Very few. That's a little sad. It's nice to have live people in the audience. Well, some is better than none. So as I said last time, so first I remind, although at least the people are here, around Zoom, obviously know that the new hours of the course for the last two weeks, so this week and next week, are Monday and Wednesday. So today and after tomorrow and the next week from 2 to, well, I'll write it. European style, 15, 14 to 15, 30. So 2 to 3, 30. And I'll talk either all four lectures or three out of the four on the same subject because it's very pretty. So the subject for either the rest of the course or almost all of the rest, if I finish early, I have one other topic, subject is the circle method in a less standard context. So the original circle method, I talked about a little bit at the very beginning of the course. And of course, it was made famous and really discovered by Hardy and Ramanujan to study the partition function. And in their famous paper, they mentioned another problem, which is more general, which is partitions into powers, like into squares. So in how many ways can you write n as a sum of squares of integers where the order doesn't matter? And they gave a very rough answer for that. So they did look at it, but not in detail. But basically, they said it would be very hard to get really detailed information because the corresponding function is not a modular form, which it is in the case of ordinary partitions. So I worked on that. Somebody asked me a question about that. And I started working on it about a year ago. And it turns out you can say a lot of nice things. But the circle method is already the second stage. As you know, the circle method means you want to get asymptotics of the coefficients of a power series. And you do that by writing a co-scienticle. But for that, you have to know the singularity. So the behavior of the function itself near maybe one on the unit disk or maybe all roots of unity. And so there are two parts. One is understanding the function near one or near all roots of unity. And the second part is the actual circle method. So all of today will only be the first part and probably most or all of Wednesday also. Because actually, that's the most fun part is understanding the behavior of this function, which is not modular. And there are a lot of nice surprises, some nice number theory, some nice asymptotics. So it fits very well with the spirit of this course. So as I say, the problem will be specifically the squares, but the circle method itself will only come into play fairly late. So maybe next time more like the next Wednesday. I should also mention the name Fabio Novias. I don't know personally. He's in Chile. But about a year ago, he sent me an email asking if I knew or if anything was known about the asymptotics of the number of representations of a number. This is a sum of squares, which came up in connection with some problem. I have to even look because I don't remember. The term sum is formulation of 3D gravity and its connection to integral systems, nothing of which I know anything about. And so first I gave a very vague answer. It's something very sloppy, but then I got hooked and started working. So I would like to thank him at this point for having pointed out this problem, which was 100 years old. It was a known problem. It's discussed in the paper of Hardian Ramanujan. But apparently nobody had ever studied it in any kind of detail. So I'll start a little historically. So the original paper of Hardian Ramanujan, the famous paper that I talked about at the Ramanujan day, Ramanujan maybe three weeks ago just before this course started. So that paper was 1918. And it's called, if I remember, asymptotic formulae in combinatorial analysis. So I'll write that because it's such a famous paper. I don't remember if it's combinatorial or combinatorial. Both words are used, but they had combinatorial. That's what I thought. And so of course the original problem is P of n is the number of partitions of n of n. So that means you write n equals, let's say, a1 plus some a new, where the a new are positive integers, but up to order. So you can put them in decreasing order. You can consider it up to order. So that's the number of partitions. And then this function was invented by Euler, who also then promptly invented generating functions and realized that the right way to do it is to look at the function that's called P of q as the generating function without any factorials, just the straight generating function of the P of nq of n.  So the sum q to the power 1 plus 1 plus 1 with the variable number of 1s is 1 over 1 minus q. And the similar thing for the sum of squares, we have q squared and so on. So what you get is the infinite product, m for 1 to infinity, so this is an Euler's original formula. And now as I've already suggested, the key point is that this function a to q, if you p of q, if you think of q as e to the 2 pi i tau with tau in the upper half plane, then up to a simple factor, q to the 1, 24th, this is a modular form of weight minus a half. And that means it transforms in a completely known way under the modular group. So the modular group tau maps to a tau plus b over c tau plus d, where a, b, c, and d are usual are integers, so we're in SL2z with determinant 1. And in particular, infinity goes to a over c, which I could call cap, it's some rational number, but it might be infinity itself. And conversely, any rational number can be written as a over c, where a and c are co-prime, then they're unique up to a common sign. And then if you have a and c, you can fill in this matrix, a, c, you can find a b and a d, not unique, making the determinant 1. And so you can move from tau equals infinity to any rational point. But that means that if you take q as always as e to the 2 pi i tau, which I will always abbreviate by e underline of tau, because it's too much trouble to write and it's easy to read the formulas, then you see that this would correspond to q equals 0, whereas this would correspond to q is e of kappa, which is, in this case, a c-th root of unity. So mu c double mu c is the set of the group of c-th roots of unity. So in other words, on the unit circle, we have q going to 1, but you can also go with q to minus 1 or e of a third, or e of 2 thirds, or e of a quarter, which is i and so on. But because of this multi-transmission behavior of p, if you know how the function looks at q equals 0, so of course when q is infinity, oh sorry, when tau goes to infinity, that would be q equals 0, which is therefore here. So this point is equivalent with all the points, all the rational points of the roots of unity on the unit circle, and since you know the expansion of the function at 0, you also know it to all orders of all roots of unity. But for the functions we'll be having, it's not going to be like that. So now, if we go back to that paper, already in the introduction, they said that one could look at other problems, not just this problem, and in particular, they use the letter s, so I'll also use the letter s. For the moment, s will be a strictly positive integer, which means n. We look at p sub s of n, so p1 of n would be ordinary partitions. It's the number of partitions, well, I'll start with p2 as they did too. p2 of n is the number of partitions of n into squares. So for instance, if I do p of five, then it'll already be seven, because you can write five as well, of course, just five, one plus four, two plus three, that's into one part, into two parts, into three parts, it could be one plus one plus three, or one plus two plus two, into four parts, I guess it could only be one plus one plus two, and finally into five parts, so one, two, three, four, five, six, seven. But if I look at p2 of five, then since the only squares less than five are four and one, either it's one plus four, or it's one plus one plus one plus one plus one. So there are many fewer of them, p2 of five is only two. And the question is, of course, how does p2 of n behave when n is large? And of course, the generating function p2 of q, which I define as the sum p2 of n, q to the n, for exactly the same reason as before, so we'd find p2 of zero is one, there's no other reasonable way, and then it's the same form as before, but with q to the m squared. And similarly, of course, I can say ps of n is the number of partitions of n, in for instance, cubes, if s is three, so into s powers, and then of course, exactly the same thing, I'll just write it. So you have the product one over one minus q to the m to the s. But as I said, only p1 has any kind of modularity properties. So specifically, and I'll come back to this of course later, eight of tau is defined as eq to the 1 24th. You can say that the 24th root of a context number is not well defined, but remember that q is defined as e to the two pi times tau, and then you simply divide the exponent by 24, that is defined and exponentiate, and then it's times p2 of q inverse. So here it's the product one minus q to the m from one to infinity, which, and this thing is of course, p2 of q inverse, so I can, putting p on the other side, p2 of q is the quotient of q to the 1 24th and eta, q to the 1 24th is an elementary function, purely exponential in tau, and eta of tau is a modular form, so we can read off the behavior at all roots of unity, but there's nothing like that for the other tau. So the problem is to do that, and already in this paper, Hardy and Ramanichand, as I said, they discussed this other problem. Actually, they did, they discussed it for a third of a page in the introduction, and gave a very weak formula, and I only realized, in fact, when my paper, when my whole investigation was finished, that they actually came back to it. 60 pages later, at the very end of the paper, with no displayed form, it's just in the text, and they gave us a more precise, some more precise table, so one of the things I'd found, they had already found, but I re-founded 100 years later, but luckily not everything I found, so it was, there was still some interest, so they asked this, and in particular, they gave the problem, well, they defined, they gave the problem with general s, so the s is their letter, because there'll be a Riemann zeta function soon, and one's used s for s equals, well, of course, squares are the most interesting case, but it's the first case, and they gave a, in the introduction, they gave a very weak formula, so let me remind again the Hardy-Ramanichand formula, p of n is the log of p of n, that's quite easy to prove, is asymptotically equal to pi times the square root of two n over three, but there's a full formula, which I'll talk about later, but the leading term, I didn't write it down, so I don't quite remember, I think it's four n squared of three, but it's the exponential of this term here, but then it's corrected, I think it's four n squared of three, but anyway, constant times n, and then there's a far, yet much more precise form, which I talked about in my talking for the Ramanichand day, which was about Ramanichand partitions, but I'll give it again next time, the time after when I come to the circle method, so that much they gave here, and the form that they gave was the weaker one, that the log of this grows like a multiple of a power of s, and maybe I don't need the name of their paper anymore, because the form is a little long, so the constant is s plus one times the expression one over s, gamma of one plus one over s, I hope I wrote this correctly, but I think I did z to one plus one over s, the whole thing to the power one over s plus one times n to the one over s plus one, so it's a little bit complicated, but you see that if s is one, this would be two, if s is one, this would be two, this is one, gamma of two is one, z of two is pi squared over six, to the one half, and then the square root of n, and that's exactly pi times the square root of two n over three, which is what I said here. Okay, so this formula they already gave, and as I said, I only discovered many months later when I read the beginning of the paper, and some of the calculations, I'd never got to the discussion at the end, but they talk about where the method could be used in