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.
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+x1x0=2
Step 2 — Fixed-point iteration 1
Add fixed-point iteration 1 to the table.
n1g(x)1.5000residual−0.2500
Step 3 — Fixed-point iteration 2
Add fixed-point iteration 2 to the table.
n12g(x)1.50001.6667residual−0.2500+0.1111
Step 4 — Fixed-point iteration 3
Add fixed-point iteration 3 to the table.
n123g(x)1.50001.66671.6000residual−0.2500+0.1111−0.0400
Step 5 — Fixed point
The estimate is 1.6000 and the contraction check is less than 1.
x≈1.6000∣g′∣<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.