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Imagine breaking a cube into 64 identical small cubes. The length of the edge of the big cube is 4 times the length of the edge of a small cube and we say that the magnification factor is 4. As 64=4^3 we see that the number of small self similar pieces is equal to the magnification factor cubed. The number 3 is called the dimension of the cube.
If you break a line of length 1 into self similar bits of length {1\over m} there are m^1 bits and the dimension of the line is 1.
If you break up a square of side 1 into self similar squares with edge {1\over m} then there are m^2 smaller squares and the dimension is 2.
If you break up a cube of side 1 into self similar cubes with edge {1\over m} then there are m^3 smaller cubes and the dimension is 3.
In each case we say the magnification factor is m meaning that we have to scale the lengths by a factor of m to produce the original shape. The formula for dimension is: n = m^d where n is the number of self similar bits, and d is the dimension.
We can generalise what we know about 1, 2 and 3 dimensions to the non integer dimensions of fractals using the formula (where d is the dimension): \text{number of self similar bits} = \text{(magnification factor})^d.
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This is the program that draws the squareflake.
to flake :side :stage
repeat 4[line :side :stage rt 90]
end
to line :side :stage
if :stage = 0 [fd :side stop]
line :side /4 :stage - 1 lt 90
line :side /4 :stage - 1 rt 90
line :side /4 :stage - 1 rt 90
repeat 2 [line :side /4 :stage - 1] lt 90
line :side /4 :stage - 1 lt 90
line :side /4 :stage - 1 rt 90
line :side /4 :stage - 1
end
What is the total area of the triangles remaining in the nth stage of constructing a Sierpinski Triangle? Work out the dimension of this fractal.
Make a poster using equilateral triangles with sides 27, 9, 3 and 1 units assembled as stage 3 of the Von Koch fractal. Investigate areas & lengths when you repeat a process infinitely often.
This article gives a proof of the uncountability of the Cantor set.