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: 23



Work Step by Step

We have the Maclaurin series for $(1+x)^a$ as follows $(1+x)^a=1+ax+\dfrac{a(a-1)x^2}{2!}x^2+\dfrac{a(a-1)(a-2)}{3!}x^3+...$ Replace $\dfrac{1}{4}$ for $x$ in the above form through four terms. Therefore, we have: $(1+x)^{1/4}=1+\dfrac{x}{4}+\dfrac{\dfrac{1}{4}(\dfrac{1}{4}-1)x^2}{2!}x^2+\dfrac{\dfrac{1}{4}(\dfrac{1}{4}-1)(\dfrac{1}{4}-2)}{3!}x^3+...\\=1+\dfrac{x}{4}-\dfrac{3}{32}x^2+\dfrac{7}{128}x^3+...$
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.