thanks to John N. for the idea
We want to (hopefully) understand, then explain, fractals.
>Why?
They're fascinating and, if we're lucky, we can see their use in financial theory.
We'll start with Gaston Maurice Julia,
born in 1893. In 1928 he published a paper descibing the iterations of a function and ...
>Huh?
Consider this:
 We start with some point in the plane, say (x,y) and a couple of fixed parameters "a" and "b".
 We then generate another point (x_{1}, y_{1}) via the formulas:
[a1] x_{1} = x^{2}  y^{2} + a
[b1] y_{1} = 2xy + b
 Then we continue with formulas [a1] and [b1], but using x_{1} and y_{1} to generate another point (x_{2}, y_{2}):
[a2] x_{2} = x_{1}^{2}  y_{1}^{2} + a
[b2] y_{2} = 2x_{1}y_{1} + b
 Then we continue with formulas [a2] and [b2], but using x_{2} and y_{2} to generate (x_{3}, y_{3}):
[a3] x_{3} = x_{2}^{2}  y_{2}^{2} + a
[b3] y_{3} = 2x_{2}y_{2} + b
 Then we continue with formulas [a3] and [b3], but using x_{3} and y_{3} ...
>Okay! I get it, but what's so good about that?
Well, this all started with some point (x,y) and, depending upon your choice of (x,y), the points defined by
the iterates, namely
(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) etc.
may:
[A] Remain close to the origin (inside some square x<2, y<2 for example)
[B] Grow without bound ... becoming infinite
The boundary between the points (x,y) that produce either [A] or [B] might look like Figure 1: a Julia Set
It's even more interesting to identify the points which remain finite (and paint them black) as well as those which head off to
infinity and even to colour the latter points according to how quickly they head off to infinity.
If you do that, you'd get something like Figure 2.
>What's "a" and "b"?
For Figure 2? It's a = 0.75, b = 0.

Figure 1
Figure 2

If, for a = 0.75, b = 0, we start at various points (x,y) shown in red,
we can follow the first few iterations like so:
Note that, starting with (x,y) = (0.5, 0.5), the iterations head off to infinity ... and we'd colour the point
(0.5, 0.5) according to how rapidly it moved off. That is, how many iterations it took to leave some
predefined neighbourhood of the origin. (For example, how rapidly if moved out of a circle of radius 2 units, centred on the origin.)
>I assume there are other neat pictures if you change a and b?
Yes. Take a look at some pretty pictures
here
I should also point out that our prescription for iterating, namely 1, 2, 3, 4 etc. above, can
be simplified by writing the iteration formula as:
z_{n+1} = z_{n+1}^{2} + c
where z = x+iy and c = a+ib and i = SQRT(1) is the imaginary unit.
>Huh?
Don't worry about it.
However, whether or not the starting point (x,y) gives a sequence that stays
close to the origin or heads off to infinity will (as you might imagine!) depend upon (a,b).
Indeed, this time we'll keep the initial point (x,y) fixed at (0,0) and play with different values of (a,b).
Those points (a,b) which give rise to a bounded sequence we'll colour black. If the points become infinite,
we'll colour them according to how rapidly they move away from the origin, as in Figure 1A.
>A "bounded sequence"?
Yes. If, for some number B, all iterates lie within some circle of radius B, then they're "bounded".
Recognize the boundary of the black region in Figure 1A? (The black region is called the Mandelbrot Set.)
It's shown in Figure 1!

Figure 1A

>Mandelbrot?
Yes. Benoit Mandelbrot,
born in 1924 in Poland, is the Father of Fractals.
Okay, let's do something different.
>Et cetera!
Yes. After three more of these rituals we get 3072 sides and it looks like Figure 3B.
>You want me to count the number of sides?
At step n, the number of sides is 3(4^{n}) so the number of sides are 3, then 3(4), then 3(4^{2}), etc.
However, notice that at each step we replace a single line segment by four line segments, each 1/3 the size.
The original line segments had length 1, then (1/3), then (1/3)^{2}, etc.
>So the TOTAL length is ... uh ...
It's (number of segments)x(length of each segment) = 3(4^{n}) x (1/3)^{n} = 3(4/3)^{n}, after n steps.
It's called the Koch snowflake
(you can play with it here),
named after
Helge von Koch, born 1870 in Sweden.
After an infinite number of such rituals there are infinitely many "sides" and the length of the
"snowflake" boundary is infinite yet the entire snowflake will fit in your pocket and it encloses a finite area and ...
>So?
Don't interrupt!
And if you examine each tiny portion of the snowflake, you'd see the same diddle design
... no matter how much you amplified the snowflake.
That's called self similarity and is a feature common to Fractals.
Look again at the Mandelbrot Set
There's a small red circle. Click on the picture to enlarge this small circle.
>Very funny. You just displayed the same picture!
But you get the idea of self similarity, eh?
In fact, some of those wee bulbs sticking out, in the Mandelbrot Set, look like
this
>Wow! But when do we get to financial stuff?
Patience ...


