Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 25



Work Step by Step

We have the Maclaurin series for $ \sin x$ and $e^x$ as follows $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ $$\tan^{-1} x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+...$$ Now, we have the Maclaurin series of $e^x \tan^{-1} x$ as: $$e^x \tan^{-1} x=\\ =(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+)(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+)+...\\ =x+x^2+\dfrac{x^3}{2!}-(\dfrac{x^3}{3}+\dfrac{x^4}{3})+...\\=x+x^2+\dfrac{x^3}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{6}-\dfrac{x^4}{3}+...$$
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.