University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 48


$ f(x)=1+x+x^2+x^3$ and $ Error \leq 0.00017$

Work Step by Step

The Taylor series for $ f(x)$ at $ x=0$ is as follows: $ f(x)=1+x+x^2+x^3$ Consider the Remainder Estimation Theorem to find $|f^{4} | \leq M $. So, $|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$ and $|R_3(x)| =|f^{4} |\times |\dfrac{x^4}{4!}| =|\dfrac{x^4}{(1-x)^5}|$ Since, $|x| \lt 0.1$, we will find the maximum of $|R_3(x)| $, so $\space max |\dfrac{x^4}{(1-x)^5}|$ $|x| \lt 0.1 \implies -\dfrac{1}{10} \lt x \lt \dfrac{1}{10} $ or, $\dfrac{10}{11} \lt 1-x \lt \dfrac{9}{10} $ So, $\space max |\dfrac{x^4}{(1-x)^5}| \lt (0.1)^4 (\dfrac{10}{9})^5$ or, $\space max |\dfrac{x^4}{(1-x)^5}| \lt 0.00016935 \leq 0.00017$ So, $ f(x)=1+x+x^2+x^3$ and $ Error \leq 0.00017$
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