Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 637: 68

Answer

$\dfrac{1}{a}-\dfrac{(x-a)}{a^2}+\dfrac{(x-a)^2}{a^3}-\dfrac{(x-a)^3.}{a^4}....$

Work Step by Step

Given: $f(x)=\dfrac{1}{x+1}$ Taylor polynomial of order $n$ for the function $f(x)$ at the point $a$ can be defined as: $A_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$ Here, $f(a)=\dfrac{1}{a}$ and $f'(x)=-x^{-2} \implies f'(a)=-\dfrac{1}{(a)^2} \\f''(x)=2x^{-3} \implies f''(a)=\dfrac{2}{(a)^3}\\f'''(x)=-6x^{-4} \implies f'''(a)=-\dfrac{6}{(a)^4}$ Taylor polynomial of order $n$ for the function $f(x)=\dfrac{1}{x}$ is: $f(x)=\dfrac{1}{a}+(-\dfrac{1}{(a)^2})(x-a)+\dfrac{\dfrac{2}{(a)^3}}{2!}(x-a)^2+\dfrac{(-\dfrac{6}{(a)^4})}{3!}(x-a)^3....$ Hence, $f(x)=\dfrac{1}{a}-\dfrac{(x-a)}{a^2}+\dfrac{(x-a)^2}{a^3}-\dfrac{(x-a)^3.}{a^4}....$
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