University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 2



Work Step by Step

Since, we know that the Maclaurin Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Now, $e^{-x/2}=\Sigma_{n=0}^\infty \dfrac{(-x/2)^n}{n!}=1+(\dfrac{-x}{2})+\dfrac{(-x/2)^2}{2!}+\dfrac{(-x/2)^3}{3!}-\dfrac{(-x/2)^4}{4!}...$ or, $=1-\dfrac{x}{2}+\dfrac{1}{8}x^2-\dfrac{1}{48}x^3+\dfrac{1}{384}x^4-...$
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.