University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 49


$$\dfrac{x^3}{1+x^2}; |x| \lt 1$$

Work Step by Step

The Taylor series for $\dfrac{1}{ (1-x)}$ can be defined as: $\dfrac{1}{ (1-x)}=1+x+x^2+.....x^n+....; |x| \lt 1$ Consider the given series: $\\=x^3-x^5+x^7-x^9-....\\=x^3(1-x^2+x^4-x^6+....)\\=x^3(1-x^2+(x^2)^2-(x^2)^3+.......) \\ = x^3 (\dfrac{1}{1+x^2}) \\=\dfrac{x^3}{1+x^2}; |x| \lt 1$
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.