I have always been amazed by things like the Mandelbrot set. I share the view of most that it and the Koch snowflake are absolutely beautiful. I decided to get a deeper more mathematical knowledge of this, but sadly Wikipedia hasn't been of much help. I have a few questions to get me started on this then.

  • What does $z_{n+1} = z_n^2 + c$ mean? Do we take an initial value of $z_n$ and calculate successive points and go on plotting them on the complex plane?

  • What is its historical significance? I feel that knowing where something came from helps us appreciate it even more. Where did this equation first come from? Why was it required to be studied?

Those are my specific questions for the moment, but considering the fact that I am only trying to learn about fractals, I may not be very well equipped to be asking the right questions, in which case you could tell me anything else you think is worth mentioning.

EDIT: I have one specific concern about the equation. It says that $z$ and $c$ are complex numbers. All very good, apart from the fact that $\mathbb R$ is a subset of $\mathbb C$ so apparently the starting values can be real too. But wouldn't that lead to all points falling on the real line, and leading to a plain old line instead of the Mandelbrot Set that we know? Although I couldn't find any references(hence this question) wouldn't simply defining the initial values to be non-real be a little... arbitrary? I think I might have a wrong idea of what equation actually means then.

  • $\begingroup$ when you find values of an equation in loop starting from initial point, which means that you go on putting the previous value again into the equation and so on in loop. The plot gives you mandelbrot set. Thats the rough idea i have $\endgroup$ – ketan Mar 8 '14 at 18:13
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    $\begingroup$ @ketan yes I understood that. That would give us a series of $z_i$ for different starting points, presumably all have something fundamentally similar when plotted on the complex plane. My question is how exactly are they plotted? $\endgroup$ – Guy Mar 8 '14 at 18:18
  • $\begingroup$ with the help of computers..most probably $\endgroup$ – ketan Mar 8 '14 at 18:20
  • $\begingroup$ you would be surprised to know that, with the study of fractals it was found that there could be fractional dimensions too.like we have 3 dimensions, could there be 2.5? or like that $\endgroup$ – ketan Mar 8 '14 at 18:25
  • $\begingroup$ @ketan Are you talking about this? For instance I know that the Mandelbrot set has $\log_3(4)$ "dimensions" whatever that means $\endgroup$ – Guy Mar 8 '14 at 18:35

First, a distinction should be made: a fractal is one thing, and certain methods for constructing particular fractals are another.

Loosely, a fractal can be described as an object which is self similar at different scales, that is, "zooming in" repeatedly leads to the same curve. An interesting property of fractals which is sometimes used to define them, is that one can assign a non integer dimension to them. For instance, a smooth curve has dimension 1, but a Koch snowflake is in a certain sense closer to being a two dimensional object, and we can assign it a non integer dimension of $\sim1.26$. Intuitively, A Sierpinski carpet is even closer to a 2D object, and indeed we assign it a higher fractal dimension, of $\sim1.89$.

As to Mandelbrot's famous set, the idea is as follows: To check whether or not a (complex valued) point $c$ is in the set, start with $z_0=0$, and iterate. When the series stays bound, $c$ is in the set. When the series diverges, $c$ is not in the set. (try $c=-1,0,1$ for yourself, and see what you get). For instance, is the point $i$, i.e. $(0,1)$ in the Mandelbrot set? $$z_1 = 0^2 + i = i,\quad z_2 = i^2 + i = i - 1,\quad z_3 = (i-1)^2+i= -i$$ $$z_4 = (-i)^2+i = i-1$$ Thus, the point $i$ leads to a bound repeated loop, and is therefore in the Mandelbrot set (i.e. the black area in most drawings).

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    $\begingroup$ Just to add to the answer, the Mandelbrot set is usually colored by counting how many iterations it took to converge. To render a Mandelbrot, we assign each pixel on the screen a complex number, and do the procedure explained in the answer. If it diverges, paint it black; else, pick a color based on how many iterations it took to converge. That's it. $\endgroup$ – NothingsImpossible Mar 9 '14 at 2:23
  • $\begingroup$ ah that makes a lot of sense. and sounds very cool. thanks :) $\endgroup$ – Guy Mar 9 '14 at 7:22
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    $\begingroup$ @NothingsImpossible Your comment is correct, except we normally colour the region outside the Mandelbrot set, leaving the set itself black. I.e. If it does not diverge, paint it black; else pick a colour based upon how many iterations to took to diverge (normally $|z^2|>4$). $\endgroup$ – Mark Hurd Mar 11 '14 at 1:46
  • $\begingroup$ @MarkHurd You are right, I accidentally the conditions :p $\endgroup$ – NothingsImpossible Mar 11 '14 at 10:13
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    $\begingroup$ To clarify, the color/grey scale of the Mandelbrot is technically representative of the conditional rate of convergence. This means that we iterate a point by the Mandelbrot Function N times and assume that any point reaching N iterations without diverging converges. Points/pixels are colored/detailed based on how many iterations (up to N) it took for them to definitively diverge. Definitive divergence occurs when the (complex) output value of the Mandelbrot Function has a distance greater than 2 from the origin on the complex plane. @NothingsImpossible $\endgroup$ – Seth Taddiken Jul 16 '17 at 6:42

A fractal is perhaps best defined as a set that is complex at an arbitrarily small scale. Whereas a smooth curve looks like a line if you zoom in close enough, a fractal never looks like anything so simple no matter how far you zoom in. See, for instance, this video.

