University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 34


$ x-\dfrac{x^3}{2}+\dfrac{3x^5}{8}-\dfrac{5x^7}{16}+....$

Work Step by Step

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ and the Taylor series for $\tan^{-1} x $ at $ x=0$ can be defined as: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....$ We have $\sin (\tan^{-1} x) = x-\dfrac{(\tan^{-1} x)^3}{3!}+\dfrac{ (\tan^{-1} x)^5}{5!}-....$ or, $=(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-....) -\dfrac{1}{3!} (x-\dfrac{x^3}{6}+\dfrac{x^5}{5}-....)^3 +....$ or, $=x-\dfrac{x^3}{2}+\dfrac{3x^5}{8}-\dfrac{5x^7}{16}+....$
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