Notes · Further MathematicsUK · A-Levels

Numerical methods

Numerical methods give approximate answers when an exact one is unavailable. This topic extends the trapezium rule and Newton-Raphson method of A-Level Mathematics with the mid-ordinate and Simpson's rules for integration, Euler's step-by-step methods for differential equations, and iterative root-finding. Throughout, the emphasis is on applying a method carefully and commenting honestly on its accuracy.

4 sections~12 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·101010 / 13

Exam profile

AO1 · Apply numerical integration rules and iterative methods accuratelyAO2 · Justify the choice of method and comment on accuracy and errorAO3 · Use numerical methods to solve problems that have no closed-form solution

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basic level

AS Further Mathematics covers the mid-ordinate rule and simple iterative root-finding.

higher level

The full A-Level adds Simpson's rule, Euler and improved-Euler methods for differential equations, and a fuller treatment of error and accuracy.

Depth

Text

Contents · 4 sections

Numerical methods

Numerical integration: mid-ordinate and Simpson's rules#

StandardAQA 7367 (Numerical methods)DfE Further Mathematics

Mid-ordinate heights under a curve

Fig. 1The mid-ordinate rule uses the heights of the curve at the strip midpoints (drop lines) as the rectangle heights.

When a definite integral cannot be evaluated exactly, its value is estimated by replacing the curve with simple shapes over each of

n

equal strips of width

h = \dfrac{b-a}{n}

. The trapezium rule (recap from A-Level Mathematics) joins the tops of the ordinates with straight lines; the mid-ordinate rule uses rectangles whose heights are the values of the function at the midpoints of the strips. The mid-ordinate rule is

\displaystyle\int_a^b y\,\mathrm{d}x \approx h\big(y_{1/2} + y_{3/2} + \cdots + y_{n-1/2}\big)

, summing the mid-ordinates.

Simpson's rule is more accurate again: it fits a parabola through each consecutive triple of ordinates, which follows the curvature of the graph rather than approximating it by straight lines. It requires an even number of strips and weights the ordinates in the pattern

1, 4, 2, 4, \dots, 4, 1

\displaystyle\int_a^b y\,\mathrm{d}x \approx \dfrac{h}{3}\big(y_0 + y_n + 4(y_1 + y_3 + \cdots) + 2(y_2 + y_4 + \cdots)\big)

. The end ordinates carry weight

1

, the odd-indexed ones weight

4

, and the even interior ones weight

2

Applying any of these rules is a matter of careful tabulation: compute the strip width

h

, list the ordinates in order, and combine them with the correct weights. The commonest arithmetic error is a miscount of ordinates or a misplacement of the

4

s and

2

s in Simpson's rule, so it is worth setting the values out in a table with the weights written above them.

Which rule to use depends on the accuracy required and the data available. Simpson's rule is far more accurate for a smooth curve and should be preferred when the number of strips is even; the mid-ordinate rule is simple and often better than the trapezium rule; the trapezium rule is the crudest. All three improve as

n

increases and

h

shrinks, and the choice, together with a comment on likely accuracy, is itself examined.

\int_a^b y\,\mathrm{d}x \approx h\big(y_{1/2} + y_{3/2} + \cdots + y_{n-1/2}\big) \quad (\text{mid-ordinate rule})

Mid-ordinate rule

Rectangles with heights taken at the midpoints of the strips.

\int_a^b y\,\mathrm{d}x \approx \frac{h}{3}\Big(y_0 + y_n + 4\!\sum_{\text{odd}} y_i + 2\!\sum_{\text{even}} y_i\Big) \quad (\text{Simpson's rule}, n \text{ even})

Simpson's rule

Fits parabolas; requires an even number of strips.

Worked example

Simpson's rule with four strips

Estimate $\displaystyle\int_0^2 \sqrt{1 + x^3}\,\mathrm{d}x$ using Simpson's rule with four strips.

