Numerical Methods

This lesson covers Numerical Methods as required by the A-Level Mathematics Pure 1 specification. Numerical methods are techniques for finding approximate solutions to equations that cannot be solved algebraically. These methods are particularly useful for equations where no exact analytical solution exists.

Spec Mapping — AQA 7357 Section I Numerical methods. This lesson covers change-of-sign root location, fixed-point iteration, Newton-Raphson, and the trapezium rule in Section I. Refer to the official AQA specification document for exact wording.

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Locating Roots

The Change of Sign Method

If f(x) is a continuous function and f(a) and f(b) have opposite signs (i.e., f(a) × f(b) < 0), then there is at least one root of f(x) = 0 in the interval (a, b).

This is a consequence of the Intermediate Value Theorem.

Example: Show that f(x) = x³ − 3x − 1 has a root between x = 1 and x = 2.

$$\begin{aligned} f(1) &= 1 - 3 - 1 = -3 \quad \text{(negative)} \\ f(2) &= 8 - 6 - 1 = 1 \quad \text{(positive)} \end{aligned}$$f(1)f(2)​=1−3−1=−3(negative)=8−6−1=1(positive)​

Since f(1) < 0 and f(2) > 0, and f(x) is continuous, there is a root in the interval (1, 2) by the change of sign rule. ∎

Exam Tip: When using change of sign, you must: (1) evaluate f(x) at both endpoints, (2) state that one is positive and one is negative, and (3) conclude using "change of sign, therefore a root exists." You must also state (or it must be obvious) that the function is continuous.

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Limitations of the Change of Sign Method

The change of sign method can fail in certain situations:

Situation	Problem
Even number of roots in the interval	Signs are the same at both endpoints
Root at the endpoint	f(a) = 0 or f(b) = 0
Discontinuity in the interval	The function is not continuous
Interval too large	Multiple roots are missed

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Iteration — Fixed Point Iteration

To solve f(x) = 0, rearrange into the form x = g(x). Then use the iterative formula:

$$x_{n+1} = g(x_{n})$$xn+1​=g(xn​)

Starting from an initial estimate x₀, repeatedly substitute to generate x₁, x₂, x₃, ...

If the sequence converges, it approaches a root of f(x) = 0.

Example: Use the iteration xₙ₊₁ = (xₙ³ + 1)/3 with x₀ = 1 to find a root of 3x − x³ = 1.

x₀ = 1
x₁ = (1 + 1)/3 = 0.6667
x₂ = (0.2963 + 1)/3 = 0.4321
x₃ = (0.0807 + 1)/3 = 0.3602
x₄ = (0.0468 + 1)/3 = 0.3489
x₅ = (0.0425 + 1)/3 = 0.3475

The values are converging. The root is approximately x ≈ 0.347.

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Convergence and Divergence

An iteration xₙ₊₁ = g(xₙ) will converge to a root if |g'(x)| < 1 near the root:

Condition	Behaviour
	g'(x)
	g'(x)

If an iteration diverges, you need to rearrange f(x) = 0 into a different form x = g(x) and try again.

Staircase and Cobweb Diagrams

Iterative sequences can be illustrated graphically by plotting y = g(x) and y = x on the same axes:

• Staircase diagram: When the iteration converges by stepping in one direction (from one side).
• Cobweb diagram: When the iteration converges by oscillating around the root (alternating sides).

The root is where these two curves intersect.

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The Newton-Raphson Method

The Newton-Raphson method is a more sophisticated iterative technique:

$$x_{n+1} = x_{n} - f(x_{n}) / f'(x_{n})$$xn+1​=xn​−f(xn​)/f′(xn​)

This method uses the tangent to the curve at each iteration to find a better approximation to the root. It generally converges much faster than fixed-point iteration.

