## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 44

#### Answer

$$\ln (\dfrac{3}{2})$$

#### 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 $\\=\dfrac{1}{2} - \dfrac{1}{2 \cdot 2^2} + \dfrac{1}{3 \cdot 2^3}-....\\=( \dfrac{1}{2})-( \dfrac{1}{2}) ( \dfrac{1}{2})^2+( \dfrac{1}{3})( \dfrac{1}{2})^3-( \dfrac{1}{4})( \dfrac{1}{2})^4+....\\= \ln (1+\dfrac{1}{2}); -1 \leq x \leq 1\\=\ln (\dfrac{3}{2})$

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