A20: Iteration
Find approximate solutions using systematic iteration; use suffix notation in recursive formulae
📋 Key Concepts
Iteration: A method for finding approximate solutions to equations by repeatedly applying a formula.
Recursive formula: Uses previous values to find new values. Written with suffix notation: x₁, x₂, x₃, etc.
General form: xn+1 = f(xn)
Start with x₁ and find x₂, x₃, x₄, ...
Start with x₁ and find x₂, x₃, x₄, ...
📝 Setting Up Iteration
Method:
- Rearrange the equation to x = f(x) form
- Choose a starting value x₁
- Apply the formula repeatedly
- Stop when values converge (stop changing)
Example 1
Find an iterative formula for x² - 3x - 5 = 0
Solution:
Rearrange: x² = 3x + 5
x = √(3x + 5)
Iterative formula: xn+1 = √(3xn + 5)
Example 2
Find another iterative formula for x² - 3x - 5 = 0
Solution:
Alternative rearrangement: x² - 3x = 5
x(x - 3) = 5
x = 5⁄x - 3
Iterative formula: xn+1 = 5⁄xn - 3
📝 Applying Iteration
Example 3
Use xn+1 = √(3xn + 5) with x₁ = 4 to find x₂, x₃, x₄, x₅.
Solution:
x₂ = √(3(4) + 5) = √17 = 4.123...
x₃ = √(3(4.123) + 5) = √17.369 = 4.168...
x₄ = √(3(4.168) + 5) = √17.504 = 4.184...
x₅ = √(3(4.184) + 5) = √17.552 = 4.189...
📝 Finding Solutions to Given Accuracy
Example 4
Find a solution to x² - x - 2 = 0 correct to 2 decimal places, using xn+1 = √(xn + 2) with x₁ = 2.
Solution:
x₂ = √(2 + 2) = 2
x₃ = √(2 + 2) = 2
The values converge to x = 2
Check: 2² - 2 - 2 = 0 ✓
Solution: x = 2.00
Example 5
Solve x³ + x = 5 correct to 3 d.p. using xn+1 = 5⁄xn² + 1 with x₁ = 2.
Solution:
x₂ = 5⁄4 + 1 = 2.25
x₃ = 5⁄5.0625 + 1 = 1.988...
x₄ = 5⁄3.953 + 1 = 2.264...
x₅ = 5⁄5.124 + 1 = 1.976...
This formula is diverging. Try a different rearrangement.
xn+1 = ³√(5 - xn) with x₁ = 2:
x₂ = ³√(3) = 1.442
x₃ = ³√(3.558) = 1.527
x₄ = ³√(3.473) = 1.514
x₅ = ³√(3.486) = 1.516
x₆ = ³√(3.484) = 1.516
Solution: x = 1.516
📝 Convergence and Divergence
Convergence: Values get closer to the solution (good!).
Divergence: Values get further apart (try a different rearrangement).
Divergence: Values get further apart (try a different rearrangement).
Tip: If iteration diverges, try rearranging the equation differently.
🧮 Worked Examples — Exam Style
Iteration feeds an answer back into the formula repeatedly. Set the calculator to Ans memory: type x₁, =, then the formula with Ans in place of xₙ, and press = repeatedly.
Use xₙ₊₁ = √(xₙ + 6) with x₁ = 1. Find x₂, x₃ and x₄.
x₂ = √(1 + 6) = √7 ≈ 2.6458
x₃ = √(2.6458 + 6) ≈ 2.9404
x₄ = √(2.9404 + 6) ≈ 2.9900 (4 d.p.)
M1 for substituting, A1 for the sequence of values. Keep 4+ decimal places in the working; round only when asked.
Exam wording to expect: "Use the iterative formula … with x₁ = … to find x₂ and x₃", "show that the equation has a root between a and b".
❓ Practice Questions
Q1: Rearrange x² + 2x - 4 = 0 into the form x = √(4 - 2x).
Q2: Use xn+1 = 4⁄xn - 2 with x₁ = 1. Find x₂ and x₃.
Q3: Use xn+1 = √(xn + 2) with x₁ = 2. Find x₂, x₃, x₄.
Q4: Find a root of x³ - 2x - 3 = 0 using xn+1 = ³√(2xn + 3), x₁ = 2, to 2 d.p.
Q5: Write down an iterative formula from x = 5 - x³.
Q6: A sequence converges to 1.618. What equation might it be solving?
Q7: Use xₙ₊₁ = √(5 − xₙ) with x₁ = 1. Find x₂ and x₃
Q8: Why does xₙ₊₁ = √(5 − xₙ) fail if x₃ were 6?
