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: 2


$p_{0}(x)=0 \\ p_{1}(x)=x \\p_{2}(x)=x \\p_{3}(x)=x-\dfrac{x^3}{6}$

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