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 \( x^3 - 3x + 1 = 0 \) has a root between \( x = 1 \) and \( x = 2 \). [2]

Nova's hint:

This is a change of sign method question — you need to show the equation changes from positive to negative (or vice versa) in the given interval.

Ask yourself: 'What happens when I substitute both \( x = 1 \) and \( x = 2 \) into the expression — do I get one positive and one negative value?'

For example, to show a root between \( x = 0 \) and \( x = 1 \) for \( f(x) = x^2 - x - 0.5 \), calculate \( f(0) \) and \( f(1) \) and check the signs differ.

Try to Solve it with Nova?