Numerical Methods

When equations cannot be solved algebraically, numerical methods provide approximate solutions. This lesson covers the change-of-sign method, iteration (fixed-point), Newton-Raphson, and the trapezium rule.

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Locating Roots — Change of Sign

If f(x) is continuous on [a, b] and f(a) and f(b) have opposite signs, then there is at least one root in the interval (a, b).

Worked Example 1

Show that x³ − 3x − 5 = 0 has a root between x = 2 and x = 3.

f(2) = 8 − 6 − 5 = −3 (negative)
f(3) = 27 − 9 − 5 = 13 (positive)

Sign change → root exists in (2, 3). ∎

When It Fails

The method fails if:

• The function is discontinuous in the interval.
• There are multiple roots and the sign changes an even number of times.
• The root is exactly at the boundary.

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Iteration (Fixed-Point Method)

Rearrange f(x) = 0 into the form x = g(x), then use the iteration:

xₙ₊₁ = g(xₙ)

starting from an initial estimate x₀.

Convergence Condition

The iteration converges if |g’(x)| < 1 near the root.

Worked Example 2

Solve x³ − 3x − 5 = 0 using the rearrangement x = ∛(3x + 5).

x₀ = 2

x₁ = ∛(6 + 5) = ∛11 ≈ 2.2239
x₂ = ∛(3(2.2239) + 5) = ∛(11.6718) ≈ 2.2636
x₃ = ∛(3(2.2636) + 5) = ∛(11.7907) ≈ 2.2696
x₄ = ∛(3(2.2696) + 5) = ∛(11.8088) ≈ 2.2705

Converging to approximately x ≈ 2.2706.

Cobweb and Staircase Diagrams

• A staircase diagram arises when the iteration converges monotonically (always approaching from one side).
• A cobweb diagram arises when successive values alternate above and below the root.
• If |g’(x)| > 1 near the root, the iteration diverges.

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Newton-Raphson Method

Newton-Raphson is generally faster than fixed-point iteration:

xₙ₊₁ = xₙ − f(xₙ)/f’(xₙ)

Worked Example 3

Solve x³ − 3x − 5 = 0 using Newton-Raphson with x₀ = 2.

f(x) = x³ − 3x − 5, f’(x) = 3x² − 3

x₀ = 2: f(2) = −3, f’(2) = 9
x₁ = 2 − (−3)/9 = 2 + 1/3 = 2.3333

x₁ = 2.3333: f(2.3333) = 12.7037 − 7 − 5 = 0.7037, f’(2.3333) = 16.3333 − 3 = 13.3333
x₂ = 2.3333 − 0.7037/13.3333 ≈ 2.2806

x₂ = 2.2806: f(2.2806) ≈ 11.8647 − 6.8417 − 5 = 0.0230, f’(2.2806) ≈ 15.6033 − 3 = 12.6033
x₃ = 2.2806 − 0.0230/12.6033

≈ 2.2788

Rapid convergence in just 3 iterations.

When Newton-Raphson Fails

• The tangent is horizontal (f’(xₙ) = 0) — division by zero.
• The starting point is too far from the root.
• There is a discontinuity or oscillation near the root.

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The Trapezium Rule

Approximate ∫ₐᵇ f(x) dx by dividing [a, b] into n strips of width h = (b − a)/n:

∫ₐᵇ f(x) dx ≈ h/2 [y₀ + 2(y₁ + y₂ + … + yₙ₋₁) + yₙ]

Worked Example 4

Estimate ∫₁³ 1/x dx using the trapezium rule with 4 strips.

h = (3 − 1)/4 = 0.5

x	1	1.5	2	2.5	3
1/x	1	0.6667	0.5	0.4	0.3333

≈ 0.5/2 × [1 + 2(0.6667 + 0.5 + 0.4) + 0.3333]

= 0.25 × [1 + 2(1.5667) + 0.3333]

= 0.25 × [1 + 3.1334 + 0.3333]

= 0.25 × 4.4667

≈ 1.1167

(Exact answer: ln 3 ≈ 1.0986. Overestimate because 1/x is convex on [1, 3].)

