Iteration formulae

Objective. Generate approximations from iterative rules.

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2. Worked example
3. Practice
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Learn: the key idea

Iteration formulae generate a sequence of increasingly accurate approximations to a solution by repeatedly substituting an output back in as the next input, written xₙ₊₁ = g(xₙ), starting from a given value x₀. Each iteration applies the rearranged formula to the previous approximation, and after several iterations the values often settle towards a fixed decimal, which approximates a root of the original equation. These formulae are derived by rearranging an equation f(x)=0 into the form x=g(x). For example, iterating x_{n+1}=√(3xₙ+4) from x₀=2 gives x₁=√10≈3.162, then x₂=√(3×3.162+4)≈3.606, converging towards a root.

Worked example

Use the iteration formula x_{n+1} = (x_n² + 6)/5 with x₀ = 1 to find x₂, giving your answer to 3 decimal places.

1 $Find x₁ using x₀=1: x₁ = (1²+6)/5 = 7/5 = 1.4$
2 $Use x₁=1.4 to find x₂: x₂ = (1.4²+6)/5$
3 $Compute 1.4² = 1.96, so x₂ = (1.96+6)/5 = 7.96/5$
4 $Divide: 7.96/5 = 1.592$

Answer: x₂ = 1.592

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Common mistake. Pupils sometimes substitute the original starting value x₀ into every iteration instead of using the previous answer, repeatedly calculating g(x₀) instead of g(x₁), g(x₂); each new term must use the immediately preceding result.

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