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: 39


Error $\lt 1.67 \times 10^{-10}$

Work Step by Step

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ and $\sin x= x-\dfrac{x^3}{6}+\dfrac{ x^5}{120}-....$ We need to find $ x $. $|\dfrac{x^3}{6}| \lt |\dfrac{(10^{-3})^3}{6}| $ or, $|\dfrac{x^3}{6}| \lt |\dfrac{(10^{-3})^3}{6}| = \dfrac{10^{-9}}{6}$ or, Error $\lt |\dfrac{(10^{-3})^3}{3 !}| \approx 1.67 \times 10^{-10}$ or, Error $\lt 1.67 \times 10^{-10}$
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