Thank you so much for the introduction and thank you to the three organizers for the invitation. I'm very happy to be speaking here on February 2nd or 3rd because I think it's February 3rd somewhere in the world right now. What I'm going to talk about is joint work with Vivian Cooper. Although when I finished preparing my slides, I realized that the first half is really introducing the topic. Okay, so the object of the study today is the K device or function which is defined as a decay on N, N positive integer is the number of ways of writing N as a product of K factors. Now we can see, so this decay device or function shows up as an numerator of the K power of the Riemann-Zeta function. Let me see. Okay, just a minute. I want to, okay. Okay, so as I was saying, the decay device or function shows up as an numerator of the K power of the Riemann-Zeta function. And so one motivation for studying is, is obviously if you are interested in the Riemann-Zeta function, but also it's a natural function to consider especially the case K equals 2. So the K equals 2 is just the number of divisors of N, right? The number of positive divisors. And so I made the first two slides with pictures so that to kind of introduce the topic. So here's the first one. So if we consider the device or function and we want to study its behavior, well, one thing we can do is to try to kind of graph it or look at what happens and you can see it's pretty chaotic, right? So it's perhaps not a very good idea to just look at values of it. So what a natural thing to do then is to try to consider some average that maybe can give us a better idea of how it behaves. And so this is what actually did I say did. And so here I'm talking about D, which is D2, okay, from K equals 2 from the previous slide. So if you look at the sum of the N for N less or equal than X divided by X, okay, so the average, you can approximate this with log of X plus 2 gamma minus 1, where gamma is the Euler-Macheron constant. And, okay, so I promised another picture, okay? So if you want to compare these things, actually you can graph them and it's fun. So here's on the left hand side, so it's a graph of this average of D up to, with X up to 250. And on the right hand side, I just graph this log of X plus 2 gamma minus 1. And if you put the two graphs together, the two graphs together, you can see that they are pretty close to each other. So this is fun. Obviously, if we want to understand this better, okay, why stop there, we want to understand the error term. Okay, so now let's go back to the general case of the K-device or function. So if we try to estimate this sum of the KN for N less or equal than X, we can, for example, use Perron's formula and that will give a main term given by some residue of the K-power of the set of function and some error term. And this main term, actually you can compute it and is given by X times a polynomial of degree K minus 1 evaluated in log of X. And then this error term that is supposed to be, it should be smaller. And in fact, one can see that it's smaller. So it's a smaller power of X.  Okay, X to the 1 half minus 1 over 2K, but this is actually hard to prove. So an approach to study this is again, try to do some average, okay, but this error term could be positive or negative. So an average that makes sense to make is to try to take some sort of, so this, the average of the square, so the, which in other words is the variance. Okay, so if we square it, then we take into account, I mean, we forget about this, the minus signs that could cancel. And so we take this average and then it is known that this has the right asymptotic size for K equals 2 due to Cramer, 4K equals 3 due to Tongue, and then 4K greater or equal than 4 under the Riemann hypothesis due to Tongue. So like I said, this is sort of the expected size that this error term, okay, so this is twice because I'm averaging the square, but this is the expected size that this error term should have. Now, what I wanted to talk about today are variants of this question. So the first variance that people normally study has to do with distribution in intervals, and in particular, ensuring intervals. Okay, so what do I mean by that is that instead of taking the sum over all the n, this or equal to x, I take the sum on n moving in some interval between x and x plus h, and the fun thing is when h is small. So this is what is hard and what is interesting. Okay, so and then the error term, well, we are interested in this difference. Now, in this context, the variance was studied by various people. So I just mentioned names here, and I give you I give you an statement in that is written a more general statement because it's in the context of a conjecture in the next slide. So there are results of Leicester that 4K that are under our age 4K greater or equal than 3 for relatively wide intervals.  