other situations. They actually do give an asymptotic formula corresponding to this one for P s of n for any s, so that's much, much weaker than what we're going to do, is this just the leading term, and this comes only from the contribution near q equals one, but it's in fact only a tiny part of that. This is still off by an error of the order of one over n, or one over square root of n, so it's quite poor, n is 200, this is only correct of five digits. Then if you give the full calculation of q equals one, you have to add three correct terms to this. You get an elementary thing like this, but with other terms, and now it's suddenly very accurate, and it gives you, now the error is exponentially small in the square root of n, and then if you can look at the other roots of unity, you can get them all, but they say fairly clearly in their text that it's, you can't really say anything at other roots of unity, so you could at most improve this by looking at q equals one, because you don't have multidarity, and that's what, as I say, it turns out in fact to work surprisingly well. So again, just to emphasize the multidarity, I said that this function eight of tau, which I defined here, is modular, and so this is very well known, I'll just write down the corresponding things, but they're not going to be true. So here, if I, the modular group is generated by tau goes to tau plus one, which corresponds to this matrix, and then this is trivial. If you change tau to tau plus one, then from this definition, q doesn't change, because q is e to the two pi tau, but q to the 124 changes by the 24th root of unit, the standard one, I'll always use the notation if n is a positive integer, this means the standard root of unity, e to the two pi i times one over n, so the one at angle two pi over n. So this is trivial, but what's less trivial and there are many, many proofs, is that if you take minus one over tau, which corresponds to this matrix, zero minus one, one, zero, or it's negative, then you get the square root of tau over i times eight of tau. And of course, you have to say which square root, but since tau's in the upper half plane, tau over i is in the right half plane, and then there's a canonical square root, the one that is a deformation of the square roots of positive numbers. So if you're in the right half plane, then the square root is in this corner, you don't take the one on the left. So these are the two famous forms, and more generally, if you have any element, the a, b, c, d of s, l, two, z, then it follows from this that up to sum 24th root of unity, so something divided by 24, there's always a factor of square root of c tau plus d, it's a multiple form of weight, a half. And here you do have to specify which square root you take, but if you take the other one, you're changing by minus one, which is also 24th root of unity. So this will always be true for whichever square root you take, but you have to be careful. And there's a famous form of this due to, so this is essentially Riemann and Dedekind. So it's called the Dedekind's data function. This formula was published by Dedekind, but in a paper on a fragment of Riemann, he was asked here, and to Vibra, I think, to go through the papers of Riemann after his death, and he had left many unfinished things behind, and he had analyzed this function to some extent, and Dedekind finished it and worked out the details and published it, but Riemann had actually started studying this function, so let's say the Riemann Dedekind function. So this is exactly what we don't have for the higher P2, and they will have to somehow get around. But it turns out that, in fact, you do have something. And so I'll say the first problem is no modularity, no modular transformation equation for S bigger than one, but before I say for S bigger than one, I should perhaps say transformation behavior of what? Because already for P of Q, remember P of Q is just the product, well, it's the sum P of N Q to the N, it's the product one over one minus Q to the M. But that doesn't transform, you can't see anything here, because when you send T to T plus one, you just aren't changing Q at all, so P of Q is just equal to P of Q, and here you get P of Q is some crazy thing involved in the square root of the log of Q times P of a Q star, you can't read Q star off from Q, because you have to take the log inverted. So you must introduce the function eta of tau even to make sense, and so we'll do that also for higher tau, and we'll do it the same way. But now I want to get the function right, so remember E underlined as E to the two pi i times the argument, it will always be a multiple in the argument, and actually a rational multiple, so remember S so far has been a natural integer, all that will change soon, a natural number, so positive integer, so it's eta minus S to rational number, so this is a rational power of Q, which if S is two, the next interesting case, the squares is actually zero, so there isn't a correction factor, but if S was one, you get minus a half of minus a 12, which is plus 24th, which is about just what we saw, and then of course multiplied by PS of Q inverse, just as it was when S was one. So the PS of Q inverse, it looks artificial because in the definition of eta, there is no, whereas the definition there is no inverse, but in P2 I put in the inverse because it's the generating function of P2, so the natural multiple form is eta, and P2 is actually reciprocal, so it turns out that this is always a good correction factor, so here the notation will be the same as before, tau is in the upper half plane, which means complex numbers with positive measuring parts, and Q I won't repeat it again, Q is always E to the two pi i tau, but strange in a sense is why should you introduce tau at all? The point of introducing it here was that the multiple group acts by tau plus one minus one over tau and A tau plus B over C tau plus D here. Here it's not going to act, so you might think what's the point, but as you'll see there's very much a point in having the tau variable. So okay, so there's no modularity, but the generalized Euler-McLoran formula, formula, so generalized means with a shift and so on, as discussed earlier in this course, so when I need it soon I'll write it out again, so that gives the asymptotics of PS or A to S, or actually it's log, but it gives the asymptotics also these functions to all orders in epsilon when Q is equal to E to some root of unity, so Z I'll just say it's a generic root of unity, so mu without a letter of mu, infinity means all roots of unity, and then we put let's say E to the minus epsilon, or equivalently, well here I'll actually take two pi epsilon, now I'll put epsilon for the moment, I don't like either one, equitably tau goes to a rational number in Q. Okay, so as tau descends in the upper half plane towards a rational value here, that means remember that Q goes radially, let's say if this is vertical to some root of unity, and then we could do that for anything, and the Euler-McLoran formula gives that entire expansion, not just the leading term, but the entire asymptotic series, but now you can say well, but this was way, way better, because here as tau goes to infinity, that means that this argument is very small. This thing is equal to Q to the 124th times one plus O of Q, but Q is exponentially small, and therefore this Q to the 124th is a terminating series, it's true to all orders. Here the series we write down will be a bit different, but it turns out that you can improve that, and that's the first big surprise, which I'll talk about, well I'll prove next time or even next Monday. So in fact, even though it's not not there, this Euler-McLoran formula can be lifted to an exact formula. So the exact formula will describe the behavior of this function with any S as tau goes vertically to a rational pointer, as Q goes radially to root of unity, that to me was a surprise, but this is the thing that I discovered actually hard in Ramanuj and gave in one line with no details, but they said you can get that. So EG if I take the easiest case, this is when cap is zero. So remember we're interested in tau tending to a number kappa in the upper half plane, but if tau is zero then Q is tending to one, that's the easiest case to study, and then the formula for eta is the one that I wrote here. It's simply eta of minus one over tau is a power of tau, tau to the one half up to a constant, and then times eta of tau. So now here's the exact formula for eta two, and it's really kind of surprising that there is an exact formula. So the asymptotic formula to all the words that I say is quite easy, but here's the exact formula, eta two of minus one over tau is this time it's still got a square root of tau times a constant, it's a different constant from before, before it was just tau over i, now it's the square root of two pi tau over i, and then it's the product of two functions, see before it was a multiple form, so we just had eta tau, so if here at the same function eta two of tau, that would be multiple, but instead it's eta half, and if you look at this definition, well here there's no problem having S to be half, it won't be a rational number anymore, this will be some transcendental number times tau, you can still exponentiate, but PS, this definition at no place that I use, that S was an integer, and this part is, sorry, I didn't even write the definition, the sum PS of n of q to the n is of course this, that was the generating function, if S is not an integer, let's say S is 1.3, you cannot write as an integer as a sum of 1.3 powers, so there is no PS of n, q of n, but of course this function makes perfectly good sense for any real number bigger than one, actually bigger than zero I guess, right, I mean it's a perfectly convergent series, so we have an excellent function of tau, it's just that it no longer has the powers, is that there's no P one half of n, but there is a capital P one half of q, and therefore with this definition, so but this is now for any S in C actually, but I'll only use real values actually, the reciprocals of integers and integers, so for any point in the upper half, in the right half plane, this is a perfectly good function, and so now the surprise is that a to two of minus one over tau, it's not there, it doesn't matter to itself, it goes to a to half, well that would already be very nice, so it'd be kind of a reflection, S goes to one over S tau, but it's not true, first for two reasons, first of all this is the square root of tau, so it's not symmetric at all because from tau you can go to the square root of tau, but there are two square roots of tau, but you have to take the one in the upper half plane, so remember tau is in the upper half plane, so square root of tau is, I guess in the right half plane, in the right quadrant, but then there's the second term, which is not minus the square root of tau, which is the other square root, that would have been lower half plane, but i times the square root of tau, which is in the left quadrant. So it's a very, and this is an exact formula, so I can even call it theorem, we'll prove it quite a bit later, probably only next Monday, so age two of minus one over tau is exactly, and this is not approximate, this is an exact equality, that I found first purely numerically with a lot of work, and that I'll talk about today, that's actually the most fun part of the whole thing, and more generally I can put it in the same theorem, a to s of minus one over tau for any integer, but s is still an integer here, a to s of minus one over tau is again, always the square root of tau over i, but there's a power, well you can guess, here for s equals one, it was just the square root of tau over i, here it's two pi to the one half, so in general it's two pi to the power s minus one over two, times the square root of tau over i, and then it's the product for all z in the upper half plane, such that z to the s is plus or minus tau, so if you take all complex numbers, whose s power is tau, they're exactly s of them, if you take all complex numbers, s power is either tau or minus tau, there are two s of them, but they come in pairs, z and minus z, so exactly