Example: Use Newton-Raphson to find a root of f(x) = x² − 3 starting from x₀ = 2.

$$\begin{aligned} f(x) &= x^{2} - 3, \quad f'(x) = 2x \\ x_{0} &= 2 \\ x_{1} &= 2 - \dfrac{4 - 3}{4} = 2 - 0.25 = 1.75 \\ x_{2} &= 1.75 - \dfrac{3.0625 - 3}{3.5} = 1.75 - 0.01786 = 1.73214 \\ x_{3} &= 1.73205 \quad \text{(converges rapidly to } \sqrt{3}\text{)} \end{aligned}$$f(x)x0​x1​x2​x3​​=x2−3,f′(x)=2x=2=2−44−3​=2−0.25=1.75=1.75−3.53.0625−3​=1.75−0.01786=1.73214=1.73205(converges rapidly to 3​)​

When Newton-Raphson Fails

The method can fail if:

• The starting value is too far from the root.
• f'(xₙ) = 0 at some iteration (tangent is horizontal).
• The tangent overshoots to a different part of the curve.

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Trapezia for Numerical Integration

When an integral cannot be evaluated analytically, the trapezium rule provides an approximation:

∫(from a to b) f(x) dx ≈ h/2 × [y₀ + yₙ + 2(y₁ + y₂ + ... + yₙ₋₁)]

where h = (b − a)/n is the strip width, and y₀, y₁, ..., yₙ are the function values at the equally spaced x-values.

Example: Use 4 strips to approximate ∫₀² √(1 + x²) dx.

h = (2 − 0)/4 = 0.5

x:    0      0.5      1.0      1.5      2.0
y:    1    1.1180   1.4142   1.8028   2.2361

Area ≈ 0.5/2 × [1 + 2.2361 + 2(1.1180 + 1.4142 + 1.8028)]
     = 0.25 × [3.2361 + 2(4.3350)]
     = 0.25 × [3.2361 + 8.6700]
     = 0.25 × 11.9061
     ≈ 2.977

Improving Accuracy

• Increase the number of strips (larger n) for a better approximation.
• The trapezium rule overestimates the area under a convex curve and underestimates under a concave curve.

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Synoptic Links

This topic connects to:

• Differentiation for Newton-Raphson (aqa-alevel-maths-pure-1 / differentiation) — the iteration formula requires $f'(x)$f′(x) at each step.
• Numerical integration in modelling (aqa-alevel-maths-calculus-applications / applications-of-integration) — when an integrand has no elementary antiderivative, the trapezium rule is the go-to tool.
• Convergence of geometric series (aqa-alevel-maths-pure-1 / sequences-and-series) — the condition $|g'(\alpha)| < 1$∣g′(α)∣<1 for fixed-point convergence mirrors $|r| < 1$∣r∣<1 for series convergence.

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Summary

• The change of sign method locates roots: if f(a) and f(b) have opposite signs and f is continuous, there is a root in (a, b).
• Fixed-point iteration uses xₙ₊₁ = g(xₙ) to approximate roots — convergence requires |g'(x)| < 1.
• The Newton-Raphson method uses xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) and converges faster.
• The trapezium rule approximates definite integrals using trapezia.
• Numerical methods give approximate solutions — always state the degree of accuracy.

Exam Tip: When using iterative methods, show all iterations clearly with at least 4 decimal places. When using the trapezium rule, set up a clear table of x and y values. State the strip width h explicitly. Always give your final answer to the required degree of accuracy, and where possible, comment on whether your approximation is an overestimate or underestimate.

A-Level Deep Dive: Numerical Methods

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section L: Numerical methods. This section covers the location of roots of $f(x) = 0$f(x)=0 by considering changes of sign, the use of iteration of the form $x_{n+1} = g(x_n)$xn+1​=g(xn​) to find approximations to roots, and the trapezium rule for numerical integration. (The Newton–Raphson method formally sits in Pure 2 / Section M for AQA 7357 and is examined in Paper 2.) Numerical methods is intrinsically synoptic: every question reaches into Section H (integration) via the trapezium rule, into Section J (differentiation) for analysing convergence, and into Section D (functions / graphs) for sketching cobweb diagrams. The AQA formula booklet provides the trapezium rule statement; the iteration formula $x_{n+1} = g(x_n)$xn+1​=g(xn​) is given case-by-case within each question.

