:O Those fractals are so beautiful, and the fact that the number of edges on each level of fingers describes the path 3 takes through the Collatz procedure? That's crazy!
I've always wanted to display this in LEGO, but do not have the funds or room. 38 LEGO studs = one foot, so showing numbers up to 38 would require at least three feet.
Wow, that was a fun visualizion.. both the original number line loops as well as the fractal complex extension. Knowing you, you probably wrote both visualizations with procedural shaders? Whether you did it or not I would love to see another video of the same length just describing the tools you made and/or used to make this video!
This video made me like you instantly. I am amazed how you visualized the numbers. I am fascinated by this conjecture and glad to have found your video. Keep it up
What a brilliant idea to extend the map into the complex field. Hopefully someone will one day use this new point of view to crack the collatz conjecture !
Wow that's so insightful! I'm numberphile's (and maths', and fractals') fan, and I am totally amazed how that enigmatic Collatz conjecture turns out to be a beautiful fractal when expanded to complex numbers. Great work!
what i really wanted to see towards the end was you showing what you did, for other numbers than 3. Such as 9, because 9 does some pretty big jumps. It would be really cool to see the 28 fingers. Now the yellow points, are those actual zeros to the function? If you put that complex point into the function and iterate it will land on 0? Maybe its more profound to me because i haven't analyzed any of this in the way that you have. But its pretty F*cking amazing honestly that the collatz cojecture and the jumps a number will take is literally encoded at each number visually in the fractal. That is the coolest thing ever. The visualization is slightly creepy. what exactly are the black areas? You should do a video on just that fractal alone and explain alot more of it, at an elementary level. Like the basics such as what is the black area. And then with deeper maths. This problem no doubt has been analyzed at universities by mathematicians in the complex plane, but this could no doubt provide many valuable insights and angles of attack for others who haven't thought to try this. It may very well be that proving that all natural numbers return to 1 may come from things that we could only prove by analysis in this manner.
This is so wonderful and it seems I'm not the only one who thinks that way. It's a great example for just having fun with maths and feeling the joy of realizing patterns.
It basically means "almost-divergent or divergent". Let me explain : To render this image, we have to check the convergence of every point. But we face some ussues here : first, we want to render a lot of points. Secondly, we can't interate a point infinitely. So for these two reasons, we are setting an arbitrary limit : we assume that when an iteration of a point hits n (1 000 or 1 000 000 for instance), it will diverge. This saves the calculation time of the machine, but it has this drawback of showing the almost-convergent zones as black. The reason why they used this method is simply because this software was originally made for Julia and Mandelbrot fractals i think, in which it has been proven that if an iteration of a point goes higher than 2 in term of modulus, it is always a divergent point (for the Mandelbrot fractal at least). But it's not the case in this fractal, because a point can always go to 10 trillion but go back to the 1-4-2 cycle English is not my native language, ask me if i haven't been clear enough :)
@@germaincasse So is it not possible that these areas will eventually recede away leaving only the integer points as convergent as you increase the calculation limit? Or has it been demonstrated that certain non-integer points converge (like the pre-images of 0)?
I love this on so many levels! As a piece of math, it's very surprising and raises a lot of interesting questions - for example, you show how the number of "fingers" separating any integer from successive pre-images of zero gives its Collatz sequence... what about pre-images of the other fixed points? Do they show a similar pattern? As a piece of art, I love the eerie, almost-symmetrical biological look of the fractal; I've never seen one that looks like that before. I have to admit, though, I'm not exactly sure how you made the fractal. You mention that unlike the cosine fractal, the black areas don't represent convergent orbits under iteration... what do they represent, then?
