The Boundary of the Mandelbrot Set

[@Lttmtf]

It's rather interesting how the finite series..es (seri?) seem to be non-self-intersecting, but truncations of the infinite series do have self-intersections.

[@Number_Cruncher]

Thank you, for bringing up this issue. There are two different kinds of curves in this video. In the first part there are the "contour lines". They originate from the Taylor expansion and inversion. There $\Phi$ is a circle (larger than the unit circle, but shrinking for higher contours) that is mapped by the the Laurent series (polynomials, including negative powers). The shrinking of the radius makes them lying inside each other. This is a consequence of the "barrier of two". Whenever it is crossed by a sequence, there is no turning back. The authors of the papers claim that the outside of the Mandelbrot is simply connected. Self-intersections seem to conflict with this statement, but I'm not sure on this. Maybe someone knows more about this and can comment on this. I'm also confused by the seemingly isolated points of the Mandelbrot set, like the point i for instance. If it is true, the inside of the set would not be simply-connected? Is it possible, that the outside is and the inside is not? In the second part, truncations (100 and 20000 terms) of the final series are shown. It does not match with any finite contour line (simply because the unit circle is used as domain of hte mapping, turning it into a standard inverse Fourier transformation). I guess that the self-intersections disappear, when "all" terms are included or turn into isolated points. Also the limit of an infinite Fourier sum does not need to be smooth, maybe not even continuous.

[@Lttmtf]

@@Number_Cruncher Any truncation of the infinite series can also be obtained from truncating a sufficiently complicated taylor-derived series, so the difference between intersecting and non-intersecting is caused solely by having or not having last boundary terms of a finite series. The infinite series has no last 'boundary' terms, which confuses things. The way I understand it is that there's an 'outside' non-intersecting sequence of functions and an 'inside' intersecting sequence that both converge the same shape (the boundary of the mandelbrot set), so the limit is sandwiched between intersecting and not, and the result is the really spindly thread-like structures that bridge the otherwise disconnected points. I believe that the infinitely thin boundary of the set is all that connects some of the more solid islands, and also the seemingly disconnected points like i that you mentioned. I'm no expert on the topic, so these are just my second thoughts about the topic after being prompted to think further by your comment, and they hold every possibility of being wrong. I send my thanks you for allowing me and others to explore aspects of this subject I had never considered before.

[@Number_Cruncher]

The coefficients of the Taylor-derived series do not correspond to Fourier coefficients. They are not used to map the unit circle but a little bit larger circle instead. For instance in the first non-trivial case the contour is plotted by: $\sqrt{2} \exp{2\pi i}-1/2+1/8/\sqrt{2}\exp{2\pi i}+...$. So each coefficient gets multiplied by a power of \sqrt{2}, when treated as a Fourier coefficient. There are two limits happening at the same time, the convergence of the Taylor expansion and the shrinking of the circle towards the unit circle.

[@viliml2763]

The plural of "series" is "series". Like "sheep". It was that way in Latin too. It could only be "serii" if you started with "serius".

[@ethanJ496]

@@viliml2763 I am serius. And don't call me serii.

[@WhattheHectogon]

Excellent work! Really a joy to see Fourier series show up in this topic.

[@Number_Cruncher]

It's also interesting that you obtain the coefficients without a Fourier transform. They are the primary result of a straightforward computation and they would in principle give birth to the boundary of the Mandelbrot set if you considered them all.

[@denelson83]

​@@Number_CruncherSomething also tells me that if you run this series of generating polynomials out to infinity, the coefficients of the terms of increasing exponents end up being exactly the Catalan numbers.

[@Number_Cruncher]

@denelson83 It looks very much like it, although the convergences is very slow. The fourteenth polynomial starts with the coefficients 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440 followed by 9678461, which is not a Catalan number anymore. So the first 15 numbers agree. Which seems to generalize. The n-th polynomial agrees on the first n+1 coefficients (out of 2^n total coefficients).

