Quadrature — where it appears
Named by 12 essays across 4 fields — each of them below, with the objects they name alongside it.
A problem with infinitely many eigenvalues
Let the matrix depend on λ through something that is not a polynomial and three things stop being true at once. There is no linearisation, there is no characteristic polynomial, and "compute the spectrum" is not a request that can be granted — the only finite question is how many eigenvalues are inside this circle.
The last digit is the cheapest
Every cost curve measured here has the same shape: the first digits are cheap and the last ones are not. One method inverts it. Doubling the work buys twice as many digits as the previous doubling did, so the price of a digit halves every time it is paid.
The grid was the first filter
A continuous deconvolution discretised on n points and solved with no regularisation at all is not unregularised. Its error against the continuous signal is least at 24, 26 and 34 points for noise of 1%, 0.1% and 0.01% per sample — beside best truncations of 24, 28 and 32 components on a 64-point grid — and within 4 to 16 per cent of their error. The grid's own filter factors sum to n exactly and fall through a half at k = n. Choosing the grid was choosing a truncation, before anybody chose a λ.
Counting what is inside a circle
A trace of a matrix nobody wants to form, integrated around a contour, gives an integer — how many eigenvalues are inside. It converges exponentially, it is estimated with random probes, and the probe block is a ceiling that the answer does not mention.
A better discretisation is a weaker filter
A coarse grid's error has two sources — how well the discrete operator approximates the integral, and how well the grid's function represents the answer — and the grid essay could not separate them. Changed one at a time they separate: integrating the kernel against the hat functions takes a fifth off the 12-point error, reading the answer as a cubic spline takes 15 per cent more, and both roughly double the condition number on every grid. At 0.1% noise the spline discretisation's unregularised solve on 26 points reaches the best truncation of a 64-point grid to 0.3%. At 1% it is worse than the crude grid.
A ceiling with a knob on it
A contour method returns at most as many eigenvalues as its probe block has columns, and the object that comes back does not distinguish that from having found everything. One line of the derivation multiplies the ceiling by a number the caller chooses, and it costs no extra solves at all.
The conditioning that rises with the ceiling
Higher moments multiply a contour method's ceiling by K and grade its block Hankel over ρ to the 2K, so the two knobs are the same knob. One division per quadrature point separates them, and the measurement of what it is worth grows from twenty to twenty thousand.
The grid on which the discretisation stops mattering
Without regularisation, integrating the blur's kernel against cubic splines beat sampling it at 0.1% noise and lost to it at 1%. Give each discretisation its own best λ on every grid and the difference shrinks to nothing where grids are fine — 0.17, 0.28 and 0.20 per cent apart on 96 points at the three noise levels, with every discretisation choosing the same λ — and stays at 17 to 18 per cent on 16 points. The choice between them is a choice of how coarse a grid can be: at 0.1% noise the integrated discretisations reach the fine-grid answer on 26 points and the sampled one needs 40.
Where a contour's budget should go
A contour method's ceiling is the number of probes times the number of moments, and the moments are free in solves while the probes are not. Four ways of reaching one ceiling come out four orders apart, the ordering is not monotone, and what separates the best two is not the usual draw but the unlucky one.
The circle between two eigenvalues
A contour count's error is not approximately governed by the nearest eigenvalue; it is exactly one closed-form term per eigenvalue, and summing those terms reproduces the quadrature to a millionth at 1,720 radius and point pairs. Three things follow. The rate is the ratio of the two moduli the circle sits between, not a distance, so two circles 0.2 from their nearest eigenvalue converge three times apart. Ten digits cost about 21 points divided by log₁₀ of that ratio, 71 points with twelve eigenvalues inside and 3,476 with six. And the best circle is not halfway: at the geometric mean of two eigenvalues on one ray their two terms are equal and opposite, and ten digits cost 27 points where the midpoint needs 84.
A centre that chooses which pair cancels
A contour count's error is one closed-form term per eigenvalue, and two real eigenvalues on one ray cancel at their geometric mean — about the origin. Moved along the real axis, the centre was predicted to buy a further factor of two by putting real eigenvalues on two rays, the far one cancelling with a sign flip. The flip is parity: an eigenvalue behind the centre changes sign only at odd point counts. The factor is not two. Counting the delay problem's first two and three real eigenvalues, the best centre costs 22 and 21 points against the origin's 41 and 78; counting all four, 32 against 341, because about the origin the circle has to pass between a real eigenvalue and a complex pair and about 2.3 it does not. The centre does not add a cancellation. It chooses which pair cancels and pushes everything else away.
Count the upper half, read the lower
Moving a contour's centre off the real axis was predicted never to pay for a count closed under conjugation: one member of each pair comes closer, the other goes away, and the slower one always gets slower. For one circle the prediction holds at every offset measured, from 52 points on the axis to 567 three units above it. But one circle is not the only way to count a set like that. On the delay problem, the four real eigenvalues on a circle of their own and each band's upper cluster on a small circle of its own cost 55 points against 52 for one band, 68 against 97 for two, and 115 against 275 for six. What pays is the mirror: for a problem with real coefficients the count about the conjugate centre is the conjugate of the count already made, equal to 1.6·10⁻¹⁵, so the lower half of the plane costs nothing.
Named alongside it
The objects these essays reach for when they reach for this one.
Contour integralNonlinear eigenvalue problemContour eigensolverExact ground truthCondition numberArithmetic costConvergence rateRankComplex arithmeticDeconvolutionDiscretisationDiscretisation error