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: 9



Work Step by Step

We are given $f(x)=\frac{x^{2}}{4-x}=\frac{\frac{x^{2}}{4}}{1-\frac{x}{4}}$ with $c=\frac{x^{2}}{4}$ for $r$ in $(-4,4)$ The Taylor series for $f(x)=\frac{x^{2}}{4-x}$ is $f(x)=\frac{x^{2}}{4}+\frac{x^{2}}{4}(\frac{x}{4})+\frac{x^{2}}{4}(\frac{x}{4})^{2}+\frac{x^{2}}{4}(\frac{x}{4})^3+...+\frac{x^{2}}{4}(\frac{x}{4})^n+...$ $=\frac{x^{2}}{4}+\frac{x^{3}}{16}+\frac{x^{4}}{64}+...+\frac{x^{2+n}}{4^{n+1}}+...$
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