Because I'm wearing a mask. So as usual, we're a few seconds late. And I ask the few people here or the also fewer than it was people on Zoom, if anyone has questions or remarks about last time or any other time. Otherwise, I'll just continue. So let me remind you very briefly what our aim is in this section of the course, which is going to be three lectures last time, Monday, this time, and next Monday. And what I do on Wednesday, I don't know. It depends how much of my notes on this very solid convergence series I can find. So I don't make any promises. But for these three times, we were studying the problem going back to Hardy and Ramanujan in their famous paper on partitions of 1918. They mentioned, or they actually discussed in some detail, more than I'd originally realized, the problem of the powered partitions, the square or more generally power, let's say, s-th power, partitions, so p, s of n for n, well, both s and n are natural numbers, is the number of partitions, which means splitting up as a sum without regard to the order, partitions of n into s-th powers of natural numbers. So what that means is that if you define the generating function with a product m from 1, 1 over 1 minus q to the power not m is for Hardy and Ramanujan or rather for Euler, but m to the s, which is what they also wrote, that defines the function capital equal colon, and then we expand that as a power series q and call the coefficient of q to the nps of n. So that was the problem. And so there are two parts. But actually, the two parts became three parts. So part one, which was last Monday, already done, is discover the all orders. So discover the asymptotics, first the asymptotics of ps of, let's say, root of unity e to the minus x, first to all orders, all powers of x in x as x goes to 0. So here, say, does there any root of unity, not just 1? And guess so that you can do, so this is not discover, actually. This is find, but I did do it numerically, but you can do that by Ola McClourness. I'll explain today. Discover that and guess an exact formula. So this will be an exact formula describing the behavior of ps near root of unity. Then part two, which is today, fairly clear, you have to prove it. So I'll indicate the proof of the exact formula, how we first do the approximate formula. Last time, I only showed the numerical methods and said very briefly that you can use Ola McClourness. I won't give more details on that. And then part three, which will be next Monday, apply it. Well, apply to get the coefficients. So apply the famous circle method of Hardin-Ramonuchan. So remember, I mentioned that is long ago, in the first lecture and the second lecture, you can take, for instance, q is a constant where r is less than 1, or you could take some other contour which circles around 0, 1. So it doesn't have to be an exact circle of radius r less than 1. And then you just take ps of q, q to the minus n. Well, q to the minus n minus 1 dq. OK, so by Cauchy's formula, that is, of course, the formula for ps of n. And then the ideas, if you go around the circle, which has radius very near 1, then you'll come near all of the roots of unity. And there's a contribution from each one I will not attempt. And I mean, not because I don't want to give it to you, but because I don't know how to do it. I have not attempted to do anything like an exact analysis, the famous major and minor arcs of Hardin-Ramonuchan, Littlewood, Vienneproth, and so on. I don't even know if it can be done. It's probably a very non-trivial problem, maybe, of the level of a PhD thesis. It's certainly not something I think anybody could do in an afternoon or two. OK, but anyway, I haven't done it. So this will be what I'll do is find the exact contribution to all orders of each root of unity and add them up and see how far we get. But that's for next time. So today I'm not talking about the coefficients. I'm therefore not talking about partitions. I'm simply talking about this function. And of course, also it's logarithm, which is a sum. So first let me remind you from last time that just as in the case when s is 1, if s is 1, of course, then p1 of q, which is the generating function that Euler found of partitions, which is the product 1 over 1 minus q to the n, the nice way to write this is q to the 1 24th divided by 8 of tau. Where 8 of tau is, I'm not going to write it again. I'm going to write 8s in a second. And q is always E of tau, E underlined of tau, which always means E to the 2 pi i tau. And q to the lambda, even if lambda is not integral, doesn't mean you take the complex power. You take the complex number q to some random lambda power, but you take E of lambda times tau. You take that particular lambda power. So if s is arbitrary, we do the same. We'll write ps of q. Well, I've already written ps of q. But let 8s of tau in general be E to the minus half times zeta of minus s times tau. So if you want q to the minus a half zeta of minus s. So if s is 1, zeta of minus 1 is minus 12. And so minus a half of zeta of minus 1 is plus of 24, which is just what it should be. So this is, well, I can remind you, it's 1 over 24, product 1 minus q to the n. And this part, of course, is p1 of q inverse. So here, in the same way, we take ps of q. Well, I can write it else. ps of q inverse, which will therefore be the sum 1 minus q to the m to the s. Now, one amusing thing that will happen, I actually think I gave the result. Yes, last time I gave a formula that would imply it. But I think I didn't mention. The function 8 of tau is extremely small as you approach any root of unity. And the reason is it's a multiform. And so every root of unity corresponds to tau tending to a rational number. Every rational number can be moved to infinity. Infinity, the function, is exponentially small. Therefore, ps of tau, which is essentially 1 over 8, is exponentially big at every root of unity. And the size, the speed with which it grows, is something at q equals 1. Roughly the square root of that at q is minus 1. Roughly the cube root of that if q tends to a third root of unity, et cetera. But here, it's completely different. The behavior at each root of unity is different. And so already we did it last time, and I'll write it down in a minute. I did the experimental things when s was 2, the next case, and kappa was a number with denominator 5, a over 5. And zeta was e to the 2 pi i kappa, so a non-trivial fifth root of unity. Then what happens in that case, and I'll give the form this again, I'll remind I gave them last time, if I take the five fifth roots of unity, so this is e of a fifth, e of two fifths, and then, of course, three fifths and four fifths, then here, eta of p2 is exponentially big, if I remember correctly, and here, it's exponentially small. But maybe I got them reversed. I'll look it up in a second. But the point is, and in other roots of unity, maybe they're both big or both small, but with completely different orders of growth. And not only the different s roots of unity for a fixed c, so a over c, where a is prime to c, range is multi, not that they have different orders from each other, but they also have orders that you have no idea how to compare with the previous one. In the Hardy-Bramannutian situation, as I said, q equals one gives some exponential term, and therefore, some exponential term at the end in p of n, which is e to the pi squared of two n over three. It's something exponentially big and squared of n. But then minus one gives something e to the 1 half of the same x one, is half as many digits. And the two q roots of unity both give a third. And all of the 11th roots of unity give an 11th. So at each stage for each c, you add up all the contributions of that denominator. They're all the same order of magnitude, they're all big. But