Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 18

$x = 0.647765$

$$\eqalign{ & \cos 2x = {x^3} \cr & {\text{Let }}y = \cos 2x,{\text{ }}y = {x^3} \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & \cos 2x = {x^3} \cr & {\text{Subtract }}x - 1{\text{ from both sides of equation to write the functions}} \cr & {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr & f\left( x \right) = 0 \cr & \cos 2x - {x^3} = 0 \cr & \cr & {\text{Let }}f\left( x \right) = \cos 2x - {x^3},{\text{ and differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos 2x - {x^3}} \right] \cr & f'\left( x \right) = - 2\sin 2x - 3{x^2} \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 352}}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr & {\text{form}} \cr & {x_{n + 1}} = {x_n} - \frac{{\cos 2{x_n} - {x_n}^2}}{{ - 2\sin 2{x_n} - 3{x_n}^2}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = 0.5 \cr & {\text{If we choose }}{x_1} = 2{\text{ as the initial approximation}},{\text{then we obtain}} \cr & {x_2} \approx 0.5 - \frac{{\cos \left( {2\left( {0.5} \right)} \right) - {{\left( {0.5} \right)}^3}}}{{ - 2\sin \left( {2\left( {0.5} \right)} \right) - 3{{\left( {0.5} \right)}^2}}} \cr & {x_2} \approx 0.670699 \cr & {x_3} \approx 0.648160 \cr & {x_4} \approx 0.647765 \cr & {x_5} \approx 0.647765 \cr & {x_4}{\text{ and }}{x_5}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the solution of the equation }}\cos 2x = {x^3}{\text{ is}} \cr & x = 0.647765 \cr} $$