Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
NUMERICAL SOLUTIONS LAB
Learn it by playing it
Four short missions on trapping roots with sign changes and chasing them with iterations. In each one you read the picture, predict the method, and slide your answer onto the exact value - when a mission glows green, read why it worked, then take the next one.
1234
Mission 1 of 4
Trap the sign change
x^3 - x - 3 = 0 has no tidy algebra, so hunt its root instead. A continuous curve that starts below the axis and ends above it must cross in between. The white curve is f(x) = x^3 - x - 3; the shaded strip is your tenth-wide bracket [a, a + 0.1]. Slide the bracket until it traps the sign change.
Before you slide: a sign change between f(a) and f(b) tells you a root lies between a and b because...
Your bracket starts at 1.0 - the sign only flips between 1.6 and 1.7, so slide until the strip straddles the crossing.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Locating a root by a sign change 3.6
If f is continuous and f(a) and f(b) have opposite signs, then f(x) = 0 has a root between a and b - an unbroken curve cannot get from below the axis to above it without crossing.
Worked example: for f(x) = x^3 - x - 3, f(1) = -3 and f(2) = 3, so a root lies in (1, 2); tightening to tenths, f(1.6) = -0.504 and f(1.7) = 0.213 (each rounded to 3 d.p.) pin it in (1.6, 1.7).
The converse fails quietly: no sign change does not prove no root - a curve can touch the axis, or cross it twice inside one interval.
The continuity condition matters: 1/x changes sign between -1 and 1 but has no root, because it is not continuous at 0.
Decimal search: tightening the bracket 3.6
Once a root is trapped between two integers, search the tenths inside the bracket for the next sign change, then the hundredths inside that, until the root is pinned to the accuracy asked for.
Worked example: x^3 - x - 3 = 0 gives f(1.6) < 0 < f(1.7), then f(1.67) = -0.0125 and f(1.68) = 0.0616 (each rounded to 4 d.p.), so the root is 1.67 to 2 d.p.
To state a root correct to n decimal places, show the sign change across the n-d.p. bracket - stating the bracket endpoints and their signs is the working the mark scheme looks for.
A sketch or a table of values locates the integer bracket first; the decimal search then does the refining mechanically.
Fixed-point iteration x_(n+1) = F(x_n) 3.6
Rearrange f(x) = 0 into x = F(x); then a starting value x_0 generates x_1 = F(x_0), x_2 = F(x_1), and so on. If the sequence settles, it settles on a fixed point x = F(x), which is a root of the original equation.
Worked example: x^3 = 2x + 5 gives x_(n+1) = (2x_n + 5)^(1/3); from x_0 = 2 the terms are 2.0801, 2.0924, 2.0942, 2.0945 (each rounded to 4 d.p.), settling at 2.0946 (4 d.p.).
Another worked example: x_(n+1) = cos(x_n) from x_0 = 1 gives 0.5403, 0.8576, 0.6543, 0.7935, 0.7014 (each rounded to 4 d.p.), settling at 0.7391 (4 d.p.) - and cos(0.7391) = 0.7391 (4 d.p.) closes the loop.
On a diagram with y = x and y = F(x), the iteration draws a staircase or cobweb from x_0 to the intersection; the meeting point is the root.
When iteration converges 3.6
Near a root alpha, each error is roughly multiplied by F'(alpha): the iteration converges when |F'(alpha)| < 1 and diverges when |F'(alpha)| > 1.
A positive F'(alpha) below 1 gives a staircase that approaches from one side; a negative F'(alpha) above -1 gives a cobweb that spirals in from alternating sides.
Worked example: for F(x) = (2x + 5)^(1/3), F'(x) = 2/(3(2x + 5)^(2/3)) and |F'(2.0946)| = 0.1520 (4 d.p.) - under 1, so the iteration converges; for the rival rearrangement F(x) = (x^3 - 5)/2, |F'(2.0946)| = 6.58 (2 d.p.) - over 1, so it diverges.
One equation can rearrange into iterations with opposite fates, so the choice of rearrangement is the real decision - always check the derivative's size near the root.
Exam craft for numerical methods 3.6
Show every iterate to at least one more decimal place than the answer demands, and stop only when two successive terms agree to the required accuracy.
Keep radians for any iteration involving trig - x_(n+1) = cos(x_n) in degrees settles somewhere else entirely.
If a question supplies the rearrangement, use it as given and tabulate x_1, x_2, ...; if it asks you to verify a root to n d.p., use the sign-change bracket, not the iteration.
State the conclusion in words: 'the sequence converges to 2.0946, so the root is 2.095 to 3 d.p.' - the final rounding is a separate step from the iteration.
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