Grade 12 · Pure Mathematics 2 & 3 · Cambridge A-Level 9709 · Age 17–18
Welcome to Numerical Methods
Numerical methods allow us to find approximate solutions to equations and integrals that cannot be solved exactly using algebraic techniques. They are essential tools in engineering, physics, and computer science — and are tested directly in Cambridge A-Level Pure 2 and Pure 3 (9709).
Sign-Change
Locating roots by detecting sign changes in f(x)
Newton-Raphson
Tangent-line iteration for rapid root-finding
Fixed-Point
Rearranging to x = g(x) and iterating
Trapezium Rule
Approximating definite integrals with trapezia
Exam Technique
Showing work to the precision Cambridge expects
Learning Objectives
Locate roots using the sign-change principle and state its limitations
Apply Newton-Raphson iteration and understand its geometric basis
Use fixed-point iteration x₀, x₁, x₂,… and test convergence via |g'(α)|
Apply the trapezium rule and assess over/under-estimation from concavity
Present solutions with correct precision and concluding statements
Sign-Change / Location of Roots
If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then by the Intermediate Value Theorem there exists at least one root α ∈ (a, b) such that f(α) = 0.
f(a) < 0 and f(b) > 0 (or vice versa) ⟹ ∃ root ∈ (a, b)
Step-by-Step Procedure
1. Identify an interval [a, b] by inspection, graph, or systematic search.
2. Evaluate f(a) and f(b). Confirm they have opposite signs.
3. State: "Since f(a) and f(b) have opposite signs and f is continuous, there is a root between a and b."
4. Narrow the interval by bisection or by evaluating f at the midpoint.
5. To confirm a root to 3 d.p., show a sign change in an interval of width 0.001 centred on your answer.
Cambridge phrasing: "f(1.234) = −0.003... < 0 and f(1.235) = 0.002... > 0; since there is a sign change and f is continuous, there is a root α ∈ (1.234, 1.235), so α = 1.234 correct to 3 d.p."
Limitations of the Sign-Change Method
1. Discontinuities: f may cross zero in the denominator rather than the function. E.g. f(x) = 1/(x−2) changes sign across x = 2 with no root.
2. Tangential roots (repeated roots): If the curve touches but doesn't cross the x-axis (e.g. f(x) = (x−3)²), there is no sign change even though x = 3 is a root.
3. Multiple roots in one interval: If there are two roots in [a, b], the sign change may cancel out. You must check that only one root exists in the interval.
Tip: Always sketch f or check intermediate values before applying the sign-change test. A graph confirms there is exactly one root in your chosen interval.
Newton-Raphson Method
The Newton-Raphson method finds successive approximations to a root by following the tangent line to the curve at each point.
xₙₖ₁ = xₙ − f(xₙ) / f'(xₙ)
Geometric Meaning
At the point (xₙ, f(xₙ)) on the curve, draw the tangent line. This tangent has gradient f'(xₙ). The tangent crosses the x-axis at xₙₖ₁, which is closer to the root α than xₙ (usually).
Derivation from tangent line:
Tangent at (xₙ, f(xₙ)): y − f(xₙ) = f'(xₙ)(x − xₙ)
Set y = 0: −f(xₙ) = f'(xₙ)(x − xₙ)
Solve for x: x = xₙ − f(xₙ)/f'(xₙ) = xₙₖ₁
Procedure
1. Identify f(x) and compute f'(x) analytically.
2. Choose a starting value x₀ close to the root (given in exam or from sign-change).
3. Apply the formula repeatedly: x₁ = x₀ − f(x₀)/f'(x₀), then x₂, x₃, …
4. Stop when successive values agree to the required precision (usually 5+ d.p.).
5. Verify with a sign-change check at the stated precision.
When Newton-Raphson Fails
1. f'(xₙ) ≈ 0: The tangent is nearly horizontal; it meets the x-axis very far away, diverging from the root.
2. Poor starting value x₀: If x₀ is on the wrong side of a turning point, the iteration may converge to a different root or diverge entirely.
3. Oscillation: The iteration can flip between two values if the function has a particular shape near the root.
Exam tip: Always show at least 4 iterations to 5 decimal places. Conclude: "The sequence converges to α = 1.23456 (5 d.p.)."
Fixed-Point Iteration
Any equation f(x) = 0 can be rearranged into the form x = g(x). Starting from x₀, the sequence xₙₖ₁ = g(xₙ) may converge to a root α where α = g(α).
xₙₖ₁ = g(xₙ) | Convergence requires |g'(α)| < 1
Convergence Criterion
Near the fixed point α:
• If |g'(α)| < 1: the iteration converges to α.
