HookA six-hour forecast that took six weeks
In 1922 the Quaker mathematician Lewis Fry Richardson published the first serious attempt to compute the weather. He took the observed state of the atmosphere over central Europe for one morning in May 1910, divided the map into cells, and stepped the governing differential equations forward by hand — around six weeks of pencil arithmetic to produce a six-hour 'forecast'. The output was famously wrong (he predicted a 145-millibar surge in surface pressure while the real atmosphere sat almost still), but the failure lay in his rough starting data, not in the idea. Richardson imagined a hall of 64,000 human computers stepping the sky forward faster than it evolved; in 1950 the ENIAC machine ran essentially his scheme and produced the first useful 24-hour numerical forecast in roughly 24 hours of computing. Every forecast on your phone is still made this way — not by solving the equations, but by stepping them.
That is the honest confession at the heart of this section: most integrals have no antiderivative you can write down, and most differential equations no closed-form solution, so we approximate — and, crucially, we keep a running estimate of how wrong we might be. You will estimate integrals with the mid-ordinate rule and Simpson's rule, then step first-order differential equations forward with Euler's method \(y_{r+1}=y_r+hf(x_r,y_r)\) and the improved Euler method \(y_{r+1}=y_{r-1}+2hf(x_r,y_r)\). The benchmark throughout is \(\int_1^2\frac{1}{x}\,dx=\ln 2=0.693147\ldots\), an integral we can do exactly — which lets us catch each rule in the act of being wrong by a measurable amount.
ModelThe mid-ordinate rule — sample the middle of every strip
Split \([a,b]\) into \(n\) strips of equal width \(h=\frac{b-a}{n}\). The trapezium rule you met in A-level Maths reads the curve at the strip edges; the mid-ordinate rule reads it once per strip, at the midpoint, and replaces each strip with a rectangle of that height:
\[\int_a^b f(x)\,dx\approx h\big[f(m_1)+f(m_2)+\cdots+f(m_n)\big],\]
where \(m_1,m_2,\dots,m_n\) are the midpoints of the strips. The midpoint is a genuinely clever place to sample: the rectangle overshoots the curve on one side of the midpoint and undershoots on the other, and the two wedges nearly cancel — the rectangle has exactly the same area as the trapezium formed by the tangent at the midpoint. That cancellation makes the rule far better than its crude look suggests, with an error typically about half the trapezium rule's and in the opposite direction: where the curve is convex (\(f''>0\)) the midpoint tangent lies below the curve, so the rule underestimates; where it is concave, the rule overestimates. Halving \(h\) roughly quarters the error. And one counting discipline before anything else: \(n\) strips have \(n\) mid-ordinates but \(n+1\) ordinary ordinates — muddling those two counts is the classic way to wreck \(h\) and everything after it.
Estimate \(\int_1^2\frac{1}{x}\,dx\) with four strips. Here \(h=0.25\) and the midpoints are \(1.125,\ 1.375,\ 1.625,\ 1.875\). The mid-ordinates are \(\frac{1}{1.125}=0.888889\), \(\frac{1}{1.375}=0.727273\), \(\frac{1}{1.625}=0.615385\) and \(\frac{1}{1.875}=0.533333\), which sum to \(2.764880\). The estimate is \(0.25\times 2.764880=0.691220\). The true value is \(\ln 2=0.693147\), so the error is \(-0.001927\) — an underestimate, exactly as predicted, because \(f(x)=\frac{1}{x}\) has \(f''(x)=\frac{2}{x^3}>0\) on \([1,2]\): the curve is convex, so every midpoint tangent (and with it every rectangle) sits below it.
