The area of the Mandelbrot set, from two sides

The shape in every picture of itself listed what is known about the Mandelbrot set — the complex numbers $c$ for which the orbit of nought under $z \mapsto z^2 + c$ stays bounded — and one line of the list was blunt: its area is unknown. The set is closed and bounded, so it certainly has an area, a definite number of square units. Estimates put it near 1.5066. The error bars have not closed, and no method that would close them is known.

That sentence hides two quite different ways of attacking the number, and they behave in opposite ways. One counts: cover the plane with small squares and decide, for each, whether it belongs to the set. The other is a formula: the map that carries the outside of a disc onto the outside of the set has coefficients, and a classical theorem turns them into the area exactly. The count converges quickly to a number it cannot certify. The formula gives certified bounds that converge so slowly they are nearly useless. This essay runs both, and measures how far apart they stay.

Deciding a point is harder than drawing it #

The familiar pictures of the set are drawn by iterating $z \mapsto z^2 + c$ from nought, at each pixel’s $c$, and stopping either when $|z|$ passes 2 — after which the orbit certainly escapes to infinity — or when an iteration cap is reached. A pixel that reaches the cap is coloured as inside. For a picture that is harmless. For a measurement it is a guess: a point near the boundary can take thousands or millions of steps to escape, and a point exactly on the boundary never escapes and never settles.

So a measurement needs a test for inside as well as one for outside. Two regions can be recognised by formula. The big heart-shaped region is the main cardioid, the $c$ for which the map has an attracting fixed point; a short inequality decides membership. The round region to its left is the disc of radius $\tfrac14$ about $-1$, where the map has an attracting cycle of period 2. Everywhere else, a point is called inside if its orbit can be seen to settle onto a cycle: the orbit is compared, at each step, with a saved point, and the save is renewed at steps 2, 4, 8, 16, … so that a cycle of any length is eventually caught repeating. A point is outside if its orbit leaves the disc of radius 2, and undecided if neither has happened by step $N$.

The hero figure classifies cells 1/38 of a unit wide, with $N = 60$. The dark cells, certainly inside, total 1.379 square units; the orange, undecided, 0.197. The orange cells are not scattered: they trace the boundary and the antennae of the set, and the small bulbs where the orbit takes a long time to choose. Everything about the measurement problem is in that orange band.

Two estimates that close in #

Make the cells smaller and the wait longer, and the dark and orange totals change.

The certainly-inside estimate climbs from 1.4256 to 1.4968; the inside-or-undecided estimate falls from 1.5104 to 1.5069. They straddle 1.5066, the value that much larger computations of this kind have settled on, and they approach it from opposite sides. That looks like a two-sided bound, and it is not one. Each cell is judged by a single point, its centre, and a cell whose centre escapes can still contain part of the set, while a cell whose centre settles can contain points that escape. The two lines are estimates, not bounds, and the 1.5066 they close on is a statistical conclusion: careful versions of this count, with cells a few millionths wide and corrections for the cells along the boundary, agree on 1.50659 to several more decimal places, but none comes with a guarantee.

The obstacle is the boundary itself. Mitsuhiro Shishikura proved in 1998 that the boundary of the Mandelbrot set has Hausdorff dimension 2 — the largest a curve in the plane can have — so in the sense of a dimension that is not a whole number, it is as crinkled as a set of its size can be. Whether it has positive area, so that the set’s area would include a contribution from points that are neither inside nor outside in any visible sense, is not known. If it did, no count of cells could ever settle the question by refinement.

Waiting is not enough either #

At a fixed cell size the undecided share can be cut by waiting longer, and the rate at which it falls says how stubborn the boundary is.

Over the range measured, the undecided area falls from 0.165 at 100 steps to 0.0026 at 20,000, roughly in proportion to one over the wait. Points that escape slowly do so because they sit close to the set, and the closer a point is, the longer its orbit lingers near the set before leaving; points inside near the boundary converge to their cycle slowly for the same reason. Halving the undecided area costs doubling the wait. And there is a floor: a cell whose centre happens to lie exactly on the boundary is undecided for ever. Refining the grid puts more cells on the boundary, so the two refinements have to be pushed together, and the cost of a count accurate to $\varepsilon$ grows much faster than one over $\varepsilon$.

That is the practical shape of counting. It is the method of adding up rectangles applied to the indicator of the set, and like any such sum over a set with a rough edge, the error is controlled by what happens at the edge. A smooth boundary would contribute an error proportional to its length times the cell size; this boundary has no finite length, and its contribution has to be beaten down by brute force.

