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: 16


$\dfrac{x}{2}-\dfrac{x^3}{2^3 \cdot 3!}+\dfrac{x^5}{2^{5} \cdot 5!}-$

Work Step by Step

Since, we know that the Maclaurin Series for $\sin{x}$ is defined as: $ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ We need to replace in the above series $x$ with $x/2$. we have $ \sin \dfrac{x}{2}=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\dfrac{x}{2})^{2n+1}}{(2n+1)!}=\dfrac{x}{2}-\dfrac{x^3}{2^3 \cdot 3!}+\dfrac{x^5}{2^{5} \cdot 5!}-$
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