Iteration Method –

The iteration method is a Higher-tier technique for finding approximate solutions to equations that cannot be solved algebraically. By applying a formula repeatedly, each answer gets closer to the true solution. This guide focuses on applying the iterative formula, demonstrating rearrangements, and determining solutions to a given number of decimal places.

What Is the Iteration Method?

Iteration means "repeated application". You start with an initial estimate x₀ and feed it into a formula to produce x₁, then feed x₁ back in to get x₂, and so on. Each value is a better approximation of the solution.

Key Formulas

The iterative formula comes from rearranging the original equation into the form x = f(x). For example, x³ − 5x − 3 = 0 can be rearranged to x = ∛(5x + 3), giving the iterative formula xₙ₊₁ = ∛(5xₙ + 3).

Step-by-Step Method

1. Write down the iterative formula and the starting value x₀.
2. Substitute x₀ into the formula to calculate x₁. Write the full calculator display.
3. Substitute x₁ to find x₂. Use the unrounded value from the previous step.
4. Repeat until the question is answered — either a set number of iterations or until values agree to the required decimal places.
5. Never round intermediate values. Only round your final answer.
6. State the solution clearly, to the accuracy requested.

Worked Example 1 — Foundation Level

Question: Use the iterative formula xₙ₊₁ = (xₙ² + 3) / 5 with x₀ = 1 to find x₁, x₂ and x₃ to 4 decimal places.

Working:

x₁ = (1² + 3) / 5 = 4/5 = 0.8000

x₂ = (0.8² + 3) / 5 = (0.64 + 3) / 5 = 3.64/5 = 0.7280

x₃ = (0.728² + 3) / 5 = (0.529984 + 3) / 5 = 3.529984/5 = 0.7060

Answer: x₁ = 0.8000, x₂ = 0.7280, x₃ = 0.7060

Worked Example 2 — Higher Level

Question: Show that x³ + 2x − 9 = 0 can be rearranged to x = (9 − x³) / 2. Using x₀ = 1.5, find the solution correct to 1 decimal place.

Working:

Rearrangement: x³ + 2x = 9 → 2x = 9 − x³ → x = (9 − x³) / 2 ✓

x₁ = (9 − 1.5³) / 2 = (9 − 3.375) / 2 = 5.625 / 2 = 2.8125

x₂ = (9 − 2.8125³) / 2 = (9 − 22.24121...) / 2 = −13.24121... / 2 = −6.620605...

This diverges. The rearrangement x = ∛(9 − 2x) is more stable.

Using xₙ₊₁ = ∛(9 − 2xₙ) with x₀ = 1.5:

x₁ = ∛(9 − 3) = ∛6 = 1.817120...

x₂ = ∛(9 − 3.634241...) = ∛(5.365759...) = 1.750690...

x₃ = ∛(9 − 3.501380...) = ∛(5.498620...) = 1.765030...

x₄ = ∛(9 − 3.530061...) = ∛(5.469939...) = 1.761955...

x₅ = ∛(9 − 3.523910...) = ∛(5.476090...) = 1.762616...

x₄ and x₅ both round to 1.8 (to 1 d.p.).

Answer: x = 1.8 (to 1 d.p.)

Worked Example 3 — Exam Style

Question: The equation x³ − 4x − 1 = 0 has a solution near x = 2. Use the iterative formula xₙ₊₁ = ∛(4xₙ + 1) with x₀ = 2 to find this solution correct to 2 decimal places.

Working:

x₁ = ∛(4(2) + 1) = ∛9 = 2.080084...

x₂ = ∛(4(2.080084...) + 1) = ∛(9.320336...) = 2.104585...

x₃ = ∛(4(2.104585...) + 1) = ∛(9.418340...) = 2.112060...

x₄ = ∛(4(2.112060...) + 1) = ∛(9.448241...) = 2.114346...

x₅ = ∛(4(2.114346...) + 1) = ∛(9.457384...) = 2.115046...

x₄ and x₅ both round to 2.11 (to 2 d.p.).

Answer: x = 2.11 (to 2 d.p.)

Common Mistakes

• Rounding intermediate values. Using a rounded value in the next step compounds errors. Always use the full unrounded calculator display and only round the final answer.
• Not showing enough iterations. You must continue until two consecutive values agree to the required number of decimal places. Stopping early loses accuracy marks.
• Substituting incorrectly into cube roots. Be careful with brackets on your calculator. ∛(4x + 1) is different from ∛4 × x + 1.

Exam Tips

• Use the ANS button on your calculator: type the formula once using ANS in place of xₙ, then press = repeatedly to generate successive values.
• Write every iteration value to at least 6 decimal places to demonstrate you are not rounding early.
• If asked to show a rearrangement leads to the iterative formula, start from the original equation and work algebraically — do not work backwards.

Practice Questions

Q1 (Higher): Use xₙ₊₁ = ∛(2xₙ + 7) with x₀ = 2 to find x₁, x₂ and x₃ to 4 decimal places.

Q2 (Higher): Show that x³ − 6x + 2 = 0 can be rearranged to x = (x³ + 2) / 6.

Q3 (Higher): Use xₙ₊₁ = ∛(6xₙ − 2) with x₀ = 2 to find the solution of x³ − 6x + 2 = 0 correct to 2 decimal places.

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Related Topics

• Iteration
• Solving Quadratic Equations — Quadratic Formula
• Functions and Function Notation

Summary

• Iteration finds approximate solutions by repeatedly applying a formula xₙ₊₁ = f(xₙ).
• Start with x₀ and substitute to generate x₁, x₂, x₃, and so on.
• The iterative formula comes from rearranging the equation into the form x = f(x).
• Never round intermediate values — only round the final answer.
• The solution is confirmed when two consecutive iterations agree to the required number of decimal places.