# Numerical methods

> CIE A-Level Mathematics · 9709 Pure Mathematics 2 & 3
> Source: https://www.owlsprep.com/study/cie-9709-u2-numerical-methods/

This subtopic covers techniques for finding approximate values of roots of equations that cannot be solved algebraically. We cover locating a root by a change of sign, and using a given iterative formula to approximate a root, including the fact that an iteration may fail to converge.

**Prerequisites:** [Differentiation of basic functions](https://www.owlsprep.com/study/cie-9709-u1-differentiation/); Properties of continuous functions; Basic sequence convergence

## Learning objectives

- Locate roots of continuous functions using the sign change principle
- Apply a given iterative formula $x_{n+1} = F(x_n)$ to find an approximate root
- Understand that an iteration may fail to converge, depending on the rearrangement used
- Give approximations to a specified degree of accuracy

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P2 past papers (2016–2025): **Numerical methods** appears **126 times** in the last 10 years — **14.3%** of all P2 questions (126 of 455).

Most-tested forms: Iteration: locating root + applying iterative formula (70), Newton-Raphson iteration (57).

## Locating Roots by Sign Change

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.

**Root of an equation** — A value $x = \alpha$ such that $f(\alpha) = 0$, also called a zero of the function $f$.

**Worked example:** Show that $e^x - 3x^2 = 0$ 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 $(3, 4)$.

> **warning**
>
> A sign change confirms at least one root exists, but no sign change does not confirm no root exists. Roots that touch the x-axis (tangent roots) or pairs of roots do not produce a sign change.

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

## Fixed Point Iteration

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 $x^3 - 3x + 1 = 0$ 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.

## When an Iteration Fails to Converge

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 $\alpha = \frac{1+\sqrt{5}}{2} \approx 1.618$. Compare two rearrangements, both starting from $x_0 = 2$.

1. Rearrangement A: $x^2 = x + 1 \implies x = \sqrt{1 + x}$, giving $x_{n+1} = \sqrt{1 + x_n}$. 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: $x^2 - 1 = x \implies x = x^2 - 1$, 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.

> **tip**
>
> A cobweb or staircase diagram (drawing the steps between the line $y = x$ and the curve $y = g(x)$) shows this visually: for a convergent rearrangement the steps close in on the root, while for a divergent one they step away from it. In practice, calculating the first few iterates is enough to see whether they are settling towards a value or spreading apart.

> **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.

## Common pitfalls

- **Wrong:** Forgetting to state the function is continuous when justifying a root in an interval
  - Why it fails: CIE explicitly allocates one mark for this statement, which is often lost
  - Correct: Always add the line 'f(x) is continuous on [a, b]' after calculating f(a) and f(b)
- **Wrong:** Stopping after one iteration when asked for a given accuracy
  - Why it fails: Examiners require you to confirm your approximation is correct to the required accuracy
  - Correct: Calculate at least two successive approximations that round to the same value
- **Wrong:** Claiming no root exists when there is no sign change in an interval
  - Why it fails: Tangent roots and pairs of roots do not produce a sign change but still exist
  - Correct: Only use sign change to confirm a root exists, never to confirm a root does not exist

## 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 |

## 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.

- [Differentiation](https://www.owlsprep.com/study/cie-9709-u1-differentiation/)
- [Functions](https://www.owlsprep.com/study/cie-9709-u1-functions/)
- [3D Vector Geometry (Lines and Planes)](https://www.owlsprep.com/study/cie-9709-u2-vectors/)

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