half run the upper half plane, and none are on the real line because then their power would be real, and so they're exactly s factors here, and now you can guess the last term's actually shorter because I used this notation, then it was for s equals two, you have a to sub one over s of those z's, so that's the kind of one of the main theorems, but actually the main theorem on a to will be more complicated, it will be if you put in arbitrary a tau plus b over c tau plus d, or more relevant, we will probably write it usually as I'll just put for the moment i epsilon, but usually later I'll have another normalization, so we look at the formulas, you come very close to a rational number from above, first asymptotically, but then on the nose, so okay, so now the next stage is, and now I can erase almost everything, the next stage is to do the, to describe the asymptotics, to discover the asymptotics, so that's been kind of a theme of this course, that even if you're doing quite theoretical mathematics, you need computations very often to simply know what the truth is, you have a sequence of numbers, like we had recently with Manuel, with Papel who's not here today, or with many other people, even here in ICTP, you find some problem, you get a bunch of numbers, you want to recognize them, you want to understand them, you have to A, know how to calculate reasonably, and B, you have to know how to recognize asymptotics who do asymptotics, and that turned out actually here to be the most fun part of the whole project, was figuring out what is the correct exact asymptotics, so let me take, I'll concentrate most on S equals two, because the ideas are the same, and it's easier, occasionally I'll give the formulas for general S, so let me assume, so the whole problem is, so we want numerical results, and the problem is, find the asymptotics of, once again, A to S of cap, so S is always going to be an integer, greater than or equal to one, greater than or equal to two, because we know it for one, and then plus I epsilon, cap is always going to be a rational number, and epsilon, let's say, goes down to zero, it could come also in a cone, but let's say vertically for the moment, and so sorry, here, as epsilon goes to zero, but first to all orders, that'll be more Euler-Mclaurin, and then exactly, and it turns out that the answer's quite fun, and as I say, fits in well with the methods of this course, because there's a lot of, first there's some Euler-Mclaurin, then there's some guesswork, so let me do the step one, so I'll concentrate, first on the case when S is two, the very easiest case after, and cap is zero, so now we're talking about P2 as Q goes to one, so then Q goes to one, I can call it E to the minus X, where X goes to zero, okay, so we look at P2 of E to the minus X, and for this you just use Euler-Mclaurin, and I'm going to actually do that quickly because we did that at the beginning of the course, you may have forgotten, and if you haven't forgotten, it's nice and relaxing, it takes only a couple of minutes, so what will happen when you do it, is just as for the a function, you will have a constant over the square root of X, well, there it was one over X, but here it'll be one over the square root of X, for higher S would be one over the S root of X, here there's a power which is always the square root of X, but in principle, there would be a whole infinite power series, C zero plus C one epsilon, well, X plus C two X squared, but in fact, the error term is O of X to the N for all N as X goes to infinity, so in other words then first, but that we'll find from straight Euler-Mclaurin, and it's a bit of a surprise because in the one that I gave in my course as an exercise at the very beginning, I also I think called it P two for which I apologize, but here I really don't wanna change it, that was a different instead of putting Q to the N squared, which is N times N, you take it to the Nth power here, and then this one, you could take the log, but it had an asymptotic expansion that did not terminate even for the log because it had products of Bernoulli number times the Bernoulli number two later, B N times P N plus two, if N is even, N plus two is also even and both Bernoulli numbers are non-zero, whereas for P one or P three or P five, you'd have B N times B N plus one or B N plus three, they have opposite parities, and remember the odd Bernoulli numbers are zero, so this has a non-terminating, the asymptotics, asymptotic expansion, let me just say this, non-terminate asymptotics, but this one turns out as I've just written for S, but it's the same for every S, it's a slightly different form, but this is the same I wrote before, gang of one plus one over S eight upon plus one over S, X to the one over minus one over S, but it's always, there's just a finite expression and then it's to all orders, so because that's just, as I say, it's relaxing because we already know how Euler-McLoran works, let me remind you, briefed out that goes, that's the thing I did much earlier in the course, which was the asymptotics of functions of the form F of Nt, so I'll do it quite quickly, so of course, since we want a sum and this was a product, I first take the sum, so this is minus the sum, M from one to infinity, log of one minus two to the M, which is now E to the M squared or M to the, actually, now I'm doing S, why did I do it for general S, I wrote it for S equals two, it makes no difference in the calculation and it's fun to see that the fact that it terminates will always happen, whether S is even or odd, so this would be the function and so if I replace X by T to the S, then this thing is equal to the sum M from one to infinity, F sub S of Mt, where F S of T is equal to minus log of one minus E to the minus T to the S. Do you see, and we've had that before in the other function, we have to make the small change of variables to get it in the form of what I discussed towards the beginning of the course, infinite sums of the form sum of Mt, where M is a nice function, nice when small to infinity, this is certainly small at infinity if T is large, this is exponentially close to one and so it's log exponentially close to zero, so it's rapidly convergent, it isn't quite analytic at zero, so remember, well, you can expand this, this function is of course just whatever it was, so that expansion I gave when we were talking then because we had an example, the function of log of one minus E to the minus X is equal now I have to get it right that I don't make any mistakes, it's log one over X plus X over two minus X squared over 24 plus X to the fourth over 280 and so on, it's an alternating sum and the general term is Br over R factorial X to the R over R, except for the log one over X term, okay? So here you do the same X as T to the S, this in fact already was, so here we'll have the same thing but it'll therefore be S times the log one over T, T to the S plus the sum R from one to infinity Br over R factorial times T to the R over R but it's now T to the SR. Now that means for this we gave a general form that this will always be some integral which is the integral of FS divided by T plus and if it were just a power series, then remember that was Riemann and Euler, you just multiplied T to the N by Zeta of minus N but when we did that there, when you have F, the sum F of NT, you start with the integral of F divided by T but then any term T to the lambda goes to Zeta minus lambda T to the lambda so it's kind of the Euler point of view but log of one over T goes to log one over T I think times minus a half because it's Zeta half so one half plus roughly Euler's constant I'll write it in a second but there was a formula for what you have to do if there's a log term. So but the first thing is we need to know the integral so this is standard but I might as well quickly do it. So we first need the integral so I of FS is the integral from zero to infinity FS of TDT so I think the quickest way to do it that there are several ways you can do it by parts but if we just take this thing then of course using the power series of log it's this log minus log of one minus X is the sum one over K X to the K so it's minus KT to the S so this with the sum K from one to infinity one over K times the integral from zero to infinity and then it's E what did I just write? It's E to the minus K S and here you just make a change of variables you replace T again by X is one over S so this will become one over S times the integral one over S minus one DX that's the DT and this will become E to the minus K X and this of course is exactly the integral so this integral is one over S times that but that is one over S times the gamma function of one over S times K to the minus one over S and therefore this whole thing one over S gamma one over S is gamma one plus one over S and the sum over K one over K K to the minus S is Zeta of one plus one over S so that's the value of this integral it was easy and fun so therefore if I use this formula now and I use this one let me leave that there because I need it so this one will be equal to what I just wrote gamma one plus one over S Zeta of one plus one over S divided by T which is remember X to the one over S and now the log term well it's S log one over T and that's the one that I wrote before here I'll just write the answer oh it doesn't involve gamma I was stupid when you do log it involves square root of T square root of two pi over T so here you get the same so what you get is the log of T over two pi times S but if I want it in terms of X it becomes this and then the next term this is plus and then there's one more term but then that's the whole joke there's only one more term because if I look at this when R is one then for R is one I'll have minus a half because that's B one one factor all is one and I have T to the S but remember T to the S always goes to Zeta of minus S so what I get for some reason my notes say plus a half so one of them is wrong probably called I checked on the computer this morning so it should be correct so T to the S is X but now comes the fun part for all other terms if R is bigger than one bigger than or equal to two then we'll have B R over R times R factorial T to the R S will become X to the R so it's a power series in X but the coefficient here is Zeta of minus R S but if R is odd and bigger than one B R is always zero so R is even but if R is even then whether S is even or odd R S is even and Zeta negative even is zero and so it just stops to all orders so that's the little calculation I want to show you and I got some sign writer I didn't so now if you exponentiate that of course you immediately get P S of E to the minus X is the exponential it's the same except this is gamma of one plus one of S Zeta of one plus one of S X to the minus one of S here S was two then there's always a term squared of X over two pi so actually I might as well just do it this will become two pi can you put it inside X over two pi to the S this will become what I just said gamma of one over one plus S Zeta of one plus one over S X to the minus one over S and then finally there's one more exponential term which is plus half if I got the sign right for S equals two you don't see it so I can't check against what I just wrote but it's that in the exponent so this is just Euler Maclaurin and you can do the same at roots of unity I'll probably do that and I'll just give the answer later with the shifted Euler Maclaurin and the same thing happens it terminates if you do it right so that's very very nice okay but now of course we want to know can we find out anything about this epsilon of X so let me write it not as approximately that but as I did before it's that times one plus epsilon out epsilon S of X and this is very small so very small means actually it'll be exponentially small as we'll see in either X or T so it's but it's faster it's going to zero faster than any power of X as X goes to zero so now the question is since we're hoping to find an exact formula though at the beginning I didn't know there would be one but I certainly want to know the asymptotics to high order can you actually find this so again now I'll go back to the simplest case which