University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 11

Answer

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

Work Step by Step

Since, we know that the Maclaurin Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}$ We need to replace in the above series $x$ with $-x$. we have $e^{-x}=\Sigma_{n=0}^\infty \dfrac{(-x)^n}{n!}$ or, $e^{-x}=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^n}{n!}$
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.