Newton's Method — Question 4

Question 4

Use Newton’s Method to approximate a root of the equation: $$x^3 - x - 1 = 0$$ Start with an initial guess of $x_1 = 1$, and compute the next three iterations: $x_2$, $x_3$, and $x_4$, rounding answers to four decimal places.

Question 4 - Solution

Newton’s iteration is

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

Keeping full precision internally gives

$$\begin{aligned}x_{2}&=1.0000000000-\frac{-1.0000000000}{2.0000000000}\approx\boxed{1.5000}\\[6pt]x_{3}&=1.5000000000-\frac{0.8750000000}{5.7500000000}\approx\boxed{1.3478}\\[6pt]x_{4}&=1.3478260870-\frac{0.1006821731}{4.4499054820}\approx\boxed{1.3252}\end{aligned}$$