here, some of them are big, some are actually when the function is small, the contribution is actually zero to the order that we're interested in. So some don't contribute anything, but even the ones that do contribute, some contribute something huge and something very small, and they don't come in the natural order. What does happen is that c equals one, q equals one, is always the main term. But for instance, q equals four, I think, for s equals two is more important than c equals four. So i and minus i turn out to give bigger contributions than the two predecessors, minus one and the q-th root, q root of unity. That we'll see next time. But the effect on the pn, on the little p s of n, but the effect on big p we'll see already today that the growth is not at all of the same order and it's not even always big. So the function a to s can be big, even though in the classical case it's a cost point, it's exponentially small near every root of unity in q or near every rational point in tau. Here it can be very big, which means that p s can be very small and you get no contribution. So that's, so now let me recall what we did last time. So we always had this normalization and it was only partially justified. Last time today when I write on more explicit formulas, we will see that it helps a little, but even last time I said that for the final formula to see that exactly this a to minus s is the right thing to simplify the formula, you have to do some work with the function equation of the Riemann's a to function. So don't worry about it too much, it's just a harmless constant, but if you dump it in that constant, the formulas would be even worse. So what we had was, of course, in the case of the Dedekind a to function, we had a to minus one over tau, is the squared of tau over i, a to tau. But then I told you that this would become, and this is, I think I called it a theorem last time, a to s of minus one over tau will be two pi. So these are the two of the results for the main, two results I told last time, a to s of minus one over tau will turn out to be exacted. This is not asymptotic, but in equality, even though there's no modularity, it's exactly two pi to the s minus one over, two times squared of tau over i, tau over i, before s was one, so you didn't see this factor. And then it's the product, but now it's not the same formula, because the s becomes one over s. Of course, when s was one, that was the same formula. And also, it's not just minus one over tau and tau. Here, z is indeed appointed the upper half plane, but it's not tau, it's an f's root of either tau or minus tau. So in the case when s is one, this says z is either tau or minus tau. But since tau's in the upper half plane and z is in the upper half plane, it can't be minus tau, so it's plus tau. And so there's only one term, this gives the modularity. But in general, we'll have exactly two s. So this is the same, what I wrote here is the same as z to the two s is tau squared, but it's uglier to write it that way. So you see they're exactly two s zeros, but exactly half of them are in the upper half plane, exactly half of them are lower. So this is a product of, so they're s factors contributing. So it's not a modularity, because first of all, the function doesn't relate to itself at a different point, which a modular function does. Instead, a to s of tau relates to a to one over s of z, where z and, well, minus one over tau are related in some way. And secondly, it's not one value on the right, but several. So you can't go back, you cannot write a to one half of some unknown z in terms of a to two, because you'd have to combine it with all of its partners. So it's not at all the symmetrical relation, it's not at all the group action like here, and there's nothing like modularity. So the modular group does not behave nicely on a to s of the function. So that was one thing that I explained a little, how one did it numerically, first at zero, and then for other s. But then I explained a much more complicated example, which was an s was two, and a was an integer, but only multiple of five, and prime to five, so you could take a, to be one, two, three, or four, or more intelligently, plus or minus one, or plus or minus two, it's actually even. And so it doesn't, there are only two cases. And then I went into, I explained how you can numerically study at this, so this is a fifth root of unity, non-trivial fifth root of unity, and here x is going to zero, let's say, it's a real number going from above, you can also come at an angle, and remember that for the numerics, it was actually quite important to come in not just to zero horizontally, so here's x, x, and we're going to zero, but sometimes we wanted to come in on some line in order to separate the various eigenvalues numerically, but the final result, which was first found by a lot of work we're getting to 300 terms of the asymptotic, well first the asymptotic expansion was easy to all powers in x, that part was easy, but then when you subtracted that you got a second term which was again easy, it was the second expansion, and the third, and in this case, it was only after I had 300 of those terms numerically, which took like many hundreds of digits of precision and several days of work that I could recognize the rule, and the rule was in fact quite complicated, but I'll write it out again as I did last time slightly differently from last time, but very, very slightly. So there are five terms, the first is a constant, square root of five over two pi, the second is the power of x, which is x to the one half, I'm only doing s is two. Okay, so we have a constant with the square root of x, then the big term is the exponential times c A over five, so c A over five, sorry, A over five, it should be, and later this will be c two comma A over five, and this number, I don't have my notes, I have to go back and look at the notes from last time, and I want maybe, so last time I gave the formula for this c, so c A of five was the square root of pi over two times the square root of five, so the square root of pi over 20 if you want, times one-fifth times the read and minus A to function at three halves plus A over five times the value of the L series at A over five, and of course the L series is positive, this is positive, so if A is positive, so if it's one over five, I mean if A is a quadratic residue, so A over five is positive, then this is positive, but when you compute the numbers, maybe even have them written down, I certainly have them in my computer to a hundred digits, I don't actually see where it's written down so I won't bother, it's an even function so there are only two values, but c plus or minus two-fifths is actually negative, it is the opposite sign because the L series, the L series is not bigger than say it can't be because it's the sum with govish plus or minus one, but there's a one-fifth here, and so this is bigger than one-fifth of that, and so when you have a minus sign, it becomes negative, and so the result is that therefore here, that's just what I wrote here, that here P2 is exponentially big as I wrote, here of course P2 is very exponentially big, that's clear, but here it's much smaller because you have this factor, but that factor is less than one, but it's still positive, but here it's exponentially small because it's the exponential of something negative, so that was kind of a surprise that even the nature of the growth, whether it's growing or shrinking, even that changed, and then there were two terms, which were, well, I can put them just to save time, I'll put it as the sum, it's the same term twice, it's G of eight pi cubed i to the plus or minus one, so it's either i or minus i divided by, and now if I could only read my handwriting, I'd be happy, I think it's 