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: 21



Work Step by Step

Given: $f(x)=x^4-2x^3-5x+4$ Maclaurin Series of order $n$ for the function $f(x)$ at the point $k=0$ can be defined as: $f(x)=f(0)+\Sigma_{n=1}^\infty \dfrac{f^{n}(0)}{n!}(x-0)^n$ Here, $f(0)=4 \\ f'(x)=4x^3-6x^2-5 \implies f'(0)=-5 \\ f''(x)=12x^2-12x \implies f''(0)=0\\ f'''(x)=24x-12 \implies f'''(0)=-12\\f^{4}(x)=24 \implies f^{4}(0)=24 \\f^{n}(x)=0 \implies f^{n}(0)=0$ Thus, the given series follows the pattern of $f(x)=f(0)+\Sigma_{n=1}^\infty \dfrac{f^{n}(0)}{n!}(x-0)^n=4-5x-(12)\dfrac{x^3}{3!}+(24)\dfrac{x^4}{4!}=x^4-2x^3-5x+4$
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