Thomas' Calculus 13th Edition

Published by Pearson

Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 20

Answer

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

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, $\sinh x=\dfrac{e^x -e^{-x}}{2}$ $\implies \dfrac{1}{2}[(1+x+\dfrac{x^2}{2!}+...)-(1-x+\dfrac{x^2}{2!}+...)]=x+\dfrac{1}{3!}x^3+\dfrac{1}{5!} x^5=\Sigma_{n=0}^\infty \dfrac{x^{(2n+1)}}{(2n+1)!}$

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