Here we start with a square, then we remove the middle ninth, leaving eight squares.
Then for each of the remaining eight squares we remove ...
>Remove the middle ninth, etc. etc., right?
Right.
>But I don't see eight remaining squares, do you?
Like so 

Figure 3c

You can see several iterations
here.
It's a fractal, named after
Waclaw Sierpinski
(18821969).
>How do you know it's a fractal?
Well, its HausdorffBesicovitch dimension exceeds its topological dimension.
>Aw ... c'mon.
Okay, let's talk about dimension.
>Like 47 metres or 59 acres or ...?
Not exactly. Consider this:
 A curve is a 1dimensional object because any point on the curve can be
identified by giving just 1 number:
the distance along the curve from some origin.
 The surface of a sphere is a 2dimensional object because any point on the sphere
can be identified by giving just 2 numbers:
a latitude and a longitude.
 The interior of a cube is a 3dimensional object because any point
can be identified by giving just 3 numbers:
x, y and z.
>We're talking length, area and volume, right?
Well, yes, we associate with 1dimensional objects a length and
2dimensional objects an area ...and so on.
But what we do now is to prescribe a somewhat different way of defining "dimension" so that we can
calculate the dimension of, say, the Koch snowflake.
>But it's a curve so it has dimension 1.
Not exactly. Pay attention. We're going to talk about objects that have self similarity.
Objects that are made up of a bunch of smaller objects of the same shape ... like our snowflake.

Figure 4

Consider some object, like a cube (which can be considered to consist of smaller cubes each of which consists of smaller cubes etc.)
Figure 5

 Suppose its measure is S. (For a line, that'd be its length, for a square its area, for our cube its the volume.)
 For our cube, if its side length is a, its volume is S = a^{3}.
 We then subdivide the cube into smaller cubes whose side length is a/n.
 The measure of each of the smaller cubeobjects is then (a/n)^{3}.
(If n = 3, we'd divide our cube into smaller cubes whose side length is 1/3 of the original as in Figure 5.)
 If there are N of these smaller cubes that make up the original cube, then their total measure is
N (a/n)^{3}.
 But the sum of these must be equal to the original, so
S = N (a/n)^{3} = a^{3}.

We then have, from N (a/n)^{3} = a^{3},
that N = n^{3} or 3 = log(N)/log(n) = the dimension of our cube.
>But we already knew that! A cube is 3dimensional!
Sure, but it's a somewhat different determination of dimension, eh?
>You call it "measure". Why not just plain "volume"?
We'll use this procedure to calculate the dimension of objects that aren't as simple as a cube.
The general rule for selfsimilar objects is:
fractal dimension = log(number of selfsimilar pieces) / log(reduction factor)

>For a square it'd be ... uh ...
We subdivide a square into smaller squares with side length 1/10, so (reduction factor) = 10.
Each square would become 10x10 = 100 of these smaller squares, so (number of selfsimilar pieces) = 100.
>So the dimension of a square is log(100)/log(10) = log(10^{2})/log(10) = 2log(10)/log(10) = 2, right?


