X3 máis lexas, maybe we'll do one, two, three, this up to you, because I can explain many, many things, right? But at least this lecture I would like to finish the part of the technical calculation for the spectral density of grandos rengi graphs to get the sub point equation for this density of the deltas and then I'm going to show you a trick of how to solve this thing numerically by an algorithm called population dynamics. The main goal today is to obtain this equation for this w of the delta and then how to introduce the idea of something called population dynamics. This is a technique that appears in spin glasses to solve, you know, self-consistency equations of diluted spin glasses in replicacimetric answers or replicacimetric breaking, etc., etc., right? So, where were we? We were, again, calculating what? So we have the spectral density rho sub c of lambda over h over this order and we thought that this was equal to what? Minus 2 divided by pi divided by m, the imaginary part of the relative with respect to set of the limit of n going to 0 of 1 over n, the logarithm of the partition function and replicas average over the disorder. And set is equal to lambda minus i eta, right? And here somewhere I have to have this limit. Here I have the limit of eta going to 0 plus. An after a tedious derivation we saw once the member DC is connectivity matrices for the ensemble of connectivity for adjacency matrices related to for Erdos-Renegraff and the matrices are symmetric, equal to cgi, they take for all i different than j. Then for simplicity we take the diagonal elements of the matrix equal to 0 and remember that they take value of 0, 1. And the probability rule is the probability for two nodes ij being connected, that means the probability of cij equal to 1 is equal to d divided by n where d is the average connectivity. So after a not difficult but tedious calculation we arrived at the following note that the average over the partition function, average over the nth power of the partition function, this could be written as a path integral dp hat of the exponential of this two functions where s-pen of pp hat was equal to i times the integral dx sub bar p hat x bar plus d divided by two integral over dx dy dy times the exponential of the scalar product of x and y minus one plus the logarithm of the integral over respect to x of the exponential of minus z dy by two x square minus ip hat x where in this notation we have that x with the bar below is a vector x1 to xn x alpha belonging to r And you'd remember what was the goal of all these things we did. The goal is to arrive to a situation like this because this is very cool, that means that if I'm interested in the typical properties of very large matrices or what in physics we call the thermonetic limit this integral when n becomes very large goes like the exponential of n and then this function s evaluated at p0, p0 hat where p0 and p0 hat obey the saddle point equations. Well and then we end up in what? We end up in saying or putting that at the saddle point or for n when n is very large the average of the empirical spectral density can be written as follows, right? We can be written as the imaginary part of the limit and going to zero one divided by n of the integral with the limit going to zero, with the limit of eta going to zero plus, that I'm not going to write it anymore. And I use this formula to go to the second step of the replica method to explain how to make the analytical continuation of the interest to the real. And then I said ok, so now this would be step two of the replica method that I need to answer for this guy and the answer was that p0 of x in replica symmetric answers can be written as an integral over a parameter, complex parameter, real and imaginary part of some kind of density regarding this parameter of the product for alpha from one to n of the exponential of minus x alpha squared divided by two. Delta divided by the square root of two p delta. Yeah, we were like that there, right? And then I said well and then I argue why this thing should be consistent with this idea of replica symmetric answers and then I said if you plug this thing into here, then you obtain that the average of the empirical spectral density was equal to the imaginary part. Of the integral over delta omega delta delta. And we left it there, yes? So far so good, as a reminder. So what remained, that actually wrote down the questions about we and the IPs, what on earth or how on earth I determined this object here. And the way to determine is to remember that this order parameter function obeis a cyberpone equation and the idea is to transform that cyberpone equation into an equation for this omega of delta. And this from all the directions we have done maybe this is the most difficult one, but again it's not difficult, it's annoying. Ok, so let me show you the trick. Or let's do together the derivation to obtain a consistency or a close equation for this omega of delta. And once we get that equation, I'll show you how to solve it numerically by population dynamics. Ok? Very good, so let me write down again the two cyberpone equations that we have. So one cyberpone equation was that minus IP0x must be equal to what? 2 and P, the integral with respect to Y, P0y, exponential of the scalar part of X, Y minus 1. So this is correct, yes? And the second one is that P0x is equal to exponential of minus Z divided by 2 X square minus I P0 hat X divided by the integral with respect to Y of what the half in the enumerator. And then the idea is to use the leplica symmetric ansatz to get to transform this couple of equations into an equation of four, again this object here. So notice that I can combine the two equations so I can get rid of P hat by just simply plug in here P hat. Ok, so let me do that and what you get is the following. So I get that P0x is equal to exponential of minus Z divided by 2 X vector square plus D, the integral with respect to Y, P0y, exponential of the scalar part of X and Y minus 1. And for the time being, let me put the denominator as the denominator. We put here simply the denominator. And then I have to plug the ansatz into this expression here and here. And I have to try to convert this equation to an equation again for this omega of delta. Is it clear what we have to do? Is it clear? Well, so where is the trick? Because you look at this expression, it's very threatening somehow. It's just scaring you. Why? Because if I look on the right hand side, sorry, left hand side, I have that this is a convolution of Gaussians. So forget if you want this piece so this is simply a product of Gaussians. So here I have an exponential of Gaussian for this variable. But this variable X is here as a free variable in this part which is Gaussian. But this X is inside an exponential, inside an exponential. So in order to have to equate the left hand side and the right hand side, I must be able to somehow take out this X into the first exponential. That's the trick. And the way to do this is to simply do a Taylor expansion of this exponential. Of the exponential of the exponential. So let's do that. So let me take that. And for the moment I'm going to ignore the denominator. Let's focus on the numerator. So what I have, let me put the denominator, our numerator. This is equal to what? This is equal to the exponential of a minus set divided by 2 X vector squared. And then I have the exponential of minus d. That I take outside, exponential of minus d. And then I have, you see again, exponential of this order