What condition on g' near a root guarantees convergence? [1 mark]

If |g'(α)| = 1.5 at a root α, what happens to the iteration? [1 mark]

SEAB Original-style practice: The equation x^3 + x - 1 = 0 has a root near x = 0.7. Show it can be rearranged as x = 11 + x^2, and use the iteration x_n+1 = 11 + x_n^2 with x_0 = 0.7 to find the root to three decimal places.

Rearrange x^3 + x - 1 = 0: factor x from the first two terms is not direct, so instead write x(x^2 + 1) = 1, giving x = 11 + x^2, the required form. Iterate x_n+1 = 11 + x_n^2 from x_0 = 0.7: $x_1 = (1)/(1 + 0.49) = 0.67114, x_2 = (1)/(1 + 0.45043) = 0.68945,$ $x_3 = (1)/(1 + 0.47534) = 0.67782, x_4 = (1)/(1 + 0.45944) = 0.68517,$ continuing until the values settle. They converge to x ≈ 0.682 (3 d.p.). Markers reward the rearrangement to x = 11 + x^2, correct iterates, and the converged root 0.682.

SingaporeFurther MathsSyllabus dot point

How does rearranging an equation into the form x = g(x) give an iterative method, and what controls whether it converges?

Solve an equation by fixed-point iteration of the form x = g(x), and use the derivative condition to decide convergence

A focused answer to the H2 Further Mathematics outcome on fixed-point iteration. Rearranging f(x) = 0 into x = g(x), the iteration x_{n+1} = g(x_n), the staircase and cobweb diagrams, and the convergence condition that the magnitude of g prime is less than one.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to solve an equation by fixed-point iteration: rearrange $\mathrm{f}(x) = 0$ into the form $x = \mathrm{g}(x)$ , iterate $x_{n+1} = \mathrm{g}(x_n)$ from a starting value, and use the derivative condition $|\mathrm{g}'(x)| < 1$ near the root to decide whether the iteration converges. You should also recognise the staircase and cobweb diagrams that picture the process.

The answer

Rearranging into x = g(x)

A root of $\mathrm{f}(x) = 0$ is a fixed point of some function $\mathrm{g}$ , that is a value $\alpha$ with $\alpha = \mathrm{g}(\alpha)$ . There are usually several ways to rearrange an equation into $x = \mathrm{g}(x)$ , and they are not equally good: some converge and some do not.

The iteration

Starting from $x_0$ , generate the sequence

x_{n+1} = \mathrm{g}(x_n).

If it converges, the limit $\alpha$ satisfies $\alpha = \mathrm{g}(\alpha)$ and so is a root of the original equation.

The convergence condition

The iteration converges to a fixed point $\alpha$ provided, near $\alpha$ ,

|\mathrm{g}'(x)| < 1.

The reason: each step multiplies the error $x_n - \alpha$ by approximately $\mathrm{g}'(\alpha)$ , so a magnitude below $1$ shrinks the error geometrically, while a magnitude above $1$ enlarges it and the iteration diverges. The smaller $|\mathrm{g}'(\alpha)|$ is, the faster the convergence.

Staircase and cobweb diagrams

Plotting $y = x$ and $y = \mathrm{g}(x)$ , the iteration is read by bouncing between the two curves. When $0 < \mathrm{g}'(\alpha) < 1$ the path is a staircase converging monotonically; when $-1 < \mathrm{g}'(\alpha) < 0$ it is a cobweb spiralling in with alternating sides. If $|\mathrm{g}'| > 1$ the staircase or cobweb moves away from the root.

Worked example

The equation $x^2 - 5x + 3 = 0$ has a root near $x = 0.7$ . Rearrange it as $x = \dfrac{x^2 + 3}{5}$ , test the convergence condition, and iterate from $x_0 = 0.7$ .

Step 1: Rearrange to x = g(x)

From $5x = x^2 + 3$ , $x = \dfrac{x^2 + 3}{5}$ , so $\mathrm{g}(x) = \dfrac{x^2 + 3}{5}$ .

Step 2: Check the convergence condition

$\mathrm{g}'(x) = \dfrac{2x}{5}$ . Near $x = 0.7$ , $|\mathrm{g}'(0.7)| = \dfrac{1.4}{5} = 0.28 < 1$ , so the iteration converges.

Step 3: First iterations

x_1 = \frac{0.49 + 3}{5} = 0.698, \quad x_2 = \frac{0.487 + 3}{5} = 0.69749.

Step 4: Continue to settle

x_3 = \frac{0.69749^2 + 3}{5} = \frac{0.48649 + 3}{5} = 0.697298,

and further iterates change only in later decimals.

