I wrote a bunch of papers on infinite (that is iterated) electrical networks, where the same types of issues come up. You can think of some of these issues as boundary values at infinity (an electrical network being a discrete analogue of a PDE.)
I started picking random examples of the repeated quadratic to do in my head, when the process started feeling familiar. That's when I realized that by studying this iterated integral, we've accidentally ended up at a Julia set.
@@gamerpedia1535 A Julia set is the set of all points that exhibit chaotic behaviour under function iteration with respect to the other points in it's neighborhood. The actual function can be pretty much whatever you want.
That's cool. Did you notice that this is related to the Mandelbrot Set, at least its restriction to the real line? Once you define the functions a_n, the analysis of the simple convergence of the sequence a_n (n>=0) can be seen as studying the dynamics of z -> z² + c for c = -1. Furthermore, once you've found fixed points of z -> z² - 1 as you did, you can check whether these are attractive or repulsive fixed points. There are many results and questions that arise from this kind of study. I had never realized that the stacking integral could lead to such questions. Thank you for that!
So to give a definite value z to an integral stack of f(x) with lower bound a, we would need the function g(b)=F(b)-F(a) to have a "global fixpoint" z in the sense that repeatedly applying g converges to z starting from any real starting number b.
Nah mate I find that boring. I'd rather start with an intuitive approach and build up from it. Which is of course why the channel distinguishes itself from everyone else on RUclips 😎 (I just unlocked the humility skin 🤣). How are you doing mate?
Why is the 2nd method more rigorous than the 1st one? It doesn't seem obvious to me that one can say the problem as originally defined is equal to taking the limit of the sequence of the 2nd method, because in a vague sense, I consider that the way we approach a limit matters, meaning the result may not be the same if we approach the result from a limit in that "direction", compared to some other way to define the limit. I didn't understand if there is actually an answer. Does that integral have a unique result then? It can't have multiple, as you said and I understand.
Yeah, the video end felt abrupt, the motivation of the second method was "we got two results with the first method, but it cannot have two results, so what does it really converge to?". But then the second method doesn't solve that question either.
sqrt(2+sqrt(3+sqrt(5+... is the limit of the truncated expressions' values, if it exists. The integral towers needs you to insert a variable every time you truncate. You also have to integrate to turn the constant function into a real number.
If you start at 1.618, the next number is 1.617924; it's moving away from φ. Whereas if you start at -0.618, the next number is -0.618076; it's moving toward -1/φ. So the answer is -1/φ.
I originated this my first semester at USM. I got it from reading Chungwu's paper. I was supposed to go to the PhD program at Michigan but I didn't make it there.
I’m a little intrigued, but also disappointed. This definitely is something that seems interesting to explore further, but it’s lacking a proper conclusion. It definitely makes for a cool video and title to get people in the door, but if you are just getting into the door then you won’t know how to use the room properly, metaphorically speaking, if I go to a pool and I spend my whole year going into the pool, but I didn’t know that there was a hot tub, then maybe I would have used the pool building more during the winter. I think showing a few examples of fixed points and some eventual fixed points is cool, but I feel it needs a bit more analysis on the exact form of each fixed point and each oscillation point and so forth. Like you mention |t| < phi, but the only things you really mention in that section are examples: 0, phi_bar, -phi_bar, but you never fully conclude the behavior of each interval or point within this range and it is a little underwhelming. I absolutely love the videos, but I was hoping for something a bit deeper.
the first method feels what you imagine at first and then go "but WAT?", then the second method it would be kinda hard to figure out a value to start testing and finding out the convergence point could be very hard depending on t. BUT! you did the first method where you weirdely found phi, which gives you a hint on what to try for the 2nd method. The 2nd without the 1st feels kinda out of nothing.
Well we don't exactly need the first method. We can analysis the problem completely in accordance with the proper method but that would've been incredibly boring.
Isn't the domain of your I(t) function < 1, inf >. because the upper bound of an integral can not be less than the lower bound? So I(0) and I(-phi) are actually undefined?
The problem with these limits, as shown by the second method, is that they depend on what you choose as the "last" integral. In your case, getting 1=1 means that any point is a fixed point, there are no restrictions, the integral could converge to anything. Assuming you chose 0 as the lower limit of integration, the series would go like this: t, t, t, t, t, ... So a_n(t) = t for any n.
I thought you will say the answer is phi as phi conjugate (as you named it) is a negative number (exactly equal to 1-phi) and an integration is always positive, so you would have cancelled out the negative answer and say phi is the answer
Computability problem: is the integral computable? If it's not, then the reason you can't compute it is that it has an incomputable bound. But isn't it computable as long as the bound is? Yes, it is. So the computability of the intergral is irrelevant.
Cool edit: can be easily solved mentally and, o no, again the golden ratio... please revive lemniscate constant again xdd (amazing video as always just kiddin)
what an insane way to write phi! 2:07 HAHA I THOUGHT THE SAME THING ABOUT CALLING IT PHI BAR! 2:28 can't we cancel one of the solutions? that at least one of the solutions should be impossible? I see, so we can't attribute one value to I, rather discuss it as a convergence of a function depending on the value we input. Interesting way to analyze the problem!
