Newton's Method — Question 9

Question 9

Approximate a solution to the equation: $$x^3 + x - 1 = 0$$ using Newton’s Method. Start with $x_1 = 0.5$, and compute $x_2$, $x_3$, and $x_4$, rounding each to four decimal places.

Question 9 - 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

$$\begin{aligned}x_{2}&=0.5000000000-\frac{-0.3750000000}{1.7500000000}\approx\boxed{0.7143}\\[6pt]x_{3}&=0.7142857143-\frac{0.0787172012}{2.5306122449}\approx\boxed{0.6832}\\[6pt]x_{4}&=0.6831797235-\frac{0.0020432938}{2.4002036038}\approx\boxed{0.6823}\end{aligned}$$