In reality the fingers are all infinitely long, but it takes a lot of computing power to extend them. Every finger is mande out of smaller fingers, which are in turn made out of smaller fingers and so on. If you zoom in on a random point, it is certain, that you are eventually not going to be inside one of the increasingly tiny fingers. Thus almost no point is inside a truly black region
@@denyrawif you zoom in on the thick part of the fingers arent you always going to be in a black region? I think your explanation only happens in case of the perimeter of these finger like fractals isnt it so? And i also didnt understand like op what do thse black regions mean if they dont represent the converging points of the complex plane. The cosine func one i was able to understand
The cosine-based fractal from the alternative odd -> (3x+1)/2 is (in my opinion) prettier than the cosine-based fractal for the original rules; Part of the structure of the former even resembles a Mandelbrot set. Its iteration simplifies to z -> z - ((2z+1)cos(πz)-1)/4. For a non-Collatz but prettier still fractal, changing the rule to z -> z - ((2z+1)cos(πz)-7/6)/4. seems to hit a critical value, and the dark areas of the pseudo-Mandelbrot sets spring into life with further detail. The exponential-based fractals for the above aren't as nice as the above, or as neat as the exponential-based fractal for the original rules.
At 6:10 if the black areas are numbers where the iteration are divergent yet, what is differentiates them from the other areas? This is not standard way to create graphs of fractals like those used in rational functions. You off course can do that, but please explain the criteria used to color a point black.
I checked the shader he wrote for this on Shadertoy. When you zoom out you get a mostly flat plane at about y = -3, and the "finger" structures appear to go on to infinity in both directions. Interestingly, I found a very fine stripe pattern on the fractal near that -3 plane, but when I zoomed in I got blocks. I think the stripe pattern is just an artifact of precision limits, and not actually part of the fractal as it's defined in pure math. You can look at the shader here: www.shadertoy.com/view/lssfDs On line 50, you can adjust the scale variable, called 'sc' On line 51, you can adjust the graph's center, a 2D vector called 'ce' To disable the gridlines, change line 91 to '#if 0'
Great video! But is anyone else really confused? He presented the limit of a sequence as the formula for the fixed points, but the sequence definitely diverges (it has a term of 2n in it). And I'm not sure how the fact about the fractal representing the dynamics of the number under iteration of the Collatz formula is derived; why is the argument of f(z) approximately pi/2?
Inigo, could you please publish the first part of this video (with the cos function and the beautiful blue to brown color palette) to Shadertoy ? There's a lot of examples with the exp function but just one with the cos function on the site, and its color palette is not so good... plus the method seems to be different as yours, there's some artifacts in the example...
2:36 The formula is written wrong. K is acting a multiplier to 5n + 2, it's not taking 5n + 2 as input. So it should be written as k(n)(5n + 2). I was so confused until I figured that out.
i wonder if he did it on purpose to see if anyone would notice, hes clearly good at math, and thats kind of a dumb mistake for someone who probably speaks math and code as their second languages
N= positive odd number. N changes to (3N+1)/2 (3N+1)/2 could be: 1- (3N+1)/2 = positive odd integer 2- (3N+1)/2 = positive even integer 1- assume (3N+1)/2 = positive odd integer. Since N = positive odd integer, it could be a value for (3N+1)/2 So, (3N+1)/2 = N 3N + 1 = 2N 3N - 2N = -1 N = -1, which contradicts with N = positive odd integer. So, the assumption (3N+1)/2 = positive odd integer is false. 2- assume (3N+1)/2 = positive even integer. Since N + 1 = positive even integer, it could be a value for (3N+1)/2 So, (3N+1)/2 = N + 1 3N + 1 = 2N + 2 3N - 2N = 2 - 1 N = 1, which does not contradict with N = positive odd integer. So, the assumption (3N+1)/2 = positive even integer is true, and (3N+1)/2 will change to smaller value (3N+1)/4 < N, getting toward the destination 1. If N = 1, (3N+1)/4 will equal 1, which is a term within the destination loop 1 → 2 → 1. So, N = positive odd integer, just changes to (3N+1)/2 = positive even integer, which changes to (3N+1)/4 < N, getting toward the destination 1. So, Collatz conjecture is true. Eng. Mahmoud Attalla. WhatsApp: +20 1112669096.
