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: 21


The Taylor series for $f(x)=\ln(1+2x^4)$ for $r$ in $(-\infty,\infty)$ is $=x-2x^8+\frac{8}{3}x^{12}-4x^{16}+...+\frac{(-1)^n(2x^4)^{n+1}}{n+1}+...$

Work Step by Step

We are given $f(x)=\ln(1+2x^4)$ for $r$ in $(-\infty,\infty)$ 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 $f(x)=\ln(1+2x^4)$ is $f(x)=x-\frac{1}{2}(2x^4)^2+\frac{1}{3}(2x^4)^3-\frac{1}{4}(2x^4)^4+....+\frac{(-1)^n(2x^4)^{n+1}}{n+1}+...$ $=x-2x^8+\frac{8}{3}x^{12}-4x^{16}+...+\frac{(-1)^n(2x^4)^{n+1}}{n+1}+...$
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