Yes, and what's the dimension of the Koch snowflake?
>Okay, each step reduces the side length by 1/3 so (reduction factor) = 3.
Yes, and each side would become 4 sides, so the (number of selfsimilar pieces) = 4.
>So the dimension of the snowflake is log(4)/log(3) ... uh, logarithms to what base?
It doesn't matter.
To base 10 it'd be 0.60206/0.47712 = 1.26.
To base e it'd be 1.38629/1.09861 = 1.26.
>And since it's a curve, then ... hey! A dimension of 1.26? Is that possible?
It's not your ordinary, garden variety dimension, eh? It's called the (are you ready for this?) the
HausdorffBesicovitch
dimension.
It started as an ordinary, garden variety curve but became an infinitely detailed object.
This infinite detail actually increased its dimension a schnitzel ... from 1 to 1.26.
>This is weird.
Isn't it, especially when we're so familiar with dimensions of "1" or "2" or ...
>They're the "ordinary" dimensions, eh?
Actually, the basic, gardenvariety dimension is called the "topological" dimension
whereas the "fractal" or "Hausdorff" dimension may (or may not) be equal the topological dimension.
In fact, in the words of Mandelbrot:
A fractal is by definition a set for which the HausdorffBesicovitch dimension strictly exceeds
the topological dimension.
Let's do the following:
 Start with a line segment of length "1".
 Cut out the middle third. (What's left has length 2/3.)
 For each of the pieces left, cut out the middle third.
 For each of the pieces left, cut out the middle third.
 For each of the pieces left, cut out ...


>Okay, I get it. So?
So, at each step, we increase the number of pieces: 1 then 2 then 4 then 8 then ...
>That's 2^{n}?
Yes, but the pieces get smaller. They're 1/3 of the earlier length.
At each step we multiply the number pieces by 2.
At each step the reduction factor is 3.
The Hausdorff dimension ...
>Let me do it!
(reduction factor) = 3
(number of selfsimilar pieces) = 2.
fractal dimension = log(2)/log(3)
Very good ... and log(2)/log(3) = 0.631 which is greater than the topological dimension which is 0 so ...
>Huh?
How much of the original line has been cut out?
We'll add all the segment lengths that have been removed.
That's the sum of terms like: (number of segments removed)(length of each removed segment)
and that's: 1/3 + 2(1/3^{2}) + 2^{2}(1/3^{2}) + 2^{3}(1/3^{3}) + ...
or (1/3)[1 + 2/3 + (2/3)^{2} + (2/3)^{3} + ...]
and that infinite series adds up to (1/3)/(1  2/3) = 1.
>That's the entire original line!
Yes, so what's left is ... uh, nothing.
>I'd say the topological dimension is zero.
Yes, and since the Hausdorff dimension, namely 0.631,
is greater than 0, the Cantor Set is a fractal.
>The what?
Named after Georg Cantor (18451918).
Okay, now compute the Hausdorff dimension of the Sierpinski carpet.
>Gimme a hint.
It's 1.893.
When we considered the Koch snowflake, we had a "replacement rule", namely: Replace each line segment by the
Now we'll consider another curve ("one of the most remarkable ...", says Hausdorff), invented by
Giuseppe Peano, born
in 1858 in Italy.
We start with a single line segment, then replace it with
Although it looks like a line replaced by something involving a rectangle, it's actually replacing
one line segment by nine line segments
>I'll calculate its dimension, okay?
We replace one line by nine, so (number of selfsimilar pieces) = 9.
It looks like each of these nine is 1/3 of the original line segment, so (reduction factor) = 3.
The dimension of peano's curve is then log(9)/log(3) = log(3^{2})/log(3) = 2.
Does that surprise you?
>Nothing surprises me any more.
But that's the dimension of a plane or a square ... a 2dimensional object.
>So?
So, in 1890, Peano published this curve. It's a space filling curve. After an infinite number of "replacements"
Peano's curve actually completely fills a square! (You can see a picture
here.)
>Which explains why its dimension is 2, eh?
You got it!
Indeed, before Peano, it wasn't thought possible to define a "curve" that entirely filled a
square! Now everybuddy has their own version of spacefilling curves
(See, for example,
Hilbert's version.)
Now the interesting / exciting thing about fractals is that they can be used to model real live objects, not
just squares and curves and such. For example, the Koch snowflake may look like a shoreline, where, when we
look from 1000 metres then 10 meters then .1 meters ... the shoreline looks similar. Indeed, one can use
fractals to generate things that are quite recognizable, like clouds and
trees
(where each branch and twig looks like a smaller version of the tree).
>And financial stuff? What about financial stuff?
Patience. Aren't we having fun?
>Well ... yes, as a matter of fact. I'm enjoying this!
Me too! Have you seen
these, especially
the one that looks like a fern?
>That's nice, but what about financial stuff?
Take a look at the
Mandelbrot
article which appeared in the scientific American.
for Part II