[@alexarnold8461]

Great video! The Fourier series coming out of nowhere was really elegant

[@Number_Cruncher]

I'm glad that you can share my excitement about this little miracle.

[@davidgillies620]

I remember how stunned I was when I learnt that the level sets of the generating polynomials converged on the boundary. I couldn't wait to get home to play with it in Mathematica.

[@denelson83]

The reason the points _c_ = -2 and _c_ = _i_ have this kind of visual "isolation" has to do with the behaviour of their orbits under z² + c. You can see that they suddenly snap into periodic cycles that do not include zero. Points _c_ with this property are called _Misiurewicz points,_ and fall right on the boundary of the Mandelbrot set. As you zoom into a Misiurewicz point, the detail you see around it will converge toward exact self-similarity.

[@Number_Cruncher]

Thank you for this insight. Is there a sequence of points inside the set that converges to the Misiurewicz points?

[@denelson83]

@@Number_Cruncher No. You find Misiurewicz points by solving for when the output of one iteration exactly equals the output of another, with the condition that said output is not zero. Misiurewicz points are also landing points for specific rational external rays of the set. In the case of the two examples above, the 1/6 external ray lands on the point _c_ = _i,_ and the 1/2 external ray lands on the point _c_ = -2.

[@potatokiller4188]

I love videos like this, next year I'm going to college and I chose mathematics just because I want to be able to fully understand concepts like this. I probably wouldn't be as into maths as I am if it wasn't for videos like this showing just how interesting it can be.❤

[@Number_Cruncher]

Yeah, if it resonates - go for it. Math is hard, but it is beautiful at the same time.

[@unknowntimelord9557]

Strongest German accent ever. Love it

[@Number_Cruncher]

I've tried hard for years. But I've given up eventually. I started late learning it.

[@DanielIntense]

Wow, what a wonderul visualization at 14:48 I was already looking for this perspective.

[@Number_Cruncher]

Nice to hear, that people get that far:-) I like this part too, and also created a short for it kzbin.info/www/bejne/eGq2fYOma5l4aassi=9jislmQY8_JcACLv

[@pourtoukist]

Some mathematical objects are truly fascinating 😳

[@moodangelatx6580]

Beautiful cinematic math moment

[@HectaSpyrit]

This is a wonderful video! I'm not knowledgeable enough about the Mandelbrot set or about analysis more broadly (especially Fourier series) to have understood everything, but you have given me a strong urge to delve into the intricacies of this fascinating entity. Not only is there a lot to unpack in the video, the comment section seems to be rich in insights and related lines of inquiry as well. I can't wait to learn more!

[@Number_Cruncher]

I think that the Mandelbrot set is a great motivation to learn about complex numbers and a bit of programming. With these two tools at hand, progress is made very easily.

[@HectaSpyrit]

​@@Number_Cruncher Complex numbers, I'm very familiar with. I'm in my second year of master in mathematics, so I already have a fair amount of maths under my belt, and last semester I even had a course on complex analysis. But analysis was never my forte, I've always had a higher propensity towards Algebra, and the Mandelbrot set is definitely not something you often encounter in uni lectures. Still, I really want to dig into it some day! I'm a very poor programmer, so brushing up on that could be very interesting to learn how to practically compute solutions to problems like this.

[@xenontesla122]

I used this method to display the Mandelbrot set on an oscilloscope a while back. At the time I just pasted coefficients I found online. It's really cool seeing the actual logic behind it and this video may help me with making a more detailed version! Thanks!

[@Number_Cruncher]

This sounds like a fun thing to do. How do you generate the voltage signals? I guess you could use an audio program like audacity?

[@xenontesla122]

@@Number_Cruncher There's actually a whole community of people making "oscilloscope music". Some have even made apps!