A formula from the outside #

The other approach starts from a theorem of Adrien Douady and John Hubbard, who proved in 1982 that the Mandelbrot set is connected. The proof constructs a map $\Phi$ from the outside of the set onto the outside of the unit disc that is conformal — it preserves angles — and has an inverse $\psi$ going the other way, with a series expansion

$$\psi(w) = w + b_0 + \frac{b_1}{w} + \frac{b_2}{w^2} + \cdots$$

valid for every $|w| > 1$ — every point of the outside of the disc, which on the sphere the complex numbers live on is a cap around the point at infinity. The circles $|w| = r$ are carried to curves that wrap ever more tightly around the set as $r$ shrinks towards 1. A conformal map keeps angles but not areas, as angles survive and areas do not explained for maps of the globe, and it is exactly the change in area that makes the coefficients useful.

Thomas Gronwall’s area theorem of 1914 computes the area enclosed by the image of the circle $|w| = r$ from the coefficients — it is the area found by walking round a curve, applied to the image curve and written in terms of the series — and letting $r$ fall to 1 gives the area of the set:

$$\text{area} = \pi \Big( 1 - \sum_{m \ge 1} m\, b_m^2 \Big).$$

The coefficients here are real, because the set is symmetric about the real axis. Every term $m b_m^2$ is positive, so every partial sum of the series gives an upper bound on the area — a certified one, with no statistics in it — and the bound can only decrease as terms are added.

The coefficients can be computed exactly, one at a time. Write $p_0 = c$ and $p_{k+1} = p_k^2 + c$: these are the points of the critical orbit, as polynomials in $c$. Substituting $c = \psi(w)$ into $p_n$ must give $w^{2^n}$ plus terms smaller than $w^{-2^n}$, because that is what it means for $\Phi$ to straighten the iteration into squaring — and squaring, as multiplying is turning showed, doubles every angle, so the straightened picture is a circle whose points run round it twice as fast at each step. Expanding, the coefficient of each power of $w$ below the top must vanish, and the $m$-th one involves $b_m$ with weight $2^n$ and otherwise only earlier coefficients. So each $b_m$ is determined by the ones before it, and they are rational numbers with powers of 2 in their denominators: $b_0 = -\tfrac12$, $b_1 = \tfrac18$, $b_2 = -\tfrac14$, $b_3 = \tfrac{15}{128}$, $b_4 = 0$, $b_5 = -\tfrac{47}{1024}$. This recursion is due in essence to Irwin Jungreis, in 1985.

Coefficients that shrink too slowly #

The coefficients shrink, but slowly: a line fitted through their sizes falls like one over $m^{1.17}$, with swings of a factor of ten either side of it. The area series sums $m b_m^2$, whose typical size is then about $m^{-1.3}$, and a sum of terms shrinking like $m^{-1.3}$ converges — barely. Its tail after $k$ terms is of order $k^{-0.3}$ by that crude count, and in fact the measured tail behaves even worse.

The figure also shows something unexpected: 115 of the first 4,095 coefficients are exactly nought, and they sit in a pattern.

Every vanishing coefficient has an index divisible by 4. At first it is every multiple of 4 — $b_4$, $b_8$, $b_{12}$, $b_{16}$ — then only the multiples of 8 between 16 and 64, then sparser multiples of higher powers of 2. The pattern was seen in the first computations of these coefficients in the 1980s, and many of the zeros have since been shown to be forced by the way the set’s outside is built from preimages under squaring. The sizes of the coefficients that do not vanish have no known formula, and their erratic swings reflect the set’s erratic boundary: a coefficient is an average of the boundary’s shape at a particular frequency, and the boundary has structure at every frequency.

The bound, term by term #

The bound starts at 2.273 with 16 coefficients and falls to 1.786 with 16,384. Each doubling of the number of terms buys only a few hundredths, and over the last few doublings the distance to 1.5066 shrinks like $k^{-0.12}$. If it continued at that rate, a bound within a thousandth of the counting value would need of the order of $10^{25}$ coefficients. John Ewing and Glenn Schober computed 240,000 of them in 1992, and their bound was still far above the counting value.

So the two methods agree in principle — the series converges to the true area, whatever it is — and nowhere near in practice. The count gives a number to five or six decimal places that nobody can prove. The series gives a proof of a number that is too large by more than a sixth. There is no theorem that the counting value 1.5066 is the series’ limit; it is only overwhelmingly likely, and the gap between “overwhelmingly likely” and “proved” is the whole of the open problem.

