Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 349: 16

Answer

$$ 3 \sin x=x, $$ so $$ f(x)=3 \sin x-x $$ $ \Rightarrow$ $$ f^{\prime}(x)=3 \cos x-1 $$ Equation 2 becomes $$ \begin{aligned} x_{n+1}&=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\ &=x_{n}-\frac{3 \sin x_{n}-x_{n}}{3 \cos x_{n}-1} \end{aligned} $$ From the figure, the positive root of $ f(x)=3 \sin x-x $ is near $2$ So we assume that, $\quad x_{1}=2$ $ \Rightarrow$ $$ \begin{aligned} x_{2}&=x_{1}-\frac{3 \sin x_{1}-x_{1}}{3 \cos x_{1}-1}\\ &=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\ &\approx 2.323732 \end{aligned} $$ and $$ \begin{aligned} x_{3}&=x_{2}-\frac{3 \sin x_{2}-x_{2}}{3 \cos x_{2}-1}\\ &=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\ &\approx 2.279595 \end{aligned} $$ and so on $$x_{4} \approx 2.278863 \approx x_{5}$$ . So the positive root is $2.278863,$ to six decimal places.

Work Step by Step

$$ 3 \sin x=x, $$ so $$ f(x)=3 \sin x-x $$ $ \Rightarrow$ $$ f^{\prime}(x)=3 \cos x-1 $$ Equation 2 becomes $$ \begin{aligned} x_{n+1}&=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\ &=x_{n}-\frac{3 \sin x_{n}-x_{n}}{3 \cos x_{n}-1} \end{aligned} $$ From the figure, the positive root of $ f(x)=3 \sin x-x $ is near $2$ So we assume that, $\quad x_{1}=2$ $ \Rightarrow$ $$ \begin{aligned} x_{2}&=x_{1}-\frac{3 \sin x_{1}-x_{1}}{3 \cos x_{1}-1}\\ &=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\ &\approx 2.323732 \end{aligned} $$ and $$ \begin{aligned} x_{3}&=x_{2}-\frac{3 \sin x_{2}-x_{2}}{3 \cos x_{2}-1}\\ &=2-\frac{3 \sin (2)-(2)}{3 \cos (2)-1}\\ &\approx 2.279595 \end{aligned} $$ and so on $$x_{4} \approx 2.278863 \approx x_{5}$$ . So the positive root is $2.278863,$ to six decimal places.
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