Fixed-Point Iteration
Solve f(x) = 0 by rearranging to x = g(x) — elegant but convergence depends on the rearrangement.
Try Fixed-Point — No Sign-in NeededWhat is the Fixed-Point Iteration?
Fixed-point iteration solves equations of the form x = g(x) by repeatedly applying the function g: starting from an initial guess x₀, compute x₁ = g(x₀), then x₂ = g(x₁), and so on. A value x* where x* = g(x*) is called a fixed point. To use this method for a root-finding problem f(x) = 0, you rearrange the equation into the form x = g(x). For example, x² − 2 = 0 can be rewritten as x = √(x + 2) by adding x + 2 to both sides (with some algebra), giving g(x) = √(x + 2).
The elegance of fixed-point iteration lies in its simplicity — the update rule is just x ← g(x), requiring only function evaluations. Its key weakness is that convergence is not automatic: it depends entirely on the choice of g(x). The Banach Fixed-Point Theorem gives the fundamental condition: the iteration converges to a fixed point x* if and only if |g′(x*)| < 1. If this condition fails (|g′| > 1), the iterates diverge.
This sensitivity to rearrangement distinguishes fixed-point iteration from methods like Bisection or Newton-Raphson. The same equation can converge with one choice of g and diverge with another. Understanding this makes fixed-point analysis a key topic in numerical analysis courses.
The Formula
The iteration is self-correcting when |g′(x*)| < 1: small errors shrink at each step by a factor of |g′(x*)|. The closer |g′(x*)| is to 0, the faster the convergence.
Step-by-Step Algorithm
- 1
Rearrange f(x) = 0 into the form x = g(x).
- 2
Verify that |g′(x)| < 1 near the expected root (differentiating g gives the convergence criterion).
- 3
Choose a starting point x₀.
- 4
Compute xₙ₊₁ = g(xₙ).
- 5
Check convergence: if |xₙ₊₁ − xₙ| < tolerance, return xₙ₊₁ as the root.
- 6
Otherwise repeat from step 4.
Convergence & Behaviour
Convergence is linear with rate |g′(x*)|: after each step the error is multiplied by approximately |g′(x*)|. A rate of 0.1 means each iteration removes 90% of the remaining error; a rate of 0.9 means each iteration removes only 10%. If |g′(x*)| > 1 near the fixed point, small perturbations amplify each step and the iterates diverge. If |g′(x*)| = 1 exactly, convergence is possible but very slow and not guaranteed.
Fixed-point iteration is useful when a natural rearrangement of the equation produces a contractive g(x), or when the update formula has physical meaning (e.g., iterative solvers in physics simulations, Picard iteration for differential equations).
Avoid it if |g′(x)| ≥ 1 near the root — the iteration will diverge. In that case, try a different rearrangement or switch to Newton-Raphson or Bisection.
Worked Example
Find the fixed point of g(x) = √(x + 2) starting from x₀ = 1.
- 1.g′(x) = 1 / (2√(x+2)). At x* ≈ 2: g′(2) = 1/(2√4) = 0.25. Since 0.25 < 1, convergence is guaranteed.
- 2.x₁ = √(1 + 2) = √3 ≈ 1.7321
- 3.x₂ = √(1.7321 + 2) ≈ 1.9319
- 4.x₃ ≈ 1.9829, x₄ ≈ 1.9957, x₅ ≈ 1.9989
- 5.After ~15 iterations: x* ≈ 2.0000
Result: Fixed point at x = 2 (satisfying x² − x − 2 = 0). Converged in ~15 iterations with rate ≈ 0.25.
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