### Borderline fractals

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#### pauldelbrot #### Borderline fractals

« on: March 01, 2018, 02:31:47 PM »
Pursuant to recent Shoutbox activity, I thought up something pathological.

Consider this construction. We start with a line segment. At each step n = 0 ... ∞, for each line segment we:

a) Divide it into two equal halves.
b) Erect at the new vertex two perpendicular line segments whose lengths are equal to the lengths of those halves, times f(n).

If f(n) ≡ 1, we get a fairly familiar fractal. On the other hand, if f(n) = 0.5n, the segments should get so small so fast that the resulting shape may converge on Hausdorff dimension 1, boundary dimension 1, and a finite perimeter.  (If not, we just need a faster growing function for f. The TREE function would certainly suffice, in swatting a fly with a hydrogen bomb fashion.)

Is this thing a fractal? On the one hand, dimension is 1, perimeter is finite, etc. On the other hand, branch points will necessarily be dense along any segment of the end product. It will nowhere have a normal or be differentiable, and it will have many of the other pathological traits typical of fractal plane curves like the Sierpinski gasket. It also belongs to a continuously parametrized family that includes unquestionable fractals and undeniable non-fractals. (Consider f(n) = 0.5g(n), with g(n) = ns. If s = 0 we get a fractal similar to the earlier example, but if s = 1 we get our borderline thingy, and if s → ∞ we get a straight line segment with no branches, which is indubitably a non-fractal!)

So, what the heck is it?

This may be relevant to our friend the humble Mandelbrot set. Because there's a similar parametrization there. The Julia set of -2 is a straight line segment and that of 0 is a circle, both non-fractals. The rest are fractals ... or are there borderliners in there too lurking somewhere?

#### gerrit #### Re: Borderline fractals

« Reply #1 on: March 04, 2018, 08:25:23 PM »
Can't say I fully understand it.

He says that H-dim(J(c)) is "analytic" in c outside M-set, but this is a function from complex to reals. Maybe he means it's harmonic, so it has a complex extension which is analytic?

Other interesting result is if you continuously move from $$c_0 \rightarrow c$$, for example left direction on the real axis towards -2, H-dim is not continuous (it jumps to 2 whenever you hit a hyperbolic component) but you can find an (infinite) sequence $$c_0 c_1 \ldots c_n c$$ so that the H-dim changes "smoothly". I guess those points are Misiurewicz (which he mentions explicitly) and Feigenbaum points (which he does not mention).

Another interesting result is you can always find a (small) $$\epsilon$$ such that Hdim(-2+$$\epsilon$$) is as close to 1 as you wish.

None of this proves that -2 an 0 are the only points with H-dim of Julia set = 1, but that seems most likely.

#### claude #### Re: Borderline fractals

« Reply #2 on: March 04, 2018, 11:05:15 PM »
None of this proves that -2 an 0 are the only points with H-dim of Julia set = 1, but that seems most likely.

Falconer (Fractal Geometry, 2nd ed) Theorem 14.15
Quote
... when |c| is large, $$\dim_H J(f_c) \approx 2 \log 2 / \log 4 |c|$$
so there is a large c with dim < 1.  But $$\dim_H J(f_{0.25 + \epsilon}) > 1$$ for small $$\epsilon$$.  By continuity there is a c outside M with dim_H J(f_c) = 1.

Or do you mean points in M?  I think that is known, but I don't have a reference (or proof) handy - if c in M then J(f_c) is connected, so dim_H J(f_c) >= 1, the strict inequaiity for subset of M is interesting however.  I asked here https://math.stackexchange.com/questions/2676932/bounds-on-dimension-of-julia-sets-inside-mandelbrot-set

#### gerrit #### Re: Borderline fractals

« Reply #3 on: March 04, 2018, 11:51:13 PM »
But $$\dim_H J(f_{0.25 + \epsilon}) > 1$$ for small $$\epsilon$$.
Huh? 0.25+ is outside M so the Julia set is completely disconnected and Hdim<1, no?

#### claude #### Re: Borderline fractals

« Reply #4 on: March 05, 2018, 12:21:09 AM »
You can have completely disconnected sets with dimension > 1.  For example (Cartesian) product A x B of two Cantor sets A B satisfies dim (A x B) >= dim A + dim B.  The standard middle-thirds Cantor set has dim = log 2 / log 3, twice which is log 4 / log 3 > 1

#### gerrit #### Re: Borderline fractals

« Reply #5 on: March 05, 2018, 01:11:06 AM »
You can have completely disconnected sets with dimension > 1.  For example (Cartesian) product A x B of two Cantor sets A B satisfies dim (A x B) >= dim A + dim B.  The standard middle-thirds Cantor set has dim = log 2 / log 3, twice which is log 4 / log 3 > 1
Yes but that product lives in 4-d space. In 2d space I think any totally disconnected set has dim between 0 and 1.
Why do you think J(0.25+eps) has dim>1?
Maybe this is relevant: https://web.warwick.ac.uk/cody/Documents/UK_workshop_1/preiss_notes.pdf
Edit This paper http://www.ams.org/journals/ecgd/2001-05-07/S1088-4173-01-00070-4/S1088-4173-01-00070-4.pdf answers my question I think (formula 1.1).

So all we have to do is find the magical eps and then we have a Julia fractal that's not a fractal in some sense.

Learned something!
« Last Edit: March 05, 2018, 02:18:32 AM by gerrit »

#### claude #### Re: Borderline fractals

« Reply #6 on: March 05, 2018, 02:42:55 AM »
Yes but that product lives in 4-d space. In 2d space I think any totally disconnected set has dim between 0 and 1.

huh?  the middle-thirds cantor set is a subset of R = R^1, so the product lives in R^{1+1} = R^2 . you can in fact have any dimension in 0<= thru <n (in nD space) and still be completely disconnected (proof: construct a dim target/n cantor set in 1D, form the cartesian product of n copies).
see https://en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension (scroll down to log_3 4)

#### gerrit #### Re: Borderline fractals

« Reply #7 on: March 05, 2018, 02:49:37 AM »
Yes I see now, thanks.
Looks like the magical eps is around 0.25, do you know? Nothing visually obvious happens there as far as I can see from the Julias.

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