✅ Answers
- x² + 2x = 4 → x² = 4 - 2x → x = √(4 - 2x) (when x > 0)
- x₂ = 2, x₃ = 0
- x₂ = 2, x₃ = 2, x₄ = 2 (converges)
- x ≈ 1.89
- xn+1 = 5 - xn³
- x² = x + 1 (golden ratio) or similar
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. x² + 2x − 4 = 0 → x² = 4 − 2x → x = √(4 − 2x) for x > 0
Q2. x₂ = 4/1 − 2 = 2; x₃ = 4/2 − 2 = 0
Q3. x₂ = √4 = 2; x₃ = √4 = 2; x₄ = 2 — it has converged immediately
Q4. x₂ = ³√7 ≈ 1.913; x₃ = ³√(6.826) ≈ 1.897; x₄ ≈ 1.894 → x ≈ 1.89 (2 d.p.)
Q5. x = 5 − x³ → xₙ₊₁ = 5 − xₙ³
Q6. A limit L satisfies L = f(L) — e.g. 1.618 solves x² = x + 1
Q7. x₂ = √4 = 2; x₃ = √3 ≈ 1.732
Q8. You cannot square-root a negative: 5 − 6 = −1 → √(−1) is not real — the iteration leaves the valid domain
📌 Key Facts to Memorise
- xₙ₊₁ = f(xₙ): substitute the PREVIOUS value, keep full decimals until the end
- Ans button: x₁ then =, formula with Ans, then = repeatedly
- Converge → values settle towards a limit (a root); diverge → values blow up or jump
- "To 2 d.p." → iterate until consecutive values agree at 2 d.p.
- Rearranging the same equation differently gives different iterations — some converge, some do not
🗺️ In the Exam
Iteration is Higher-tier content in this folder: 3–4 marks on Higher Paper 2/3, calculator essential. The sign-change/interval version ("show a root lies between a and b") pairs with it.
Links to: solving quadratics (A18), rearranging (A5), graphs (A11).
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always show your working with each iteration step
- Use ANS button on calculator for accuracy
- Stop when values are the same to required accuracy
- If diverging, try a different rearrangement
- Check your answer in the original equation
🧠 Problem-Solving Strategies
Problem-Solving
Rearrange the equation into x = f(x) form. Choose a starting value near the expected root. Use the ANS button on your calculator for accuracy. Keep going until the value stabilises to the required accuracy. If values diverge, try a different rearrangement.
Multi-Step Problem
Find a root of x³ - x - 3 = 0 correct to 3 decimal places. (a) Find an iterative formula. (b) Use x₁ = 1.5 to find x₂, x₃, x₄, x₅. (c) State the root.
Solution:
(a) x³ = x + 3 → x = ³√(x + 3). So xn+1 = ³√(xn + 3)
(b) x₂ = ³√(4.5) = 1.651... x₃ = ³√(4.651) = 1.669... x₄ = ³√(4.669) = 1.671... x₅ = ³√(4.671) = 1.672...
(c) Converging to x = 1.672 (3 d.p.)
⚠️ Common Errors
Watch Out!
1. Wrong: Rounding intermediate values instead of using full calculator display Correct: Use the ANS button to keep full accuracy between iterations
2. Wrong: Stopping too early before values have converged Correct: Continue until consecutive values are the same to the required decimal places
3. Wrong: Assuming all rearrangements will converge Correct: Some rearrangements diverge — if values oscillate wildly, try a different form
✍️ 6-Mark Exam Question
Extended Answer
6 marks: The equation x² + 2x - 5 = 0 has a root between 1 and 2. (a) Show that x = √(5 - 2x) is a rearrangement. (b) Use the iterative formula xn+1 = √(5 - 2xn) with x₁ = 1.5 to find x₂, x₃, x₄, x₅. (c) Find the root correct to 2 decimal places.
(a) x² + 2x - 5 = 0 → x² = 5 - 2x → x = √(5 - 2x) ✓ (positive root between 1 and 2)
(b) x₂ = √(5 - 3) = √2 = 1.414... x₃ = √(5 - 2.828) = √2.172 = 1.474... x₄ = √(5 - 2.948) = √2.052 = 1.433... x₅ = √(5 - 2.866) = √2.134 = 1.461...
(c) Continuing: the values converge to approximately x = 1.45 (2 d.p.)
Mark scheme: (a) 1 mark. (b) 3 marks for iterations (allow small rounding differences). (c) 2 marks for correct root.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A population is modelled by Pn+1 = 0.8Pn + 100, where Pn is population in year n.
(a) If P₁ = 200, find P₂, P₃, P₄.
(b) What does the 100 represent in context?
(c) The population approaches a stable value. Find this value.
Answers: (a) P₂ = 0.8(200) + 100 = 260. P₃ = 0.8(260) + 100 = 308. P₄ = 0.8(308) + 100 = 346.4. (b) 100 represents annual immigration/new arrivals. (c) When Pn+1 = Pn = P: P = 0.8P + 100 → 0.2P = 100 → P = 500.
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