term. So I'm going to do a Taylor expansion of this exponential. So this would be equal to what? It would be equal to the series from k, from 0 to infinity, or 1 divided by k factorial. And then I have, let's do it step by step, d, the integral over y, p0, rs, y. The exponential of X squared product with y to the power k. It's clear what I've done. I've simply done a Taylor expansion of this exponential with argon being all this weird thing. Yeah? Clear? Exponential of what? I don't hear you, sorry, you have to speak up, man. It's e to the minus d because here there is d with a minus. What I have is, you see, I have this piece and this is normalized. Because this is a distribution, so I have here a minus d that I take out. So I have the exponential of minus d and then the exponential of this piece and the exponential of this piece I do a Taylor expansion. Better? Yeah. Món questions? Good. So then the other thing I have to do now is to rewrite this thing, right? So let us do that. So then I have that this is the exponential of minus c divided by 2, X vector square. And then I write this thing as well as sum for k from 0 to infinity of exponential of minus d, d to the k divided by k factorial. And then these are the same, it's the same integral k times, right? So what I do is I put an extra index to label those integrals and the other thing that remains equal in all these integrals is the value of X, ok? Because the integration variable is Y. So these I'll write it as follows. I'll write it as the integral for the product for L from 1 to k, d, y, L, p0, rs, y, L. You can do this thing piece by piece. Product L from 1 to k, sorry, p0, rs, y, L. And then I have here the exponential of X, the product with the sum for L from 1 to k, y, L. The derivatives? What do you mean? Yeah, ok. So what I have is a, so what I do is a exponential of something, I write, I use the Taylor expansion of r on A equals serial A, being this, yeah? Do you have more questions? Now what I do is to put the other sunsets here. So remember, let me delete here. Remember that there are sunsets for p nodes, integral over d, delta, omega, delta, the product of alpha from 1 to n of the exponential of minus X, alpha squared divided by 2 delta, divided by the square root of 2. Right, so I plug this thing into here. And then I have to be a bit careful with having the integrals and new integrals have to label them appropriately, etc., etc., right? So now I will have here k extra integrals, so this would be the following. I have here the exponential of minus z divided by 2, X squared, the sum k from 0 to infinity, exponential of minus d, d to the k, divided by k factorial. And then I have the following, I have the integral for the product of kl from 1 to k, d, delta, l, omega, delta, l. Right, this comes from this part here. And then I have the following. I have the product of alpha from 1 to n of what? Of the 1 to l, give me a second, the product of k of l from 1 to k, the product of alpha. No, this part is difficult. So let's do it step by step better. And I have here the product of l from 1 to k, d, y, l. And then I have here the product of alpha from 1 to n, the product of k of l from 1 to k of the exponential of minus y, square alpha l divided by 2 delta l, divided by the square root of 2 pi delta l. Right? And then I have this guy over here, exponential of the scalar product X, the sum l from 1 to k of y vector l. Let me see this thing from a distance. Very good. That makes sense. And again, so the idea is, I have to rewrite this thing in such a way that I will have a Gaussian measure for X. So after doing the integral over y, what remains has to be X to the square with something multiplied by something. There is one parent X, this one here. Yeah? Very good. So now, at replica indexes, where do we appear? Because they appear here as a product. Here they are hidden here. Also as a product I can put explicitly. And here in the scalar product. And they also appear here as well. So I do the following now. This is equal to something, is equal to the sum of l or k, sorry, from 0 to infinity of the exponential of minus d, d to the k divided by k factorial, the multiple integral for l from 1 to k of the product of d delta l omega delta l. And now, I do the following, times the product of alpha from 1 to n. Of what? The first part that has the replica index alpha with X is this one here. So I have then the exponential of minus z divided by 2 X alpha square. And then I have the integrals over the ys, no? Here I have now the integral half, sorry, the product with respect to the l from 1 to k. Of the integral of dy alpha l, the exponential of minus ys square alpha l divided by 2 delta l, divided by the square root of 2 by delta l. And then I have this factor here, no? Which I can put here, plus X alpha y alpha l. So again, the trick here is to put things in the right order. You'll see why I'm doing this. You'll see just in the following step. Clear so far? Now, this integral everybody should be able to do because it's a Gaussian integral. So this give us the following, the sum of k from 0 to infinity of exponential of minus e d to the k divided by k factorial. The integral, multiple integral for the deltas, d delta l omega delta l. And then I have here the following. The product alpha from 1 to n I have the exponential of minus z divided by 2 X alpha square. And this integral will give what? Well, it would be simply X alpha square multiplied by delta l and then the sum of l from 1 to k. With the one half, plus one half the sum over l from 1 to k of delta l and this multiplies X alpha square. Yeah? And if you want, I can put this thing together. Y alpha that comes from this scalar product. So this is a Howard Straton's transformation, simply the typical Gaussian integral. So let me put this thing in a compact way with a half is what? A half, one half, one half or alpha square half of z minus the sum of l from 1 to k of delta. Right? And that's the end of the numerator in the derivation. So far so good, I have not done anything fancy. Just annoying. Yeah? Now notice the following. So I want this thing. Now let's remember that I have the Saipan equations and this is the right hand side of the Saipan equations. It's actually the numerator. This must be equal or proportional to p0 rs at x and this remember that is equal to what? To the interval with respect to delta omega delta the product alpha from 1 to n of the exponential of minus X alpha square divided by 2 delta divided by the square root of 2 by delta. So it's actually proportional because for the moment I'm forgetting about the denominator. Clear? So I have to combine somehow so now I have to find a way to rewrite this guy in this way so I can identify with an equation this omega that appears here with this bunch of omegas that appear here. Again so what I have to do is to rewrite this part this result in such a way that resembles this one so I can identify left hand side with right hand side. So how can I do this thing with the beautiful magic of Dirac deltas? So the idea is the following. You mean Taylor's function? I understand the confusion. So I have an equation what is called a Saipan equation for p0 of x. So this is a function of... I can write just one equation. I have an equation for this function. And then based on an observation related to replicas or the symmetry of replicas I say this object has to have a certain form and this form depends on something that still needs to be determined which is this omega of delta. So what I'm trying