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: 49


a) $ Q(x)=1+kx+\dfrac{k(k-1)}{2}x^2$ b) $0 \lt x \lt 0.21544$

Work Step by Step

a) The Taylor series for $ f(x)$ at $ x=0$ is as follows: $ Q(x)=f(0)+f^{,}(x-0)+\dfrac{f^{,}(0)}{2!}(x-0)^2=1+kx+\dfrac{k(k-1)}{2}x^2$ b) Consider the Remainder Estimation Theorem to find the Error. So, $|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$ and $|R_2(x)| =|\dfrac{ 3! x^3}{3 !}|=|x^3|$ Since, $|x| \lt \dfrac{1}{100}$, thus, $|x|^3 \lt \dfrac{1}{100}$ or, $\dfrac{10}{11} \lt 1-x \lt \dfrac{9}{10} $ or, $0 \lt x \lt \dfrac{1}{(100)^{1/3}}$ or, $0 \lt x \lt 0.21544$ Thus, the required values of $ x $ are $0 \lt x \lt 0.21544$ . a) $ Q(x)=1+kx+\dfrac{k(k-1)}{2}x^2$ b) $0 \lt x \lt 0.21544$
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