University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 52


$$-\ln \dfrac{(1-x)}{x}; -1 \leq x \leq 1$$

Work Step by Step

The Taylor series for $\ln (1+x)$ can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$ We have $\\=1+\dfrac{x}{2}+ \dfrac{x^2}{3} + \dfrac{x^3}{4}+....\\=(- \dfrac{1}{x})(-x- \dfrac{x^2}{2}- \dfrac{x^3}{3} - \dfrac{x^4}{4}-......)\\= (-1/x) \ln (1-x) \\=-\ln \dfrac{(1-x)}{x}; -1 \leq x \leq 1$
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