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: 39



Work Step by Step

$xcos(\frac{1}{2}x^{2})=x\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\frac{1}{2}x^{2})^{2n}}{2n!}$ $=x\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\frac{1}{2})^{2n}x^{4n}}{2n!}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{4n+1}}{2^{2n}(2n)!}$
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