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: 20


\begin{aligned} T_{n}(x) =\frac{1}{3}-\frac{1}{9}(x-4)+\frac{1}{27}(x-4)^{2}-\frac{1}{81}(x-4)^{3}+\cdots+\frac{(-1)^n}{3^{n+1}}(x-4)^{n} \end{aligned}

Work Step by Step

Since \begin{aligned} f(x) &=\frac{1}{x-1}, & f(4) &=\frac{1}{3} \\ f^{\prime}(x) &=-\frac{1}{(x-1)^{2}}, & f^{\prime}(4) &=-\frac{1}{9} \\ f^{\prime \prime}(x) &=\frac{2}{(x-1)^{2}}, & f^{\prime \prime \prime}(4) &=-\frac{2}{27} \\ \vdots & & & \vdots \\ f^{(n)}(x) &=\frac{(-1)^{n} \cdot n !}{(x-1)^{n+1}} & & f^{(n)}(4)=\frac{(-1)^{n} \cdot n !}{3^{n+1}} \end{aligned} Then \begin{aligned} T_{n}(x) &=f(4)+\frac{f^{\prime}(4)}{1 !}(x-4)+\frac{f^{\prime \prime}(4)}{2 !}(x-4)^{2}+\cdots+(x-4)^{n} \frac{f^{(n)}(4)}{n !} \\ &=\frac{1}{3}-\frac{1}{9}(x-4)+\frac{1}{27}(x-4)^{2}-\frac{1}{81}(x-4)^{3}+\cdots+\frac{(-1)^n}{3^{n+1}}(x-4)^{n} \end{aligned}
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