University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 15


$x-\dfrac{\pi^2 x^3}{2!}+\dfrac{\pi^4 x^5}{4!}-\dfrac{\pi^6 x^7}{6!}+..$

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)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ Now, $ x \cos \pi x=x(1-\dfrac{(\pi x)^2}{2!}+\dfrac{(\pi x)^4}{4!}-\dfrac{(\pi x)^6}{6!}+...)=x-\dfrac{\pi^2 x^3}{2!}+\dfrac{\pi^4 x^5}{4!}-\dfrac{\pi^6 x^7}{6!}+..$
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