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: 4


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

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, $f(0)=0 ; f'(x)=(1+x)^{-1} \implies f'(0)=1; f''(x)=-(1+x)^{-2} \implies f''(0)=-1; f'''(x)=2(1+x)^{-3}\implies f'''(0)=2$ Thus, $p_{0}(x)=0 \\ p_{1}(x)=x \\p_{2}(x)=x-\dfrac{x^2}{2} \\p_{3}(x)=x-\dfrac{x^2}{2}+\dfrac{2}{6}(x-0)^3 =x-\dfrac{x^2}{2} +\dfrac{x^3}{3}$
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