100, and I'm sure it's 125x, so there are two correction terms and those are the ones that were a huge sum of exponentials that it took a lot of work to find, and so G of x is a pure sum of some coefficient, that was the part that was relatively easy to recognize, remember, as long as we tried to look at P itself, it was very hard to recognize because it was a mixture of different exponentials, so once you took the log, then these terms were the form of pure exponentials with square root of n times either this eight pi cubed instead of square root of i or square root of minus i, that was relatively easy to see, but the coefficient is something of a mess, and so here's what the coefficient is, this coefficient is one plus or minus the square root of five over two, so either the golden ratio or its algebraic conjugate times the sum d squared divides m of one over d if m is concurrent to plus or minus a, mod five, remember a is either plus or minus one or plus or minus two, mod five, so if m is also not zero, mod five is the same, the genre is similar to this a, that's the contribution that you get, okay? If m is also not zero, mod five, but has the opposite, the genre symbol, so a is a quadratic question, m not or vice versa, m is concurrent to a plus or minus two a, mod five, then you get zero as the contribution, and finally, if m is zero, mod five, then what you get, if I'm reading this from my notes and I hope I didn't make, I mean I even slightly transcribed the form, but I think it's correct, you take the sum d squared divides m, and then it's one over d, and so that's just like it is here times one except that this one plus or minus squared of five over two has become minus one, which is kind of good because it means the sum of all these five numbers will be zero. If I take this part, this would have a minus one, two zeros, and a one plus squared of five over two, one minus squared of five over two add up to one, the whole thing adds up to zero, so that part, so I think I'll call this science right, but then you have another Legendre symbol, which is a times m over d squared over five divided by five. So if I write it like that, I didn't write it quite like that last time, then you see that since m is zero, mod five, if d is prime to five, which it might be, then this thing is prime to five, m over d squared is an integer, but if m is zero, mod five, and d does have a five, then this will be zero, and it won't contribute this term. But in order for this to contribute since m is zero, mod five, m over d squared has to be non-zero, mod five, so d has to be divisible by five, and therefore m has to be divisible by five squared. So this is only, this only occurs if 25 divides m, and that's why when we wrote it out, yesterday the formula was kind of a mess because it was one thing if m was prime to five, but actually not one thing, but four different things depending on whether m was one, two, three, or four, mod five, and then to make things worse, if m was zero, mod five, there was a term which was just the same, except a different coefficient in front, but a new term which only happened at all if m was zero, mod 25, and that's why the things, you know, when you write it like that, it looks halfway reasonable when you're trying to do experimentally, it's impossible to figure anything out. Now if you take this thing and then take the last part and put in the sum, you know, q to the m and sum you can write it in terms of the log, and that's what I did last time, but the way I wrote it now, I think it's a little better. Okay, so those are the things that we found last time experimentally, and now the idea is that we want to prove, to show why these things aren't true, I won't certainly give every detail, this would be, there's a lot of calculation and it's not very interesting which takes too long, but I'll give, I'll try to give all of the main ideas. So let me start with the case that's well known, because that's when we understand everything, and then we can see exactly what we have to do and then we can do some of it for every s. So if s is zero, I'm going to actually number a couple of formulas so that I can refer to them. First of all, it's modular. The remember eight, it's probably still written here, eight of tau is q to the 124th, probably one minus q to the n. It's a modular form and that means that for any element gamma is A, B, C, D. In SL2Z, we have an equation, eight of A tau plus B over C tau plus D is the 12th root of unity, which I'll write as E remember, E to the 2 pi i times an integer depending on gamma over 24. This thing is an integer. Times C tau plus D to the 1 half, and then you have to make some conventions for which square root you take, but if I don't tell you what n is, then of course you take the other, you have to make a fixed convention, which is the same as the whole upper half plan where C tau plus D is never zero. So once you've chosen at one point, it's a constant, and then there's an n and if you take the other choice, you have to change that n by 12. I'm not going to write down every formula anyway, so it doesn't matter. Now let's write that if I define the appropriate logarithm. So I'm going to take a logarithm of eight of tau, but that's not actually correct. The logarithm of eight of tau, the logarithm of a non-zero complex number is not well-defined. It has inf the many changed by two pi. The only function that's well-defined is log of Z, which is the one that I think everybody knows if you start with the log, which is positive, real, on the positive axis, and then you extend to the complement of the negative axis, and if it makes you happy, you can do something on the negative axis. Take the argument of your number to be either i pi or minus i pi, we won't need that. So log is the principal branch, and that's a well-defined function, but of course it's not continuous. The analytic, as you cross the line, it jumps by two pi i. However, here, so this is going to be some log, some log, but that's not well-defined of eight of tau. As I move eight, if I go sufficiently close to the real axis, but this formula will not be, it's not going to be the standard branch, it's a particular branch. So it is a log, its exponential will be eight of, and of course it's clear what I have to do. Here I take the log of one minus q to the m, but since q is in the unit d, so is q to the m, and so one minus q to the m is always in the circle of radius one around one, so it's always at the right half plane, there's no problem. So we have the standard branch, and so we have a way to choose. And now if you take the statement that we actually want, of course, let me call that one, is it's the same as the first, but at the level of the log, then we get the same integer over 24, plus the log, sorry, times two pi i, undoubtedly pi i over 12 if you want, and then the capital log of c tau plus d, now I can make a choice plus h of tau, and so if I've made some reasonable choice, people actually take the log of minus c tau plus d squared to make sure, I don't want to worry about details here. So this is what we know because it's modular, of course that's exactly what we can use later, but now I want t to tend to a rational number, and the rational number I can always write as a over c, but I'm not going to actually, I'm gonna write this minus d over c, where c will always be positive, and d will be a number multiple of c squared, or actually let's say it's just primed, it's an integer primed to c, because it's not exactly periodic period one, might be period 24. So I'm going to let k, I'm not sure if this is capital, let me just say that I let capital, I think it won't be in a second, I just let it tend to minus d over c. So how will I do that? Sorry, there certainly could be and should