to do is to rewrite this equation into an equation for omega of delta because this part is still undetermined. The only thing I did when I imposed replicasimetric answers is some form for this but there is an unknown function still too that has to obey the Saipan equation. Better? The rest is just massaging the expression. More questions? Ok, so let us use the magic of the Dirac delta in the following. So I continue in this here and I write. So this is equal. I introduce an integral delta of ok, let's do it maybe step by step. So this is equal to the sum of k from 0 to infinity exponential of minus b d to the k divided by k factorial of this multiple integral for the omegas product of L from 1 to k delta L omega delta L and then I introduce an integral over a delta or a variable I introduce of the product of alpha from 1 to n of exponential of minus x alpha squared divided by 2 delta divided by a square root of 2 p delta of a Dirac delta delta minus 1 divided by z minus the sum for L from 1 to k of delta L and there is a half a termism but I will put the term. What on earth am I doing? I am doing nothing in the sense that if I were to integrate over delta you know this delta is substituted here and I obtain back this expression right? I'm not doing anything however I'm actually doing something because I'm isolating the right pieces to go to the place I want to go so still this expression is not correct it's not exactly that expression over there because here I have a normalization factor that it doesn't appear here so I have to multiply by a a square root of 2 by delta to the power n ok? Another expression is unequality but remember since we are doing the replica limit when n goes to zero this term will not be important for this limit but in other cases it is important very good now what I can do is to say I can take this integral at the beginning I can put things in a different order and then I can identify with the left hand side so I can write this thing as follows I do first the integral over this delta and then I do this is equal to the sum k from zero to infinity exponential of minus delta delta to the k divided by k factorial of the multiple integral containing the product for L from 1 to k d delta L omega delta L I put this thing here Dirac delta of delta minus 1 divided by z minus the sum of L from 1 to k d delta I close the bracket if you want let me see I can put this factor here in front and then I have the product or actually let me put it like this no, like this one the product of alpha from 1 to n of the exponential of minus x alpha squared divided by 2 delta divided by the square root of 2 pi delta you see what I have done I simply rearrange the terms the differential of what? here I put it in the first place why I am doing this because on the left hand side of the Salmon equation the order parameter in RS has this form so now I know that this that comes from the derivation of my numerator must be equal or proportional to this so what I have here I see so I have this term here which is well so I have the differential of delta this one here which is this one here then I have the product for this Gaussian functions which are this one here and the left hand side must be equal to the right hand side because they must obey the Salmon equation so that means now you can proceed as you wish you put this thing on the other side equated to zero, you have an integral equated to zero, this means that the integral must be zero so therefore this must be to use the word equal must be equal to precisely this part times this now it's not equal because I'm forgetting the denominator now if the denominator you do the same trick I've done and then you integrate you have something to the power n so when you also include the denominator and you equate left hand side and right hand side of the equations what is the following you get that omega of delta is equal to this here you have something like and then you have something the denominator also to the power n of the sum k to infinity exponential of minus d d to the k divided by k factorial and then this multiple integral for L from 1 to k d delta L omega delta L and the direct delta of omega sorry delta minus 1 divided by z minus the sum 1 to k delta L no no because I you integrate over over landas this landa appears here instead yes it will depend but you have to still make the replica limit so whatever you have there when you send n going to 0 is 1 now you make the replica limit and this goes this gives you bless you this gives you 1 and this day now the self-consistency equation that you have for the omega of delta yeah so actually if you think about the difference between numerator and denominator is that in the denominator is equal to the numerator but you are integrating over the free variable so if you look at the direction we did this equivalent of I think I'm going to break it down again don't worry so if you look at with it in the denominator denominator is equal to this part here but I have to integrate now over x right when I integrate over x this gives 1 and then this integral ok disappears right this disappears because it's normalized and you have something to a given power that when n goes to 0 it gives you 1 yeah because if you look at the self-consistency equation since p0 must be normalized the denominator is the numerator integrated do you remember the self-consistency equation so the self-consistency equation in terms of p0 p0 of x was equal to right so when you do this derivation the denominator is exactly the same derivation but then you integrate over x no no no it's another thing this denominator is the denominator of the framework is the denominator left in the self-consistency equation because I was only analyzing the numerator yeah more questions ok so you have to do this derivation it's the only proper way to learn it ok but anyway so suppose that you do derivation or you believe me the next problem is how on earth I solve numerically this because suppose now I want to simply do a plot of the spectral density for Poissonian graphs so what do I have at this point I have that within a replica symmetric ansatz I have that the spectral density is equal to taking the imaginary part of the the delta of omega delta delta where the omega of delta obeis this self-consistency equation omega of delta is equal to k from 0 to infinity exponential of minus d d to the k divided by k factorial integral of the product of L from 1 to k d delta L omega delta L and then I direct delta of delta minus divided by set minus the sum of L from 1 to k delta L so the question is now what can I solve this in exactly y any note so then how I evaluate this numerically because the idea would be the following so in principle remember that set is equal to lambda minus a eta and then I want to plot the spectral density for certain value of d for certain value of the hours connectivity so fixing the value suppose I take a small value of eta and I fix the value of lambda I have to put it here I have to solve this equation somehow and when I have the solution for this omega delta I have to do this where you have to remember that delta is a complex number so let's take that the real part of delta let's call it a the imaginary part of delta let's call it b so when I put here like omega of delta is a density of two variables a and b and then the spectral density is simply the integral with respect to the the second the spectral density would be actually in this notation would be this would be the integral over dA dB