So this would be considered a relatively wide interval. There are results for K equals 2 that are like for for small intervals by Utila, Coppola, and Salerno, Ibi, Leicester, and Yesha. There are also results with small intervals that are like upper bounds. So it's some upper bounds. So it's not an asymptotic of the error term by Milinovic and Ternache, Butterwell. So to summarize something that we need to keep in mind is that the interest will be short intervals. And so we are interested in h less than x to the 1 minus 1 over k. Now, in this regime, a few years ago, a Keating-Rodger-Rodger de Gershon and Rudnick made a conjecture. So here we take h to be x to the delta and this delta exponent is strictly less than 1 minus 1 over k. And the conjecture is that as x goes to infinity, the asymptotics of these variants is given by the following. So it's a constant, an arithmetic constant given by this formula. So this is an Euler product. So over all the primes, and this term, but also there is some sum actually coming from the gamma function. Then there is this pk of delta term. So that's a piecewise polynomial function evaluated in delta. So delta is my exponent of x of degree k square minus 1. So what it means is that this is going to be a continuous function that is defined by pieces in intervals. And whenever it is defined, it's defined by a polynomial that has degree k square minus 1. And then, so those are coefficients and then there is h, or x to the delta if you want, and then log of x to the k square minus 1. Let me tell you a bit more about this pk of delta. So this pk of delta more precisely is obtained by, okay, so the polynomial function originally, okay, I explain this why it's written in this way. So there's some inversion of the variable, but this is the gamma k is a polynomial function, this piecewise polynomial function, and it actually shows up as a very interesting integral. So this is an integral of the delta. So this delta c here is the delta distribution and I apologize that there are two deltas here that have nothing to do with each other, okay, but this is how the notation is coming. And so this will indicate, okay, so we do, this is a multivariable integral in this simplex, okay, so there are k variables and each of them moves between 0 and 1. And we take the delta, so they take the sum of the variables when it's equal to c, okay, so that we do 1 and 0 otherwise, and there is some random bounded type factor there. Now, okay, so it turns out that this integral is actually this piecewise polynomial function. So for example, to give you a clear example, here's a case k equals 2, okay, so you get there a polynomial of degree k square minus 1, in this case it's 3, okay, and so you get a certain polynomial in the interval between 0 and 1, and another polynomial in the interval 1, 2, outside that is 0, and it's continuous, right, because if you plug c equals 1, you are going to get the same value in both places. Now, let me talk about the other problem. So the other typical problem that one may consider when talking about this k-device function has to do with the sum over arithmetic progressions, okay, and in this context, again, there are various results. And, okay, I just wanted to mention results on the variance by Motohashi, Blomer, Lohanshaw, Kowalski, Rikota, and also in the same paper, Keating, Rothschild, Roddy, Tegershon, and Rudnik, formulated conjecture. So again, this is for the variance, this a k is the same a k that I have in the previous slide, the gamma k is the same gamma k that I've been talking about, okay, and again, we have, okay, so the gamma k now is being evaluated in log of x over log of q, so q is the modulus of my arithmetic progression, and we are summing up to x. Okay, now the question is how they come up with these conjectures, and the answer is, they come up with these conjectures by studying the analogous problem over function fields. Okay, so what's the analogous problem over function fields? Well, if we have f, a monic polynomial over with coefficients over fq, q is an odd prime power, okay, q is not a prime in this talk. Then we consider the k of f defined analogously as over the monic polynomials as the number of ways of writing f as a product of k monic polynomial. And again, this dk of f shows up as the numerator of the set of function over fqt, so if we take the set of function over fqt, which is given by the sum of 1 over absolute value of f to ds, with f going over all the monic polynomials, then when we take the k power, we get that. And here before I go on, let me just mention that the absolute value of f here is q to the degree of f, so it's the number of elements when you take the residue field of fqt by the idea generated by f. Okay, so to give a bit more context, before presenting the results that they prove and our results, et cetera, I want to talk a bit about L-functions, our function fields. So a main object that we use all the time in this context is Dirichlet L-functions. So let me remind people a Dirichlet character is somehow defined over some question. So we have fqt, question by an idea generated by dt, say ds, a polynomial. So the character