be, thank you very much, yeah, there certainly should be half. I mean, I also didn't tell you which log, but that's not important, but that is important, thanks a lot. Okay, it's written wrong in my notes too, so at least I'm reading correctly, but it's not much of an excuse. So I'm gonna tend to a point that I call minus pi over c, and I'm going to write the number as minus d over c, and then plus, it has to be the upper half thing, i over x over c, where x is gonna go to infinity. Okay, so then tell will go to minus d over c, and gamma of tell, which is the same as it was, a tau plus b over c, tau plus d, will now be on the nose equal to a plus ix over c. But I'm not, well of course it doesn't matter if x is going to infinity, you know, but I'm doing it for the moment, it's exact. So now I suddenly can't remember whether this x, because I kept changing x and one over x, if you're supposed to think that it's big or small, no, this x is supposed to be small actually, because in the end, I want to be tending to a over c, which will be my capital plus, actually I'm gonna have x going to zero. So let's not even worry about tau tend to minus d over c, this is just a statement, if I make a change of variables, and I write any tau having fixed my gamma as minus d plus i over x over c, then gamma of tau is a plus i times x, rather than x inverse over the same c. So if I do that, then this equation becomes, this is exactly the same, it's now a plus ix over c, but now I want to write it slightly differently. So first there's this same pi i over 12 that we already had, times a plus ix over c, but now something has happened with the log now, there must be a log of x somewhere, c tau plus d, not able to do this, c tau plus d is i over x, so there's going to be a log of i over x, I'm not gonna worry too much, maybe half, don't worry about that, I'm interested, the interesting part, but I have lost the term, sorry, copying, and now what you get is of course the sum of the logs of this thing at the corresponding point on the other side, but that point will have a c through divinity. Sorry, no, but I'm right, I'm completely wrong, there is no, I'm not applying yet the functional equation, that will come later, I'm just multiplying it out, I'm alerting x is going to zero, so tau is tending to a number kappa, and the other part is that I'll write that kappa is a tau plus b some other tau here, so if I take this, then this is just the definition, h of something is pi i tau over 12 plus the sum, so I'm not doing anything, so I'm just taking this, but now q to the m will be e to the two pi m times this thing, as that's very close to e to the two pi i m times a over c, but that will then depend on m mod c, so what I get is minus the sum l mod a of a function that I'll define in one second of a over c, sorry, l over c, comma al over c, and two pi i x, this is not from the multilayer, this is just from the definition, from the original definition of h, but I'm breaking up the sum according to n will be congruent to sum number l mod a, and so I have to tell you what l is, so l of alpha, beta, and variable t, and I should tell you where these are, these are real numbers, but multiloseum, and here they're rational, but in principle they could be real, they could even be complex, but that's the choice of roots, anyway they will be real, so if I bother, and t will be the positive real number, or it could be the real part of t is positive, I think would be good enough, everything I do, or certainly if it's near the real axis, positive real axis. So l of alpha, beta, t is now going to be the minus the sum, so this is the important definition, alpha, beta, two numbers, multiple of one, I take the sum over all numbers, which are in the congruent class z plus alpha, so that indeed only depends on alpha z, but positive, we did exactly the same with Euler-Mclaurin, I'll remind you in a minute how Euler-Mclaurin shifted, Euler-Mclaurin looked, and then it's one minus e to the two pi i, and the application, the a l over c, so beta would be a l over c, and then here it's e to the minus u t. So that's the basic, the first step is you split it up into little pieces, exactly c pieces, where c is the denominator of the rational number you're approaching. Okay, but now, now we use Euler-Mclaurin, it's a shifted Euler-Mclaurin, which I talked about earlier and even proved in an earlier course, lecture of this course. So remember how it went, if f is some nice function, let's say it's smoothed at zero and small to infinity, then the sum, so f is more or less arbitrary, this is f of t and this is t, then the sum f of nu t, originally it was nu as a positive integer, but more generally it could be shifted, this was shifted Euler-Mclaurin, if f is again in z, r, z, or q, it doesn't matter, and I take only the positive numbers, so if f is zero, this is somewhere positive edges, otherwise it's shifted, and what you got to all orders was i f of t, where i f is simply the integral f of x dx, and then the other terms where there are two ways of writing them, one is you put the orbit z function, but already put it with the Bernoulli function, b r plus one bar of alpha, remember alpha is only multiple of one, this is the periodic Bernoulli number, which I'll remind you in a second what it is, and then you take the original Fourier expansion of f, it's zero, so f r of zero over r factorial, you multiply by minus b r plus one of alpha over r plus one factorial, which is the value of orbit z function at minus r, and then here it's t to the r, and that's to all orders, and so b r plus one of b n bar of x is b n of the fractional part of x, I make it periodic, unless n is one, remember if n is one, it's a sawtooth function, and then at the jumping points you take the middle, and this b n of x in all cases is n minus one factorial over two pi i to the n times, since it's periodic, it better has a Fourier expansion, some e of m x, and the constant term is zero if n is positive, so it's m different from zero, and the denominator of m to the n, actually you have two pi i m to the n. Okay, so this is the Euler-Mclaurin summation formula, so now we can use that, and so if I do that, then this thing here will be, by Euler-Mclaurin, one over t times the integral, that's an exercise, I won't do it, it's the dialog rhythm, remember, li two of x is the sum x to the n over n squared, if x is less than or equal to one, then it converts, if x is bigger, there's an issue of analytic continuation, but we don't care here, because this will be e of beta, or e of minus beta, I forget, e of beta, so remember, beta is also multiple of one, so the constant term, of course, doesn't care about alpha ever, because that only is the integral of f. I mean, here my f, I didn't say, f of t in this example will, of course, be minus log of one minus e of beta, so beta is fixed e to the minus t, and then I'm summing replacing t by nu t, so this thing only depends on e of beta, and if you just expand the log as sum one over k e of beta k e to the minus tk, you've got this formula trivially, so it's that, plus this sum r from zero to infinity, b r plus one bar of minus alpha, multiple mistakes, maybe it was minus before, over r plus one, it probably was, because it was the, I really don't care, I'm not gonna get every form of the right anyway. Do you have an entire expansion, and the coefficients here of t to the r involve the periodic renewing number, as they have to, according to this general Euler-McLean summation form, as I say, probably there's a minus alpha, I'd have to look, but it's almost the same, it's a question of a sign. I'm not sure if it's right or wrong, but it's that, but then there's still a part that depends on beta, and that's the negative index, l i, so l i, I can put l ij of