Fixed-Point Iteration - Numerical Methods Step by Step

A root of f(x)=0 can sometimes be found as the fixed point of a rearrangement g(x)=x. Here g(x)=1+1/x has the golden ratio as its fixed point (g(x)=x ↔ x²-x-1=0). Starting from x0=2, three iterates yield x3=1.6000, with |f(x3)|=0.04 < 0.1.

fixed-point-iteration Rewrite f(x)=0 as x=g(x). If |g'(x)| < 1 near the fixed point, the iteration x_{n+1}=g(x_n) converges. For g(x)=1+1/x near the golden ratio ≈1.618, |g'|≈0.38 < 1, guaranteeing convergence.

Example

Iterate x = g(x) and track the residual.

highlighted = computed this step

Step 1 — Set up

Start fixed-point iteration from x 0 equals 2.

g(x)= 1+\frac{1}{x} \quad x_{ 0 }= 2

Step 2 — Fixed-point iteration 1

Add fixed-point iteration 1 to the table.

\begin{array}{ccc}\text{n} & \text{g(x)} & \text{residual} \\ \hlmath{1} & \hlmath{1.5000} & \hlmath{-0.2500}\end{array}

Step 3 — Fixed-point iteration 2

Add fixed-point iteration 2 to the table.

\begin{array}{ccc}\text{n} & \text{g(x)} & \text{residual} \\ 1 & 1.5000 & -0.2500 \\ \hlmath{2} & \hlmath{1.6667} & \hlmath{+0.1111}\end{array}

Step 4 — Fixed-point iteration 3

Add fixed-point iteration 3 to the table.

\begin{array}{ccc}\text{n} & \text{g(x)} & \text{residual} \\ 1 & 1.5000 & -0.2500 \\ 2 & 1.6667 & +0.1111 \\ \hlmath{3} & \hlmath{1.6000} & \hlmath{-0.0400}\end{array}

Step 5 — Fixed point

The estimate is 1.6000 and the contraction check is less than 1.

x\approx \hlmath{1.6000} \quad |g'|< 1