Im happy that I've gotten far enough in my journey of mathematics that I can undertsand maybe 60% of this video. A year or two ago I would have no idea what's happening! Hopefully one day I can understand 100%, or even be the one making awesome and fascinating videos like this.
Hi Cruncher. I am happy to see that you computed both an “outer measure” and an “inner measure” for the area, because ever since Archimedes, this has been the accepted method for integration in complicated situations. Cheers.
Beautifully done! I remember when I studied advanced calculus and tried to sketch a proof of the Stokes theorem (or just Green-Stokes) by rectangular approximation. Apparently the difficulty lies in the observation that we can't approximate the length of a line, say f(x)= -x, by a rectangular staircase curve.
I like to apply Stokes' theorem in all possible situations and I'm always amazed that it works so beautifully. I can imagine that it's not so easy to proof it for the general case.
I've never used differential forms before but your color coding of particular elements of calculations and formulae is amazing and takes what would've been completely impossible for me to understand to something i can chunk through with a couple google searches. I've used stokes in an E&M class but never seen it in terms of a differential form. I've got some reading to do! Great video!
Thanks for your pleasant feedback. It's never easy to find the right amount of detail. But if you really want to learn it, you have to sit down and work on it with additional sources anyways. So I'm glad that is motivation for you.
This was absolutely mind blowing. Thank you for sharing such an interesting fact about differential forms. I have worked with them for many years and it never occured to me that ther should exist a canonical n- form whose surface integral will always be the encosed volume of a n+1 compact object of any shape in any geometry. Specially on the unit 2 sphere, if you find a closed curve whose enclosed area is 2pi then you know you are dividing the sphere in 2 equal areas and that therefore if you reverse the direction in which you do the line integral, by symetry, it should be the same 2pi result? Beause what is exterior and interior changes, but I might be wrong on that intuitive explanation.
You are absolutely right with your intuition. In the simplest case, when your curve is the equator, the integral reads \int_0^{2\pi} (1-\cos\theta)d\phi. At the equator \theta={\pi\over 2}, which makes the cosine disappear and you get 2\pi as area for the northern hemisphere. If you go the other way round, the enclosed area will be the southern hemisphere with the same area. But this should be true for any curve that splits the sphere into two halves. Also, if you do not take the equator, but any other latitude \theta you get the area of the enclosed part of the sphere as 2\pi(1-\cos\theta). It is zero for \theta =0 (which corresponds to a collapsed curve on the north pole) and it approaches 4\pi for a tiny curve around the south pole that basically encloses the entire sphere.
13:45 ".. better estimates roughly converting to 1.96" That may hold for that curve, but it is further off the value that I find on various online sources for the area of the Mandelbrot set, the most accurate (claimed) seems to be 1.5065918857±0.0000000085 So it seems the "tendrils" remain too fat, and counting smaller pixels does not solve the accuracy bottleneck.
Yes of course, this is not an approximation for the area of the Mandelbrot set but only for the area inside this particular curve. The point is, that the pixel counting matches the value obtained from the sum over the coefficients. The computation of the area for the Mandelbrot set was done at 1:58 in the video. This upper bound with 1.78 is still far away from your value. When 5 million coefficients are summed the upper bound is lowered to 1.68. There are some discussions, whether there is a non-vanishing contribution to the area from the boundary (since it has a fractal dimension of 2). But I personally believe, that you would reach 1.506..., if you could sum infinitely many coefficients. The curve that I show at 13:45 is a curve in the contour plot outside of the Mandelbrot set (have a look at ruclips.net/video/Oh1AiiPpoTo/видео.html for more details). It includes all points, where the iteration |z_n|
@@Number_Cruncher Sure, understood. I just got the impression that the goal was to estimate the area of the Mandelbrot, and then this circumferential method seems less effective than e.g. Monte Carlo integration. Also, it is biased to overestimate the result, giving a systematic error rather than a random error. But maybe that wasn't the goal after all..? 😉
@@Number_Cruncher Just thinking out loud, it might be interesting to see if Euler summation (or some other method that improves series convergence, like the Shanks transform) would lead to more accurate values for that truncated infinite sum...
I will give it a try. The smallest outcome will be that I learn something about the Euler transform. However, I'm not sure, whether it will be much of an acceleration. When I look at the way the curve moves closer to the boundary the more coefficients are introduced (1:55), I get the impression that the high-frequency details introduced by the later coefficients are desperately needed to get closer to fractal boundary. The truncated curves do not oscillate around the true boundary but rather approach it steadily from the outside.