ModelSimpson's rule — a parabola through every triple of points
Simpson's rule spends its readings differently: take an even number of strips \(n\), evaluate the curve at all \(n+1\) strip edges \(y_0,y_1,\dots,y_n\), and through each consecutive triple of points fit a quadratic. A parabola can be integrated exactly, and doing so hands each triple the weights \(1,4,1\); stitching the triples together gives
\[\int_a^b y\,dx\approx\frac{h}{3}\Big[y_0+y_n+4(y_1+y_3+\cdots+y_{n-1})+2(y_2+y_4+\cdots+y_{n-2})\Big]\]
— the ends, plus four times the odd-numbered ordinates, plus twice the interior even ones. The idea is older than its name: Johannes Kepler published the end–middle–end version in 1615 to check how much wine Austrian barrels really held, and German textbooks still call it the Fassregel, the barrel rule; Thomas Simpson's name got attached about 120 years later. Its accuracy is out of all proportion to its cost. By a symmetry fluke the parabola's leading error cancels, so the rule is exact not just for quadratics but for cubics too, and its error shrinks like \(h^4\): halve the strip width and the error divides by roughly sixteen. The price of that power is the eligibility rule — an even number of strips (equivalently an odd number of ordinates), or the 4–2 pairing collapses and the formula simply does not apply.
Same integral, \(\int_1^2\frac{1}{x}\,dx\), four strips — so five ordinates at \(x=1,1.25,1.5,1.75,2\): \(y_0=1\), \(y_1=0.8\), \(y_2=0.666667\), \(y_3=0.571429\), \(y_4=0.5\). Then
\[\frac{0.25}{3}\Big[1+0.5+4(0.8+0.571429)+2(0.666667)\Big]=\frac{0.25}{3}\big[1.5+5.485714+1.333333\big]=0.693254.\]
Against \(\ln 2=0.693147\) the error is \(+0.000107\) — eighteen times smaller than the mid-ordinate estimate built from almost the same information (five curve readings against four). That ratio is the whole sales pitch: comparable arithmetic, an extra order of accuracy, provided you respect the even-strip rule.
MechanismEuler's method — walk the tangent
Now for differential equations. Given \(\frac{dy}{dx}=f(x,y)\) with a start point \(y(x_0)=y_0\), the equation tells you the gradient of the solution curve at any point you happen to be standing on — but not the curve itself. Euler's method walks: from \((x_r,y_r)\), compute the gradient \(f(x_r,y_r)\), freeze it, and slide along the tangent for one step of width \(h\):
\[x_{r+1}=x_r+h,\qquad y_{r+1}=y_r+hf(x_r,y_r).\]
Then recompute the gradient where you land, and repeat. The error is baked into the freeze: across each step the true gradient changes and the tangent does not, so each step commits a small truncation error of order \(h^2\), and after the \(\frac{b-a}{h}\) steps needed to cross a fixed interval the accumulated error is of order \(h\). That makes Euler a first-order method: halve the step and you halve the final error — but double the arithmetic. This is precisely Richardson's move on the atmosphere, and the reason his imagined forecast factory needed 64,000 clerks: with a first-order method, accuracy is bought with steps.
For \(\frac{dy}{dx}=x+y\) with \(y(0)=1\) and \(h=0.1\), estimate \(y(0.2)\). Step 1: \(f(0,1)=0+1=1\), so \(y_1=1+0.1\times 1=1.1\) at \(x_1=0.1\). Step 2: \(f(0.1,1.1)=0.1+1.1=1.2\), so \(y_2=1.1+0.1\times 1.2=1.22\). This particular equation happens to have the exact solution \(y=2e^x-x-1\) (check: it satisfies both the equation and \(y(0)=1\)), giving \(y(0.2)=2e^{0.2}-1.2=1.242806\). Euler's \(1.22\) is adrift by \(0.0228\), about \(1.8\%\) — respectable for two steps of crude tangent-walking, and the deficit has a predictable direction: this solution curve is convex (\(y''=2e^x>0\)), so every frozen tangent undershoots it.
