University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 64


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

Work Step by Step

Since, we know that the Taylor Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Now, $e^{(-x^2)}=\Sigma_{n=0}^\infty \dfrac{(-x^2)^n}{n!}=1+(-x^2)+\dfrac{(-x^2)^2}{2!}+\dfrac{(-x^2)^3}{3!}+\dfrac{(-x^2)^4}{4!}+...$ or, $\Sigma_{n=0}^\infty (-1)^n\dfrac{(x)^{2n}}{n!}$
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