omega AB B which is what you were asking the order what was delta so delta is a complex number so therefore omega of delta is a density of for two variables what? no idea en cancel it so there were different denominators as far as I recall and what happened is like then you have to take the limit n going to zero so you have something to the power n in the denominator and something to the power n in the denominator so both terms go to zero sorry go to one where n goes to zero so they are not cancelled they disappear in the limit delta of delta sorry because this is real so this is a density of the two independent variables of the complex numbers ok and this you are taking the imaginary part of delta which is B so I am doing the double integral of A and B of this density with respect to B so delta is A plus IB so therefore if I put this thing here the imaginary part is the imaginary part of delta is B ah ok so again this notation what this yeah this what is this notation what it means is dA dB omega so this just compare the notation because otherwise you know the derivation would be a a bit more annoying which I understand that is annoying enough that maybe one should be more careful so this you understand it or the proper way to write it from the beginning should have been the product of L from 1 to K of dA L dB L omega A L B L and then you take here and this direct delta means you have to take the real delta for the real and the imaginary part of this so you have to take this object split it into the real and the imaginary part and this is you have to direct delta 1, 4 the relationship between real to real and the other one imaginary to imaginary better? sorry what is your doubt no please no it's not an integral over the complex plane this is notation that's why it's not a contour integral yeah more questions ok so let's go to the ok so again so I want to plot a spectral density of this so you see the problem the problem is fixing an eta that actually in this case you can take the limit to 0 explicitly yeah speaking and there is no problem I have to fix the value of lambda I put it here I have to solve this somehow and then I have to do this integral then how do I do this well so this kind of an integral equation which is very annoying no so it's very difficult to see what to do with this to solve it so the way to do it is by realizing that all the pieces that appear here can be understood in a probabilistic manner and you can use some kind of a updating that keeps this probabilistic interpretation of any of these pieces what is called population dynamics so what is the idea let me delete this part so the idea is to use what is called the fixed point iteration method you know what the fixed point iteration method is right fixed point iteration method is you have a fixed point equation and you solve it by doing the fixed point iteration method if x's point fixed point equation let's remind this fix point is a point é un qüesión de este tipo. É? O que é? O que é o que é? Não, é que foi complexo, é que o método de cavité é complexo. Ok. Eu entendo, eu entendo que eu uso a mesma notación para significar diferentes coisas e é o meu erro. Ok. Então, é que... Desculpe-me clarificar. É complexo normal, é que significa contúr integral, o que é? Talvez eu should have written something like this. Ok. Mas o que é isso significa? O que é o início? Sim. Sim, não, mas é que isso é só a notación compacta para representar isso. Ok. Eu entendo. O problema é que se eu coloco a real e a parte imaginaria a derivación se torna um pouco mais anónimo. Sim, sim. É o que eu doo. Ok. E também para anónimo. Ok. Mais questões? Sim, é a densidade. Sim. É algo... Desculpe-me. Eu não menciono, mas eu menciono a nossa última lectura. Em as réplicas e metricas ansatz, é a densidade, no sentido da densidade real, a distribución da parametra que aparece na distribución gaussian. Sim. É a función real, que tem dois argumentos. O que é a parte real da delta e o que é a parte imaginaria da delta. O que é real e que tem o sentido da densidade. Ok. E se você ola na dispersión da parametra da orda na arre, a parametra da orda é a distribución, então therefore a densidade deve ser positiva definitiva e deve ser... e é normal. Porque a parametra da orda é normal. Se esse objeto non é a densidade, eu não posso usar a tricón, eu vou usar, não faz sentido. É uma boa questão. Dê-me. Que delta? Delta ou delta capital? Delta capital delta é simplemente um número complexo que aparece na santa arre da orda da parametra. É o que é. Ok? Então lembra que na santa arre da orda da parametra eu escrevo assim que é a exponencia da x alfa squared dividida por 2 delta dividida por squared root of... Na realidade eu deveria ter escrito isto, mas isto é anónimo, não é? É igual ao inter sobre da db omega a db product alfa 1 e a exponencia e eu quero curar o meu already, ok? Então isto é 2a plus i b dividida. Blá, blá, blá, blá, e senão isso deve ser anónimo, tenho que dizer que senão isso é un número complexo dê-me cuidado porque o real par deve ser todo o que eu imagino. Sim, é a variante. É a variante complexa. Dê-me. Este é o... O que? É, o... O que? Não, a solución para o sálpão. É, o sálpão é a mesura estacionaria que é... Sim, mas eu me lembro que eu chamo as probabilidades porque estou usando... Eu estou seguindo o vocabulario de mecánics, mas estas não é a probabilidades porque estas são complexas blá, blá, blá, blá isto não é a funcione de partículas eu não quero. Ok, eu não quero because in any part, in no part of this derivation, I have to use probability identities or certain properties of probabilities to the derivation, no way. In the only place that is going to be important is here, and here I know that P is a normalized quantity and therefore this density. This is the only place where the idea comes for probabilities are important, not in the rest of the derivation. More questions? Tell me. And now I'm going to show you a very cool trick, it was introduced by George, no it was introduced by an article by a German author that I forget, I apologize, and then George Athen popularized it in diluted spin glasses, which is called population dynamics. Ok, and then for RS is very easy, and then if you have one step of the application of the breaking, this is the result in the population dynamics called survey propagation in certain limit for the temperature, etc. But I'm going to show you a very cool trick to solve this, it's very cool. And if you understand this trick, then you can do, you can then apply to study the case of, you know, the statistics of the number of eigenvalues in a given interval. More questions? Ok, so let me go back to this, this is a small reminder, I'm pretty sure that you know, ok? So a fixed point is a point x that obeis an equation of this sort, and it's anything in life or in mathematics, you find you want to introduce ways to solve equations of this sort, there is something called fixed point iteration method. Fix point is fixed point iteration method, that of course is an numerical way of approximating a fixed point to a given fixed point equation, ok? So suppose, fixed point iteration method consists in the following. Suppose I have a guess, let me call the fixed point, I call the fixed point p, it's going to be my fixed point. It's the point that solves this equation. So suppose I have a guess for a fixed point, let me call this thing p0, so this is a guess for the fixed point. What's up? It's introduced as sum with the function of Dirac delta, I'm telling that this p, p of x must be 1 divided by n, the sum over there, yes. But these are empirical spectral densities, so this still has the property of being a density, no, a distribution. One, I'm not sure what you mean by imaginary. Ok, continue, I think I know what you want to ask, but go ahead. So first of all, when you say that the Dirac delta cannot have imaginary parts, I'm not sure what you are referring to. If you refer to the argument of a Dirac delta, it can have imaginary parts. If you understand, if the argument of a Dirac delta is a complex number, and you understand that the Dirac delta applies to the imaginary part. And for the thing you mentioned before, so let me see. So this p of x came from what? Came first for defining this, right? So here there is no issue with, yeah? And then, what? Ah, yeah, and then this goes to the derivation and then we go, we get a side point and an integral over, a path integral over the p and p hat, yeah?  