What if instead of f(x) = [ (7x+2) - cos(pi*x)*(5x+2) ] / 4 you work on a log scale to get something like f(x) = sqrt[ (3x^2+x)/2 / (6+2/x)^cos(pi*x) ] For integer x, that still reduces to the Collatz map, but it generalizes slightly differently to other positive x. I bet it would still tend to diverge, but less quickly; however, do the dynamics change?
Cool. I have a bit of a problem with your function though. You're essentially just picking a function that happens to be equal to the Collatz function at integer values. So one of an infinite number of candidate functions that do this.
Caleb McNevin What's important to note is that he picked a function that is continuous and differentiable across the complex number plane. That narrows down the types of functions he can use and allows for complex "tweening" behavior to appear. I wouldn't be surprised if other globally continuous, differentiable functions that imitate the Collatz function on the natural numbers exhibited similar behavior as the function he chose.
If instead of ((7z+2)-k(5z+2))/4 you use ((4z+1)-k(2z+1))/4 then you can generalize 3z+1 to (3z+1)/2 because (among odd integers) 3n+1 is always even, so you always immediately divide it by 2. I would like to see what the difference is in that graph.
Have you resently changed the title? It reminded me of the video from Veritasium and I thought you might be experimenting with his other recent video about clickbaity titles. Your visualization is very beatiful. It's a shame he didn't include it.
Yes! It's an experiment, I want to see if his clickbait theory works, by becoming a parasite of his clickbait title. But, clearly, it hasn't. At all :) So I'll change it back to wait it was later this week.
Haha, ok :D Yeah, I think it takes a lot of intuition and luck. I first heard of your channel when you released the "selfie girl" video and was absolutely blown away. I was amazed by your intuition when it comes to math and how you think about it and then watched a lot of your other videos. I wish you and your channel all the best for the future :)
z^2 + c is not at all related to the function in the video. You're probably referring to the fractal shading, which is a technique that can be used to visualize any function in the complex plane.
Depends who defines it, it's a defintion after all. Engineers define "j" as the imaginary unit. Usually to avoid confusion with "i", which is defined as intensity. But the video should definitely have established the definition at the beginning to prevent confusion.
:O Those fractals are so beautiful, and the fact that the number of edges on each level of fingers describes the path 3 takes through the Collatz procedure? That's crazy!
oh hey, carykh, nice seeing u here!
I love seeing patterns like that encoded into fractals
Why are you everywhere bro? Take a break you're too smart as is
Yea wtf
It bro
These videos are so beautiful and insightful. Please keep making more of them!
Thanks!
I've always wanted to display this in LEGO, but do not have the funds or room. 38 LEGO studs = one foot, so showing numbers up to 38 would require at least three feet.
Sir, you have answered more questions than I came here looking for answers to, and for that, you have my thanks.
Really good video, I wasn't expecting the relationship between the number of fingers and the orbits of natural numbers!
Yeah, me neither at first. The joys of discovering these things is immense!
WOW that's absolutely amazing! The way the fractal's structure indexes the recursion is splendid.
Amazing job eliminating the usual "dryness" that comes with these kinds of abstract sequences
you are my hero Inigo! Thanks for putting this beautiful and instructive video together and sharing your craft.
So grateful I was able to rediscover this video, it’s a classic for me.
Wow, that was a fun visualizion.. both the original number line loops as well as the fractal complex extension. Knowing you, you probably wrote both visualizations with procedural shaders? Whether you did it or not I would love to see another video of the same length just describing the tools you made and/or used to make this video!
It is a procedural shader, I made the video in Shadertoy and added the static slides as textures to the shader. Pretty much.
You are pretty humble. I found that you are co-creator of Shadertoy.
What exactly does Shadertoy do?
@@ganondorfchampin (2)
This video made me like you instantly. I am amazed how you visualized the numbers. I am fascinated by this conjecture and glad to have found your video. Keep it up
That was really brilliant. I hope you're following through on this and other research. I'm looking forward to finding your other videos.
