# Calculate/Estimate the fractal dimention of the logistic map

This is the logistic map:
. It is a fractal, as some might know here.
It has a Hausdorff fractal dimension of 0.538.
Is it possible to calculate/measure its fractal dimension using the box counting method?
A "hand waving calculation" is good enough.

Update: I understand there is another way to calculate the logistic map using the Kaplan-Yorke Conjecture. Can anyone explain that and how it can help calculating the fractal dimension of the logistic map?

Update2: Seems like the way to go around this is not Kaplan-Yorke Conjecture (which is a unproven Conjecture still), but use the correlation dimension. There is a paper with the solution here, hope I'll know more as I read it.

• @KaziarafatAhmed Surely you have heard about chaos theory? Well, one great importance to physics is that fractals are very closely related to chaos in many cases. So this question is very much on topic. – Wouter May 25 '13 at 17:23
• @KaziarafatAhmed Fractals are important in physics. Look up the Hofstadter Butterfly. First fractal found in physics :) I do however motion a move of this post to mathematics. Cheers. – John M May 26 '13 at 0:40
• @KaziarafatAhmed fractal dimensions are used in some intermittency models in turbulence theory, where the Hausdorff dimensions appear too. – Dilaton May 26 '13 at 0:47
• I just spoke to my physics professor today. Apparently you can do this using the correlation dimension. I am reading about it and hope to post an answer. There is a paper on how to do this here – GuySoft May 26 '13 at 21:54
• @KaziarafatAhmed: enwp.org/Causal_dynamical_triangulation – Abhimanyu Pallavi Sudhir Sep 6 '13 at 1:50

Yes you can estimate the dimension by box counting. It is not quite hand waving but the idea has an advantage to be intuitive.

1) You consider the logistic map attractor like an analogy to a Cantor set whose dimension you can compute by box counting.

2) You remember that when the chaotic bands double, their sizes scale like $1/a$ and $1/a^2$ where $a$ is the second Feigenbaum constant, $a \approx 2.5029$. So the procedure looks like the Cantor set production because at each doubling you "remove" a part of the previous band. The difference being that the new smaller 2 parts have not the same size like in the Cantor set.

3) You suppose that at the Nth doubling you need $2^n$ boxes of size $R_n$ to cover the bands. Then at the $N+1$ stage you will need $2^{n+1}$ boxes of average size $$R_{n+1} = \frac{R_n}{2}\left(\frac{1}{a} + \frac{1}{a²}\right).$$ The hand waving part is the arithmetical average because there is no rigorous reason to use it but one sees the idea of approximation.

4) The box counting dimension is then $$D_b = - \frac{\log 2}{\log(1/2(1/a + 1/a²))} = 0.544$$

You will admit that the hand waving was not so bad because one is not far from the much more rigorously derived value of $0.538$.

As a particular remark, the box counting is not very practical when the attractor is not strictly self similar like f.ex the Cantor set because the results depend then on the particularities of the covering method chosen

• The Feigenbaum constant is 4.669... Your calculations don't work. $$1/2 \cdot (1/a+1/a^2)=0.130...$$ This results in a value of $0.339...$ It's better to just use $$1/a$$ and get an estimate of $0.449$. – Zach466920 Aug 14 '15 at 1:42

As the number of bifurcations goes to infinity, we can make some various claims.

$$d_f={{\ln(N)} \over {\ln(s)}}$$ Where $d_f$ is the fractal dimension, N is the number of boxes, and s is the scale.

(Examine $N\sim s^{d_f}$ to get the above)

$$N=2^n$$ And $$s \sim {\delta_f}^n$$ Where $\delta_f$ is the Feigenbaum constant.

We obtain, in the limit...

$$d_f \sim { {\ln(2)} \over {\ln(\delta_f)}}=0.4498...$$

Which has about 16% error from the empirical value of 0.538...

• 0.538... is not empirical, its calculated. Its the Hausdorff dimension. This not even a good approximation when you consider box counting. – GuySoft Aug 16 '15 at 8:07
• @GuySoft odd, that you don't have an exact value then...either way, my approximation is better than the above. – Zach466920 Aug 16 '15 at 14:08
• @GuySoft btw the very link you give cites the dimension as approximate and calculated rather than exact... – Zach466920 Aug 16 '15 at 21:21
• Its an approximate to any degree of accuracy you want. You could do the calculation and get a finer result if you want. – GuySoft Aug 18 '15 at 10:32
• @GuySoft but you said it was exact. It's not, so there's a contradiction. Either way, the above answer gets a more accurate answer based on a mathematical mistake, so you should take that into consideration too. – Zach466920 Aug 18 '15 at 14:17