetic lattice and is usually positive, periodic, and of the order of ∼ρ/Q2 x. Second, the radius of curvature and orbit path-length C vary with the momentum and, to ﬁrst order, are characterized by the momentum compaction factor αc, αc ≡∆C/C ∆p/p = 1 C I Dx(s) ρ(s) ds , (31.11) which typically is of order 1/Q2 x. Energy deviations also result in changes of machine focusing lattice properties and variations of the particle tunes, characterized by the chromaticity Q′ x,y ≡ ∆Qx,y/(∆p/p). The natural chromaticity due to momentum dependence of the quadrupole focusing is negative and large ∼−Qx,y. Corresponding chromatic tune variations can, therefore, become unacceptably large even for relatively small energy deviations (∆p/p) ∼(0.1 −1) · 10−3. To assure transverse particle stability, usually, the chromaticity is partially or fully compensated by additional sextupole magnets placed at locations of non-zero dispersion. Radiofrequency electric ﬁelds in s direction provide a longitudinal focusing eﬀect, allowing a stable increase of particle energy. The frequency fs of such longitudinal synchrotron oscillations is (expressed in units of revolution frequency f0, to become the synchrotron tune Qs) Qs ≡fs f0 = s (αc −1/γ2)hZeVRF sin(φs) 2πβcp , (31.12) where h = fRF /f0 denotes the RF harmonic number, VRF the RF voltage, and φs = cos−1(∆E/ZeVRF) the synchronous phase, with ∆E the average energy loss per turn (e.g. due to synchrotron radiation and impedance). The synchrotron tune Qs determines the amplitude of longitudinal oscillations for a particle with an initial momentum oﬀset, e.g., the rms bunch length σz relates to the rms momentum spread δp/p as: σz = c(αc −1/γ2) 2πQsf0 δp p . (31.13) Similarly to the case of transverse oscillations, the area of the longitudinal phase space {∆E, ∆t}, or {γβδp/p = (1/β)∆γ, z = βc∆t}, encircled by a moving particle is an adiabatic invariant, and the corresponding normalized longitudinal emittance εn,L = βγmcσz(δp/p) is a generally conserved quantity in hadron accelerators and also in linear accelerators. In the case of lepton storage rings, synchrotron radiation determines the equilibrium relative momentum spread, which grows linearly with beam energy [7,8], and the corresponding bunch length follows from Eq. (31.13). In hadron synchrotrons, the longitudinal emittance sometimes is intentionally blown up during acceleration, so as to preserve longitudinal beam stability. Longitudinal oscillations are the slowest of all the periodic processes which take place in the accelerators. For example, in the LHC, the frequency of synchrotron oscillations at the top energy of 7 TeV is about fs = 23 Hz, the revolution frequency is f0 =11.3 kHz, the frequency of betatron oscillations is about Qx,yfrev ≃700 kHz and the RF frequency is fRF = 400.8 MHz (h = 35640). It should be noted that longitudinal motion is practically absent in linacs. In the absence of bending dipoles, dispersion Dx(s) is zero and so are the momentum compaction factor αc and the synchrotron tune Qs. As a result, ultrarelativistic particles in a linac barely change their relative positions during acceleration despite signiﬁcant energy spread. 31st May, 2024 6 31. Accelerator Physics of Colliders Highest-energy circular colliders face a serious impediment in the form of synchrotron radiation (SR) that causes an energy loss per turn of ∆ESR = 1 3ε0 Z2e2β3γ4 ρ , (31.14) here, ε0 is the permittivity of vacuum. For electrons and positrons it is equal to ∆ESR = 88.5 [keV/turn] E4 b[GeV]/ρ[m] and requires correspondingly high total RF voltage per turn to replenish the loss. Above a few hundred GeV, the SR energy loss becomes comparable to beam energy ∆ESR ∼Eb, which makes circular e+e−colliders impractical for c.m.e. above ∼500 GeV. Dynamics of the particle spin and sophisticated methods to maintain beam polarization along the acceleration chain, from the polarized sources to collisions, dedicated spin matching procedures to enable self polarization in e+/e−storage rings and the resonant depolarization method of ultra-precise c.m.e. calibration are described in . 31.2.2 High Intensity Beams Ultimate collider luminosity calls for high beam currents Ib = Zef0nbN. Three related major diﬃculties include growing RF demands to compensate the synchrotron-radiation power loss P = Ib∆ESR in e+/e−beams, the advent of so-called coherent (or collective) beam instabilities, and growing demands for minimization of radiation due to inevitable particle losses. Many types of single- and multi-bunch instabilities are caused by beam interactions with electromagnetic ﬁelds induced by the beam itself due to the impedance of the vacuum chambers and RF cavities , or caused by unstable clouds of secondary particles, like electrons or ions, which are formed around the circulating beams . These instabilities can develop as quickly as within tens to thousands of turns and need to be controlled. Mechanisms that are routinely employed to avoid coherent instabilities include the use of nonlinear magnets to generate suﬃcient spread of the tunes and therefore, provide Landau damping, fast beam-based transverse and longitudinal feedback systems, and electron/ion clearing (either by weak magnetic or electric ﬁelds or by modulation of the primary beam current proﬁle rendering secondaries unstable, or by reducing the yield of secondary electrons via either a special coating or extensive beam scrubbing of the vacuum chamber walls). High current beam operation is sensitive to even minuscule fractional intensity losses caused by particles’ scattering at a large angle or with a large energy loss, suﬃcient for either the particle amplitudes q 2Jx,yβx,y(s), or their dispersive position deviations ∆x = Dx(s)(δp/p) to exceed the available transverse aperture, usually set by collimators (otherwise, by the vacuum chamber and magnet apertures). This can be due to residual vacuum molecules near the beam orbit or Compton scattering oﬀthermal photons, due to Coulomb scattering oﬀother particles within the same bunch (Touschek eﬀect), or due to collisions with opposite beam particles and ﬁelds, such as inelastic interaction of protons, Bhabha scattering e+e−→e+e−, or radiative Bhabha scattering e+e−→e+e−γ (see corresponding chapters in ). Particles can also get lost on the aperture as a result of much slower mechanisms of diﬀusion caused either by the above processes with smaller scattering amplitudes, but stochastically repeated many times, such as intensity-dependent multiple Coulomb intrabeam scattering , by external noises such as ground motion or magnetic ﬁeld ﬂuctuations, or via chaotic mechanisms like Arnold diﬀusion, modulational diﬀusion, or resonance streaming in nonlinear ﬁelds, enhanced by minor tune modulations. Diﬀusion leads to a slow evolution of the beam distribution function and appearance of highly unwanted large-amplitude tails and beam emittance growth. The only way to counteract it is to arrange beam cooling (damping of particle oscillations). The cooling requires a reaction force opposite to particle momentum arranged such that, on average, the corresponding dissipative particle energy loss is compensated for by external power [13,14]. 31st May, 2024 7 31. Accelerator Physics of Colliders In the case of electron or positron storage rings, such cooling occurs naturally due to synchrotron radiation and provides an automatic route to achieve small equilibrium emittances through a bal-ance between radiation damping and excitation of oscillations by random radiation of individual photons. Fast radiation damping allows top-up injection of new particles without removing existing ones, a useful method to maximize the integrated luminosity of circular e+e−colliders. Synchrotron radiation damping will also be an important cooling mechanism for future energy-frontier hadron colliders, like the proposed FCC-hh and SppC (see below). Four other methods of beam cooling have been developed and successfully employed to attain low emittances, namely electron cooling and stochastic cooling of heavy particles (ions and antiprotons), laser cooling of ion beams, and the ionization cooling of muons. To avoid damage or excessive irradiation of accelerator components so that these remain acces-sible for maintenance in the tunnel, sophisticated collimation systems are utilized. These systems usually employ a series of targets or primary collimators which scatter the halo particles, and numerous absorbers (sometimes as many as a hundred, which intercept particles in dedicated loca-tions) [15, Ch.9.7]. In the highest energy modern and future colliders, extreme total beam energies nbNEb ranging from MJs to GJs and impacting surface energy densities reaching many GJ/mm2 pose one of the biggest challenges for high eﬃciency and robust particle collimation. 31.2.3 High Luminosity Collisions Eq. (31.3) for luminosity can be recast in terms of normalized transverse emittances Eq. (31.10) and the beta-functions β∗at the IP as: L = f0γnb N2 4π q εnxβ∗ xεnyβ∗ y F . (31.15) Here, equal bunch populations N are assumed in two Gaussian beams with the same emittances. Naturally, to achieve a high luminosity, one has to maximize the total beam populations nbN within the lowest possible emittances, and collide the beams at high frequency at locations where the focusing beam optics provides the lowest possible values of the amplitude functions β∗, the so-called low-beta insertion. The latter requires sophisticated systems of strong focusing elements, sometimes occupying quite a signiﬁcant fraction of the collider’s total length. The lowest β∗ x,y is determined by the maximum ﬁeld gradients and apertures in the interaction region (IR) magnets and the eﬀectiveness of compensation of chromatic and nonlinear aberrations. The typical geometric reduction factor is F ≈1, and it rarely drops below 0.5 for the majority of colliders, unless this is speciﬁcally required by physics processes under study. The reduction due to the hourglass eﬀect is caused by the increase in transverse beam sizes as one proceeds away from the IP, where β(s) grows parabolically, as in Eq. (31.8). For long round bunches the eﬀect scales as F ≈√πA exp(A2)erfc(A) , where A = β∗/σz. Nonzero beam crossing angles θc in the horizontal plane and long bunches (rms bunch length σz) will reduce the luminosity, too, by a factor F ≈1/(1 + Φ2)1/2, where the parameter Φ = σz tan(θc/2)/σ∗ x is known as the Piwinski angle. One of the most common limits to producing high luminosity arises from electric and magnetic forces of the opposite bunch at the IPs, characterized by a dimensionless beam-beam parameter : ξx,y = r0Nβ∗ x,y 2πγσ∗ x,y(σ∗ x + σ∗ y) , (31.16) where r0 = Z2e2/(4πε0mc2) is the classical radius of the colliding particle (with charge Ze and mass m). From Eqs. (31.3) or (31.15) and (31.16), one can note that the path to higher luminosity via higher beam intensity and smaller beam sizes almost automatically calls for a higher beam-beam 31st May, 2024 8 31. Accelerator Physics of Colliders parameter as L ∝ξ. Cited in the Tables, the beam-beam parameter is roughly equal to the betatron tune shift experienced by small-amplitude particles – positive in the case of opposite charge beams, like e+e−, and negative for same charge beams as in pp collisions. Beam-beam forces can lead to coherent eﬀects, such as unstable beam oscillations or blow-up of one beam’s size while the other beam remains small or even shrinks (ﬂip-ﬂop eﬀect). The tune spread arising from ξ and the nonlinear nature of beam-beam interactions results in strong diﬀusion along high-order transverse resonances kQx + lQy = m and, ultimately, in beam size growth and beam losses. Operational experience indicates that the aforementioned eﬀects are tolerable below certain beam-beam limit of ξx,y ≈0.003 −0.012 in hadron colliders , and – due to strong synchrotron radiation damping – an order of magnitude higher one in e+e−colliders, with maximum ξx,y ≈0.03 −0.12 [17,18]. The accessible beam-beam parameter range can also be restricted by coherent beam-beam instabilities. These various limits translate into a maximum allowed single bunch intensity N and call for an increase of the number of bunches nb to achieve higher luminosities. In linear colliders, where each bunch collides only once, with typically much smaller beam size and experiencing much stronger forces, the strength of the collision is measured by the ratio of the rms bunch length σz to the beam-beam