Newton on the Bratu problem: analytic Jacobian, differenced Jacobian at ε = 10^-6, and matrix-free
At its defaults it draws newton on the bratu problem: analytic jacobian, differenced jacobian at ε = 10^-6, and matrix-free. Three residual sequences, all starting from zero. The analytic Jacobian gives 8, 0.052, 3·10⁻⁶, 5.2·10⁻¹³; a Jacobian differenced at ε = 10^-6, whose entries are correct to about 1.3·10⁻¹⁰, gives 8, 0.052, 3·10⁻⁶, 4.6·10⁻¹³. They stop at the same residual. The matrix-free run, where GMRES sees only a closure and no entry exists anywhere, lands 8.1·10⁻¹⁶ from the analytic answer in 438 products.
newton-trails is one function in lib/figures/matfree.js —
matrix-free — what survives when the matrix is a subroutine, and what it costs. 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.
Three residual sequences, all starting from zero. The analytic Jacobian gives 8, 0.052, 3·10⁻⁶, 5.2·10⁻¹³; a Jacobian differenced at ε = 10^-6, whose entries are correct to about 1.3·10⁻¹⁰, gives 8, 0.052, 3·10⁻⁶, 4.6·10⁻¹³. They stop at the same residual. The matrix-free run, where GMRES sees only a closure and no entry exists anywhere, lands 8.1·10⁻¹⁶ from the analytic answer in 438 products.
logEps: -6, n: 64
The arguments are the ones The accuracy that is thrown away passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
Three residual sequences, all starting from zero. The analytic Jacobian gives 8, 0.052, 3·10⁻⁶, 5.2·10⁻¹³; a Jacobian differenced at ε = 10^-6, whose entries are correct to about 1.3·10⁻¹⁰, gives 8, 0.052, 3·10⁻⁶, 4.6·10⁻¹³. They stop at the same residual. The matrix-free run, where GMRES sees only a closure and no entry exists anywhere, lands 8.1·10⁻¹⁶ from the analytic answer in 438 products.
logEps: -6
The arguments are the ones The part of a solver that may be rounded passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
Three residual sequences, all starting from zero. The analytic Jacobian gives 8, 0.052, 3·10⁻⁶, 5.2·10⁻¹³; a Jacobian differenced at ε = 10^-6, whose entries are correct to about 1.3·10⁻¹⁰, gives 8, 0.052, 3·10⁻⁶, 4.6·10⁻¹³. They stop at the same residual. The matrix-free run, where GMRES sees only a closure and no entry exists anywhere, lands 8.1·10⁻¹⁶ from the analytic answer in 438 products.
What it checked while drawing
Every figure above checked its own claims on the way to being drawn, and a claim that failed
would have stopped the picture rather than shipped a wrong one. Those checks 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.
8 distinct claims across 3 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 differencing step inside the range the sweep covers
a power of ten rather than an exponent literal
a size the repeated Jacobian assemblies can afford
and the matrix-free solve lands on the same point
in the same number of steps, or at most one more
LU is for square matrices
matmul shapes agree
the differenced Jacobian stops at the same residual as the analytic one
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 217
of 397 generators —
199 print a residual and
18 are exempt with a published reason;
180 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 accuracy that is thrown away
A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.
Methods that were designed apartThe part of a solver that may be rounded
A preconditioner computed and applied with a three-bit significand still returns thirteen correct digits — it costs seventeen extra iterations and nothing else. Round the working arithmetic instead and the step count barely moves while the answer loses exactly the digits the format dropped.