x, is this over n to the j, so l i1 of x is just the function that we've been using, minus log one minus x, so this is simply l i1 of e to the two pi beta e to the minus t, but l i zero of x is just a geometric series, and it's x over one minus x, and similarly l i minus one of x is x over one minus x squared, they're just polynomials, so this is basically just a trigonometric function, it's a rational function of e to the two pi beta, as long as, as soon as r is strictly bigger than one, even equal to one, even equal to one, yeah, so only the term r equals zero involves a log. Okay, so that's that, but now, I mean, so far this has been straightforward, it's just what you'd expect, but now comes the key point, that is the reason that it's gonna work for eta, but you would think it only works for eta because eta's multiple, but it's going to work for every s, so that's the nice surprise, but first I'll do it for the case we know. The asymptotic series for l alpha beta of t, so alpha and beta are fixed numbers multiple of one, and t is going to zero, is infinite and it's factorially divergent, so there are infinitely many non-zero terms and they blow up and never converges, and the reason it's kind of clear from this, this thing grows, this one also grows like r factorial, this grows like r factorial, we're dividing by only one r factorial, it's not enough, believe me, it blows up like r factorial, so it's no good, but the asymptotic series with the symmetrized function, so let's call this l sim of alpha beta and t, by definition is the sum of what it was before, plus it's really, it's actually twice the symmetrization, I mean it's the, well it's the symmetrization, it's twice the even part, so if you change the sign to both alpha and beta and you add l of alpha beta t and l of minus alpha minus beta t, then when you do this what you use is that bk bar of minus alpha picks up a sign minus one to the k, bk bar of alpha, but li k of one over x, after analytic continuation, picks up a sign minus one to the k minus one, times li k of x, you have to do analytic continuation outside of the circle except on the negative cut, and so because of that in this product, now something has gone wrong because now it looks like it's at the same symmetry, so I must have shifted something by one, I've been feeling that the one minus r, where this term would be r is minus one, so one minus r is two, that's very good, actually don't see what's wrong, but these formulas also look right, so li one is the log, but of course that isn't true for k equals one, but for k at least two, li two is essentially an odd function under inversion, that looks like, that's right, r plus one and one minus r have the same parity, because of the two, I got confused, they both have the opposite parity part, so I have two k's with the same parity, and these two things add, so this is opposite parity, and so when I change the sign to both alpha and beta, everything cancels, and so what you get is much simpler, and while to know typing mistakes, which are certainly possible, it will be the periodic version B2 bar of beta over T, then there'll be a constant term, which is the periodic version of alpha, times the periodic version of beta, so B1 bar of alpha, B1 bar of beta, and then there'll be the last term, which is linear, and that will be minus B2 bar of alpha over two times T, but then that's plus to all orders, so that's the very nice thing, that what was a divergent asymptotic series now gives us some of two divergent series, but every term except the linear term, the constant term and the one over T term cancel, and you're left with the terminating asymptotic series, so now you have something exact, and in fact, so now if you put this formula back, into this one, then the important point, otherwise it wouldn't have helped us know that the symmetrized function was so nice, but of course here for free, I can put, well, symmetrics, the symmetric should be the even parts, I'd have to divide by two, because when I've L, I can replace L by minus L, then I change both signs, so they always come in pairs, alpha beta and minus alpha minus beta, and therefore the whole thing symmetrizes, and if you put that in, what you'll get, I mean you'd say the exercise takes a few minutes, we'll be, remember the C tells us T was log one over X, the log one over X is some extra term that we had somewhere, I can't trace it all through, but anyway, when you now plug it in, you get exactly the function equation of the eta function, but only asymptotic series to all terms, but it actually lifts, but in fact, if you think about it, we've got much more, because the eta function, and I've actually never seen this proof, I think of the transformation of the eta function in this form, if you look at A to a very near root of unity, then the product, one minus Q to the N, to a higher order, well Q to the N will be very near, and it's A to the N, and so it'll depend on what's A to, so it factors into a lot of things, and we're taking the product of all of them, but what I've said now is if you take the product where N is okay, zero mod C, let's forget that, but N congruent plus or minus L mod C, then I'm saying that the smaller products, let's say I'm zeroing on the fifth root of unity, but I take L congruent to two mod five, and L congruent to minus two mod five, I have to symmetrize, then those two together will also give a terminating series, and that if you know your multiple forms at all, you'll recognize immediately the product, N congruent to plus or minus two mod five, let's say, and also zero, so you always have to include zero and plus or minus two, and that's the famous Jacobi triple product, and that's some theta series, so some Jacobi theta, and then by the transformation equation because it's a Jacobi form, it's some other theta, and I'll just put one over x dot dot dot, so roughly you invert the argument, and so indeed you get something much stronger from this argument asymptotically, but afterwards exactly, it's not just that the whole A to function has a nice form, but near a C through the unity, it's a product of roughly C over two terms, each other over every pair, L and minus L multiple of C, ignoring zero, and each of those separately has a terminating series, you get something very nice, and that will happen, it's not so important to us, but it will happen exactly like that for A to S. So now I do the same for A to S, and probably I shouldn't even erase what I said, because a lot of it's very similar, so the multilayer, of course we don't have, H S of tau we already had, remember that this pi I over 12 will become minus A to minus S over two, times two pi I tau, so I guess it's minus pi I, is A to minus S tau, and this will be the same with M to the S, so that was the definition of the A to function, A to just the exponential of this, E to the minus pi I is A to minus S tau times the product one minus Q, and here this is M to the S, not M times S. Okay, so this is the log, some log, but a well-defined log, but I called A to S, and so now we can do exactly the same thing here. Maybe I, it's hopeless to try to, oh no, I think I can actually do it, so the pi I over 12, I think it's completely fine, will become minus two pi I times A to minus S, times A plus I X over C, and this will be L S, and the only thing that changes is that it's now A L to the S over C, so that's not the only thing that changes, maybe I will write it again because it's getting impossible to read at the bottom of the board, so the statement is that H S of A plus I X over C will be minus two pi I, Z of minus S times the same, that's the boring part, X over C minus the sum L mod to the C, but the mod each time L sub S of L over C, comma A to the, A L to the S over C, and then the argument