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: 18



Work Step by Step

We are given $f(x)=\frac{6}{3+x^2}=\frac{2}{1-(-\frac{x^2}{3})}$ for $r$ in $(-\infty,\infty)$ The Taylor series for $\frac{1}{1-x}$ is $1+x+x^2+x^3+...+x^n+...$ The Taylor series for $f(x)=\frac{2}{1-(-\frac{x^2}{3})}$ is $f(x)=2.1+2(-\frac{x^2}{3})+2(-\frac{x^2}{3})^2+2(-\frac{x^2}{3})^3...+2(-\frac{x^2}{3})^n+...$ $=2-\frac{2x^2}{3}+\frac{2x^4}{9}-\frac{2x^6}{27}+...+\frac{2(-x^{2+n})}{3^n}$
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