Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 2: Pure Mathematics 2 · official syllabus
NUMERICAL METHODS LAB
Learn it by playing it
Four short missions on bracketing a root with a sign change, climbing a fixed-point iteration, running a given rule to a prescribed accuracy, and spotting the rearrangement that actually converges. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Hunt the sign change
The curve is f(x) = x^3 - 2x - 5, with its value marked at every integer. Somewhere it crosses the x-axis. Slide k to an integer where f(k) and f(k + 1) have opposite signs - that brackets a root.
Before you slide: f is a continuous curve and f(a) and f(b) have opposite signs. That means...
f(0) = -5 and f(1) = -6 - same sign, so this interval proves nothing. Keep hunting.
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 sign change 2.6
If f is continuous and f(a) and f(b) have opposite signs, the curve crosses the x-axis somewhere between a and b, so a root lies in the interval (a, b). This is the intermediate value idea, and it is the standard way to locate a root approximately.
Worked example: for f(x) = x^3 - 2x - 5, try integers: f(2) = 8 - 4 - 5 = -1 and f(3) = 27 - 6 - 5 = 16. The sign change brackets a root in (2, 3), and no other consecutive integer pair in 0 to 5 changes sign, so this is the unit interval to work in.
A sketch does the same job graphically: where the curve crosses the axis you can read off roughly where the root sits, which tells you which integers to test.
A warning on the converse: no sign change does not prove there is no root - a curve can touch the axis or cross it twice inside one interval. The sign change proves existence; its absence proves nothing.
Iteration notation and fixed points 2.6
An iteration writes a sequence of approximations as x(n+1) = F(x(n)): each new estimate is the rule F applied to the previous one, starting from an initial value x0.
If the sequence settles at a value x*, then x* = F(x*): it is a fixed point of F. When F came from rearranging an equation into the form x = F(x), that fixed point is a root of the original equation.
Worked example: x^3 - 2x - 5 = 0 rearranges to x^3 = 2x + 5 and then x = (2x + 5)^(1/3). A fixed point of F(x) = (2x + 5)^(1/3) satisfies x^3 = 2x + 5, which is the original equation.
The same equation always has several rearrangements, and the choice matters - one F can converge while another fails, which the next sections show.
Using a given iteration to a prescribed accuracy 2.6
Worked example: solve x^3 - 2x - 5 = 0 with x(n+1) = (2x(n) + 5)^(1/3) starting from x0 = 2. The values run x1 = 2.0801, x2 = 2.0924, x3 = 2.0942, x4 = 2.0945, x5 = 2.0945 (4 decimal places).
Stop when two consecutive values agree at the accuracy asked for: x4 and x5 both read 2.0945 to 4 decimal places, so the root is 2.09 to 2 decimal places. The root itself is 2.094551..., a non-terminating decimal - every answer here is a rounded value, so state the rounding.
A sign change confirms the rounded answer brackets the true root: f(2.085) and f(2.095) have opposite signs, which traps the root in (2.085, 2.095) and justifies 2.09 to 2 decimal places.
Exam questions usually hand you the iteration and the starting value; your job is to iterate, show enough terms to justify the accuracy, and conclude with a correctly rounded statement.
Rearrangements can fail to converge 2.6
The syllabus does not require the convergence condition, but it does expect you to know that an iteration may fail to converge - and the worked examples make that concrete.
All three of these are honest rearrangements of x^3 - 2x - 5 = 0: x = (2x + 5)^(1/3), x = (x^3 - 5)/2 and x = x^3 - x - 5. Only the first converges.
From x0 = 2, the iteration x(n+1) = (x(n)^3 - 5)/2 runs 2 -> 1.5 -> -0.8125 -> -2.768 -> -13.1 -> -1128, sprinting away from the root, and x(n+1) = x(n)^3 - x(n) - 5 runs 2 -> 1 -> -5 -> -125 -> -1953005. Same equation, same start, total failure.
So the rearrangement is a real choice, not a cosmetic one. When a question gives you the iteration to use it has done this work for you; when it asks you to rearrange, pick the form that isolates x gently (a cube root, a reciprocal) rather than one that leaves x^3 on the right.
Worked example: cos x = x 2.6
The equation cos x = x has no exact solution, but the graphs y = cos x and y = x cross once, so a single root exists between 0 and 1 (cos 0 = 1 > 0 and cos 1 = 0.5403 < 1, a sign change for g(x) = cos x - x).
Use the given iteration x(n+1) = cos(x(n)) with x0 = 1 (radians): the values run 1, 0.5403, 0.8576, 0.6543, 0.7935, 0.7014, 0.7640, 0.7221, 0.7504, ... zigzagging inward on either side of the answer.
The sequence settles at 0.739085..., the fixed point where cos x = x. To 2 decimal places the root is 0.74.
The zigzag is typical of converging iterations: successive estimates alternate above and below the root while the gap shrinks. Watching two consecutive terms agree at the required accuracy remains the stopping rule.
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