qep-backward-error
At its defaults it draws one quadratic eigenvalue problem in 9 systems of units: what the solver reports and what the answer is worth. An overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 7.59·10⁻¹³ at γ = 108 — it moves by a factor of 1928 while the other two move by 1.68·10¹⁰. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 8.76·10¹⁰ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 0.00126. Nothing went wrong with the solver at any stop.
qep-backward-error is one function in lib/figures/qepbe.js —
the price of linearising — a residual that stays flat while the answer loses eleven orders. 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.
An overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 7.59·10⁻¹³ at γ = 108 — it moves by a factor of 1928 while the other two move by 1.68·10¹⁰. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 8.76·10¹⁰ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 0.00126. Nothing went wrong with the solver at any stop.
top: 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.
An overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 7.59·10⁻¹³ at γ = 108 — it moves by a factor of 1928 while the other two move by 1.68·10¹⁰. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 8.76·10¹⁰ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 0.00126. Nothing went wrong with the solver at any stop.
top: 4
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.
An overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 7.81·10⁻¹⁵ at γ = 104 — it moves by a factor of 19.8 while the other two move by 1.14·10⁵. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 1.02·10⁵ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 8.57·10⁻⁹. Nothing went wrong with the solver at any stop.
top: 10
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.
An overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 2.2·10⁻¹⁰ at γ = 1010 — it moves by a factor of 5.59·10⁵ while the other two move by 3.99·10¹⁵. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 2.44·10¹⁴ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 299. Nothing went wrong with the solver at any stop.
top: 6
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.
An overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 1.26·10⁻¹³ at γ = 106 — it moves by a factor of 320 while the other two move by 7.31·10⁷. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 1.86·10⁸ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 5.47·10⁻⁶. Nothing went wrong with the solver at any stop.
n: 6
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.
An overdamped chain of 6 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 7.1·10⁻¹⁶ at γ = 1 and 2.57·10⁻¹² at γ = 108 — it moves by a factor of 7022 while the other two move by 7.37·10¹¹. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 7.03·10¹¹ across the same sweep, and the forward error follows it: 9.14·10⁻¹⁵ to 0.00674. Nothing went wrong with the solver at any stop.
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.
18 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 change of units
a positive mass
a reduction this file knows
a size the eigenvectors can be computed at
a sweep that spans at least four decades of units
an inverse iterate that is a vector
an overdamped chain, whose spectrum is entirely real
and the quadratic's grows about an order per decade of units
damping that removes energy rather than adding it
every computed eigenvalue is matched to an unused exact one
LU is for square matrices
matmul shapes agree
so the two part company by half an order per decade
some real eigenpair to measure
the linearisation's backward error stays near the rounding level at every stop
two spectra of the same size
Against the rule
The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.
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 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 factorisingA 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 linearA 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 itA 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 linearA 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 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.
Two errors, and whose fault they areThe roots are not the coefficients
A polynomial whose roots are the integers one to twenty, expanded exactly, handed to the routine every library uses. The computed roots are wrong in the third digit, the computation is backward stable for the matrix it factorised, and above degree eighteen the coefficients are not double-precision numbers at all.
The eigenvalue problem that is not linearThe 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.
The arithmetic underneathThe units that overflow before the answer does
A change of variable that is exact in the algebra requires γ² times a matrix to be a number the format can hold. In binary64 that is a bound nobody meets by accident. In binary32 it arrives at 10¹⁹ and in fp16 at 256, and past it there is no answer rather than a poor one.