Fractals are not selfsimilar
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by Super User, 4 years ago
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One technical note: It's possible to have fractals with an integer dimension. The example to have in mind is some *very* rough curve, which just so happens to achieve roughness level exactly 2. Slightly rough might be around 1.1dimension; quite rough could be 1.5; but a very rough curve could get up to 2.0 (or more). A classic example of this is the boundary of the Mandelbrot set. The Sierpinski pyramid also has dimension 2 (try computing it!).
The proper definition of a fractal, at least as Mandelbrot wrote it, is a shape whose "Hausdorff dimension" is greater than its "topological dimension". Hausdorff dimension is similar to the boxcounting one I showed in this video, in some sense counting using balls instead of boxes, and it coincides with boxcounting dimension in many cases. But it's more general, at the cost of being a bit harder to describe.
Topological dimension is something that's always an integer, wherein (loosely speaking) curveish things are 1dimensional, surfaceish things are twodimensional, etc. For example, a Koch Curve has topological dimension 1, and Hausdorff dimension 1.262. A rough surfaces might have topological dimension 2, but fractal dimension 2.3. And if a curve with topological dimension 1 has a Hausdorff dimension that *happens* to be exactly 2, or 3, or 4, etc., it would be considered a fractal, even though it's fractal dimension is an integer. Pick a random fractal from a hat, though, and it will almost certainly have a noninteger dimension.
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4 years ago