MechanismThe improved Euler method — measure the gradient in the middle
Euler's bias comes from measuring the gradient at the start of each step — always the lopsided place. The improved Euler method on this specification,
\[y_{r+1}=y_{r-1}+2hf(x_r,y_r),\]
fixes the lopsidedness geometrically: it takes a double-width stride, from \(y_{r-1}\) clear across to \(y_{r+1}\), using the gradient measured at \(x_r\) — the exact midpoint of that double step. A centrally measured gradient overshoots on one half of the stride and undershoots on the other, the two halves largely cancel, and the method becomes second-order: halving \(h\) quarters the accumulated error rather than merely halving it. The price is a memory requirement. Because the formula reaches back two rows, it cannot start itself: the standard recipe is one ordinary Euler step to manufacture \(y_1\), after which the midpoint formula takes over — \(y_2\) built from \(y_0\), \(y_3\) from \(y_1\), leapfrogging forward with three columns of book-keeping (\(x_r\), \(y_r\), \(f(x_r,y_r)\)) live at all times.
Continue the same problem: \(\frac{dy}{dx}=x+y\), \(y(0)=1\), \(h=0.1\). The Euler bootstrap gave \(y_1=1.1\). Now leap: \(y_2=y_0+2hf(x_1,y_1)=1+0.2\times f(0.1,1.1)=1+0.2\times 1.2=1.24\). Against the exact \(y(0.2)=1.242806\) the error is \(0.0028\) — eight times closer than plain Euler's \(1.22\), built from exactly the same two gradient evaluations. The march then continues from two rows back each time: \(y_3=y_1+2hf(x_2,y_2)=1.1+0.2\times(0.2+1.24)=1.388\), and so on. One honest footnote: whatever error the crude bootstrap planted in \(y_1\) is carried along the odd-numbered rows — the midpoint formula stops error growing quickly, but it cannot repair a bad start.
CaseWhat accuracy costs — from ENIAC to the exam hall
Put the four methods on one scoreboard and the pattern is the point. Halve \(h\), and: Euler's error halves; improved Euler's quarters; the mid-ordinate rule's quarters; Simpson's divides by about sixteen. Every halving doubles the arithmetic, so a method's order decides what accuracy costs — with a first-order method, one extra decimal place costs roughly ten times the work; with Simpson's rule, less than a doubling. That trade is modern numerical analysis in a sentence: the professional tools (the Runge–Kutta family, adaptive step-size solvers) are direct descendants of improved Euler that sample the gradient at cleverer points inside each step, and a global weather model still crosses its forecast interval in time steps of a few minutes, exactly as Richardson dreamed.
In the exam hall the same cost–benefit thinking earns actual marks. Asked how to improve an estimate, the answer is always the same two-parter: more strips, or a smaller step — at the stated cost of more computation. Asked whether an estimate is too big or too small, argue from shape: concavity for the mid-ordinate rule, the direction of tangent drift for Euler. And treat every number this section produces as what it is — an approximation, quoted to a stated accuracy, never presented as exact. Precision of presentation is not precision of knowledge; the method's order and step size set the real accuracy, whatever the calculator display implies.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
The numerical-methods formulae are printed in the AQA formulae booklet, so nothing here is a memory test: the marks are method marks for organised deployment. The standard grammar of an answer: state the rule and \(h\), tabulate the sampling points and their function values (for Simpson's rule, label which ordinates take the 4 and which the 2), show the substitution into the quoted formula, then give the final value to the accuracy demanded. Work in radians for any trigonometric integrand, carry full calculator precision between steps, and round only at the end — a rounded intermediate value is the commonest way to drop the final accuracy mark in this topic.
For the step-by-step methods, a three-column table (\(x_r\), \(y_r\), \(f(x_r,y_r)\)) is the examiner-proof format: each row is independently checkable, so one slip costs one accuracy mark rather than the whole question. With improved Euler, show the Euler bootstrap for \(y_1\) explicitly, then make the leapfrog structure visible — \(y_2\) built from \(y_0\), \(y_3\) from \(y_1\). Around 60% of further maths marks sit in AO1, routine procedures, and this section is where they are cheapest; the AO2 garnish is the comment question — 'is this an overestimate?', 'how could the estimate be improved?' — where the creditworthy answers are, respectively, an argument from concavity or tangent drift, and 'more strips / a smaller step, at the cost of more computation'.