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: 16


\begin{aligned} T_{n}(x) &= \frac{1}{2}-\frac{(x-1)}{4}+\frac{(x-1)^{2}}{8}+\cdots+(-1)^{n} \frac{(x-1)^{n}}{2^{n+1}} \end{aligned}

Work Step by Step

Since \begin{aligned} f(x) &=\frac{1}{1+x} & & f(1)=\frac{1}{2} \\ f^{\prime}(x) &=\frac{-1}{(1+x)^{2}} & & f^{\prime}(1)=\frac{-1}{4} \\ f^{\prime \prime}(x) &=\frac{2}{(1+x)^{3}} & & f^{\prime \prime}(1)= \frac{1}{4} \\ f^{\prime \prime \prime}(x) &=\frac{-6}{(1+x)^{4}} & & f^{\prime \prime \prime}(1) =\frac{-3}{8} \\ f^{(4)}(x) &=\frac{24}{(1+x)^{5}} & & f^{(4)}(1) =\frac{3}{4} \end{aligned} Then \begin{aligned} T_{n}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+\ldots \ldots+\frac{f^{(n)}(a)}{n !}(x-a)^{n} \\ &=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime \prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime \prime}(1)}{3 !}(x-1)^{3}+\ldots \ldots+\frac{f^{(n)}(1)}{n !}(x-1)^{n} \\ &=\frac{1}{2}-\frac{1}{4}(x-1)+\frac{1}{8}(x-2)^{2}-\frac{1}{16}(x-2)^{3}+\frac{3}{32}(x-3)^{4}+\cdots \\ &=\frac{1}{2}-\frac{(x-1)}{4}+\frac{(x-1)^{2}}{8}+\cdots+(-1)^{n} \frac{(x-1)^{n}}{2^{n+1}} \end{aligned}
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