University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 27


$f(x)=1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+...=\Sigma_{n=0}^{\infty} (-1)^n (n+1)(x-1)^n$

Work Step by Step

We have $f'(x)=\dfrac{-2}{x^3} \implies f'(1)=1$; $f''(x)=\dfrac{-6}{x^4} \implies f''(1)=-2\\ f'''(x)=\dfrac{-24}{x^5}\implies f'''(1)=-24$ and $f^{4}(1)=-120$ and so on. Therefore, the Taylor's series at $x=1$ is as follows: $f(x)=f(1)+f'(-1) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+\dfrac{f''(1)(x-1)^3 }{3!}+\dfrac{f''(1)(x-1)^4 }{4!} \space on$ or, $f(x)=1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+...=\Sigma_{n=0}^{\infty} (-1)^n (n+1)(x-1)^n$
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.