OxfordAQA International AS and A-level Mathematics: Numerical methods: roots, iteration, mid-ordinate rule and Simpson's rule (9660) – Revision Notes

These notes condense section P2.9 (Numerical methods) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification lists this content under International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS. For full explanations and worked examples, read the study guide, then test yourself with the practice questions. Course pages: hub and checklist. Not sure what to revise first? Take a free diagnostic.

Key facts at a glance

Idea	Rule
Sign change	f continuous on [a, b], with f(a) and f(b) differing in sign → at least one root in (a, b)
Iteration	Rearrange f(x) = 0 to x = g(x); use xₙ₊₁ = g(xₙ) from a starting value x₁
Convergence	Converges near α when
Staircase	0 < g′(α) < 1: one-sided approach
Cobweb	−1 < g′(α) < 0: alternates either side of α
Strip width	h = (b − a)/n for n strips
Mid-ordinate	∫ₐᵇ y dx ≈ h × (sum of the n midpoint heights)
Simpson	(h/3)[first + last + 4(odd ordinates) + 2(even interior ordinates)], n even

Locating roots

Method.

1. Write f(x) = 0 in the form “expression = 0”.
2. Evaluate f at both ends of the interval and record the signs.
3. State that f is continuous (name the type, for example “polynomial” or “exponential”).
4. Conclude: opposite signs, so a root lies in the interval.

To show a root is correct to a given number of decimal places, evaluate f at the two values that bracket it. For a root of 1.24 to 2 d.p., test 1.235 and 1.245.

Where a sign change misleads

• A jump or asymptote inside the interval, for example tan x between 1 and 2, can change sign without a root.
• A root where the graph touches the axis, for example f(x) = (2x − 5)², gives no sign change.
• Two roots in one interval can leave the end signs equal.

Iteration

Method.

1. Rearrange to x = g(x) (several rearrangements usually exist).
2. Substitute x₁ to get x₂, and repeat.
3. Carry extra decimal places, stop when successive values agree to the accuracy asked, and state the result.

Example. From x³ + 2x² − 5 = 0, the rearrangement xₙ₊₁ = √(5/(xₙ + 2)) from x₁ = 1 gives 1.2910, 1.2326, 1.2437, 1.2416, 1.2420, 1.2419. The root is about 1.2419.

Must-know distinctions

Staircase	Cobweb
g′(α) positive	g′(α) negative
Values move steadily in one direction	Values alternate above and below α
Converges if g′(α) < 1	Converges if g′(α) > −1

Drawing a diagram. Plot y = x and y = g(x). From x₁ go up or down to the curve, across to y = x, and repeat. In an exam sketch, mark α where the curves cross, label x₁, x₂, x₃ on the axis, and use arrows. If the arrows head towards α, the iteration converges; if away, it diverges.

A rearrangement that fails. For the same equation, xₙ₊₁ = 5/xₙ² − 2 has g′(α) ≈ −5.22, so the cobweb spirals outwards. From 1.3 it gives 0.9586, 3.4414, −1.5778, 0.0084, then 70 754.76.

Mid-ordinate rule

Method.

1. Find h = (b − a)/n.
2. List the midpoints: a + h/2, a + 3h/2, and so on.
3. Evaluate y at each midpoint.
4. Add the heights and multiply by h.

Geometry. Each rectangle matches the trapezium under the tangent at the midpoint. Convex curve: tangent below the curve, estimate too low. Concave curve: estimate too high.

Reminder. ∫₁³ (1/x) dx with 4 strips: h = 0.5, midpoints 1.25, 1.75, 2.25, 2.75, heights 0.8, 0.5714, 0.4444, 0.3636. Estimate = 0.5 × 2.2194 = 1.0898, below the exact value ln 3 = 1.0986 because 1/x is convex.

Simpson’s rule

Method.

1. Check the number of strips is even (odd number of ordinates).
2. Find h and list ordinates y₀ to yₙ.
3. Ends: weight 1. Odd positions: weight 4. Even interior positions: weight 2.
4. Multiply the weighted total by h/3.

Reminder. ∫₀² x/(1 + x²) dx with 4 strips: ordinates 0, 0.4, 0.5, 0.46154, 0.4. Estimate = (0.5/3)[0 + 0.4 + 4(0.4 + 0.46154) + 2(0.5)] = 0.8077.

Simpson’s rule is exact for polynomials up to cubic, so a quadratic or cubic is a good way to check your set-up.

