Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.9 Representations of Functions as Power Series - 11.9 Exercises - Page 798: 34



Work Step by Step

$f(x)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{2n}$, $f'(x)=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2nx^{2n-1}}{(2n)!}$ $f''(x)=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n(2n-1)x^{2n-2}}{(2n)!}$ Let $f(x)+f''(x)=I$ Then $I=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{2n}+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n(2n-1)x^{2n-2}}{(2n)!}$ After simplification $f''(x)=-f(x)$ Thus, $f(x)+(-f(x))=0$ or Hence, $f(x)+f''(x)=0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.