Numerical methods

Numerical methods produce controlled approximations when an exact answer is unavailable, unnecessary or too difficult to obtain. At A-level, you use them to:

• locate and approximate roots of equations;
• understand when an iterative process converges or fails;
• approximate a definite integral using the trapezium rule.

An approximation is not a guess. A complete numerical solution identifies the method, shows enough working, controls rounding and justifies the stated accuracy.

Prerequisites

You should be able to:

• evaluate functions and rearrange equations;
• interpret graphs, roots and intersections;
• differentiate standard functions;
• evaluate definite integrals;
• round to decimal places and significant figures.

Review functions, function graphs, differentiation basics or definite integrals and areas if any of these are uncertain.

The central principle: approximation with evidence

Suppose the equation

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

has no convenient exact solution. Define

$$f(x)=x^3+x-1.$$

There are several different questions you might ask.

1. Does a root exist in a given interval? Use continuity and a change of sign.
2. What is the root approximately? Use bisection, fixed point iteration or Newton Raphson iteration.
3. Will the chosen iteration work? Examine the iteration function, its derivative and the starting value.
4. How accurate is the result? Give a bracket, test rounding boundaries or apply an appropriate stopping criterion.

These questions are related, but they are not interchangeable. A calculator display such as $0.6823278038$ does not by itself prove that a root exists, that it is unique or that the displayed digits are correct.

Locating a root by a change of sign

If $f$ is continuous on $[a,b]$ and

$$f(a)f(b)<0,$$

then $f(a)$ and $f(b)$ have opposite signs. The graph must cross the $x$-axis at least once, so there is a root in $(a,b)$.

Worked example: establish a bracket

Show that $x^3+x-1=0$ has a root between $0.6$ and $0.7$.

The function $f(x)=x^3+x-1$ is a polynomial, so it is continuous. Also,

$$f(0.6)=0.6^3+0.6-1=-0.184$$

and

$$f(0.7)=0.7^3+0.7-1=0.043.$$

Thus

$$f(0.6)f(0.7)<0.$$

By continuity, at least one root lies in $(0.6,0.7)$.

Notice the careful conclusion: a sign change proves at least one root, not exactly one. Here,

$$f'(x)=3x^2+1>0,$$

so $f$ is strictly increasing and the root is unique. Learn how to halve a bracket repeatedly in locating roots using a change of sign.

Fixed point iteration

Rearrange an equation into the form

$$x=g(x),$$

choose $x_0$, then generate

$$\boxed{x_{n+1}=g(x_n)}.$$

If $x_n$ converges to a limit $\alpha$ and $g$ is continuous there, then

$$\alpha=g(\alpha),$$

so $\alpha$ is a fixed point and hence a solution of the rearranged equation.

Worked example: carry out an iteration

Use

$$x_{n+1}=\sqrt[3]{1-x_n}, \qquad x_0=0.7,$$

to find three further approximations to the root of $x^3+x-1=0$.

Substitute each value into the formula and retain full calculator precision:

$$\begin{aligned} x_1&=\sqrt[3]{1-0.7}=0.669432\ldots,\\ x_2&=\sqrt[3]{1-0.669432\ldots}=0.691541\ldots,\\ x_3&=\sqrt[3]{1-0.691541\ldots}=0.675771\ldots. \end{aligned}$$

The values alternate around the root. That is expected because $g'(x)<0$. Near a fixed point $\alpha$, errors approximately obey

$$x_{n+1}-\alpha\approx g'(\alpha)(x_n-\alpha).$$

Thus $|g'(\alpha)|<1$ usually makes nearby errors shrink, while $|g'(\alpha)|>1$ usually makes them grow. The rearrangement matters: algebraically equivalent formulae can have completely different numerical behaviour. Study fixed point iteration before when numerical iteration fails.

Newton Raphson iteration

Newton Raphson uses the tangent to $y=f(x)$ at $x=x_n$. The tangent meets the $x$-axis at the next approximation:

$$\boxed{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}}.$$

It is often much faster than bisection or simple fixed point iteration when the starting value is suitable and $f'(x)$ is not close to zero.

Worked example: one Newton Raphson step

For $f(x)=x^3+x-1$, take $x_0=0.7$. Since

$$f'(x)=3x^2+1,$$

the next approximation is

$$\begin{aligned} x_1 &=0.7-\frac{0.7^3+0.7-1}{3(0.7)^2+1}\\ &=0.7-\frac{0.043}{2.47}\\ &=0.682591\ldots. \end{aligned}$$

A second step gives $x_2=0.682328\ldots$, already close to the root. Fast convergence is not a guarantee: a poor starting value, a nearly horizontal tangent or a complicated graph can send the sequence away from the intended root. Continue with the Newton Raphson method.

Proving accuracy

Do not confuse successive approximations agreeing with a proof of accuracy. A robust way to prove a rounded root is to test its rounding boundaries.