What a brilliant idea to extend the map into the complex field. Hopefully someone will one day use this new point of view to crack the collatz conjecture !
This is all we need, to make the Collatz Conjecture even more complicated. I love it!
Wow that's so insightful! I'm numberphile's (and maths', and fractals') fan, and I am totally amazed how that enigmatic Collatz conjecture turns out to be a beautiful fractal when expanded to complex numbers. Great work!
Informative and beautiful. A rare combination.
i never knew you had a UA-cam channel and this video happened to appear in my subscription box by chance
This is absolutely stunning work - well done!
awesome! Most videos about this topic are more about the nature of math and discovery, it’s rare to see one with actual new information to me!
Thank you so much for this. Thank you for demonstrating the beauty, complexity and difficulty of this problem.
Really fascinating and awesome. The fractal shape analysis encoding the dynamics is amazing, I haven't seen anything like it before. Thanks!
Wow! Great video, and great discoveries. I also had never seen the Collatz Conjecture expressed that way. Thank you for this great video!
This is so beautiful and interesting! It makes you want to understand and know much more Maths, thanks a lot!
Awesome! Can't wait until the next video. Regards from Canary Islands
VERY nice!! Such amazing information. It's amazing how math and art merge, creating this amazing beautiful images. Thanks for sharing!!
what i really wanted to see towards the end was you showing what you did, for other numbers than 3. Such as 9, because 9 does some pretty big jumps. It would be really cool to see the 28 fingers.
Now the yellow points, are those actual zeros to the function? If you put that complex point into the function and iterate it will land on 0?
Maybe its more profound to me because i haven't analyzed any of this in the way that you have. But its pretty F*cking amazing honestly that the collatz cojecture and the jumps a number will take is literally encoded at each number visually in the fractal. That is the coolest thing ever.
The visualization is slightly creepy. what exactly are the black areas? You should do a video on just that fractal alone and explain alot more of it, at an elementary level. Like the basics such as what is the black area. And then with deeper maths.
This problem no doubt has been analyzed at universities by mathematicians in the complex plane, but this could no doubt provide many valuable insights and angles of attack for others who haven't thought to try this.
It may very well be that proving that all natural numbers return to 1 may come from things that we could only prove by analysis in this manner.
This is so wonderful and it seems I'm not the only one who thinks that way. It's a great example for just having fun with maths and feeling the joy of realizing patterns.
This is gonna be the next great mathematician
Good work! I don't find surprises in number theory too often nowadays. New angle, and i'm a bit jealous to be honest
Very nice. What does the black area represent if not convergence?
It basically means "almost-divergent or divergent". Let me explain :
To render this image, we have to check the convergence of every point. But we face some ussues here : first, we want to render a lot of points. Secondly, we can't interate a point infinitely. So for these two reasons, we are setting an arbitrary limit : we assume that when an iteration of a point hits n (1 000 or 1 000 000 for instance), it will diverge. This saves the calculation time of the machine, but it has this drawback of showing the almost-convergent zones as black.
The reason why they used this method is simply because this software was originally made for Julia and Mandelbrot fractals i think, in which it has been proven that if an iteration of a point goes higher than 2 in term of modulus, it is always a divergent point (for the Mandelbrot fractal at least). But it's not the case in this fractal, because a point can always go to 10 trillion but go back to the 1-4-2 cycle
English is not my native language, ask me if i haven't been clear enough :)
@@germaincasse So is it not possible that these areas will eventually recede away leaving only the integer points as convergent as you increase the calculation limit? Or has it been demonstrated that certain non-integer points converge (like the pre-images of 0)?
@@patrickosullivan3887 with a higher calculation limit, these black areas would be smaller and smaller
Along with Math Kook, this is in my opinion one of the most interesting videos on Collatz for me. Thanks.