As a special case, for many fractals, zooming in on a small region (and clipping it) will produce a set identical to the original fractal, and this holds true no matter how far you zoom in. Such fractals are called self-similar.

Concerning the definition of the Mandelbrot set: For functional notation, let $f^n(z)$ denote $$\underbrace{f \circ f \circ \cdots \circ f}_{n\text{ times}}$$ For a fixed point $c \in \mathbb C$, let $f_c$ be the function $f_c(z) = z^2 + c$. Then $c$ is a member of the Mandelbrot set if and only if $\lvert f_c^n(0)\rvert\not\to \infty$ as $n \to \infty$. If $c$ happens to be a real number, then it is true that $f^n_c(0)$ is real for all values of $n$. There are, roughly, two reasons to bring complex numbers into the picture:

  1. You care about what happens to complex numbers.
  2. You want to get a pretty picture. (In less glib language, you find the Mandelbrot set intrinsically interesting, even if it is considered only as a subset of $\mathbb R^2$, and complex numbers are the easiest way to define it.)

Update: Certain geometric figures are differentiable (i.e., $C^1$) and yet seem to be complex at arbitrarily small scales: see the bottom of the page in the Hevea project. Should these be considered fractals? If one defines fractals based on the fractal dimension, then the answer is "no." But I'm inclined to think this highlights a deficiency in the fractal dimension-based definition. The authors of the Hevea project call these objects "$C^1$ fractals."

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    $\begingroup$ "Whereas a smooth curve looks like a line if you zoom in close enough, a fractal never looks like anything so simple no matter how far you zoom in." +1 for that. It was so obvious from the statement that fractals are self-similar yet I don't know why I overlooked it. $\endgroup$ – Guy Mar 9 '14 at 7:39

Mathematically, the term is "fractal set". A fractal set is a set with a non-integer Hausdorff dimension, which is a generalization of what you normally think of "dimension" (e.g. the Hausdorff dimension of a line is 1; of a plane is 2).

The most simple example of a fractal set is the Cantor set:

Construction of the Cantor set

The image represents the construction of the set and the set is defined in the limit as this construction goes to infinity (i.e. as it goes downwards). It is possible to compute the dimension of this set, which is $\log(2)/\log(3)\notin \mathbb{N}$.

Any other way to define a fractal is ambiguous and prone to magicality. Given the amount of subtleties in this topic, sticking to some mathematical rigor helps.

Relation with chaos

The equation you wrote, more generally written as $x_{n+1}=F(x_n)$, is a dynamical system in discrete time (also called a map).

The reason why they normally appear related with fractal sets is that chaotic dynamical systems generally have fractal sets. Skipping a full undergrad book, the most notable relation between fractals and chaotic systems is that the dimension of the attractor is related with the Lyapunov exponent of the system, i.e. a topology property (dimension) is related with a dynamical property (Lyapunov exponent).

To better illustrate this point, consider the following chaotic map:

$$x_{n+1} = 3 x_n\ \ for\ \ 0 < x_n < 1/2;\ \ \ \ \ \ x_{n+1} = 3 (1 - x_n)\ \ for\ \ 1/2 < x_n < 1$$

On each iteration of the map there are some points that stay inside the interval [0,1], and some points that leave the interval.

The points that stay up to time $n^*$ are exactly the construction of Cantor set in the $n^*$ iteration. Those that never leave are the Cantor set.

This and much more can be found in the fascinating (and old) topic of Chaos... ^_^

