Concept

Estimate

A bound or approximate value for a quantity that cannot conveniently be computed exactly, with a reason to trust it. A good estimate says how far off it can be, so that it can settle the questions the exact value would settle.

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

p(n) to 60, against the Hardy–Ramanujan estimate. The number of partitions of each number up to sixty on a logarithmic scale, with the asymptotic estimate drawn over it and the ratio of the two tabulated.

The size of a number with no formula

There is no closed expression for the number of partitions of n. There is an expression for how large it is — with a square root in the exponent and a π in front — and it is accurate enough that rounding a few terms of its refinement gives the exact count.

number · Partitions

Discs from the rows, and the eigenvalues inside them. The complex plane with one disc per row of a 3×3 matrix — centred on its diagonal entry, with radius the sum of the sizes of the rest of the row — and the 3 eigenvalues marked: 4.14, −5.09, −1.06.

Discs that fence in the eigenvalues

Draw one disc for each row of a square matrix, centred on the diagonal entry, with a radius equal to the sum of the sizes of everything else in that row. Every eigenvalue lies inside one of the discs, and a group of discs set apart from the rest holds exactly as many eigenvalues as it has discs. Nothing is solved to find them.

algebra · Eigenvectors

Latin squares of order 4, and the ones that are also Sudoku grids. Two 4×4 Latin squares with their 2×2 boxes outlined: one in which every box also holds 1 to 4, and the cyclic square, whose top-left box repeats a symbol. Of all 576 Latin squares of order 4, 288 pass the box test.

A Latin square with boxes

A finished Sudoku is a Latin square of order nine with one extra rule: each 3×3 box holds every digit once. At order four the extra rule keeps exactly half of the 576 Latin squares, the 288 survivors are two grids in disguise, and no puzzle can be pinned down by fewer than four clues. At order nine every one of those questions needed a computer, and the answers are 6.67 × 10²¹ grids, 5.47 billion essentially different ones, and seventeen clues.

computation · Latin squares

One shared birthday, as a measurement of the year. Histogram over 20000 rooms of the estimate T(T − 1)/2 from the first shared birthday, averaging 361 against 365, spread widely.

Counting a population by its repeats

The birthday problem runs forwards from a known year to the chance of a match. Run it backwards and the matches measure the year: count how many pairs among the draws came out the same, and the number of kinds is about the number of pairs divided by that. It costs the square root of the population rather than a census of it — and when the kinds are unevenly common, what the repeats measure is not how many kinds there are but how many they behave like.

probability · Birthday problem

Named alongside it

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

Counting Analytic continuation Approximation Asymptotics Birthday problem Collision Complex numbers Continuity Convergence Diagonal dominance Eigenvalue Exhaustive search

All concepts