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: 39


$P'_n(a)=f'(a) ; P''_n(a)=f''(a) ; P^{n}_{n}(a)=f^{n}(a)$

Work Step by Step

The Taylor's series for $f(x)$ at $x=a$ is as follows: $P_n(x)=\Sigma_{n=0}^{\infty} \dfrac{f^{k}(a)}{k!}(x-a)^k=f(a)+f'(a)(x-a)+.....+\dfrac{f^{n}(a)}{n!}(x-a)^n$ Therefore, $P'_n(x)=\Sigma_{n=0}^{\infty} \dfrac{f^{k}(a)}{k!}(x-a)^k=f'(a)+2(x-a) \dfrac{f^{n}(a)}{2!}(x-a)^n+....+n(x-a)^{n-1} \dfrac{f^n(a)}{n!}$ and $P^{n}_n(x)=[n(n-1)(n-2).....1]\dfrac{f^n}{n!}$ So, $P'_n(a)=f'(a) ; P''_n(a)=f''(a) ; P^{n}_{n}(a)=f^{n}(a)$
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