Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 24

Answer

$cosx=\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\pi)^{2n}}{(2n)!}$ The radius of convergence is $R=\infty$.

Work Step by Step

Taylor series centered at $a= \pi$ is $-1+\frac{1}{2!}(x-\pi)^{2}-\frac{1}{4!}(x-\pi)^{4}+....$ $cosx=\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\pi)^{2n}}{(2n)!}$ The radius of convergence is $R=\infty$.
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