focal length. This ratio, called disruption parameter Dy, is related to ξy via Dy = 4πσzξy/β∗ y. Stronger beam-beam focusing and higher Dy lead to eﬀectively smaller beam sizes and a resulting luminosity enhancement; but it also makes the collisions more sensitive to small oﬀsets, resulting in a kink instability. Additional beam-beam eﬀects arising in the collisions at linear colliders are the emission of beamstrahlung (synchrotron radiation in the ﬁeld of the opposing beam), along with e+e−pair creation, and depolarization by various mechanisms. Beamstrahlung is relevant for both linear colliders, where it may signiﬁcantly degrade the luminosity spectrum, and for future highest-energy circular colliders, where it may limit the beam lifetime, and also increases the energy spread and bunch length of the stored beam. For both types of colliders, the beamstrahlung is mitigated by making the colliding beams as ﬂat as possible at the interaction point (σ∗ x ≫σ∗ y) to lower the beams’ EM ﬁelds. The photon energy spectrum of the beamstrahlung is characterized by the parameter Υ = (2/3)ℏωc/Eb , with ℏωc denoting the critical photon energy. The spectrum strongly deviates from the classical synchrotron radiation spectrum for Υ approaching or exceeding 1. For hadron colliders, two fundamental luminosity limits are the beam lifetime, determined by burn-oﬀin the collisions due to inelastic pp interaction dN/dt = −Lσin, and the radiation from the collision debris, which may induce “quenches” (transitions to the normal-conducting state) of the superconducting ﬁnal quadrupole magnets, and, in the long term, aﬀect the equipment lifetime. Another limit on the achievable integrated luminosity in circular colliders is set by the minimum or average turnaround time (the time between the beam abort at the end of a physics ﬁll and the start of the next physics collisions). Achieving practical ﬁlling times with many bunches in the collider requires either fast cycling injector machines and/or the top-up injection operation. The latter makes the average luminosity of circular electron-positron colliders approximately equal to the peak luminosity. 31.3 Recent High Energy Colliders In this and the following section, elaboration is made on various issues associated with some of the recently operating colliders, particularly factors which impact integrated luminosity. Only general references are provided, where further information can be obtained. A more complete list of recent colliders and their parameters can be found in the High-Energy Collider Parameters tables. 31.3.1 Tevatron The ﬁrst superconducting synchrotron in history, the Tevatron was converted into a proton-antiproton collider in 1985. Its 4.4 T dipole magnets employed Nb-Ti superconducting cable op-31st May, 2024 9 31. Accelerator Physics of Colliders erating at 4.5 K, requiring what was then the world’s largest cryogenic system. With √s up to 1.96 TeV it was the highest energy collider for 25 years and delivered more than 12 fb−1 of the integrated luminosity to each p¯ p detector experiments (CDF and D0) before being shut oﬀin 2011. The route to high integrated luminosity in the Tevatron was governed by the antiproton produc-tion rate, the turn-around time to produce another store, and the resulting optimization of store time . The antiproton production complex consisted of three 8 GeV ¯ p accelerators (the Ac-cumulator, Debuncher, and Recycler – the latter was the ﬁrst high energy accelerator built with permanent magnets), and employed 25 independent stochastic cooling systems and one pioneering high-energy electron cooling set-up to accumulate up to a record high value of 25·1010 ¯ p per hour. Despite severe parasitic long-range interactions of the two beams, each consisting of 36 bunches placed on helical orbits by two dozen ±150 kV high-voltage (HV) separators, a total beam-beam tune shift parameter of nIPξ ≈0.025-0.03 was achieved, a record for hadron beams, with nIP = 2 primary collision points. Other notable advances in the accelerator science and technology included advanced longitudinal beam manipulation techniques of slip-stacking and momentum mining and the ﬁrst operational use of electron lenses for beam collimation and for compensation of long-range beam-beam eﬀects . The Tevatron ultimately achieved luminosities a factor of 430 higher than the original design speciﬁcation. 31.3.2 HERA The ﬁrst lepton-proton collider, the 6.4 km long Hadron-Elektron-Ring-Anlage (HERA) at DESY in Germany , operated between 1992 and 2007 and delivered nearly 1 fb−1 of integrated luminosity at √s of about 320 GeV to the electron-proton collider experiments H1 and ZEUS [24, Ch.10.5]. It was the ﬁrst facility to employ both applications of superconductivity: 5 T magnets in the 920 GeV proton ring and SRF accelerating structures to provide about 12 MW of RF power to compensate for synchrotron radiation losses of 30 GeV lepton beams (positrons or electrons, in a conventional-magnet ring). With proper orbit and optics control, the HERA lepton beam would naturally become transversely polarized to about 60% (within about 40 minutes) thanks to the Sokolov-Ternov eﬀect. Special magnets called spin rotators were implemented on either side of the collider IPs to produce 30–45% longitudinal polarization at the experiments. 31.3.3 LEP Installed in a tunnel of 26.7 km circumference, LEP was the largest circular e+e−collider built so far. LEP was operated from 1989 to 2000 with beam energies ranging from 45.6 to 104.5 GeV. The synchrotron radiation loss per turn reached some 3% of beam energy and it was the total available RF voltage and power, respectively, that determined LEP maximum energy and luminosity of 1032 cm−2s−1. At a beam energy of 98 GeV, LEP operated with a beam-beam parameter ξ ≈0.083 , i.e., the total beam-beam tune shift for four interaction points was nIPξ ≈0.33. In the last year of operation, 288 SRF cavities were powered by 36 klystrons with an average power of 0.6 MW each and provided VRF= 3.63 GV. Up to about 60 GeV, LEP used resonant depolarization to measure the beam energy with 0.001% accuracy. 31.3.4 SLC Based on an existing 3-km long 2.85 GHz warm RF linac, the SLC was the ﬁrst and only linear collider. It was operated from 1987 to 1998 with a constant beam energy of 45.6 GeV, up to about 80% electron-beam polarization, quasi-ﬂat beams, a ﬁnal-focus optics with local chromatic correction based on four interleaved sextupoles and β∗ y ≈1 mm. In its last year, SLC achieved a peak luminosity of about 3 × 1030 cm−2s−1, roughly half of the design value. The SLC had a high-eﬃciency positron source providing 5 × 1012 e+ per s for 120 Hz injection into the linac. It also employed the BNS damping to suppress the single-bunch beam break up instability, and 31st May, 2024 10 31. Accelerator Physics of Colliders also demonstrated an about 2-fold increase of luminosity from disruption enhancement due to the mutual focusing of the colliding electron and positron bunches at the interaction point, 31.4 Presently Operating Colliders 31.4.1 LHC With a beam energy of 6.8 TeV (to be raised to the design value of 7 TeV through consolida-tion and magnet training), the superconducting Large Hadron Collider presently is the world’s highest energy collider. In the latest runs, peak luminosities of up to 2.1 × 1034 cm−2s−1 have been achieved at 13 TeV cme - more than twice the design value (the current status is best checked at the Web site ). To meet its luminosity goals, the LHC operates with a high beam current of approximately 0.5 A, leading to stored energies of about 330 MJ per beam. Controlled energy deposition and component protection are given a high priority and a sophisticated highly eﬃcient system of more than 100 beam collimators is employed . At the energy of about 7 TeV per particle, synchrotron radiation poses a challenge, as the cryogenic system must remove roughly 7 kW due to synchrotron radiation, intercepted with a specially designed beamscreen inside the vacuum chamber, at a temperature of about 5–20 K, to be compared with a temperature of 1.9 K for the magnet cold bore. The elevated temperature allows for a more energy-eﬃcient removal of beam-induced heat. The beamscreen also provides an eﬀective cryo-pump for the vacuum system. When synchrotron-radiation photons hit the beamscreen, they can generate photoelectrons. These photoelectrons, and also any other electrons generated in the vacuum system, e.g. by residual-gas ionization, are accelerated in the electric ﬁeld of the beam and may multiply via secondary-electron emission, with consequent electron cloud development. To mitigate this issue, the beamscreen is regularly subjected to beam-induced surface conditioning (scrubbing), thereby lowering the sec-ondary emission yield. The two proton beams of 2556 bunches spaced by 25 ns are contained in separate pipes throughout most of the circumference and are brought together into a single 130 m long beam pipe at the interaction points. To avoid approximately 30 head-on collisions a small crossing angle of about 0.3 mrad is employed, which reduces the luminosity by about 15%. Still, the bunches moving in one direction experience multiple long-range encounters with the counter-rotating bunches and the resulting perturbations of the particle motion substantially contribute to the beam lifetime reduction. The dominant source of approximately 8 hour characteristic luminos-ity decay time is proton burn-oﬀdue to inelastic pp interaction with σin ≈81 mbarn, corresponding to pile-up of up to 50 (number of events per individual bunch crossing). In special physics runs with a few bunches and large β∗, the LHC achieved a head-on beam-beam tune shift of nIP\|ξ\| ≈0.02 with nIP = 2, about twice as high as in regular operation. The Tables also show the LHC luminosity performance in Pb-Pb collisions, which for the AT-LAS and CMS experiments well exceeded the design value, while for the ALICE experiment, the luminosity is levelled near the Pb-Pb design value of 1027 cm−2s−1. The LHC can also provide Pb-p collisions as it did in 2013 and 2016, and other ion-ion or ion-proton collisions, at diﬀerent energies. In the coming years, an ambitious upgrade program, HL-LHC , with the accompanying LHC Injectors Upgrade , has as its target an order-of-magnitude increase in integrated luminosity through doubling the proton beam current, the utilization of new larger aperture Nb3Sn supercon-ducting ﬁnal quadrupoles to allow squeezing the β∗to as low as 10 cm, superconducting compact crab cavities and luminosity leveling also for ATLAS and CMS as its key ingredients. 31.4.2 Electron-Positron Rings Asymmetric energies of the two beams allow for the enhancement of B-physics research and for interesting interaction region designs. SuperKEKB operates with 7 GeV electron and 4 GeV positron beams since 2018 and is aiming for luminosities of 8 × 1035 cm−2s−1 . By summer 31st May, 2024 11 31. Accelerator Physics of Colliders 2022, a world record luminosity of 4.7 × 1034 cm−2s−1 has been reached, still at rather low beam currents. Vertical beam-beam tune shifts of ξy ≈0.05 for the 4 GeV positron beam, and ξy ≈0.03 for the 7 GeV electron beam have been achieved. These values are still about a factor of two lower than at the previous KEKB. Since 2020 SuperKEKB operates with a virtual crab-waist collision scheme, ﬁrst developed for the FCC-ee design . The original crab-waist scheme, based on additional sextupole magnets, was earlier implemented at DAΦNE . The general crab-waist concept combines a large Piwinski angle Φ, and an extremely low β∗ y (≪σz) with the cancellation of the transverse betatron resonances which occur under conditions of kQx + lQy = n, where k, l, n are integers. The latter is achieved by means of existing or additional electromagnetic sextupoles with special betatron phase advances to the collision point . The crab-waist collision scheme has become a design choice for all proposed future e+e−circular colliders. Beside SuperKEKB and DAΦNE, three other e+e−ring colliders currently in operation are VEPP-2000 with √s up to 2.0 GeV, BEPC-II with √s up to 4.6 GeV and VEPP-4M with maximum c.m.e. of 12 GeV . 