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks):

(a) Show that the equation $x^3 - 4x + 1 = 0$x3−4x+1=0 has a root $\alpha$α in the interval $[1, 2]$[1,2]. (2)

(b) Show that this equation can be rearranged into the form $x = \sqrt[3]{4x - 1}$x=34x−1​, and use the iteration $x_{n+1} = \sqrt[3]{4x_n - 1}$xn+1​=34xn​−1​ with $x_0 = 2$x0​=2 to find $x_1$x1​, $x_2$x2​, $x_3$x3​ to four decimal places. (4)

(c) State, with justification, the value of $\alpha$α to two decimal places. (2)

Solution with mark scheme:

(a) Step 1 — evaluate at the endpoints.

Let $f(x) = x^3 - 4x + 1$f(x)=x3−4x+1. Then $f(1) = 1 - 4 + 1 = -2$f(1)=1−4+1=−2 and $f(2) = 8 - 8 + 1 = 1$f(2)=8−8+1=1.

M1 — evaluating $f$f at both endpoints. A common slip is computing only one value and asserting "sign change" without showing both.

A1 — $f(1) < 0$f(1)<0 and $f(2) > 0$f(2)>0, with the conclusion stated explicitly: "since $f$f is a polynomial it is continuous on $[1, 2]$[1,2], and $f(1)$f(1) and $f(2)$f(2) have opposite signs, so by the change-of-sign criterion there exists $\alpha \in (1, 2)$α∈(1,2) with $f(\alpha) = 0$f(α)=0." The continuity statement is essential — without it the sign-change argument is incomplete.

(b) Step 1 — derive the rearrangement.

$x^3 - 4x + 1 = 0 \implies x^3 = 4x - 1 \implies x = \sqrt[3]{4x - 1}$x3−4x+1=0⟹x3=4x−1⟹x=34x−1​.

B1 — correct rearrangement, shown algebraically.

Step 2 — iterate.

Using $x_0 = 2$x0​=2:

$x_1 = \sqrt[3]{4(2) - 1} = \sqrt[3]{7} = 1.9129...$x1​=34(2)−1​=37​=1.9129... $x_2 = \sqrt[3]{4(1.9129) - 1} = \sqrt[3]{6.6516} = 1.8806...$x2​=34(1.9129)−1​=36.6516​=1.8806... $x_3 = \sqrt[3]{4(1.8806) - 1} = \sqrt[3]{6.5224} = 1.8684...$x3​=34(1.8806)−1​=36.5224​=1.8684...

M1 — substituting $x_0$x0​ correctly into the iteration formula. A1 — $x_1 = 1.9129$x1​=1.9129, $x_2 = 1.8806$x2​=1.8806, $x_3 = 1.8684$x3​=1.8684 all to 4 d.p. A1 — at least three iterates correct (allow recoverable rounding throughout).

(c) Continuing the iteration: $x_4 \approx 1.8638$x4​≈1.8638, $x_5 \approx 1.8620$x5​≈1.8620, $x_6 \approx 1.8614$x6​≈1.8614. The iterates appear to converge to about $1.86$1.86.

M1 — taking enough further iterates (or by sign-change refinement on $[1.855, 1.865]$[1.855,1.865]) to justify two-decimal-place accuracy. To confirm, evaluate $f(1.855) \approx -0.0397$f(1.855)≈−0.0397 and $f(1.865) \approx 0.0507$f(1.865)≈0.0507 — opposite signs, so $\alpha \in (1.855, 1.865)$α∈(1.855,1.865) and hence $\alpha = 1.86$α=1.86 to 2 d.p.

A1 — $\alpha = 1.86$α=1.86 (2 d.p.) with the sign-change confirmation explicitly written.

Total: 8 marks (M3 A4 B1).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The curve $y = \ln(1 + x^2)$y=ln(1+x2) is to be integrated numerically over $[0, 2]$[0,2].

