Iteration Methods — Edexcel A-level Mathematics (Pure)

SPEC 9.2 • NUMERICAL METHODS

Fixed Point Iteration

Using iterative formulas and graphical interpretation with staircase and cobweb diagrams

What is Fixed Point Iteration?

Fixed point iteration is a method for finding roots by repeatedly applying a formula until the values converge to a solution.

Definition

A fixed point of a function g(x) is a value α where α = g(α). At this point, applying g doesn't change the value.

Key Idea

To solve f(x) = 0, we rearrange to the form x = g(x), then iterate:

x_{n + 1} = g (x_{n})

Starting from x₀, we generate x₁, x₂, x₃, ... hoping they converge to the root.

Rearranging to x = g(x)

The same equation can often be rearranged in multiple ways:

Example

For x³ − 3x − 5 = 0, we could rearrange to:

x = \frac{x ^{3} - 5}{3}

→ gives

g (x) = \frac{x ^{3} - 5}{3}

x = 3 3 x + 5

→ gives

g (x) = 3 3 x + 5

x = \frac{5}{x ^{2} - 3}

→ gives

g (x) = \frac{5}{x ^{2} - 3}

Important

Not all rearrangements will converge! The choice of g(x) determines whether iteration succeeds or fails.

The Iteration Process

Starting from an initial estimate x₀, we repeatedly apply the formula:

Process

Step 1: Choose starting value x₀ (often from sign change method)

Step 2: Calculate x₁ = g(x₀)

Step 3: Calculate x₂ = g(x₁)

Step 4: Continue until values stabilise to required accuracy

Convergence

Values get closer together:

x₀ → x₁ → x₂ → x₃ → α

SUCCESS!

Divergence

Values grow or oscillate wildly:

x₀ → x₁ → x₂ → ∞

FAILURE!

Exam Technique & Calculator

Exam Technique

Show substitution for the FIRST iteration only!

Write: x₁ = ∛(3 × 2 + 5) = ∛11 = 2.2239...

After that, you can just list values: x₂ = 2.2679..., x₃ = 2.2761...

This gets you the method mark for showing understanding!

Calculator Shortcut

Quick iterations using ANS button:

1

Type your starting value: 2 =

2

Type the formula using ANS: ³√ ( 3 × ANS + 5 )

3

Press = repeatedly to get x₁, x₂, x₃, ...

2.277929...

Each press of = gives the next iteration!

Worked Example01

The equation x³ − 3x − 5 = 0 can be rearranged to x = ∛(3x + 5). Using x₀ = 2, find the root correct to 3 decimal places.

Using

x_{n + 1} = 3 3 x_{n} + 5

with x₀ = 2:

Show first substitution: x₁ = ∛(3 × 2 + 5) = ∛11 = 2.2239...

Answerα = 2.278 (3 d.p.)

Graphical Interpretation

The solution to x = g(x) occurs where the curve y = g(x) intersects the line y = x.

Why This Works

At intersection (α, α): the y-value equals both x (from line) and g(x) (from curve), so x = g(x) ✓

Staircase Diagrams

When 0 < g'(α) < 1 (positive but shallow gradient), iterations form a staircase pattern.

How to Draw

Vertical to curve (dashed) → Horizontal to line → Repeat. Steps get smaller, approaching α from one side.

Cobweb Diagrams

When −1 < g'(α) < 0 (negative but shallow gradient), iterations form a cobweb pattern.

Pattern

Iterations alternate sides of the root, spiralling inward like a spider's web. Each "loop" is smaller.

Worked Example02

For x = ∛(3x + 5), sketch the staircase diagram showing convergence from x₀ = 2 to the root.

Step 1: Draw y = x and y = ∛(3x + 5) on same axes

Step 2: Mark x₀ = 2 on x-axis

Step 3: Draw vertical to curve, horizontal to line, repeat

Pattern: Converging staircase (g'(α) ≈ 0.26 > 0, |g'(α)| < 1)

Worked Example03

The equation x = 2 − eˣ/3 has a root near x = 1. With x₀ = 1, perform 4 iterations and sketch the cobweb diagram.

Show first: x₁ = 2 − e¹/3 = 2 − 0.906... = 1.094

Pattern: Cobweb (oscillating) — values alternate: 1, 1.09, 1.00, 1.09...

g'(x) = −eˣ/3 < 0 confirms negative gradient (cobweb)

Divergence: When |g'(α)| > 1

If the curve y = g(x) is steeper than y = x at the root, iterations diverge.