And you say that in which part it says that it's imaginary. No, no, no, no, no. In p0 there is no imaginary part, but in the way you parameterize p0, there is one parameter that can be imaginary. This delta, no? OK. Sorry, so maybe we'll ask later on those two more questions. OK, so what is the idea of the fixed point iteration method? So you have a fixed point, you have I guess for a fixed point p0, e sen xfxp, poi xfxpmxx, xfxgps yjgider e xfx. Phil Premium notito sono pairs fil, xfxane x嫁 x 제� ni destroyed norma to no como linn mining say antenna randou slammed xfx truck solidaron que a sequencia converges e significa que o limite do infinito de Pn é algo, o que se chama P tilde.  Clico? O método fixe-point é muy coo, o método é muito coo porque any root finding problem can be transformed into a fixe-point problem and fixe-point problems in numerical analysis are much easier to treat than root finding problems. Any way. Sorry? This process, you start with a guess of a fixe-point, whatever, I don't know, code comes and give it to you. And then you use the guess to generate P1, then P1 to generate P2 and then you generate a sequence of piece. Now, if the limit of the sequence converges, this limit must be equal to a fixe-point. Why? Because if I take this rule that generated the sequence and I make the limit, the limit of n going to infinity of Pn, this would be P tilde, this must be equal to the limit n going to infinity of G Pn minus 1, assuming that the function is continuous, etc, etc, this equal to G of P tilde. So P tilde is equal to G of P tilde, so therefore P tilde must be P, which is a fixe-point. Good? Are you with me? Ok, so I am going to use this stupid thing and the observation that things that appear here can be understood, different pieces can be understood as densities, to solve this thing numerically. Cool? Ok, so I do the following. Let me delete this thing, this is something that you learned in kindergarten. You see, so take that omega of delta is a density. We already discussed what it means, omega of delta is omega of A and B, a density of two variables. Now, suppose that, now forget about this problem. Suppose that you have a private distribution and you have a machine that produce random numbers according to that distribution. And then you have a collection of random numbers according to this distribution. Let's say that this collection is this one. So I have a collection of random numbers, L from 1 to N, that were generated according to this density. Right? In the sense that, if I have a collection of random numbers that are generated with this density, it is clear that I can construct a histogram, or an estimator for the density that would be what? It would be simply one over my collection of variables of the sum of A, L from 1 to N of a direct delta of delta minus delta lambda. This delta has nothing to do with other deltas, this something separate. Ok? So I have a density, whatever, and then there is a machine that produces random numbers according to this density. So therefore I can construct the histogram or an estimator for this density in the sense that if I take N going to infinity, here I should recover back to this density. So let's put this estimator in such a way that the limit of N goes to infinity of omega axon n of delta, this would be omega of delta. So far so good. I have not said anything complicated yet. Now, the idea is the following now. How I am going to use, how I am going to solve this self-consistency equation for the omegas. What I am going to do is, instead of taking this density and parameterizing it with parameters and maybe close these equations for those parameters, I am going to represent this density with a population of random variables in such a way that if I were to do the histogram, I would recover that density. So the density would be, I don't know, maybe do a Fourier transform and try to close those equations for an infinite number of parameters. But what I do is I first represent, this would be zero, I represent omega of delta with a collection of variables that if I were to construct the histogram, I would recover this. Ok, very well and in the limit of the population going to infinity precisely you will get that density. Now, when I look at this equation, what do I see? What do I see is that if I take from this expression one element of the population that represents this density, this equation is telling you, ok, this population of random variables that represent this density must be updated according to the following. Draw first a random number according to a Poissonian distribution with mean value of d. Let's say that this random number is k. Right? Go to my population that represents the density and pick k numbers completely randomly, right? Calculate this object, take another number of the population randomly and update this value according to this. Ok, so what I do is the following. So first I represent the density with a population of random variables. Now, I start with some kind of initial, I can generate this sequence randomly. Ok, and then what I do is the following. Now, I generate a Poisson number, a Poisson random number. Suppose that it's k to follow this notation, say k, and the Poisson random number has mean value of d. Ok, because this is the parameter of the Poisson distribution with mean value. These integrals seem some integrated over densities and the densities are represented by a collection of random variables. Ok, integrating over these densities is equivalent of taking one element with the same probability one over n in each of the integrals. That means to take one k elements randomly from the population. So the next step two would be pick up k elements from this population, randomly, that would be uniformly randomly, uniformly. Suppose that these elements are the following. So the other goal is delta L1 up to delta Lk. Then pick another one, pick an extra one, extra element. Let's call it delta prime or let's call it, I don't know, delta, I don't know, L tilde and do the following, do the following substitution. Do the following replacement. Delta L prime must be substituted by one divided by z minus the sum. Ok. And then you repeat. And then you go to step one. And you repeat this process any times until you go to each element of the population and you update it according to this what is called a probabilistic rule. And at the end of the day this population will have a profile which is this profile, this density that would be the solution to this equation. So the numerical solution of this equation. Ok. So this is called population