Strip width and ordinates
$h = \dfrac{2-0}{4} = 0.5$ ; the ordinates at $x = 0, 0.5, 1, 1.5, 2$ are $y_0 = 1$ , $y_1 = \sqrt{1.125} = 1.0607$ , $y_2 = \sqrt{2} = 1.4142$ , $y_3 = \sqrt{4.375} = 2.0917$ , $y_4 = \sqrt{9} = 3$ .
Apply the weights
$\displaystyle\int \approx \frac{0.5}{3}\big(y_0 + y_4 + 4(y_1 + y_3) + 2y_2\big) = \frac{1}{6}\big(1 + 3 + 4(1.0607 + 2.0917) + 2(1.4142)\big)$ .
$\int \approx \frac{1}{6}\big(4 + 4(3.1524) + 2.8284\big)$
Evaluate
$= \dfrac{1}{6}\big(4 + 12.6096 + 2.8284\big) = \dfrac{19.438}{6} = 3.2397$ .
State to 3 d.p.
The estimate is $3.240$ (3 d.p.).

Result: $\displaystyle\int_0^2 \sqrt{1 + x^3}\,\mathrm{d}x \approx 3.240$ .

Typical mistakes

Using Simpson's rule with an odd number of strips, or misplacing the $4$ and $2$ weights.
Confusing the number of strips with the number of ordinates (there is always one more ordinate than strips).

Active revision

Use Simpson's rule with four strips to estimate $\displaystyle\int_0^2 \sqrt{1 + x^3}\,\mathrm{d}x$ , giving your answer to three decimal places.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Accuracy and error of numerical integration#

AdvancedAQA 7367 (Numerical methods)DfE Further Mathematics

Every numerical integration rule gives an approximation, and understanding the error is part of using it responsibly. As a general principle, halving the strip width improves the estimate, and the rules differ in how fast: the trapezium and mid-ordinate rules have errors that scale roughly with

h^2

, while Simpson's rule has an error scaling with

h^4

, so Simpson's rule improves dramatically faster as the strips are refined.

The sign of the error can often be predicted from the curvature. On a curve that is concave (bending downwards), the trapezium rule underestimates the area because its chords lie below the curve, while the mid-ordinate rule overestimates; on a convex (upward-bending) curve the signs reverse. Interestingly, the trapezium and mid-ordinate errors tend to be of opposite sign and roughly in the ratio

2:1

, which is the fact that makes Simpson's rule — a weighted blend of the two — so accurate.

Simpson's rule is exact for any cubic (and hence for quadratics and linear functions), because a parabola through three points already reproduces a cubic's integral over a double strip. This is why it is the method of choice for smooth curves: for polynomials of degree three or less it makes no error at all, and for smoother functions its error is very small. Knowing this exactness result lets you argue about accuracy rather than merely quoting a number.

In an exam you may be asked to increase the number of strips and comment on how the estimate changes, or to state which rule is likely to be the most accurate and why. A good answer refers to the curvature of the graph and to the order of the error (

h^2

versus

h^4

), and recognises that the true value lies between an over- and an under-estimate when one of each is available. Numerical answers should always be given to an accuracy consistent with the method — quoting ten digits from a four-strip estimate is dishonest.

\text{error}_{\text{trap}},\ \text{error}_{\text{mid}} \sim h^2, \qquad \text{error}_{\text{Simpson}} \sim h^4

Orders of error

Simpson's rule improves far faster as the strips are refined.

Worked example

Trapezium and mid-ordinate as bounds

Estimate $\displaystyle\int_0^2 \frac{1}{1+x}\,\mathrm{d}x$ using the trapezium rule with two strips, and comment (the exact value is $\ln 3 = 1.0986$ ).

Ordinates
$h = 1$ ; ordinates at $x = 0, 1, 2$ are $y_0 = 1$ , $y_1 = \tfrac12$ , $y_2 = \tfrac13$ .
Trapezium estimate
$\displaystyle\int \approx \frac{h}{2}\big(y_0 + y_2 + 2y_1\big) = \frac{1}{2}\big(1 + \tfrac13 + 2\cdot\tfrac12\big) = \frac{1}{2}\cdot\frac{7}{3} = \frac{7}{6} \approx 1.1667$ .
$\int \approx \frac{7}{6} \approx 1.1667$
Compare with the exact value
The curve $y = \tfrac{1}{1+x}$ is convex (bends upward), so the trapezium chords lie above the curve and the estimate $1.1667$ is an overestimate of the true $1.0986$ .
Comment
A finer partition or Simpson's rule would reduce the error; the mid-ordinate rule would give an underestimate, bracketing the true value.

