# 8.7 Webwork.pdf - William Forbes Assignment Section 8.7 due...

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William ForbesSharmaMAT266ONLINEASpring2018Assignment Section8.7 due 02/22/2018 at 11:59pm MST1.(1 point)The functionf(x) =sin(7x)has a Maclaurin series. Find thefirst 4 nonzero terms in the series, that is write down the Taylorpolynomial with 4 nonzero terms.Solution:SOLUTIONWe use the known Maclaurin series for sin(x), whose firstfour nonzero terms aresin(x)x-x33!+x55!-x77!So the first four nonzero terms off(x) =sin(7x)aresin(7x)7x-(7x)33!+(7x)55!-(7x)77!2.(1 point) Find the Maclaurin series of the functionf(x) = (3x2)e-3x(f(x) =n=0cnxn)c1=c2=c3=c4=c5=Solution:SOLUTIONWe use the known Maclaurin series forex, whose first fournonzero terms areex1+x+x22!+x33!So the first four nonzero terms off(x) =e-3xaree-3x1-3x+(-3x)22!+(-3x)33!And so the first four nonzero terms off(x) =3x2e-3xare3x2e-3x3x21-3x+32x22!-33x33!=3x2-9x3+3(32)2x4-3(33)6x5Equating this tof(x) =n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+c5x5+···,we getc0=0c1=0c2=3c3=-9c4=3(32)2=272c5=-3(33)6=-272for the coefficients up to thex5term.3.(1 point) Match the series with the right expression. (Usethe Maclaurin series.)1.n=0(13)nn!2.n=0(-1)n(13)2n(2n)!3.n=0(-1)n(13)2n+12n+14.n=0(-1)n(13)2n+1(2n+1)!A. sin(13)B. arctan(13)C. cos(13)D.e1/3Solution:SOLUTIONAnalyzing the answers:A.sin(13),sin(x) =n=0(-1)nx2n+1(2n+1)!, so sin13=n=0(-1)n(13)2n+1(2n+1)!B.arctan(13),arctan(x) =n=0(-1)nx2n+12n+1, so arctan13=n=0(-1)n(13)2n+12n+1C.cos(13),cos(x) =n=0(-1)nx2n(2n)!, so cos13=n=0(-1)n(13)2n(2n)!D.e1/3,ex=n=0xnn!, soe1/3=n=0(13)nn!1
So1. D,becausen=0(13)nn!=e1/32. C,becausen=0(-1)n(13)2n(2n)!=cos(13)3. B,becausen=0(-1)n(13)2n+12n+1=arctan(13)4. A,becausen=0(-1)n(13)2n+1(2n+1)!=sin(13)4.(1 point) Match each of the Maclaurin series with rightfunction.1.n=02nxnn!2.n=0(-1)n22nx2n(2n)!3.n=0(-1)n2x2n+1(2n+1)!4.n=0(-1)n2x2n+12n+1A. cos(2x)B. 2arctan(x)C.e2xD. 2sin(x)Solution:SOLUTIONAnalyzing the answers:A.cos(2x),cos(x) =n=0(-1)nx2n(2n)!, socos(2x) =n=0(-1)n(2x)2n(2n)!=n=0(-1)n22nx2n(2n)!B.2arctan(x),arctan(x) =n=0(-1)nx2n+12n+1, so2arctan(x) =2n=0(-1)nx2n+12n+1=n=0(-1)n2x2n+12n+1C.e2x,ex=n=0xnn!, soe2x=n=0(2x)nn!=n=02nxnn!D.2sin(x),sin(x) =n=0(-1)nx2n+1(2n+1)!, so2sin(x) =2n=0(-1)nx2n+1(2n+1)!=n=0(-1)n2x2n+1(2n+1)!So1. C,becausen=02nxnn!=e2x2. A,becausen=0(-1)n22nx2n(2n)!=cos(2x)3. D,becausen=0(-1)n2x2n+1(2n+1)!=2sin(x)4. B,becausen=0(-1)n2x2n+12n+1=2arctan(x)5.(1 point)Write out the first four terms of the Maclaurin series off(x)iff(0) =7,f0(0) =-1,f00(0) =-11,f000(0) =-15f(x)=+···Solution:Solution:The first four terms of the Maclaurin series off(x)aref(0)+f0(0)x+f00(0)2!