regularisation-tradeoff

At its defaults it draws what the regularisation costs, and what 6 steps of refinement take back. Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

regularisation-tradeoff is one function in lib/figures/quasidef.js — quasi-definite — every ordering legal, and the fill a symbolic phase can promise exactly. 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.

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σ_min(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σ_min and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

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.

43 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.

It draws a decomposition and prints its residual. It calls svd, regularisationSweep, 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 165 of 306 generators — 150 print a residual and 15 are exempt with a published reason; 141 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.