I've learned that many fractals can be described like the below:

$$z_n \in F \text{ if } \limsup_{n\to\infty} |z_{n+1}| \leq i$$

Where $i$ is some number, usually I see 2, and $z$ is a member of the set. This is paired with some iteration on $z_n$ to make $z_{n+1}$. Classic example is the Mandelbrot set: $z_{n+1} = z_n^2 + c$, where $c$ is a complex number.

But some of the things I've read about fractals is they are self-similar, self-symmetric, have a finite area yet infinite perimeter.

I have a few examples (color scheme = more red as approaches the set, magenta very close to the set, blue touching the set, black inside the set):

A "tri-wing" vein set := $z_{n+1} = (-z_n^{-1})^2 + z_n$

enter image description here

A Julia set (k = -0.835 - 0.232i) := $z_{n+1} = z_n^2 + k$

enter image description here

These all are symmetric in some way, have finite area and infinite perimeter. But then I consider fractals like the Burning Ship ($z_{n+1} = |z_n^2| + c$) which have no symmetry. Usually I can define a line somewhere on the graph intersecting the origin and it will be a mirror image on both sides, but not for everything I can define.

Consider a basic example like $z_{n+1} = z_n^0 + c$:

enter image description here

This has symmetry but it's just a circle. This has a finite area as well as a finite perimeter. Or am I wrong? Only $\{0\} \in F$ for this example, so it's not a fractal.

I'll disclaim I've only learned about fractals in the context of $z_{n+1} = z_n + c$, with $z_n$ to the power of an integer, or adding extra terms. Is it safe to assume this is not a particularly good means of defining fractals?

  • $\begingroup$ Your last image is actually a resolution artifact - you didn't do enough iterations of the map. The bifurcation locus is a single point, namely, it is just $\{ 0 \}$. Colors on the outside are not the locus but rather highlighting used to pick up fine features that otherwise could not be drawn by computer. E.g. your first two images would look like nothing without it because the actual sets are infinitely thin. $\endgroup$ – The_Sympathizer May 5 at 5:49
  • $\begingroup$ (add: Not sure about your first pic btw, those black regions could be artifact, but could also be areas of positive measure [area] [hence not infinitely thin]. Is that $z^{-1}$ actually conjugate, as it looks kind of like the "tribar"?) $\endgroup$ – The_Sympathizer May 5 at 5:53
  • $\begingroup$ @The_Sympathizer because of limitations of my algorithm, I set $i$ in the first equation to $5$ and instead of $n\to\infty$ it's $n\to 64$ because much higher takes 30-40s to render. $\endgroup$ – gator May 5 at 5:59
  • $\begingroup$ Yes. That's because that map causes $z$ to move with a constant "velocity" given by $c$ - in each step, you add the same $c$ to the previous addition thereof, which is just incremental, linear motion. When $c$ is very small, i.e. near the origin, $z$ moves very, very slowly and thus takes a long time to reach the bound. But from that reasoning, you can easily see then that the BL must be only the point $0$, because only $c = 0$ produces no motion at all and thus no (eventual) escape. $\endgroup$ – The_Sympathizer May 5 at 6:03
  • $\begingroup$ In one textbook a fractal in $\Bbb R^n$ was defined as a subset of $\Bbb R^n$ whose $n$-Hausdorff-Besicovich dimension is larger than its Small Inductive Dimension. Such a set, even a compact one, need not be self-similar nor have any ordinary type of symmetry. $\endgroup$ – DanielWainfleet May 5 at 10:22

Your last picture, actually, answers your title question!

What the "thing" you are asking about that the Mandelbrot set is an example of is called a bifurcation locus of a self-map on the complex plane. To put this in some cleaner notation, if you have a complex self-map $f_c: \mathbb{C} \mapsto \mathbb{C}$ with one parameter $c$, then the orbits of the map with a starting point $z_0$ are the sequences $O(f_c, z_0)$ defined by

$$[O(f_c, z_0)](n) := f_c^n(z_0)$$

where the exponent represents repeated application of $f_c$. At each $n$ the orbit gives the result of $n$ self-applications of $f_c$ to the starting point $z_0$. The bifurcation locus developed with the starting point $z_0$ is

$$BL[f, z_0] := \{\ c \in \mathbb{C} : \mbox{$O(f_c, z_0)$ is a bounded sequence}\ \}$$

The reason for the name "bifurcation locus" is that this is related to the notion of a bifurcation diagram like that commonly used to represent a real self-map, most popularly seen in the real logistic (Verhulst) map. The difference is the bifurcation locus is simply the points where you get a bounded orbit subject to bifurcation, while the bifurcation diagram is a plot of the whole orbit at every single parameter value, or at least the limiting behaviors of the orbits. For complex functions, the latter actually can't be done so faithfully because it would require a four-dimensional space.

In the last picture, your map $f_c(z) := z + c$ has a bifurcation locus consisting only of the point $0$. This can be understood by physical reasoning: repeated action of it is like a discretized motion with steady velocity, which here is $c$ - one step of $c$ for each "unit" of time. If you keep up a steady motion with nonzero speed long enough, eventually it will cross any boundary you put in its way. As a result, unless $c = 0$, all starting points $z_0$ go to infinity.

For the mathematical proof, it is easy to see that

$$f_c^n(z_0) = \underbrace{(\cdots((z_0 + c) + c) + c)}_{\mbox{$n$ nestings}} = z_0 + nc$$

and thus $\lim_{n \rightarrow \infty} f_c^n(z_0) = \infty$ unless $c = 0$, hence $BL[f, z_0] = \{0\}$ no matter what $z_0$ is. A single point set is not a fractal.

Your picture almost shows this, except that you didn't iterate enough times. As you mentioned, it gets very slow to plot because, effectively, the "speed" of the point gets arbitrarily slow as you get close to the origin. Thus for drawing near there, the computer has to do many, many, many iterations to reach your approximated escape boundary. You could get an accurate picture if you increased the iterations more and left it to cook, but the above should illustrate to you that you don't need to. Nonetheless, you can imagine what the final plot would look like: it will just be a kaleidoscope of colors converging at that one point.

  • $\begingroup$ This is a good answer. $\endgroup$ – Alexander Gruber May 5 at 6:22
  • $\begingroup$ Interesting, very good answer! I noticed a larger maximum depth does slow it down but also "shrinks" the black portion of the image; by this I mean it can render with higher fidelity the set. It makes sense: $\lim_{n\to 64}$ is adequate enough to visualize some sets but $\lim_{n\to\infty}$ would be ideal. Interestingly, using the Mandelbrot set, a depth of 1 shows a completely black image. Of 2 shows a large black circle. Of 4 it looks like a black bicycle seat. And so on. It realizes the actual shape of the set better the larger the maximal depth is. $\endgroup$ – gator May 5 at 6:49
  • 1
    $\begingroup$ @gator : Yep, because at a low depth of recursion, there is not enough "time" for points that will eventually escape the boundary circle to actually do so and thus they will be mistakenly counted as being within the set when they aren't. Using these algorithms is kind of like doing scientific laboratory measurements: if something looks funny, you should tweak parameters specific to the algorithm's assumptions to see if it's not an artifact thereof. $\endgroup$ – The_Sympathizer May 5 at 6:54

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