To show that a root is $0.682$ correct to three decimal places, test

$$0.6815 \quad\text{and}\quad 0.6825.$$

For $f(x)=x^3+x-1$,

$$f(0.6815)=-0.002005\ldots, \qquad f(0.6825)=0.000416\ldots.$$

The continuous function changes sign between the rounding boundaries, so the unique root lies in $(0.6815,0.6825)$ and therefore rounds to

$$\boxed{0.682}$$

to three decimal places.

The trapezium rule

Numerical methods also approximate integrals. Divide $[a,b]$ into $n$ equal strips of width

$$h=\frac{b-a}{n}.$$

If $y_r=f(a+rh)$, the trapezium rule is

$$\boxed{ \int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0+2(y_1+y_2+\cdots+y_{n-1})+y_n\right] }.$$

The end ordinates occur once and every interior ordinate occurs twice.

Worked example: approximate an integral

Use four strips to approximate

$$\int_0^2 e^{-x^2}\,dx.$$

Here $a=0$, $b=2$, $n=4$, so

$$h=\frac{2-0}{4}=0.5.$$

The ordinates are:

$x$	$0$	$0.5$	$1.0$	$1.5$	$2.0$
$e^{-x^2}$	$1$	$0.778801$	$0.367879$	$0.105399$	$0.018316$

Therefore

$$\begin{aligned} \int_0^2e^{-x^2}\,dx &\approx \frac{0.5}{2}\left[1+2(0.778801+0.367879+0.105399)+0.018316\right]\\ &=0.880619\ldots\\ &\approx 0.881. \end{aligned}$$

The exact antiderivative is not an elementary A-level function, but the numerical rule still gives a useful estimate. Smaller strip widths usually improve the approximation. Whether the rule overestimates or underestimates depends on the curve’s concavity. Study the trapezium rule for unequal data presentation, error interpretation and exam technique.

Choosing a method

Task	Natural method	Main strength	Main caution
Prove a continuous function has a root in an interval	Change of sign	Reliable existence test	Does not prove uniqueness
Narrow a known sign changing bracket	Bisection	Guaranteed to retain a bracket	Converges relatively slowly
Generate roots from $x=g(x)$	Fixed point iteration	Simple and often calculator friendly	Choice of rearrangement controls convergence
Find a root rapidly using $f$ and $f'$	Newton Raphson	Usually fast near a simple root	Can fail or approach another root
Approximate a definite integral	Trapezium rule	Works from tabulated values	It is an approximation, not exact integration

Common misconceptions

• “Opposite signs mean exactly one root.” They prove at least one root, assuming continuity.
• “No sign change means no root.” A graph can touch the axis and turn around, as $f(x)=x^2$ does at $x=0$.
• “Every rearrangement converges.” The condition $x=g(x)$ identifies fixed points, but does not ensure iteration reaches one.
• “Matching decimal displays prove accuracy.” Agreement can be misleading, especially for slowly converging or cyclic sequences.
• “Newton Raphson always finds the nearest root.” Its tangent steps depend on the shape of the whole graph and the initial value.
• “The trapezium rule formula uses every ordinate twice.” Only interior ordinates have coefficient $2$.
• “Round at every step.” Keep full precision during calculations and round only the final answer unless instructed otherwise.

Diagnostic self-check

1. Interpret a sign change

A continuous function satisfies $f(2)=-3$ and $f(3)=5$. What can you conclude?

Answer

There is at least one root in $(2,3)$. You cannot conclude that there is exactly one without further information, such as strict monotonicity.

2. Generate an iterate

Given

$$x_{n+1}=\ln(4+x_n), \qquad x_0=1,$$

find $x_1$ and $x_2$ to four decimal places.

Answer

$$x_1=\ln 5=1.6094,$$

then, using the unrounded value,

$$x_2=\ln(4+1.609437\ldots)=1.7245.$$

3. Diagnose convergence

An iteration has a fixed point $\alpha$ with $g'(\alpha)=-0.4$. What local behaviour do you expect?

Answer

Since $|g'(\alpha)|=0.4<1$, nearby values should converge. The negative sign means successive errors tend to have opposite signs, so the approximations alternate across the fixed point.

4. Set up a trapezium rule calculation

Five equally spaced ordinates cover the interval $[1,3]$. Find $n$ and $h$, and state the coefficient pattern.

Answer

Five ordinates define four strips, so

$$n=4, \qquad h=\frac{3-1}{4}=0.5.$$

The coefficients are

$$1,2,2,2,1.$$

Recommended learning pathway

1. Start with locating roots using a change of sign to understand continuity, brackets and justified rounding.
2. Learn fixed point iteration and connect algebraic rearrangements with graphical fixed points.
3. Study the Newton Raphson method for tangent based iteration and rapid root finding.
4. Consolidate convergence, divergence, cycles and unsuitable starting values in when numerical iteration fails.
5. Learn the trapezium rule after definite integration, since it approximates area and accumulation rather than solving equations.

For mixed questions, practise choosing a method and checking answers.