is slightly, this is just all keeping track of the notation, it's completely boring, but when you do it, it's this. Okay, so it's this and L S of alpha, beta, and T is now minus the sum U in C plus alpha, and again, positive of log, it's exactly the same as before, the only difference here is of course that here we have E to the minus U T to the S, but the T, if I want that to be T, I have to put this two pi to the C to the S minus one times X to the power of one over S, otherwise it doesn't come out by right. Okay, so it's the same story, and so this again, gives some divergent asymptotics by shifted Euler McLaurin, it's a complete exercise to do, and then this gives an exact terminating, so this gives me terminating asymptotics, actually will turn out to be exact, terminating asymptotics for, and now the important thing is the definition L S symmetrized of alpha, beta, and T, but now it's clear what you have to do, it's not quite the same as before, you take the value at alpha, beta, and T, that would correspond to some L in our infinite product, but now we have L over C, L goes to minus L, this becomes minus L, but if S is even, that one doesn't change, so therefore the alpha is always minus alpha, but the beta either changes or doesn't change depending whether S is even or odd. So this symmetrization is slightly different, and that will mean that all the later formless are slightly different when S is even and S is odd, and so this again gives terminating asymptotics, which actually turned out then to be exact. So what you get, finally, as an asymptotic series, I'll just give the whole answer, because I've given essentially all the steps of the calculation with Euler McLaurin, but it takes certainly several hours to do by yourself, and even if I wrote it, you copy it, and it's tedious and there's absolutely nothing happening, so the proposition says, well, it'll be easier if S is even, so I'll write it for S even and then I'll mostly put the modifications if S is odd, so remember, alpha and beta are real numbers, multiple of Z, they don't have to be rational, T is going to zero, let's say a positive real number, 10 to zero, then what we'll get is that remember the first term is a one over T, and that's the integral, and that means it's very easy to compute, and it's two gamma of one plus one over S times Li one plus, before remember it was Li two, now it's Li one plus one over S of E of beta, so that's the main term, and then there's a very simple correction term if S is even, which depends whether alpha and beta are zero or not, remember it's only multiple of one, so when I say they're not zero, I mean they're non-interal. If they're both non-zero, then it's really simple, it's just zero, and that's the entire expansion, but if alpha is zero, but beta isn't, then the formula is log of one minus E to the beta, and if alpha is not zero, but beta is zero, then it's minus S times log, now I don't need absolute value because it's the real part, so it's actually absolute value of one minus E of alpha, and if alpha equals zero equals beta, if they're both zero, then it's S, and in this case there's a log term, which is why in the final expansion when you add everything else there'll be one log left over, because somewhere this product you have L equals zero, and then both L over C and A L to the S surface are both zero, so you always have that term in our application. So it's not that bad, there's a very, very triple correction term which is a constant or a constant plus a constant times log T, and the main term is just one over T times a constant, the constants depend on S, alpha, and beta, so it couldn't be simpler. If S is odd, then the answer is slightly different, it's still two gamma one plus one over S over T, but now it's the real part of Li one plus one over S of E of beta, so you have to symmetrize differently, and so you symmetrize with beta and minus beta, so you've got the real part, and then there are two more terms that you didn't see at all before, one of them surprisingly is the same as it was for the dedicated zeta function, so it's a product of two periodic versions of B one bar, number B one bar is the software function, so it's B one bar of alpha, B one bar of A times T to the zero, it's a constant, that's independent of what S is, it's always the same, okay, and then there's another term that you don't see if S is even because it's zero, no sorry, this is simply new, it's zeta bar of S alpha, I'm not gonna redefine it, in an earlier course I said this is the orbit zeta function, but made periodic, so the same sum, U congruence alpha mod one and U positive of one over new to the S, and then that's a negative interest given by the same Bernoulli number, made periodically we had, so these terms are new, and then we again have four terms, this is the same, the first one, so it's zero, the third is the same, and the fourth is the same, so that part doesn't change, these three, but this one becomes again the real part, so it becomes the log of the absolute value of one minus E of A, in the second case out of the four, so here's the complete asymptotic expansion of this thing, symmetrized, and so that's kind of a pain in the neck, but now I can finish, oh yeah, fused up a lot of the time, already just for the asymptotics, I wanted to get to the exact part, but I'll do that more, so the corollary of this proposition is the actual behavior of P S, and so now cap is going to be a rational number, cap will be A over C, so denominator C, and then here I'll put E to the minus one over T to the S, so T is going to infinity, so this is something you can attend to zero, and this will be a product of several terms, and the first one involves a generalized dataconsum, so this is one half, the usual dataconsum, which is the N that I had there, that N multiple of 24, it's a famous form to do the dataconsum, two or three, a trivial additive term, and then a multiple of the dataconsum, and the dataconsum is defined as the sum L multiple of C of the periodic Bernoulli polynomial B one of L over C times B one bar of A L over C, that's the classical dataconsum. Now here remember I told you that this extra term, which was only true when S was even, but here it doesn't matter, I said that part doesn't change, alpha and beta, but the beta changes, because it's L over C and it's A L to the S over C, so this is a modified, it's a generalized dataconsum, so that's kind of amusing, and you might say wait a second, that term is only here if S is odd, it's not there if S is even, but it's also not here if S is even, because then if S is odd, if S is even, then this is an even function, that's an odd function, the whole sum vanishes, so this is only non-zero if S is odd, so there's that multiplicative factor, then what you'd expect, which is a constant over T to the S over two, but actually the constant isn't just one, it's an integer N S of C, and this is the smallest integer N, positive integer N, such that it's S power is divisible by N, so when S was one it was just C, and if C is square free, then it's also just, I guess just C, but otherwise if C has some higher powers, it can be a little smaller, anyway it's an integer, and then we'll have a purely exponential number, that's the one that I wrote before when S was two and K was plus or minus fifth, I'll write it in a second, so that will be a term that's exponentially big, so that's the one that really counts, and then we still have this stupid one half Z of minus S that you wouldn't see when S is two, which is the case we care most, because Z of minus two is zero, but you see it for the dedicated data function, that's the Q to the 124th, and then we'll have again one plus O of T to the minus N for all N, so that's like what we found experimentally, that