Fractal Dimension

English: Sierpinski triangle, a fractal having...
English: Sierpinski triangle, a fractal having a Hausdorff dimension of ln3 / ln2 (which is approximately 1.58). (Photo credit: Wikipedia)

Today I sat in a lecture about Hausdorff dimension which is pretty technical and was very nice. However, what I am going to write is my own intuitive view on Hausdorff dimension. First, let us assume the fact that we have “n-dimensional volume” that are length for n=1, area for n=2 and well, volume, for n=3 and something that you can imagine for yourself for higher n. Now one characteristic of n-dimensional object is when you you dilate/magnify them by a factor, say K, then their new volume is K^n times the old volume. For example, take “n-dimensional cube” of volume 1 which are a line segment of length one for n=1, a square of sides one for n=2 and a cube of sides one for n=3. Note that when you dilate these “n-dim cube” by 2, their volume are 2 for n=1, 4=2 x 2 for n=2 and 8= 2 x 2 x2 for n=3. By this simple character, we can try to explain objects of non-integer dimension (regardless of whether they exist or not). For instance, if I want to say an object is a half-dimensional object, then when I dilate the object by a factor of 2 then I want the new volume to be 2^(1/2) of the old volume. One example for this object with non-integer dimension is Sierpinski triangle (the one on the picture). If you magnify the triangle by a factor of 2, then the “volume” of the triangle is 3 times the old volume. This suggests that the Hausdorff dimension of this triangle, is A where 2^A=3, that is A=ln3/ln2. This is of course not a proof, because the proof is somewhat much more complicated. But my point is that you can see this characterization is very intuitive.

The problem is, I wrote the above argument as if I do have the “volume” of this triangle. But as you may probably try to compute yourself, defining the “volume” of this triangle is not a trivial matter.  Hausdorff does this by introducing the so called Hausdorff measure, which is parametrized by dimension. And although the definition of this measure is not trivial at all, the usage of this measure to prove a certain object has a certain dimension is really intuitive. You say a dimension of an object is D when its “A-dimensional volume” is infinite when A<D and its “A-dim volume” is 0 when A>D. Why I say this very intuitive is because when we compare to the discrete dimension again, a 2-dim object has infinite 1-dim volume (length) and 0 3-dim volume (volume). And my point is that with these measure and dimension being so intuitive, it will be easy to remember the definition and although it may hard to prove that a certain object has a certain dimension, computing that dimension is a much easier task.


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