Generator

overdamped-gap

One function in the hyperbolic library, called 14 times across 8 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 13 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws the 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between them. A chain of 8 masses with C = 8K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.6698 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

overdamped-gap is one function in lib/figures/hyperbolic.js — real by class — a cholesky that certifies a spectrum, and a double root that costs half the digits. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 8K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.6698 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-33-28.2505-23.5009-18.7514-14.0018-9.25229-4.502750eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap0.67critical β for this n5.8the spectrum is real by classnot by outcome

A chain of 8 masses with C = 8K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.6698 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

beta: 8

The arguments are the ones A backward-stable answer to a problem nobody asked passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 8K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.6698 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-33-28.2505-23.5009-18.7514-14.0018-9.25229-4.502750eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap0.67critical β for this n5.8the spectrum is real by classnot by outcome

A chain of 8 masses with C = 8K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.6698 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

beta: 7

The arguments are the ones Every eigenvalue real, and a test that says so passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 7K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.48 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-29-24.8546-20.7092-16.5637-12.4183-8.2729-4.127480eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap0.48critical β for this n5.8the spectrum is real by classnot by outcome

A chain of 8 masses with C = 7K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.48 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

beta: 20

The arguments are the ones Every eigenvalue real, and a test that says so passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 20K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 2.31 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-83-71.0448-59.0895-47.1343-35.1791-23.2238-11.26860eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap2.3critical β for this n5.8the spectrum is real by classnot by outcome

A chain of 8 masses with C = 20K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 2.31 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

beta: 10

The arguments are the ones Every eigenvalue real, and a test that says so passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 10K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.9861 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-41-35.0456-29.0912-23.1368-17.1824-11.228-5.27360eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap0.99critical β for this n5.8the spectrum is real by classnot by outcome

A chain of 8 masses with C = 10K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.9861 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

beta: 14

The arguments are the ones Every eigenvalue real, and a test that says so passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 14K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 1.539 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-58-49.6426-41.2852-32.9278-24.5704-16.213-7.855590eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap1.5critical β for this n5.8the spectrum is real by classnot by outcome

A chain of 8 masses with C = 14K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 1.539 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.

What it checked while drawing

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

13 distinct claims across 6 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

a chain long enough to have a spectrum and short enough to draw

a hyperbolic quadratic to split

a linearisation has as many eigenvalues as it has rows

a positive mass

a reduction this file knows

a spectrum that fits along the axis

an entirely real spectrum

damping above the critical value, so the spectrum is real

damping that removes energy rather than adding it

Jacobi needs a symmetric matrix

LU is for square matrices

matmul shapes agree

n eigenvalues each side of the certificate

Against the rule

It draws a decomposition and prints its residual. It calls splitAtCertificate, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 173 of 325 generators — 158 print a residual and 15 are exempt with a published reason; 152 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

The eigenvalue problem that is not linear

A backward-stable answer to a problem nobody asked

One quadratic eigenvalue problem, in nine systems of units, with a change of variable that is exact in both directions. The residual the solver prints stays at the rounding level at every stop. The answer loses eleven orders of magnitude, and the two facts are consistent.

Iterating, instead of factorising

A Krylov space for a problem that is not linear

A quadratic eigenvalue problem has no matrix to build a Krylov space out of. The recurrence that builds one anyway stores half as many numbers, returns twice as many Ritz values — and stops being a basis at twenty vectors while the answer it gives keeps improving.

The eigenvalue problem that is not linear

A matrix that depends on its own eigenvalue

A damped structure does not produce Ax = λx. It produces (λ²M + λC + K)x = 0, where the matrix whose null vector is wanted is a function of the number being solved for — so there is nothing to factorise, an n × n problem has 2n answers, and the eigenvectors cannot be a basis.

Structure, and the solver that cannot see it

A perturbation that keeps the symmetry

The smallest perturbation that makes a computed answer exact is the backward error. Ask for the smallest one that also keeps the problem's structure and the number can only go up — and measured on a palindromic quadratic it goes up by 1.17, while the structure the computed spectrum has lost is not in either number.

The eigenvalue problem that is not linear

A spectrum that comes in reciprocal pairs

A palindromic quadratic reads the same backwards, so λ is an eigenvalue exactly when 1/λ is. A general solver discards that, computes the large half of the spectrum perfectly and the small half to seven digits — and the small half is a division away from being perfect too.

The eigenvalue problem that is not linear

Every eigenvalue real, and a test that says so

A quadratic eigenvalue problem has no reason to have real eigenvalues. One class does, as a property rather than an outcome, and the proof is a Cholesky that completes. The boundary of the class has a closed form, and at the boundary the arithmetic loses half its digits with nothing ill conditioned anywhere.

The eigenvalue problem that is not linear

Six routes to one spectrum

Three linearisations of one quadratic, each reduced to a standard eigenvalue problem two ways. All six have exactly the same eigenvalues in exact arithmetic. On a well-scaled problem they differ by noise; on a badly scaled one by a factor of forty; and two of the six are the same matrix.

The eigenvalue problem that is not linear

The scaling that buys ten orders

Two lines computed from three norms, a change of variable that is exact in both directions, and the whole of the loss the previous essay measured comes back — flat, at every stop, because after scaling every stop is the same problem.

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