Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 578: 43

Answer

$$\sum_{n=0}^{\infty}(-1)^{n+1}(x-5)^{n}$$ $ |x-5|<1 $

Work Step by Step

Given $$\frac{1}{4-x}$$ Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$ For $c=5 $, by using (1), we get \begin{aligned} \frac{1}{4-x}&=\frac{1}{-1-(x-5)}\\ & =\frac{-1}{1+(x-5)}\\ & =\frac{-1}{1-(-(x-5))}\\ &=(-1) \sum_{n=0}^{\infty}(-1(x-5))^{n}\\ &=\sum_{n=0}^{\infty}(-1)^{n+1}(x-5)^{n} \end{aligned} Such that $ |x-5|<1 $
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