Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 34


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

Work Step by Step

Recall the Taylor series for $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ and $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....$ $\sin (\tan^{-1} x) = x-\dfrac{(\tan^{-1} x)^3}{3!}+\dfrac{ (\tan^{-1} x)^5}{5!}-....$ or, $\sin (\tan^{-1} x)=(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-....) -\dfrac{1}{3!} (x-\dfrac{x^3}{6}+\dfrac{x^5}{5}-....)^3 +....$ or, $\sin (\tan^{-1} x) =x-\dfrac{x^3}{2}+\dfrac{3x^5}{8}-\dfrac{5x^7}{16}+....$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.