The Mandelbrot set extends infinitely along the x-axis. Although it seems intuitive that the area contributions far from the origin are insignificant, does this have to be proven?
I've seen it De Rham's cohomology in connection with Hodge duality the context of classifying Calabi-Yau manifolds. It's definitely something that I would like understanding better :-)
@@Number_Cruncher well it's not too far off what you treated in this video, it is just the duality between differential forms and homology groups of a manifold, with the inner product given by the integral in Stoke's Theorem. The domain of integration is a simplex of the homology, so you're really treating possible domains of integration as elements of a vector space whose dual is the space of differential forms Edit: Nakahara's "Geometry, Topology and Physics" has a great treatment of the subject
You can map any image that you have created in a plane onto a sphere with stereographic projection. For each point of the (x,y) there is a unique point on the sphere. You need to use a program that can perform the point-wise mapping. If you want to do it in real life, it should be difficult, because the plane gets stretched and squeezed during the mapping. Blender is a very powerful program for such tasks. I showed more details for the work with blender in one of my older videos (Riemann spheres in blender)
One thing I've always thought was confusing was that in this presentation of Stoke's theorem, you end up with an integrand that is just f. That looks like it should be a function, not a one-form or any other kind of differential form, so it takes a leap of imagination to realise that actually it just means it's a differential form of degree one less than df. In other words notationally it hides the fact that it's a differential form, kinda - I'm sure courses in maths cover that, but when you show that to a bunch of people who have some knowledge of integration but haven't yet met differential forms, it's very confusing.
I see the point you are making. On the wiki-page, they use a small omega, which maybe doesn't lead to the association with a pure function. I decided to use f to make the connection with the zero-dimensional case, when it is a pure function as easy as possible.
@@Number_Cruncher In that case it's likely I am entirely wrong! I had no idea that you could actually integrate a one-form. At my basic level of calculus knowledge, you'd have to integrate with respect to something - I couldn't integrate f, but I could integrate f(x)dx. Without an explicit differential in the integrand, what are you integrating with respect to? Thank you very much for replying, and please keep producing these excellent videos
When you write f(x)dx --- this actually is a one-form and you can integrate it over a one-dimensional interval. This integration can be related to the integration of a zero-form along the boundary of the interval. The corresponding zero-form is a function F(x), with the property that F'(x) = f(x). The boundary of the interval are just two points a and b. And the zero-dimensional integral reduces to F(a)-F(b). The second term is negative, because the "outside" of the boundary points towards the negative direction. A two-dimensional integral has to be performed over a two-form. In three dimensions, the most general two-form is written as f(x,y,z) dx∧dy+g(x,y,z) dx∧dz + h(x,y,z) dy∧dz, this is usually denoted as v•dF (v is a vector function and dF is the surface area element, cf. en.wikipedia.org/wiki/Surface_integral) . In a two-dimensional world the generic two-form looks like f(x,y) dx∧dy, which is then just written as f(x,y)dxdy for simplicity.
@@Number_Cruncher that was extremely clear! So, the integral of a zero form F is just F(b) - F(a) where b and a are the integration limits. You have a skill for teaching complicated maths, thank you for sharing it
Noone knows wether the value of the area is a rational or an irrational number, let alone algebraic or transcendental. I think it is irrational, but I think that is not proovable.
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Im happy that I've gotten far enough in my journey of mathematics that I can undertsand maybe 60% of this video. A year or two ago I would have no idea what's happening! Hopefully one day I can understand 100%, or even be the one making awesome and fascinating videos like this.
I wish you a joyfully continuing journey.
Hi Cruncher. I am happy to see that you computed both an “outer measure” and an “inner measure” for the area, because ever since Archimedes, this has been the accepted method for integration in complicated situations. Cheers.
THIS VIDEO IS SIMPLY AMAZING. I have learnt so much!!!! I'm going to share this video with all my math freak friends. Beautiful, phenomenal job!
@@itz_RavjiBoy THX!!! I really appreciate it.
Beautifully done! I remember when I studied advanced calculus and tried to sketch a proof of the Stokes theorem (or just Green-Stokes) by rectangular approximation. Apparently the difficulty lies in the observation that we can't approximate the length of a line, say f(x)= -x, by a rectangular staircase curve.