31.4.3 RHIC The Relativistic Heavy Ion Collider employs 3.45 T Nb-Ti superconducting magnets, and collides combinations of fully-stripped ions such as H-H (p −p), p-Al, p-Au, d-Au, h-Au, Cu-Cu, Cu-Au, Zr-Zr, Ru-Ru, Au-Au, and U-U over a wide energy range. The high charge per particle (+79 for gold, for instance) makes intra-beam scattering of particles within the bunch a special concern, even for seemingly moderate bunch intensities. In 2012, 3-D stochastic cooling was successfully implemented in RHIC and is now routinely used. With stochastic cooling, steady increases in the bunch intensity, and numerous other upgrades, RHIC now operates at 44 times the Au-Au design average luminosity. Unique among high energy colliders, RHIC heavy ions beams cross the transition energy γrmt = 1/√αc during acceleration – see Eqs.(31.11, 31.12) – a point where the derivative with respect to momentum of the revolution period is zero. This period of time is kept as short as allowed by the magnet ramp rate and must be dealt with carefully. RHIC is also unique in its ability to accelerate and collide polarized proton beams. As proton beam polarization must be maintained from its low-energy source, successful acceleration through the myriad of depolarizing resonance conditions in high energy circular accelerators has taken years to accomplish . An energy of 255 GeV per proton with 60% ﬁnal polarization per beam has been realized. As part of a scheme to compensate the head-on beam-beam eﬀect, electron lenses operated routinely during the record high beam-beam parameter polarized proton operation at 100 GeV energy in 2015 . 31.5 Future High Energy Colliders and Prospects Modern nuclear physics and high energy particle physics face critical questions which require next-generation high-energy and high-intensity experiments using hadron-hadron, lepton-lepton, and lepton-proton colliding-beam facilities. Understanding the structure of the proton and neutron directly from the dynamics of their quarks and gluons governed by the quantum chromodynamics calls for new ion-ion and electron-ion colliders. Two types of colliders are generally aspired by the HEP community [40, 41]: i) Higgs factories with a c.m.e. of 240–250 GeV in e+e−collisions for precision studies of the Higgs boson (mH = 125 GeV) and exploration of the Higgs sector in greater detail, including measurements of Higgs couplings to fermions and vector bosons, self-coupling, rare decays, mass and width, that can also deliver other electroweak precision physics, e.g. on the Z-pole (91 GeV c.m.e.), at the W-pair threshold (about 160 GeV), and when run as a top quark factory (365–380 GeV); and ii) colliders with c.m.e. levels signiﬁcantly beyond those of the LHC to explore the energy frontier for potential discoveries through direct searches in pp, µµ, and e+e−interactions. In addition, precision physics at future high-luminosity factories operating at 31st May, 2024 12 31. Accelerator Physics of Colliders the τ-charm energy also provides sensitivity to new physics at multi-TeV energies and beyond. A comprehensive review of the future colliders’ projects, ideas, and R&D activities can be found in Ref. ; Ref. presents comparative analysis of implementation challenges of numerous future HEP collider proposals. Below we only brieﬂy summarize leading collider proposals for construction over the next several decades which rely mostly on currently available technologies, such as normal-conducting (NC) or SC RF and/or NC or SC magnets, some of them requiring reasonable scope and duration mission-oriented development programs, as well as advanced schemes based on plasma acceleration and other innovative ideas. Tentative parameters of some of the colliders discussed, or mentioned, in this section are summarized in Table 31.1 and Table 31.2. 31.5.1 Ion-Ion and Electron-Ion Colliders NICA (Nuclotron-based Ion Collider fAcility) is a new accelerator complex under construction at the Joint Institute for Nuclear Research (JINR, Dubna, Russia) to study properties of hot and dense baryonic matter, strong interactions between quarks and gluons, and spin physics . NICA will provide a variety of beam species, ranging from protons and polarized deuterons to massive gold ions. The 500 m circumference SC magnet based collider is designed for average luminosity in heavy ion and light ion interactions at √sNN=4–11 GeV of 1×1027cm−2s−1 for a variety of nuclei up to 197Au79+, and for polarized proton and deuteron collisions at √s=12–27 GeV with L=(1–10)×1031cm−2s−1. NICA major accelerator challenges include strong intrabeam scattering and space-charge eﬀects which will be mitigated by extensive use of electron and stochastic cooling systems. The recently announced Electron-Ion Collider (EIC) for nuclear physics research will be built at Brookhaven National Laboratory in the US and arrange collisions between an the reconﬁgured RHIC with a 41–275 GeV proton beam and a 5-18 GeV electron beam stored in a new ring (eRHIC) . The EIC physics requirements , include highly polarized (Pe,n ∼70%) electron and nucleon beams (as the precision of measurements of interest scales as LP 2 e P 2 n), a spectrum of ion beams from deuterons to the heaviest nuclei (U or Pb), variable c.m.e. values from √s =20 GeV to 140 GeV, high luminosities of up to 1033−34 cm−2s−1, as well as possibilities of having more than one interaction region. Main accelerator design challenges on the path to the required energy, luminosity, and polarization, include the development of SRF crab-cavities and advanced SC magnets for interaction region focusing, energy-recovery linac (ERL) based electron cooling of hadron beams, essential to attain luminosities two orders of magnitude beyond the predecessor HERA ep collider, and high intensity polarized particle sources, augmented by the development of special magnets and operational techniques to preserve the polarization through the acceleration process to the collisions, including swap-out injection. 31.5.2 Higgs/Electroweak Factories Higgs factory proposals generally aim at improving the precision of coupling measurements of Higgs boson, top quark, W and Z by an order of magnitude or more compared with previous studies. Two proposals for ∼100 km circumference circular e+e−colliders have recently gained momentum: the Future Circular Collider (FCC-ee) at CERN and the Circular Electron-Positron Collider (CEPC) in China . Design philosophy of these machines assumes use of the maximum RF power available to compensate O(100 MW) synchrotron radiation losses PSR = 2I · ∆ESR and operation at the beam-beam limit ξy that yields peak luminosity: L = 3 16πr2 0(mec2) PSRξyρ β∗ yγ3 , (31.17) that scales approximately as 1/E3.5 b for practical limits on P, ξy and β∗ y. The short beam lifetime at the high target luminosity, due to radiative Bhabha scattering, requires these machines to be 31st May, 2024 13 31. Accelerator Physics of Colliders Table 31.1: Tentative parameters of selected future e+e−high-energy colliders. Parameters associated with diﬀerent beam energy scenarios are comma-separated; H and V indicate horizontal and vertical directions. FCC-ee CEPC ILC CLIC Species e+e− e+e− e+e− e+e− Beam energy Eb (GeV) 46, 120, 183 46, 120, 180 125, 250 190, 1500 Circumference or length (km) 90.66 100 20.5, 31 11, 50 Interaction regions 4 2 1 1 Est. integrated luminosity per experiment (ab−1/year) 17, 0.6, 0.15 15, 0.65, 0.07 0.2, 0.3 0.1, 0.6 Peak lumi. L/IP (1034cm−2s−1) 140, 5.0, 1.25 115, 5.0, 0.5 1.4, 1.8 1.5, 6 Rep.rate (Hz, frev for rings) 3307 3000 5 50 Polarization (%) ≥10, 0, 0 5–10, 0, 0 80/30 (e−/e+) 80/0 (e−/e+) Time between collisions (µs) 0.025, 0.3, 2.5 0.025, 0.68, 2.6 0.55 0.0005 Energy spread (rms, 10−3) 1.09, 1.43, 1.92 1.3, 1.7, 2.0 e−: 1.9, 1.2 e+: 1.5, 0.7 3.5 Bunch length σz (rms, mm) 15.5, 4.7, 2.2 8.7, 3.9, 2.9 0.3 0.07, 0.044 IP beam size σ∗(rms, µm) H: 9, 13, 40 V: 0.04, 0.04, 0.05 H: 5.9, 14, 38 V: 0.04, 0.04, 0.11 H: 0.52, 0.47 V: 0.008, 0.006 H: 0.15, 0.04 V: 0.003, 0.001 Emittance, εn (rms, µm) H: 63, 167, 568 V: 0.17, 0.3, 0.6 H: 24, 150, 493 V: 0.12, 0.3, 1.7 H: 5, 10 V: 0.035, 0.035 H: 0.95, 0.66 V: 0.03, 0.02 β∗at interaction point (cm) H: 11, 24, 100 V: 0.07, 0.1, 0.16 H: 13, 33, 104 V: 0.09, 0.1, 0.27 H: 1.3, 1.1 V: 0.041, 0.048 H: 0.8, 0.69 V: 0.01, 0.0068 Full crossing angle θc (mrad) 30 33 14 20 Crossing scheme crab waist crab waist crab crossing crab crossing Piwinski angle Φ = σzθc/(2σ∗ x) 26, 5.4, 0.8 24, 4.8, 1.3 0 0 Beam-beam param. ξy (10−3) 97, 88, 134 110, 127, 100 n/a n/a Disruption parameter Dy 1.3, 0.9, 2.0 0.6, 1.3, 0.8 35, 25 13, 8 Average Upsilon Υ (10−2) 0.01, 0.04, 0.06 0.02, 0.04, 0.05 3, 6 17, 500 RF frequency fRF (MHz) 400, 400, 800 650 1300 11994 Particles per bunch N (1010) 21, 12, 16 14, 13, 20 2 0.52, 0.37 Bunches per beam nb 11200, 440, 60 11951, 249, 35 1312 (pulse) 352, 312 (trains at 50 Hz) Average beam current Ib (mA) 1270, 27, 4.9 804, 16.7, 3.3 0.021 0.014, 0.009 Injection energy (GeV) on Eb (top oﬀ) on Eb (top oﬀ) 5.0 (linac) 9.0 (linac) RF gradient G (MV/m) 5.7, 10.6, 20.1 17.4, 24.9, 27.6 31.5 72, 100 Total SR power loss (MW) 100 60 n/a n/a Total beam power (MW) n/a n/a 5.3, 10.5 5.6, 28 Key technology — — high grad. SC RF two-beam accel. constructed with a full-energy injector ring installed in the same tunnel to top oﬀthe electron and positron currents in the collider rings operating at constant energy. Beamstrahlung introduces an additional beam lifetime limitation depending on momentum acceptance (so that achieving suﬃcient oﬀ-momentum dynamic aperture becomes one of the design challenges), as well as some bunch lengthening. These ambitious, large-scale projects based on well-established technologies are not extendable to TeV or multi-TeV energies, but they oﬀer several important advantages that include the potential for much higher luminosities, and, thus, higher precision, the ability to operate multiple experiments simultaneously, and their ∼100 km circular tunnels that could later house O(100 TeV) hadron colliders. The high energy eﬃciency of circular e+e−colliders is further boosted by advances in RF power sources, by improved SC cavities, and by innovative low-power magnet systems including ones based on high-temperature superconductors (HTS) at moderate magnetic ﬁeld. 31st May, 2024 14 31. Accelerator Physics of Colliders Table 31.2: Tentative parameters of selected future high-energy hadronic and muon colliders. Parameters associated with diﬀerent particle species for NICA and EIC, and diﬀerent beam-energy scenarios for a muon collider, are comma-separated. Quantities are, where appropriate, r.m.s.; unless noted otherwise, energies refer to beam energy; H and V indicate horizontal and vertical directions. Parameters of HL-LHC can be found in the High-Energy Collider Parameters review tables. NICA EIC FCC-hh SPPC µµ collider Species ion-ion, pp ep, e-ion pp pp µ+µ− Beam energy Eb (TeV) 10−3·(4.5/u, 13) 0.01(e),0.275(p) 40-58 62.5 0.063, 5 Circumference C (km) 0.503 3.834 90.66 100 0.3, 10 Interaction regions 2 1(2) 4 2 1, 2 Est. integr. luminosity per exp. (ab−1/year) 10−8,−3 (ii, pp) 0.1 0.2–1.0 0.6 0.001, 2.0 Peak luminosity L (1034cm−2s−1) 10−7,−2 (ii, pp) 1.05 5–30 4.3 0.008, 20 Rep.rate (Hz, frev for rings) 5.9·105 7.8·104 3307 3000 15, 5 Time between collisions (µs) 0.077 0.009 0.025 0.025 1, 33 Energy spread (rms, 10−3) 1.6 (Au) 0.6 (e), 0.7 (p) 0.1 0.1 0.04, 1 Bunch length σz (rms, mm) 600 7 (e), 60 (p) 80 60 63, 1.5 IP beam size σ∗(H/V rms, µm) 360 95/8.5 6.7-3.5 (init.) 3.0 (init.) 75, 0.9 Emittance εn (H/V rms, mm mrad) 1.1 11.3/1.0 (e), 9.2/1.6 (p) 2.2 (init.) 1.2 (init.) 200, 25 Beta function at IP β∗(H/V cm) 60 45/5.6 (e), 80/7.2 (p) 110–30 50 1.7, 0.15 Beam-beam param. ξ (10−3 H/V) 25 72/100 (e),12 (p) 5–15 15 22, 78 RF frequency fRF (MHz) 13/39 591 400 400/200 805/1300 Particles per bunch N (1010) 0.23 17.2(e), 6.9(p) 10 4 400, 180 Bunches per beam nb 22 1160 9648 10082 1 Average beam current Ib (mA) 480 2500(e),1000 (p) 500 190 640, 9 (peak) Injection energy (GeV) 1-3.8 on Eb (e), 25 (p) ≥1000 3200 on Eb Peak magnetic ﬁeld B (T) 1.8 0.248 (e), 3.80 (p) 14-20 20 10 Polarization (%) 0(i), >50(p) > 70(e), >70(p) 0 0 0 SR power loss/beam (MW) 10−6 10(e), < 10−6(p) 1.0-4.25 2.2 10−3, 0.16 Key technology electron and stoch. cooling strong hadron cooling Nb3Sn/HTS magnets HTS magnets muon prod. & cooling For more than four decades, eﬀorts have been devoted to developing high-gradient RF technol-ogy linear e+e−colliders in order to overcome the synchrotron radiation limitations of circular e+e− machines. The International Linear Collider (ILC), with a c.m.e. of 250 GeV in e+e−collisions, has been under consideration for more than two decades and could potentially be upgraded to √s=500 GeV and further to 1 TeV. CERN’s Compact Linear Collider (CLIC) design, developed since the mid-1980s, also includes possible upgrades, from an initial 380 GeV c.m.e. to ultimately 3 TeV, which would enable searches for new particles of signiﬁcantly higher masses. The primary challenge confronting a high energy, high luminosity single pass collider design is the beam power requirement, so that measures must be taken to keep the demand within bounds as illustrated in a transformed Eq.