Newton's Method — Question 10

Question 10

Consider the equation $$x^3-3x+1=0.$$

• Explain why Newton’s method fails with the initial guess $x_1=1$.

• Restart with $x_1=0$, and compute $x_2$, $x_3$, and $x_4$. Round the answers to four decimal places.

Question 10 - Solution

At $x_1=1$, $f(1)=-1$ and $f\prime(1)=0$, so the Newton step is undefined. Restart at $x_1=0$. Newton’s iteration is

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

Keeping full precision internally gives

$$\begin{aligned}x_{2}&=0.0000000000-\frac{1.0000000000}{-3.0000000000}\approx\boxed{0.3333}\\[6pt]x_{3}&=0.3333333333-\frac{0.0370370370}{-2.6666666667}\approx\boxed{0.3472}\\[6pt]x_{4}&=0.3472222222-\frac{0.0001955804}{-2.6383101852}\approx\boxed{0.3473}\end{aligned}$$