• If |g'(α)| > 1: the iteration diverges away from α.
• If |g'(α)| = 1: the test is inconclusive (higher-order analysis needed).
Staircase Diagrams
When 0 < g'(α) < 1, successive iterates approach α in a staircase pattern — always on the same side, stepping closer monotonically.
Cobweb Diagrams
When −1 < g'(α) < 0, the iterates alternate sides of α, spiralling inward in a cobweb pattern.
How to sketch: Draw y = g(x) and y = x on the same axes. From (x₀, x₀) go vertically to the curve y = g(x) to get (x₀, x₁), then horizontally to y = x to get (x₁, x₁), then repeat. The path traces the staircase or cobweb.
Choosing a Good Rearrangement
Not every rearrangement converges. For example, x³ − 3x − 1 = 0 can be rearranged as:
(a) x = (x³ − 1)/3 → g'(x) = x² diverges near x = 1.88
(b) x = ∛(3x + 1) → g'(x) = 1/[3(3x+1)^(2/3)] < 1 near x = 1.88 → converges
The Trapezium Rule
The trapezium rule approximates a definite integral by dividing the area under the curve into n trapezia of equal width h = (b − a)/n.
1. State n (number of strips) and compute h = (b − a)/n.
2. List the x-values: x₀ = a, x₁ = a+h, …, xₙ = b.
3. Evaluate y₀, y₁, …, yₙ (usually to 4–5 d.p.).
4. Apply the formula: ½h[y₀ + 2(y₁+…+yₙ₋₁) + yₙ].
5. State the answer to the required accuracy.
Over- or Under-Estimate?
• If the curve is concave up (f''(x) > 0, curves upward), the trapezia lie above the curve → overestimate.
• If the curve is concave down (f''(x) < 0, curves downward), the trapezia lie below the curve → underestimate.
• For a mixed curve, you cannot state simply — must say the answer may be either.
Increasing Accuracy
Doubling n (halving h) approximately halves the error. Using more strips always gives a better approximation. In the exam, if asked "how could the estimate be improved?", answer: "Increase the number of strips (decrease h)."
Error ≈ O(h²) — error is proportional to h² (halving h quarters the error for smooth functions)
Write: "Since there is a sign change and f is continuous on [a,b], there is a root α ∈ (a, b)."
To verify to 3 d.p.: evaluate f at [root − 0.0005, root + 0.0005] and show a sign change.
Newton-Raphson in Exam
State f(x) and f'(x) explicitly.
Write the formula: xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ).
Show x₀, x₁, x₂, x₃, x₄ to at least 5 d.p.
Conclude: "The values converge to α = [value] to [precision]."
Fixed-Point Iteration in Exam
Show the rearrangement f(x) = 0 → x = g(x) algebraically.
If asked to show convergence: compute g'(x) and show |g'(α)| < 1.
List x₀, x₁, x₂, … until two successive values agree to required d.p.
Trapezium Rule in Exam
Draw a clear table of x and y values.
Apply the formula, showing the bracket expansion.
If asked over/under-estimate: comment on concavity with reason.
Common mark allocations (9709 style):
M1 — setting up the method correctly (table, formula, iteration formula stated)
A1 — correct numerical values to required accuracy
B1 — correct statement about sign change or over/under-estimate
Golden rule: Never round intermediate values. Keep full calculator precision until the final answer, then round only once.
Example 1 — Sign Change to Locate Root
Show that f(x) = x³ − 3x − 1 has a root between x = 1 and x = 2.
f(1) = 1 − 3 − 1 = −3 < 0 M1
f(2) = 8 − 6 − 1 = 1 > 0 A1
Since f(1) < 0 and f(2) > 0 and f is continuous, there is a root α ∈ (1, 2). B1
Example 2 — Newton-Raphson for x³ − 3x − 1 = 0
Use Newton-Raphson with x₀ = 2 to find the root correct to 3 d.p.
If |g'(α)| < 1, then |eₙ₊₁| < |eₙ|, so the error decreases geometrically → convergence.
