Learn9MA0Paper 1 · Pure Mathematics 1Numerical methodsFixed-point iteration

Numerical methods · Subtopic 2 of 4

Fixed-Point Iteration

Feed a number into a formula, feed the answer back in, and repeat. When the values settle down they settle on a root. This lesson is the engine of every P2 and P3 numerical question.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $x_{n+1}=F(x_n)$ from a given starting value, recording each iterate to enough decimal places.
Decide when to stop, and give the root to the accuracy asked for.
Justify the accuracy of the answer with a change of sign.
Use the calculator's answer key so that no accuracy is lost between steps.

Pause and think

Put your calculator in radians, type $1$ , then press $\cos$ over and over. What happens?

Section 2 of 14

Core idea

Start hereFeed the answer back in

Type any number into a calculator and press $\sqrt{\ }$ then $+2$ then $\sqrt{\ }$ then $+2$ ... repeatedly. Whatever you start with, the display settles on $2$ . You've just solved $x^2-x-2=0$ by iteration: using each answer as the next input.

The big idea

Rearrange $f(x)=0$ as $x=g(x)$ and iterate $x_{n+1}=g(x_n)$ . If the sequence converges to $\alpha$ , then $\alpha=g(\alpha)$ , so $\alpha$ is a root.

1. The idea

Why the limit is a root

$x_n\to\alpha\ \text{and}\ x_{n+1}\to\alpha$
Both sides approach the same limit.
$x_{n+1}=g(x_n)\Rightarrow\alpha=g(\alpha)$
Take limits ( $g$ continuous).
$\alpha=g(\alpha)\iff f(\alpha)=0$
$x=g(x)$ was a rearrangement of $f(x)=0$ .

Staircase for x(n+1) = √(x(n) + 2), starting at 0

Go up to the curve (that is

g(x_n)

), across to the line

y=x

(that makes it the new

x

), up to the curve again. The steps climb towards where the two graphs meet.

Worked example $x_{n+1}=\sqrt{x_n+2}$ , $x_0=0$

Iterate and identify the equation solved.

Goal: Limit.
Strategy: Calculator with ANS.

$x_1=1.4142,\ x_2=1.8478,\ x_3=1.9616,\ x_4=1.9904,\ x_5=1.9976$
Iterate.
$\alpha=\sqrt{\alpha+2}\Rightarrow\alpha^2=\alpha+2$
Limit equation.

AnswerConverges to

2

, a root of

x^2-x-2=0

✓ Check

4-2-2=0

✓.

Worked example $x^3-2x-5=0$ via $x_{n+1}=\sqrt[3]{2x_n+5}$

Find the root to 3 d.p. from $x_0=2$ .

Goal: Root.
Strategy: Check the rearrangement, iterate until stable.

$x^3=2x+5\Rightarrow x=\sqrt[3]{2x+5}$
Valid rearrangement.
$2.0801,\ 2.0924,\ 2.0942,\ 2.0945,\ 2.0945$
Iterate.

Answer

2.095

(3 d.p.).

✓ Check

f(2.0945)\approx-0.0006\approx0

✓.

Common misconceptionNot every rearrangement works Some rearrangements diverge (next lesson). If the values run away, try a different

g

Pause and think

What equation does $x_{n+1}=\frac{x_n}2+\frac3{x_n}$ solve?

Summary

$x=g(x)$ ; iterate.
Limit satisfies $\alpha=g(\alpha)$ .

Iteration: Repeatedly applying a formula to its own output.

HistoryTwo thousand years of iteration. Heron of Alexandria, writing in the first century AD, found square roots like this: to find

\sqrt{720}

, guess

27

, then average

27

and

\tfrac{720}{27}

, and repeat. Babylonian clay tablets suggest the method may be far older. It is the iteration

x_{n+1}=\tfrac12\left(x_n+\tfrac{N}{x_n}\right)

, which you will meet again in this topic.

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