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: 31


$f(x)=-1+4 \dfrac{(x-\pi/4)^2}{2!}-16 \dfrac{(x-\pi/4)^4}{4!}+....=\Sigma_{n=0}^{\infty} (-1)^{n+1} (2)^{2n} \dfrac{(x-\dfrac{\pi}{4})^{2n}}{(2n)!}$

Work Step by Step

We have $f'(x)=-2 \cos 2x \implies f'(\pi/4)=0 $; $f''(x)=4 \sin 2x \implies f''(\pi/4)=4 \\ f'''(x)=8 \cos 2x \implies f'''(1)=0$ and $f^{4}(1)=-16$ and so on. Therefore, the Taylor's series at $x=\pi/4$ is as follows: $f(x)=f(\pi/4)+f'(\pi/4) (x-\pi/4)+\dfrac{f''(\pi/4)(x-\pi/4)^2 }{2!} \space on$ or, $f(x)=-1+4 \dfrac{(x-\pi/4)^2}{2!}-16 \dfrac{(x-\pi/4)^4}{4!}+....=\Sigma_{n=0}^{\infty} (-1)^{n+1} (2)^{2n} \dfrac{(x-\dfrac{\pi}{4})^{2n}}{(2n)!}$
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