Iteration — Closing in on a Solution

Grade 7

The idea behind iteration

Iteration is a method for solving equations that you cannot rearrange exactly. You start with an initial estimate, feed it into a recurrence relation (a formula of the form xₙ₊₁ = f(xₙ)), and the outputs get closer and closer to the true solution with each step.

The iteration converges when the values stop changing significantly — your estimate has homed in on the solution.

Worked example

EXAMPLE

Use the iteration xₙ₊₁ = √(xₙ + 5) with x₀ = 2 to find a solution to 2 decimal places.

x₀ = 2. x₁ = √(2 + 5) = √7 ≈ 2.6458. x₂ = √(2.6458 + 5) = √7.6458 ≈ 2.7651. x₃ = √(2.7651 + 5) = √7.7651 ≈ 2.7866. x₄ = √(2.7866 + 5) ≈ 2.7905. x₅ ≈ 2.7912. Both x₄ and x₅ round to 2.79, so the solution is x ≈ 2.79.

How many decimal places to carry

Carry more decimal places in intermediate steps than you need in the final answer. Round only at the very end — rounding too early introduces errors that compound with each iteration.

What equation does the iteration solve?

The recurrence relation xₙ₊₁ = √(xₙ + 5) converges to a fixed point where x = √(x + 5). Squaring: x² = x + 5, so x² − x − 5 = 0. This is the equation being solved. Examiners often ask you to show this connection by rearranging the given equation to match the iteration formula.

Try this one

Show that the equation \( x^3 - 6x + 2 = 0 \) can be rearranged to give the iterative formula

\[ x = \sqrt[3]{6x - 2} \]

Nova's hint:

Get the cubed term on its own first.

Ask yourself: 'what do I move across so only \( x^3 \) remains on one side?' Then reverse the cubing.

Try to Solve it with Nova?