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: 9


$p_{0}(x)=2\\ p_{1}(x)=2+\dfrac{(x-4)}{4} \\p_{2}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}\\p_{3}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{1}{512}(x-4)^3$

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, $f(4)=2 \\ f'(4)=\dfrac{1}{4}\\f''(4)=-\dfrac{1}{32}\\ f'''(4)=\dfrac{3}{256}$ Thus, $p_{0}(x)=2\\ p_{1}(x)=2+\dfrac{(x-4)}{4} \\p_{2}(x)= 2+\dfrac{(x-4)}{4}+(-1/32)\dfrac{(x-4)^2}{2}=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}\\p_{3}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{(3/256)}{6}(x-4)^3=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{1}{512}(x-4)^3$
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