Newton's Method

1 min read Last updated Sat Jun 27 2026 08:46:52 GMT+0000 (Coordinated Universal Time)

A special case of fixed point method.

Given a root-finding problem f(x)=0f(x) = 0, define:

g(x)=xf(x)f(x)g(x) = x - \frac{f(x)}{f'(x)}

Start from p0p_0 and iterate: pn+1=g(pn)p_{n+1} = g(p_n). The sequence converges to a root of ff.

Order of convergence is 2 for simple roots (quadratic convergence).

Failure Cases

  • f(pn)=0f'(p_n) = 0: division by zero; method fails at that step.
  • p0p_0 far from root: sequence may diverge or cycle between values.
  • Multiple root (order >1> 1): convergence degrades to linear (order 1).

Example

Find a root of f(x)=x22f(x) = x^2 - 2 starting from p0=1p_0 = 1.

f(x)=2xf'(x) = 2x, so g(x)=x(x22)/(2x)g(x) = x - (x^2-2)/(2x).

nnpnp_n
01.000000
11.500000
21.416667
31.414216
41.414214

Converges to 21.414214\sqrt{2} \approx 1.414214 in 4 steps.

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