Accuracy of the Trapezium Rule

• More strips → more accurate.
• Overestimate for convex (concave-up) curves.
• Underestimate for concave (concave-down) curves.

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Bounds for Roots

After locating a root in an interval, you can narrow it down:

Worked Example 5

Show that x³ − 3x − 5 = 0 has a root correct to 1 decimal place as x = 2.3.

f(2.25) = 11.3906 − 6.75 − 5 = −0.3594 (negative)
f(2.35) = 12.978 − 7.05 − 5 = 0.928 (positive)

Root lies in (2.25, 2.35), so the root rounds to 2.3 to 1 d.p. ∎

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Exam Tips

• For change-of-sign questions, always evaluate f at both endpoints and state the sign change clearly.
• Newton-Raphson converges faster than fixed-point iteration — but requires the derivative.
• Show iterative steps to the accuracy requested — usually 4 or 5 decimal places.
• The trapezium rule: remember the formula pattern — “first + last + 2 × middle terms, all times h/2”.
• State whether the trapezium rule gives an overestimate or underestimate and justify using the shape of the curve.
• If an iteration diverges, try a different rearrangement x = g(x) where |g’(x)| < 1 near the root.

A-Level Deep Dive: Numerical Methods

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section L (Numerical Methods). AS Pure 1 introduced locating roots via the change-of-sign rule on a continuous function and fixed-point iteration of the form $x_{n+1} = g(x_n)$xn+1​=g(xn​). Year 2 (Pure 2) extends this with the Newton-Raphson method $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$xn+1​=xn​−f′(xn​)f(xn​)​, including its failure modes: stationary point at the iterate ($f'(x_n) = 0$f′(xn​)=0), oscillation between values, divergence away from the root, and convergence to a different root than the one sought. The AQA formula booklet gives the Newton-Raphson formula but does not describe the conditions under which it fails — that reasoning must be supplied by the candidate.

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Worked example with full mark scheme

Question (8 marks):

The equation $f(x) = x^3 - 2x - 5 = 0$f(x)=x3−2x−5=0 is known to have a root $\alpha$α in the interval $[2, 3]$[2,3].

(a) Apply the Newton-Raphson method with $x_0 = 2$x0​=2 to find $x_1$x1​, $x_2$x2​ and $x_3$x3​, giving each to 5 decimal places. (5)

(b) Verify that $\alpha$α lies in the interval $[2.0945, 2.0946]$[2.0945,2.0946]. (2)

(c) Explain why Newton-Raphson would fail if $x_0 = \sqrt{2/3}$x0​=2/3​ were used. (1)

Solution with mark scheme:

(a) Step 1 — differentiate. $f'(x) = 3x^2 - 2$f′(x)=3x2−2.

B1 — correct derivative.

Step 2 — apply the iteration.

$x_{n+1} = x_n - \dfrac{x_n^3 - 2x_n - 5}{3x_n^2 - 2}$xn+1​=xn​−3xn2​−2xn3​−2xn​−5​

With $x_0 = 2$x0​=2: $f(2) = 8 - 4 - 5 = -1$f(2)=8−4−5=−1, $f'(2) = 12 - 2 = 10$f′(2)=12−2=10.

$x_1 = 2 - \dfrac{-1}{10} = 2.1$x1​=2−10−1​=2.1

M1 — correct substitution into the Newton-Raphson formula. A1 — $x_1 = 2.10000$x1​=2.10000.

Step 3 — iterate. $f(2.1) = 9.261 - 4.2 - 5 = 0.061$f(2.1)=9.261−4.2−5=0.061, $f'(2.1) = 13.23 - 2 = 11.23$f′(2.1)=13.23−2=11.23.

$x_2 = 2.1 - \dfrac{0.061}{11.23} = 2.09457\ldots$x2​=2.1−11.230.061​=2.09457…

$f(2.09457) \approx 0.000186$f(2.09457)≈0.000186, $f'(2.09457) \approx 11.16187$f′(2.09457)≈11.16187.

