hyperbolic-boundary
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.
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.
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.
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.
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.
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.
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.
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 linearA 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 linearEvery 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 linearSix 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.