Keywords 020 100 points find a power series

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keywords:02010.0pointsFind a power series representation for thefunctionf(z) =15 + 16z2.1.f(z) =summationdisplayn=0(-1)n42nz2n5n2.f(z) =summationdisplayn=04nzn5n+1
choice (hac762) – HW Quest Week 8 – cepparo – (53850)103.f(z) =summationdisplayn=0(-1)n42nz2n5n+1correct4.f(z) =summationdisplayn=042nz2n5n+15.f(z) =summationdisplayn=0(-1)n4nzn5n+1Explanation:After simplification,f(z) =15 + 16z2=15parenleftbigg11 +165z2parenrightbigg=15parenleftbigg11-(-165z2)parenrightbigg.On the other hand,15summationdisplayn=0parenleftbigg-165z2parenrightbiggn=summationdisplayn=0(-1)n42nz2n5n+1keywords:02110.0pointsFind a power series representation centeredat the origin for the functionf(y) =y3(6-y)2.1.f(y) =summationdisplayn=3n-26n-1yncorrect2.f(y) =summationdisplayn=3n6nyn3.f(y) =summationdisplayn=216n-1yn4.f(y) =summationdisplayn=316n-3yn5.f(y) =summationdisplayn=2n-16nynExplanation:By the known result for geometric series,16-y=16parenleftBig1-y6parenrightBig=16summationdisplayn=0parenleftBigy6parenrightBign=summationdisplayn=016n+1yn.This series converges on (-6,6).On the other hand,1(6-y)2=ddyparenleftBig16-yparenrightBig,and so on (-6,6),1(6-y)2=ddyparenleftBiggsummationdisplayn=0yn6n+1parenrightBigg=summationdisplayn=1n6n+1yn-1=summationdisplayn=0n+ 16n+2yn.Thusf(y) =y3summationdisplayn=0n+ 16n+2yn=summationdisplayn=0n+ 16n+2yn+3.Consequently,f(y) =summationdisplayn=3n-26n-1yn.02210.0pointsFind a power series representation for thefunctionf(x) = tan-1(2x).1.f(x) =summationdisplayn=012n+ 1x2n+12.f(x) =summationdisplayn=022n+12n+ 1x2n+1
choice (hac762) – HW Quest Week 8 – cepparo – (53850)113.f(x) =summationdisplayn=0(-1)n22n+12n+ 1x2n+1correct4.f(x) =summationdisplayn=0(-1)n22n2n+ 1x2n5.f(x) =summationdisplayn=0(-1)n12n+ 1x2n+16.f(x) =summationdisplayn=012n+ 1x2nExplanation:We know thattan-1(x) =integraldisplayx011 +t2dt .On the other hand,11-x= 1 +x+x2+. . .on the interval (-1,1). Replacingxby-x2we thus see that11 +x2= 1-x2+x4-x6+. . .=summationdisplayn=0(-1)nx2non (-1,1). But thenf(x) =integraldisplayx0parenleftBigsummationdisplayn=0(-1)nt2nparenrightBigdt=summationdisplayn=0parenleftBig(-1)nintegraldisplayx0t2ndtparenrightBig=summationdisplayn=0(-1)n2n+ 1x2n+1on (-1,1). Consequently,tan-1(2x) =summationdisplayn=0(-1)n22n+12n+ 1x2n+1is a power series representation for tan-1(2x)onparenleftbigg-12,12parenrightbigg.02310.0pointsEvaluate the integralf(t) =integraldisplayt0s1-s4ds .as a power series.1.f(t) =summationdisplayn=0t4n+24n+ 2correct2.f(t) =summationdisplayn=4t4n4n+ 23.f(t) =summationdisplayn=0t4n4n4.f(t) =summationdisplayn=0(-1)nt4n+24n+ 25.f(t) =summationdisplayn=0(-1)nt4n4nExplanation:By the geometric series representation,11-s=summationdisplayn=0sn,and sos1-s4=summationdisplayn=0s4n+1.But thenf(t) =integraldisplayt0parenleftBigsummationdisplayn=0s4n+1parenrightBigds=summationdisplayn=0parenleftBigintegraldisplayt0s4n+1dsparenrightBig.Consequently,f(t) =summationdisplayn=0t4n+24n+ 2.

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