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Some Fractals and Fractal Dimensions

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The Cantor set: we take a line segment, and remove the middle third. For each remaining piece, we again remove the middle third, and continue indefinitely.

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To calculate the fractal / Hausdorff / capacity / box-counting dimension, we see how many boxes (circles) of diameter 1/r^n we need to cover the set (in this case, we will use r = 3, since it fits nicely).

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D = Lim(log(N r )/log(1/r)) = log(2) / log(3) r N r 1 1 1/3 2 1/3^2 2^2 1/3^3 2^3 1/3^n 2^n
The Koch snowflake: We start with an equilateral triangle. We duplicate the middle third of each side, forming a smaller equilateral triangle. We repeat the process.

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To calculate the fractal / Hausdorff / capacity / box-counting dimension, we again see how many boxes (circles) of diameter (again)1/3^n we need to cover the set.

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D = Lim(log(N r )/log(1/r)) = log(4) / log(3) r N r 1 3 1/3 3 * 4 1/3^2 3 * 4^2 1/3^3 3 * 4^3 1/3^n 3 * 4^n

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Unformatted text preview: The Sierpinski carpet: We start with a square. We remove the middle square with side one third. For each of the remaining squares of side one third, remove the central square. We repeat the process. D = Lim(log(N r )/log(1/r)) = log(8) / log(3) r N r 1 1 1/3 8 1/3^2 8^2 1/3^3 8^3 1/3^n 8^n The Sierpinski gasket: we do a similar process with an equilateral triangle, removing a central triangle. (Note: we could also do a similar thing taking cubes out of a larger cube -- the Sierpinski sponge -- but its hard to draw :-) D = Lim(log(N r )/log(1/r)) = log(3) / log(2) r N r 1 1 1/2 3 1/2^2 3^2 1/2^3 3^3 1/2^n 3^n We can also remove other shapes. D = Lim(log(N r )/log(1/r)) = log(4) / log(4) = 1 r N r 1 1 1/4 4 1/4^2 4^2 1/4^3 4^3 1/4^n 4^n D = Lim(log(N r )/log(1/r)) = log(3) / log(3) = 1 r N r 1 1 1/3 3 1/3^2 3^2 1/3^3 3^3 1/3^n 3^n...
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