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: 29


$f(x)=e^2+e^2(x-2)+e^2\dfrac{(x-2)^2}{2!}+....=\Sigma_{n=0}^{\infty} e^2\dfrac{(x-2)^n}{n!}$

Work Step by Step

We have $f'(x)=e^x \implies f'(2)=e^2$; $f''(x)=e^x \implies f''(0)=3\\ f'''(x)=e^2 \implies f'''(0)=e^2$ and $f^{4}(1)=e^2$ and so on. Therefore, the Taylor's series at $x=2$ is as follows: $f(x)=f(2)+f'(2) (x-2)+\dfrac{f''(2)(x-2)^2 }{2!} \space on$ or, $f(x)=e^2+e^2(x-2)+e^2\dfrac{(x-2)^2}{2!}+....=\Sigma_{n=0}^{\infty} e^2\dfrac{(x-2)^n}{n!}$
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