Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 30


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

Work Step by Step

Differentiate the given function $f(x)$ as follows: $f'(x)=2^x \ln 2 \implies f'(1)=2 \ln 2 \\f''(x)=2^x (\ln 2)^2 \implies f''(1)=2 (\ln 2)^2 \\ f'''(x)=2^x (\ln 2)^3 \implies f'''(1)=2 (\ln 2)^3$ The Taylor series at $x=1$ can be written as: $f(x)=f(1)+f'(2) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+....\\=2+2 \ln 2(x-1)+2 \dfrac{(\ln 2)^2 (x-1)^2}{2!}+.... \\=\Sigma_{n=0}^{\infty} 2 (\ln 2)^n \dfrac{(x-1)^n}{n!}$
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