(31.15): L = 1 8παr0 Pwall √s η σ∗ y Nγ HD . (31.18) Here, Pwall is the total wall-plug power of the collider, to be converted into beam power Pb = 2f0NEb with eﬃciency η, Nγ ≈2αr0N/σ∗ x is the number of beamstrahlung photons emitted per 31st May, 2024 15 31. Accelerator Physics of Colliders e± (α ≈1/137 denotes the ﬁne-structure constant), and the last factor HD, typically between 1 and 2, represents the enhancement of luminosity due to the pinch eﬀect, the additional focusing occurring during the collision of oppositely charged bunches. The management of Pwall leads to an upward push on the bunch population N with an attendant rise in the energy radiated due to the electromagnetic ﬁeld of one bunch acting on the particles of the other (beamstrahlung). Keeping a signiﬁcant fraction of the luminosity close to the nominal energy represents a design goal, which is met if Nγ does not exceed a value of about 1. A consequence is the use of ﬂat beams, where Nγ is managed by the beam width, and luminosity adjusted by the beam height, thus the explicit appearance of the vertical beam size σ∗ y. The ILC [47, 48] is based on 1.3 GHz standing-wave superconducting accelerating structures operating at 2 K with 31.5 MV/m average gradient, up to 8 nm vertical beam size at the IP, and luminosity comparable to the LHC. Progress toward higher ﬁeld gradients and Q values continues to be made, with nitrogen-doping techniques being a recent example . CLIC is based on a novel two-beam acceleration scheme . Here, room-temperature NC copper high-gradient 12 GHz accelerating structures are powered by a high-current 1.9 GeV drive beam to eﬃciently enable accelerating gradients of up to 100 MV/m (though optimal G=70 MV/m for the ﬁrst CLIC stage at √s=380 GeV, and for this stage an alternative RF power drive option with 12 GHz klystrons powering is also being considered). To reach their design luminosities, both CLIC and ILC require unprecedented rates of positron production about 40 times the world record set by the SLC positron source, and very tight control of imperfections, such as O(10 µm) accuracy of pre-alignment of the main linac and beam delivery system components at the level, suppression of fast vibrations of the quadrupoles due to ground motion to O(1 nm) level at frequencies above 1 Hz, advanced beam-based trajectory tuning, and mitigation of the eﬀect of wakeﬁelds . Recent developments of the RF technologies show promise of more compact linear e+e−Hig-gs/Electroweak factories based on either ∼70 MV/m travelling-wave 1.3 GHz SRF structures or on 70–120 MV/m high-eﬃciency, distributed RF coupling, normal-conducting cold copper 5.7 GHz accelerating cavities at liquid nitrogen temperature . There are a number of alternative ideas proposed for studies of the Higgs/Electroweak physics, such as high-energy, high-luminosity e+e−collider in a 100 km tunnel using ERLs to accelerate particles to collision energy in 4 to 6 passes and return up to 81% of the energy back into the SRF cavities on deceleration turns, thus, lowering the required facility power several-fold ; similar power recovery in one pass can greatly improve eﬃciency of linear colliders [55–57]; an arrangement of γγ collision through near-IP conversion of high energy electron beams into intense photon beams by backward Compton scattering oﬀa high-power laser [58,59] or oﬀan FEL photon pulse [60,61]; µµ Higgs factory with unprecedented 0.004% energy resolution , and a high-energy lepton-hadron collider bringing into collision a 60-GeV electron beam from an ERL with the 7 TeV protons circulating in the LHC (LHeC) . At lower energies, Super Tau-Charm Factory proposals aim at the production and precise study of charmonium states and of the tau lepton [64,65]. 31.5.3 Energy Frontier Circular Colliders Several hadron and lepton colliders have been proposed to extend the energy reach beyond the LHC. As noted above, ambitious plans have been proposed to upgrade the FCC and CEPC to hadron colliders – FCC-hh at CERN and Super Proton Proton Collider (SPPC) in China, respectively – by means of next- or next-next generation SC magnets installed in the arc sections of the 100 km rings, so as to enable √s of the order 100 TeV or above [66,67]. Comparable discovery reach is expected for a circular 10–14 TeV muon collider , signiﬁcantly beyond that of practical e+e−linear colliders. The maximum beam energy Eq.(31.1) is directly proportional to the magnetic ﬁeld and to the 31st May, 2024 16 31. Accelerator Physics of Colliders ring circumference, hence,future hadron colliders like FCC-hh and SppC rely on the development of the technology of 14–16 T Nb3Sn dipole magnets or dipole magnets based on high-temperature superconductors (HTS) with a ﬁeld of up to 18-20 T, including iron-based HTS . Though higher ﬁelds are possible with HTS, more cost-eﬀective might be hybrid magnet designs incorporating Nb-Ti, Nb3Sn, and an inner layer of HTS and providing ﬁelds of about 20 T. Another important technology is the cryogenic beam vacuum system, which has to cope with unusually high levels of synchrotron radiation (up to 15 MW in total, for FCC-hh) in a cold environment. The beam-screen intercepting the radiation inside the cold bore of the magnets should operate at or above 50 K — signiﬁcantly higher temperature than in the LHC. Design luminosities of these hadron colliders O(1035 cm−2s−1) will result in a pile-up of events per crossing O(500) (from up to 50 in LHC) and fast intensity drop due to burn-oﬀ. Signiﬁcant radiation damping of beam emittances will naturally level luminosity evolution, though the total beam-beam tune shift nIPξ ∼(0.01 −0.03) might need a special control as it will increase during the store . Future hadron colliders are characterized by record high stored beam energy, e.g. 6–9 GJ in FCC-hh, rendering machine protection a paramount concern. A very challenging multi-stage col-limation system is needed to avoid local beam loss spikes near cold magnets, which would induce magnet quenches. The primary and secondary collimators of the LHC are based on carbon-carbon composite material. For the future hadron colliders, ever stronger materials are being developed and examined, which also feature higher conductivity and, hence, lower impedance. More advanced options include the use of short bent crystals as primary collimators and the deployment of hollow electron-beam lenses as non-destructible collimators . It is noteworthy that machines like FCC-hh or SPPC can additionally be used for ion-ion/ion-proton collisions; their high-energy proton beams can also be collided with high-intensity O(60 GeV) electrons from an ERL resulting in c.m.e. of 3.5 TeV. The lifetime of the muon, γτ0 where τ0=2.2µs, is suﬃcient to allow fast acceleration to high energy before most, or all, of the muons decay, and storage for some 300B turns in a ring with an average bending magnets ﬁeld of B (in units of Tesla). The muon to electron mass ratio of 210 implies removal of the synchrotron radiation barrier and possibility of a muon collider facility scale to a level compatible with on-site placement at existing accelerator laboratories. High-energy muon colliders, as presently conceived, are predicted to be more compact, more power-eﬃcient and signiﬁcantly less expensive than the equivalent energy-frontier hadron or e+e−machines , and a neutrino factory could potentially be realized in the course of their construction . The Higgs production cross-section in the s-channel is enhanced by a factor of (mµ/me)2 compared to that in e+e−collisions. The average luminosity of a muon collider, ⟨L⟩= f0γ2 cτ0 2C nbN2 4πεnβ∗F = BPb Nr0 4πεn γ β∗ cτ0F 8πe , (31.19) scales with B, the total beam power Pb, and the beam brightness – the third factor above is nothing but the muon beam-beam tune shift Eq. (31.16). There is an obvious incentive to have all the particles in just one bunch per beam. The beta-function at the two IPs β∗≈σz scales as 1/γ within certain range of energies, giving overall scaling ⟨L⟩∝γ2 with other limiting parameters ﬁxed. The main challenges to luminosity achievement with decaying particles are related to production and fast cooling and acceleration of O(1012) muons per bunch without emittance degradation. A multi-TeV c.m.e. high luminosity O(1034 cm−2s−1) muon collider would consist of : (i) a high power proton driver (e.g., 8 GeV 2-4 MW H−SRF linac), (ii) pre-target accumulation 31st May, 2024 17 31. Accelerator Physics of Colliders and compressor rings, in which high intensity 1-3 ns long proton bunches are formed, (iii) a liquid-mercury or other high-power target for converting the proton beam into a tertiary muon beam with energy of about 200 MeV, (iv) a multi-stage ionization cooling section that reduces transverse and longitudinal emittances by several orders of magnitude and creates a low emittance beam, similarly to that recently demonstrated , (v) a multistage acceleration system, possibly employing either rapid cycling synchrotrons or RLAs to accelerate muons in a modest number of turns up to the ﬁnal energy using superconducting RF technology, and, ﬁnally, (vi) a 3–14 km diameter collider ring, where counter-propagating muon beams are stored and collide over the roughly 3000 turns corresponding to the muon lifetime. The intense neutrino ﬂux originating from the multi-TeV muon beams decaying in the collider poses another challenge — the need to minimize the environmental impact. The collider complex is usually located underground and when the produced neutrinos emerge at the surface, a small fraction interacts with the rock (and other material) and produces ionizing radiation dose scaling as γ3. The impact of this neutrino-induced radiation can be mitigated, for example, by continually adjusting the orbits of the beams to spread them out on a wider area, by deeper collider tunnels or with a further reduced emittance of the muon beam so that the required luminosity could be obtained using a substantially smaller number of muons. 31.5.4 Plasma Acceleration and Other Advanced Concepts Since about the mid-1950s, it has been understood that collective plasma-based accelerators promise extremely large accelerating gradients, approximately three orders of magnitude greater than ∼100 MV/m obtained in conventional breakdown limited RF structures . Ionized plasmas can sustain electron plasma density waves with electric ﬁelds in excess of E0 = cmeωp/e or E0 ≈96 [V/m] q n0[cm−3], (31.20) where n0 denotes the ambient electron number density and ωp = p e2n0/(meε0) is the electron plasma frequency . Such gradients can be eﬀectively excited by either powerful external pulses of laser light or by electron bunches if they are shorter than the plasma wavelength λp = c/ωp ≈1 mm× p 1015 cm−3/n0, or by longer beams of protons if their charge density is modulated with the period of λp. In the past decade, we have seen impressive progress in the plasma wakeﬁeld acceleration of high-quality beams. Laser-driven electron energy gain of about 8 GeV over 20 cm of plasma with density 3×1017cm−3 has been demonstrated at the BELLA facility at the Lawrence Berkeley National Laboratory (LBNL) . Short electron bunches were used to boost the energy of externally injected electron bunches by 9 GeV over 1.3 m of ∼1017cm−3 plasma at the FACET facility in SLAC . The AWAKE experiment at CERN used self-modulating long 450 GeV proton bunches to accelerate electrons to 2 GeV over 10 m of 1015 cm−3 plasma . Whether plasma acceleration will ﬁnd application in an HEP facility is not yet clear, given the necessity of staging and phase-locking acceleration in multiple plasma chambers . Another critical issue is the power eﬃciency η for a collider based on plasma acceleration, whose luminosity would still be described by Eq. (31.18). Positrons can be accelerated as well but, contrary to electrons, plasma-wakeﬁelds will naturally defocus positively charged particles. Such fundamental deﬁciency can be addressed, e.g., in more complicated plasma acceleration schemes for e+ ; or by acceleration of electrons in both linacs with conversion into a γγ collider at the IP. In addition, many novel approaches have been proposed to extend the energy or physics reach of future particle colliders, reduce their cost, and improve their luminosity or energy eﬃciency. Those include: i) dielectric wakeﬁeld accelerators in which resonant dielectric accelerating structures are fed by ultra-short RF pulses of wakeﬁelds driven by either collinear or preceding high charge 31st May, 2024 18 31. 