AQA A-Level Mathematics: I: Numerical methods (7357) – Revision Notes

These notes condense Section I: Numerical methods (I1 to I4) of the AQA A-level Mathematics (7357) specification, version 1.3, for A-level exams from June 2018 onwards. Section I is Paper 1 content in the specification, and Papers 2 and 3 can also assess any Paper 1 content. A calculator is required in every 7357 paper. For full explanations and longer worked examples, read the Numerical methods study guide first.

Then test yourself with the Numerical methods practice questions. The 7357 course hub, the printable checklist and the free 10-minute diagnostics help you plan what to revise next.

The four outcomes at a glance

Ref	In one line
I1	A change of sign of a continuous f on [a, b] shows a root of f(x) = 0 in (a, b); know how this fails
I2	Iterate x_(n+1) = g(x_n), draw staircase and cobweb diagrams, use Newton-Raphson; know how these fail
I3	Trapezium rule; area estimates and the limits the area must lie between
I4	Apply all of the above to problems in context

Key formulas

Method	Formula or rule
Change of sign	f continuous on [a, b] and f(a) × f(b) < 0 ⇒ a root in (a, b)
Fixed-point iteration	x_(n+1) = g(x_n), where f(x) = 0 has been rearranged to x = g(x)
Newton-Raphson	x_(n+1) = x_n − f(x_n) / f′(x_n)
Trapezium rule	∫ from a to b of y dx ≈ (h/2)[y₀ + y_n + 2(y₁ + … + y_(n−1))], h = (b − a)/n
Convergence guide	from a start close to α, iteration converges when −1 < g′(x) < 1 near α

I1: Change of sign

Method in steps: show a root lies in an interval

1. Evaluate f(a) and f(b). Write the values, not just the signs.
2. State that f is continuous on [a, b] (for example, “f is a polynomial”).
3. State: “change of sign, so there is a root between a and b”.

Method in steps: show α = 1.52 to 2 d.p.

1. Evaluate f(1.515) and f(1.525), the rounding bounds.
2. Show the change of sign.
3. Conclude: 1.515 < α < 1.525, so α = 1.52 to 2 d.p.

When it fails

• A vertical asymptote: f(x) = 1/(x − a) changes sign across x = a with no root.
• A repeated root: (x − k)² touches the axis with no sign change.
• An even number of roots in the interval: f(a) and f(b) have the same sign.
• No sign change does not prove there is no root.

I2: Iteration

Method in steps

1. Rearrange f(x) = 0 to x = g(x), showing every algebraic step if asked to “show that”.
2. Enter x₁, press =, type g(Ans), press = repeatedly.
3. Write each requested iterate to the required accuracy. Keep full values in the calculator.

Must-know distinction: staircase vs cobweb

Diagram	Path of iterates	g′ near the root
Staircase (converging)	from one side only	0 < g′ < 1
Cobweb (converging)	alternately above and below	−1 < g′ < 0
Diverging staircase or cobweb	moves away from the root	g′ > 1 or g′ < −1

To draw either: sketch y = x and y = g(x). From x₁ go up (or down) to the curve, across to y = x, then to the curve again. Label x₁, x₂, x₃ on the x-axis.

When it fails: |g′| > 1 near the root, so the iterates move away. The sequence may then diverge, or converge to a different root from the one you need. A different rearrangement of the same equation may work.

I2: Newton-Raphson

Method in steps

1. Differentiate: write f′(x) clearly.
2. Substitute into x_(n+1) = x_n − f(x_n)/f′(x_n).
3. Iterate until two values agree to the required accuracy.
4. If asked, confirm with a change of sign.

Small reminder: for f(x) = x² − 7 and x₁ = 3, x₂ = 3 − 2/6 = 2.6667. Since √7 = 2.6458, one step has already moved most of the way.

When it fails

• f′(x_n) = 0: a horizontal tangent, so division by zero.
• x_n near a stationary point: f′ is small, the step is huge, and the next value may be far away or near a different root.
• A poor start can cause oscillation or divergence.

I3: Trapezium rule and bounds

Method in steps

1. Find h = (b − a)/n. Strips n, ordinates n + 1.
2. Make a table of x and y values (4 d.p. or more).
3. Ends once, middle ordinates twice, multiply by h/2.

