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: 7



Work Step by Step

We are given $f(x)=\frac{8x}{1+3x}=\frac{8x}{1-(-3x)}$ with $c=8x$ for $r$ in $\frac{-1}{3}\lt x \lt \frac{1}{3}$ The Taylor series for $f(x)=\frac{8x}{1+3x}$ is $f(x)=5x+5x(-3x)+5x(-3x)^{2}+5x(-3x)^3+...+5x(-3x)^n+...$ $=5x-15x^2+45x^3-135x^4+...+5x(-3x)^n+...$
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