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

This guide teaches Section I: Numerical methods (content references 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 listed under Paper 1 in the specification, and Papers 2 and 3 can each assess any Paper 1 content as well. A calculator is required in every 7357 paper, and the specification says it must have an iterative function, so the methods here are always done with a calculator to hand.

Use it with the Numerical methods revision notes and the Numerical methods practice questions. The 7357 course hub lists every topic, the printable checklist lets you tick off outcomes, and the free 10-minute diagnostics show where to start.

What Section I covers

Ref	What you must be able to do
I1	Locate roots of f(x) = 0 by a change of sign of f(x) over an interval where f is sufficiently well-behaved; understand how change of sign methods can fail
I2	Solve equations approximately with simple iterative methods; draw cobweb and staircase diagrams; use Newton-Raphson and other recurrence relations x_(n+1) = g(x_n); understand how these methods can fail
I3	Understand and use numerical integration, including the trapezium rule, estimating the area under a curve and the limits it must lie between
I4	Use numerical methods to solve problems in context

Why bother? Many equations, such as e^(−x) = x or ln x + x = 3, cannot be solved with algebra. The overarching theme OT2.4 says you should understand that numerical methods give a solution to a required level of accuracy instead. OT2.5 adds that you should judge how accurate such a solution is. The overarching themes guide covers those themes across the course.

I1: Change of sign

If f is continuous on the interval [a, b], and f(a) and f(b) have opposite signs, the graph must cross the x-axis somewhere between a and b. So there is at least one root of f(x) = 0 in (a, b).

Worked example 1

Show that x³ − 4x − 2 = 0 has a root α between 2.2 and 2.3, then show that α = 2.21 to 2 decimal places.

f(x) = x³ − 4x − 2
f(2.2) = 10.648 − 8.8 − 2 = −0.152    (negative)
f(2.3) = 12.167 − 9.2 − 2 =  0.967    (positive)

f is a polynomial, so it is continuous. There is a change of sign, so 2.2 < α < 2.3.

To show α = 2.21 to 2 d.p., test the bounds of that rounding interval, 2.205 and 2.215:

f(2.205) = −0.0992   (negative)
f(2.215) =  0.0073   (positive)

The change of sign means 2.205 < α < 2.215, so α rounds to 2.21. Always finish with a conclusion sentence of this kind.

How change of sign can fail

• Discontinuity. f(x) = 1/(x − 2) is negative at x = 1 and positive at x = 3, but there is no root. The graph jumps across an asymptote at x = 2. This is why the function must be well-behaved (continuous) on the interval.
• Repeated root or touching. f(x) = (x − 1)² has a root at x = 1, but f is positive on both sides, so no sign change shows it.
• Two roots close together. If the interval holds two roots, f(a) and f(b) can have the same sign. No change of sign does not prove there is no root.
• Interval too wide. A wide interval can hide several roots, or roots that cancel in sign. Use narrow intervals and a sketch.

I2: Iteration x_(n+1) = g(x_n)

Rearrange f(x) = 0 into the form x = g(x). A root α satisfies α = g(α). Start with a value x₁ (or x₀) and feed each output back in. If the sequence converges, its limit is a root.

Worked example 2

The same equation x³ − 4x − 2 = 0 can be written x³ = 4x + 2, so x = ∛(4x + 2). Use x_(n+1) = ∛(4x_n + 2) with x₀ = 2.

x₁ = ∛(10)       = 2.15443
x₂ = ∛(10.6177)  = 2.19791
x₃ = 2.20985
x₄ = 2.21310
x₅ = 2.21399

On a calculator, type 2, press =, then enter ∛(4 × Ans + 2) and press = repeatedly. Write down each value the question asks for. The sequence climbs steadily towards 2.214.

Staircase and cobweb diagrams

Draw y = x and y = g(x) on the same axes. The root is where they meet. From x₁, go vertically to the curve y = g(x), then horizontally to y = x, then vertically to the curve again.

• If the iterates approach the root from one side, the path looks like a staircase. This happens when 0 < g′(x) < 1 near the root. In example 2, g′(2.214) ≈ 0.27.
• If the iterates alternate either side of the root, the path spirals in like a cobweb. This happens when −1 < g′(x) < 0 near the root.

Example of a cobweb: x = e^(−x) with x₀ = 0.5 gives 0.6065, 0.5452, 0.5797, 0.5601, 0.5712, … above and below the root 0.5671 in turn. Here g′(x) = −e^(−x) ≈ −0.57 near the root.

How iteration can fail

When |g′(x)| > 1 near the root, the iterates move away from it. Try the other rearrangement of example 2, x = (x³ − 2)/4, with x₀ = 2:

x₁ = 1.5,  x₂ = 0.34375,  x₃ = −0.48985,  x₄ = −0.52938,  x₅ = −0.53709

Near 2.214, g′(x) = 3x²/4 ≈ 3.7, so the sequence leaves that root. It settles on a different root, −0.539. The same equation can give a useful or a useless iteration depending on the rearrangement. Check your answer against the root you were asked for.