(a) Use the trapezium rule with four strips (five ordinates) to estimate $\displaystyle \int_0^2 \ln(1 + x^2)\,dx$∫02​ln(1+x2)dx, giving your answer to three decimal places. (4)

(b) State, with a reason referring to the concavity of the integrand, whether the trapezium-rule estimate is an over-estimate or an under-estimate of the true value of the integral. (2)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — strip width $h = (2 - 0)/4 = 0.5$h=(2−0)/4=0.5 stated.
• M1 (AO1.1b) — evaluating the five ordinates: $y_0 = \ln(1) = 0$y0​=ln(1)=0, $y_1 = \ln(1.25) \approx 0.2231$y1​=ln(1.25)≈0.2231, $y_2 = \ln(2) \approx 0.6931$y2​=ln(2)≈0.6931, $y_3 = \ln(3.25) \approx 1.1787$y3​=ln(3.25)≈1.1787, $y_4 = \ln(5) \approx 1.6094$y4​=ln(5)≈1.6094.
• M1 (AO1.1b) — applying the trapezium rule $T = \tfrac{h}{2}\left[(y_0 + y_4) + 2(y_1 + y_2 + y_3)\right]$T=2h​[(y0​+y4​)+2(y1​+y2​+y3​)] correctly: $T = 0.25\left[1.6094 + 2(2.0949)\right] = 0.25(5.7992) \approx 1.4498$T=0.25[1.6094+2(2.0949)]=0.25(5.7992)≈1.4498.
• A1 (AO1.1b) — $T \approx 1.450$T≈1.450 (3 d.p.).

(b)

• M1 (AO2.4) — observing that $y = \ln(1 + x^2)$y=ln(1+x2) has second derivative changing sign on $[0, 2]$[0,2], or specifically that the curve is convex (concave up) for small $x$x and concave down for larger $x$x, so the direction of the trapezium-rule error depends on which behaviour dominates.
• A1 (AO2.4) — a clean argument is: on $[0, 1]$[0,1] the curve is concave up so trapeziums lie above the curve; on $[1, 2]$[1,2] the curve is concave down so trapeziums lie below. Awarding the mark requires a clear, reasoned response — either "over-estimate, because the convex region dominates" or "no simple judgement, the regions act in opposite directions" with justification.

Total: 6 marks split AO1 = 4, AO2 = 2. Part (a) is procedural; part (b) tests genuine reasoning about geometry of the trapezium error.

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Synoptic links

1. Section H — Integration: the trapezium rule is an explicit numerical alternative to the antiderivative method. When $\int e^{-x^2}\,dx$∫e−x2dx has no elementary closed form, trapezium estimates are how A-Level students access the integral. The link runs both ways: numerical estimates can be checked against an antiderivative when one exists, building confidence in the algorithm.

2. Section J — Differentiation: convergence of fixed-point iteration $x_{n+1} = g(x_n)$xn+1​=g(xn​) near a root $\alpha$α is governed by $|g'(\alpha)| < 1$∣g′(α)∣<1. This is a direct application of the chain rule and the mean-value theorem — the iteration shrinks errors by a factor of approximately $|g'(\alpha)|$∣g′(α)∣ each step.

3. Section D — Graphs and functions: cobweb (and staircase) diagrams sketched on the lines $y = x$y=x and $y = g(x)$y=g(x) are visual proofs of convergence or divergence. Drawing them requires fluent transformation between Cartesian curves — pure Section D content.

4. Section F — Sequences and series: the iterates $x_0, x_1, x_2, \ldots$x0​,x1​,x2​,… form a recursively defined sequence. The language of "monotonic", "bounded", "convergent" from Section F applies directly, and the formal idea of a limit is the fixed point.

5. Section M — Newton–Raphson (Pure 2): Newton–Raphson is fixed-point iteration with the specific choice $g(x) = x - f(x)/f'(x)$g(x)=x−f(x)/f′(x). The $|g'(\alpha)| < 1$∣g′(α)∣<1 convergence criterion specialises (since $g'(\alpha) = 0$g′(α)=0 when $f(\alpha) = 0$f(α)=0, $f'(\alpha) \neq 0$f′(α)=0) to give the famous quadratic convergence of Newton's method.

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Mark-scheme literacy

Numerical-methods questions on AQA 7357 split AO marks roughly as follows: