The Mandelbrot Set, a Map of Julia Worlds, math that surprises you lesson

A Julia set picture keeps the rule $z^2 + c$ fixed and lets the starting point change from pixel to pixel. What changes if the starting point stays fixed at $0$ and the parameter $c$ varies instead? Then every pixel is a different rule, and the picture becomes a map of the whole family of rules. This page defines the Mandelbrot set, works out its exact cases, explains the radius $2$ test and the limits of escape-time pictures, and states the theorem that makes the set a map of Julia worlds, with connected Julia sets inside it and dust outside.

Key terms

Mandelbrot set

Definition The Mandelbrot set $M$ is the set of complex numbers $c$ for which the orbit of $0$ under the rule $z^2 + c$ , which is $z_0 = 0$ , $z_1 = c$ , $z_2 = c^2 + c$ and so on, stays bounded. Each point of the set is a parameter, so each point stands for a whole rule. The radius $2$ decides membership for every $c$ . One point of the orbit with $\lvert z_n \rvert > 2$ forces escape, so $c$ belongs to $M$ exactly when $\lvert z_n \rvert \le 2$ for every $n$ . When $\lvert c \rvert \le 2$ this is the escape radius $\max(2, \lvert c \rvert) = 2$ from Julia Sets, One Rule and Many Starting Points. When $\lvert c \rvert > 2$ the orbit escapes anyway. It begins with $z_1 = c$ , and whenever $\lvert z_n \rvert \ge \lvert c \rvert$ ,

\begin{aligned} \lvert z_{n+1} \rvert &\ge \lvert z_n \rvert^2 - \lvert c \rvert \\[0.4em] &\ge \lvert z_n \rvert^2 - \lvert z_n \rvert \\[0.4em] &\ge \lvert z_n \rvert \bigl(\lvert c \rvert - 1\bigr). \end{aligned}

The factor $\lvert c \rvert - 1$ is greater than $1$ , so every later point is again at least as far out as $c$ , and the distances grow at least like the powers of $\lvert c \rvert - 1$ . Hence $M$ lies in the closed disk $\lvert c \rvert \le 2$ , and on the real axis it is exactly the segment from $-2$ to $\tfrac14$ , as the third question below shows. The start $0$ is not arbitrary. Every value $w$ other than $c$ is reached in one step from exactly two starts, the two square roots of $w - c$ , while $c$ is reached only from $0$ , so $0$ is the one point where the two starts merge. The orbit of this special point decides whether the Julia set is in one piece, and the connectedness card below gives the theorem. The set is named after Benoit Mandelbrot, who published computer pictures of it in 1980.

In plain words Each point of the Mandelbrot picture is a rule. Run that rule from $0$ , and the point belongs to the set when the orbit never gets farther than $2$ from $0$ . Belonging has nothing to do with $c$ itself being small.

Example Three exact cases come first. For $c = 0$ the orbit is $0, 0, 0, \dots$ , a fixed point, and for $c = -1$ it is $0, -1, 0, -1, \dots$ , a cycle of length $2$ , so both values lie in $M$ . For $c = 1$ the orbit is $0, 1, 2, 5, 26, \dots$ . The value $2$ at the second step lies exactly on the radius and decides nothing, but the value $5$ at the third step proves escape, so $1$ is not in $M$ . The parameter $c = -2$ shows why the test must be strict. Its orbit is $0, -2, 2, 2, 2, \dots$ , which reaches distance exactly $2$ and stays there, so $-2$ belongs to $M$ and is its leftmost point. A test that removed every orbit with some $\lvert z_n \rvert \ge 2$ would wrongly throw it out. Size alone does not decide, since $-2$ is far from $0$ and belongs, while $0.3$ is close to $0$ and does not, as the second question below shows.

Parameter plane

Definition For the family of rules $z^2 + c$ , the parameter plane is the complex plane of values of $c$ , in which each point stands for one rule. It differs from the dynamical plane, the complex plane of starting points $z_0$ for one fixed rule, where filled Julia sets and Julia sets are drawn. A Julia picture fixes $c$ and lets $z_0$ vary over the dynamical plane. The Mandelbrot picture fixes $z_0 = 0$ and lets $c$ vary over the parameter plane. Both pictures use the same rules and the same kind of escape test, but their coordinates answer different questions, so a point of one picture is not a point of the other.

In plain words Moving across a Mandelbrot picture changes the rule. Moving across a Julia picture changes only the start. Behind every point of the parameter plane lies a whole dynamical plane with its own Julia set.

