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: 53


$$2(x+\dfrac{x^3}{3}+ \dfrac{x^5}{5} +....); -1 \lt x \lt 1$$

Work Step by Step

The Taylor series for $\ln (1+x)$ can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$ and $\ln (1-x)=-x-\dfrac{ x^2}{2}-\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$ We have $\ln (1+x) -\ln (1-x)=(x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....) - (-x-\dfrac{ x^2}{2}-\dfrac{x^3}{3}-....) $ or, $\ln (\dfrac{1+x}{1-x})=2(x+\dfrac{x^3}{3}+ \dfrac{x^5}{5} +....); -1 \lt x \lt 1$
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