Concept

Invariant measure

A way of assigning sizes to sets that a map leaves unchanged: the preimage of every set has the same size as the set. It is what turns long-run questions about an orbit into questions about areas, and for a chaotic map it gives the share of time spent in each region.

Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.

How long four orbits of the doubling map spend in each quarter. Grouped bars for four starting points of the doubling map, each group showing the fraction of time the orbit spends in each quarter of the unit interval, against a dashed line at one quarter.

Almost every orbit is fair

Double a number and keep the fractional part, and do it again, and again. Where the orbit goes is written in the number's binary digits, and for almost every starting point it spends exactly a quarter of its time in each quarter of the interval. The proof is a short argument about Fourier coefficients being pushed to infinity — and it leaves room for a set of exceptions with no length at all, which includes every fraction and, for all anyone can prove, every number anyone has ever named.

dynamics · Iteration
The Chebyshev maps T₂, T₃, T₄ and T₅. Four small square plots of the Chebyshev polynomials of degrees two to five on the interval from minus one to one, each with the diagonal drawn; the graph of degree n sweeps between the bottom and top of the square n times.

A solvable chaos of every degree

The logistic map at four is chaotic and, through a change of coordinates, completely solvable: its orbits are cosines of doubling angles. The trick is not a one-off. For every whole number n there is a polynomial of degree n that multiplies angles by n instead of 2, and every one of them is exactly as solvable, has exactly nᵏ points of period k, and preserves the same distribution — and any two of them commute, which almost no two polynomials do.

dynamics · Iteration
Where x ↦ βx mod 1 spends its time, the golden ratio φ. A histogram of 2000000 orbit points of the β-map for the golden ratio φ in 50 bins, with Parry's staircase density drawn over it; they agree, and the density steps down at the orbit of 1.

Where a base-β orbit spends its time

Follow x ↦ βx mod 1 for a million steps and count how often the orbit visits each part of the interval. For the doubling map the answer is evenly; for the golden ratio it is a staircase with one step, spending 1.17 times the average near 0 and 0.72 times it near 1. The step sits exactly where the orbit of the number 1 lands, and in a base whose 1 never has a finite expansion the staircase has infinitely many steps.

dynamics · Symbolic dynamics
Where one Hénon orbit spends a million steps. A density map of 1000000 steps of one Hénon orbit counted into 420 × 140 cells: 2542 cells visited, half the time spent in 608 of them.

Where the time goes on an attractor

A chaotic orbit's position is unpredictable within a few dozen steps. How it divides its time is not — start anywhere, follow long enough, and the fraction of time spent in each region comes out the same. That distribution lives on a set of no area, and among the infinitely many ways an orbit could spend its time, typical starts pick exactly one.

dynamics · Strange attractor
Five triangles between the same two circles, every one closing. An outer circle of radius 1 and an inner of radius 0.38 at Euler's distance 0.4899; five triangles inscribed in the first and circumscribed about the second, started from different points.

A triangle that fits once fits everywhere

Put one circle inside another and try to fit a triangle between them, its corners on the outer circle and its sides touching the inner. Usually no triangle fits. But if one does, then one fits starting from every point of the outer circle — and whether it does is decided by a single equation in the two radii and the distance between the centres.

geometry · Triangle centres
The cosines of the powers of two. cos(2^k) for k < 600, with a histogram of 20,000 points against the arcsine density.

The cosines of the powers of two

Start at cos 1 and apply x ↦ 2x² − 1 over and over. The map is chaotic and exactly solvable: the k-th point is cos(2ᵏ), the cosine of two to the k radians. Iterate it in floating point and the computed orbit leaves the true one after 51 steps. Read the binary digits of 1/(2π) instead and the true orbit can be followed for as long as digits can be computed — here 262,080 steps — and at that length it is as fair as any orbit can look. Whether it stays fair is a question about those digits, and nobody can answer it.

dynamics · Iteration

Named alongside it

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

Binary expansionChebyshev polynomialErgodic theoremErgodicityNormal numberPeriodic orbitTopological entropyChaosCircumcircleConjugacyDensityDoubling map

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