chi, when we define on the multiplicative group, they are, it gives you complex values. And you can extend it to fqt by periodicity. And with the condition that chi of a is zero when a is not compliant to d. Then the Dirichlet L-function associated to chi will be defined as the sum of chi of f over f over absolute value of f to ds. And here, yes, sorry, the notation m means monic. I was trying to prevent this notation, to delete this notation from my size, but I forgot here. So I just want to say the sum is over f monic, just like in the definition of the Riemann-Zeta function over fqt. Now, okay, so this is one way of writing the Dirichlet L-function. However, we can organize this a bit better because the absolute value of f only depends on the degree of f. So you could do the sum by degree. Okay, so just from zero to infinity. So the absolute value of f would be q to the n. So you get one over q to the ns. And then inside, so you get a sum, like an inner sum of chi f over all the f that had a harmonic of degree n. Now, this inner sum, okay, so when the degree of f is bigger than the degree of d, this is a full sum on the character. So if the character is not trivial, this will give you zero. And so what happens is that this term becomes zero when n is sufficiently large. Okay, so at the end of the day, my L-function that was defined as an infinite series is actually a finite sum. Okay. Well, this is one elementary way of seeing this, but it's actually a manifestation of the big conjecture. So this is one of the, yes, so this is one of the consequences of the big conjecture. Now, the next slide, okay, before I go on to the next scene, I just wanted to say we sometimes it's convenient to make this change of variables where we call q to the minus s, we call it u.  And then the L-function becomes a polynomial. Okay, so it's convenient to think of this digital function as a polynomial. Okay, so when we think of it as a polynomial, then, okay, so here I'm writing some lies, this is not precise at all, but I just wanted to give an idea. So it will be a polynomial of degree, essentially the degree of d. Okay, remember d was the modulus or, well, the conductor of my character. This is not, I'm not being precise here, but say if you want an example that is more concrete, if we take a quadratic character, then like the ones I'm going to use in my results, then to be concrete, we will take d to be a polynomial of degree to g plus one, and then the L-function will have degree to g. And g will correspond to the genus of the hyper elliptic curve defined by y square equals d. But anyway, so the picture given by the vague constructions is that this polynomial, we can write it as a product of linear factors where the reciprocal of the roots, this thing that I call pi j of chi, will have absolute value q to the one half, the square root of q. And so in this context, one can think of this product as a characteristic polynomial of some matrix that after normalization, we can view it as a unitary matrix. Okay, so this theta of chi. And so this to give you an idea that this, so make these models that show up in this context. And so, and this is part of a bigger, so this is part of what is known as the Katz and Sarnab philosophy. And the Katz and Sarnab philosophy says that the statistics for the zeros and families of L-function should follow the distribution laws of classical random matrices. And I wrote here as d goes to infinity. So what I have in mind is as the genus goes to infinity. Now, some, some results that I'm going to use, take a different approach, so a different limit, and they, they take q going to infinity. So instead of d going to infinity, q going to infinity. And so they're, so these are typically, so this a key distribution results, they have a shape, they have the following shape. So we have a function f defined, so it's a c-value central continuous function on some matrix group. Okay, so what I mean by that is a function that is defined on controversy classes in this matrix group. And the matrix group, you have to think of it as unitary or some group like symplectic, orthogonal, etc. And then the results, so it will be determined by, by the family of L-functions. So my family of L-functions, I'm going to call it in, in this very general context, I'm going to call it fd. And then the, the, the general result is that when q goes to infinity, the average of f over the elements in fd will approach the integral of fudu over my, my group. Okay, so this is very a general thing. Okay, so it's not precise, but this is a general statement that then for some specific cases, there will be a theorem that we will use. Okay, so now I'm going to go back to this question that Keating, Keating, Rogers, Rolyte Gershon, and Rudnik were looking at. And basically, okay, so the first question has to do with model, the, this aberration short intervals. And so a way to take a short interval over the function fields has to do with you fix a polynomial, okay, of degrees say n, and then you take