Mini-Lesson

Numerical methods

When an equation cannot be solved exactly, we approximate. This mini-lesson covers root location by change of sign, fixed-point iteration x = g(x) (with staircase and cobweb diagrams), the Newton–Raphson method, and the trapezium rule — including when it over- or under-estimates.

Where this sits: Unit A2 1: Pure Mathematics — numerical methods: location of roots, iterative methods including Newton–Raphson, and numerical integration by the trapezium rule.

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.

Root location

Change of sign

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

Worked example — f(x) = eˣ − 3x

f(1) = e − 3 = 2.718 − 3 = −0.282 (negative)

f(2) = e² − 6 = 7.389 − 6 = +1.389 (positive)

Sign change ⇒ a root lies between 1 and 2. ∎

Two traps: (1) the method fails if f is discontinuous (e.g. 1/(x − 1.5) changes sign across an asymptote with no root); (2) an even number of roots in the interval produces no sign change, so "no sign change" does not prove "no root".

Calculate

Change of sign

1f(x) = eˣ − 3x. Calculate f(1) to 3 decimal places.

Hint: f(1) = e¹ − 3(1) = 2.71828 − 3 = −0.28172.

Iteration

Fixed-point iteration x = g(x)

Rearrange f(x) = 0 into the form x = g(x), then iterate xₙ₊₁ = g(xₙ) from a starting value x₀.

Worked example — x³ − x − 3 = 0

Rearrange: x³ = x + 3 ⇒ x = (x + 3)^(1/3)

x₀ = 1 ⇒ x₁ = 4^(1/3) = 1.587

x₂ = (4.587)^(1/3) = 1.661 · x₃ = 1.671 · … converging to 1.6717

Convergence: the iteration converges near a root if |g′(x)| < 1 there, and diverges if |g′(x)| > 1. A different rearrangement of the same equation can diverge — that is why the exam gives you the rearrangement to use. Diagrams: a staircase when g′ > 0, a cobweb when g′ < 0.

Calculate

One iteration

2Use xₙ₊₁ = (xₙ + 3)^(1/3) with x₀ = 1 to find x₁, to 3 decimal places.

Hint: x₁ = (1 + 3)^(1/3) = 4^(1/3) = 1.5874.

Quick check

Why no sign change?

?f is continuous and f(2) and f(3) are both positive. What can you conclude about roots in (2, 3)?

Newton–Raphson

The tangent method

xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ)follow the tangent at xₙ down to the x-axis — that landing point is the next estimate

Worked example — f(x) = x³ − 2x − 5, x₀ = 2

f(2) = 8 − 4 − 5 = −1 · f′(x) = 3x² − 2 ⇒ f′(2) = 10

x₁ = 2 − (−1)/10 = 2.1

f(2.1) = 9.261 − 4.2 − 5 = 0.061 · f′(2.1) = 13.23 − 2 = 11.23

x₂ = 2.1 − 0.061/11.23 = 2.0946 (4 d.p.)

When it fails: if f′(x₀) = 0 the tangent is horizontal and never meets the axis (division by zero); a starting value near a turning point can throw the iteration far away.

Calculate

Newton–Raphson, first step

3f(x) = x³ − 2x − 5 and x₀ = 2. Use Newton–Raphson to find x₁.

Hint: f(2) = −1 and f′(2) = 3(4) − 2 = 10. x₁ = 2 − (−1)/10 = 2 + 0.1.

Calculate

Newton–Raphson, second step

4Continuing from x₁ = 2.1 for f(x) = x³ − 2x − 5, find x₂ to 4 decimal places.

Hint: f(2.1) = 9.261 − 4.2 − 5 = 0.061; f′(2.1) = 3(4.41) − 2 = 11.23. x₂ = 2.1 − 0.061/11.23 = 2.09457.

Numerical integration

The trapezium rule

∫ₐᵇ y dx ≈ h/2 [y₀ + 2(y₁ + y₂ + … + yₙ₋₁) + yₙ], h = (b − a)/nn strips means n + 1 ordinates — count them carefully

Worked example — ∫₀⁴ x² dx with 4 strips

h = 1, ordinates y = 0, 1, 4, 9, 16

T = ½[0 + 2(1 + 4 + 9) + 16] = ½[0 + 28 + 16] = 22

The exact value is [x³/3]₀⁴ = 64/3 = 21.33, so the trapezium rule over-estimates here.

Which way? The chords lie above a convex curve (f″ > 0), so the rule over-estimates; they lie below a concave curve, so it under-estimates. For a straight line it is exact. Increasing n improves accuracy.

Calculate

Trapezium rule

5Use the trapezium rule with 4 strips (h = 1) to estimate ∫₀⁴ x² dx.

Hint: Ordinates: 0, 1, 4, 9, 16. T = (1/2)[0 + 2(1 + 4 + 9) + 16] = 0.5 × 44.

Sort it

Over-estimate or under-estimate?

Tap a trapezium-rule estimate, then tap what it does compared with the true value.

⬆️ Over-estimate

⬇️ Under-estimate

🎯 Exact

Quick check

Newton–Raphson failure

?Newton–Raphson fails if the starting value x₀ makes f′(x₀) = 0. Why?

Quick check

Improving the trapezium estimate

?How do you make a trapezium-rule estimate more accurate?

Match it

Match the numerical method

Tap the idea on the left, then its meaning on the right.

Statement

Answer

Quick check

Reading an iteration

?An iteration gives x₁ = 1.587, x₂ = 1.661, x₃ = 1.671, x₄ = 1.6717, x₅ = 1.6717. What does this show?

Examiner traps

Numerical method pitfalls

"No sign change" does not mean "no root" — there could be two.
Sign change with a discontinuity (like 1/(x − 2)) does not imply a root. State that f is continuous.
Confirming a root to 3 d.p. requires a sign change over the interval (x − 0.0005, x + 0.0005) — not just a converged iteration.
Newton–Raphson fails if f′(x₀) = 0, or if x₀ is close to a turning point.
Trapezium ordinates: n strips means n + 1 ordinates, and h = (b − a)/n.
Over/under-estimate: justify it by the curve's convexity, not by guessing.

Recap

The big ideas to know

Change of sign: f continuous and f(a)f(b) < 0 ⇒ a root in (a, b) — beware discontinuities and even numbers of roots

Iteration: xₙ₊₁ = g(xₙ); converges if |g′| < 1 near the root (staircase or cobweb)

Newton–Raphson: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ) — fails when f′(xₙ) = 0

Trapezium rule: h/2[y₀ + 2(y₁ + … + yₙ₋₁) + yₙ], h = (b − a)/n

Convex → over-estimate · concave → under-estimate · straight line → exact

Numerical methods are the tools for the equations and integrals algebra cannot crack. Press Finish to see your score.

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