you get a terminating asymptotic expansion, the only thing I tell you is what this constant is, and that's also quite pretty, it's gamma of, well in the case of S equals two, it's gamma of three halves over C, I mean I wrote earlier today, last time when C was five, L mod five, and then it's the one over S E of L to the S Kappa, so multiple mistakes, none of this matters the exact formulas, except that you see that they're complicated, they have many pieces, but that there's, it's still a finite expression, so if you only want to all orders, then you're left with something finite, and this is to all orders, and the reason is because we're the terminating thing. Okay, so now that's the thing to all orders, but now the claim, I think that's, the claim, I'm not sure if I'm saying something right, it's not quite that it's exact, no, this obviously can't be exact, this is going to become a product of one minus something exponentially small, so we will lift this to an exact formula, that's the point, I mean, it's still, even for the dedicated data function, that thing which is one to all orders is still a product of one minus some powers of Q tilde, where Q tilde is E to the two pi I of, for instance, minus one over tau or gamma of tau, and that's very small, but I mean, so the, you'll have something exponentially small and powers out, but you'll still have it, so they'll be here exponentially small in some explicit way to the power T to be actually one over S, so we have to lift this to an exact formula, so now comes the important philosophical point, I mean, the thing to take home with you out of this complicated calculation, I made that point in a very early lecture of this course, actually there was a question, in fact, most of the questions by Emanuel, when I talked about the Euler-McClure information form, he said, well, couldn't you in some situations use Poisson summation, wouldn't it be better? And the answer is yes, when you can use it, it's way better, but you can't always use it, and the point is, Euler-McClure, remember, is the sum in the K form I did, a function f of n, or to keep track of how things were doing asymptotically, I put f of n t with t very small, and it's a sum over half of that, sir. You can, of course, include, if f of zero is finite, you can include this, but the Poisson summation formula is a formula if f is a function which is small in both directions, and you take the sum f of n t, or in both directions, and this one is an asymptotic formula, this is the one that we know, i over t plus something plus something times t plus something times t squared, but it's in general, it's an infinitely long formula, there's no kind of, it doesn't terminate, and it's certainly not exact. You can't even sum it, because the terms are essentially always factorally big, so it never converges, but this one, when you have the sum n in z, for any smooth function like this which is small at both plus minus infinity, that's the sum, this is Poisson summation, it's the sum f tilde of, well, one over t, it's really without the t, it's just f tilde of n, but when you rescale, it's n over t, where f tilde of s is the integral for minus infinity, e to the two pi i, s t, f of t dt. So it's the Fourier Transformer. So if you think of what that looks like, if this function is very rapidly decaying, let's say it's even more than rapidly decaying, let's say that it actually is extremely rapidly decaying, then f tilde will become an analytic function in some strip, it's a very nice function, it will be very, very, and if this function is also c infinity, or even better, omega, if it's c infinity, this function will decay faster than any negative power of s at infinity, so it's very, very small, so therefore this sum is convergent faster than any power sum. But if f is actually analytic, this thing will actually be exponentially small. And so now when I do this, the first term is f tilde of zero, but f tilde of zero is of course what I called i of f before, it's simply the integral. So the main term, which is the same one we had before, well now it's the integral, it's the integral from minus infinity to infinity, before it was zero to infinity, and then typically it's exponentially small. But it's convergent, it's a sum, but it's a convergent and explicit sum of exponentially small things which are getting exponentially smaller. So it's way, way better. So it's not that it isn't this weird after you, when you need an exact form, it becomes easier than the original asymptotic, it always becomes easier. However, you're not done, because when you do this, if you just want the asymptotic things, you get i of protein, all these other coefficients, and indeed they will cancel when you symmetrize, as they just did. But now you only have the i of t, you still have to compute the other terms. But the point is there is a form that it's exact and will be exponentially small. And so based on, I finished today's lecture, even if the time runs out completely, because when you do that, you will get all of the states, everything comes out and the things that were found experimentally that at the end were on the nose, turned out to be exactly that. And the reason is exactly the symmetrization. So I won't show you in a little detail how that works here. And I'll only actually carried out in the case when s is even, remember there was already case distinctions for s even and s odd, in the proposition with most of which, no, I think I even kept it, it was easier when s was even, there was only one main term, and then the constant may be a log t. But when s was odd, there was a second constant, another constant times t to the s, and also these formulas changed a little bit. So I mean, I just to save time because anyway the most interesting case is s equals two, let's say the case when s is even. But now if you remember what ls was, ls was, I hope it's still here, here's the definition of ls, it's the sum over all nu, which are strictly positive, but congruent to alpha multiple of one, of log of this thing. But now if s is even, then if I send t to minus t, I change nothing when I change nu to minus nu, I don't change the s power, I do however change alpha to minus alpha. But I don't change beta, but that's good because remember the symmetrization, which is still written here somewhere, when s is even, you just symmetrize in alpha, alpha and minus alpha, you don't change beta. And so you see that what you have is that this one, if s is even, is simply the sum and nu is simply in z plus alpha and it's either positive or it's negative, and then it will be the same thing, log of one minus e of beta, e to the minus nu t to the s, which I can write as the sum nu in z plus alpha of some f, which is an f alpha beta, I usually just call it f of nu t, where f is the same thing that you see here, minus capital log of one minus e to the two pi beta, e to the minus t to the s without the nu. And here it's either positive or negative, but every real number is either positive or negative, unless it happens to be zero. And so if it's zero, that will only happen if zero occurs, but zero could only occur if alpha is zero mod one, and then you'll have that contribution, but then t is just zero, so you'll get log of one minus e of beta.  