Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 33

$${x_1} \approx - 0.335408,{\text{ }}{x_2} \approx 1.333057$$

$$\eqalign{ & f\left( x \right) = \cos 2x - {x^2} + 2x \cr & f\left( x \right) = 0 \cr & \cos 2x - {x^2} + 2x = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos 2x - {x^2} + 2x} \right] \cr & f'\left( x \right) = - \sin 2x - 2x + 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 311}}} \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 \left( {2{x_n}} \right) - x_n^2 + 2{x_n}}}{{ - \sin \left( {2{x_n}} \right) - 2{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{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = - 0.5 \cr & {x_1} \approx - 0.5 - \frac{{\cos \left( {2\left( { - 0.5} \right)} \right) - {{\left( { - 0.5} \right)}^2} + 2\left( { - 0.5} \right)}}{{ - \sin \left( {2\left( { - 0.5} \right)} \right) - 2\left( { - 0.5} \right) + 2}} \approx - 0.325253 \cr & n = 1,{\text{ }}{x_1} = - 0.325253 \cr & {x_2} \approx - 0.325253 \cr & - \frac{{\cos \left( {2\left( { - 0.325253} \right)} \right) - {{\left( { - 0.325253} \right)}^2} + 2\left( { - 0.325253} \right)}}{{ - \sin \left( {2\left( { - 0.325253} \right)} \right) - 2\left( { - 0.325253} \right) + 2}} \cr & {x_2} \approx - 0.337378 \cr & {\text{Computing the same calculation we obtain:}} \cr & {x_3} \approx - 0.335038 \cr & {x_4} \approx - 0.335477 \cr & {x_5} \approx - 0.335395 \cr & {x_6} \approx - 0.335410 \cr & {x_7} \approx - 0.335407 \cr & {x_8} \approx - 0.335408 \cr & {x_9} \approx - 0.335408 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 1.5 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = - 0.5 \cr & {x_1} \approx 1.5 - \frac{{\cos \left( {2\left( {1.5} \right)} \right) - {{\left( {1.5} \right)}^2} + 2\left( {1.5} \right)}}{{ - \sin \left( {2\left( {1.5} \right)} \right) - 2\left( {1.5} \right) + 2}} \approx 1.289686 \cr & {\text{Computing the same calculation we obtain:}} \cr & {x_2} \approx 1.352618 \cr & {x_3} \approx 1.325454 \cr & {x_4} \approx 1.336209 \cr & {x_5} \approx 1.331782 \cr & {x_6} \approx 1.333577 \cr & {x_7} \approx 1.332844 \cr & {x_8} \approx 1.333143 \cr & {x_9} \approx 1.333021 \cr & {x_{10}} \approx 1.333071 \cr & {x_{11}} \approx 1.333057 \cr & {x_{12}} \approx 1.333057 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} \approx - 0.335408,{\text{ }}{x_2} \approx 1.333057 \cr} $$