Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.4 Taylor Polynomials - Exercises - Page 493: 18

Answer

\begin{aligned} \ln (1+x) =x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+(-1)^{n-1} \frac{x^{n}}{n} \end{aligned}

Work Step by Step

Since \begin{aligned} f(x) &=\ln (x+1) & & f(0)=0 \\ f^{\prime}(x) &=\frac{1}{1+x} & & f^{\prime}(0) =1 \\ f^{\prime \prime}(x) &=\frac{-1}{(1+x)^{2}} & & f^{\prime \prime}(0) =-1 \\ f^{\prime \prime \prime}(x) &=\frac{2}{(1+x)^{3}} & & f^{\prime \prime \prime}(0) =2 \\ f^{(4)}(x) &=\frac{-6}{(1+x)^{4}} & & f^{(4)}(0) =-6 \end{aligned} Then \begin{aligned} \ln (1+x) &=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\ldots \ldots+\frac{f^{(n)}(0)}{n !} x^{n} \\ &=\frac{1}{1 !} x-\frac{1}{2 !} x^{2}+\frac{2}{3 !} x^{3}-\frac{6}{4 !} x^{4} \ldots \\ &=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+(-1)^{n-1} \frac{x^{n}}{n} \end{aligned}
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