I2: Newton-Raphson

Newton-Raphson follows the tangent at x_n down to the x-axis and uses that crossing as the next approximation:

x_(n+1) = x_n − f(x_n) / f′(x_n)

It needs f′, so you must differentiate first (see Section G: Differentiation).

Worked example 3

Use Newton-Raphson with x₁ = 2 to find a root of ln x + x − 3 = 0 to 4 decimal places.

f(x)  = ln x + x − 3
f′(x) = 1/x + 1

f(2)  = ln 2 − 1 = −0.30685
f′(2) = 1.5
x₂ = 2 − (−0.30685)/1.5 = 2.204569
x₃ = 2.207939
x₄ = 2.207940

x₃ and x₄ agree to 4 d.p., so the root is 2.2079 (4 d.p.). A change of sign check on f(2.20785) and f(2.20795) would confirm it.

How Newton-Raphson can fail

• f′(x_n) = 0. The tangent is horizontal and never meets the x-axis, so the formula divides by zero. For f(x) = x³ − 3x + 1, f′(1) = 0, so x₁ = 1 cannot be used.
• Starting near a turning point. f′ is small, so the step f/f′ is huge. For the same f, x₁ = 0.9 gives f(0.9) = −0.971 and f′(0.9) = −0.57, so x₂ = 0.9 − 1.7035 = −0.8035. The sequence then runs to the root 1.532, not the nearer root 0.347.
• Poor starting value or discontinuity. The iterates may jump to another root, cycle, or diverge.

I3: The trapezium rule

Split [a, b] into n strips of equal width h = (b − a)/n. Join the tops of the ordinates y₀, y₁, …, y_n with straight lines and add the trapezium areas:

∫ from a to b of y dx ≈ (h/2) [ y₀ + y_n + 2(y₁ + y₂ + ... + y_(n−1)) ]

There are n strips but n + 1 ordinates.

Worked example 4

Estimate ∫ from 0 to 1 of e^(x²) dx using 4 strips. Say whether it is an over- or underestimate, and find limits the integral must lie between.

h = 0.25.

x	0	0.25	0.5	0.75	1
y = e^(x²)	1	1.0645	1.2840	1.7551	2.7183

T = 0.125 × [1 + 2.7183 + 2(1.0645 + 1.2840 + 1.7551)]
  = 0.125 × 11.9255
  = 1.4907 (4 d.p.)

Over or under? f″(x) = (2 + 4x²)e^(x²) > 0, so the curve is convex (bends upwards). Each straight chord lies above the curve, so the trapezium rule gives an overestimate. For a concave curve (f″ < 0) it gives an underestimate.

Limits. f is increasing on [0, 1], so rectangles using the left-hand heights sit under the curve and rectangles using the right-hand heights sit over it:

Lower sum = 0.25 × (1 + 1.0645 + 1.2840 + 1.7551) = 1.2759
Upper sum = 0.25 × (1.0645 + 1.2840 + 1.7551 + 2.7183) = 1.7055

So 1.2759 < I < 1.7055. Because the trapezium value is an overestimate, the tighter statement is 1.2759 < I < 1.4907. (A calculator’s numerical integration gives 1.4627, which agrees.) More strips narrow the gap.

I4: Numerical methods in context

A drug’s concentration is modelled by C = 12t e^(−0.4t) mg per litre, t hours after a dose. It is effective while C ≥ 8. Find when it stops being effective.

C peaks at t = 2.5 (C ≈ 11.04), so the second crossing of C = 8 is after that. Solve f(t) = 12t e^(−0.4t) − 8 = 0, which has no algebraic solution.

f′(t) = 12e^(−0.4t)(1 − 0.4t)
t₁ = 6:  f(6) = −1.46831,  f′(6) = −1.52406
t₂ = 6 − (−1.46831)/(−1.52406) = 5.0366
t₃ = 5.0740

Check to 2 d.p.: f(5.065) = 0.0146 > 0 and f(5.075) = −0.0017 < 0, so t = 5.07. The drug stops being effective about 5.07 hours (about 5 hours 4 minutes) after the dose.

Interpret the answer: the model ignores differences between patients, so a precision of seconds would be meaningless. State an answer to a sensible accuracy and note the model’s limits. The trapezium rule appears in context too, for example estimating distance from a table of velocity readings.

Common errors

• Showing a change of sign but giving no conclusion, or not saying f is continuous.
• Testing 2.21 and 2.22 to “show α = 2.21 to 2 d.p.”. Test 2.205 and 2.215.
• Calculator in degrees for an equation involving sin or cos. Use radians.
• Writing only the final iterate when the question asks for x₂, x₃ and x₄.
• Using n + 1 as the number of strips, or the wrong h.
• Calling a trapezium estimate an overestimate without a reason based on the curve’s shape.
• Rounding iterates early, which drifts the final answer.

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.