[@djsmeguk]

13:20 or so. The location of the apparent poles seems to correlate with the known features of the set around that point- specifically the number of points in the cycle. As you increase the degree of the polynomial I wonder if the new poles that appear correspond to the cycle count of the associated iteration.. fascinating stuff.

[@Number_Cruncher]

I hope that I understand everything correctly. The zeros are indeed points of perfect cycles. The sequence restarts whenever it reaches zero. The zeros of a polynomial of degree n can give rise to cycles upto order n. Each polynomial trivially has 0 as a zero. This point will always be a one-cycle. The last example at 3:30 has such a zero as initial value that creates a five-cycle. Once the iteration hits zero, it returns to the initial value.

[@dsagman]

Fascinating. I studied fractals in the late 1980s and did not know this. I wonder how this translates to Julia sets.

[@Number_Cruncher]

I know that this method has been applied to other fractals as well. But I'm not sure about the Julia set, which can be disconnected in general.

[@landsgevaer]

Not sure what you mean. But for c that is inside the mandelbrot set, the Julia set is a "fat" fractal (i.e. local patches are part of the set); for c outside the set, the Julia set contains only "dust" (i.e. isolated points).

[@harriehausenman8623]

Some more random engagement! This thing has to go viral. 🤗

[@gigaprofisi]

babe wake up new number cruncher video

[@GrandMoffTarkinsTeaDispenser]

This is absolutely amazing. I can't thank you enough. I look forward to doing my own research.

[@ilikemitchhedberg]

Thank you! this went at a good pace for me, not too slow, not too fast. never saw Fourier thingamabobs in this context!! I love it!!! p. s. what would we see if we zoomed all the way in? 😉

[@laziz193]

Such an amazing presentation! You answered several questions I had specifically about the Mandelbrot Set. I was wondering if you were aware of the relationship between the logistic map and orbits of points inside the Mandelbrot set - the latter actually contains the former! I am also wondering if given your insight through the Taylor expansion one could sweep out the output space for a given complex number (in M or outside), thus generating the corresponding Julia set.

[@Number_Cruncher]

No, I ddn"t know about it. Can you please share references for the logistic map relationship?

[@denelson83]

There is actually no relationship between the logistic map and the Mandelbrot set. See this video to find out why. kzbin.info/www/bejne/Z6a6Xqtofd2Ahas

[@Number_Cruncher]

That's a great reference.

[@JavSusLar]

4:18 the simplest point with this property is (-2+0i). The Mandelbrot set is connected, i.e. it is possible to find a path between any two points in the set, a path that never leaves the set. In the case of the extreme point (-2+0i), it is easy to follow the path because it is a straight line, but in the case of (0+i), the path is a fractal line. In the case of (-2+0i), any infinitesimal deviation from the point, except in the direction towards (0+0i), lays outside the set. In the case of (0+i), it is even weirder, because ANY infinitesimal deviation from the point lays outside the set, but there is some limit, a spiral limit or something like this I suppose, that belongs in the set, so that the point (0+i) is connected to the set.

[@Number_Cruncher]

This fills in nicely with the question raised in the video. I checked infinitesimal deviations from (0+i) and didn't find any. If the path is fractal, it might explain, why an infinitesimal deviation cannot be found. And also, if it is just a line, it won't appear no matter how far you zoom in. But I guess, the line is located inside the fractal structure that builds a bridge from the set to the point (0+i).

[@Uuugggg]

I also dug into this. The parts of the Mandelbrot set with an area (ie the cardioid and bulbs) are created because they are near a stable fixed point that passes through 0. e.g. (-1)^2 - 1 = 0, (0)^2 - 1 = -1. It loops back to -1 through 0. Whereas the point i (and all ends of all the fractal lines) are Misiurewicz points : i doesn't return to back i: (i)^2+i = -1+i => (-1+i)^2 + i = -i => (-i)^2 + i = -1+i. It bounces between two points. Any movement from these points creates big changes, and is unstable, so you don't get an area around i that is stable. A movement around -1 creates small changes, because the points near 0 are squared to being even closer to 0, so there is an area around -1 that some movement stays stable. ( I don't have a great rationale as to why "close enough" is close enough, but certainly passing by 0 is better than -1+i )

[@michaelhiebert2867]

Could the Fourier series "arm" be used to draw for us factorizations of high degree polynomials or at least some approximation of these? Could we work backwards from the boundary of the Mandelbrot set to understand high degree polynomials better?

