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: 18


$5-\dfrac{5 \pi^2x^2}{2!}+\dfrac{5 \pi^4x^4}{4!}-\dfrac{5 \pi^6x^6}{6!}+..$

Work Step by Step

Since, we know that the Maclaurin Series for $\cos{x}$ is defined as: $ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$ Now, $5 \cos (\pi x)=5 \Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi x)^{2n}}{(2n)!}$ or, $ =5-\dfrac{5 \pi^2x^2}{2!}+\dfrac{5 \pi^4x^4}{4!}-\dfrac{5 \pi^6x^6}{6!}+..$
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