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: 41

Answer

$\lt 1.87 \times 10^{-4}$

Work Step by Step

The Taylor series for $ e^x $ can be defined as: $ e^x=1+x +\dfrac{x^2}{2}+\dfrac{ x^3}{6}-....$ 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)| \leq |\dfrac{e^ex^3}{3!}| $ So, Error $ \lt \dfrac{3^{0.1} \cdot (01)^{3} }{3!} \lt 1.87 \times 10^{-4}$
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