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Stage | Length | Area |
0 | 4 | 1 |
1 | $4\times2=8$ | 1 |
2 | $4\times2^2=16$ | 1 |
I have made the following observations:
The length of the fractal at each stage is twice its length at
the preceding stage.
The area of the fractal is constant because the two colored
areas are equal (the blue one is added, the yellow one
subtracted)
For the $n^{th}$ stage the length of the curve is $4 \times 2^n =2^{n+2}$ and the area is $1$ square unit |
The length of the fractal is $$\lim_{n\to \infty} 2^{n+2}$$ and hence the fractal curve has infinite length. This is an example of a curve of infinite length surrounding a finite area.
What is the dimension of the fractal? Solution: $d = 1.5$
Explanation: the dimension $d$ is given by the formula $n = m^d$.
$n = 8$ because each segment is broken up into 8 self-similar segments.
$m = 4$ because each of the $8$ new segments is $1/4$ the length of the original segment, and thus must be multiplied by $4$ to be the length of the original segment.
So: $8 = 4^d$ and hence $d = log_4 8 = 1.5$.
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.