all the polynomials that are close to this polynomial closer than certain distance. Okay, so in this case, say closer than q to the h. So what it means here, if you think about the definition of the norm, means that the highest coefficients, they coincide. Okay, so up to degree, okay, anything higher than degree h, they will have the same coefficient as a, and then the smaller coefficients can be, are free to be anything. Okay, and in this context, the rest is that, okay, so the average can be computed exactly. And if we give the variance by these, okay, just, you know, the return with the average and sum of absolute value square, then the result is that the variance can be written as an integral over the group of unitary matrices with respect to the harm measure. And the integral is of the following. So the integral is of this sum. So we take the sum of products of what are called the security coefficients. These coefficients here, they are just the coefficients of a certain way of writing the characteristic polynomial. So if we take the characteristic polynomial by taking one plus xu, then those are the coefficients. And then we take the sum on some homogeneous weight n, and we take the absolute value square of the sum, and we integrate that, okay, times q to the h, okay, times the size of the interval. The general strategy for proving such a result actually is using even characteristic characters to pick up actually to restrict to the interval. I'm not going to go into the details on how this is done, but basically, one can relate, so one can recover the sum of the kf by considering twisted sums by the Dirichlet characters. And so we take the k power, we get the sum with the k, but twisted by a Dirichlet character. And then playing with that, well, I mean, I shall say playing, but it's difficult sums, but anyway, one can approximate the variance, one can approximate with this sum, and then use some a solution result that like the general result that I mentioned to turn this sum into an actual integral over all the unitary matrices. Let me also mention that for the other version of the problem, the other problems, so the arithmetic progressions, there is also a result that is very similar, okay, due to the same authors, where again, basically, they take the variance and using similar methods is not exactly the same story, because now for arithmetic progressions, they take odd characters, but again, using characters, they recover this type of result. Now, this doesn't answer the question of how they get the conjectures, because, okay, so you can see more or less the shape, but I mean, one has to understand this integral. And so they actually work with this integral a lot. So the general integral that one has to understand is the sum of secure coefficients to the square over the unitary matrices, and they do, they take various approaches to understand this. So one thing that can be done is to relate to the count of lattice points in z to the k square power lying in certain convex polytop ones also to extract this gamma kc that I mentioned before, okay, so what's the relationship between this and gamma kc? Well, this has to do with the main coefficient of the asymptotic. So the size of this integral is like n to the k square minus one, and this gamma kc that I keep popping up in all these results is basically the coefficient of this. So we take c n, small n over capital N, and basically that's what we get, and this gamma kc is what I like I said before, this piece why continue polynomial function given by this integral. Now, our project started out of a working group at AIM with various people. We wanted to, so that everybody was sort of interested in this gamma kc coefficient, and the main question was, okay, so let's study a symplectic version of it. And then the question that Cooper, Verano and I started with was, okay, so can we get, let's go back and let's try to get the problem that gives you a symplectic distribution. So let's go backwards, okay, can we get symplectic distribution? So the reason, I mean, the both results that I presented by Keaton, Roger, Roddy, Tegershon, and Rudnik, they are in relation with unitary distributions. And so they come from different sides, so one is even primitive characters, the other is all primitive characters. If we want to get symplectic, the natural set of characters would be quadratic characters. And so our first thing was quadratic character with conductor D, and D you take monic and square free. The problem with this is that it's a bit hard to detect squares with this. I mean, if this not prime, okay, detecting squares is not so easy. And so we decided to, okay, so let's restrict to primes. Okay, so what about we study the sum of dkf over f monic of degree n with f congruent to square mod p? And so this is what we do, okay. And so to be concrete here, we are going to sub p prime in the context of function field means that p is a monic reducer polynomial. And I'm going to fix here degree to g plus one. This is so that