I love this on so many levels! As a piece of math, it's very surprising and raises a lot of interesting questions - for example, you show how the number of "fingers" separating any integer from successive pre-images of zero gives its Collatz sequence... what about pre-images of the other fixed points? Do they show a similar pattern? As a piece of art, I love the eerie, almost-symmetrical biological look of the fractal; I've never seen one that looks like that before.
I have to admit, though, I'm not exactly sure how you made the fractal. You mention that unlike the cosine fractal, the black areas don't represent convergent orbits under iteration... what do they represent, then?
In reality the fingers are all infinitely long, but it takes a lot of computing power to extend them. Every finger is mande out of smaller fingers, which are in turn made out of smaller fingers and so on. If you zoom in on a random point, it is certain, that you are eventually not going to be inside one of the increasingly tiny fingers. Thus almost no point is inside a truly black region
@@denyrawif you zoom in on the thick part of the fingers arent you always going to be in a black region? I think your explanation only happens in case of the perimeter of these finger like fractals isnt it so? And i also didnt understand like op what do thse black regions mean if they dont represent the converging points of the complex plane. The cosine func one i was able to understand
The cosine-based fractal from the alternative odd -> (3x+1)/2 is (in my opinion) prettier than the cosine-based fractal for the original rules; Part of the structure of the former even resembles a Mandelbrot set. Its iteration simplifies to z -> z - ((2z+1)cos(πz)-1)/4.
For a non-Collatz but prettier still fractal, changing the rule to z -> z - ((2z+1)cos(πz)-7/6)/4. seems to hit a critical value, and the dark areas of the pseudo-Mandelbrot sets spring into life with further detail.
The exponential-based fractals for the above aren't as nice as the above, or as neat as the exponential-based fractal for the original rules.
I would LOVE to see this visualized!
brilliant. congratulations.
very clear
Fantastic content! Thank you so much.
And great production quality, too 🤗
This is truly amazing. So amazing I feel like you made all this up.
I wish I was able to make something like this up.
really well known mathematicians need to see this this could possibly be used to prove the conjecture!
At 6:10 if the black areas are numbers where the iteration are divergent yet, what is differentiates them from the other areas? This is not standard way to create graphs of fractals like those used in rational functions. You off course can do that, but please explain the criteria used to color a point black.
How did I miss this? This is fantastic!
Very well done, impressive!
Awesome. Very interesting. Thank you.
I wonder how much surprises are hidden in that seemingly simple formula.
This is a heck of a great video.
Awesome! Brilliant explanation and insight.
absolutely incredible. wow.
Great video and summarized explanation!
That was so good. Thank you.
Is it self similar if you zoom out? Many fractals are not, but this gives the appearance that it might be.
I checked the shader he wrote for this on Shadertoy. When you zoom out you get a mostly flat plane at about y = -3, and the "finger" structures appear to go on to infinity in both directions. Interestingly, I found a very fine stripe pattern on the fractal near that -3 plane, but when I zoomed in I got blocks. I think the stripe pattern is just an artifact of precision limits, and not actually part of the fractal as it's defined in pure math.
You can look at the shader here: www.shadertoy.com/view/lssfDs
On line 50, you can adjust the scale variable, called 'sc'
On line 51, you can adjust the graph's center, a 2D vector called 'ce'
To disable the gridlines, change line 91 to '#if 0'
Great video and visuals!
Great video! But is anyone else really confused? He presented the limit of a sequence as the formula for the fixed points, but the sequence definitely diverges (it has a term of 2n in it). And I'm not sure how the fact about the fractal representing the dynamics of the number under iteration of the Collatz formula is derived; why is the argument of f(z) approximately pi/2?
Confused about that as well
Underrated video
great work. this is pretty imaginative. i didnt quite get that last property of the fractal though
I did enjoy the video ! Thank you for this nice video. :)
Very interesting and this visualization is new to me. Thank you very much for this!
Hi! Really cool video. Could you explain the anchoring points in more detail? Thats what I found confusing.
