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 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 633: 46

Answer

$$\tan^{-1} (\dfrac{2}{3})$$

Work Step by Step

Recall the Taylor series for $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....; |x| \leq 1$ OR, $=\dfrac{2}{3} - \dfrac{2^3}{3^3 \cdot 3!} + \dfrac{2^5}{3^5 \cdot 5!}-....$ OR, $=\dfrac{2}{3}-( \dfrac{1}{3}) (\dfrac{2}{3})^3+( \dfrac{1}{5}) (\dfrac{2}{3})^5-...$ Or, $=\tan^{-1} (\dfrac{2}{3})$
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.