Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 37



Work Step by Step

$f(x)=xcos(2x)$ Since, $sinx=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{(2n)}}{(2n)!}$ Change $x$ to $2x$ $cos(2x)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(2x)^{2n}}{(2n)!}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(2)^{2n}x^{2n}}{(2n)!}$ Hence, $xcos(2x)=x \cdot \Sigma_{n=0}^{\infty}(-1)^{n}\frac{(2x)^{2n}}{(2n)!}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(2)^{2n}x^{2n}}{(2n)!}$ $xcos(2x)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(2)^{2n}x^{2n+1}}{(2n)!}$
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