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94	https://www.youtube.com/watch?v=uqCgXpd9JGo	Clayden l Chapter 21 (Part-1) Electrophilic Aromatic Substitution Reactions l Aromatic substitution Rpb Resonance Chemistry 3870 subscribers 6 likes Description 419 views Posted: 21 Oct 2020 Telegram group link : Chapter 19 playlist link : 1 comments Transcript: [Music] hi everyone welcome to rpp resonance chemistry now let's continue our organic chemistry tried in series of videos in this session i'm going to discuss about um chapter number 21 that is electrophilic aromatic substitution reaction simply aromatic electrophilic substitution reaction now onwards i'll convert the thoughts of author to like csr net gate and iatj the iit john aspirants okay now so the topic is electrophilic aromatic substitution reaction electrophilic aromatic substitution reaction okay simply it is the electrophilic aromatic substitution reactions okay so now generally aromatic compounds is a kind of a alkenes that is easy kind of unsaturated molecules so generally unsaturated compounds a simple alkenes are all kinds it undergoes the addition reactions but uh aromatic compounds cannot undergoes the addition reactions the first thing okay so general unsaturated compounds it undergoes the addition reactions but it's same the case even though aromatic compounds are unsaturated but it cannot use the addition reactions the first like this is the first question for us okay so now the second one some of the aromatic compounds if some of the aromatic substituted compounds some of the compounds will directly like uh outdoor and para position okay like a first and fourth second and fourth position but some of the groups it uh directs the third position that is nothing but a meta position okay now based on the substituent in nature it is deviated from like either ortho para direction or meta direction okay so it is a different different cases are there in this in this chapter they are well explained how it is happen okay so uh which type of substituents that will use the ardu para products under which type of substituent will give the meta product they are clearly mentioned so some of the exceptions also some of the exceptions also so don't waste our time now let's move on to the topic that is a aromatic electrophilic substitution reaction this is the like initial video of the chapter 21. so now generally so before going to the topic we are aware about a simple thing that is keto in all total medicine okay so keto enough total medicine this is the compound that is nothing but a three pentanone okay so generally it it easily undergoes the ketone ketogenization simply ketone formation so here the keto analyzation in presence of a deuterium so it produces the d plus okay now it undergoes the keto in all tautomerization body okay so this is the od this is the od so whenever this bonded to two electrons that is a lone pair of electrons will shift it towards the like a carbon oxygen single bond then it becomes a negative charge that negative charge will obstruct the electrophile will captures the electrophile and it forms the like a deuterium so here it is one hydrogen and one deuterium so this is the simplest compound btm plus okay so whenever the loss of a d plus a deteriorated plus so now we will get the like uh alpha electrophilic alpha electrophilic substitution reaction simple one okay so whenever we will repeat the these kind of steps in four times now we will get the like up four substitutions so this is the product okay so now here this is the keto form here it is the anal form the ketone and the enol form keto and the unknown forms so they both are equilibrium to each other so general aliphatic compounds ketone content is more some of the exceptions are there okay so generally aliphatic compounds keto content is more but some of the aromatic compounds low down some of the aromatic compounds so it exists only in all form that is the the aromatic compound which exists only stabilized enol form that is nothing but a phenol that is nothing but a phenol so the same thing so it is the general aliphatic one which is treated with deuterium it is substituted the deteriorated plus that that is a d plus ion at the alpha position so that means the ketogenic atomization position okay now what will happen the phenol treated with a d3o plus that is a deuterated water heavy water so what will happen okay so we don't know whether it produces the like a deuterated phenol or like a 2 4 6 a day tri-substitute a tri-butylated phenol or all the proteins are replaced we don't know okay so but so arthur was considered now with the help of arthur was gives the complete information with the help of nmr spectroscopy okay now let's move on to the those nmr spectroscopy details okay now the phenol treated with deuterium what will happen we don't know okay so but uh so that that spectrum that penile this this is the simplest funeral for phenol so the spectrum of phenol at a 5.2 so again here it is a 6.228 it appears two types of peaks okay this is the two hundred megahertz spectrum that's why we cannot able to uh completely those uh triplet under multiplied those splitting clearly okay this is the 200 megahertz spectrum okay so whenever this is treated with deuterium this is treated with deuterium so now we will get the like a product like this od d d b okay so now we will get the product like this so that you that was confirmed by the our that was confirmed by the our nmr spectroscopy that was confirmed by the spectroscopy here all the peaks already disappeared here the aromatic peaks are disappear so these aromatic peaks related to the like the second proton fourth proton sixth proton okay so when you go through the anomal study you will easily confirm that so second fourth sixth protons are disappeared when it is treated with deuterium so that indicates the second fourth sixth position was substituted by the deuterium ion was substituted by the deuterium so this is the nmr study confirms that it undergoes the deterioration it it will produces a second four six okay two four six deuterated uh funeral try deteriorated free note okay so now this was how it is happen okay whether it undergoes the which type of mechanism whether it follows the which type of mechanism now let's move on to the mechanism i think it is clear whenever phenol treated with deuterium it produces a two 246 a deteriorated phenome deteriorated final okay now here so this is the phenol so in the chapter they are clearly mentioned that how to draw the beijing rings okay so this is also been using a representation okay this is the best better representation in case of naphthalenes so this is the naphthalene beta representation instead of instead of this so this is the best representation now here benzene it is better but whenever it is having the substituent will go through the like a complete mechanism so this is the very best representation of aromatic things okay now let's move on to the phenol which is treated with deuterium will happen heavy water deteriorated water deuterated what so simply here it is the phenol phenol okay so simply deteriorated phenol deteriorated phenol which is treated with simply which undergoes the like a ketone catalyzation which i'm proposing like a ketone automorization electrons will shift it towards the uh like a position carbon atom or the position carbon atom now we will get so here it is the negative charge that negative charge will obstruct the deuterium d plus simply it is the deuterium so now here deuterium so one hydrogen is hydrated now here o d o d okay so this is the like a keto form not the inorder form now here it is the you know form enol form is stabilized by the its aromatic nature stabilized by the its aromatic nature so here it is the keto form it is the keto form so that keto form was destabilized due to the lossing of the aromaticity okay so this is the keto form but it is unstable why because it lost the aromaticity nature it lost the aromatic nature that's why it is unstable okay so that's why it undergoes the like reversal of reaction now whether it is it here two types of reactions are there whether it undergoes the like a simple d elimination that is a d plus elimination it will produces now here so simple d plus elimination it produces deuterium compound okay this is the common general one or instead of deuterium so here it undergoes the like keto in all cautionarization ketone for harmonization now it produces od substituent at a substituent at upper position substituent at a order position this is the simplest aromatic substitution reactions in phenol compound so this is the mono substitution like we follow this similar mechanism we will get the like a dye substitution also by the substitution so this is the simplest mechanism okay now the third one how the aromat like how the electrophile that is nothing but a deep plus is substituted at a para position we don't know now let's move on to the that mechanism also so that is the final one of a phenol compound okay now here this is the simplest phenol od okay so now here deuterium deuterium okay so now the lone pairs of oxygen will shift it towards the towards the ring okay so then now i'll write the these double bonds like this for the sake of our simplicity okay so now here it creates the excess of counter then bonded to two electrons will shift it towards the another atom so then these bonded two electrons will shift at the para position that is the fourth position then we'll obstruct the d plus then we'll abstract the d plus now here it produces the double bond shifted over here okay now here it creates the deuterium deuterium and h double bond over here so this is the od along with the positive charge here deuterium deuterium here is nothing but a hydrogens here it is the hydrogens here it is the hydrogen then it undergoes the again para ketone catamarasm this is simply it is nothing but a parabolism okay so parata medicine trautomalism now here it undergoes the like parathyroidism here the loss of h plus loss of h plus so it regains the aromaticity it regains its aromaticity deuterium od deuterium d so here these two are protons only these two are protons only this is the simplest substitution reactions whenever phenol treated with the deuterium okay this may be like asked in like this may be possible to ask this is an examination also cser examination this is the initiation of chapter 21 okay so from this thoughts they are extrapolating this object of electrophilic aromatic substitution reaction with binging rings so that is our first doubt what is that so unsaturated compounds will undergoes the addition reactions but aromatic compounds cannot gives the addition so instead of addition reaction it will use this substitution so how it is possible now let's move on to the second one our first doubt okay so i think it is clear so the phenol treated with deuterium it produces the two four six two four six deteriorated phenol so three and fifth position third and fifth position will not deteriorated so due to the its electron releasing capacity okay now laptop gases will explain the why it is not deteriorated okay now let's move on to the aromatic electrophilic substitution reactions in aromatic compounds i think it is clear ok so now so in generally ok so in generally so this is the only thing simple only thing will undergoes the like bromination so it gives the anti-addition we are discussed we are very familiar with this type of anti-addition in chapter 19 okay so i'll upload the chapter 19 and i'll give you uh link in description marks also so which is treated with the acids so here also it it produces the epoxy so this is also our familiar chemistry this is also not familiar epoxy chemistry so here olivines will undergo like a bromination as well as epoxidation but at the same time