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: 14


$2+\Sigma_{n=1}^\infty 3x^n$

Work Step by Step

Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as: $p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$ Now, $f'(x)=3(1-x)^{-2}\\f''(x)=6(1-x)^{-3}\\f'''(x)=18(1-x)^{-4}$ and $f^{n}(x)=3(n!)(1-x)^{-n-1}$ Here, $f(0)=2 \\ f'(0)=3=3 \cdot 1! \\f''(0)=6=3 \cdot 2!\\ f'''(0)=18=3 \cdot 3 !$ Thus, the given series follows the pattern of $2+3x+3x^2+3x^3+....=2+\Sigma_{n=1}^\infty 3x^n$
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