Holy cow. If you have any sort of control over how many ads are inserted into your videos, pleaseeee turn it down. Your explanation was interrupted on me 6 times. 😢
I wrote a bunch of papers on infinite (that is iterated) electrical networks, where the same types of issues come up. You can think of some of these issues as boundary values at infinity (an electrical network being a discrete analogue of a PDE.)
Would you mind sharing a link if they are public papers ?
I started picking random examples of the repeated quadratic to do in my head, when the process started feeling familiar. That's when I realized that by studying this iterated integral, we've accidentally ended up at a Julia set.
A Julia set is z²+c, or more aptly z_(n+1) = (z_n)² + c
While close, this is not actually a Julia set.
nice pfp
@@RichConnerGMNlmaooo
@@gamerpedia1535 I don't see why it wouldn't be, if we take z_0=t and c=-1.
@@gamerpedia1535 A Julia set is the set of all points that exhibit chaotic behaviour under function iteration with respect to the other points in it's neighborhood. The actual function can be pretty much whatever you want.
That's cool. Did you notice that this is related to the Mandelbrot Set, at least its restriction to the real line? Once you define the functions a_n, the analysis of the simple convergence of the sequence a_n (n>=0) can be seen as studying the dynamics of z -> z² + c for c = -1. Furthermore, once you've found fixed points of z -> z² - 1 as you did, you can check whether these are attractive or repulsive fixed points. There are many results and questions that arise from this kind of study. I had never realized that the stacking integral could lead to such questions. Thank you for that!
Love this, I wonder what would happen for functions other than 2x 🤔
This is literally my first time seeing the solve of these type of integral ,didn't know they even existed wow
So to give a definite value z to an integral stack of f(x) with lower bound a, we would need the function g(b)=F(b)-F(a) to have a "global fixpoint" z in the sense that repeatedly applying g converges to z starting from any real starting number b.
Method 2 starts to show what's going on, but still missing that Michael Penn level of rigor to prove convergence.
Nah mate I find that boring. I'd rather start with an intuitive approach and build up from it. Which is of course why the channel distinguishes itself from everyone else on RUclips 😎 (I just unlocked the humility skin 🤣). How are you doing mate?
@@maths_505 are you a graduate student by chance?
Why is the 2nd method more rigorous than the 1st one? It doesn't seem obvious to me that one can say the problem as originally defined is equal to taking the limit of the sequence of the 2nd method, because in a vague sense, I consider that the way we approach a limit matters, meaning the result may not be the same if we approach the result from a limit in that "direction", compared to some other way to define the limit.
I didn't understand if there is actually an answer. Does that integral have a unique result then? It can't have multiple, as you said and I understand.
just presenting things. Uniqueness wasn't the main concern
Yeah, the video end felt abrupt, the motivation of the second method was "we got two results with the first method, but it cannot have two results, so what does it really converge to?". But then the second method doesn't solve that question either.
The "extra examples" you showed at 10:10 aren't actually new: +-sqrt(1+phi)=+-phi, so the exact same value already found.
bro used clickbait on a math videos title and actually held up to it oh my god
this actually was insane thank you for making this video >:0
sir, the stack is not infinte, at least in my system it's only 8 megabytes.
sqrt(2+sqrt(3+sqrt(5+... is the limit of the truncated expressions' values, if it exists. The integral towers needs you to insert a variable every time you truncate. You also have to integrate to turn the constant function into a real number.
If you start at 1.618, the next number is 1.617924; it's moving away from φ. Whereas if you start at -0.618, the next number is -0.618076; it's moving toward -1/φ. So the answer is -1/φ.
Very good Analysis. I'm proud to be a subscriber for this channel.
That's one of the most heartwarming comments I've ever read. Thank you so much professor, it really means alot.
The title should be: The GOLDEN story of the infinite stack integral
That was so cool! I got interested into a comment here that wondered what would happen for other functions other than 2x. Lets see where that leads
An interesting way to get a Julia set.
So it doesn't converge? Im uncertain of what the answer was meant to be.
Yes, strictly speaking, it doesn't converge
DELICIOUS !!!! SCRUMPTIOUS THOUGHTS ARE ENTERING MY BRAIN MEATS !!!!! I LOVE IT !!!
Great vid . What software u used 4 ur vids
I originated this my first semester at USM. I got it from reading Chungwu's paper. I was supposed to go to the PhD program at Michigan but I didn't make it there.
What was wrong with the first method?