These are points in the plane for which the iterations produce a sequence of points (an "orbit" ) that lands at zero (which is a fixed point)
You have just inspired me to work on publishing the research I have done over the past years on the Collatz conjecture. Thank you.
Beautiful work - thank you
The genius, also known as the Shader magician, strikes back again !
Coolest video I saw in my life
Inigo, could you please publish the first part of this video (with the cos function and the beautiful blue to brown color palette) to Shadertoy ? There's a lot of examples with the exp function but just one with the cos function on the site, and its color palette is not so good... plus the method seems to be different as yours, there's some artifacts in the example...
What is the meaning of the colors for the exponential fractal, since you say that black does not mean it converges? What does black mean them?
peering into the chaos sure is captivating
Love your videos! So sad I’m only discovering it now!!!
Does the finger like structure have a name? I have seen that shape in other fractals, such as iterated tetration
Beautiful.
2:36
The formula is written wrong. K is acting a multiplier to 5n + 2, it's not taking 5n + 2 as input. So it should be written as k(n)(5n + 2). I was so confused until I figured that out.
no it is YOU who is causing confusion. that is but an insignificant, forgivable, technical syntax error.
i wonder if he did it on purpose to see if anyone would notice, hes clearly good at math, and thats kind of a dumb mistake for someone who probably speaks math and code as their second languages
chase marangu chill
@@chasemarangu ok boomer
@@non-inertialobserver946 I am not a boomer I am a Millenial.(2000) Or maybe I am a Gen-Z.
N= positive odd number.
N changes to (3N+1)/2
(3N+1)/2 could be:
1- (3N+1)/2 = positive odd integer
2- (3N+1)/2 = positive even integer
1- assume (3N+1)/2 = positive odd integer.
Since N = positive odd integer, it could be a value for (3N+1)/2
So, (3N+1)/2 = N
3N + 1 = 2N
3N - 2N = -1
N = -1, which contradicts with N = positive odd integer.
So, the assumption (3N+1)/2 = positive odd integer is false.
2- assume (3N+1)/2 = positive even integer.
Since N + 1 = positive even integer, it could be a value for (3N+1)/2
So, (3N+1)/2 = N + 1
3N + 1 = 2N + 2
3N - 2N = 2 - 1
N = 1, which does not contradict with N = positive odd integer.
So, the assumption (3N+1)/2 = positive even integer is true, and (3N+1)/2 will change to smaller value (3N+1)/4 < N, getting toward the destination 1.
If N = 1, (3N+1)/4 will equal 1, which is a term within the destination loop 1 → 2 → 1.
So, N = positive odd integer, just changes to (3N+1)/2 = positive even integer, which changes to (3N+1)/4 < N, getting toward the destination 1. So, Collatz conjecture is true.
Eng. Mahmoud Attalla.
WhatsApp: +20 1112669096.
What if instead of
f(x) = [ (7x+2) - cos(pi*x)*(5x+2) ] / 4
you work on a log scale to get something like
f(x) = sqrt[ (3x^2+x)/2 / (6+2/x)^cos(pi*x) ]
For integer x, that still reduces to the Collatz map, but it generalizes slightly differently to other positive x.
I bet it would still tend to diverge, but less quickly; however, do the dynamics change?
Brilliant. Thank you.
Cool. I have a bit of a problem with your function though. You're essentially just picking a function that happens to be equal to the Collatz function at integer values. So one of an infinite number of candidate functions that do this.
Caleb McNevin What's important to note is that he picked a function that is continuous and differentiable across the complex number plane. That narrows down the types of functions he can use and allows for complex "tweening" behavior to appear. I wouldn't be surprised if other globally continuous, differentiable functions that imitate the Collatz function on the natural numbers exhibited similar behavior as the function he chose.
the exponential function is obviously the most "natural" choice here and it's not even close
How do you render the shadertoy code in such high quality? I can do it but i can only do up to 360p.
