Cynthia Lanius

Fractal Properties

Fractional Dimension

 
A point has no dimensions - no length, no width, no height.
That dot is obviously way too big to really represent a point. But we'll live with it, if we all just agree what a point really is.
 
A line has one dimension - length. It has no width and no height, but infinite length.

Again, this model of a line is really not very good, but until we learn how to draw a line with 0 width and infinite length, it'll have to do.
 
A plane has two dimensions - length and width, no depth.

It's an absolutely flat tabletop extending out both ways to infinity.

Space, a huge empty box, has three dimensions, length, width, and depth, extending to infinity in all three directions.
 
Obviously this isn't a good representation of 3-D. Besides its size, it's just a hexagon drawn to fool you into thinking it's a box.

Fractals can have fractional dimension. A fractal might have dimension of 1.6 or 2.4. How could that be? Let's investigate.
 

Again this isn't a great picture of a fractal. It's really just an approximation of one. Fractals really are formed by infinitely many steps, not just the three of this one. So we have to remember that there are infinitely many smaller and smaller triangles inside the real fractal, and infinitely many holes (the black triangles) at the same time..

In order to see how fractals could have dimension of a fraction, let's see what we mean by dimension in general. Take a self-similar figure like a line segment, and double its length. Doubling the length gives two copies of the original segment.

 

Take another self-similar figure, this time a square 1 unit by 1 unit. Now multiply the length and width by 2. How many copies of the original size square do you get? Doubling the sides gives four copies.

Take a 1 by 1 by 1 cube and double its length, width, and height. How many copies of the original size cube do you get? Doubling the side gives eight copies.
Let's organize our information into a table.

 Figure   Dimension   No. of Copies 
Line segment
1
2 = 2 1
Square
2
4 = 2 2
Cube
3
8 = 2 3

Do you see a pattern? It appears that the dimension is the exponent - and it is! So if we know the number of copies, we write it as a power of 2 and the exponent will be the dimension.

Let's add that as a row to the table.

 Figure   Dimension   No. of Copies 
Line segment
1
2 = 2 1
Square
2
4 = 2 2
Cube
3
8 = 2 3
 Any Self-Similar Figure 
d
n = 2 d

Now let's use this to figure out the dimension of a fractal called Sierpinski Triangle.
  

Start with a Sierpinski triangle of 1-inch sides. Double the length of the sides. Now how many copies of the original triangle do you have? Remember that the black triangles are not a part of the Sierpinski triangle

Doubling the sides gives us three copies, so 3 = 2 d , where d = the dimension.

But wait, 2 = 2 1 , and 4 = 2 2 , so what number could this be? It has to be somewhere between 1 and 2, right? Let's add this to our table.

 Figure   Dimension   No. of Copies 
Line segment
1
2 = 2 1
Sierpinsi's Triangle
?
3 = 2 ?
Square
2
4 = 2 2
Cube
3
8 = 2 3
 Any Self-Similar Figure 
d
n = 2 d

So the dimension of Sierpinski's Triangle is between 1 and 2. Do you think you could find a better answer? Use a calculator with an exponent key (the key usually looks like this ^ ). Use 2 as a base and experiment with different exponents between 1 and 2 to see how close you can come. For example, try 1.1. Type 2^1.1 and you get 2.143547. I'll bet you can get closer to 3 than that. Try 2^1.2 and you get 2.2974. That's closer to 3, but you can do better.

Okay, I got you started; now find the exponent that gets you closest to 3, and that's its dimension.

That's how fractals can have fractional dimension.

Note 1: For those of you that know about logarithms: Yes, you could use logs to solve this, but remember, this is primarily designed for students in elementary and middle school who haven't studied them yet. If you have studied logarithms, then yes, use them to solve this equation.

Note 2: Is there anything special about us doubling the lengths here? Could you have tripled them and derived the formula as well? Why don't you investigate by trying it.

Copyright 1997-2007 Cynthia Lanius URL http://math.rice.edu/~lanius/fractals/dimpr.html