Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 19


$\Sigma_{n=0}^\infty \dfrac{x^{(2n)}}{(2n)!}$

Work Step by Step

Consider the Maclaurin Series for $\cos x$ and $e^x$ are as follows: $ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$ and $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=(1+x+\dfrac{x^2}{2!}+...$ Then, $ \cosh x=\dfrac{e^x+e^{-x}}{2}$ $ \implies \dfrac{1}{2}[(1+x+\dfrac{x^2}{2!}+...)+(1-x+\dfrac{x^2}{2!}+...)]=1+\dfrac{1}{2!}x^2+\dfrac{1}{4!} x^4=\Sigma_{n=0}^\infty \dfrac{x^{(2n)}}{(2n)!}$
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.