Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 17

$x = 1.934563$

$$\eqalign{ & \sin x = x - 1 \cr & {\text{Let }}y = \sin x,{\text{ }}y = x - 1 \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & \sin x = x - 1 \cr & {\text{Subtract }}x - 1{\text{ from both sides of equation to write the functions}} \cr & {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr & f\left( x \right) = 0 \cr & \sin x - x + 1 = 0 \cr & \cr & {\text{Let }}f\left( x \right) = \sin x - x + 1,{\text{ and differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x - x + 1} \right] \cr & f'\left( x \right) = \cos x - 1 \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 352}}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr & {\text{form}} \cr & {x_{n + 1}} = {x_n} - \frac{{\sin {x_n} - {x_n} + 1}}{{\cos {x_n} - 1}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = 2 \cr & {\text{If we choose }}{x_1} = 2{\text{ as the initial approximation}},{\text{then we obtain}} \cr & {x_2} \approx 2 - \frac{{\sin \left( 2 \right) - \left( 2 \right) + 1}}{{\cos \left( 2 \right) - 1}} \cr & {x_2} \approx 1.93595115 \cr & {x_3} \approx 1.93456387 \cr & {x_4} \approx 1.93456321 \cr & {x_3}{\text{ and }}{x_4}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the solution of the equation }}\sin x = x - 1{\text{ is}} \cr & x = 1.934563 \cr} $$