Example The point $c = -1$ of the parameter plane is a single point of $M$ , and behind it lies the dynamical plane of $z^2 - 1$ , where the filled Julia set is the basilica. The large shapes of $M$ have exact meanings. The big heart-shaped region, the main cardioid, consists of the values $c$ for which $z^2 + c$ has an attracting fixed point $p$ , one with $\lvert 2p \rvert < 1$ . For a fixed point, $p^2 + c = p$ , so

\begin{aligned} (z^2 + c) - p &= z^2 - p^2 \\[0.4em] &= (z - p)(z + p), \end{aligned}

and near $p$ each step multiplies the distance to $p$ by about $\lvert 2p \rvert$ . Nearby orbits are pulled in, and in fact the orbit of $0$ is drawn to $p$ . For $c = -\tfrac12$ the attracting fixed point is $p = \tfrac{1 - \sqrt{3}}{2} \approx -0.366$ , with $\lvert 2p \rvert \approx 0.732$ . The disk of radius $\tfrac14$ centered at $-1$ consists of the values $c$ for which the rule has an attracting cycle of length $2$ , and it touches the cardioid at $c = -\tfrac34$ . For $c = -1.2$ , inside that disk, the orbit of $0$ approaches the cycle of the two numbers $-\tfrac12 \pm \tfrac{3\sqrt{5}}{10}$ , about $0.171$ and $-1.171$ . The region just above the meeting point, around $-0.75 + 0.1i$ , is known as Seahorse Valley.

Escape-time algorithm

Definition The escape-time algorithm draws a finite approximation of a set defined by bounded orbits. It needs a rectangle of the plane, a grid of sample points with one point for each pixel, an escape radius and a budget of $N$ steps. For the Mandelbrot set the sample point is the parameter $c$ , the orbit starts at $0$ and the radius is $2$ . Each orbit is computed for at most $N$ steps. If some $n \le N$ has $\lvert z_n \rvert > 2$ , the smallest such $n$ is the escape time of that pixel, and the pixel is colored by it. Otherwise the pixel gets a separate color, often black, which means only that no escape was observed within $N$ steps. An escape proves that the sample point lies outside $M$ , provided the rounding errors of the computer arithmetic are small enough. A non-escape proves nothing, because points just outside $M$ can take very many steps to escape. The picture also depends on the grid, since a pixel shows one sample point and thin parts of the set can fall between samples. Zooming in means recomputing the orbits at new, closer sample points, usually with a larger budget, and enlarging an old picture only enlarges its pixels. Ordinary double-precision arithmetic carries about $16$ significant digits, so very deep zooms need more precise arithmetic.

In plain words Run the orbit of every pixel for a fixed number of steps and color the pixel by how soon the orbit crossed radius $2$ . The colors outside the set are measurements, while the dark region holds the sampled points that have not escaped yet, which can include points just outside the set.

Example For $c = 1$ the escape time is $3$ , since $\lvert z_2 \rvert = 2$ is not beyond the radius and $\lvert z_3 \rvert = 5$ is. Real parameters just to the right of $\tfrac14$ , the right end of $M$ on the real axis, escape slowly.

$c$	Escape time
$0.26$	$30$
$0.2501$	$312$
$0.250001$	$3140$

With a budget of $1000$ steps, the pixel at $0.250001$ would be painted as if it belonged to $M$ , although its orbit escapes at step $3140$ . The counts follow a curious pattern, since multiplying each by $\sqrt{c - \tfrac14}$ gives $3.0$ , $3.12$ and $3.14$ , close to $\pi$ . Parameters just above the meeting point $-\tfrac34$ escape slowly too. The parameter $-0.75 + 0.01i$ escapes at step $315$ and $-0.75 + 0.001i$ at step $3143$ , and multiplying by the heights $0.01$ and $0.001$ again gives about $\pi$ . The colors are a choice of palette applied to these counts, so a different palette shows the same computation with a different look.

Connectedness

Definition A closed and bounded set in the plane is connected when it cannot be split into two nonempty parts that are a positive distance apart, in other words when it is all in one piece. It is totally disconnected when none of its connected parts has more than one point, like a dust. For the quadratic family, the following theorem goes back to the work of Pierre Fatou and Gaston Julia around 1919. If the orbit of $0$ under $z^2 + c$ is bounded, the filled Julia set $K_c$ and the Julia set $J_c$ are connected. If the orbit of $0$ escapes, $K_c$ and $J_c$ are the same set, and that set is totally disconnected, a dust with the same structure as the Cantor set built by removing middle thirds. So the Mandelbrot set is exactly the set of parameters whose Julia sets are connected, which is what makes it a map of Julia worlds. This all-or-nothing choice relies on the rule having just one special point like $0$ . For polynomial rules of higher degree with several such points, a Julia set can be disconnected without being a dust. The Mandelbrot set itself is connected, as Adrien Douady and John Hubbard proved in 1982.

