University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 5


$p_{0}(x)=\dfrac{1}{2} \\ p_{1}(x)=\dfrac{1}{2}-\dfrac{(x-2)}{4} \\p_{2}(x)=\dfrac{1}{2}-\dfrac{(x-2)}{4}+\dfrac{(x-2)^2}{8} \\p_{3}(x)=\dfrac{1}{2}-\dfrac{(x-2)}{4}+\dfrac{(x-2)^2}{8}-\dfrac{(x-2)^2}{16}$

Work Step by Step

Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as: $p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$ Here, $f(2)=\dfrac{1}{2} ; f'(x)=-x^{-2} \implies f'(2)=\dfrac{-1}{4}; f''(x)=2x^{-3} \implies f''(2)=\dfrac{1}{4}; f'''(x)=-6x^{-4}\implies f'''(2)=\dfrac{-3}{8}$ Thus, $p_{0}(x)=\dfrac{1}{2} \\ p_{1}(x)=\dfrac{1}{2}-\dfrac{(x-2)}{4} \\p_{2}(x)=\dfrac{1}{2}-\dfrac{(x-2)}{4}+\dfrac{(x-2)^2}{8} \\p_{3}(x)=\dfrac{1}{2}-\dfrac{(x-2)}{4}+\dfrac{(x-2)^2}{8}+(\dfrac{-3}{8})\dfrac{(x-2)^2}{6} =\dfrac{1}{2}-\dfrac{(x-2)}{4}+\dfrac{(x-2)^2}{8}-\dfrac{(x-2)^2}{16}$
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