Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 637: 57

Answer

$\Sigma_{n=0}^\infty 2^{(n)} x^{(n)} $ for $|x| \lt \dfrac{1}{2}$

Work Step by Step

Consider the Taylor Series for $\dfrac{1}{1-x}$ as follows: $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$ and $\dfrac{1}{1-2x}=1+(2x)+(2x)^2+....+(2x)^n=\Sigma_{n=0}^\infty 2^n x^n $ As we can see that $|2x| \lt 1 \implies |x| \lt \dfrac{1}{2} $ Thus, we have $\dfrac{1}{1-2x}=\Sigma_{n=0}^\infty 2^{(n)} x^{(n)} $ for $|x| \lt \dfrac{1}{2} $
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