Generator

hyperbolic-boundary

One function in the hyperbolic library, called 8 times across 4 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 where a quadratic stops being hyperbolic, located by a cholesky and by a sine. The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 8 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 5.75877048314, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 5.75877048314, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

hyperbolic-boundary 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.

Where a quadratic stops being hyperbolic, located by a Cholesky and by a sineThe overdamping margin min over modes of (βκ)² − 4κ, for a chain of 8 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 5.75877048314, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 5.75877048314, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.44.95985.919596.879397.839188.7989810⁻¹110¹10²stiffness damping βoverdamping marginβ* = 5.75877two routes to a boundaryclosed form β*5.8by certificate5.8difference3.6·10⁻¹³bisection steps44a factorisation that completesand a sine, agreeing to twelve digits

The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 8 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 5.75877048314, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 5.75877048314, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

n: 8

The arguments are the ones A matrix that depends on its own eigenvalue 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.

Where a quadratic stops being hyperbolic, located by a Cholesky and by a sineThe overdamping margin min over modes of (βκ)² − 4κ, for a chain of 8 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 5.75877048314, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 5.75877048314, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.44.95985.919596.879397.839188.7989810⁻¹110¹10²stiffness damping βoverdamping marginβ* = 5.75877two routes to a boundaryclosed form β*5.8by certificate5.8difference3.6·10⁻¹³bisection steps44a factorisation that completesand a sine, agreeing to twelve digits

The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 8 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 5.75877048314, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 5.75877048314, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

n: 4

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.

Where a quadratic stops being hyperbolic, located by a Cholesky and by a sineThe overdamping margin min over modes of (βκ)² − 4κ, for a chain of 4 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 3.2360679775, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 3.2360679775, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.22.539343.078693.618034.157384.6967210⁻¹110¹10²stiffness damping βoverdamping marginβ* = 3.23607two routes to a boundaryclosed form β*3.2by certificate3.2difference4.7·10⁻¹³bisection steps44a factorisation that completesand a sine, agreeing to twelve digits

The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 4 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 3.2360679775, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 3.2360679775, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

n: 12

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.

Where a quadratic stops being hyperbolic, located by a Cholesky and by a sineThe overdamping margin min over modes of (βκ)² − 4κ, for a chain of 12 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 8.29622981056, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 8.29622981056, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.56.38277.765419.1481110.530811.913510⁻¹110¹10²stiffness damping βoverdamping marginβ* = 8.29623two routes to a boundaryclosed form β*8.3by certificate8.3difference3.5·10⁻¹³bisection steps44a factorisation that completesand a sine, agreeing to twelve digits

The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 12 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 8.29622981056, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 8.29622981056, agreeing to 13 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

n: 6

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.

Where a quadratic stops being hyperbolic, located by a Cholesky and by a sineThe overdamping margin min over modes of (βκ)² − 4κ, for a chain of 6 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 4.49395920743, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 4.49395920743, agreeing to 15 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.33.748994.497995.246985.995976.7449710⁻¹110¹10²stiffness damping βoverdamping marginβ* = 4.49396two routes to a boundaryclosed form β*4.5by certificate4.5difference8·10⁻¹⁵bisection steps44a factorisation that completesand a sine, agreeing to twelve digits

The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 6 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 4.49395920743, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 4.49395920743, agreeing to 15 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

n: 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.

Where a quadratic stops being hyperbolic, located by a Cholesky and by a sineThe overdamping margin min over modes of (βκ)² − 4κ, for a chain of 10 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 7.02667418333, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 7.02667418333, agreeing to 14 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.56.171117.342228.513349.6844510.855610⁻¹110¹10²stiffness damping βoverdamping marginβ* = 7.02667two routes to a boundaryclosed form β*7by certificate7difference1.9·10⁻¹³bisection steps44a factorisation that completesand a sine, agreeing to twelve digits

The overdamping margin min over modes of (βκ)² − 4κ, for a chain of 10 masses, against β. It reaches zero at β* = 1/sin(π/2(n+1)) = 7.02667418333, which is the closed form. Bisecting on a completely different question — does a Cholesky of −Q(μ) complete for some μ — gives 7.02667418333, agreeing to 14 digits. Neither route computes an eigenvalue. The marks below the axis are the largest imaginary part in the computed spectrum, which is zero to the rounding level above β* and not below it, so a third route agrees with the other two about where the boundary is.

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 linearisation has as many eigenvalues as it has rows

a positive mass

a reduction this file knows

a size the bisection can afford

and the lower end is not

damping that removes energy rather than adding it

Jacobi needs a symmetric matrix

LU is for square matrices

matmul shapes agree

some stop of the sweep is entirely real

the two routes to the boundary agree

the upper end of the bracket is overdamped

Against the rule

It draws a decomposition and prints its residual. It calls qepEigen, criticalByCertificate, 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 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.

The eigenvalue problem that is not linear

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 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.

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