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: 61


$\Sigma_{n=0}^\infty \dfrac{(-1)^n (x)^{10n/3}}{(2n)!}$

Work Step by Step

Since, we know that the Taylor 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!}-...$ Now, $ \cos (x^{5/3})=\Sigma_{n=0}^\infty \dfrac{(-1)^n (x^{5/3})^{2n}}{(2n)!}=1-\dfrac{(x^{5/3})^2}{2!}+\dfrac{(x^{5/3})^4}{4!}-...$ or, $=\Sigma_{n=0}^\infty \dfrac{(-1)^n (x)^{10n/3}}{(2n)!}$
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