[@Number_Cruncher]

I've seen a paper, where they try to find all the zeros of such polynomial of high degree numerically. Already this is quite a challenge. They used regular Newton"s method. I wanted to have a closer look at this. I'm not sure there will be any help with factorization. But I'll check out the Galois groups for the first few polynomials.

[@Number_Cruncher]

But, another issue, the Taylor-series approach will break down inside the Mandelbrot set. So, I guess, it will not be helpful with the zeros.

[@FrancisDuranx]

Great video!

[@cosimobaldi03]

Beautiful video! But does the length of the border converge or not? It should be able to calculate the length from the Taylor expansion, right?

[@Number_Cruncher]

Nice question! Here are the first curve lengths (100 terms of expansion included): 12.6 (4pi), 8.89, 7.74, 7.44, 7.69, 8.13, 8.88, 9.66, 10.3, 10.7, 10.9, which clearly indicates my suspicion, that it will not converge. When I run the computation with 200 terms, the first six values are the approximately the same, the remaining ones are even larger.

[@denelson83]

The length of each curve diverges as it gets closer to the border of the set. As the Mandelbrot set is a fractal, the length of its boundary is infinite. This is reflected in the fact that the boundary's Hausdorff dimension is 2.

[@jeremy.N]

okay I now need to know the area of the mandelbrot set! iirc the hausdorf dimension of the boudary is 2, and the area is like 1.56...? but that's years ago since I got into that. Im exited for future videos, this one was amazing!

[@Number_Cruncher]

As far as I know, there is still a discrepancy between the area obtained from pixel-counting and the area obtained from the coefficients. The first result is about 1.50 and the later including 5 million coefficients is about 1.68 ( arxiv.org/abs/1410.1212 ). Where do you get your number from?

[@jeremy.N]

@@Number_Cruncher Again, this is like several years ago, but i think I remember it was from counting boxes or linear interpolations + triangulation of a certain finite depth/ iteration.

[@swamihuman9395]

- I can't like this enough!!! - Thx :)

[@harriehausenman8623]

Liked everything there is to like, shared with everyone I know, anything I forgot? Oh yes, and obviously subscribed 😄

[@Number_Cruncher]

Thank you, I really appreciate your efforts.

[@trevorclinton2573]

The boundary of the Mandelbrot set is topologically a circle and has topological dimension of 2. It is not known if the boundary has positive area.

[@Number_Cruncher]

When the area is calculated from the coefficients, it is larger than from pixel counting. It's kind of obvious, because a finite number of coefficients always leads to a curve that overestimates the area. But as far I know, it is not obvious, whether it is slow convergence or whether there is a contribution from the boundary. May I ask for references that explore in these directions?

[@EmissaryOfSmeagol]

12:31 you say here that we "compute the Taylor expansion of the inverse function." What happens on screen is what I might call 'Series Reversion,' but I'm not sure. I had to look up the term for this, since the equation shown is not for _c^(-1)_ but rather _c_ and _Phi_ have more or less swapped places. Am I misunderstanding? Also, great video overall, I am glad KZbin recommended this video!

