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: 11



Work Step by Step

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