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A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.6 (also Paper 2) · ~6 min read

Numerical solution of equations

Locating a root by a change of sign or by sketching two graphs, rearranging an equation into x = F(x), running the iteration xₙ₊₁ = F(xₙ) to a stated accuracy, and understanding why some rearrangements converge and others do not.

Change of signSketching to count rootsRearranging to x = F(x)IterationStaircase & cobweb diagramsConvergenceStating accuracy
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Change of sign: if f(x) is continuous and f(a) and f(b) have opposite signs, there is a root between a and b.
  • Iteration: rearrange f(x) = 0 into x = F(x), pick x₁, and keep calculating xₙ₊₁ = F(xₙ). If the values settle, the limit is a root.
  • Convergence needs ǀF′(x)ǀ < 1 near the root. A different rearrangement of the same equation may run away instead.
  • Accuracy: to give a root to 2 d.p., iterate until two successive values agree to 2 d.p. (show 4 d.p. in working), or check the sign change across the interval ±0.005.

By the end you can

  • Locate roots graphically or by a sign change, and know when a sign change can mislead.
  • Show that a root satisfies a given rearranged equation.
  • Use an iterative formula to find a root to a given accuracy, showing all iterates.
  • Understand convergence and divergence using staircase and cobweb diagrams.

1Locating roots

Sign change and sketches

  • Sketch two graphs: to count the roots of eˣ = 3 − x, sketch y = eˣ and y = 3 − x. Each crossing is a root. Questions often say "by sketching a suitable pair of graphs".
  • Sign change: evaluate f at both ends and state the signs clearly, with a conclusion sentence: "the sign changes and f is continuous, so there is a root in (a, b)".
  • When it fails: a vertical asymptote (e.g. tan x or 1/(x − 2)) gives a sign change with no root; two roots close together give no sign change at all.
Worked example

Show that x³ + 2x − 5 = 0 has a root between 1 and 2.

  1. f(x) = x³ + 2x − 5; f(1) = −2 < 0, f(2) = 7 > 0 M1
  2. sign change and f is continuous, so a root lies between 1 and 2 A1
Common mistakeMissing the conclusion, or writing "f(1) = −2, f(2) = 7" with no comment. The mark needs the sign change and a statement.

2Iteration

xₙ₊₁ = F(xₙ)

  • Show that a root satisfies x = F(x): start from the original equation and rearrange step by step to the given form. For example x³ + 2x − 5 = 0 ⇒ x³ = 5 − 2x ⇒ x = ∛(5 − 2x).
  • Calculator routine: type x₁, press =, then type F(Ans) and press = repeatedly. Write down every iterate to 4 or more d.p.
Worked example

Use xₙ₊₁ = ∛(5 − 2xₙ) with x₁ = 1.3 to find the root to 3 d.p.

  1. x₂ = 1.33887, x₃ = 1.32425, x₄ = 1.32978, x₅ = 1.32770, x₆ = 1.32849, x₇ = 1.32819, x₈ = 1.32830, x₉ = 1.32826 M1 A1
  2. values settle at 1.328 (3 d.p.) A1
Worked example

The stationary point of y = x cos x for 0 < x < π/2 has x-coordinate p. Show that p = tan⁻¹(1/p), and use an iteration to find p to 2 d.p.

  1. dy/dx = cos x − x sin x = 0 ⇒ tan x = 1/x ⇒ x = tan⁻¹(1/x) M1 A1
  2. x₁ = 1: 0.7854, 0.9050, 0.8352, 0.8749, 0.8520, 0.8651, 0.8576, 0.8619, 0.8594, 0.8609, … M1
  3. p = 0.86 A1 (check: f(x) = tan x − 1/x changes sign between 0.855 and 0.865)
Common mistakeStopping too early. Two iterates that round to the same value to the required accuracy are the minimum evidence. The 2024 Paper 3 report noted iterates given to only 2 d.p.

3Why iterations converge

Staircases and cobwebs

xy0.20.40.60.81.00.20.40.60.81.0y = e⁻ˣy = xx₁ = 0.1root 0.567xₙ₊₁ = e^(−xₙ)0.1 → 0.905 → 0.405→ 0.667 → 0.513→ … → 0.567Vertical to the curve, across to y = x, repeat. Here the path spirals (a cobweb) in to the root.
For xₙ₊₁ = e−xₙ the gradient of F is negative, so the iterates spiral in (a cobweb) towards 0.567.
  • Picture: draw y = x and y = F(x). Go up to the curve, across to y = x, up to the curve again. The root is where they cross.
  • Staircase: when 0 < F′(x) < 1, the values approach the root from one side. Cobweb: when −1 < F′(x) < 0, they alternate either side.
  • Divergence: when ǀF′(x)ǀ > 1 near the root, the iterates move away. Try a different rearrangement.
Worked example

The equation x = 2 + ln x has a root near 3. Compare the iterations xₙ₊₁ = 2 + ln xₙ and xₙ₊₁ = exₙ − 2.

  1. first: F′(x) = 1/x ≈ 0.32 near 3.15, so it converges: 3, 3.0986, 3.1310, 3.1413, 3.1446, 3.1457, 3.1460, … → 3.146 M1 A1
  2. second: F′(x) = ex − 2 ≈ 3.15 > 1, so it diverges: 3.1, 3.0042, 2.7296, 2.0743, 1.0771, … moving away A1
Common mistakeClaiming an iteration "doesn't work" because the root is wrong. Both forms have the same root; only the convergence differs.
Linking questionThe functions you iterate come from Logarithms and exponentials and Further trigonometry; stationary-point equations come from Further differentiation; and the trapezium rule is in Further integration.

Test yourself — cover the answers first

Quick check

  1. f(2) = −0.3 and f(3) = 1.4. What can you conclude if f is continuous?
  2. Why might f(x) = 1/(x − 2) give a sign change on (1, 3)?
  3. What condition makes x = F(x) converge?
  4. Iterates alternate either side of the root. What diagram is this?
  5. Rearrange x² − 3x − 1 = 0 into x = F(x) with a square root.

Exam-style

  1. Use xₙ₊₁ = e−xₙ with x₁ = 0.5 to find the root of x = e−x to 2 d.p.
  2. Show that x = 2 + ln x has a root between 3 and 4.
  3. Explain why xₙ₊₁ = exₙ − 2 cannot find the root near 3.15 of x = 2 + ln x.
Show answers
  1. there is a root between 2 and 3
  2. asymptote at x = 2, not a root
  3. ǀF′(x)ǀ < 1 near the root
  4. a cobweb
  5. x = √(3x + 1)
  6. 0.6065, 0.5452, 0.5797, 0.5601, 0.5712, 0.5649, … → 0.567, so 0.57
  7. f(x) = 2 + ln x − x: f(3) = 0.099 > 0, f(4) = −0.614 < 0; sign change, continuous, so a root
  8. F′(x) = ex − 2 ≈ 3.1 > 1 there, so the iterates diverge
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.