I really have a concrete set that I can relay to the, to the modular space corresponding to the curves, the hyperactive curves of genus G. And then the result that we have is that while we can compute the, the average, and then if we compute the variance, then we get, we get a very similar integral to the ones I presented before, but now I'm integrating over the symplectic matrices of dimension 2g. Okay, well, again, I repeat the p has degree 2g plus one. Okay, here maybe I should mention I put a star here because the variance is not really I'm taking, I'm not taking the difference with the average, I'm taking the difference with the main term of the average. So this is why I put the star. Okay, so the ingredients in the proof, okay, as I mentioned, the idea is that while we use this, the quadratic character, this Legendre symbol to the text square module of p, and we use an equidivision results of cuts. Now, the original plan was to use the whole family of hyperelectric curves. Here, however, when we restricted to primes, we actually have a question about this, because then you're looking into y squared equals p of x. And so it wasn't clear that this was symplectic, and then actually we went and asked cuts about it. And yeah, and he thought for some time for a few hours, and then he had a good answer about it. So it turns out that if we fix certain factorization type, any factorization type within the square free polynomials of degree to g plus one, we actually get symplectic. So basically, we consider this set to be the set of a square free, sorry, I forgot to mention this H means a square free, the square free polynomials of degree to g plus one, that have certain factorization time meaning that they can be written as a product of money polynomial that have degree to di. Okay, so we prescribe the eyes. And here, when I write this, I don't mean to say that the FDI are irreducible. Okay, so they could be, you could factorize more. Okay, they could be even thinner factorizations of this. But the whole point is that the monotomy loop of this on the one hand is included in the symplectic. On the other hand, it contains the monotomy of having a just one linear factor. And this can be seen to be the symplectic. So basically, you sandwich this between two and you get symplectic for all these type of factorizations. And in particular, if you take you, you can recover the price by doing some inclusion exclusion principle. So it's actually a very nice argument, which is why I kind of spending some time on it. Okay, so with this result, one can go back and formulate the conjecture over the number FIM case. Okay, so over the rational. So the conjecture will be that if we have P prime, and we define, we are looking at the sum of DKN, where N is less or equal than X and it goes over all the Ns that are congruent to the square mod P, okay, with P that doesn't divide N. Then we picked a variance computed, okay, over the P in some the other key interval. So between Y and 2Y to be asymptotically given by a formula that is contained some arithmetic constant X over 4. So X, we are summing up to X, okay, that's where the X is coming from. Then some gamma. Okay, so this gamma is going to be like the gamma that I have before, but now it's a symplectic gamma. So it's a piece white polynomial function of degree, okay, so the degree is different, 2K squared plus K minus 2. And then evaluated in log X over log Y minus 1 to the log Y minus 1 multiplied by log Y minus 1 to the 2K squared plus K minus 2 power. So now, okay, so this is what we have. We also look at a different context given by, so it's a different frame set, different problem that will also give symplectic. So this context has to do with Gaussian integers and it has to do with constructing an analog of Gaussian integers over function fields. And so this, so this construction is found in some paper of Barry Sorokers, Milanski and Wolff, and it's also used by Rudnik and Guassman to study distribution of Gaussian primes again by modeling what happens over function fields. And the starting idea here is that if we have p a monarchy reducible polynomial, then we can write it as a square plus, okay, so t my variable, so we can write it as a square plus t b if and only if p of zero is a square over sq. And this problem, I mean, this equation should remind you of the primes that are sum of two squares, okay. And so it's sort of an analog of that. And to make it an analog, we should make t to be the square of minus the square of i, right? So basically what we do is we take a new variable s to be the square root of minus t. And then we see the polynomials in t inside the polynomials in s. And so this s will play the role of i. And so in this context, you can build the Gaussian integers, okay. So this complex conjugation that is defined as you will expect something that changes to minus s over formal power c s is a norm defined also as one would expect. And then in this context, you can build a unit circle. And so the unit circle in question