Result: Trapezium estimate $\tfrac{7}{6} \approx 1.167$ , an overestimate of $\ln 3 = 1.099$ .

Typical mistakes

Claiming more accuracy than the method warrants by quoting too many decimal places.
Getting the direction of the trapezium error wrong by not considering whether the curve is concave or convex.

Active revision

For $\displaystyle\int_0^1 \mathrm{e}^{-x^2}\,\mathrm{d}x$ , estimate the integral with the trapezium rule and with the mid-ordinate rule using two strips, and explain why the true value lies between them.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Step-by-step solution of differential equations#

AdvancedAQA 7367 (Numerical methods)DfE Further Mathematics

Euler's method drifts from the true curve

Fig. 2For

\tfrac{\mathrm{d}y}{\mathrm{d}x}=y

y(0)=1

, the Euler polyline (step

0.5

) falls below the true solution

y=\mathrm{e}^x

When a first-order differential equation

\dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x,y)

cannot be solved exactly, its solution can be built up numerically from a starting point. Euler's method uses the gradient at the current point to step forward a small distance

h

y_{n+1} = y_n + h\,f(x_n, y_n)

, with

x_{n+1} = x_n + h

. Geometrically, it follows the tangent line for one step, then re-evaluates the gradient and repeats — approximating the solution curve by a chain of short straight segments.

Because Euler's method uses the gradient only at the start of each step, it systematically drifts away from the true curve, and the error accumulates as the steps proceed. The figure of the true curve against the Euler polyline shows this drift clearly: the polyline lies below a convex solution because each tangent step undershoots the curving path. Reducing the step size

h

reduces the error but requires more steps and more arithmetic.

The improved Euler (or mid-point) method is more accurate for the same step size. It first takes a trial half-step to estimate the gradient at the middle of the interval, then uses that better gradient to take the full step:

y_{n+1} = y_n + h\,f\!\left(x_n + \tfrac{h}{2},\ y_n + \tfrac{h}{2}f(x_n, y_n)\right)

. Sampling the gradient nearer the middle of the step captures the curvature better and roughly squares the accuracy gain, at the cost of two function evaluations per step.

Carrying out either method is a disciplined, repetitive calculation, and setting the work out in a table of

x_n

y_n

and

f(x_n, y_n)

keeps it accurate. The examinable understanding is that these are approximations whose error depends on the step size and on the method, and that a smaller

h

or the improved method buys more accuracy — a trade-off between effort and precision that mirrors the choice of integration rule in the earlier sections.

y_{n+1} = y_n + h\,f(x_n, y_n) \quad (\text{Euler})

Euler's method

Step forward along the tangent using the gradient at the current point.

y_{n+1} = y_n + h\,f\!\left(x_n + \tfrac{h}{2},\ y_n + \tfrac{h}{2}f(x_n, y_n)\right) \quad (\text{improved Euler})

Improved Euler (mid-point) method

Uses the gradient estimated at the middle of the step for greater accuracy.

Worked example

Three steps of Euler's method

For $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x + y$ with $y(0) = 1$ , use $h = 0.1$ to estimate $y(0.3)$ .

Step 1 (x = 0 to 0.1)
$f(0,1) = 0 + 1 = 1$ , so $y_1 = 1 + 0.1(1) = 1.1$ at $x = 0.1$ .
Step 2 (x = 0.1 to 0.2)
$f(0.1, 1.1) = 0.1 + 1.1 = 1.2$ , so $y_2 = 1.1 + 0.1(1.2) = 1.22$ at $x = 0.2$ .
Step 3 (x = 0.2 to 0.3)
$f(0.2, 1.22) = 0.2 + 1.22 = 1.42$ , so $y_3 = 1.22 + 0.1(1.42) = 1.362$ .
$y(0.3) \approx 1.362$
Comment
The exact solution is $y = 2\mathrm{e}^x - x - 1$ , giving $y(0.3) = 1.3997$ ; Euler underestimates, and a smaller $h$ would close the gap.

Result: $y(0.3) \approx 1.362$ (exact $1.400$ ).