Diverging Staircase

g'(α) > 1

Diverging Cobweb

g'(α) < −1

Failure

Each iteration amplifies the error. This rearrangement is unsuitable — try a different one!

The Convergence Condition

Key Condition

∣ g^{'} (α) ∣ < 1

For iteration to converge, the magnitude of the gradient at the root must be less than 1.

ConditionPatternResult

0 < g'(α) < 1Staircase✓ Converges

−1 < g'(α) < 0Cobweb✓ Converges

g'(α) > 1Staircase✗ Diverges

g'(α) < −1Cobweb✗ Diverges

Exam Tip

To check convergence: find g'(x), substitute approximate root, check if |answer| < 1

Worked Example04

The equation x³ − 3x − 5 = 0 can be written as x = (x³ − 5)/3. Show that this rearrangement does NOT converge near x = 2.

Step 1: Identify g(x) = (x³ − 5)/3

Step 2: Differentiate: g'(x) = 3x²/3 = x²

Step 3: At x ≈ 2.28: g'(2.28) = (2.28)² ≈ 5.2

Step 4: Since |g'(α)| ≈ 5.2 > 1, this rearrangement diverges.

Conclusion: This rearrangement is unsuitable. The gradient is too steep!

Compare: x = ∛(3x + 5) gives g'(x) = 1/(∛(3x+5)²) ≈ 0.26 < 1 ✓

Worked Example05

For the equation eˣ = 3x, compare the rearrangements x = eˣ/3 and x = ln(3x). Which converges to the root near x = 1.5?

Rearrangement 1: x = eˣ/3

g₁'(x) = eˣ/3. At x ≈ 1.5: g₁'(1.5) = e¹·⁵/3 ≈ 1.49 > 1 ✗

Rearrangement 2: x = ln(3x)

g₂'(x) = 1/x. At x ≈ 1.5: g₂'(1.5) = 1/1.5 ≈ 0.67 < 1 ✓

AnswerUse x = ln(3x) because |g'(α)| < 1 near the root.

Practice

Question 1: The equation x³ + x − 1 = 0 can be written as x = 1/(x² + 1). Starting with x₀ = 0.5, find the root correct to 3 decimal places.

Show first: x₁ = 1/(0.5² + 1) = 1/1.25 = 0.8

Continue iterations: x₂ = 0.610, x₃ = 0.729, x₄ = 0.653, x₅ = 0.701...

Values oscillate: 0.5, 0.8, 0.61, 0.73, 0.65, 0.70, 0.67, 0.69, 0.68...

Pattern: Cobweb — g'(x) = −2x/(x²+1)² < 0 at root

Answerα = 0.683 (3 d.p.)

Practice

Question 2: For the iteration xₙ₊₁ = 2 + 1/xₙ with x₀ = 2, find x₄. State whether this is a staircase or cobweb pattern and justify your answer.

Show first: x₁ = 2 + 1/2 = 2.5

x₂ = 2 + 1/2.5 = 2.4

x₃ = 2 + 1/2.4 = 2.4167

x₄ = 2 + 1/2.4167 = 2.414

Pattern check: g'(x) = −1/x²

At x ≈ 2.4: g'(2.4) = −1/5.76 ≈ −0.17

Since g'(α) < 0 and |g'(α)| < 1 → Converging cobweb

Practice

Question 3: The equation x = cos x has a root α ≈ 0.739. For the iteration xₙ₊₁ = cos(xₙ), determine whether the iteration converges and identify the diagram type.

Step 1: g(x) = cos x

Step 2: g'(x) = −sin x

Step 3: At α ≈ 0.739 (radians): g'(0.739) = −sin(0.739) ≈ −0.674

Step 4: |g'(α)| = 0.674 < 1 ✓

Since g'(α) < 0, this forms a cobweb diagram.

Try it: Press cos repeatedly on your calculator — it always converges to 0.739!

Summary

Fixed Point Iteration

ConceptKey Point

Iteration Formula

x_{n + 1} = g (x_{n})

Convergence Condition

∣ g^{'} (α) ∣ < 1

near the root

Staircase DiagramWhen 0 < g'(α) < 1

Cobweb DiagramWhen −1 < g'(α) < 0

DivergenceWhen |g'(α)| > 1

Remember for Exams

Show first substitution for method marks, then list values

Use ANS button for quick calculator iterations

Draw diagrams: vertical to curve, horizontal to line

Check convergence: find g'(x), evaluate at root, check |g'| < 1