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: 50


$\frac{1}{\sqrt[10] e}=e^{-0.1}\approx 0.90483$

Work Step by Step

Given:$\frac{1}{\sqrt[10] e}=e^{-0.1}$ The series for $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$ Put $x=-0.1$ $e^{x}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(0.1)^{n}}{n!}$ Here, $a_{n}=(-1)^{n}\frac{(0.1)^{n}}{n!}$ So, $s_{2}=a_{0}+a_{1}+a_{2}+a_{3} \approx 0.90483$ Thus, $\frac{1}{\sqrt[10] e}=e^{-0.1}\approx 0.90483$
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