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: 20


$\Sigma_{n=0}^\infty \dfrac{x^{(2n+1)}}{(2n+1)!}$

Work Step by Step

Consider the Maclaurin Series for $\sin x$ and $e^x$ are as follows: $ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ and $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+...$ Then, $ \sinh x=\dfrac{e^x -e^{-x}}{2}$ $\implies \dfrac{1}{2}[(1+x+\dfrac{x^2}{2!}+...)-(1-x+\dfrac{x^2}{2!}+...)]=x+\dfrac{1}{3!}x^3+\dfrac{1}{5!} x^5=\Sigma_{n=0}^\infty \dfrac{x^{(2n+1)}}{(2n+1)!}$
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.