Typical mistakes

Using the gradient at the wrong point, or forgetting to update $x_n$ as well as $y_n$ at each step.
In the improved-Euler method, using the initial gradient for the full step instead of the mid-point gradient.

Active revision

Given $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x + y$ with $y = 1$ at $x = 0$ , use Euler's method with step $h = 0.1$ to estimate $y$ at $x = 0.3$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Numerical solution of equations#

StandardAQA 7367 (Numerical methods)DfE Further Mathematics

Newton-Raphson: the tangent locates the next estimate

Fig. 3The tangent to

y = x^3 - 2x - 5

x = 2

meets the axis near the root

x \approx 2.0946

A root of

f(x) = 0

that cannot be found algebraically is located numerically. The starting point is a change of sign: if

f

is continuous and

f(a)

and

f(b)

have opposite signs, a root lies between

a

and

b

. Interval bisection then repeatedly halves the interval, keeping the half in which the sign change occurs, so the root is trapped in an ever-shorter interval — reliable but slow.

Linear interpolation improves on bisection by using the actual function values: it joins the points

(a, f(a))

and

(b, f(b))

with a straight line and takes the point where that line crosses the axis as the next estimate, which is usually closer to the root than the midpoint. It is a modest refinement of the sign-change idea, converging faster than bisection while remaining robust.

The Newton-Raphson method (recap and extension from A-Level Mathematics) is the fastest of the standard methods when it works. From an estimate

x_n

it follows the tangent to the curve down to the axis, giving

x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}

. Geometrically, the tangent at

(x_n, f(x_n))

meets the

x

-axis at the next estimate; near a simple root the method roughly doubles the number of correct digits each step.

Newton-Raphson can fail, and knowing when is examinable. If the initial estimate is poor, if the tangent is nearly horizontal (

f'(x_n) \approx 0

), or if there is a stationary point or discontinuity nearby, the iteration may diverge, oscillate, or converge to the wrong root. A good numerical answer states the method used, shows the iterations converging, and — where relevant — comments on the conditions needed for convergence. Comparing the robustness of bisection with the speed of Newton-Raphson is the kind of evaluation the paper rewards.

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \quad (\text{Newton-Raphson})

Newton-Raphson iteration

Follows the tangent at $x_n$ to the axis; fast near a simple root.

Worked example

Newton-Raphson for a cubic root

Solve $x^3 - 2x - 5 = 0$ near $x = 2$ by Newton-Raphson, to three decimal places.

Set up the iteration
$f(x) = x^3 - 2x - 5$ , $f'(x) = 3x^2 - 2$ , so $x_{n+1} = x_n - \dfrac{x_n^3 - 2x_n - 5}{3x_n^2 - 2}$ .
First iteration
$f(2) = 8 - 4 - 5 = -1$ , $f'(2) = 12 - 2 = 10$ , so $x_1 = 2 - \dfrac{-1}{10} = 2.1$ .
Second iteration
$f(2.1) = 9.261 - 4.2 - 5 = 0.061$ , $f'(2.1) = 13.23 - 2 = 11.23$ , so $x_2 = 2.1 - \dfrac{0.061}{11.23} = 2.09457$ .
$x_2 = 2.09457$
Converge
A further iteration gives $x_3 = 2.09455$ ; the value is stable to three decimal places.

Result: The root is $x \approx 2.095$ (3 d.p.).

Typical mistakes

Forgetting to check for a sign change before iterating, or misreading which subinterval contains the root.
Dividing by a very small $f'(x_n)$ in Newton-Raphson and not recognising the resulting instability.

Active revision

The equation $x^3 - 2x - 5 = 0$ has a root near $x = 2$ . Use the Newton-Raphson method with $x_0 = 2$ to find the root correct to three decimal places.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Numerical methods

4 sections~12 min reading time3 competenciesLevel Standard 2 · Advanced 2

Numerical methods

Key points

Simpson's rule with four strips

Key points

Trapezium and mid-ordinate as bounds

Key points

Three steps of Euler's method

Key points

Newton-Raphson for a cubic root

Sources

Numerical methods

Key points

Simpson's rule with four strips

Key points

Trapezium and mid-ordinate as bounds

Key points

Three steps of Euler's method

Key points

Newton-Raphson for a cubic root

Sources