Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Numerical Solution of Equations – Revision Notes

These notes condense section 2.6, Numerical solution of equations, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.6 belongs to Pure Mathematics 2 and is examined on Paper 2, which is taken only in the AS Level Pure Mathematics route (Paper 1 plus Paper 2). The same outcomes appear as section 3.6 for Paper 3. A scientific calculator is allowed, but unsupported calculator answers earn no marks.

For full explanations and longer worked examples, use the study guide. Then test yourself with the practice questions. Other links: the whole-paper Pure Mathematics 2 revision notes, the Pure Mathematics 3 revision notes (section 3.6 is the same content), the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

The three outcomes in one line each

2.6 outcome	In short
Locate a root	Sketch graphs and/or find a sign change; e.g. name two consecutive integers either side of the root
Sequences of approximations	x₁, x₂, x₃, … converging to α; write xₙ → α
Iterative formula xₙ₊₁ = F(xₙ)	The limit satisfies α = F(α), so it is a root of x = F(x). Use it to reach a stated accuracy. It may fail to converge

Not required: the condition for convergence. Required: recognising from the numbers that an iteration is not converging.

Key facts

Idea	What to remember
Sign-change rule	f continuous on [a, b] and f(a), f(b) of opposite sign ⇒ f(x) = 0 has a root between a and b
Graphs	Roots of f(x) = g(x) are the x-coordinates where y = f(x) and y = g(x) meet; the number of meeting points is the number of roots
Limit of an iteration	If xₙ → α then xₙ₊₁ → α too, so α = F(α)
From equation to formula	Rearrange to x = F(x), then write xₙ₊₁ = F(xₙ)
From formula to equation	Put xₙ₊₁ = xₙ = x and rearrange
Stopping rule	Stop when two successive iterates agree to the accuracy required
Accuracy check	α = c to 2 d.p. is confirmed by a sign change on [c − 0.005, c + 0.005]

Method: show that a root lies in an interval

1. Write the equation as f(x) = 0 (everything on one side).
2. Work out f(a) and f(b); write both values, to 2 or more s.f.
3. State: "sign change, and f is continuous, so a root lies between a and b".

Worked reminder. Show that 3 sin x = x has a root between 2 and 3.

f(x) = 3 sin x - x   (radians)
f(2) = 3 sin 2 - 2 =  0.7279   (positive)
f(3) = 3 sin 3 - 3 = -2.5766   (negative)

Sign change and f is continuous, so a root lies between 2 and 3. (It is about 2.2789.) In degree mode you would get nonsense, so check the mode first.

Method: iterate to a prescribed accuracy

1. Write down x1 exactly as given.
2. Calculate each iterate using the stored value (Ans), not a rounded one.
3. Write each iterate to the places asked for (usually 4 d.p.).
4. Stop when two successive iterates round to the same value.
5. State the root to the accuracy asked for.

Worked reminder. The equation x² + 3 ln x = 10 can be rearranged as x = √(10 − 3 ln x). Use xₙ₊₁ = √(10 − 3 ln xₙ) with x₁ = 3 to find the root to 2 d.p.

x1 = 3
x2 = 2.5892
x3 = 2.6732
x4 = 2.6552
x5 = 2.6590

x₄ and x₅ both round to 2.66, so the root is 2.66 (2 d.p.). Check: with f(x) = x² + 3 ln x − 10, f(2.655) = −0.0216 and f(2.665) = 0.0428, a sign change.

Method: find the exact limit

1. Replace x(n+1) and x(n) by alpha.
2. Rearrange to an equation with no fractions.
3. Solve exactly; reject any value the sequence cannot reach.

Worked reminder. xₙ₊₁ = 6/(xₙ + 1), x₁ = 1, converges to α.

alpha = 6/(alpha + 1)
alpha^2 + alpha - 6 = 0
(alpha + 3)(alpha - 2) = 0

All iterates are positive (1, 3, 1.5, 2.4, 1.7647, …), so α = −3 is rejected and α = 2.