is S equals two and I'm already just taking X going to zero so cap is zero so Q goes to one I'm not looking at other points so what is epsilon two of X so of course you can graph it so here is X it starts at let's say one and when you graph this I won't even try what you get is that first of all it's extremely small when you make a graph it's exponentially small and secondly it oscillates so it's very hard to see what's happening so what do you do numerically so let's say the thing I want to emphasize now is that it oscillates and therefore it's not easy to recognize so I'll tell you the zeroth step which is how to recognize it a little bit but then there are three simplifications and without all three you couldn't do the higher case at all even this one would be extremely difficult but with that it becomes quite sorry so this epsilon of X epsilon two of X is both exponentially small which makes it hard to recognize but of course you can increase the size through axis you don't care that it's a small number I mean you're adding up lots of contributions in Paris you can easily work with 1000 or 2000 digits and this power series for PS remember that this was the product one minus E to the minus M squared X so it's converged in the Mth term it's not just exponentially small it's exponentially small than M squared so if you take you know if you go to 1000 you're taking about E to the million X even if X is very small so there's no problem calculating this to very high precision these numbers are exact you divide, you subtract one you get epsilon of X to all the precision you need the problem is not to compute it it's to recognize it so it's exponentially small and highly oscillatory as X goes to zero and so that makes it a priori not easy to recognize however even a quick look shows that the amplitudes of the size is roughly I mean of the order of E to some constant in this case over the square root of X well minus the square root of X so it's blowing up P2 is getting very big sorry P2 is getting very big 8 is getting very small so it's a constant over the square root of X okay as you tend to infinity so that already tells you we should make a change of variables as we already did before remember X was T to the S and so T was the S root of X and also the oscillations the period of the oscillations so how far you have to go as you go in it oscillates more and more but if you look at the difference roughly between the troughs you see from the difference that the period is growing roughly like X to the three halves so you know if X is a thousandth then a thousandth to the three well let's say ten thousandth it's easier then it would be a million then if you take ten thousandth minus a millionth it'll be the next trough so that suggests the first simplification so we're going to do this in three steps I have to follow my notes because it's a complicated thing so this implies the first simplification I change variables from X to a new T so X is going to zero T will be going to infinity and they're related that X is a constant over T squared or in general it'll be two to the S but it turns out that it's I could put one but it would rescale them if the forms will come out nicer if I take all this two pi over T to the S and that's logical if you remember the Q because now I'm writing the thing as E to the minus X but then Q to E to the two pi times something remember Q deep down is E to the two pi I tell so the two pi is reasonable but anyway that's just normalization and then it's not so hard to recognize for epsilon two that if I take X which is now two pi over T squared so now as I said it's exponentially small than T purely exponentially in T as T goes to infinity times an oscillatory function but now the oscillatory function has a constant period because I've made this change of variables right it's T is one over the square root of X but if I take the nth position the nth trough it'll be the difference one of the constants one over the square root of n minus one over the square root of n plus one that's like n to the minus three halves so the period between two maximum and minimum that was like X to the three halves but it means that the period itself I mean the variable has to be something like sine or cosine of square root of X so therefore you get that and now how do you do that in practice so you have a function now T is going to infinity and you have something whose graph looks like this it's exponentially small but it is roughly constant period so how do you find so it looks roughly like E to the minus lambda T times cosine of let's say beta T plus a phase something like that so how do you recognize it how do you find those three numbers well first you make a guess and then you look if you can recognize the numbers so for the maximum what you do is you take the maxima of the absolute value so you graph it first and then you can zero in on the computer very quickly in Paris it takes a split second to compute individual values and you find that if you take the maxima and you take their ratio then that tells you exactly what lambda is and so you quickly find lambda just from the graph to let's say three decimals and then you recognize so I'll write it down in a second and now once you've recognized you divide your function by E to the minus lambda T now just the order of one it's a pure sine function and then what you do is you look at the successive crossings to a high thing and they're extremely regular so you take that distance and you recognize that and it's very easy in fact it's the same lambda and beta do not be the same it makes it even easier and then once you have that you have to work out the phase and here that's even easier so that turns out not at all difficult so after this change of variables you easily recognize the leading this should have been in a box because I'm giving the formula this thing is roughly two so there's a concept but it's two well two lots of digits and once you've guessed it then you just divide by this or you subtract this and you see that it's equal to exponentially high order so the exponential turns out to be e to the minus square root of two times pi times T that's not a hard number if you have an exponential or cosine you want to write it as something times two pi anyway and then recognizing square root of two is not very hard so this is the exact formula let me call this function and you'll see why in a second a one of T so here we have a one of T which is a pure exponential multiplied by cosine function or if you like complex numbers it's a sum of two pure exponentials e to the minus lambda one T and lambda two T where lambda one and lambda two have real parts squared of two pi but they've imagined a part also the square root of two pi both signs so that's the first the beginning and now you continue and you say okay I won't write that formula again because I haven't erased it it's right there so epsilon two of x which is two pi over T squared is very nearly epsilon one so now that we know that I mean this is experimental everything's experimental you subtract that and you make a graph of this next function and the next function again is exponentially small but it's exactly the same thing you look at the highest points you take the ratio you find the new lambda and then you take the other part the cosine and it turns out it's really easy so this thing is just as easy as what we just did it turns out that it's a new function and it's particularly easy to recognize because it's just the original function stretched by factor squared of two so let's call this one a two of tau and now if I take epsilon I won't keep writing this minus a one of T minus a two then still just as easy you just do the same thing and what you find well you could almost guess it's T times the square root of three which I'll now call a three of tau a three of T so this would be something with e to the minus pi times I guess the square root of six e to the minus square root of six times pi T but now so now of course everybody can see you know that's how it goes on I'll put some l from one to three a l of tau but it's not a four what you would now guess is of course two T T times the square root of four so that would be if it were that in fact sorry a one of two T so it's not a one of two T it's something which you can recognize and I'll say it so let's call this a four of T but it's no longer given by quite the same formula let me get it right a four of T it has exacted the exponential you would have expected namely the previous exponential e to the square root of two pi T times two which is the square root of four but the pre-factors no longer two cosine it's one plus four times the same cosine so it's not you know wildly difficult to recognize but it's quite easy to recognize but it's not easy to predict and above all now what's the next one going to be okay so already there's a bit of a surprise that the you know there's a slight irregularity so now we continue I'll go I'm going to the left because I didn't plan the use of the board correctly so now when you do the next one now epsilon minus the sum one to the four of a i of tau and now I'm using i I think I used i is indeed a five of tau which is just what you would expect so this is defined as a one of square to five T so here there's no anomaly but now the next one is not just not right when you subtract that epsilon of this minus the sum one to five a five so at least when we subtracted one to three it wasn't quite what we expected but it was the right order magnitude it was the same exponential it was just a different cosine but this thing is much bigger than what you would have expected which is e to the minus squared of six times two so twelve times pi times t I mean it's exponentially much bigger it's the wrong answer so we have a new term which we didn't expect at all and so you do the same again you stared at it for a while and it's harder to recognize because when you can guess the exponential after a while you divide by the exponential but now you see a function that's not a pure cosine curve it's periodic it does I can't draw it but it's got two cosine terms so it's a sum of four pure exponentials so it's already considerably harder to recognize so in fact this sum when I subtract from one to five the next term is something completely different it's again two cosine but this time it's two plus the squared of two times pi t plus two times the cosine of the complex algebraic conjugate two minus the squared of two pi t times e to the minus two plus squared of two pi t which is not at all the squared of twelve pi t it's a it's bigger well I mean it's a smaller number but negative so this is a bigger sorry this is all yeah everything is small I mean it's extremely small but this is not as small as squared of twelve so let's call this one b one of t so now we have a one up to a five where a one is this rather simple function a pure cosine function where's a one even gone I hope it's still did I raise it if I raised I should put it back so no uh yeah a one of t maybe I should put back was is it still here somewhere where ah it just didn't say oh yeah here a one of t thank you so a one is this pure exponential a two a three and a five we're just this where you replace t by t times squared of m a four had little anomaly but now we've got this b one which looks completely different so now we continue and then after a while you see if we keep continuing that way we have more and more guesswork and at some point you just it wouldn't work that way so but let me give you the next little bit of these are I'm telling the actual truth these are the experiments in the same order that I did it basically it's once you have to do as you study it but it it takes several days then you at each stage you find the next bit it keeps coming out a little at a time so when you continue uh then well I'm going to stop for that but for the moment because that is where I stopped instead at this point we have to think a little this was already quite a nasty thing to recognize it turns out when you subtract this the next term is fairly easy it's exactly the a six of t you would have expected and so is a seven but then again