Most equations cannot be solved exactly. OCR section 1.09 gives you three tools to close in on a root — change of sign, fixed-point iteration and Newton–Raphson — and, just as important, asks you to explain when each one fails.
Throughout we hunt the roots of f(x) = x³ − 5x + 1. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then f must cross zero somewhere between them — so a root lies in the interval.
f(2) = 8 − 10 + 1 = −1 (negative)
f(2.5) = 15.625 − 12.5 + 1 = +4.125 (positive)
f is a polynomial, so it is continuous. The sign changes, so there is a root between 2 and 2.5.
Two ways it fails. (1) A discontinuity: f(x) = 1/(x − 2) changes sign across x = 2 but has an asymptote there, not a root. (2) An even number of roots in the interval: the signs at the ends match, yet roots are hiding inside. Always state that f is continuous — it is a marked point.
Rearrange f(x) = 0 into the form x = g(x), then iterate: xn+1 = g(xn). If it converges, the limit is a root.
Rearrange: x³ = 5x − 1, so x = ∛(5x − 1).
Start at x₀ = 2:
x₁ = ∛(5 × 2 − 1) = ∛9 = 2.0801 (4 d.p.)
x₂ = ∛(5 × 2.0801 − 1) = ∛9.4005 = 2.1105
x₃ = ∛(5 × 2.1105 − 1) = ∛9.5525 = 2.1218 … creeping towards the root 2.1284.
Staircase or cobweb? Sketch y = x and y = g(x). The iteration is drawn by bouncing between the two curves. It produces a staircase when g′ is positive, and a cobweb (spiralling in) when g′ is negative. It converges only when |g′(x)| < 1 near the root — a different rearrangement of the same equation can diverge.
Take the tangent to the curve at your current estimate, and see where it cuts the x-axis. That is your next estimate. It converges very fast when it works.
f′(x) = 3x² − 5.
f(2) = −1 and f′(2) = 12 − 5 = 7.
x₁ = 2 − (−1)/7 = 2 + 0.142857 = 2.1429 (4 d.p.).
Next: f(2.142857) = 0.12537, f′(2.142857) = 8.77551.
x₂ = 2.142857 − 0.12537/8.77551 = 2.142857 − 0.014286 = 2.1286 (4 d.p.).
Compare the iteration method, which needed many more steps to reach 4-figure accuracy. Newton–Raphson roughly doubles the number of correct digits each step.
How it fails. If f′(xn) = 0 the tangent is horizontal and never meets the x-axis — the formula divides by zero. A starting value near a turning point can also fling the next estimate far away, or towards a completely different root.
The same idea applies to areas: when you cannot integrate exactly, approximate.
h = (3 − 1)/2 = 1. Ordinates: y(1) = 1, y(2) = 0.5, y(3) = 0.33333.
Estimate = (1/2)[1 + 0.33333 + 2(0.5)] = 0.5 × 2.33333 = 1.167 (3 d.p.).
Compare the exact value: ∫1/x dx = ln 3 = 1.099. Our estimate is too big — as expected, because 1/x is convex, so the chords lie above the curve.
Improving it: doubling the number of strips roughly quarters the error of the trapezium rule. It never becomes exact for a curved graph.
Tap a description, then tap the method it belongs to.
Tap a description on the left, then the formula or condition it names.
Change of sign: f continuous and f(a), f(b) opposite signs ⇒ a root lies between
Its failures: discontinuities (asymptotes) and an even number of roots in the interval
Iteration: xn+1 = g(xn); staircase (g′ positive) or cobweb (g′ negative); needs |g′| < 1
Newton–Raphson: xn+1 = xn − f(xn)/f′(xn) — fast, but fails when f′ = 0
Trapezium rule: (h/2)[y₀ + yn + 2(middles)] with h = (b − a)/n
Accuracy: the rule overestimates a convex curve; more strips means less error
That is OCR 1.09 — and in the exam the explanation of a failure is worth as much as the calculation. Press Finish to see your score.
You've worked through Numerical methods for OCR A-level Mathematics A. 🎉
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