Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 317: 36

$${x_1} = - 1,{\text{ }}{x_2} \approx - 0.683550,{\text{ }}{x_3} \approx 1.183555$$

$$\eqalign{ & f\left( x \right) = \frac{{{x^5}}}{5} - \frac{{{x^3}}}{4} - \frac{1}{{20}} \cr & {\text{Let }}f\left( x \right) = 0 \cr & \frac{{{x^5}}}{5} - \frac{{{x^3}}}{4} - \frac{1}{{20}} = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^5}}}{5} - \frac{{{x^3}}}{4} - \frac{1}{{20}}} \right] \cr & f'\left( x \right) = {x^4} - \frac{3}{4}{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 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{{\frac{{x_n^5}}{5} - \frac{{x_n^3}}{4} - \frac{1}{{20}}}}{{x_n^4 - \frac{3}{4}x_n^2}} \cr & {\text{Multiply the numerator and denominator by 20}} \cr & {x_{n + 1}} = {x_n} - \frac{{4x_n^5 - 5x_n^3 - 1}}{{20x_n^4 - 15x_n^2}} \cr & {\text{From the graph we can see that the first exact root is}} \cr & {x_1} = - 1 \cr & \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = - 0.7 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = - 0.7 \cr & {x_1} = \left( { - 0.7} \right) - \frac{{4{{\left( { - 0.7} \right)}^5} - 5{{\left( { - 0.7} \right)}^3} - 1}}{{20{{\left( { - 0.7} \right)}^4} - 15{{\left( { - 0.7} \right)}^2}}} = - 0.683233 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx - 0.683550 \cr & {x_3} \approx - 0.683550 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 1.2 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 1.2 \cr & {x_1} = \left( {1.2} \right) - \frac{{4{{\left( {1.2} \right)}^5} - 5{{\left( {1.2} \right)}^3} - 1}}{{20{{\left( {1.2} \right)}^4} - 15{{\left( {1.2} \right)}^2}}} = 1.184235 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 1.183551 \cr & {x_3} \approx 1.183555 \cr & {x_4} \approx 1.183555 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} = - 1,{\text{ }}{x_2} \approx - 0.683550,{\text{ }}{x_3} \approx 1.183555 \cr} $$