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: 30


$f(x)=2+2 \ln 2(x-1)+2(\ln 2)^2 \dfrac{(x-1)^2}{2!}+....=\Sigma_{n=0}^{\infty} 2 (\ln 2)^n \dfrac{(x-1)^n}{n!}$

Work Step by Step

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