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 - Practice Exercises - Page 637: 64

Answer

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

Work Step by Step

Consider the Taylor Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Then, we have $e^{(-x^2)}=\Sigma_{n=0}^\infty \dfrac{(-x^2)^n}{n!}$ Hence, $1+(-x^2)+\dfrac{(-x^2)^2}{2!}+\dfrac{(-x^2)^3}{3!}+\dfrac{(-x^2)^4}{4!}+...=\Sigma_{n=0}^\infty (-1)^n\dfrac{(x)^{(2n)}}{n!}$
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