dynamics algorithm. A step four, you see, this is telling you that this element, so this delta, that would be one element of this density, or this delta, which is the argument of this density that according to this representation would be an element of this population must be updated by the value given on the right hand side. Yeah. Yes. So this is also a famous population dynamics algorithm. And you can do this thing and it converges very fast actually. Tell me. Yes. Yes, yes. And what happens is the following, right? So the reason, so first you draw a person number according to that has mean value d. Let's say this value is k. So this part is going to select one of the possible weights that will appear in this interval. Because you have an infinite, here you have an infinite sum of integrals. Each integral has a weight to appear. And this weight is a Poissonian weight. So what happens is like, ok, so from here this infinite number of integrals and then the number of integrations in each integral is growing, you know, will have a weight according to a Poisson distribution. So first I draw a number according to a Poisson distribution. So first I draw a number according to a Poisson distribution. So first I draw a number according to a Poisson distribution. Let's say that this is k. Right? Yes. Because I'm using fixed point iteration. I understand this. I'm solving this using fixed point iteration method. And the fact that each of these pieces in this equation has the meaning of density. And I understand the integrals in the sense of a Monte Carlo integral. In the sense of a Monte Carlo integral. In the sense of a Monte Carlo integral. So let me put this thing in the sense of a fixed point problem. Right? So this fixed point equation is an unusual fixed point equation because it's given in an integral way. But now I want to solve it iteratively by using fixed point iteration method. So I start with some guess of the density. This guess is omega zero delta. I put it here and then I get omega one delta. Then omega one delta. I put it here. Omega two delta etc. So the first thing you have to do is to understand, to realize or to give up your hope of solving this in any other possible way by using fixed point iteration method because you don't know anything else. Right? So first I put this thing as a fixed point iteration method and it's weird because it's an integral equation. You know? The second step is to say I could, since these are densities I could parameterize them with a certain number of parameters. For instance you could say I'm going to do a Gaussian approximation. You could do that. And then you write down equations for the Gaussian parameters that will result from this equation. Yeah? Or you can say instead of approximating this density by some functional form, let's say Gaussian or whatever what I'm going to do is to approximate the density by a population of random numbers in such a way that if I were to do the histogram this would represent the density. Yeah? So then you start at time zero well at step zero in your algorithm which is the initial condition you start with some initial values let's put it that way some initial values for this collection of random variables that would represent your initial case at T equals zero. Right? And then what this guy the left, sorry, the right hand side what is telling you is the following because here again you have an infinite number of some of interas. You know this is a pluralistic weight. It's telling you that from a more possible infinite number of interas that here appear you know they are coupled to a Poissonian weight and then the result of this integral is conditioned to that value of the Poissonian weight. So first you draw a random number according to a Poisson distribution and this will allow you then to estimate the integral coupled to that weight of the Poissonian distribution. Yes, yeah, yeah, yeah. Yeah? At step one I generate a Poisson random number with parameter D and let's say this is K, let's say first time is two then is seven, then is six five etc etc fixed K I go here and I interpret this thing as a Monte Carlo integral where what I'm doing is parametraxing these densities with their population so the way to understand now how this is used to integrate the integral which is a Dirac delta is to say well means you know this density is given in terms of a population the probability that one element of the density appears is taking one of the elements uniformly randomly from the population with probability one over n so you have to take them uniformly uniformly. How many of them? Well, in this condition now for a given value of the Poisson number you have to take K, yeah? Once you have taken K uniformly randomly you calculate this object you calculate one divided by set for this K delta that you got from the population and then what is left is to decide what to do with this number this number here so for this you realize that this number is equal to delta inside the delta and this delta is the argument of the density on the other side but the density is parameterized by its population so the probability that one value appears in the population is one over n because the population is represented by collection of random numbers and the probability of appearing is one over n so you take one you take one element in the population uniformly and randomly you take one element that would be step 3 by the value you have calculated and then you repeat and then you repeat and then you repeat you go from 0 to 1, 3, 4 etc etc and at the end of the day in principle this will converge to a population whose profile is a solution to the this self-consistency equation now it's a distribution so the idea of a fixed point when we learn it we learn it for a real number but it's any kind of mathematical object that obeys an equation of that sort and of course you can apply it here uniformly randomly and then you substitute that value for the value you calculated with the other elements you choose from the population and what well and now you have to know with this thing you have to be careful but it must be big enough so that it would represent well a density and if happens that the population that you have is smaller than the value of d maybe you take replacement or something like this you have to be a bit careful with this but normally for the numerics you do for an average connectivity let's say 5, 7, 10 1000, 10, 1000 and this is more than enough so you are not going to have that problem more questions the ones I chose so I chose for this one I chose K randomly and this could be L1 and L2 and LK just to be simple and this would be as opposite I couldn't put here delta 1, delta K because you could misunderstand that this would be the first element of the population and I am taking K elements of the population randomly the first one is L1 L1 goes from 1 to N up to LK that goes from 1 to N tell me you can use the same criteria that you use in the fixed point iteration method which are called stopping criteria you can use any of them applied now not to a real number but to a density for instance what you could do is to take the expectation value the expectation value of this of some object that relate to this one and see how the expectation value changes when you are using this algorithm and as soon as it reaches a stationary point you are happy any kind of stopping criteria well you have to adapt the stopping criteria from real numbers to densities but the idea is the same that is what sorry this