Must-know distinctions

• Sign change vs root. A sign change on an interval where f is continuous proves a root. No sign change does not prove there is no root: a touching root such as (x − 2)² = 0, or two roots close together, gives no sign change.
• Discontinuity. f(x) = 1/x changes sign between −1 and 1, but 1/x = 0 has no root. The graph jumps across an asymptote.
• Converging vs diverging. Iterates that settle down converge. Iterates that move away from the root, or oscillate with growing size, do not converge. A different rearrangement of the same equation may converge.
• Monotonic vs oscillating convergence. Iterates may approach from one side (a “staircase” of values) or jump either side of α on alternate steps. Both are fine.
• Iterates vs answer. Iterates are usually to 4 d.p.; the final answer is to the accuracy the question asks for, often 2 or 3 d.p.
• Exact vs approximate. “Find the exact value of α” needs algebra from α = F(α), not a calculator decimal.

Quick self-test

1. Show that x³ + x − 3 = 0 has a root between 1 and 2.
2. Find the pair of consecutive integers between which the root of eˣ = 10 − x lies.
3. f(x) = 1/x. Does f(−1) < 0 < f(1) prove that f(x) = 0 has a root? Explain.
4. f(x) = (x − 2)². Is there a sign change on [1, 3]? Is there a root?
5. Using xₙ₊₁ = √(2xₙ + 5) with x₁ = 3, find x₂ and x₃ to 4 d.p.
6. The sequence in question 5 converges to α. Find the exact value of α.
7. Which cubic equation, in the form f(x) = 0, does xₙ₊₁ = (xₙ³ + 4)/5 solve?
8. The iteration xₙ₊₁ = (xₙ + 12)/(xₙ + 1), x₁ = 3, converges to β. Find β exactly.
9. Four successive iterates are 2.4136, 2.4178, 2.4165, 2.4169. State the root to 2 d.p. and justify.
10. Which interval would you test for a sign change to confirm that a root is 1.79 to 2 d.p.?
11. Rearrange eˣ = 4 − x² into the form x = F(x) using a logarithm.

Answers

1. f(1) = 1 + 1 − 3 = −1, f(2) = 8 + 2 − 3 = 7. Sign change and f continuous, so a root lies in (1, 2).
2. f(x) = eˣ + x − 10: f(2) = −0.611, f(3) = 13.086. Between 2 and 3 (the root is about 2.07).
3. No. f is not continuous at x = 0; its graph jumps across the asymptote, and 1/x is never 0.
4. f(1) = 1 and f(3) = 1, so no sign change, but x = 2 is a root (the graph touches the axis).
5. x₂ = √11 = 3.3166, x₃ = √(2 × 3.3166 + 5) = 3.4108.
6. α² = 2α + 5, so α² − 2α − 5 = 0 and α = 1 ± √6. The iterates are positive, so α = 1 + √6.
7. x = (x³ + 4)/5 ⇒ 5x = x³ + 4 ⇒ x³ − 5x + 4 = 0.
8. β(β + 1) = β + 12 ⇒ β² = 12 ⇒ β = ±2√3. Iterates are positive, so β = 2√3.
9. The last three iterates all round to 2.42, so the root is 2.42 (2 d.p.). The first, 2.4136, rounds to 2.41 and is ignored.
10. [1.785, 1.795]: a sign change there confirms the root rounds to 1.79.
11. eˣ = 4 − x² ⇒ x = ln(4 − x²) (valid for −2 < x < 2). Its positive root is about 1.058.

Where marks are usually lost

• Values of f written without a statement that there is a sign change, or without “so a root lies between a and b”.
• A sign change used across an asymptote, e.g. for a function containing 1/(x − k) or tan x.
• Only one iterate shown to round to the answer; two successive iterates must agree.
• Iterates retyped from their rounded values, so later iterates are slightly wrong.
• Iterates given to 3 d.p. when the question says “give each iteration to 4 decimal places”.
• Calculator left in degree mode when F(x) involves sin, cos or tan.
• In “find the exact value of α”, a decimal given instead of the exact form, or the wrong root of the quadratic kept.
• An iteration that is clearly not converging carried on for many lines instead of being described as failing to converge.
• In “show that α satisfies x = F(x)”, steps skipped between the original equation and the given form.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.6 Numerical solution of equations.