I’m a little intrigued, but also disappointed. This definitely is something that seems interesting to explore further, but it’s lacking a proper conclusion. It definitely makes for a cool video and title to get people in the door, but if you are just getting into the door then you won’t know how to use the room properly, metaphorically speaking, if I go to a pool and I spend my whole year going into the pool, but I didn’t know that there was a hot tub, then maybe I would have used the pool building more during the winter. I think showing a few examples of fixed points and some eventual fixed points is cool, but I feel it needs a bit more analysis on the exact form of each fixed point and each oscillation point and so forth. Like you mention |t| < phi, but the only things you really mention in that section are examples: 0, phi_bar, -phi_bar, but you never fully conclude the behavior of each interval or point within this range and it is a little underwhelming. I absolutely love the videos, but I was hoping for something a bit deeper.
the first method feels what you imagine at first and then go "but WAT?", then the second method it would be kinda hard to figure out a value to start testing and finding out the convergence point could be very hard depending on t. BUT! you did the first method where you weirdely found phi, which gives you a hint on what to try for the 2nd method.
The 2nd without the 1st feels kinda out of nothing.
Well we don't exactly need the first method. We can analysis the problem completely in accordance with the proper method but that would've been incredibly boring.
@@maths_505 that's my point the first problem justify you nailing the fixed points on the second, so skipping the boring part is fine.
So what does the integral equal to then?
Unfortunately it doesn't converge. However when reframed as a series problem, it converges for specific starting values of the parameter.
which whiteboard are u using?
Is there some things we can prove with the Sequence u_0=0 and u_{n+1}=(n+1)^(-u_n) ?
I do not like why phi pops up every time I deal with infinite nested functions.
wouldn't +/-sqrt(1+phi)=+/-sqrt(phi^2)=+/-phi?
However, +/-sqrt(1/phi+1) is a new value that is eventually fixed as is +/-sqrt(1/phi)
You should try the integral from -inf to +inf of ((x^2-x+pi)/(x^4-x^2+1))^2 dx
Isn't the domain of your I(t) function < 1, inf >. because the upper bound of an integral can not be less than the lower bound? So I(0) and I(-phi) are actually undefined?
Ofcourse it can be less than the upper bound. That would just result in a negative sign when we flip the limits.
Damn bro I was doing this the other day. I was trying it with just dx instead of 2xdx. I just got 1=1.
Any idea what that means?
The problem with these limits, as shown by the second method, is that they depend on what you choose as the "last" integral.
In your case, getting 1=1 means that any point is a fixed point, there are no restrictions, the integral could converge to anything.
Assuming you chose 0 as the lower limit of integration, the series would go like this: t, t, t, t, t, ... So a_n(t) = t for any n.
@@kikones34 nvm i just resaw and got it. Thanks for the explanation
10:16 --- t = √(1+φ) = ±φ, cuz φ² = φ+1
You could have at least given credit to Dr. Barker for the question
I didn't like his solution development so I added the missing elements.
why do your think, that this integral is actually equal to lim(I_n(t))?
Nope
Wait, but sqrt(1+phi)=phi? That doesn't add new solution
I=[x^2](1...I)=I^2-1...I^2-I-1=0...I=(1+√5)/2
Can you make a respective vidéo about the intégrale from 0 to 1 for x^x
Already made one bro
So the solutions should be contained in the Julia Set for c=-1+0i. However the set also contains points that aren't solutions like 0+0i
At 1:31 I think you mistakenly wrote -1 instead of 1
Nope
Isn’t there a little bit of begging the question going on here? “Suppose this expression equals a number - then it equals a number.”
Strictly speaking the integral doesn't converge as I indicated in the first part of the video.
im taking real analysis next semester!
I thought you will say the answer is phi as phi conjugate (as you named it) is a negative number (exactly equal to 1-phi) and an integration is always positive, so you would have cancelled out the negative answer and say phi is the answer
Always positive????
@@maths_505 wait my bad i confused with something else i think
Didn't watch the solution, didn't read the comments, liked the video for a good measure. Golden ratio is the answer.
Well....it's not exactly the answer.
@@maths_505 Fo some reason I thought you can't integrate backwards. I guess you can though.
I was hoping the mandelbrot face would show its face 🥸
I got 1/2 as my answer
its something plus c
I wanna know the asymptotic expansion of this sequence😂
Computability problem: is the integral computable? If it's not, then the reason you can't compute it is that it has an incomputable bound. But isn't it computable as long as the bound is? Yes, it is. So the computability of the intergral is irrelevant.
Cool
edit: can be easily solved mentally and, o no, again the golden ratio... please revive lemniscate constant again xdd (amazing video as always just kiddin)
This format is a JEE examiner's pet😅.
what an insane way to write phi!
2:07 HAHA I THOUGHT THE SAME THING ABOUT CALLING IT PHI BAR!
2:28 can't we cancel one of the solutions? that at least one of the solutions should be impossible?
I see, so we can't attribute one value to I, rather discuss it as a convergence of a function depending on the value we input. Interesting way to analyze the problem!
Im a ninja🥷
Holy cow. If you have any sort of control over how many ads are inserted into your videos, pleaseeee turn it down. Your explanation was interrupted on me 6 times. 😢
NICE
The question was never really answered.
Useless.
Love everything about this, BUT PLEASE call it like y'all English speakers "fee" and not "py" it's so annoying
☝️🤓
amazin