would be the power series whose constant coefficient is one and who have norm one. And then if we have this, okay, we can build sectors in this unit circle. And a sector will be elements in the unit circle that are closer to some point by some distance, okay. And my distance is going to be the norm infinity of f is going to be q to the minus order of f. And order is the highest power of s, that divides f. Now, in another way of understanding this sector is the power series in the circle that are congruent to a modular high power of s, okay. So basically you pick some point in the circle and you look at things that the difference with it have a divisible by high power of s. Okay. Now, with this one can study the divisible function asking, okay, so I didn't define this u, but this u has to do with the angle of f. So obviously there is a way, but not obviously, but there is a way of seeing what the argument of f is defining an argument. And then we're going to ask that the argument of f is in some particular sector of the unit circle. And then in that context, we come through a similar result. Basically we can compute the mean and then we can compute the variance and the variance again will be some integral similar to what I was describing before, but now over the symplectic matrices. Okay. Now, I should hurry up a bit. To prove this, so one important ingredient is again working with characters. And the characters that we need in this case are called super even characters. To be concrete, they are defined over the polynomials modulo sum power of s. Okay. And super even means, okay, so in the even characters in general on fq are the ones fqt are the ones that are trivial over the constant. So trivial or fq. Here we are asking more. We're asking that they are trivial on the analog of the real numbers, which means they are trivial on the, essentially on the polynomials that the only powers of s are even. Okay. So they are super even characters. And there's a result of cats that says that these characters, okay, they are associated, they become uniformly distributed in the symplectic loop. Now, one, there is another, okay, so there's another character that one can define on the elements on fqs that is just detect, detect the square of f0. Okay. So this is a kichu that is one, if f0 is a square and minus one if not. Now, here's a twist and this is pun intended. So the twist of the story is also that if you twist your character by this kichu, there is another result by cats that says that we get distribution on the orthogonal group. Okay. So then we can go back and cook up a problem that, okay, I have to admit it's a bit more artificial than the one I had before. Okay. But we can cook up a problem that the distribution now is over the orthogonal group. Okay. Now, one thing I forgot to mention before I should have mentioned is that the first problem I showed you was had to do with congruence of a square modulo p. So it's some sort of congruence condition. And the second problem I showed you on the symplectic is, you know, something lying on some segment in the, okay, so in some sector of the unit circle, which you can think of it as some interval somewhere. So these two problems that I presented, they have a spirit that is associated to the original two problems that we look at before. Now, this is a bit harder to formulate because this is a bit harder to translate. Before, okay, so before translating, one has to study what these integrals give you, okay. So in the same way as the other authors, they study the integrals over the unitary. Here we study the integrals over the symplectic and orthogonal. And because we are computing variants, we actually have a square associated to these integrals. We make things quite complicated in this case. Let me just mention some elements that show up in this study so one thing, sorry, a big, yeah, a big central point, central results that we use are coming from symmetric function theory that basically there are these formulas that relate these integrals over these products of characteristic polynomials to some of sure functions. So sure functions are essentially generating polynomials that count semi-standard young tableau of some sort. And so there are very beautiful results such as this one, okay, that allows to basically you can think of these as a generating function for the integrals that we want to find. And so basically by looking at the right hand side, we may be able to to find the integral. So basically in our case, we have this square so we have to specialize half of the variables in one way and the other half in the other way and look at diagonal terms in the generative function. And these then the counting over young tableau semi-standard young tableau can be related to counting points in a lattice, okay, inside certain polytope, okay, and then using error theory one can prove that then here I should say the asymptotic it's not really the the inter is a inter integral but the asymptotic when n goes to infinity ends up being