Concept

Bisection — where it appears

Halving an interval repeatedly, keeping the half a monotone test says the answer is in. For eigenvalues the test is a count of negative pivots, so the invariant is a statement about integers and the interval it returns is a bracket rather than an estimate.

Named by 4 essays across 4 fields — each of them below, with the objects they name alongside it.

-101-101(x − 1) ÷ 0.001p(x) ÷ 2.68·10⁻¹⁵a square, drawn negativepoints sampled401negative, unfused179negative, fused196signs disagree at98widest gap1.4·10⁻¹⁵unfusedfusedthe true valuethe true curve is the flat oneeverything else is rounding

A square that evaluates negative

(x − 1)⁶ evaluated near x = 1 comes out negative at 179 of 401 points on one build and 196 on another, and the two disagree about the sign at 98 of them. Neither is nearer the truth: both traces are made entirely of rounding.

machine · Fma contraction
grey: proved empty · filled: proved to contain exactly one roota covered squareboxes proved empty42boxes proved unique2undecided0operator evaluations87every rectangle carries a proofand the two crosshairs are where the roots are

Where the box is cut

A branch-and-bound with an interval operator settles a whole square — two roots proved unique, forty-two regions proved empty, nothing left undecided, in 87 evaluations. Move the roots so one lands on the first bisection and it proves nothing at all, at any depth. Cutting at 0.485 instead of 0.5 finds both, in a quarter of the work.

arithmetic · Interval
-4-2.35371-0.7074290.9388572.585144.231435.8777102468101214shift σν(σ)an answer that is an integershifts2000disagreements0eigenvalues14steps14the marks are a Jacobi decompositionand the staircase never saw one

An eigenvalue count that cannot be slightly wrong

Every spectral computation here returns floats with errors in them. Counting eigenvalues below a shift by the signs of an unpivoted elimination returns an integer, and an integer cannot be 6.9999999997 — so the answer is exactly right, or wrong by a whole eigenvalue, and where the second happens is a band of measurable width.

spectra · Inertia
110¹00.30.60.91.2trust-region radius Δshare of the exact model decreasethe exact subproblem7 radii, n = 60share at the smallest radius0.95share at the largest0.3products, at most8radii stopped by the curvature4a few products against an eigendecompositionand most of the decrease

The certificate that arrives soonest is worth least

The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth. At λₘᵢₙ = −10 the step arrives after two products and gets 39.6 per cent; at −10⁻³ the same two products get 89.8, and the whole sweep costs eight.

iterative · Trust-region

Named alongside it

The objects these essays reach for when they reach for this one.

Exact ground truthBitwise reproducibilityBranch and boundCatastrophic cancellationCertificateCondition numberConjugate gradientsEigenvalue bracketEigenvaluesError accumulationExistence and uniquenessExpression contraction

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