1. May 7, 2008

### Verblun

Unfortunately my level of maths does not allow me to fully understand fractals.
I am at a tenth grade level of mathematics and my understanding of fractals is minimal. Very minimal. After searching the net tirelessly I have realised that any information available to my level of understanding will simply tell me that a fractal is a geometric "shape." Anything more complicated than that is out of my league.
Please, all I need to know is WHAT IS A FRACTAL. Nothing complicated, just a basic explanation.
Thank You.

2. May 7, 2008

### HallsofIvy

Staff Emeritus
As simply as possible, a "fractal" is a geometric figure that has fractional (that's where the name "fractal" comes from) dimension, where the dimension can be defined in a number of ways. Most commonly used is the "Hausdorff" dimension.

For example, you know, I am sure, the area of a 2 dimensional figure depends on a length squared, the volume of a 3 dimensional figure depends on a length cubed, etc. In general the "size" of a d dimensional figure depends on a length to the d power.

Now, consider the "Cantor ternary set": start with the unit line segment, [0, 1]. Remove the middel third (1/2, 2/3) (leaving the end points). Now remove the middle third of the two remaining intervals. remove the middle third of the (4) remaining intervals. The "Cantor ternary set" is what you get by continuing to remove "middle thirds" infinitely. Notice that at each step we take out the middle third so have twice as many intervals, each 1/3 as long as the intervals at the previous step. That is, we start with one interval of length 1, remove its middle third so we have 2 intervals of length 1/3 (total length 2/3), remove the middle third of each of those so we have 4 intevals, each of length 1/9 (total length 4/9= (2/3)^2), etc. In the limit the total length is the limit (2/3)^n= 0. But there are many good reasons, the fact that the Cantor set is uncountable is one of them, why we would not want to consider the Cantor set as have "length" 1. If we treat the Cantor set as, instead, a set of Hausdorf dimension d, we need to include the d power of a length so we would be taking the limit of (2/3^d)^n. That will have a finite, non-zero, limit if and only if 2/3^d= 1 or 2= 3^d. Taking the logarithm of both sides log(2)= dlog(3) so d= log(2)/log(3) or about 0.6309. The Cantor ternary set has fractional Hausdorf dimension 0.6309 and so is a "fractal" set.