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    $\begingroup$ Your definition of a fractal is almost correct, except you can have a fractal curve of Hausdorff dimension, say, 2 which is an integer. The correct definition is that Hausdorff dimension is strictly larger than topological dimension. $\endgroup$ – Moishe Kohan Mar 9 '14 at 2:28
  • $\begingroup$ @studiosus, ok, I was not aware of it. Can you provide a reference of that please? $\endgroup$ – Jorge Leitao Mar 9 '14 at 7:36
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    $\begingroup$ I agree that this definition is the one to use if you want to prove a theorem that "such-and-such is (or is not) a fractal set." But I don't think it's a good definition for explanations of "fractals," which are more general and have no formal definition. For instance, the Mandelbrot set is (informally, for the noun "fractal" cannot be used otherwise) a fractal, but not a fractal set; its boundary is a fractal set. $\endgroup$ – Charles Staats Mar 9 '14 at 12:56
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    $\begingroup$ @CharlesStaats, I see your point, but notice the question is "What exactly are fractals", which includes the sentence "I decided to get a deeper more mathematical knowledge of this, but sadly wikipedia hasn't been of much help". I myself am from Physics, and I know that there are situations for "intuitive ideas" and other for more formalism; I believe a proper answer to this particular question deserves the latter. $\endgroup$ – Jorge Leitao Mar 9 '14 at 16:34
  • $\begingroup$ @J.C.Leitão A generalized Cantor set can be constructed like above but have something arbitraty rather than 1/3 removed each step. My example would involve $C_2$, where you remove the middle half on each step: $$\scriptstyle..\ \ \ \ \ ..\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..\ \ \ \ \ ..\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..\ \ \ \ \ ..\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..\ \ \ \ \ ..$$ Fractal dimension of $C_2\times C_2$ is 1 (topological 0). $\endgroup$ – Heimdall Jun 14 '16 at 22:58

Where does $z_+=z^2+c$ come from?

Actually, this story began with the study of Newtons method. If you apply it to a polynomial equation, you get an iteration of a rational function.

From that the topic was generalized to arbitrary rational functions with no connection to Newton's method and then specialized to the "easy" cases.


The first being linear functions. $z_+=ax+b$. From the solution theory of linear recurrences on knows that one can multiply by $(a-1)$ and distribute $b$ to obtain $$w_+=(a-1)z_++b=a((a-1)z+b)=aw$$ to get a normal form and to know that for $|a|<1$ it is a contraction to zero and for $|a|>1$ it diverges to infinity.


Next simple are quadratic iterations $z_+=az^2+bz+c$. Again, shifting and scaling of the sequence will not change the quality of the picture, so one may first try to reduce the number of free coefficients. Multiplying by $a$ gives $$az_+=(az)^2+b(az)+ac,$$ so wlog. one may set $a=1$. Now complete the square, $$ az_++\tfrac b2=\left(az+\tfrac b2\right)^2+\tfrac b2-\tfrac{b^2}4+ac $$ So set $w=az+\tfrac b2$ and $\tilde c=\tfrac b2-\tfrac{b^2}4+ac$, and the reduced form of the quadratic iteration is $$ w_+=w^2+\tilde c, $$ the iteration of the Mandelbrot fractals.

There is another normal form of this iteration, the Feigenbaum iteration, that is usually only considered on the real line, more specifically on the interval $[0,2]$, with a parameter $λ\in[0,4]$, \begin{align} x_+=λ x(1-x) &\iff (-λx_+)=(-λx)^2+λ(-λx) \\ &\iff (-λx_++\tfrac λ2)=(-λx+\tfrac λ2)^2+\tfrac14-\tfrac{(λ-1)^2}4 \end{align} so it covers the real line in the Mandelbrot diagram from $-\infty$ to $\tfrac14$, with the $λ\in[0,4]$ covering the range starting from $-2$, the point of the antenna of the Mandelbrot set.


One can now continue with cubic iterations. There one can again normalize the leading coefficient to $1$ and the quadratic to $0$, so the reduced form is $$ z_+=z^3+sz+t $$ but now we have 4 real degrees of freedom and a symmetry, in that changing $z$ to $-z$ results in the equally reduced iteration $$ z_+=z^3+sz-t $$

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  • $\begingroup$ I remember 20-odd years ago reading about 4-dimensional mandelbrot set within $\Bbb C\times\Bbb C$, defined as $(z,c)$ being an element if $w_0=z$, $w_{n+1}=w_n^2+c$ doesn't escape to infinity. (So you don't necessarily start the sequence with 0.) Intersect it with the plane $z=0$ and you get normal Mandelbrot set. Alternatively, intersect it with the plane $c=\rm constant$ and you'll get various Julia sets. It also mentioned intersecting with other planes, but I don't remember much about those. $\endgroup$ – Heimdall Jun 15 '16 at 0:39

This is a good documentary about fractals; and here a TED presentation from the father of fractals - Benoit Mandelbrot.

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    $\begingroup$ It is a little sad that right now I am on a terribly slow connection, but I am bookmarking it. I'll see it as soon as possible. :) $\endgroup$ – Guy Mar 8 '14 at 18:14
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    $\begingroup$ Just a little suggestion though, do you know how to use Markup to embed the links like this? Looks more readable I think $\endgroup$ – Guy Mar 8 '14 at 18:17
  • $\begingroup$ Here is the link. $\endgroup$ – Guy Mar 8 '14 at 18:19

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