[@Number_Cruncher]

Thank you for giving me the chance to expand on this as well. I was thinking to give more details in the video but I didn't want to make it too heavy. The idea is simple: After the Taylor expansion, we have a series for Phi(c). But what we are actually interested in is the inverse function c(Phi), which maps the circle to the contour line. Finding the inverse function is normally a straight forward procedure. For instance, if you want to find the inverse function for y(x)=x^2, you just solve for x in terms of y to find x(y) = \sqrt{y}. The inverse series is found by a general ansatz c(Phi), which is a Laurent series with coefficients b_{m} for m\in \mathbb{Z}. Now you impose the condition c(Phi(c)) =c. Comparing coefficients for powers of c, yields the values for the b_{m}. It is tedious but more or less straight forward and easily performed with computer programs.

[@realcygnus]

Nifty AF

[@PrancerFrisco]

How do you do the animations?

[@Number_Cruncher]

The animations are rendered in blender and the blender files are created with a python library that I work on. It is inspired by the manim library. For instance, if you check out the video kzbin.info/www/bejne/nGfPdp6laMSSZq8 , the shader created manually there, is just one function call in the library.

[@wcdeich4]

Can we plot the edges of the Barnsley fern with this method?

[@Number_Cruncher]

I'm afraid, we can't. The Barnsley fern is generated by an affine transformation of R2. (x,y)-> A.(x,y) +(e,f), where A is a real-valued 2x2 matrix. If you wanted to translate this into the complex plane, it would look like z->α z+β z*+c, where α and β are related to the coefficients of the matrix ( math.stackexchange.com/questions/3331261/complex-numbers-and-linear-transformation ). Therefore, the iteration function is not holomorphic. The complex conjugate z* appears in the iteration. Taylor series expansion only works for functions that can be purely written in terms of the complex variable z. Also, holomorphic transformations preserve angles. Linear transformations do not have this property in general.

[@wcdeich4]

@@Number_Cruncher Thanks

[@dsagman]

By the end, did you essentially outline a proof of the connected nature of the Mandelbrot set?

[@Number_Cruncher]

The authors referenced in the video description claim, that this proofs the connectedness of the outside of the Mandelbrot set including the point at infinity. I'm not certain, whether it also implies the connectedness of the set itself.

[@denelson83]

@@Number_Cruncher Connectedness outside a set does not imply connectedness inside, as can be demonstrated with a set consisting of two disjoint areas.

[@Number_Cruncher]

@@denelson83 yes, this makes sense. If I remember correctly, the authors even claim that the outside is simply connected. I guess, this poses a stronger constraint on the inside, doesn't it? But I wonder how this would be compatible with the Misiurewicz points? If a loop goes around such a point, it could not be shrunk to a point that is part of the outside.

[@michaelyork4554]

How Transformative. Have you used the Mandelbulb 3D application? You can create fractal movies, and zoom into them,

[@Number_Cruncher]

I use blender for these animations. Infact, I made another video, where I explained the creation of some of the animations: kzbin.info/www/bejne/nGfPdp6laMSSZq8

[@C0MPLEXITY]

great video and the animations were breathtaking!

[@JwalinBhatt]

Could this method be used to find the fourier series for the boundary of the littlewood fractal?

[@Number_Cruncher]

Honestly, I don't know. My first guess: I doubt it. At least, I don't find any similarities. But I must confess that I just started reading about Littlewood fractals, having read your question. For the Mandelbrot set, we have algebraically defined subsets s_n, each of which is simply connected and s_{n+1} is a subset of s_{n}. Each of the set's boundaries can be described by a function with a Fourier spectrum. If I understand it correctly, each subset of the the Littlewood fractal is just a point set, collecting the roots of a set of polynomials. Only in the limit of infinitely many iterations the set becomes dense and developes a continuous boundary. I could be completely wrong, though.

[@brotherWesley]

One definitive answer to the question "Are there criteria by which an objective assessment of the beauty of iterative mathematical processes may be judged?"can be understood through your explanations of the boundaries of the Mandlebrot set.

[@doraemon402]

Yo. Great vid. It's awesome and I'm serious about it.

[@paolarei4418]

No way doemoram

[@JwalinBhatt]

Awesome video! Is there a reason for taylor expansion specifically at infinity? Would it rather not be expanded at 2? And would it work if it was expanded at 2?

