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: 62


$\Sigma_{n=0}^\infty \dfrac{(-1)^n (x)^{6n}}{5^n(2n)!}$

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 (\dfrac{x^3}{\sqrt 5})=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\dfrac{x^3}{\sqrt 5})^{2n}}{(2n)!}=1-\dfrac{(\dfrac{x^3}{\sqrt 5})^2}{2!}+\dfrac{(\dfrac{x^3}{\sqrt 5})^4}{4!}-...$ or, $=\Sigma_{n=0}^\infty \dfrac{(-1)^n (x)^{6n}}{5^n(2n)!}$
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