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: 46


The Taylor series for $f(x)=\ln(1+\frac{1}{3}x)$ for $r$ in $(-3,3]$ is $x-\frac{1}{18}x^2+\frac{1}{81}x^3-\frac{1}{324}x^4+...+\frac{(-1)^n(\frac{1}{3}x)^{n+1}}{n+1}+...$

Work Step by Step

We are given $f(x)=\ln(1+\frac{1}{3}x)$ for $r$ in $(-3,3]$ The Taylor series for $\ln(1+x)$ is $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{(-1)^nx^{n+1}}{n+1}+...$ The Taylor series for $f(x)=\ln(1+\frac{1}{3}x)$ is $f(x)=x-\frac{1}{2}(\frac{1}{3}x)^2+\frac{1}{3}(\frac{1}{3}x)^3-\frac{1}{4}(\frac{1}{3}x)^4+....+\frac{(-1)^n(\frac{1}{3}x)^{n+1}}{n+1}+...$ $=x-\frac{1}{18}x^2+\frac{1}{81}x^3-\frac{1}{324}x^4+...+\frac{(-1)^n(\frac{1}{3}x)^{n+1}}{n+1}+...$
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