Step 5: State the root

The iteration converges to $x \approx 0.6972$ (4 d.p.), the smaller root of the quadratic, consistent with the rapid convergence implied by the small $|\mathrm{g}'| = 0.28$ .

Common traps

Choosing a divergent rearrangement: Not every $x = \mathrm{g}(x)$ converges; check $|\mathrm{g}'|$ at the root and switch form if it exceeds $1$ .
Testing $\mathrm{g}'$ at the wrong place: The condition is on $|\mathrm{g}'|$ near the root, not at the starting value far away.
Confusing the fixed point with a root of $\mathrm{g}$: The root is where $x = \mathrm{g}(x)$ (intersection with $y = x$ ), not where $\mathrm{g}(x) = 0$ .
Stopping after a fixed count: Iterate until successive values agree to the required accuracy, not after an arbitrary number of steps.
Ignoring the sign of $\mathrm{g}'$: A negative derivative gives alternating (cobweb) iterates; this is normal and still converges if $|\mathrm{g}'| < 1$ .

Examples in context

Example 1. Iterating a population map. The discrete logistic map $x_{n+1} = rx_n(1 - x_n)$ is a fixed-point iteration; its fixed points and their stability (governed by $|\mathrm{g}'|$ ) determine whether a population settles, oscillates or behaves chaotically, a gateway to dynamical systems.

Example 2. Solving a transcendental equation. An equation like $x = \cos x$ is already in the form $x = \mathrm{g}(x)$ ; iterating $x_{n+1} = \cos x_n$ from any start converges to the Dottie number, since $|\mathrm{g}'| = |\sin x| < 1$ there.

Try this

Q1. State the iteration used in fixed-point iteration once the equation is in the form $x = \mathrm{g}(x)$ . [1 mark]

Cue. $x_{n+1} = \mathrm{g}(x_n)$ .

Q2. What condition on $\mathrm{g}'$ near a root guarantees convergence? [1 mark]

Cue. $|\mathrm{g}'(x)| < 1$ .

Q3. If $|\mathrm{g}'(\alpha)| = 1.5$ at a root $\alpha$ , what happens to the iteration? [1 mark]

Cue. It diverges, because the error is multiplied by more than $1$ each step.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksThe equation

x^3 + x - 1 = 0

has a root near

x = 0.7

. Show it can be rearranged as

x = \dfrac{1}{1 + x^2}

, and use the iteration

x_{n+1} = \dfrac{1}{1 + x_n^2}

with

x_0 = 0.7

to find the root to three decimal places.

Show worked answer →

Rearrange $x^3 + x - 1 = 0$ : factor $x$ from the first two terms is not direct, so instead write $x(x^2 + 1) = 1$ , giving $x = \dfrac{1}{1 + x^2}$ , the required form.

Iterate $x_{n+1} = \dfrac{1}{1 + x_n^2}$ from $x_0 = 0.7$ :

x_1 = \frac{1}{1 + 0.49} = 0.67114, \quad x_2 = \frac{1}{1 + 0.45043} = 0.68945,

x_3 = \frac{1}{1 + 0.47534} = 0.67782, \quad x_4 = \frac{1}{1 + 0.45944} = 0.68517,

continuing until the values settle. They converge to

x \approx 0.682

(3 d.p.).

Markers reward the rearrangement to $x = \dfrac{1}{1 + x^2}$ , correct iterates, and the converged root $0.682$ .

Original6 marksAn equation is rearranged into the form

x = \mathrm{g}(x)

. State the condition on

\mathrm{g}'(x)

near the root that guarantees the iteration

x_{n+1} = \mathrm{g}(x_n)

converges, and explain what happens when this condition fails.

Show worked answer →

The iteration $x_{n+1} = \mathrm{g}(x_n)$ converges to a root $\alpha$ (a fixed point, $\alpha = \mathrm{g}(\alpha)$ ), provided that near the root

|\mathrm{g}'(x)| < 1.

The factor

\mathrm{g}'(\alpha)

controls how the error shrinks: each step multiplies the error by approximately

\mathrm{g}'(\alpha)

, so

|\mathrm{g}'| < 1

makes the error decrease and the iteration converge.

If $|\mathrm{g}'(x)| > 1$ near the root, each step multiplies the error by a factor larger than one, so the error grows and the iteration diverges away from the root, no matter how close the start. (If $\mathrm{g}'$ is negative the iterates alternate sides of the root, spiralling in a cobweb if $|\mathrm{g}'| < 1$ , or out if $|\mathrm{g}'| > 1$ .)

Markers reward the condition $|\mathrm{g}'(x)| < 1$ for convergence, the error-multiplier explanation, and the statement that $|\mathrm{g}'| > 1$ causes divergence.