Must-know distinction: over or under

• Convex curve (f″ > 0, bends upwards): chords sit above the curve, so overestimate.
• Concave curve (f″ < 0, bends downwards): chords sit below, so underestimate.
• If f″ changes sign on the interval, you cannot say in general.

Limits the area lies between. For an increasing function, rectangles of height taken at the left of each strip give a lower bound; heights taken at the right give an upper bound. For a decreasing function, swap them. A trapezium estimate known to be an over- or underestimate can replace one of these bounds and give a narrower interval.

Small reminder: ∫ from 0 to 1 of √(1 + x) dx with 2 strips. h = 0.5, y = 1, 1.2247, 1.4142. T = 0.25 × (1 + 1.4142 + 2.4495) = 1.2159. The curve is concave, so this is an underestimate (the exact value is 1.2190).

I4: In context

• Say what the variable means and give units in the final answer.
• Give a sensible accuracy for the context, not every decimal place.
• Comment on limitations of the model or method, for example “the model may not hold for large t”, or “more strips would give a better estimate”.

Quick self-test

1. Show that x³ + x − 5 = 0 has a root between 1.5 and 1.6.
2. f(x) = (x − 2)² has a root at x = 2. Why does a change of sign test on [1, 3] miss it?
3. Use x_(n+1) = ∛(x_n + 5) with x₁ = 2 to find x₂, x₃ and x₄ to 4 d.p.
4. Near a root, g′(x) ≈ −0.6. Describe the diagram and whether the iteration converges.
5. Use one step of Newton-Raphson on f(x) = x² − 7 from x₁ = 3.
6. Use one step of Newton-Raphson on f(x) = cos x − x from x₁ = 1 (radians), to 4 d.p.
7. How wide is each strip when ∫ from 0 to 3 uses 6 strips, and how many ordinates are needed?
8. A trapezium rule estimate is made for a curve with f″(x) < 0 throughout. Over or under?
9. f is increasing on [a, b]. Which rectangles give a lower bound for the area?
10. Why can Newton-Raphson not start at x₁ = 1 for f(x) = x³ − 3x + 1?
11. Show that the root in question 1 is 1.52 to 2 d.p.

Answers

1. f(1.5) = −0.125 < 0, f(1.6) = 0.696 > 0. f is continuous (a polynomial), so there is a root between 1.5 and 1.6.
2. f(x) ≥ 0 everywhere, so f(1) and f(3) are both positive: the graph touches the axis without crossing it.
3. x₂ = 1.9129, x₃ = 1.9050, x₄ = 1.9042.
4. A cobweb: iterates alternate either side of the root. |g′| < 1, so it converges.
5. f(3) = 2, f′(3) = 6, so x₂ = 3 − 2/6 = 2.6667 (4 d.p.).
6. f′(x) = −sin x − 1. x₂ = 1 − (cos 1 − 1)/(−sin 1 − 1) = 0.7504.
7. h = 3/6 = 0.5, with 7 ordinates.
8. An underestimate, because the curve is concave and the chords lie below it.
9. Rectangles with heights taken at the left-hand end of each strip.
10. f′(x) = 3x² − 3, so f′(1) = 0. The tangent is horizontal and the formula divides by zero.
11. f(1.515) = −0.0077 < 0, f(1.525) = 0.0716 > 0. Change of sign, so 1.515 < α < 1.525 and α = 1.52 to 2 d.p.

Where marks are usually lost

• Stating “change of sign so root” with no values, or no word about continuity.
• Testing the wrong interval to confirm a rounded root (1.52 and 1.53 instead of 1.515 and 1.525).
• Leaving the calculator in degrees for trigonometric equations.
• In a “show that” rearrangement, skipping a line of algebra or not reaching the exact printed form.
• Copying only the last iterate when x₂, x₃ and x₄ were each asked for.
• Cobweb or staircase diagrams with no labels on the x-axis, or lines drawn to the wrong graph.
• Confusing n strips with n ordinates, so h is wrong.
• Saying “overestimate” with no reason based on the shape of the curve.
• Giving a context answer to six decimal places with no units.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams June 2018 onwards, published by AQA. Section 3.10, I: Numerical methods.