Newton's Method — Question 6

Question 6

Use Newton’s Method to approximate a root of the equation: $$x^3 - 2x - 5 = 0$$ Start with an initial guess of $x_1 = 2$, and compute the next three approximations: $x_2$, $x_3$, and $x_4$. Round answers to four decimal places.

Question 6 - Solution

Newton’s iteration is

$$x_{n+1}=x_n-\frac{f(x_n)}{f\prime(x_n)},\qquad f(x)=x^3-2x-5,\quad f\prime(x)=3x^2-2.$$

Keeping full precision internally gives

$$\begin{aligned}x_{2}&=2.0000000000-\frac{-1.0000000000}{10.0000000000}\approx\boxed{2.1000}\\[6pt]x_{3}&=2.1000000000-\frac{0.0610000000}{11.2300000000}\approx\boxed{2.0946}\\[6pt]x_{4}&=2.0945681211-\frac{0.0001857232}{11.1616468418}\approx\boxed{2.0946}\end{aligned}$$