Why the formula is so slow #

The slowness has a geometric reason. The image of the circle $|w| = r$ under $\psi$ is a smooth curve around the set, and Gronwall’s formula computes the area inside it exactly from the coefficients; truncating the series is, roughly, measuring the area inside the image of a circle that does not quite reach radius 1. For the Mandelbrot set those curves approach the boundary extremely slowly where the set has its antennae and thin filaments. A curve at $r = 1.001$ still encloses a great deal of empty plane around each filament, and the series has to resolve every one of them. The filaments are long and thin, and there are infinitely many of them; the area they hold is tiny, but the area around them that a smooth curve must include is not.

The same structure makes the count fast where it is fast. A grid cell either falls in the solid body of the set, where it is decided in a few steps, or in the empty plane, where it escapes quickly; only the cells along filaments and boundary are hard. The count measures the solid body precisely and leaves a band of uncertainty around the edge; the series measures everything at once and is dominated by the edge.

The comparison is a familiar one. The error that does not care how many dimensions contrasted a slow method whose error can be estimated with fast methods whose error cannot; here the roles are split differently, between a fast method that cannot certify and a certifying method that cannot finish. What would close the problem is a lower bound: a certified area that the set definitely contains, which is easy — the cardioid and the period-2 disc already give 1.374 square units — but whose improvement requires certifying that whole regions are inside, and that is possible only for the hyperbolic components, whose total area is itself an infinite sum nobody has evaluated.

Lower bounds from the components inside #

Inside the set, every point that the cycle test catches lies in a hyperbolic component — a region over which, as one c, one picture showed for the Julia sets, the dynamics keeps the same attracting cycle and the same shape of picture: a region where the map has an attracting cycle of a fixed period. The main cardioid is the component of period 1, the disc at $-1$ that of period 2, and there are components of every period, budding off the cardioid and off each other. Each has an explicit description, and its area can be computed. Adding them up gives certified lower bounds: the cardioid’s area is $3\pi/8 \approx 1.178$ and the disc’s $\pi/16 \approx 0.196$, together 1.374, and each further component raises the total — but there are infinitely many, their number grows exponentially with the period, and the higher-period ones are tiny and must each be measured numerically.

It is conjectured that the hyperbolic components fill the interior of the set — that every interior point has an attracting cycle. That is the density of hyperbolicity conjecture, one of the central open problems of the subject, and by work of Douady and Hubbard it would follow from the conjecture that the set is locally connected. If it is true, the area is the sum of the areas of the components plus the area of the boundary; if the boundary also has zero area, the area is exactly the sum over components. Neither part is known, and the sum over components is itself an infinite series nobody has evaluated precisely enough. Both certified approaches, from inside and from outside, run into the same unproved facts about the boundary.

What the pictures cannot show #

None of the figures shows the boundary itself. The cell picture shows a band of undecided cells, but the true boundary, of dimension 2, lies inside that band in a form no finite grid can draw. The coefficient pictures show finitely many coefficients of an infinite series. The bound figure shows a curve that is still falling when it stops, and nothing on the page says where it would end; the 1.5066 on every figure is drawn from the counts, and the claim that the series reaches it is a belief, not something the computation established.

The counts, too, rest on choices: the centre of each cell, the cardioid and disc tests, the cycle test with its tolerance. A different choice of sample point in each cell would give slightly different totals at each grid size, converging to the same limit if the boundary has zero area, and possibly to different limits if it does not. Careful published counts use many points per cell and statistical models of the boundary cells, and they agree with the simple centres used here to the precision shown.

Still open: the area, and the boundary’s area #

The area of the Mandelbrot set is not known. Counting estimates agree on about 1.50659, and the simplest certified bounds — the cardioid and disc from inside, the series from outside — leave an interval from 1.374 to above 1.7, and narrowing it by certified means requires measuring infinitely many components or summing a series that barely converges. Whether the set’s boundary has positive area is open; it has the largest possible dimension, which is consistent with either answer. And the density of hyperbolicity — whether every interior point has an attracting cycle — is open, though it is known to follow from the conjecture that the set is locally connected, which has resisted proof since the 1980s.

There is a computational question as well. The coefficients $b_m$ can be computed exactly, and their pattern of zeros is partly understood, but the asymptotic size of the nonzero ones is not known. A proven rate for how fast $\sum m b_m^2$ converges would turn the series into an estimate with an error bar, and the measured rate here — the tail shrinking roughly like $k^{-0.12}$ over a factor of a thousand in $k$ — suggests a convergence so slow that even a precise error bar would be of little use. The two ways of measuring the set will meet, if they meet, at a number nobody has yet been able to prove.