[@Number_Cruncher]

The expansion of the expression \sqrt{1+a_1 z^{-1}+a_2 z|^{-2}+...} is particularly simple at z=\infty. Because all function evaluations and evaluations of the derivatives are equal to 1 at this point, since all the terms with z^{-m} are zero. Another issue is the radius of convergence. The expansion will certainly break at z=0. Which means that the radius of convergence around 2 would be at most 2. If you expand around z=\infty, it seems to converge everywhere apart from z=0.

[@JwalinBhatt]

@@Number_Cruncher Thank you for the response and the explanation, I see your point. But as I understand, the series would be most accurate near the point at which it is expanded. And since the set lies inside a unit disk of radius 2 should that not play a role in the value we chose?

[@Number_Cruncher]

Yes, the set lies inside the disk of radius 2 centered around the origin. The most natural point to expand would be at zero. But we can't expand there, since it is a singular point. It gets even worse. I mention it quickly in the video that there are lots of zeros inside the Mandelbrot set for the polynomials p_n (more precisely 2^n zeros). When the (2^n)th root is taken from this polynomial, there are branch cuts between the zeros. The function will be multivalued, since the root function is not unique. But all these branches are located inside the Mandelbrot set connecting two zeros. It is a very unpleasant place for any kind of expansion. In contrast, the outside of the Mandelbrot set is like a paradise, free of zeros, free of branch cuts. The root is single-valued and the series expansion converges nicely.

[@JwalinBhatt]

@@Number_CruncherAch so, ich verstehe. Danke dir :)

[@moodangelatx6580]

Cruncher explaining to mere mortals “how many angels dance on head of pin” 😂 🎉 ❤

[@michaelnash5542]

Just to say that your link to the Jyptiter notebook isn't working. Excellent content though!

[@Number_Cruncher]

Thank you, I fixed the link.

[@harriehausenman8623]

Mind sufficiently blown! 🤯 Unfortunately, I can't replicate it in Wolfram 😥

[@Number_Cruncher]

I will link a notebook to this reply, once I'm back at the end of January.

[@harriehausenman8623]

@@Number_Cruncher 🙏 thx

[@Number_Cruncher]

Here is a link to my notebook: www.dropbox.com/scl/fi/dk72u1i8jo0rc37b784rz/MandelbrotSetCurves.nb?rlkey=mh7x3b5qhajh1izo6df2hk8mv&dl=0 Let me know, whether there are still problems. I added a bonus feature in the notebook, where a Manipulate draws the boundary with rotating arrows as shown in the video.

[@harriehausenman8623]

@@Number_Cruncherthank you! 🙏

[@MichaelDarrow-tr1mn]

wait are the coefficients converging to the catalan numbers

[@Number_Cruncher]

You are the second person raising this suspicion. I don't understand. The coefficients alll are fractions with powers of two as denominator. Where do the Catalan numbers appear?

[@MichaelDarrow-tr1mn]

The polynomials near the start. c, c+c², c+c²+2c³+c⁴, 1c+1c²+2c³+5c⁴+6c⁵+6c⁶+4c⁷+c⁸. Look at the coefficients up to c⁴: 1c+1c²+2c³+5c⁴. 1, 1, 2, 5. Those are the catalan numbers.

[@Number_Cruncher]

You are right, the fourteenth polynomial starts with the coefficients 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440 followed by 9678461, which is not a Catalan number anymore. One can show by induction that the first n+1 coefficients for the polynomial p_n are the Catalan numbers C_0, C_1, ..., C_n. Furthermore the last coefficient is 1 and the last before last coefficient is 2^(n-1). The inductive step relies on the recursion relation for Catalan numbers $C_n = \Sigma_{i=0}^n (C_i C_{n-i})$, which is precisely generated by the recursion relation z_{n+1}=z_n^2+c.

[@moodangelatx6580]

❤