How to find a fractal with a predetermined Hausdorff dimension?

Question

For many patterns that display self-similarity, the Hausdorff dimension can be found. Sometimes the dimension is calculated and approximate - as is the case with the Feigenbaum attractor - but often its closed form in terms of known mathematical constants be obtained - for instance, the Hausdorff dimension of the Cantor set is $ \log_{3}(2)$.

I am interested in the inverse problem: suppose we consider a mathematical constant like $\pi^{-1}$ or $\gamma e$. Can we always find and describe a fractal whose Hausdorff dimension is equal to this preset number?

Here's how to ask a good question. Specifically, remember to show your own work on the problem. — jjagmath, May 22 '22 at 11:54
https://math.stackexchange.com/questions/1713514/is-there-a-general-metod-to-construct-a-fractal — Claude, May 22 '22 at 12:27

Klaas van Aarsen · Answer 1 · 2022-05-22T13:11:04.533

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We can construct a fractal similar to the Koch curve with arbitrary dimension $1<D<2$.

Consider starting with a line segment of length 1. We divide it into 2 line segments at a slight angle. Then replace each of those segments again with the opposite angle. And so on.

If we pick the relative length of the line segments such that after 2 iterations we get a equilateral triangle, we have the "normal" Koch curve with dimension $D=\log_3 4$. Otherwise we get a curve with an arbitrary dimension $1 <D <2$.

If we pick $f$ as the length fraction of the original line segment that we divide into 2 segments, then the length of the curve with respect to dimension $D$ after $n$ iterations is: $$L_n = 2^n (f^n)^D = (2f^D)^n$$ It means that the dimension $D$ of the curve is such that $$2f^D=1 \implies f=e^{-\ln 2/D}$$ Fill in $D=\gamma e$ to get the corresponding curve.

Similarly we can construct a Cantor set with $D=\pi^{-1}$ as @Claude pointed out.

edited May 22 '22 at 13:11

answered May 22 '22 at 12:57

Klaas van Aarsen

5,908

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And here's a picture - though, I don' think it really works for $\pi^{-1}$. As Claude pointed out in his comment, his answer to the duplicate question is more complete. – Mark McClure May 22 '22 at 13:07
Ah right @MarkMcClure, $\pi^{-1}<1$. I've removed it from the post. – Klaas van Aarsen May 22 '22 at 13:11
Thank you @KlaasvanAarsen and @Claude! Now the question remains: can a similar approach be formulated for fractals with Hausdorff dimension greater than or equal to two? – Max Muller May 22 '22 at 13:23
@MaxMuller, we can divide a plane into 4 sub planes in the same fashion and repeat. Or a cube in 8 sub cubes in hyperspace. And so on. – Klaas van Aarsen May 22 '22 at 13:30
@MarkMcClure, I've just realized that this Koch approach actually also works for $\pi^{-1}$. If $f<\frac 12$, then we get disconnected line segments resulting in a Cantor set. – Klaas van Aarsen May 22 '22 at 13:34
@MaxMuller If you read the answer that Claude points to in his comment, you'll find find that he addresses the higher dimensional case quite completely via the Cartesian product. His picture in 2D space is really just an illustration of a process that can be continued to any dimension. – Mark McClure May 22 '22 at 14:11
@MarkMcClure Ah I see, you're right – Max Muller May 22 '22 at 14:19
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Unlike the answer to the duplicate question, this shows that any dimension $d\geq1$ has a fractal realization that is (path-)connected. – Jacob Manaker Aug 05 '22 at 07:13

How to find a fractal with a predetermined Hausdorff dimension?

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