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: 60

Answer

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

Work Step by Step

Since, we know that the Taylor Series for $\sin x$ is defined as: $ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ Now, $ \sin (\dfrac{2x}{3})=\Sigma_{n=0}^\infty \dfrac{(-1)^n ((\dfrac{2x}{3}))^{2n+1}}{(2n+1)!}=(\dfrac{2x}{3})-\dfrac{(\dfrac{2x}{3})^3}{3!}+\dfrac{(\dfrac{2x}{3})^5}{5!}-\dfrac{(\dfrac{2x}{3})^7}{7!}+...$ or, $=\Sigma_{n=0}^\infty \dfrac{(-1)^n (2)^{2n+1} (x)^{2n+1}}{3^{2n+1}(2n+1)!}$
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