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: 19


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

Work Step by Step

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