Improving estimates

• Double the number of strips and recompute. Stop when two successive estimates agree to the accuracy you want.
• For ∫₁³ (1/x) dx the mid-ordinate estimates for 2, 4, 8 strips are 1.0667, 1.0898, 1.0963; the error roughly quarters each time.
• The same integral with Simpson’s rule gives 1.1111, 1.1000, 1.0987, so it settles faster.
• You can reuse earlier function values: doubling the strips for Simpson’s rule keeps all the old ordinates.

Choosing and presenting

Task says	Do this
“Show that f(x) = 0 has a root in [a, b]”	Evaluate f(a) and f(b), state signs, continuity, conclusion
“Show that the root is 1.24 to 2 d.p.”	Evaluate f at 1.235 and 1.245 and state the sign change
“Use the iteration to find x₂, x₃ …”	Show each substitution, give 4 or more decimal places
“Sketch a diagram to illustrate”	Both curves, the root, and arrows from x₁ through x₂, x₃
“Estimate using n strips”	Write h first, then the table of heights, then the sum
“Comment on your estimate”	Refer to convex or concave, or to a further estimate with more strips

Write the table of y-values every time. It makes your working easy to follow, and it lets you spot a wrong weighting before the final answer.

Quick self-test

1. For f(x) = 2x³ − x − 9, find f(1) and f(2) and state what they show.
2. Show that f(x) = 2x³ − x − 9 has a root between 1.7 and 1.8.
3. Show that 2x³ − x − 9 = 0 can be written x = ∛((x + 9)/2). With x₁ = 2, find x₂ and x₃ to 4 d.p.
4. xₙ₊₁ = √(2xₙ + 3) with x₁ = 1. Find x₂ and x₃ to 3 d.p., then find the limit α exactly.
5. Is the iteration in question 4 a staircase or a cobweb? Justify with g′(α).
6. An iteration has g′(α) = −1.3. Does it converge? Describe the diagram.
7. Use the mid-ordinate rule with 2 strips to estimate ∫₀⁴ √x dx to 3 d.p. Is it an over- or underestimate?
8. Use Simpson’s rule with 2 strips for ∫₀⁴ √x dx, to 3 d.p.
9. Use Simpson’s rule on ∫₀³ x² dx with 2 strips. What do you notice?
10. Ordinates 2.0, 3.1, 3.6, 3.0, 2.2 are taken at x = 0, 0.5, 1, 1.5, 2. Estimate the area using Simpson’s rule.

Answers

1. f(1) = −8, f(2) = 5. The sign changes and f is continuous (polynomial), so a root lies between 1 and 2.
2. f(1.7) = −0.874 and f(1.8) = 0.864. Opposite signs, f continuous, so a root lies in (1.7, 1.8).
3. 2x³ = x + 9, so x³ = (x + 9)/2. Then x₂ = ∛5.5 = 1.7652 and x₃ = ∛5.3826 = 1.7525.
4. x₂ = √5 = 2.236 and x₃ = 2.734. At the limit α = √(2α + 3), so α² − 2α − 3 = 0, (α − 3)(α + 1) = 0. Since the iterates are positive, α = 3.
5. g(x) = √(2x + 3), g′(x) = 1/√(2x + 3), so g′(3) = 1/3. It is positive and less than 1: a staircase, converging.
6. −1.3 is below −1, so it diverges: the cobweb spirals outwards away from α.
7. h = 2, midpoints 1 and 3: 2(1 + 1.732) = 5.464. √x is concave, so this is an overestimate (exact value 16/3 = 5.333).
8. h = 2, ordinates 0, 1.414, 2: (2/3)(0 + 2 + 4 × 1.41421) = 5.105.
9. h = 1.5, ordinates 0, 2.25, 9: (1.5/3)(0 + 9 + 9) = 9, which is exact, as expected for a quadratic.
10. (0.5/3)[2.0 + 2.2 + 4(3.1 + 3.0) + 2(3.6)] = (1/6)(35.8) = 5.967.

Where marks are usually lost

• A “show that” root argument with the numbers but no mention of continuity or the sign change.
• Accepting a sign change over an asymptote.
• Rounding each iterate to 2 or 3 decimal places and letting the error build up.
• Naming a convergent iteration “staircase” or “cobweb” without checking the sign of g′(α).
• Using strip edges in the mid-ordinate rule instead of midpoints.
• Applying Simpson’s rule to an odd number of strips.
• Weighting 4 and 2 the wrong way round, or forgetting to multiply by h/3.
• Saying “over-estimate” with no reference to convexity or concavity.
• Giving more or fewer decimal places than the question asks.

Examiner insights on how answers gain or lose credit are in the practice set’s answers.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. This content is listed under P2.9 in International A-level Unit P2 (Pure Maths).