I like to apply Stokes' theorem in all possible situations and I'm always amazed that it works so beautifully. I can imagine that it's not so easy to proof it for the general case.
I've never used differential forms before but your color coding of particular elements of calculations and formulae is amazing and takes what would've been completely impossible for me to understand to something i can chunk through with a couple google searches. I've used stokes in an E&M class but never seen it in terms of a differential form. I've got some reading to do! Great video!
Thanks for your pleasant feedback. It's never easy to find the right amount of detail. But if you really want to learn it, you have to sit down and work on it with additional sources anyways. So I'm glad that is motivation for you.
fantastic. mandelbrot. stokes. and fourier transform rotations all in one.
This was absolutely mind blowing. Thank you for sharing such an interesting fact about differential forms. I have worked with them for many years and it never occured to me that ther should exist a canonical n- form whose surface integral will always be the encosed volume of a n+1 compact object of any shape in any geometry. Specially on the unit 2 sphere, if you find a closed curve whose enclosed area is 2pi then you know you are dividing the sphere in 2 equal areas and that therefore if you reverse the direction in which you do the line integral, by symetry, it should be the same 2pi result? Beause what is exterior and interior changes, but I might be wrong on that intuitive explanation.
You are absolutely right with your intuition. In the simplest case, when your curve is the equator, the integral reads \int_0^{2\pi} (1-\cos\theta)d\phi. At the equator \theta={\pi\over 2}, which makes the cosine disappear and you get 2\pi as area for the northern hemisphere. If you go the other way round, the enclosed area will be the southern hemisphere with the same area. But this should be true for any curve that splits the sphere into two halves. Also, if you do not take the equator, but any other latitude \theta you get the area of the enclosed part of the sphere as 2\pi(1-\cos\theta). It is zero for \theta =0 (which corresponds to a collapsed curve on the north pole) and it approaches 4\pi for a tiny curve around the south pole that basically encloses the entire sphere.
13:45 ".. better estimates roughly converting to 1.96"
That may hold for that curve, but it is further off the value that I find on various online sources for the area of the Mandelbrot set, the most accurate (claimed) seems to be 1.5065918857±0.0000000085
So it seems the "tendrils" remain too fat, and counting smaller pixels does not solve the accuracy bottleneck.
Yes of course, this is not an approximation for the area of the Mandelbrot set but only for the area inside this particular curve. The point is, that the pixel counting matches the value obtained from the sum over the coefficients.
The computation of the area for the Mandelbrot set was done at 1:58 in the video. This upper bound with 1.78 is still far away from your value. When 5 million coefficients are summed the upper bound is lowered to 1.68. There are some discussions, whether there is a non-vanishing contribution to the area from the boundary (since it has a fractal dimension of 2). But I personally believe, that you would reach 1.506..., if you could sum infinitely many coefficients. The curve that I show at 13:45 is a curve in the contour plot outside of the Mandelbrot set (have a look at ruclips.net/video/Oh1AiiPpoTo/видео.html for more details). It includes all points, where the iteration |z_n|
@@Number_Cruncher Sure, understood.
I just got the impression that the goal was to estimate the area of the Mandelbrot, and then this circumferential method seems less effective than e.g. Monte Carlo integration. Also, it is biased to overestimate the result, giving a systematic error rather than a random error.
But maybe that wasn't the goal after all..? 😉
I very much agree on the inefficiency. For me it is fascinating that it is possible at all and that it is not too complicated to understand.
@@Number_Cruncher Just thinking out loud, it might be interesting to see if Euler summation (or some other method that improves series convergence, like the Shanks transform) would lead to more accurate values for that truncated infinite sum...
I will give it a try. The smallest outcome will be that I learn something about the Euler transform. However, I'm not sure, whether it will be much of an acceleration. When I look at the way the curve moves closer to the boundary the more coefficients are introduced (1:55), I get the impression that the high-frequency details introduced by the later coefficients are desperately needed to get closer to fractal boundary. The truncated curves do not oscillate around the true boundary but rather approach it steadily from the outside.
This is Green's theorem which is the 2 dimensional case of Stokes' theorem.
Nice i wasn't expecting it!
I love this so much!
I tire of watching font glyphs get drawn super slowly. It was cool at first but now it's like the tag
I'll keep that in mind.