the aromatic compound simple benzene which treated with the bromine it cannot use the like additional reaction it cannot gives the substitution reaction so similarly here also that benzene which treated with a rcoh so here also it does not gives the it does not use the addition reactions it does not use the addition here it is the anti-addition here it is the scene addition so both the cases we cannot get the we can we can substitute the aromatic the aromatic reactant it cannot gives the product it cannot gives the product that indicates aromatic compounds cannot give the addition reactions and obviously oxygen reactions you can write the that particular statement aromatic compounds cannot give the addition reactions okay so now so if addition is not possible then we have looked out substitution reactions how it is possible now let's move on to the substitution reaction of aromatic compounds so this is the aromatic compound okay simply this is the aromatic compound so the aromatic compound will treat it with bromine so with the help of some of the lewis acids like a alcl3 or if if bromine is taken if you br three if chlorine is taken if yes f is like a cl3 those kind of catalysts are used catalysts are used so whenever the aromatic compounds treated with the bromine here one of the hydrogen atom one of the hydrogen atom was substituted by the incoming electrophile it was substituted by the incoming electrophile that indicates aromatic compounds cannot undergoes the addition reactions but it always undergoes the electrophilic substitution reactions electrophilic substitution reactions it undergoes esr reactions okay now how it is possible let's move on to the mechanism of the electrophilic substitution reaction of the aromatic rings okay so simply here it is the benzene ring so the benzene ring will treat it with the electrophile okay so it it attracts the electrophile so if there is any double bond is that the bonded or two electrons will shift it towards the like a top side of the carbon atom that indicates here top one becomes a negative bottom one becomes positive okay so then negative will abstract the electrophile so the other other end of the carbon atom other end of the olefin will get the positive charge only okay these two types simply it is represented as this kind of arrow these two setups simply it is represented as this kind of arrow now here the this arrow indicates so the bonded two electrons will shift it towards the top of the carbon atom so then that top of the carbon atom will abstract to the electrophile now it creates the electrophile okay another end is positive charge another end is positive charge here it is the intermediate okay so but this intermediate is very unstable okay so whenever compared to the reactant so the reactant is aromatic compound but our intermediate is very unstable one due to the its lossing of the aromaticity okay so this is the cationic intermediate cationic intermediate but it is very unstable okay so the unstability was uh stabilized by the delocalization now i'll explain the delocalization of those cations also okay so then finally these hydrogen atom will shift it will share the pair of electrons with a carbocation and regains and it regains the aromaticity and its regains their aromaticity so now it it substituted the electrophile simply here the loss of h which gives the its aromatic nature return back which gives the its aromatic region in return back so this is the simplest electrophilic substitution reactions in the aromatic compounds okay now so this is the very unstable intermediate but it was stabilized by the cations now let's discuss those stabilization of cation okay positive charge here it is the electrophile okay so here it is the simple electrophile now here double bonds okay so these double bonds will undergoes the delocalization now the bonded two electrons will shift towards this carbon atom then it becomes a double bond so here loss of electrons it is the positive charge this is the positive charge so again these double bond also similarly are delocalized the same way now here the positive charge will shifted over here okay so simply if this is the delocalization of aromatic electrophilic substitution reactions simply these kind of intermediates is called sigma complex this is the one of the type of intermediate in electrophilic aromatic substitution reactions okay i think it is clear so aromatic compounds treated with electrophile it produces the like a electrophile substituted so that means a electrophile which is a substituted the hydrogen one of the its aromatic hydrogens and it produces the electrophilic electrophilic aromatic compounds electrophilic aromatic compounds okay now here this is the cationic intermediate that cationic intermediate was stabilized by the its delocalising stabilized breathing it's a delocalization so in generally this is the concept of first chapter initiation okay so but whenever you're selected for the like a bulk or yngc those kind of like a highly competitive competitive field so they are asked in the interviews in this chapter particularly they may be asked in the in this chapter ask the questions in this chapter so that is the first thing so here aromatic electrophilic aromatic substitution reactions will produces the cationic intermediate we know that so that's that's all we know that okay so but how it is confirmed okay aromatic compounds will undergoes the aromatic electrophilic substitution reactions via cationic intermediate so how do you confirm that cationic intermediate okay so then our faces are written okay so how do you confirms the like our cationic intermediate so this was confirmed by the spectroscopic studies simply now i'll explain those confirmation of those confirmation of cationic intermediate okay now let's discuss those cationic intermediate so whenever this is the aromatic ring so which is treated with a like a non-basic under non-nucleophilic reagent so that is nothing but a sbf6 minus so this is a non-basic and a non-nucleophilic reagent non nucleophilic reagent so which is in the form of sb fo5 plus fs43h so this is the combination these two combinations will produce as the sbf6 okay so that will produces the like a cationic intermediate so that it produces the cationic intermediate so that cation was delocalized in the uh like all the five carbon atoms that was the d localized all the five carbon atoms so in this case here it is the electrophile so the positive charge was delocalized around the five carbon atoms we know that okay so now here it is the cation so that cation again here f minus will shift it towards the usb f5 so then it gets the after hydraulic environment after cutting environment this is the sbf fighter structure whenever f minus will attach to the usb file that negative charge distributed over all the compo that negative charge distributed over all the counter so this is the formation of cationic intermediate this is the formation of cationic intermediate so that was studied by the proton and the nmr studies as well as x-ray spectroscopic study it will it will it it will use the information it forms the cationic intermediate okay so now here so this is the cationic that's why this is the hydrogen okay now here positive charge positive charge okay the positive charge was delocalized around the five carbon atoms one two three four five six this is the numbering okay so now so according to data of proton and the c-13 nmr data three phi on fourth position okay so proton data so the first position the proton data is simply it is a 5.6 now second and sixth position that is 9.7 third and fifth position 8.6 and fourth position that is nothing but uh like a 9.3 that is nothing but a 9.3 again similarly c 13 nmr data that is 52.2 now second and the sixth position carbon atoms are that is 186 okay so third and fifth position that is approximately 170 okay now here 178 it is approximately 160 around 160 ppm around 160 ppm so that indicates second the sixth and the fourth position was uh delocalized with the positive charge but the third and fifth cannot uh cannot replaces the positive charge okay so third and third and fifth position so the positive charge cannot appear so that's why it gives the lesser less chemical shift values less chemical shifty values it was uh also confirmed by the like uh x-ray studies here positive charge is 0.26 your positive charge is 0.09 the fourth position positive charge is 0.3 so this is the like a positive charges of delocalized brain okay so in this way we are confirmed cationic intermediate is found okay so this is the very simple introduction video of chapter 21 chapter 21. now the next video i have to covered so the nitration sulfonation freedom craft alkylation friedel craft acceleration again the total summary of the reaction total summary of the electrophilic aromatic substitution reaction simply that aromatic compound is binging so whenever benzene will undergoes the nitration what will happen sulfonation freedom craft alkylation friedel craft silish so these four topics was covered in the next video okay i think it is very helpful for you guys if you like this video you can share to your friends okay that is very helpful for them okay thank you so much for watching
95	https://www.youtube.com/watch?v=3D_xCkzegew	Problem 5.9 \| Introduction to Electrodynamics (Griffiths) Hayashi Manabu 9270 subscribers 134 likes Description 11126 views Posted: 18 Jan 2020 The trick is to utilize past results by breaking up the wire into smaller, manageable pieces. 11 comments Transcript: Introduction certified their magnetic field at Point P we're going to do two things so this is the magnetic field at the center of a so of a circle and there's curds going around the side of the circle so this is going to be the Manitou field at the center and for this this is a line of water and you have current going along the wire and the magnetic field at this point will be equal to this so I'm not going to prove these results you can find them many examples in the book so using using these two results we can solve for the magnetic field at Point P for these two cases so let's start with Part A so for Part A we have first of Part a all this is a quarter of a circle so first we have a quarter of a circle and then we have another quarter of a circle but this time this length here is B so essentially we can add up these two contributions well the contributions from these sides they contribute nothing to point P because using the right hand rule these to the vector is if you just point your fingers back to this direction you'll find that these two these two components over here don't contribute anything to that magnetic field so I think my another field is only coming from these two sides so for this side the amount of magnetic field that's coming from this side so first of all we can treat it like a circle so if it's a full circle then by this result this is the magnetic field right but now we have a quarter of a circle so we multiply by 1/4 and then the contribution from this side it's going to go in the opposite direction of wherever this is going so if you use your the right hand rule you see that the contribution from this side actually points outwards and from this side it's going to point inwards so it's going to be a minus and then once again we use the same formula which we don't like a circle and then when we divide by 4 so pulling out the like terms you get this result yeah this is going to be a vector it's going to point outside so for Part B we Part b have a semi circle this is a center and then we have two lines that extend to infinity so there's a current because you guys are round then leaves this way so first of all we can add the contribution from this half circle the radius is going to be R so just like before you have two R so this half a circle we multiply that won't have and here we have a straight current and this is only when we have a straight current as opposed to what we have here this is going to extend to infinity in both directions well in this case is own it only extends to infinity in one direction but then we have to actually we have to so both of these add up together to form a complete line of current that extends from infinity to infinity so we can actually add up these two contributions and treat them as one complete line line of current so we can just add this result to the electric to that magnetic field so all I have to do is to add this to cut off in attention I would have it this is our answer so we can pull out the like terms so that's gonna be a 1 plus 2 over PI so this is the answer and then if you use a right hand if you your right hand rule you see that the magnetic field points inside of this surface so that inside inside of the screen