$x_3 = 2.09457 - \dfrac{0.000186}{11.16187} \approx 2.09455$x3​=2.09457−11.161870.000186​≈2.09455

M1 — at least one further correct iteration. A1 — $x_2 = 2.09457$x2​=2.09457, $x_3 = 2.09455$x3​=2.09455 (5 d.p.).

(b) Change of sign: $f(2.0945) \approx -0.0001\ldots < 0$f(2.0945)≈−0.0001…<0 and $f(2.0946) \approx 0.0010\ldots > 0$f(2.0946)≈0.0010…>0.

M1 — both function values evaluated. A1 — sign change observed and $f$f continuous on the closed interval, hence by the intermediate value theorem there is a root in $[2.0945, 2.0946]$[2.0945,2.0946].

(c) B1 — at $x = \sqrt{2/3}$x=2/3​, $f'(x) = 3 \cdot \tfrac{2}{3} - 2 = 0$f′(x)=3⋅32​−2=0, so the tangent at $x_0$x0​ is horizontal and the next iterate is undefined (division by zero).

Total: 8 marks (B1 M1 A1 M1 A1 M1 A1 B1).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): $g(x) = e^{-x} - x$g(x)=e−x−x.

(a) Show that the equation $g(x) = 0$g(x)=0 has a root $\beta$β in $[0, 1]$[0,1]. (2)

(b) Starting with $x_0 = 0.5$x0​=0.5, apply two iterations of the Newton-Raphson method to $g$g, giving $x_2$x2​ to 5 decimal places. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — evaluate $g(0) = 1$g(0)=1 and $g(1) = e^{-1} - 1 \approx -0.632$g(1)=e−1−1≈−0.632.
• A1 (AO2.1) — sign change plus a statement of continuity gives a root in the open interval by the intermediate value theorem.

(b)

• B1 (AO1.1b) — $g'(x) = -e^{-x} - 1$g′(x)=−e−x−1, so $g'(0.5) = -e^{-0.5} - 1 \approx -1.60653$g′(0.5)=−e−0.5−1≈−1.60653.
• M1 (AO1.1b) — substitute into $x_1 = 0.5 - \dfrac{e^{-0.5} - 0.5}{-e^{-0.5} - 1}$x1​=0.5−−e−0.5−1e−0.5−0.5​.
• A1 (AO1.1b) — $x_1 \approx 0.56631$x1​≈0.56631.
• A1 (AO2.5) — $x_2 \approx 0.56714$x2​≈0.56714 (5 d.p.).

Total: 6 marks split AO1 = 4, AO2 = 2. AQA Pure 2 numerical-methods questions are AO1-led but reserve A1 marks for connecting change of sign to continuity and for the disciplined rounding that distinguishes a usable iterate from a calculator readout.

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Synoptic links

Connects to:

1. Section H — Differentiation (Pure 1 and 2): Newton-Raphson is literally a tangent-line construction. The line through $(x_n, f(x_n))$(xn​,f(xn​)) with gradient $f'(x_n)$f′(xn​) is $y - f(x_n) = f'(x_n)(x - x_n)$y−f(xn​)=f′(xn​)(x−xn​); setting $y = 0$y=0 and solving for $x$x gives the iteration formula. Without confidence differentiating polynomials, exponentials, logs and trigonometric functions, candidates cannot even start the iteration.

2. Pure 1 numerical methods — locating roots and fixed-point iteration: the change-of-sign rule and $x_{n+1} = g(x_n)$xn+1​=g(xn​) form the foundation. Newton-Raphson is a specific choice of $g$g, namely $g(x) = x - f(x)/f'(x)$g(x)=x−f(x)/f′(x), that is engineered to converge faster than a generic rearrangement.

3. Section A — Proof: justifying that a root lies in an interval requires stating the intermediate value theorem condition ($f$f continuous, sign change at the endpoints). AQA marks reward the logical statement, not just the numerical observation.