you have a funny one coming in and it's again the sum of several terms like this a kind of a b2 and it will be called b2 they get very hard to recognize but now is the second simplification well I'll write it in a second but first I'll explain the reason so what is the problem all of the terms so far for epsilon all of these a and b and so on were epsilon which here was 2 pi over about epsilon 2 of 2 pi over t squared is the sum of pure exponentials I mean a linear combination sum I guess the sum with coefficient so linear combination it's got let's say c nu e to the lambda nu t minus lambda nu t where the real part of lambda nu is going to infinity but the problem is that the that there are several news with the same real part and so there's different exponentials with exactly the same growth with different oscillatory different phases so the solution now assume that there were one just one if I had a function epsilon which was just c e to the lambda t so if I have some function of t let's call it g of t and if it is this plus exponentially smaller terms then there's absolutely no problem when it's exponentially good recognizing lambda t because you just take the ratio g of t plus one over g of t when t is large t is going to infinity then the c will cancel the lambda t will cancel you have just lambda plus exponentially small in t so if t is 20 this will be 10 to the minus 10 or something so you get e to the lambda to the high precision take the log and you recognize lambda once you've got lambda you divide this function by e to the lambda it converts exponentially rapidly to c and so you're done the whole problem is we have two we have now we have let's say two terms like that before c one e to the lambda one and c two but c one is c two bar and lambda one is lambda two bar and then they get mixed up and then in the next case we had four such terms so it gets harder and harder so the trick is really simple and this is the trick I'd found I'm sure other people have thought of it in life well I should write out his name Stavros Garafolitus with whom I did the work that was discussed in last year's course we had similar asymptotic problems for something we found the trick and the trick is really really simple you let t go to infinity as before but along a line in c with a positive or negative slope you just choose the slope so instead of your t simply you go to infinity 10, 11, 12, to 15 you go at some angle let's say 20 degrees it doesn't really matter when you do that the t is now some absolute value of t times e to the i theta where theta is maybe relatively small now the different terms lambda and u of t will have completely different real parts and so one will be asymptotically dominant as long as you have different terms different real parts you just do this you take your function at t plus one and t the ratio tends to e to the lambda you divide by e to the lambda you get the c, subtract it you know? what? oh of going it, yeah but no but his real trick was taking a big sum of n terms and that surprised me no the thing of coming at the angle indeed I've been using now for three or four years thanks for the comment, yeah okay and here you should be careful if you take this way high then much later terms that you don't want to see yet will start to dominate so you just want to go ideally you just go in a three degrees it's just to separate like you do when you do spectroscopy there's a line with multiplicity three and you do something so it separates but if you separate it too little they're hard to recognize you want to separate it enough but they shouldn't bump into the next eigenvalue but when you do the experiments there's no problem and so when you do this it immediately works and you can get the next couple of terms and so what you find so this works so it works here I can be enthusiastic and put it works great it's not quite English but it's standard colloquial English it works great and up to the order of e to the minus so remember we always had if I just did the ams it was the squared of 2m times t so if I would go I'm going to include 11 but not 12 so I'm going to go beyond 22 I actually stopped at 22.3 because then the terms were still getting harder and harder and then comes the third simplification but so far if we go up to this then what you find is that epsilon m in this range epsilon 2 we're always doing s is 2 is the sum well the 11s remember e am of t was roughly e to the minus squared of 2m times pi times t so 2m should go up to 22.3 so m goes to 11 not infinity 11 am of t and then we have some 4 bm of t and that's all you get up to 11 so we already had 1b1 that I wrote here rather complicated mess and so the ams let me give I think how am I doing on time I'm kind of doing okay maybe I'll actually give the formless for these because they're sort of fun and I'm given what I prepared I think I'm I'm okay so what you find I have to tell you what the am is so am of t I have to give you 11 values but it was simply 2 times the cosine sorry it's the exponential is always the same it's the square root of 2m times pi times t times and then remember the original one was 2 cosine of the same number squared of 2m pi t and this was for all am up to 11 because that's as far as we're going but remember for m equals 4 there was an anomaly it wasn't quite like for m equals 1 I think it's still on the board it was 1 plus 4 but for 1, 2, 3 and 5 it was the same so it's also for all am up to 11 if m is square free well there are only 3 square numbers that aren't square free 2 of them are 4 and 8 so powers of 2 and if it's 4 and 8 then in both cases you get just what we had before the same cosine but this time multiplied by 4 plus 1 so we already had that for m equals 4 and it's the same for 8 but it's not quite the same for 9 which is the only others non-square free number in this particular range that one has a 6 times cosine of the square root of uh... well I can just put 2m but of course 2m will be 18 but I want to write it more or less uniformly well this is t over 3 and you can see t over 3 is t over d where d squared divides m it's non-square free so the second term has a smaller order of magnitude and here it looks odd because I didn't divide this by 2 you'll see in a minute why it comes so actually this is the sum of 2 exponentials e to the ix and e to the minus x this is the sum of 3 but one of them happens to be e to the 0x and this is the sum of 4 so in general there's a certain number of terms okay but and bm is much more interested so the interest well there are only 4 of them so I already gave it to you if m was 1 so it's on the board and I don't want to write all 4 this I didn't put on my paper because the forms are too long here I'll give you either the first one or maybe the most complicated so each one will be a sum of 2 terms I'll do the last one so if m is 2, 3 if anybody wants to see them of course I can write it out I'll just give the last one this m has no meaning I've just numbered them in order but the fourth one will be 2 cos but now it's not just a square of a integer times pi t or even like it was before the square of 2 plus 1 it's the sum of 2 square it's square root of 10 plus the square root of 2 and then there's another term which is what you'd now expect which is exactly the same so it's 2 cos and then there's a single exponential which is again what you now might expect square root of 10 plus the square root of 2 times pi times t so the b terms have a more complicated exponential which is e to the minus pi times the sum of 2 square roots of even numbers instead of a single square root but now if you keep going you'll eventually come to your first c which is three things it gets more and more complicated you wouldn't want to do this forever and as far as I know there is no reasonable closed form like this but the third simplification is actually obvious once you think of it but for some reason I didn't when I was doing this because you get hooked you know getting one more term at a time and then you said oh I can recognize that and I got you know 11 plus 4 that's 15 terms and of course it was already right to 400 digits and so one is very pleased but the third simplification is since well the reason is remember this one plus epsilon is the error term in p s of something which is some pre-factor which is something exponential times one plus epsilon but p is defined as an infinite product so surely you should be looking at the log of it which is an infinite sum and therefore this exponential in front well there's a constant power of x and exponential that'll just be something linear and powers and maybe a log term but then you won't have epsilon you'll have epsilon minus epsilon squared over two and so so we should look at the log of one plus epsilon s of x instead of epsilon s of x itself and now when you do that at least for this case which remember we're still doing the simplest case s is two that doesn't make such a big difference s equals three would go the same way but we're still only letting x 10 to zero or tell 10 to zero or q 10 to one we're not yet looking at rational numbers of root stability which I'll do next but without these three simplifications you couldn't even find the leading term there and you'd be totally lost I assure you at least I certainly was and it's really numerically very unconvincing but if you do this here it's very easy and you immediately see what you're getting numerically so let me find in the notes on what page I put this so what you find is that you find to all orders so now this was a formula that was not to all orders this was up to e to the minus squared of 22.3 there were 11 a terms and even the a terms weren't in closed form I gave you the generic one for square free but some exceptions b I didn't even give you all four it's kind of a mess but then there'll be c terms d terms and so on with more and more complicated exponents but when you take the log again I'm just doing the case of s equals two but you could do any s it comes out much the same way then this one is the sum of pure exponentials and this is now to all orders but it's not just all orders in one over t which it already was but now it's e to the minus something t so in other words any error that we still have will be smaller than any exponential so it's more than exponential decay so it's equal I could put three equal signs very, very nearly and now it's the sum and now you have pure terms it's always I can call it m again it's simply e to the exactly what we had before squared of even numbers times pi times t times the cosine of two pi of sorry if squared of two m pi t so that was exactly our a term except there's a pre factor but that's all there are no mixed terms when you have a sum of terms and you exponentiate you get one that was the one of the one plus epsilon then we have all the terms then you have the half of the sum of two at a time three at a time but since they're all different exponentials that's what gave those mixed terms so this is just a huge simplification and the only problem is to recognize the coefficient and that turns out also to be rather easy this is the minus one sum so remember sigma we've used this notation before the minus first sigma so sigma k in general of n is the sum of a positive device of n of d to the k so here it's d inverse so that means that if it's squared of n then we have to take one over d summed over all d which as algebraic numbers divide the squared of n which means we take all squared divisors of m and so that's why when m was square free it's just one and you didn't get anything you just call it two but if m isn't square free you get mixed terms and then you get the whole formula so that's once you realize that take the log then you immediately just get this up to a coefficient and the coefficient is one, one, one, three halves one, one, one again three halves then four thirds and after a while you see them just take the sum one over d over all divisors over d such that d squared divides m so that was very easy but let me first of all write this now in a nicer form so I don't want to lose my notes so if I write this thing by the way this should be a reflex if you're used to