If |g'(α)| > 1, then |eₙ₊₁| > |eₙ|, so the error grows → divergence. ∎
Proof 3 — Trapezium Rule Area Derivation
Divide [a, b] into n equal strips of width h = (b−a)/n. On the i-th strip [xᵢ₋₁, xᵢ], approximate the curve by the chord joining (xᵢ₋₁, yᵢ₋₁) and (xᵢ, yᵢ). The area of this trapezium is:
Aᵢ = ½h(yᵢ₋₁ + yᵢ)
Summing over all n strips:
∫ₐᵇ f(x) dx ≈ Σᵢ₌₁ⁿ ½h(yᵢ₋₁ + yᵢ)
Expanding the sum: y₀ appears once, y₁ appears in strips 1 and 2 (twice), …, yₙ₋₁ appears twice, yₙ appears once:
= ½h[y₀ + 2y₁ + 2y₂ + … + 2yₙ₋₁ + yₙ] ∎
The approximation error for each strip is O(h³); summing n = O(1/h) strips gives total error O(h²).
Interactive Visualiser
Visualising f(x) = x³ − 3x − 1 on [−3, 3]. Enter x₀ and step through Newton-Raphson iterations.
Enter x₀ and press Start.
n
xₙ
f(xₙ)
f'(xₙ)
xₙ₊₁
4
Exercise 1 — Sign-Change & Root Location
Exercise 2 — Newton-Raphson
Exercise 3 — Fixed-Point Iteration
Exercise 4 — Trapezium Rule
Exercise 5 — Mixed Methods
Practice — 30 Questions
Challenge — 15 Harder Questions
Exam Style Questions
Q1 [4 marks]
The equation x³ + 2x − 5 = 0 has exactly one real root. Show that this root lies in the interval (1, 2). Starting with x₀ = 1.3, perform two iterations of Newton-Raphson and state the root to 3 significant figures.
Use the trapezium rule with 5 strips to estimate ∫₁³ ln x dx. Show all ordinate values. State whether your answer is an over- or underestimate, giving a reason.
Show that x = (x³+1)/3 is a rearrangement of x³−3x−1=0. Starting with x₀ = −0.3, obtain x₁, x₂, x₃. By calculating |g'(x)| near the root, explain whether this iteration converges.
B1 x³−3x−1=0 → x³ = 3x+1 → cannot directly give x=(x³+1)/3 from 3x+1 without swap. Actually: 3x = x³−1 → x=(x³−1)/3. As stated g(x)=(x³+1)/3 is a different rearrangement: 3x=x³+1 → x³−3x+1=0 (slightly different eq). Checking the iteration itself: M1 g(x)=(x³+1)/3; x₁=(−0.027+1)/3=0.3243; x₂=(0.034+1)/3=0.3447; x₃=(0.041+1)/3=0.3470. A1 Values stabilising near 0.347. M1 g'(x)=x². At x≈0.347: |g'|≈0.120 < 1. A1 Since |g'(α)| < 1, the iteration converges.
Q4 [4 marks]
The iteration xₙ₊₁ = √(3xₙ+1) is used to find a root of x²−3x−1=0. Show algebraically that this is a valid rearrangement. Starting with x₀=3.3, find the root correct to 2 decimal places.
Explain why the sign-change method cannot be used to show that f(x) = (x−2)² has a root at x = 2 by evaluating f(1.5) and f(2.5).
M1 f(1.5) = 0.25 > 0; f(2.5) = 0.25 > 0. A1 Both values are positive — there is no sign change. B1 The root at x=2 is a repeated (tangential) root — the function touches but does not cross the x-axis, so the sign-change method fails to detect it.
Q6 [4 marks]
Use Newton-Raphson with x₀ = 0.5 to find the positive root of f(x) = 2x − tan x = 0 near x = 1, performing 3 iterations. Give your answer to 4 d.p.
B1 f'(x) = 2 − sec²x. Formula: xₙ₊₁ = xₙ − (2xₙ − tan xₙ)/(2 − sec²xₙ). M1 x₁ = 0.5 − (1−0.5463)/(2−1.2984) = 0.5 − (0.4537/0.7016)... wait, f(0.5)=1−tan0.5=1−0.5463=0.4537; f'(0.5)=2−sec²(0.5)=2−1.2984=0.7016. x₁=0.5−0.4537/0.7016=0.5−0.6467=−0.147. This converges to 0 (trivial root). For root near x=1.17: use x₀=1. M1 x₁=1−(2−1.5574)/(2−3.4255)=1−0.4426/(−1.4255)=1+0.3104=1.3104. A1 Continuing: root ≈ 1.1656 (4 d.p.) after further iterations.
Q7 [3 marks]
The trapezium rule with n strips gives an estimate T(n) for ∫₀¹ x² dx. The exact value is 1/3. If T(4) ≈ 0.34375, explain how you could obtain a better estimate without computing more ordinates.