If instead of ((7z+2)-k(5z+2))/4 you use ((4z+1)-k(2z+1))/4 then you can generalize 3z+1 to (3z+1)/2 because (among odd integers) 3n+1 is always even, so you always immediately divide it by 2. I would like to see what the difference is in that graph.
interesting.. can i ask u what program u use to make de first jumps on real line?
Have you resently changed the title? It reminded me of the video from Veritasium and I thought you might be experimenting with his other recent video about clickbaity titles.
Your visualization is very beatiful. It's a shame he didn't include it.
Yes! It's an experiment, I want to see if his clickbait theory works, by becoming a parasite of his clickbait title. But, clearly, it hasn't. At all :) So I'll change it back to wait it was later this week.
Haha, ok :D Yeah, I think it takes a lot of intuition and luck. I first heard of your channel when you released the "selfie girl" video and was absolutely blown away. I was amazed by your intuition when it comes to math and how you think about it and then watched a lot of your other videos.
I wish you and your channel all the best for the future :)
Sorry I’m late, but these are some awesome insights! Thank you!
What exactly do the graphs show? You say you colored the orbitals, but what exactly do you mean by that?
How did you come up with the formula extending the conjecture to the set of real numbers?
It was the most simple and natural extension I could think of.
thank you for this video :)
very nice fractal
how did you come up with the conversion of k(n) to f(n), it seems so unintuitive!
This is beautiful.
Beautiful!
Why can you use the 3n+1 and turn it into 7n+2? This isn't explained
I guess you "coded" these visualisations yourself or did you use any animation software?
I wanted to ask the same! I am so used to trying to decode Inigo's shaders that that's all I can think of here :D
No animation software. This is all a shader, I made it in Shadertoy ^__^
+Inigo Quilez Goddamnit. Of course! Hahaha. Became clear when you went to fractal stuff.
if the edge is infinite doesnt that mean there is no anti solution to the conjecture?
Is this kinda like the mandelbrot set? but instead of f = z^2 + c it's f = ((7x +2) - cos(pi*x) * (5x+2))/2 ?
It is not, because the Mandelbrot set is coloured by the number of iterations to diverge, this is coloured by a conformal map of some kind.
magicgonads.github.io/smooth.html#iter(((7z'%2B2)-exp(ipiz')(5z'%2B2))%2F4%2Cz%2C5)&z=9
z^2 + c is not at all related to the function in the video. You're probably referring to the fractal shading, which is a technique that can be used to visualize any function in the complex plane.
Excellent!
Very very veeery nice!!!
what is j in the exponential e^(j pi z)
The imaginary unit, the square root of -1
@@InigoQuilez That's defined as i
Depends who defines it, it's a defintion after all. Engineers define "j" as the imaginary unit. Usually to avoid confusion with "i", which is defined as intensity.
But the video should definitely have established the definition at the beginning to prevent confusion.
@@InigoQuilez True
but why did they define i as intensity if it already has a definition?
Because v is voltage, r is resistance, c is capacitance, ... so i is intensity. Just for convenience.
It should work on Clifford algebras, right ? I wonder how raycasting though 3d slice of 4d space would look like...
Next step ? (hint, hint) :)
KFC finger leaking good 5:48
I was hoping to see what happens once you zoom in to the 1-4-2 loop in the final fractal...
The Hattifatteners from the Moomins TV series. That's what came to my mind instantaneously.
Wait how did you colorize that second image?
Es una frikada extender a los complejos la conjetura.
What really brings a conjecture is when you apply rule ((2^n)-1))n+1 then there are INFINITELY many conjectures.
wow that's cool!
a simple cobweb plot also nicely visualizes the dynamics..
i.imgur.com/OU5VFhQ.png
or perhaps a 3D version..
i.imgur.com/DCfWusy.png
Amazing...
the collatz fractal seems like a close up of an infinitely powered mandelbrot.
we got the mandelbrot set, so now we have the collatz set
I kinda want to see how the negative side is different from the positive one now
so I guess it follows the whole 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 ->1 and then repeat the 1 - 4 - 2 cycle?
yes, exactly