interest is a contraction in the sense of fixed point that this thing must be a contraction otherwise there would not be a fixed point sure but this is that stays in the part that when I assume that the limit well no no it's not only the limit it's that the limit exists and you have to have certain properties of the mapping in this case I don't know how to prove mathematically where this map is a contraction or no just for the cases I've seen it always converges so what happens in spin glasses it doesn't converge but it's understood why also this is the annoying part you fix a value of lambda so you fix a value of lambda then you solve this by population dynamics then you get the spectral density for that value of lambda then you have a point in the spectral density have you seen before this kind of population dynamics algorithm I guess you have not seen it sorry youtube can I leave you an exercise because many of you have studied you have gone to lectures for spin glasses so at some point maybe you should have seen you have diluted the spin glasses for instance the vianabri model or even the ferromagnet on Poissonian graphs so if you go to the first example I used to do the replica method which was the IC model on Erdos-Renyi graphs you can do the same thing I've done here do there also the RS ansatz and what you obtain is the following you have also a density in this case it's a density of cavity fields and the density has to obey the following equation no k from 0 to infinity exponential of minus d d to the k divided by k factorial the integral for L from 1 to k dhl omega hl and then you have the direct delta of h minus the sum L from 1 to k of u hl where just this function u is the function u that appear in the cavity equations for that case that was the propagating field or message right this would be the result for the IC model on Poissonian graphs and now the magnetization of this model would be given in terms of the following would be given by the integral hyperbolic tangent of beta h omega of h so you see so what happens remember what is this u this u of x y it was 1 over beta the hyperbolic tangent of the hyperbolic tangent of beta x that is the hyperbolic tangent of beta y so I leave you this thing as an exercise it's a very cool numerics to do use a population dynamics in this case now this is a real number it's not a density of two variables you apply the same trick and then you get a solution numerical solution for this for a gain value of beta you plug it into here and then you can plot the magnetization as a function of t and you should obtain a critical value you see that actually you can derive exactly its value using bifolgación analysis but that's a different story and if you are into Monte Carlo simulations and things like that compare the result of the magnetization you obtain here with Monte Carlo simulations questions tell me yes no no I can continue doing ok this would finish this part of the spectral density but we can continue with the lectures we can discuss more mapping how do you watch how does it look that's interesting that depends on the connectivity of the graph and normally the connectivity has two parts one continuous part and one discrete part and that's actually very interesting and at the beginning it was maybe it was not understood but it's very trivial so depending on the value of d ok when d goes to infinity when the average connectivity goes to infinity but you have to be careful how to rescale something in the in the system when d goes to infinity you obtain a spectral density that looks like this it's a semicircle this is called the Wigner's semicircular law right this would be for d going to infinity and even you can make the limit properly, mathematically and you should obtain this law but you have to rescale something so that this limit is done in a non-trivial manner so you obtain the Wigner's semicircular law for values smaller than infinity or for small values of d let's say d equal to 3 4, 5 small values what you obtain and actually you can do this thing you can generate Poisonian graphs numerically and you can they analyze them and you can construct the histogram you would obtain a continuous part whatever but it has to be symmetric and you can prove that it has to be symmetric but inside the continuous part you have direct deltas also symmetric and these direct deltas they have a weight related with the topology of the network and how connected or disconnected you have different components of the network right like for instance you know you have a direct delta at zero with a given weight and this weight can be related with certain having a graph that can be you know this composed into two giant components connected to each other and also can be related to the fact that you may have isolated nodes why because if you have isolated nodes what is going to happen suppose that this is small remember that this is the average connectivity so it might happen that when you draw a Poisonian graph randomly it might happen that a node is not connected do you agree with me? so suppose that this would happen with more polarity when this is smaller and smaller suppose that you know you have a poor node here and then you have the rest of the graph I'm going to put it here so this part is connected and this node is alone by itself so that means that in the that's it that this node is not number one right this is node one the rest so that means that in the connectivity matrix so this row is all zeros because poor node number one is not connected to with anything else that means if I were to construct to calculate again values of this have to do one minus landa the identity matrix so then a landa will appear here and then here landas for the rest of the graph so that means that there is an eigenvalue which is zero so the weight here is not the only contribution one part of the weight of this drag delta is due to isolated nodes good now you might have situations like following now I don't remember the whole the whole so what you have now so let me see you have a graph and instead of having one not isolated you have a couple of nodes connected but isolated with the rest so that means you can deonize these two pieces apart and this will give a particular contribution okay I don't think to the zero but I think you know to other eigenvalues and you may have then starts okay this connected to the rest of the graph and this gives an extra contribution etc etc okay so any small part of the network which is not connected to a yajen component will contribute to the different ways of the drag delta and in some cases some of these conditions are you can calculate them exactly yeah more questions? the yajen component the giant component sorry for my broken English more questions? don't remember or I didn't write no explanation that it was like h of something like the h i is equal to that comes from the cavity equations no from the replica method in the replica so I use again the easy model for a ferlamagnet on Poissonian graphs as an example of the replica method we did the step one but we didn't do step two but now you know how to do step two so I think it's better that you do it in that case more questions? before we go give me just one second and then you decide if you want to have more lectures because now I can talk about other mappings so remember that in this case when you take the replica limit the replica is n going to zero and there are some factors that disappear in some other mappings in random matrices that are objects