B1 The error in the trapezium rule is O(h²), where h = 1/n. M1 Halving h (doubling n to 8) would reduce the error by a factor of approximately 4. A1 Alternatively, Richardson extrapolation: 2T(2n)−T(n) eliminates the leading error term and gives a more accurate estimate. (Accept: "use more strips" with explanation of why this helps.)
Q8 [5 marks]
A curve satisfies y = e^(x/2). (a) Use the trapezium rule with 4 strips to estimate ∫₀² e^(x/2) dx. (b) State the exact value and determine whether your estimate is an over- or underestimate with a reason.
The equation x + 3 ln x = 6 has one real root. Show that the root lies in the interval (3, 4). Use Newton-Raphson with x₀ = 3.5 to find the root correct to 3 decimal places.
B1 Let f(x) = x+3ln x−6. f(3)=3+3.2958−6=0.2958>0; f(4)=4+4.1589−6=2.1589... Wait: f(3)=3+3ln3−6=3+3.2958−6=0.2958>0 and f(2)=2+3ln2−6=2+2.0794−6=−1.9206<0. Root in (2,3). B1 Correcting: f(3)>0, f(2)<0 → root in (2,3). Use x₀=2.5. M1 f'(x)=1+3/x. x₁=2.5−(2.5+3ln2.5−6)/(1+3/2.5)=2.5−(2.5+2.7489−6)/2.2=2.5−(−0.7511)/2.2=2.5+0.3414=2.8414. A1 x₂≈2.9098, x₃≈2.9147, x₄≈2.9147. Root≈2.915 (3 d.p.). A1 Verify: f(2.9145)<0, f(2.9155)>0 → root=2.915. ✓
PP Q2 — 9709/32/M/J/20 style [6 marks]
Use the trapezium rule with 6 strips to find an approximation to ∫₀³ √(1+x³) dx, giving your answer to 3 significant figures. Explain, with a reason, whether the true value is greater or less than your approximation.
M1 h=0.5; x: 0, 0.5, 1, 1.5, 2, 2.5, 3. M1 y=√(1+x³): 1, 1.0308, 1.4142, 2.0916, 3.0000, 4.0927, 5.2915. A1 ≈½(0.5)[1+2(1.0308+1.4142+2.0916+3.0000+4.0927)+5.2915]
=0.25[1+2(11.6293)+5.2915]=0.25[1+23.2586+5.2915]=0.25×29.5501=7.388. A1 Approximation ≈ 7.39 (3 s.f.). B1 f(x)=√(1+x³) is concave up for x>0 (check f''>0). Trapezia overestimate. B1 True value is less than 7.39.
PP Q3 — 9709/23/O/N/21 style [5 marks]
Show that the equation x³ = 5 − 2x can be written as x = ∛(5−2x). Use the iteration xₙ₊₁ = ∛(5−2xₙ) with x₀ = 1.5 to find the root to 2 decimal places. Determine the nature of the convergence diagram.
B1 x³=5−2x → x=∛(5−2x). ✓ M1 g(x)=(5−2x)^(1/3). x₁=∛(5−3)=∛2=1.2599; x₂=∛(5−2.5198)=∛2.4802=1.2536; x₃=∛2.4928=1.2553; x₄≈1.2549. A1 Root ≈ 1.25 (2 d.p.). M1 g'(x)=−2/[3(5−2x)^(2/3)]. At x≈1.255: g'≈−2/[3×1.573]=−0.424. So g'(α)≈−0.424. A1 −1<g'(α)<0 → cobweb diagram (iterates alternate either side of the root).
PP Q4 — 9709/32/O/N/18 style [4 marks]
The sequence x₀=1, xₙ₊₁=2sin(xₙ) is used to find a root of 2sin x−x=0. Find x₄ correct to 4 decimal places. State the equation whose root is being found and explain whether the iteration converges.
M1 x₁=2sin1=1.6829; x₂=2sin(1.6829)=1.9972; x₃=2sin(1.9972)=1.8196; x₄=2sin(1.8196)=1.9367. A1 x₄≈1.9367. B1 Root of 2sin x=x, i.e., 2sin x−x=0. B1 g(x)=2sin x; g'(x)=2cos x. At x≈1.9: |g'|=|2cos1.9|=|2×(−0.3233)|=0.647<1 → converges (cobweb since g'<0).
PP Q5 — 9709/22/M/J/17 style [6 marks]
f(x) = x⁴ − 8x + 3. (a) Show there is a root near x=2. (b) Apply Newton-Raphson twice from x₀=2 to find the root correct to 3 d.p. (c) Show by sign change that your answer is correct to 3 d.p.