The Mandelbrot set extends infinitely along the x-axis. Although it seems intuitive that the area contributions far from the origin are insignificant, does this have to be proven?
I don't think so. The set is entirely contained in a disk of radius 2 around the origin. Definite upper bound on the area is 4pi.
the actual area seems to be 1.506 something
A-M-A-Z-I-N-G confirmation
De Rham's Cohomology is probably one of my favorite bits of math
I've seen it De Rham's cohomology in connection with Hodge duality the context of classifying Calabi-Yau manifolds. It's definitely something that I would like understanding better :-)
@@Number_Cruncher well it's not too far off what you treated in this video, it is just the duality between differential forms and homology groups of a manifold, with the inner product given by the integral in Stoke's Theorem. The domain of integration is a simplex of the homology, so you're really treating possible domains of integration as elements of a vector space whose dual is the space of differential forms
Edit: Nakahara's "Geometry, Topology and Physics" has a great treatment of the subject
Yeah, I've actually read parts of this book, while I was studying. I'll get it out of the shelf again.
マンデルブロを球面に貼り付けるやり方を教えて下さるとありがたいです。
特に、グラフィック、CGでその操作をしたいと思っております。
あと、マンデルブロを球面に貼り付けたら、その世界線の小説やゲームを作ることができそうですね。
You can map any image that you have created in a plane onto a sphere with stereographic projection. For each point of the (x,y) there is a unique point on the sphere. You need to use a program that can perform the point-wise mapping. If you want to do it in real life, it should be difficult, because the plane gets stretched and squeezed during the mapping.
Blender is a very powerful program for such tasks. I showed more details for the work with blender in one of my older videos (Riemann spheres in blender)
One thing I've always thought was confusing was that in this presentation of Stoke's theorem, you end up with an integrand that is just f. That looks like it should be a function, not a one-form or any other kind of differential form, so it takes a leap of imagination to realise that actually it just means it's a differential form of degree one less than df. In other words notationally it hides the fact that it's a differential form, kinda - I'm sure courses in maths cover that, but when you show that to a bunch of people who have some knowledge of integration but haven't yet met differential forms, it's very confusing.
I see the point you are making. On the wiki-page, they use a small omega, which maybe doesn't lead to the association with a pure function. I decided to use f to make the connection with the zero-dimensional case, when it is a pure function as easy as possible.
@@Number_Cruncher In that case it's likely I am entirely wrong! I had no idea that you could actually integrate a one-form. At my basic level of calculus knowledge, you'd have to integrate with respect to something - I couldn't integrate f, but I could integrate f(x)dx.
Without an explicit differential in the integrand, what are you integrating with respect to?
Thank you very much for replying, and please keep producing these excellent videos
When you write f(x)dx --- this actually is a one-form and you can integrate it over a one-dimensional interval.
This integration can be related to the integration of a zero-form along the boundary of the interval. The corresponding zero-form is a function F(x), with the property that F'(x) = f(x). The boundary of the interval are just two points a and b. And the zero-dimensional integral reduces to F(a)-F(b). The second term is negative, because the "outside" of the boundary points towards the negative direction.
A two-dimensional integral has to be performed over a two-form. In three dimensions, the most general two-form is written as f(x,y,z) dx∧dy+g(x,y,z) dx∧dz + h(x,y,z) dy∧dz, this is usually denoted as v•dF (v is a vector function and dF is the surface area element, cf. en.wikipedia.org/wiki/Surface_integral) . In a two-dimensional world the generic two-form looks like f(x,y) dx∧dy, which is then just written as f(x,y)dxdy for simplicity.
@@Number_Cruncher that was extremely clear! So, the integral of a zero form F is just F(b) - F(a) where b and a are the integration limits. You have a skill for teaching complicated maths, thank you for sharing it
ChatGPT said the area of Mandelbrot Set is unknown :/
There is no fundamental understanding for the area. There are only approximate results.
👍
Sie haben leider die letzte Stunde verpasst. Da haben wir die Mandelbrotmenge visualisiert.
Noone knows wether the value of the area is a rational or an irrational number, let alone algebraic or transcendental. I think it is irrational, but I think that is not proovable.
👍🏼
You overlooked the fact that the Mandelbrot set is a fractal and encloses areas that are not part of the set.
Can you please share a few more details?