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: 19


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

Work Step by Step

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