Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 47


For $r$ in $(-\infty,\infty)$ The Taylor series for $f(x)=e^{-2x^2}$ is $=1-2x^2+2x^4-\frac{4x^6}{3}+...+\frac{(-2x^2)^n}{n!}...$

Work Step by Step

We are given $f(x)=e^{-2x^2}$ for $r$ in $(-\infty,\infty)$ The Taylor series for $e^x$ is $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...+\frac{1}{n!}x^n+...$ The Taylor series for $f(x)=e^{-2x^2}$ is $f(x)=1+(-2x^2)+\frac{(-2x^2)^2}{2!}+\frac{(-2x^2)^3}{3!}+...+\frac{(-2x^2)^n}{n!}+...$ $=1-2x^2+2x^4-\frac{4x^6}{3}+...+\frac{(-2x^2)^n}{n!}...$
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.