University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 63


$\Sigma_{n=0}^\infty \dfrac{(\pi)^n x^n}{2^nn!}$

Work Step by Step

Since, we know that the Taylor Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Now, $e^{(\frac{\pi x}{2})}=\Sigma_{n=0}^\infty \dfrac{(\pi x/2)^n}{n!}=1+(\pi x/2)+\dfrac{(\pi x/2)^2}{2!}+\dfrac{(\pi x/2)^3}{3!}+\dfrac{(\pi x/2)^4}{4!}+...$ or, $\Sigma_{n=0}^\infty \dfrac{(\pi)^n x^n}{2^nn!}$
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