Iteration and numerical methods (Higher tier)

What is iteration?

Iteration applies a formula repeatedly. Each output becomes the next input.

The notation: xₙ₊₁ = f(xₙ).

• x₀ is the starting value (given).
• x₁ = f(x₀), x₂ = f(x₁), and so on.

This produces a sequence (x₀, x₁, x₂, …). If it converges, it tends towards a solution of x = f(x).

Why use it?

Many equations cannot be solved exactly with algebra (cubic, transcendental). Iteration approximates a root by zooming in.

Setting up an iteration formula

Start from the equation, e.g. x³ + x − 7 = 0. Rearrange to x = f(x). For example: x = (7 − x)^(1/3) or x³ = 7 − x → take cube root.

Different rearrangements give different iteration sequences — some converge, some don't.

Carrying out iterations

Use the calculator's ANS key (or a memory) to feed each output back in.

Example: xₙ₊₁ = (3 + xₙ)^(1/2), x₀ = 2.

• x₁ = √5 = 2.236067977…
• x₂ = √(3 + 2.236…) = 2.287…
• x₃ = √5.287… = 2.299…
• x₄ = 2.302…
• x₅ = 2.303…

Converging to ≈ 2.30 (3 sf).

Showing convergence to required accuracy

A standard CCEA Higher question: "Use the iteration to find a root correct to 3 sf — show enough iterations to justify your answer."

Iterate until two consecutive values agree to 1 more decimal place than required, then round.

Sign-change method (locating roots)

If f(a) and f(b) have opposite signs and f is continuous on [a, b], there is at least one root between a and b.

Example: f(x) = x³ − 7x + 1. f(2) = 8 − 14 + 1 = −5; f(3) = 27 − 21 + 1 = 7. Opposite signs → root in (2, 3).

Common CCEA exam tip

Show at least 4 iterations to justify rounding to 3 sf. Don't truncate — keep extra decimals throughout, only rounding at the final answer.