University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 10



Work Step by Step

Since, we know $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+...$ Now, the given equation can be written as: $\dfrac{1}{2-x}=\dfrac{1}{2} [\dfrac{1}{1-\dfrac{x}{2}}]$ or, $\dfrac{1}{1-\dfrac{x}{2}}=1+(\dfrac{x}{2})+(\dfrac{x}{2})^2+(\dfrac{x}{2})^3+(\dfrac{x}{2})^4+...$ or, $=\dfrac{1}{2}[1+\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^3}{8}+\dfrac{x^4}{16}+...]$ or, $=\dfrac{1}{2}+\dfrac{x}{4}+\dfrac{x^2}{8}+\dfrac{x^3}{16}+\dfrac{x^4}{32}+...$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.