In plain words Inside the Mandelbrot set every Julia world is in one piece. Outside it, the Julia world falls apart into dust. The orbit of the single point $0$ decides which.

Example For $c = 0$ , inside $M$ , the Julia set is the unit circle, which is in one piece. For $c = -1$ it is the edge of the basilica, and for $c = -2$ it is the segment from $-2$ to $2$ , both connected. For $c = 1$ the orbit of $0$ runs $0, 1, 2, 5, \dots$ and escapes, so $J_1$ is a dust. Pictures can only suggest these facts, and the theorem proves them. Pictures can also mislead about $M$ itself. At modest resolution some small pieces of $M$ look like separate islands, yet they are joined to the main body by filaments too thin for the grid to catch. Many of those islands resemble the whole set, but they are distorted versions of it, not exact scaled copies, so a zoom into such a region shows resemblance rather than exact repetition.

Check yourself

Is $c = i$ in the Mandelbrot set? Follow the orbit of $0$ for four steps.

Show the answer

Starting from $z_0 = 0$ , the orbit is

\begin{aligned} z_1 &= i, \\[0.4em] z_2 &= i^2 + i = -1 + i, \\[0.4em] z_3 &= (-1 + i)^2 + i = -i, \\[0.4em] z_4 &= (-i)^2 + i = -1 + i, \end{aligned}

using $(-1 + i)^2 = 1 - 2i + i^2 = -2i$ . From then on the orbit alternates between $-1 + i$ and $-i$ . Its distances from $0$ are $0$ , $1$ , $\sqrt{2}$ , $1$ , $\sqrt{2}$ and so on, never more than $2$ , so $i$ belongs to $M$ , and by the theorem its Julia set is connected.

For $c = 0.3$ the first nine values after the start are all below $1$ . Does that make $0.3$ a member of $M$ ?

Show the answer

No. For a real parameter $c$ , each step changes the value by

\begin{aligned} z_{n+1} - z_n &= z_n^2 - z_n + c \\[0.4em] &= \left(z_n - \tfrac12\right)^2 + c - \tfrac14, \end{aligned}

which is at least $c - \tfrac14 = 0.05$ . So the orbit climbs by at least $0.05$ at every step and must pass $2$ within $40$ steps. In fact $z_{10} \approx 1.037$ , $z_{11} \approx 1.376$ and $z_{12} \approx 2.193$ , so the escape time is $12$ , and a budget of $10$ steps would have shown $0.3$ as dark. The same identity shows that every real $c$ greater than $\tfrac14$ lies outside $M$ , however slowly its orbit climbs.

Extend the pattern. The values $-2$ and $\tfrac14$ are the ends of $M$ on the real axis. Show that a real number $c$ belongs to $M$ exactly when $-2 \le c \le \tfrac14$ .

Show the answer

There are four cases. If $c > \tfrac14$ , the previous answer shows that the orbit climbs by at least $c - \tfrac14 > 0$ at every step, so it escapes. If $c < -2$ , then $\lvert z_1 \rvert = \lvert c \rvert > 2$ , so it escapes.

If $-2 \le c \le 0$ , every $x$ with $c \le x \le -c$ has $x^2 \le c^2$ , so

\begin{aligned} c \le x^2 + c &\le c^2 + c \\[0.4em] &\le -c, \end{aligned}

where the last step holds because $c^2 + 2c = c(c + 2) \le 0$ . The rule keeps the segment from $c$ to $-c$ inside itself, and $0$ lies in it, so the orbit of $0$ stays in that segment and is bounded.

If $0 \le c \le \tfrac14$ , every $x$ with $0 \le x \le \tfrac12$ has $0 \le x^2 + c \le \tfrac14 + \tfrac14 = \tfrac12$ , so the orbit of $0$ stays between $0$ and $\tfrac12$ .

Hence the real points of $M$ are exactly the numbers from $-2$ to $\tfrac14$ , and both ends belong. The orbit for $c = -2$ is $0, -2, 2, 2, \dots$ , and the orbit for $c = \tfrac14$ climbs toward $\tfrac12$ without ever reaching it.

The Mandelbrot Set, a Map of Julia Worlds

Key terms

Mandelbrot set

Parameter plane

Escape-time algorithm

Connectedness

Check yourself

How this lesson connects

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