Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 73



Work Step by Step

Given: $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{x^{4n}}{n!}$ As we know $e^{x}=\Sigma_{n=0}^{\infty}\dfrac{x^{n}}{n!}$ and $e^{-x}=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{x^{n}}{n!}$ Thus, $e^{-x^{4}}=\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{x^{4n}}{n!}$
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