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: 16


Maclaurin series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n)!}$ and $R=\infty$

Work Step by Step

$f(x)=xcosx=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n!}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{x^{2n+3}}{(2n+2)!}}{\frac{x^{2n+1}}{(2n)!}}|$ $=\lim\limits_{n \to\infty}|\frac{x^{2}}{(2n+2)(2n+1)}|$ $=\lim\limits_{n \to\infty}|\frac{x^{2}}{4n^{2}+6n+2}|$ $=0\lt 1$ Maclaurin series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n)!}$ and $R=\infty$
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