One can calculate the dimension of a fractal in several ways; one of them is the so-called Hausdorff(-Besicovitch) dimension. From my understanding this is quite a complicated method, but I can’t seem to find out how it works in practice.
I already know the formula d=(log k)/(log n), but I suspect that this does not calculate the Hausdorff dimension. Is this the case and if so, what is the name of the dimension that is calculated with it?
Please do not spare me the mathematical details and elaborations. Thanks for your reply!
Answer
Dear Jeroen,
I’m not a specialist in fractals myself, but since your question has been asked for a while, I’ll try to answer as best I can.
I think you do indeed calculate the Hausdorff dimension that way. The reasoning is as follows: if you double an n-dimensional cube in side, then the original cube goes 2n times in the new cube. The dimension is then n= 2log(2n). This is true for any n-dimensional figure that is the union of a number of smaller copies of itself. This last property is one of the characteristic properties of fractals, and so this formula lends itself perfectly to calculate the dimension of a fractal. In this way one sometimes gets non-integer dimensions. For example, the two-dimensional case of the Sierpinski fractals: an equilateral triangle is divided into four equal triangles and you take the middle way. Do the same with the three remaining triangles, and so on. If you now take one of those remaining triangles, it will go exactly 3 times into the original triangle (because the middle triangle has been removed), but it has half the side. So the dimension is 2log 3=log 3/log 2. So the dimension is general alog b, where a is the factor of homothety to bijectively map part of the fractal onto the whole fractal, and b is the number of copies of this part contained in the whole fractal.
I hope this brings some clarity.
Regards,
Henry Van Maldeghem
Answered by
prof. Henry Van Maldeghem
Mathematics, geometry, algebra
http://www.ugent.be
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