Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 39

Answer

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

Work Step by Step

Recall that the Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ and $\sin x= x-\dfrac{x^3}{6}+\dfrac{ x^5}{120}-....$ Now, $|\dfrac{x^3}{6}| \lt |\dfrac{(10^{-3})^3}{6}| $ $ \implies |\dfrac{x^3}{6}| \lt |\dfrac{(10^{-3})^3}{6}| = \dfrac{10^{-9}}{6}$ $\implies \space Error \lt |\dfrac{(10^{-3})^3}{3 !}| \approx 1.67 \times 10^{-10}$ So, $\space Error \lt 1.67 \times 10^{-10}$
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