Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.7 Newton's Method - Exercises - Page 219: 5


\begin{aligned} x_{1}& \approx 0.2854036\\ x_{2}& \approx 0.24288009\\ x_{3}& \approx 0.24267569 \end{aligned}

Work Step by Step

Given $$ f(x)=\cos x-4 x, \quad x_{0}=1 $$ Since $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ Then \begin{align*} x_{n+1}&= x_n -\frac{\cos x_n-4 x_n}{-\sin x_n -4}\\ &=\frac{x_{n} \sin x_{n}+\cos x_{n}}{\sin x_{n}+4} \end{align*} Hence \begin{aligned} x_{1}&=\frac{x_{0} \sin x_{0}+\cos x_{0}}{\sin x_{0}+4}=\frac{(1) \sin (1)+\cos (1)}{\sin (1)+4} \approx 0.2854036\\ x_{2}&=\frac{x_{1} \sin x_{1}+\cos x_{1}}{\sin x_{1}+4}=\frac{(0.2854) \sin (0.2854)+\cos (0.2854)}{\sin (0.2854)+4} \approx 0.24288009\\ x_{3}&=\frac{x_{1} \sin x_{1}+\cos x_{1}}{\sin x_{1}+4} =\frac{(0.24288) \sin (0.24288)+\cos (0.24288)}{\sin (0.24288)+4} \approx 0.24267569 \end{aligned}
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