a polynomial of degree 2k square plus k minus 2. And also the sum over the sure functions can be used, okay, so there are formulas coming from representation theory that allows us to actually also compute these coefficients if we have enough patience. So in practice, I mean when things go big then this doesn't work very well. But say for example if we take maybe let me let me skip that but for small values of when k is a small respect to n we can compute things and when sorry when small n is small respect to big n. And so for example with that we can compute for example the okay so the case when there's only one term so the sum inside the the integral is just one term. We can even compute some cases when two terms and so on. Let me also mention that the orthogonal case is ended up being quite tricky. So there are some for the formula for generating function in terms of the sure function. We actually couldn't find it in the literature. We found a formula for dimension even but our problem has odd dimension so we actually had to recover this formula which wasn't easy to do. And again doing a sum of young tableau so using the semi-standard young tableau we can turn this count into a lattice inside some polytope and then prove that we expected polynomial of degree 2k square minus k minus 2. And then with that we can go back to the number field case and actually write conjectures. So let me maybe state everything together. So for example so for the number field case what we have is if so basically what we are going to do is we're going to do the sum of dka where a goes over the ideas that have bounded norm and whose direction vector is in certain sector of the unit circle. And what we mean by direction vector well is defined by this hake character. So basically we take a generator of the ideal and then we take the generator over the conjugate to the square and this and you look at the argument there well one for of the argument and that is well defined actually if you change your generator it's still well defined. And so basically with this argument okay so basically restricting this argument to this sector so here's a picture that shows more or less what we are doing okay. So basically we are counting the dots in here okay what happens to the device of function over those dots then we can formulate the conjecture okay. So here I have just formulas what I mean by the variance let me skip over that for the orthogonal case we can also write something. This is a bit harder to kind of see in a natural way so we need to define this hake 2 character that we used to twist our super even characters and so what we do is overprimes we are going to detect when a prime is a sum of two squares but not really okay so this is a prime in the Gaussian primes and what we write is a square plus i b square. So it's really a sum of two squares in yeah it's going to q omega 8 it's really going beyond that. So again we can now look at the device of function distributed over these ideals using this twisted by this detector okay so this chi 2 of alpha will be you know alpha is defined multiplicatively so on alpha the condition is not so nice but again we can formulate the conjecture in terms of what the variance of this sum should be okay and so basically to summarize okay so what we would like to do for the future well there is still some more to study of these gamma coefficients that show up in the symplectic and the orthogonal case. Integral expression what I mean by that is that in the beginning I showed you very nice integral expression for gamma in this in the unitary case the ones that we have for symplectic and orthogonal are not so nice there are also questions about lower terms like maybe we can study okay so here this gamma has to do with the main coefficient okay the highest degree but what about the next coefficient this should be also possible to study understanding the arithmetic coefficient this I really left them under the rug here there is some understanding that has to be done we don't have a conjecture about what they should be so there is some work that has to be done all of what I said in principle one should be able to extend it to von Mangel convolutions and we've done some work on symplectic but not an orthogonal and also not not an understanding everything in that case so so there is some work to be done there and obviously one could ask well this is kind of asking much further one could try to work with functions of higher degree so there are all these directions and with that I thank you very much for your attention


