Approximate solutions by iteration [Higher tier]
Some equations cannot be solved by factorising or the formula — they need a numerical method. Iteration repeatedly substitutes a guess into a rearranged version of the equation, generating a sequence x_0, x_1, x_2, … that converges to a root.
The recipe
- Rearrange the equation
f(x) = 0into the formx = g(x)(the iterative formula). - Choose a starting value
x_0(usually given). - Compute
x_1 = g(x_0), thenx_2 = g(x_1), and so on. - Stop when consecutive values agree to the required number of decimal places.
✦Worked example
Solve x³ - 3x - 5 = 0 using x_{n+1} = ∛(3x_n + 5) with x_0 = 2.
x_1 = ∛(3 × 2 + 5) = ∛11 = 2.224 (to 3 d.p.)
x_2 = ∛(3 × 2.224 + 5) = ∛11.671 = 2.269
x_3 = ∛(3 × 2.269 + 5) = ∛11.807 = 2.278
x_4 = ∛(3 × 2.278 + 5) = ∛11.834 = 2.279
x_5 = ∛(3 × 2.279 + 5) = ∛11.837 = 2.279
Both x_4 and x_5 round to 2.279 (3 d.p.) — converged. The root is x ≈ 2.279.
Calculator habit (essential)
Modern calculators have an answer key ANS. Once you compute the first iteration, type the formula using ANS in place of x; press = repeatedly. Each press generates the next term — you don't have to retype.
For x_{n+1} = ∛(3x_n + 5): type ∛(3 × 2 + 5) then = to get x_1. Then type ∛(3 × ANS + 5) and press = repeatedly. Watch the digits stabilise.
Showing your working
Examiners want at least 3 iterations shown explicitly, then a clear conclusion: "The root, to required accuracy, is x ≈ ___". State each x_n to enough decimal places (4-5 d.p. of working, then round to required precision).
Verifying convergence
If the sequence settles, the formula has converged. Sometimes iterations diverge (numbers grow without bound); that means your rearrangement won't work for that starting point — try a different rearrangement or starting value.
⚠Common mistakes— Common mistakes (examiner traps)
- Rounding intermediate values too aggressively. Carry 4-5 d.p. through the working; round only the final answer.
- Using x₀ = the question's hint, then computing only x₁. You usually need 3-4 iterations to demonstrate convergence.
- Mis-stating the answer's precision. Round the FINAL stabilised value to the required d.p. or s.f.
- Confusing iteration with trial and improvement. Iteration follows a formula; trial-and-improvement is guess-and-test.
- Showing only the final answer with no iterations. The marks are mostly for the working, not the answer.
➜Try this— Quick check
Use x_{n+1} = (8 + x_n)/3 with x_0 = 4 for at least three iterations.
x_1 = 12/3 = 4 (already at the fixed point). The root is exactly x = 4: substitute into 3x = 8 + x ⇒ 2x = 8 ⇒ x = 4. ✓
AI-generated · claude-opus-4-7 · v3-deep-algebra