doing any kind of number theory you would have this in very simple analytic number theory just formal, this is a double sum because I'm summing over m but here I have a sum d divides this was for any n so n here is square root of m so d squared divides m but then it becomes a separate sum I could write m as d squared times k where d and k now independently range over positive numbers and then I'll have one over d and the square root of two m will be nice the square root of k is a mess but that's just pure exponential and so I get a sum, the sum over d is the sum one over d times something to the d but that's again a log so what you get therefore is equal exactly to the sum m from one to infinity but then there are gonna be two terms because the cosine remember is e to the pi i t plus e to the minus pi t so it's an infinite sum m from one to infinity and then the sum over all just put plus or minus and then minus log of one minus e of course it has to be oscillatory and it's this two anymore in the square root i to the one half remember is the square root of i not square root of i is one plus i over the square root of two and that's why we had the square root of two with the two becomes square root of two and that's why we had that and the one plus i times a real number lambda is why we always said e to the minus lambda t times cosine lambda t with the same lambda so this is the formula but now this equality is that this infinite sum converges and is equal to that this was to all orders but actually big surprise for me except that as I discovered that Hardin-Roman-Leuchttin that come that century earlier big surprise present discovery when you now compute this new approximation to let's say a thousand decimals where t now doesn't have to be very big anymore you can take t to be one then what you find is that this is simply equal and so it's equal to this and that is I've improved it of course but that is the theorem that I gave at the beginning remember that was that a to two of minus one over tau was something with a to half the square root of tau and a to half the square root of i tau you have to figure out tau is ix and then there's some extra squares but this is exactly that statement but you have to still prove it but I've shown I haven't even shown this to every order I only showed by Orlan McClure and that the leading term was one plus zero all I knew was that epsilon is very small but then experimentally I got these successive terms but it is in fact equal and I'll be proving that a little bit later so there is a small subtlety and I want to mention it if you compare the theorem as we just got it here so now I've probably erased all the definitions of what p two originally was but remember the original thing was that the p two function itself it had a big exponential which contained gamma of one plus one of S zeta of one plus one of S and then there was a factor squared of T over some power of two pi and then there was this one plus epsilon so when you take the log I've just given you the one plus epsilon that's essentially the eta but if you remember the definition of eta which you have no reason to do so I'll write it out again eta of tau I'll just use the q notation in a q to a complex power means e to the two pi times that number times tau this was q to the minus a half zeta of minus S so for ordinary partitions zeta of minus one is minus 12 that was q to the 124th and then was the product m greater than or equal to one one minus q to the m to the S right this was our original sorry eta would be if S is one this is eta S of tau so there was this factor one half zeta of minus S but now we have this factor gamma of one plus S so the small point if you believe this thing let's call this equation star so this one we found empirically by finding log of one plus epsilon two to all orders we found the sum with the sigma minus one that I raised we rewrote it as a sum of logs this is eta and so that's and then you find out it's exactly true and I claim that's the same as the theorem I wrote at the beginning which well I could even rewrite it because I think this was square root of two I've only forgotten the constant and it's on the first page of my notes it was the square root of two pi I might as well write it again the theorem said that this was exactly true but when you compare then eta two contains this minus one half zeta of minus two but zeta of minus two is zero anyways you don't see it but eta half on the right hand side is eta of minus a half and here we had zeta of a half times eta of three halves so we need you have to use the I'll put the Euler but Riemann proved it but Euler found it the functional equation of zeta to get that the number that we actually had in some exponent which is two pi to the minus a half gamma of three halves zeta of three halves is equal to minus pi times the square root of two times eta of minus a half I mean it's it's easy if we know the functional equation but if you won't compare the two formulas it doesn't come out with the original as it comes out with the s on the wrong side so you need to use that okay so that finishes my discussion of the numerical fun how you find this expansion numerically and then the final surprises when you've so you first can't find it at all because it's oscillatory you have to do a little guesswork to even get going then you realize the oscillation of the square you should change variables from x to the square root of x which you call one over t then the next step is you let t go at an angle then you can separate the different expansion then you can write down the first 20 terms but after a while you have to give up and then you say how about if I take the log and then it's easy to recognize so we're going to do all of that at roots of unity and then we'll absolutely need it because it's still much harder to recognize even when you have that it's only barely doable I'll do that I think that's the last thing I want to do today so I can do that right now and then I'll stop for today so when you do p s I think I want no I did already write it I think for p s back you know p s did I write or didn't I e to the minus x I think I actually wrote it but if I didn't just to feel virtuous I'll put it so the full expression for p s of x is x goes 0 s is a fixed integer is squared of x over 2 pi to the s over 2 and I know I did write it I remember now it was the gamma 1 plus 1 over s zeta 1 plus 1 over s x to the minus 1 over s plus 1 half zeta minus s x and then times 1 plus epsilon of course I wrote that okay so that's what happens is x goes to 0 but now we want to know about rational points and now we're going to need all of our skill I mean it would be completely hopeless without the three tricks and it's actually it certainly took me more than a week with the tricks before I recognized all the terms I kept making every mistake in the book so I presented it as if it's really easy to do but frankly it's it's kind of a mess so I'm going to take so the final question for today is what about rational points so either we're going to look at a to s I'll usually take s equals 2 because that that's not the important thing we all behave the same either a to s as tell tends to a kappa which is a rational number I can't quite put q over z because a to s of tell has this exponential factor that maybe I wrote and maybe I didn't here and this is a rational number so if you change tau by one you don't change q but you do change q to the one twenty fourth you have to change tau by twenty fourths in general it would be periodic but that's not the problem so tell is really in q not q mod c so it's some rational number uh... that's the easy part but so so an equivalent to q which is e to the 2 pi tau tends to some z which is a root of unity okay so I'm going to choose as my example s equals 2 and the non-data of cap is 5 so I'll call it a over 5 where a is prime to 5 and I mean actually if I don't do it for eta but if I do it for q then uh... it is now periodic except now a p p2 remember is the same thing as eta without the pre-fact and with an inverse so now it's truly periodic period one it's a power series in q and so now cal is really in q mod c so here I wrote a is 5 so you could take a is 1, 2, 3 or 4 but it's periodic and so a is really an element of z mod of 5z star multiplicative so I'm going to do this that's the one that I chose if I started with the half maybe it would have been easier but maybe it wouldn't because then it's much simpler and then you wouldn't guess the next one anyway the one I chose to do on my computer was work it out at 5th root infinity as already said there's no problem getting the numbers so we're going to again write uh... you know it's a p2 remember I was writing z to n for the standard e of 1 over n so z to 5 is here and then z to 5 squared is here here and here so here I'm talking about z to 5 for the power a times e to the minus x and so what you find with Euler-McLauren so this I won't go through I did it today uh... at zero but when I talked about Euler-McLauren in the earlier thing we didn't just have the sum f of nt and from 1 to infinity but more generally we had to shift to Euler-McLauren with the rational number which here would be for instance you know two-fifths or something and then there was a similar form then you can do it so that I'm not going to go into but the result of Euler-McLauren shifted Euler-McLauren is exactly the same and the same miracle happens that it's terminating so though you would have had in the exponent when you take the log an infinite sum of Bernoulli numbers times times other Bernoulli numbers but luckily it was always two of opposite parodies and so it was a terminating sum so what you find is again maybe I should write it the constant should be separate so it's still square root of x so that's a little strange it's as if all of these things had way to half it doesn't matter what x you take it's always square root of x and then there's always a constant over x with this constant depends on a mod five but it's periodic so I'll write it as a function of a over five mod of one over the square root of x and then again there'll be an epsilon two comma a over five so that's my kappa of x which is again exponentially small but here C a is not such an obvious number C a over five is the following number it's pi when you work it out with Euler McLaren it takes on two values depending whether a is plus or minus one it's even so a is either plus or minus one mod five or it's plus or minus two mod five which happens to be quadratic residues and non residues it depends on the the genre symbol and the actual formula is we still have this eight of three halves that we had before remember there was this eight of three halves or one plus one of rest times gamma three halves but gamma three halves is just a half square root of pi so that's some square root of pi that didn't get it got lost because I didn't write it this was square root of pi over two square root of five so it's one fifths eight of three halves and then plus or minus depending whether a is a quadratic residue of five or not so plus in this case and minus in this case and then it's the value of the L series the Diraclet series again the same Diraclet series so you know L of s sum over five I remind you is the sum n over five n to the minus s it's the Diraclet series one minus one over two to the s minus one over three to the s plus one over four to the s plus one over six to the s and so on okay so and then you can compute that you know two thousand decibels he does it instantaneously you just ask for it and he gives it so this constant is known and again we know uh... first all orders in x that's what all the McLaurin gives that this is smaller than any power of x and now you do the same as before and so the kind of the same thing happens and you find now we know remember the three tricks is first you change variables from x to two p over two pi over t squared and then you change you go at an angle then you can separate the different terms and then it's a mess and you take the log so when you do all of that what you find is something that's almost easy to recognize but not quite it's not quite