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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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This note was uploaded on 02/10/2012 for the course MATH 2112 taught by Professor Carter during the Fall '09 term at University of Central Florida.
 Fall '09
 Carter
 Differential Equations, Equations, Counting

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