Ideally, a retiree receives an annuity for the rest of their life. A basic level perpetuity due a double-dot angle infinity is an annuity that pays $1 at t equals 0, 1, 2, and so on forever. Likewise, a basic level perpetuity immediate, a angle infinity, is an annuity that pays $1 at t equals 1, 2, and so on forever. Suppose the annuity earns interest rate i, so v is 1 over a, where a is 1 plus i. Then, a double-dot angle infinity at interest rate i. That's going to be the infinite sum of the powers of v. And here we use the infinite geometric series summation formula if the absolute value of our common ratio is less than 1, then the sum 1 plus r plus i squared and so on is equal to 1 divided by 1 minus r. The important thing to remember is that this infinite series is only equal to this nice simple expression if our absolute value of that common ratio is strictly less than 1. So let's think about this. Since v is the reciprocal of 1 plus i, and generally speaking we'll assume that i is positive, then the absolute value of v will be less than 1. And so this infinite series will be, but we can simplify further. This reciprocal 1 over 1 minus v, remember v is 1 over 1 plus i, so we can rewrite. And let's simplify by multiplying numerator and denominator by 1 plus i, and that gives us. Now by essentially the same argument, a angle infinity at interest rate i, again that's the sum of the discount functions. We'll remove the factor of v. We can use our infinite geometric series summation formula. And this time we can simplify further. v times 1 over 1 minus v will be, note that since a double dot angle infinity at interest rate i includes a payment of 1 at t equal to 0, the present value of this annuity is 1 more than the present value of a angle infinity at interest rate i. For example suppose you deposit 1,000 every year for 30 years at the end of which you have a perpetuity due. Assuming an interest rate of 5%, what is the payment of the perpetuity due? So the value of the account at t equals 30 will be 1,000 s double dot angle 30 at 5%, while the value of the perpetuity paying p forever will be p a double dot angle infinity at 5%. So we want to solve the value of the perpetuity equals the value of the account. Now p a double dot angle infinity at 5%, that's the payment amount times the infinite geometric series of the discount functions. And we could use our geometric series summation formula to find that value. Similarly 1,000 s double dot angle 30 at 5% is going to be, which gives us an equation we can solve for the payment value. Solving our equation gives, or since a is 1.05 and v is the reciprocal we find. So the annuity can pay out $3,321.94 every year indefinitely where we round down to make sure that the annuity never runs out of money. Now it's worth having some method of verifying our answer. Remember no computer was ever fired for mismanaging a retirement fund. So we might check on this as follows. The accounts value at t equal to 30 is 1,000 s double dot angle 30 at 5% and we can find that to be. Now at that point let's suppose it pays out $3,321.94. Over the next year it will accrue interest equal to 5% of $6,977.79 minus the $3,321.94 payment and that amount will be. And so the payments equal the interest earned leaving the principal untouched. The previous problem leads to the following generalization. Suppose we fund a perpetuity due by making n payments of q into an account bearing interest at rate i. Then the perpetuity pays out p at interest rate i where p a double dot angle infinity at interest rate i is equal to q s double dot angle n at interest rate i. Now p a double dot angle infinity at interest rate i is equal to also q s double dot angle n at interest rate i is equal to. And because our interest rates are the same we can remove this common factor of 1 minus v. And we can go one step further since v is the reciprocal of a then a to the n times 1 minus v to the n is. And this gives us a nice relationship between the payment of a perpetuity and the contributions made over a time period. Now remember don't memorize and I'll forget it. So suppose the investment earns 8% interest compounded annually. How much would you need to contribute for 15 years to obtain a perpetuity due paying $10,000 each year? If you make n payments of q then a perpetuity due pays p where p is q times a to the n minus 1 where a is 1 plus i the interest rate. So we have a equals 1.08, p equals 10,000 and so we find. Now we want to make sure we've deposited enough so we'll round up to an annual contribution of 460370. Again we can verify this by noting that at t equal to 15 the account balance will be. At that point it pays out 10,000 and over the next year will generate a little bit over 10,000 in interest which will be available for the next payment.


Kills in this movie are actually quite good. Breaks that guy's neck and it's great effects where you see the bone come out and everything. Hey! Get away from my trot! Yeah! Wait, Jamie dies on that machine, like that's pretty awesome. The mom when she gets killed. I like how you don't see it, but you see the blood. The lead up to the kill is great. The chest through all those sheets. Again, the sheets are a hark back to the first movie, right? Exactly.  How can you get any better than that? That's a great death. He uses the guy's face to open up that gate and like... He kills everybody, all the nurses and doctors in that room is crazy. Good kills and high kill counts.