which which really was kind of a mess and now we're taking the log of one plus epsilon and we get that this is the sum m from one to infinity and again this is sum over plus or minus and then there's a certain coefficient which remember before with sigma minus one of the squared of m when kappa was s is still two but kappa is now a over five and then here there's again a pure exponential and multiple writing mistakes it's one plus or minus i times the squared before it was m now m over the square of m over 125 because remember I'm looking at this kappa is a over five so that part is fine you get a pure exponential times the cosine or sine of the same number because it's one plus or minus i so that's very nice that's good but you'd expect but this coefficient is really kind of a nuisance and so I'll just tell you briefly and then give you the final formula m equals one remember before when m equals one was a one and that was just a single two cosine times the single e well here's the same so this one again is easy to recognize just as it was before now when you continue remember before a two a three and a four well a two and a three were very easy they were just a one of t squared of two a one of t squared of three a four was a little trickier but that's only because I hadn't taken the log if you took the log it was well it was slightly different the coefficient changed from one to three halves but it's very easy and so m equals two three and four are similar and therefore easy to recognize after you've recognized that when you changed a little bit in some very obvious way but m equals five is different it's not very different so when you stare at it you eventually also can be recognized when you look a while I'll write it in a second but it's already trickier so we have you know before there was once I took the log every term was the same it was just sigma one of one over squared of m just simple numerical coefficients but now five remember five is special I'm looking to nominate or five is now different but now six seven eight etc are like two three and four well one two three and four and similarly 10 15 and 20 are similar and they're a little different but once you've recognized five it's easy to change things and to guess what you do so then I thought I'm done now 25 and there are just two kinds of terms after all five is a prime but then unfortunately f equals 25 is new again and so when you go beyond 25 and you know we're talking about exponential things we're talking about now e to the minus 300 I mean it's very very you know we have a number of hundreds of decibels because you have to subtract all this stuff and then what's left is very small so m is five is new again then of course 26 27 are like one and two 30 and 35 are like five but indeed 50 is like 50 75 and 100 are similar but then you start worrying because the next one is 125 and now we have something for every part of five but luckily the next one kind of fits the pattern so it's you can write a formula which is uniform also for that but you aren't sure it's true so if you're me at least I only was sure that I was writing down the correct formula when I'd computed these once I know this you can compute each numerical copies by subtracting the previous and divide it by the next exponential which is exponentially bigger than the next term but only if t is very big which means these things are incredibly small sounding like e to the minus 1800 but you can do it and then only when I'd gone to 300 was I completely sure you can guess the answer completely but and in fact I said 125 but there was some ambiguity in how you write it and so it wasn't completely clear but the formula that worked for 25 then worked for the next few multiples up to 300 I didn't go to 625 because at some point it's so then anyway once you've guessed the same miracle will happen then it will become an exact formula I can already reveal that so let me give you the final formula and the final formula so I didn't copy it on the paper so I'll copy it from my notes so here's the final final answer so the answer but it's going to be an infinite sum but then I told you the same surprise but it's no longer surprise now happens just before that now it's simply equal and if I move all the other stuff then it's just this log of 1 plus epsilon 2 a over 5 so remember a was 1 2 3 or 4 of x this thing is the sum of two terms remember there was a sum over m and a sum of ga so this is a remember a is 1 2 3 or 4 mod 5 so it's going to be ga of 2 pi to the 3 halves times 1 plus i over the square root of 125 x and then it's it's not even or odd in a but the other term you take the conjugate here and you take minus a so it's 2 pi to the 3 halves and it's 1 minus i but because of this mixture that really made it somehow not at all easy to recognize these things so this is what comes out where and now I have to define ga of x so there are already two terms but ga of x is not just an infinite sum it's a sum of two quite different infinite sums and so that's why I needed 300 terms before I could figure out the rule this is exactly the same as it was before this is sigma minus 1 of square root of m but it's not quite true because here I have a number a inverse times m and d I'll tell you the value of d in a second it's a it's a periodic function on the integers with period 5 okay and I can actually tell you the values so since it's for integers here's n and here's d of n d of 0 is minus 1 d of plus remember it's mod 5 so there are only 5 3 value and even no it's not even but d of plus or minus 1 is the golden ratio or it's reciprocal or negative reciprocal I guess it's conjugate and d of plus or minus 2 is 0 so now I take but a inverse n makes sense because the argument of a is modular m so this one is mod of 5 m might not be prime to 5 but then I'll get 0 which is minus 1 but a is prime to 5 second invert a mod of 5 and then it's it's a pure exponential but it's not e to the minus mx it's the square root of mx so that's the main term and that one is the one that if they all were the same would be all the same but unfortunately it wasn't all like that and so then there's another term and that one is in a sense even easier but it's lower order so it takes more time to find now you take a simpler function of a inverse m mod 5 which is just the genre symbol I could also put a m but it's really a inverse m and here there's also something but now it's sigma minus 1 of integers and so you can do that internal sum in closed form but sorry no you can't it was a double sum and only when you put it together so if you expand it out there are lots of contributions so this is the final formula so as x goes well I was going to say it's x goes to infinity when you do the experiments x is very large so you can keep the terms apart but in the end this just converts for any positive x or real part of x positive and as I say this is an exact formula so that's the end of the story and of course if you now exponentiate this it will give you a formula for eta 2 of tau as tau approaches ever five but it's a complicated formula that's the one that I have not written on the board I only wrote the one that's tau approach zero so next time I'll talk more about that and finish the discussion of the behavior oh I've actually gone slightly over so next time will be the behavior of this function well also giving some idea of the proof the behavior of these functions p to an eta at roots of unit I've explained how you do it experimentally but a little more how it works and then we'll be maybe next time I begin on the search method and then the third lecture we'll be applying all of that to to p2 to get uh by the search method to get information and that turns out to be full of surprises too that didn't happen in the case of ordinary partitions so I'm sorry I went slightly over time but I hope that all this numerics was kind of fun I mean it was an amazingly complicated numerical investigation to get you know for instance final form experimentally just really remember this itself the second sum so there's a double sum and then you know it's a complete mess okay so then see you I hope some of you next week next week or next Wednesday day after tomorrow in fact at 2 does anyone have questions I'm not sure we should have questions maybe we should have them privately since I'm already over and we shouldn't keep marker and the others sorry uh you can ask that question when I finish talking so it's not on my website yet it will be as soon as the course is finished uh I don't want I spent a lot of time writing the paper and I put everything the best way I know how so now I'm kind of following the exposition so if you have it everybody will be reading and it or potentially so you you don't get the reference there is no reference yet you will get it uh but uh for the moment you have to listen or wait for another week okay so thank you and any questions can I'll be private


Moving on to Member's Statements, I bet to inform the House that the following document has been tabled. A report entitled Expanditure Monitor 2023-2024, Q3 from the Financial Accountability Office of Ontario. Member's Statement, I recognise the Member from Leeds-Granville. Today I rise to recognise International Women's Day on Friday, March 8th and want to reflect on an exciting day in my riding of Leeds-Granville we had last month. Our region is rich with women entrepreneurship and that was evident on Monday, February 12th when I welcomed the Associate Minister of Women's Social and Economic Opportunity, the Honourable Charmaine Williams, the 10th Ministerial Visit to My Riding Since Labor Day. The visit began with an informative women in business and entrepreneurship and skills development roundtable discussion. The meeting was very successful with 16 women leaders across our region sharing very meaningful feedback as women in business. We followed that up the same day. The Minister and I toured our riding presenting congratulatory scrolls to five women-owned businesses. We were welcomed by Lynn Libos at Echo Clothing, Cynthia Peters at Maison-Maitland, Brenda Visser at Flowers of the Field, Jenny Stotz at Tandem Unified Wellness and Kathy Lewis at the Bobbley Cafe where we enjoyed a delicious soup and sandwich. I'm so grateful for the women who make a difference in Leeds-Granville and communities across Ontario through their skilled work, their resiliency and their leadership. Happy International Women's Day. Member's Statement, the Member for Toronto Centre. Thank you, Speaker. Today marks 150 days since the start of the war on Gaza. We grieve with everyone hurting from the senseless violence unfolding in the Middle East. We grieve every innocent civilian life loss. Instead of seeing the freeing of hostages and the capture of Hamas, we have witnessed the brutal massacre of over 30,000 Palestinians and the wounding of over 71,000 civilians and the collective punishment of Gaza by the Israeli government. While we condemn October 7th attacks, the International Court of Justices' interim measures to prevent genocide cannot wait. This issue might be international, but our local communities' grief knows nothing of borders or jurisdictions. I hope that this House and all members can reach out to their federal representatives and all Ontarians can do the same and urge them to support the NDP motion scheduled for debate on March 18th that calls on the government to, among many things, demand an immediate ceasefire in the release of all hostages as the NDP has called for since October 11th, suspend all military trade and technology with Israel, and increase efforts to stop the illegal trade of arms to Hamas. As April is too far away, immediately reinstate funding to UNRWA and support the independent investigation, demand unimpeded humanitarian aid to Gaza, advocate for an end to the decades-long occupation of Palestinian territories, and work towards a two-state solution. Never again means never again for everyone. Thank you. Thank you, Speaker. This past Saturday I had the honour of serving as a judge at the New Market Lions Club Effective Speaking Contest. The Lions Club is a network of community-minded volunteers who continue to