Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.9 - Representations of Functions as Power Series - 11.9 Exercises - Page 757: 4

Answer

$\Sigma_{n=0}^{\infty}(5)(4x^{2})^{n}$, The interval of convergence is $(-\frac{1}{2},\frac{1}{2})$ The radius of convergence is $\frac{1}{2}$

Work Step by Step

The sum of a geometric series with initial term $a$ and common ratio $r$ is $S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$ Therefore, $f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(5)(4x^{2})^{n}$ This is the power series representation of $f(x)$. We know that the power series converges when $r=|4x^{2}|\lt 1$ $x^{2}\lt \frac{1}{4}$ Hence, the interval of convergence is $(-\frac{1}{2},\frac{1}{2})$ The radius of convergence is $\frac{1}{2}$
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