which are much more interesting that require to take the replica limit going to an imaginary number depends on how you do the mapping this in this consistency equation for the omega of delta will appear as something like this so you have omega of delta would be proportional then you have the Poisson distribution that would be this always appears for Poissonian graphs of course and then you have this multiple integral x1 then you have a direct delta that tells you that this delta here and this delta are connected somehow and now here if you don't take the replica limit going to zero to something else you have here something like this an exponential of a function that depends on these delas which is annoying because this is not doesn't have now in principle interpretation o maybe just if you manage to understand that this can be understood in somehow as a weight then you can generalize population dynamics to something that is called weighted population dynamics and then you can solve much more cooler problems in random matrix so I let you to think how you would if you would have something like this how you generalize population dynamics to account for this extra term that's it questions coffee


Hey everybody, welcome to the Waldoch Way. I'm Jessica. And today's video is going to be about our consonant boxes. So I actually made our first set of consonant boxes when Emily was in kindergarten, I think, maybe the summer of preschool into kindergarten. And I have a blog post that details everything that was in those original consonant boxes. So I will link that in the description box down below in case your kiddos are younger and you're thinking, oh my gosh, I can't put all of that in my consonant boxes. They'll swallow it or they'll put it in their mouth or they won't, you know, keep up with it. And at that point, most of what was in our consonant boxes was books, those little safari LTD like tube figures, and some laminated maps. Like it wasn't a ton of stuff. It was just enough that we would pull the boxes out and she kind of knew like what animals and what landmarks belonged on what continent. Like that was kind of the gist of it. I never had any rhythm or rules or structure as to how those were ever used. We just used them. We utilized them when we felt like it. And it was a really great way to introduce her to the different continents and what was in them. Well, since we were getting ready to start traveling the world, I decided it was time to up my consonant box game because our consonant boxes have been sitting for about six months at least unused because they were just too childish for Emily. So we are aging them up a little bit and I have mostly new stuff in them. And I want to share that with you today. So I am going to turn you around and open all of our consonant boxes and show you what is in them. But first, I wanted to let you know that all of the links in what you're gonna see today will be in a blog post that will also be in the description box down below. It'll be an updated consonant box post. And the way we're planning to use these is because traveling world follows or what we're gonna do is because all of the continents or all the countries are grouped into a continent is I'm gonna specifically pull that continent box when we're studying the countries there. And I'm just gonna kind of pull it to the front, leave it out and let her have full access to it. Now she'll have access to all of the boxes at any point in time. So she has free reign to do anything she wants with them. But I really want to make a point to kind of like shrew, if you will, the continent that we're going to be spending time in while we're spending time there. So it'll be a really great kind of introduction to it before we dive a little deeper. It'll also be a way for her to become more familiar with the other countries in that continent that maybe we're not specifically studying at that time. Before I show you guys the boxes, know that there is going to be a blog post linked in the description that will have all of the details for what is in each of these boxes. So you can go there for all of the links. Now, before I show you what's in the boxes, I wanna show you the box itself. These boxes are from treasures from Jennifer and they are absolutely amazing. Now they were my splurge item. I will link the blog post to our original constant boxes in which I used just a plastic box and they worked fine. We had them for over three years. There was nothing really wrong with them. It was just time that we upgraded and I had the ability to be able to finally purchase the box that have been kind of my dream box for ever. And I will show you why. Each of these boxes is well made wood. Like they are definitely quality. They each have the continent and the name on the lid in the corresponding Montessori color. They also have the name and the continent on the ends of the boxes. So if you have them lined up on a shelf, you can see them easily. And then under the lid, they have a full map with the continent in that color where it belongs. So the quality of them is amazing. Now I'm gonna show you inside two of the seven boxes, mainly because a lot of the items repeat and because I'm gonna link everything in that blog post. I'm just gonna show you inside Africa and Antarctica because Antarctica is a tiny bit different than the others and you'll see why in just a minute. So inside each of the boxes, I have the set from Hatching Curiosity that we absolutely love and that includes the cloth maps for each of the seven continents with the landmark pins. So you can either just lay these on the map or you can actually pin them to the map itself. And again, we have these for each continent. The control for the pin, so this shows you where each pin should be laid. And I love that she accents everyone with the corresponding Montessori color too so that everything in this box matches. If for some reason everything was to get mixed up, we would know that all of the green went back in the green Africa box. And then we also have the animal and landmark cards from Hatching Curiosity as well. They are beautiful. The animal cards include the area they live in, what animal classification they are, and then what they eat. And you can see the color even extends to these. And then the landmark cards have a fact on the back as well as some thinking questions. And again, has the continent and color. So you have tons of different resources for that continent. I also have the geo puzzle for Africa in here. And that is from this geo puzzle box. And then the last thing I have in each of the boxes is two books. Now this one, Explore Africa, the Bobby Kalman books we had in our old continent boxes and we still really enjoyed them so we moved them over. And we used to have a rookie read as well but Emily's outgrown the rookie read so we purchased the true books as an upgrade and they are more for third to fifth grade. So that is what is in the Africa box. Now what is different in the Antarctica box is because there is not a geo puzzle for Antarctica, what I did was use the world puzzle. So in this box it has the world in it because Antarctica is obviously pictured in the world puzzle and because I wanted a place to put this puzzle but it still has all of the other amazing things that the other boxes have, that is just the only difference is that's actually the world puzzle. So we still have that beautiful cloth map from Hatching Curiosity, the landmark and master here, all of those beautiful cards and the two books.