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: 52


(a) $(1+x)^{-1/4}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{^{n}}{4^{n}n!}$ (b) $\frac{1}{\sqrt[4] {1.1}}\approx 1+(-0.025)+(0.0015625)\approx 0.976$

Work Step by Step

(a) $\frac{1}{\sqrt[4] {1+x}}=(1+x)^{-1/4}$ $(1+x)^{-1/4}=\Sigma_{n=0}^{\infty}(-1/4)nx^{n}$ $\frac{1}{\sqrt[4] {1+x}}=(1+x)^{-1/4}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{^{n}}{4^{n}n!}$ (b) From part (a), we have $\frac{1}{\sqrt[4] {1+x}}=(1+x)^{-1/4}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{^{n}}{4^{n}n!}$ Here $a_{n}=1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{^{n}}{4^{n}n!}$ Thus, $\frac{1}{\sqrt[4] {1.1}}\approx 1+(-0.025)+(0.0015625)\approx 0.976$
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