4. Section J — Integration: numerical integration via the trapezium rule (Pure 1) and Simpson-style refinements (further reading) belongs to the same family of approximation by iteration. The conceptual link — exact answers replaced by sequences that converge — is the unifying theme.

5. Use of calculator (specification preamble): AQA expects candidates to use the $\text{ANS}$ANS key or table function to iterate efficiently. Marks are not awarded for tedious by-hand long division; they are awarded for evidence that the iteration was applied correctly and the final iterate is reported to the requested precision.

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Mark-scheme literacy

Newton-Raphson and root-locating questions on 7357 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Correct derivative, correct substitution into the iteration formula, accurate arithmetic to the requested precision
AO2 (reasoning / interpretation)	20–30%	Stating continuity for IVT, identifying which root the iteration is converging to, explaining failure modes in words
AO3 (problem-solving)	5–15%	Choosing a sensible starting value, deciding when to stop iterating, sketching tangent-line behaviour to predict failure

Examiner-rewarded phrasing: "since $f$f is continuous on $[a, b]$[a,b] and $f(a) \cdot f(b) < 0$f(a)⋅f(b)<0, by the intermediate value theorem there exists $\alpha \in (a, b)$α∈(a,b) with $f(\alpha) = 0$f(α)=0"; "the iteration fails because $f'(x_0) = 0$f′(x0​)=0, so the tangent is horizontal and does not meet the $x$x-axis"; "to 5 decimal places, $x_3 = 2.09455$x3​=2.09455". Phrases that lose marks: "the values are getting closer so it converges" (no quantitative criterion); "$x = 2.0945503\ldots$x=2.0945503…" copied straight from the calculator without rounding to the requested precision; omitting the continuity statement when invoking sign change.

A specific AQA pattern to watch: when the question gives an interval and asks you to verify a root, you must evaluate $f$f at both endpoints and state the sign change explicitly. Stating only one value, or saying "by trial $f \approx 0$f≈0", does not earn the M1 — sign change is a binary observation requiring two function values.

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Grade-band model answers

3-mark question

Question: Given $f(x) = x^3 + x - 3$f(x)=x3+x−3, apply one iteration of Newton-Raphson with $x_0 = 1$x0​=1 to find $x_1$x1​, giving your answer to 4 decimal places.

Grade C response (~150 words):

$f'(x) = 3x^2 + 1$f′(x)=3x2+1, so $f'(1) = 4$f′(1)=4. Also $f(1) = 1 + 1 - 3 = -1$f(1)=1+1−3=−1.

Then $x_1 = 1 - \dfrac{-1}{4} = 1 + 0.25 = 1.2500$x1​=1−4−1​=1+0.25=1.2500.

Examiner commentary: Full marks (3/3). The candidate correctly identifies the derivative, evaluates both $f(1)$f(1) and $f'(1)$f′(1), and substitutes cleanly into the Newton-Raphson formula. The final answer is presented to 4 decimal places as required. This is a textbook B1 M1 A1 split: derivative, substitution, accurate rounding. A common slip — writing $x_1 = 1 - \dfrac{1}{4}$x1​=1−41​ (wrong sign on $f(1)$f(1)) — would cost the A1.

Grade A response (~210 words):*

Differentiate: $f'(x) = 3x^2 + 1$f′(x)=3x2+1.

Evaluate at $x_0 = 1$x0​=1:

• $f(1) = 1^3 + 1 - 3 = -1$f(1)=13+1−3=−1
• $f'(1) = 3(1)^2 + 1 = 4$f′(1)=3(1)2+1=4

Apply the Newton-Raphson formula:

$x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)} = 1 - \dfrac{-1}{4} = 1 + \dfrac{1}{4} = 1.25$x1​=x0​−f′(x0​)f(x0​)​=1−4−1​=1+41​=1.25

To 4 decimal places: $x_1 = 1.2500$x1​=1.2500.