96	https://mnemonicdictionary.com/word/celerity	celerity meaning - definition of celerity by Mnemonic Dictionary MnemonicDictionary Home (current) Wordlists Contact Us Popular Wordlists GRE Word List Word of the Day RECENT SEARCHES scheol loddon pondweed execration celerity all-knowing execrate besmear karl czerny inhibit unsaid Top Searched Words xxix repudiate obsequious abate abjure contrite desiccate obdurate cogent recondite Discover more Dictionary mobile app Language learning tool tea leaves Online movie streaming services Vocabulary building course Mobile app Vocabulary building apps Oxford Picture Dictionary plumb bob Invidious word of the day celerity celerity - Dictionary definition and meaning for word celerity Definition (noun) a rate that is rapid Synonyms : quickness , rapidity , rapidness , speediness Mnemonics (Memory Aids) for celerity Previous Powered by Mnemonic Dictionary celer(ity) ~ cereal>>> if you eat cereals, you can be agile and quick 2 6 celerity -> celebrity get famous at rapid rate and speed. 1 0 CELEBRITY dances with CELERITY 0 1 Relate to: Accelerate = Ac (celerity) = which means speed up something or doing something rapidly. 89 1 celerity sounds like celebrity, and a celebrity like salman khan who always drive his car in maximum speed and do the accidents. 40 14 Celerity...Celebrity Celebrities, such as Michael Jordan and Jerry Rice, usually like driving fast and expensive sports cars such as the Porsche. 7 7 Powered by Mnemonic Dictionary celer(ity) ~ cereal>>> if you eat cereals, you can be agile and quick 2 6 celerity -> celebrity get famous at rapid rate and speed. 1 0 CELEBRITY dances with CELERITY 0 1 Relate to: Accelerate = Ac (celerity) = which means speed up something or doing something rapidly. 89 1 celerity sounds like celebrity, and a celebrity like salman khan who always drive his car in maximum speed and do the accidents. 40 14 Celerity...Celebrity Celebrities, such as Michael Jordan and Jerry Rice, usually like driving fast and expensive sports cars such as the Porsche. 7 7 Powered by Mnemonic Dictionary celer(ity) ~ cereal>>> if you eat cereals, you can be agile and quick 2 6 celerity -> celebrity get famous at rapid rate and speed. 1 0 CELEBRITY dances with CELERITY 0 1 Next 1 2 Discover more purchase language learning software buy vocabulary builder app Learning aids Study guides plumb bob find English learning resources Thesaurus books hire content writer service Memory techniques books order English Vocabulary in Use Download our Mobile App Today App StorePlay Store Connect with us on Facebook and Instagram to get a daily "Word of the Day" directly in your social media feed. Widget by EmbedSocial→ FacebookInstagram Discover more Writing ESL dictionary Educational games Educational board games tea leaves Vocabulary building apps Daily word email Audiobook subscriptions Dictionary API access 504 Essential Words We are Hiring! × We are looking for Content Writers (1-2 years experience) for our ed-tech startup based out of Gurgaon. If interested, please reach out to us at career@opencubicles.com Close Books We Recommend! × 5 Great English Vocabulary Books we Recommend! Oxford Picture Dictionary Word Power Made Easy McGraw-Hill Education Essential ESL Dictionary 504 Absolutely Essential Words 6th Edition English Vocabulary in Use Advanced with CD-ROM: Vocabulary Reference and Practice 2nd Edition Close Receive our word of the day on Whatsapp Download Mobile App Download Mobile App × Introducing "Memli" — Our New Vocabulary Building App! Embark on a journey to expand your vocabulary with "Memli". Whether you're a language learner or simply love words, "Memli" is tailored to enhance your lexicon effectively and engagingly. Download for iOSDownload for Android Join our community! Follow us on social media and receive our 'Word of the Day' directly in your feed. Engage, learn, and share your insights with a community of word lovers. WhatsAppInstagramFacebook Close
97	https://www.algebralab.org/lessons/lesson.aspx?file=algebra_onevariablewritingequations.xml	Translating Word Problems into Equations AlgebraLAB: Lessons Discover more Jelly bean Gumdrop candies gumdrops Wallet Coin jelly Calculator Common bean beans Site Navigation Discover more jelly beans Zoom candy Gelatine dessert beans Graphing Calculator gumdrops Candy Raspberry Gumdrop Translating Word Problems into Equations Most of the time when someone says “word problems” there is automatic panic. But word problems do not have to be the worst part of a math class. By setting up a system) AlgebraLAB Project Manager Catharine H. Colwell Application Programmers Jeremy R. Blawn Mark Acton Copyright © 2003-2025 All rights reserved.
98	https://heal.heuristiclab.com/research/combinatorial-and-simulation-based-optimization	Combinatorial & Simulation-Based Optimization Combinatorial problems arise in many domains and involve finding a suitable solution out of a huge, but finite solution space. Many well-known problems can be categorized into _selection_, _assignment_, _ordering_, and _grouping_ problems as depicted below. In general, combinatorial optimization involves a number of variables with a finite domain and a solution consists of a concrete element picked for each variable from the domain. Practical applications of combinatorial optimization include production planning, tour planning, truck door assignment, graph coloring, scheduling, and more. Typically, in an applied research project we start by identifying the type of problem that needs to be solved and derive a (mathematical) program to describe the objectives and constraints. Then we evaluate existing solution techniques to find optimal or near-optimal solutions, or, if existing techniques do not suffice, develop new techniques that are specific to the given problem. Assignment problems Solving assignment problems requires to find the best mapping between two finite sets of items. The solution space is mainly described by the cardinalities that are possible in the assignment. The cardinalities on both sides can indicate optionality, identity, and multiplicity. Multiple items of one set may be assigned to one or even multiple items of the other set. Soemtimes even, not all items of one or both sets need to be mapped. Assignment is a very general problem description, so that many of the other problem types could be represented as an assignment problem. For instance, selection could be represented as an assignment between a second set with two items: selected and discarded. _Well-known problems:_ linear assignment problem (LAP), quadratic assignment problem (QAP), generalized assignment problem (GAP), generalized quadratic assignment problem (GQAP) Ordering problems Solving an ordering problem requires to find the best order of a number of given finite set of items, e.g. N customers. For instance, the customers that have to be visited in a tour, or the jobs that have to be produced on a certain machine. The search space of such problems can be described in general by a permutation of the items and thus if of size N!. Solution spaces may be slightly smaller as sometimes there are symmetries, for instance when all distances between customers are equally large in both directions or when there are jobs that have the same properties. In addition, it can be the case that the order forms a cycle and that only the relative positioning of items is important and not the absolute position in the sequence. _Well-known problems:_ traveling salesman problem (TSP), single-machine scheduling problem Selection problems Solving selection problems requires to find a subset of a set of items. Thereby, variations of the problem occur whether an item may be selected multiple times or just once. The selected set of items constitutes the solution and may for instance denote a set of projects that should be invested in. _Well-known problems:_ knapsack problem (KSP) Grouping problems Solving grouping problems generally requires to find a subset of the powerset of all items. For instance, consider the task of coloring a graph where each node should be colored such that all adjacent nodes have different color. This is equivalent to identifying a partition of the nodes. Restrictions such as the minimal or maximal number of groups, or minimum and maximal number of items per group are quite common in practice. _Well-known problems:_ graph coloring problem (GCP) Combined problems Often real-world and pratical problems may also be described as a combination of e.g. ordering and grouping. Consider for instance, the delivery of goods from a central warehouse to several retailing locations. The locations can be grouped into tours and within each group a tour, i.e., a sequence of locations, needs to be identified. Identifiying high quality solutions to such problems efficiently presents many challenges. _Well-known problems:_ vehicle routing problem (VRP), location routing problem (LRP), traveling thief problem (TTP) Mathematical Programming Many optimization problems can be formulated as mathematical programs, e.g., mixed integer programming (MIP) and solved with commercial off the shelf software in reasonable time even for larger dimensions thanks to advances in solver technology and a strong increase in the availability of computational power. We investigate the use of mathematical programming, devise a corresponding mathematical description of the problem and evaluate the use of exsiting solvers. Simulation and optimization Often the problem may not be described in closed mathematical form. There may be complex interrelationships and dynamics that need to be considered. In this case, the problem is described in form of a simulation model and simulation-based optimization is applied to identify good solutions. Solving such problems is computationally very expensive. We use parallel execution of many simulation models on distributed computers as well as population-based evolutionary algorithms to identify good solutions quickly. To improve convergence we investigate in the use of surrogate models, i.e., data-based models that approximate the simulation response, as well as custom variation operators to guide the search towards better solutions faster. We distinguish between two types of simulation-based optimization as shown in the picture below. In case I), the optimization algorithm is the main driver. It computes new solutions and the simulation model is evaluating its quality, because the problem cannot be described in closed mathematical form. In case II), the simulation is the main driver. It calculates and updates a world state, but some decision need to be implemented and this needs to be provided by an optimization algorithm. Generally, such an algorithm is called policy in the second case. While in the first case the interest is to identify a good solution for a certain problem, the second case is more concerned about the dynamic effects that arise when implementing solutions from a certain policy. Relevant projects HOPL Within the K-project "Heuristic Optimization in Production and Logistics" we investigated solution of combined problems. Together with the company partners voestalpine, Rosenbauer, carvatech, we described complex problems such as the stacking and batching within the hot storage area, as well as supply chain and production planning optimization problems. We extended HeuristicLab to allow describing problems by combining multiple solution representations (various vectors (binary, integer, real-valued, permutations, etc.) into one solution representation whereby for each representation variation operators would be applied independently. In addition, we devised a framework whereby solvers of sub-problems could interact with each other through an orchestration procedure that guides the search. We could among others successfully demonstrate this approach in solving the location routing problem, as well as the traveling thief problem. LOISI Within the project "Logistics Optimization in Steel Industries" we investigated the problems of scheduling WIP among the facilities, from casting to