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: 22


\begin{aligned} T_{n}(x) =\frac{1}{4}-\frac{1}{4}(x-2)+\frac{3}{16}(x-2)^{2}-\frac{1}{8}(x-2)^{3}+(-1)^{n}(x-2)^{n} \frac{n+1}{2^{n+2}} \end{aligned}

Work Step by Step

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