Calculus with Applications (10th Edition)

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

Chapter 12 - Sequences and Series - 12.5 Taylor Series - 12.5 Exercises - Page 647: 28


$e^{-x}\geq 1-x$ for all x

Work Step by Step

The Taylor series for $e^{-x}$ with $r$ in $(-\infty,\infty)$ is $1+(-x)+\frac{1}{2!}(-x)^2+\frac{1}{3!}(-x)^3+...+\frac{1}{n!}(-x)^n+...$ $=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...+\frac{1}{n!}(-x)^n+...$ $e^{-x}=(1-x)+(\frac{x^2}{2!}-\frac{x^3}{3!})+...+\frac{1}{n!}x^n+...$ $e^{-x}-(1-x)\geq0$ (0 represents $(\frac{x^2}{2!}-\frac{x^3}{3!}+...+\frac{1}{n!}(-x)^n+...)$ $e^{-x}\geq 1-x$ Hence, we can suggest that $e^{-x}\geq 1-x$ for all x
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.