Numerical methods

Q: Do I need to prove convergence in CIE exams?

No. In 9709 you are only asked to locate a root by a change of sign, or to use an iterative formula that is given to you to reach a required accuracy. You are not required to test in advance whether an iteration will converge — you should simply understand that some rearrangements converge to the root while others fail to converge.

CIE A-Level Mathematics· 15 min read

1. Locating Roots by Sign Change★★☆☆☆⏱ 15 min

If a function $f (x)$ is continuous on the closed interval $[a, b]$ , and $f (a)$ and $f (b)$ have opposite signs, then there is at least one root of $f (x) = 0$ in the open interval $(a, b)$ . This is the foundation of all numerical root finding.

📘 Definition

Root of an equation

A value $x = α$ such that $f (α) = 0$ , also called a zero of the function $f$ .

📐 Worked Example

Show that $e^{x} - 3 x^{2} = 0$ has a root between $x = 3$ and $x = 4$ .

1
Define the function for the equation:
2
$f(x) = e^x - 3x^2$
3
Evaluate $f (x)$ at the endpoints of the interval:
4
$f(3) = e^3 - 3(3)^2 \approx 20.0855 - 27 = -6.9145 < 0$
5
$f(4) = e^4 - 3(4)^2 \approx 54.598 - 48 = 6.598 > 0$
6
$f (x)$ is continuous for all real $x$ , as a combination of exponential and polynomial functions. Since $f (3)$ and $f (4)$ have opposite signs, there is at least one root in $(3, 4)$ .

Exam tip:

Always explicitly state that the function is continuous when justifying a root in an interval, CIE allocates one mark for this step.

2. Fixed Point Iteration★★★☆☆⏱ 20 min

To find an approximate value for a root, we rearrange $f (x) = 0$ into the form $x = g (x)$ . Starting from an initial guess $x_{0}$ , we generate a sequence of approximations using the recurrence relation $x_{n + 1} = g (x_{n})$ .

📐 Worked Example

Find the root of $x^{3} - 3 x + 1 = 0$ between 0 and 1, correct to 2 decimal places, starting from $x_{0} = 0.5$ .

1
Rearrange the equation into $x = g (x)$ form:
2
$x^3 + 1 = 3x \implies x = \frac{x^3 + 1}{3}, \quad g(x) = \frac{x^3 + 1}{3}$
3
Starting from $x_{0} = 0.5$ , iterate and observe the values settling towards the root:
4
$x_1 = \frac{(0.5)^3 + 1}{3} = 0.375$
5
$x_2 = \frac{(0.375)^3 + 1}{3} \approx 0.35095$
6
$x_3 \approx 0.34774, \quad x_4 \approx 0.34730$
7
Both $x_{3}$ and $x_{4}$ round to 0.35 to 2 decimal places, so the root is approximately 0.35.

Exam tip:

Always show two successive iterations that round to the same value to confirm your accuracy is correct.

3. When an Iteration Fails to Converge★★★☆☆⏱ 10 min

An equation $f (x) = 0$ can be rearranged into the form $x = g (x)$ in more than one way, and the choice matters. Some rearrangements produce a sequence of approximations that settles onto the root, while others produce a sequence that moves away from it, so the iteration fails to converge. (A rearrangement may also settle onto a different root of the same equation.) You are only ever asked to use a formula that is given to you, but you should understand that not every rearrangement works.

📐 Worked Example

The equation $x^{2} - x - 1 = 0$ has a positive root $α = \frac{1 + 5}{2} \approx 1.618$ . Compare two rearrangements, both starting from $x_{0} = 2$ .

1
Rearrangement A: $x^{2} = x + 1 ⟹ x = 1 + x$ , giving $x_{n + 1} = 1 + x_{n}$ . Starting from $x_{0} = 2$ :
2
$x_1 = 1.7321, \quad x_2 = 1.6529, \quad x_3 = 1.6288, \quad x_4 = 1.6213, \quad x_5 = 1.6191$
3
The values are settling down towards $1.618$ : this iteration converges to the root.
4
Rearrangement B: $x^{2} - 1 = x ⟹ x = x^{2} - 1$ , giving $x_{n + 1} = x_{n}^{2} - 1$ . Starting from the same $x_{0} = 2$ :
5
$x_1 = 3, \quad x_2 = 8, \quad x_3 = 63, \quad x_4 = 3968, \quad \ldots$
6
The values grow without bound and move away from the root: this iteration diverges. The same equation, with a different rearrangement, fails to converge.

Exam tip:

If a given iteration diverges, do not assume you have made an arithmetic slip. Some rearrangements simply do not converge, and 9709 never asks you to predict this in advance — you only ever use a formula that you are given.

4. Common Pitfalls

Wrong move:

Forgetting to state the function is continuous when justifying a root in an interval

Why:

CIE explicitly allocates one mark for this statement, which is often lost

Correct move:

Always add the line 'f(x) is continuous on [a, b]' after calculating f(a) and f(b)

Wrong move:

Stopping after one iteration when asked for a given accuracy

Why:

Examiners require you to confirm your approximation is correct to the required accuracy

Correct move:

Calculate at least two successive approximations that round to the same value

Wrong move:

Claiming no root exists when there is no sign change in an interval

Why:

Tangent roots and pairs of roots do not produce a sign change but still exist

Correct move:

Only use sign change to confirm a root exists, never to confirm a root does not exist

5. Quick Reference Cheatsheet

Method	Key Rule	Accuracy Check
Sign Change	f(a)f(b) < 0 + continuous = root in (a,b)	Check interval bounds round to same value
Fixed Point Iteration	$x_{n + 1} = g (x_{n})$ from a given rearrangement	Two iterations round to same value

6. Frequently Asked

Do I need to prove convergence in CIE exams?

No. In 9709 you are only asked to locate a root by a change of sign, or to use an iterative formula that is given to you to reach a required accuracy. You are not required to test in advance whether an iteration will converge — you should simply understand that some rearrangements converge to the root while others fail to converge.

Going deeper

syllabusOfficial CIE 9709 P3 Syllabus

What's Next

Numerical methods are the foundation for solving real-world problems in engineering, data science and physics, where most equations cannot be solved algebraically. The iterative principles you learned here for root finding extend directly to numerical integration, differential equations and other advanced numerical techniques. Working carefully and systematically, and recording your successive approximations clearly, will help you master the numerical problems that appear in CIE P2/P3 exams.