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: 28



Work Step by Step

We have $f'(x)=\dfrac{3}{(1-x)^4} \implies f'(0)=1$; $f''(x)=\dfrac{12}{x^5}\dfrac{3}{(1-x)^4} \implies f''(0)=3\\ f'''(x)=\dfrac{60}{(1-x)^6}\implies f'''(0)=12$ and $f^{4}(1)=60$ and so on. Therefore, the Taylor's series at $x=1$ is as follows: $f(x)=f(0)+f'(0) (x)+\dfrac{f''(0)(x)^2 }{2!}+\dfrac{f''(0)(x)^3 }{3!}+\dfrac{f''(4)(x)^4 }{4!} \space on$ or, $f(x)=1+3x+6x^2+10x^3+15x^4+....$
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