Since the Mandelbrot fractal has infinite complexity and we can zoom inside of it infinitely long one might ask if any combination of pixels eventually emerge and thus It contains all possible images (assuming an adecuate colormap that samples all necessary colours).

I guess the answer is that we don't know yet.

A similar question asks if you can find any possible sequence of digits inside the decimal expansion of Pi. We know that Pi has a non-periodic infinite decimal expansion but still we don't really known if some sequences might be forbidden. In other words, we don't know if Pi is a normal number. Even if unproven, many matematicians think Pi is probably normal and thus contain every sequence.

It has been show that the number Pi is naturally encoded in the Mandelbrot set. This might imply that the random non-repeating pattern of Pi might translate into random non-repeating shapes in the fractal. If we eventually prove that Pi is normal, would this mean that the Mandelbrot set is also normal, and thus that It really contains all possible images?

Another way to prove It, I guess, would be finding a known normal number shaping some patterns of the fractal somewhere somehow, instead of Pi.

By the way, I'm considering pictures of the fractal where each iteration is represented and not only the boundary between the set and the non-set, which has properties that might forbid some possible shapes (since that would only allow connected shapes with infinite lenght boundaries)

Ted Shifrin
  • 101,067
  • 4
  • 80
  • 135
  • 218
  • 1
  • 6
  • 1
    Note that "normal" is stronger than "containing every finite digitsequence". The only evidence for the normality of $\pi$ (it is even assumed to be absolutely normal, that is normal in every base $b$-system, where $b>1$ is an integer) is that the enormous number of computed digits approve this assumption. – Peter Dec 22 '21 at 09:49
  • We do not know whether a given digit appears infinite many often, we cannot even rule out that eventually only two digits occur. But the link between Mandelbrot and $\pi$ is very surprising and a great result , although I keep sceptical that the normality of $\pi$ (or many other irrational numbers) can be decided. – Peter Dec 22 '21 at 09:52
  • The normality (even absolute normality) is also conjectured for every algebraic irrational real number as well as for $e$ and even for $\gamma$ , although in this case we do not even know whether its decimal expansion is non-terminating. But not a single of those numbers in a single base could be proven to be normal. – Peter Dec 22 '21 at 09:54
  • @Peter: I was just about to make this same comment (your first comment), then saw your comment (which at the time I saw it had "1 min ago" associated with it). For more about the distinction, see [Normal Numbers as members of a larger set?](https://math.stackexchange.com/q/2728446/13130). Most real numbers are normal in the sense of Lebesgue measure, but most real numbers are NOT normal in the sense of Baire Category (indeed, ["not"](https://mathoverflow.net/questions/374059/irrationality-measure-of-powers/374068#comment1046655_374068) in a very strong way). **(continued)** – Dave L. Renfro Dec 22 '21 at 09:59
  • However, most real numbers in both the Lebesgue measure sense and in the Baire category sense (and even more -- all but a $\sigma$-porous set of real numbers) have every finite digit string in their decimal expansions. – Dave L. Renfro Dec 22 '21 at 10:00
  • @DaveL.Renfro Is "almost every real number" also "absolutely normal" in this sense ? I guess this is the case. – Peter Dec 22 '21 at 10:02
  • @Peter: Yes, all but a Lebesgue set of measure zero set of real numbers has, in every base (not just base 10), every finite digit string for that base appearing with the "appropriate" frequency. On the other hand, all but a Baire category set of real numbers has, in each base, the property that each finite string of digits (e.g. $1103$ or $3141592654$ or $010011000111)$ appears so "uniformly erratically" that for each $\epsilon > 0$ (no matter how small) there are infinitely many locations in the decimal expansion in which the frequency of that string **(continued)** – Dave L. Renfro Dec 22 '21 at 10:14
  • up to that point is less $\epsilon$ and there are infinitely many other locations in the decimal expansion in which the frequency of that string up to that point is greater than $1 - \epsilon$ (i.e. the not only do we not have the appropriate limiting frequency for that digit string, but the limiting frequency for that string doesn't exist in the worst possible way -- the lim-inf for that digit string's frequency is $0$ and the lim-sup for that digit string's frequency is $1).$ – Dave L. Renfro Dec 22 '21 at 10:14
  • 1
    Incidentally, related to your question is a result proved by Peter M. Gruber ([*Your picture is everywhere*](https://zbmath.org/?q=an%3A1002.52502); 1991) that I believe shows that most compact subsets $K$ of ${\mathbb R}^n,$ in the sense of Baire category for the Hausdorff metric, have the following property: for each $\epsilon > 0$ and for each compact subset $C$ of ${\mathbb R}^n,$ the set $K$ contains a [homothetic copy](https://en.wikipedia.org/wiki/Homothety) of $C$ whose Hausdorff distance from $K$ is less than $\epsilon.$ **(continued)** – Dave L. Renfro Dec 22 '21 at 10:40
  • Thus, if the Mandelbrot set has this same property that most compact sets have, then "up to an $\epsilon$-distortion" the Mandelbrot set contains a copy of every compact set. Hence, for all practical purposes, it contains every shape. Of course, perhaps the Mandelbrot set doesn't have this property, but regardless of whether it does or does not, there are lots of sets (indeed, "most" compact sets in a certain sense) that have this property. – Dave L. Renfro Dec 22 '21 at 10:40
  • Incidentally, I don't think I've correctly stated the result Gruber proved (and I can't find my copy of his paper), but the gist of what he proved is that most compact sets essentially contain a copy of every compact set. – Dave L. Renfro Dec 22 '21 at 11:38
  • 4
    If you play with software that allows you to zoom into the Mandelbrot set, then it should be obvious that the answer is *no*, since the same patterns keep recurring - though, with growing complexity. The reason is that, if you zoom into a portion that contains part of the boundary, then there are always [Misiurewicz points](https://en.wikipedia.org/wiki/Misiurewicz_point) that dominate the appearance. If you zoom into a portion that entirely misses the boundary, then the image is guaranteed to be quite calm. – Mark McClure Dec 22 '21 at 12:18
  • https://math.stackexchange.com/questions/458400/is-the-mona-lisa-in-the-complement-of-the-mandelbrot-set – Claude Dec 22 '21 at 19:37
  • "The number pi is naturally encoded in the Mandelbrot set" ... In fact, the number pi is naturally encoded in a circle. So the quoted reasoning will not lead us to suspect that the Mandelbrot set contains all images. – GEdgar Dec 22 '21 at 22:36

0 Answers0