delivery. A mathematical model was devised based on the resource constraint project scheduling problem (RCPSP) which was formulated and compared for mixed integer programming solvers as well as for constraint programming solvers. The results showed that constraint programming was highly suitable to solve this problem. Additionally, we investigated the use of stacking problems in a further hotstorage environment. SimGenOpt2 Within this FFG production of the future project we investigated the use of heuristic optimization techniques for optimizing production planning parameters which are used for instance in material requirements planning (MRP). An NSGA-II algorithm was adapted for a discrete encoding with custom variation operators that included domain specific knowledge. The algorithm identified a good approximation of the Pareto front ranging from solutions which favored low inventory costs to those that favored low tardiness costs. In addition, a mathematical model of the production planning task was devised for a battery production industrial case which calculated the master production schedule. The effect of the production planning parameters on the KPIs (inventory and tardiness costs) was simulated using an Anylogic-based simulation model. Optimal Workforce Within this production funded by the regional government of Upper Austria we investigated methods for the optimization of qualification matrices. A simulation model was created to describe two industrial production planning cases arising out of the electronics industry. An NSGA-II algorithm with custom solution encoding was applied to approximate the Pareto front consisting of solutions with high service levels to those with low number of total qualifications. The solutions have been analyzed together with the industry experts and recommendations were given on the qualification strategy for an improvement to service levels. JRZ adaptOp The Josef Ressel center for adaptive optimization in dynamic environments considers many problems that arise when the world changes dynamically. Solutions need to be constantly updated to the changing situations and feedback between environment and solver has to be considered. While the problems solved within JRZ adaptOp are often combinatorial, the topic of dynamic optimization deserves its own section on our site. Selected publications Karder, J. A., Beham, A., Hauder, V., Altendorfer, K., & Affenzeller, M. (2021). Simulation-based optimisation for worker cross-training. _International Journal of Simulation and Process Modelling_ , 16(3), 185-194. doi:10.1504/IJSPM.2021.117309. Sybilska, A., Słonina, M., Lejba, P., Lech, G., Karder, J. A., Suchodolski, T., Wagner, S., Sybilski, P., & Mancas, A. (2021). WebPlan - web-based sensor scheduling tool. Paper presented at _8th European Conference on Space Debris_. Hauder, V., Beham, A., Leitner, S. J., Parragh, S., & Affenzeller, M. (2020). Resource-constrained multi-project scheduling with activity and time flexibility. _Computers and Industrial Engineering_, 150, . doi:10.1016/j.cie.2020.106857. Altendorfer, K., Schober, A., Karder, J., & Beham, A. (2020). Service level improvement due to worker cross training with stochastic worker absence. _International Journal of Production Research_, 59(14), 4416-4433. doi:10.1080/00207543.2020.1764126. Beham, A. (2019). Fitness Landscape Analysis and Algorithm Selection for Assignment Problems. PhD Thesis, Johannes Kepler University Linz. Hauder, V., Karder, J., Beham, A., Wagner, S., & Affenzeller, M. (2018). A general solution approach for the location routing problem. In R. Moreno-Diaz, A. Quesada-Arencibia, & F. Pichler (Eds.), _Computer Aided Systems Theory – EUROCAST 2017 - 16th International Conference, Revised Selected Papers (pp. 257-265). (Lecture Notes in Computer Science Vol. 10671 LNCS)_. Springer. doi:10.1007/978-3-319-74718-7_31. Beham, A., Wagner, S., & Affenzeller, M. (2018). Algorithm selection on generalized quadratic assignment problem landscapes. In _GECCO 2018 - Proceedings of the 2018 Genetic and Evolutionary Computation Conference_ (pp. 253-260). Association for Computing Machinery, Inc. doi:10.1145/3205455.3205585. Raggl, S. J., Beham, A., Wagner, S., & Affenzeller, M. (2018). Analysing a hybrid model-based evolutionary algorithm for a hard grouping problem. In R. Moreno-Diaz, A. Quesada-Arencibia, & F. Pichler (Eds.), _Computer Aided Systems Theory – EUROCAST 2017 - 16th International Conference, Revised Selected Papers (pp. 347-354). (Lecture Notes in Computer Science Vol. 10671 LNCS)_. Springer. doi:10.1007/978-3-319-74718-7_42. Raggl, S. J., Beham, A., Hauder, V., Wagner, S., & Affenzeller, M. (2018). Discrete real-world problems in a black-box optimization benchmark. In _GECCO 2018 Companion - Proceedings of the 2018 Genetic and Evolutionary Computation Conference Companion_ (pp. 1745-1752). Association for Computing Machinery, Inc. doi:10.1145/3205651.3208280. Tricoire, F., Scagnetti, J., & Beham, A. (2018). New insights on the block relocation problem. _Computers & Operations Research_, 89, 127-139. doi:10.1016/j.cor.2017.08.010. Raggl, S. J., Beham, A., Tricoire, F., & Affenzeller, M. (2018). Solving a Real World Steel Stacking Problem. _International Journal of Service and Computing Oriented Manufacturing_, 3(2). doi:10.1504/IJSCOM.2018.091621. Karder, J., Beham, A., Wagner, S., & Affenzeller, M. (2018). Solving the traveling thief problem using orchestration in optimization networks. In R. Moreno-Diaz, A. Quesada-Arencibia, & F. Pichler (Eds.), _Computer Aided Systems Theory – EUROCAST 2017 - 16th International Conference, Revised Selected Papers (pp. 307-315). (Lecture Notes in Computer Science Vol. 10671 LNCS)_. Springer. doi:10.1007/978-3-319-74718-7_37. Karder, J., Scheibenpflug, A., Beham, A., Wagner, S., & Affenzeller, M. (2017). Extending Sim# for simulation-based optimisation of Semi-Automated machinery. _International Journal of Simulation and Process Modelling_, 12(6), 485-497. doi:10.1504/IJSPM.2017.089634. Beham, A., Affenzeller, M., & Wagner, S. (2017). Instance-based algorithm selection on quadratic assignment problem landscapes. In _GECCO 2017 - Proceedings of the Genetic and Evolutionary Computation Conference Companion_ (pp. 1471-1478). ACM Sigevo. doi:10.1145/3067695.3082513 Hauder, V., Beham, A., Wagner, S., & Affenzeller, M. (2017). Optimization networks for real-world production and logistics problems. In _GECCO 2017 - Proceedings of the Genetic and Evolutionary Computation Conference Companion_ (pp. 1411-1414). Association for Computing Machinery, Inc. doi:10.1145/3067695.3092959. Karder, J., Scheibenpflug, A., Wagner, S., & Affenzeller, M. (2016). Generation of dispatching rules for job sequencing in singlemachine environments. In A. G. Bruzzone, E. Jimenez, L. S. Louca, L. Zhang, & F. Longo (Eds.), _28th European Modeling and Simulation Symposium, EMSS 2016_ (pp. 117-121). DIME UNIVERSITY OF GENOA. Beham, A., Scagnetti, J., Kommenda, M., Wagner, S., Winkler, S., & Affenzeller, M. (2015). Optimization Strategies for Integrated Knapsack and Traveling Salesman Problems. In F. Pichler, R. Moreno-Díaz, & A. Quesada-Arencibia (Eds.), _Computer Aided Systems Theory – EUROCAST 2015 - 15th International Conference, Revised Selected Papers (pp. 359-366). (Lecture Notes in Computer Science Vol. 9520)_. Springer. doi:10.1007/978-3-319-27340-2_45.
99	https://www.cun.es/diccionario-medico/terminos/hipotesis-cientifica	ÁREA DEL PACIENTE ÁREA DEL PACIENTE NAVARRA +34 948 255 400 MADRID +34 91 353 19 20 MADRID PACIENTES INTERNACIONALES BIENVENIDOS PACIENTES INTERNACIONALES ¿NECESITA AYUDA? ¿NECESITA AYUDA? SEGUNDA OPINIÓN A DISTANCIA PEDIR CITA Inicio > hipótesis científica DICCIONARIO MÉDICO Hipótesis científica Diccionario médico La hipótesis científica es un elemento esencial del método científico que desempeña un papel crucial en el avance de la medicina. Se define como una proposición tentativa que busca explicar un fenómeno o conjunto de observaciones y que debe ser comprobable mediante experimentación o análisis. En el ámbito médico, las hipótesis son fundamentales para el diseño de estudios clínicos, el desarrollo de nuevos tratamientos y la comprensión de enfermedades complejas. Qué es una hipótesis científica Una hipótesis científica es una suposición fundamentada que propone una explicación a un problema o fenómeno observable. Para ser considerada científica, debe cumplir con ciertas características: Comprobabilidad: Debe ser susceptible de ser probada o refutada a través de experimentos o estudios. Clara y específica: Debe expresar con precisión el fenómeno que se desea explicar. Basada en evidencia previa: Debe tener un fundamento lógico o empírico derivado de investigaciones anteriores. En la medicina, una hipótesis puede abordar preguntas como el efecto de un fármaco sobre una enfermedad o los factores de riesgo asociados a una condición médica específica. Importancia de las hipótesis en la investigación médica En el contexto médico, las hipótesis son el punto de partida para la generación de conocimiento. Ayudan a estructurar investigaciones, guiar el diseño de experimentos y establecer un marco para interpretar los resultados. Su relevancia se manifiesta en varios aspectos: Desarrollo de tratamientos: Una hipótesis bien formulada puede conducir al descubrimiento de nuevos medicamentos o terapias. Prevención de enfermedades: Permiten identificar factores de riesgo y diseñar estrategias de prevención. Avances diagnósticos: Facilitan la creación de pruebas más precisas y rápidas. Cómo se formula una hipótesis científica La formulación de una hipótesis científica sigue un proceso estructurado que incluye: Identificación del problema: Observación de un fenómeno médico o clínico que requiere explicación. Revisión de la literatura: Análisis de investigaciones previas para entender el contexto y las posibles variables involucradas. Formulación de la hipótesis: Redacción de una proposición clara y comprobable que conecte las variables observadas. Por ejemplo, una hipótesis en medicina podría ser: "El consumo regular de ácidos grasos omega-3 reduce significativamente los niveles de inflamación en pacientes con artritis reumatoide." Tipos de hipótesis científicas En la investigación médica, las hipótesis pueden clasificarse en: Hipótesis nula (H0): Plantea que no existe relación o efecto entre las variables estudiadas. Hipótesis alternativa (H1): Propone que sí existe una relación o efecto entre las variables. Hipótesis descriptiva: Busca caracterizar un fenómeno sin establecer causalidad. Hipótesis causal: Intenta establecer una relación de causa-efecto entre las variables. Aplicaciones de las hipótesis científicas en medicina Las hipótesis son esenciales en numerosos campos médicos, incluyendo: Investigación farmacológica: Probar la eficacia y seguridad de nuevos medicamentos. Epidemiología: Identificar factores de riesgo y patrones de enfermedades. Genética médica: Explorar el impacto de variantes genéticas en trastornos específicos. Cuándo acudir al médico en el contexto de la investigación médica Aunque las hipótesis científicas son elementos de investigación, su aplicación puede tener implicaciones directas en la salud del paciente. Es fundamental buscar atención médica en los siguientes casos: Cuando un tratamiento experimental puede ser una opción para condiciones graves o sin cura. Si se forma parte de un estudio clínico y surgen efectos adversos inesperados. Para discutir opciones terapéuticas basadas en investigaciones recientes. Preguntas frecuentes ¿Cuál es la diferencia entre hipótesis y teoría científica? Una hipótesis es una proposición inicial que se somete a pruebas, mientras que una teoría científica es un marco conceptual bien establecido respaldado por evidencia sólida. ¿Qué papel juega la hipótesis en un ensayo clínico? En un ensayo clínico, la hipótesis establece el efecto esperado de una intervención médica, guiando el diseño del estudio y la interpretación de los resultados. ¿Cómo se valida una hipótesis en medicina? Se valida mediante experimentos controlados, análisis estadísticos y replicación de los resultados en estudios independientes. ¿Es posible refutar una hipótesis científica? Sí, una hipótesis puede ser refutada si los datos obtenidos no apoyan la proposición inicial, lo que también contribuye al avance del conocimiento científico. © Clínica Universidad de Navarra 2023

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