live up to their motto, We Serve. The Effective Speaking Contest is the largest bilingual contest of its kind in Canada. The goal is to promote competitive speaking to stimulate independent thinking among students. I was asked to judge the speakers of the French portion of the contest. Juniors grades 4 to 6 and Intermediate grades 7 to 9. Students delivered a prepared speech followed by an intermediate student presenting an impromptu speech. Topics range from the importance of teaching history to animal companionship to the impromptu speeches on who they admire the most and the impacts of social media. Criteria for assessment included subject selection and organisation of material, how the student kept the interest of the crowd, the delivery of that speech, etc. Congratulations to the intermediate French speaker Riley and the junior French speaker Rami for their outstanding speeches, earning first place accolades. They will go on to compete in the regional contest and perhaps to the provincial contest this May. Congratulations to all the volunteers. Members statements? A member from the Schegelach James Bay. Not a week goes by without people in my writing coming to see me to talk about the Northern Travel Grant in the last year when the Hearst neonatal unit was lacking an obstetrician and several women had to then be redirected to the Kappus Kissing Clinic that also had no space. They were then sent elsewhere. They had to deliver three hours from their home. How can we help these women? The government, with this travel grant, gives people a certain maximum and it's not a realistic price. Find me a hotel that somebody can pay for this grant. That's just one example among hundreds. In Kappus Kissing there is a huge waiting list for doctors, for family doctors, about half of the population and the travel grant only applies if one is going to see a specialist when the government, the government needs to improve this grant to truly support Northerners, but rather instead they decided to play politics and abandon these people. If only the Minister of Health could go up north and see that it's a two-tier system, maybe then the Minister would push the government to change the conditions of this travel grant. Thank you Mr. Speaker. Thank you. Member statements? The Member for Kitchener, Conestoga. Well thank you very much Mr. Speaker and I want to talk to you a little bit today about nosies. A notice of security interests commonly referred to as a nosy are a tool similar to a lien that can be placed on the land registry system by a business or personal property without the individual's knowledge. This nosy placed on the title of the home creates an encumbrance on the title that they must clear before they sell or refinance their home. Scammers target vulnerable people and can leave them with massive debt. This is an issue right across the province, but Waterloo Regional Police have been dealing with it for years. Detective Adam Stover who's here joining us today, who I've had the pleasure of discussing this issue with, is an expert on the subject. Over the past few years these scams have evolved Mr. Speaker. A recent CTV News article stated Kitchener resident Ian Craig has seven nosies on his property totaling more than $150,000. He's fighting to make sure that others aren't scammed too. Ian said in the article, this is not right. For people who have invested in their homes, that somebody comes along and puts a nosy on it for more than half of it. The way they can take advantage of people is just disgusting. This is why we need protection. Speaker, it's not right and that's why we are fighting back. The Ministry of Public and Business Service delivery recently ran consultations to get to the bottom of this issue. I want to thank the Minister of Waterloo Regional Police, PA Riddell and the entire team at the Ministry for their hard work. Let's stop this unscrupulous practice. Let's say no to nosies. Member Statements, the member for London Fanshawe. Thank you Speaker. Today I would like to use my platform to amplify the concerns expressed by those in the child care sector regarding issues attracting and retaining staff and early childhood educators. I share these concerns on behalf of child care providers across the province, but specifically one child care provider in my community. They are losing 15 ECE graduates of Fanshawe College because they are leaving Ontario. They are relocating to take advantage of a better immigration pathways for ECEs and other provinces. The loss of these workers directly impacts the care and education of 120 preschoolers who would benefit from their expertise. Without immediate action we risk compromising the quality of early childhood education in our communities and hindering the future success of our children. The Ontario immigrant nominee program serves as a gateway for individuals with the skills and experience that our economy needs to obtain permanent residency in Ontario. We must recognize and take proactive measures to retain and attract skilled professionals to this vital sector. I call upon the Minister of Labor, Ministers of Labor, Immigration, Training, Skills Development, Education and College of Universities to address this urgent matter by implementing a comprehensive provincial immigration pathway tailored explicitly to early childhood educators. Investing in our educators and prioritizing quality of early childhood education is not only essential for the well-being of our children but also for the prosperity and success of our province as a whole. Together let's work to create a more equitable and prosperous future for all Ontarians. Thank you very much. Member statements, the member for Markham, Thorn Hill. Mr. Speaker, I rise today with the heavy heart full of sadness. We mourn the loss of truly remarkable leader. Brian Malroni was a visionary and principal Prime Minister of Canada. He dedicated his life pursuing Canada, national interest and humanitarian issue worldwide. Mr. Speaker, when I came to Canada in 1983, I was inspired by Prime Minister Malroni and what represented to people like me coming from a war-torn country. His compassion, his leadership and his dedication to freedom, democracy and social justice improved the life of millions of people around the world. He was a truly a dynamic leader for Canada, a transformative force for positive change. During the 1983 program against the Tamil people during the Civil War, thousands of Tamils came to Canada seeking asylum and better place to live. Prime Minister Malroni first strong pressure not to accept such refugees. As a man of principle and strong moral conviction, he simply replied that Canada is not in the business to turning away refugees. Prime Minister Malroni didn't sigh away from international pressure either. Despite the opposition he faced from his allies, Prime Minister Malroni's support of Nelson Mandela against anti-apathite government of South Africa helped bring about end of the ugly regime. He was always committed to do the right thing, not the popular thing. Mr. Speaker, leaders are never born, they are made. Prime Minister Malroni's spirit burns bright that will light the world forever. Thank you Mr. Speaker. Thank you very much. Member's statements, the member for Don Valley East. Thank you Mr. Speaker. Today I rise to discuss an issue affecting many of my constituents in Don Valley East, dem evictions. Dem evictions occur when under existing regulations landlords evict their tenants to demolish and rebuild. Under the current government's lack of oversight and tenant protections this often leaves tenants with nowhere to go and if they can ever return it's a significantly higher rent. This is the case for tenants of 10 different buildings in my riding. These tenants have a limited safety net under current regulations predicated on something called rent gaps payments. In this developers can pay the rent gap between a tenants current rate and in most cases the CMHC's 2022 average rate for a similar apartment. The problem though is rents have skyrocketed since 2022 meaning the present rate is entirely insufficient for my constituents. At 48 Grenoble Drive a low rise apartment in Flemington Park the rent gap payment falls $1,500 short of competitive pricing. That's not right. My community is made up of new immigrants, seniors, people dependent on fixed incomes. These buildings represent stability in troubled waters. We need more housing but we've got to protect tenants and lighten their loads so they're not left high and dry. That means fixing rent gap payments ensuring tenants have the right to return and making sure that no tenant is ever left behind as we fight to ensure everyone in this province has access to affordable housing. Thank you very much. Member Statements, the member for Peterborough Court. Thank you Speaker. On March 1st I had the pleasure of introducing a brand new form of hockey to Peterborough. Volt hockey is played using specially designed power wheel chairs. The stick blade is attached to the front of the chair and a big wiffle ball is used for the puck. It's played on a basketball court with pond hockey style nets. I was first introduced to it by Karen Stintz from Variety Village when I attended the North American Championships. I had the opportunity to speak to a family from Boston whose daughter was a quadriplegic. Their daughter was able to play volt hockey because the controls were adapted to work with the movement of her head. Her parents told me they never thought she'd be able to play a team sport and yet here she was at the North American Championships. They described it as life-altering. I knew instantly that I needed to find a way to bring this to my community. Speaker we worked with Variety Village, Five Counties Children's Center, the YMCA Eastern Ontario, the Quartha Comets, community alternatives, and the Council for People with Disabilities to create the team. The final hurdle was the startup funds. But Speaker, Saundra Sukumaran, Deanna Hayes, Peter Lemelin and Glenn White from the under-the-lock hockey tournament, the tournament I founded and chaired for 10 years, agreed to donate all of the proceeds from Canada's most inclusive hockey event to start volt hockey. Thank you to everyone who helped bring this vision to help kids with disabilities play Canada's favorite game. Thank you. Member Statements, Member Ford, Eglinton Lawrence. Thank you, State Day, a global annual holiday drawing attention to women's human rights. But this year it's hard to celebrate. On October 7th, Hamas terrorists made systematic sexual assault and barbaric sexualized torture and violence a weapon of war and the evidence has been available from that moment. The terrorists broadcast their depraved actions. The world saw the footage of multiple women being carried away on vehicles with telltale, bloodied sweatpants and many bodies that speak were recovered. On November 14th, a young woman reported to international media that hiding in the bushes she witnessed a horrific gang rape, mutilation and murder, corroborating physical evidence of this assault was found at the scene. After that report there was no excuse for any women's or international group to equivocate and yet many have been shamefully silent or worse. According to the Me Too movement we're supposed to believe all women who testify about sexual violence but this year we learned that apparently for some that does not apply to Jewish or Israeli women. This week a UN report found evidence of sexual violence by Hamas on October 7th that was conflict related and there are indications that sexual violence continues to be carried out against those still being held hostage by that group which includes 14 Israeli women. On October 7th, 2023 in its aftermath we're marking a very dark period in the history of the international women's right movement as women in Israel have been betrayed by many. Women's bodies are not a battleground on which to wage war.  It cannot be contextualized. There is no yes but when it comes to rape. Now more than ever women need to stand together to support all women because they are human beings. Thank you. That concludes our member statements for this morning.