I hope I got the sign right, otherwise it's the opposite log, capital log of one minus e to the two pi beta. So that's the extra contribution of alpha zero, that's really a detail, but the important thing is that this is a sum, and this one now is convergent if s is even, because as nu goes to plus infinity, you've e to something exponentially small, but as nu goes to minus infinity, you're squaring it, it's still exponentially small. If s is one or odd, then it isn't quite, and you see that already in the original eta function, or in the Jacobi triple product, you have one minus q to the n times sum x, and the triple product you would like to say n can be, here n is positive, and so you'd like to say n could also be negative, but of course this one would blow up exponentially, but then you say up to an infinite constant, which we'll do a la Euler, this would be q to the minus nx times one minus q to the plus n, x inverse, and so you take out the infinite constant, which becomes something like q to the zeta of minus one times x to the zeta of zero or minus x, you have to of course do a little fiddling to get the right Jacobi, but morally this term sending n to minus n up to this triple factor is the same as sending x to minus x, and also if you take the derivative, the logarithm derivative, it's really true. If you take the log and take the second derivative, you kill this linear term, and so then it's really true and you can apply Poisson, but at the end you have to integrate twice and it's a pain in the neck. So believe me, you can do it and it comes out, but it's considerably more work. So what I say for s even will work morally for s odd, but if s is even, you can simply do Poisson directly because the sum of all new in z plus alpha is what you want it to be. So for that reason, you're done, and now I'm certainly doing much better than I thought with time. So okay, so if we do that, then what we'll get? When I symmetrized, so I mean I wrote the general formula of Poisson with f tilde, and in our case, as I said, f of t for us, f of t will depend on two parameters, alpha and beta, which are real numbers, multiple of one, and this was minus log, capital log of one minus e to the two pi beta times e to the minus, but yes, sorry, there doesn't seem to be much of an alpha in that, does there? And indeed, there shouldn't be, this is just an f beta as far as I can see here. Ah, because I just wrote the simplest Poisson. Of course, I forgot to tell you the shifted one. If you take instead f of new t, where new is not all integers, but shifted integers, so without any condition new positive, then you get exactly the same thing, f tilde, but now it's e to the two pi i times n alpha. So you insert, and so if alpha is zero, it's just the sum, otherwise, and it's the same theorem as before, if you replace f by f times, f shifted, then it's Fourier transform changes by that exponential. So this is the function that we're going to have, and so what you get when you do it now, I'll just do the even case, but it's almost the same in the other, but I don't wanna keep making case distinctions, then you have an elementary term, so I'll use e for elementary, and epsilon everybody knows that epsilon in mathematics is always small, so there'll be an elementary term plus a small term, but this is very small, then we'll write it down in a second, but this one is elementary, and it is two parts because in the Poisson summation form that remember one term is the one when n is zero, and that's the integral of infinity from minus infinity to infinity, that's independent of alpha, and then the small terms, typically, but here remember our function had one extra term if alpha was zero, there was an extra term, so we have that one, so this one is, I mean I already actually more or less said it, it's one over t times a constant depending on s and beta, and the constant is two gamma one plus s, now I index one plus one over s of e of beta, so just in case you've forgotten what this is, this means it's the sum m from one to infinity, m to the power of one plus one over s, and here it's e to the two pi i m beta, okay, so that's the constant except that there's this slight correction term if alpha is zero then I have to add the capital of the principle log of one minus e to the beta, so that's the elementary part, and epsilon is something small, but epsilon you can read off from this, there's a thing with an e to the two pi and alpha, there's a Fourier expansion, a Fourier coefficient, so for this f I have to compute f tilde, let's call it y, so this would be integral from minus infinity to infinity, of well just this function that I just wrote, e of ty, dt, and now you integrate by parts, so basically whenever you have an integral in life that has a log, you know that log you can't deal with log, but you can deal with its derivative, but luckily a pure exponential e to the two pi ty is also the derivative of pure exponential, which is one over two pi y times the same exponential, and so you pick up, then this becomes a derivative by integration by parts, it's small that both ends, you get the same exponential times the derivative, and so multiple mistakes, it's s, t to the s, I could take out the s, e again of ty, y divided by e of minus beta, sorry dt, excuse me, and here it's e to the t to the s minus one, so this is a very, very rapidly convergent function you can expand in various ways and get what you want, so what you get, so it's now you insert that and you do this and you interchange the sum or something, I mean these are all, this was an epsilon s, epsilon s of alpha beta t, the extra bit is the following finite expression suddenly, it's the sum, no, sorry, it's an infinite expression, I sum over all points in the upper half plane if I did it right, such that they're s parts just like what we had before, but now before remember we were summing over a shifted lattice z plus alpha, but now because of this thing when you unravel it, it becomes the shifted lattice z plus beta, so alpha and beta change their roles, and there's also a second sum with only two terms, plus and minus, and then it's the principal log of one minus e of plus or minus alpha, so the, then before we had a one minus e to the beta, so beta's become alpha, and alpha's become beta, before we summed over lattice z plus alpha, and we had a factor of one minus e to the beta, and then this is simply plus x over t, so this is, I mean it's a closed expression, this convert is very, very rapidly, and it's a closed expression, so now essentially we're done, but since I still have 10 minutes, I'll give a little of the details of what we're done, we've got the answer, but now what is the answer, and so it's a little prettier than you would, it might have been, and also a little uglier than it might have been, you're kind of half lucky and half unlucky, so remember what I said in the case s equals one, that the proof of the functional equation for eta actually broke up into lots of terms, because eta near a c through the punitive here, eta when you apply, you know, a tau plus b over c tau plus d with tau big or small, it splits into c sub-products, but actually c over two, because you always have to combine l and minus l, and those two form a Jacobi triple problem, probably one minus q to the nx, that was one minus q to the nx inverse, which Jacobi gave us a theta series times some simple eta function, and each of those has a function equation, which is exact, and so actually you get something much more precise, and the same is true here, except all of the forms are out there, so I'll give them in a somewhat approximate form, but I'll more or less save the truth at the end, so let me define, I'm going to have the function capital lambda, s, but I'll have, in a second I'll modify, so I'll first put a zero, this is the first attempt, this will be simply e to the minus the previous ls, and remember there was an x, but if you remember the x was tau to the s over two pi i, so if I want to get the tau to be tau, then here I have to take the s root of two pi tau over i, so these are just stupid renormalizations, so essentially here I've done nothing, I've just exponentiated, so exponentiate l, l sub s, okay, and renamed the variable very slightly, so this if I write it out would be the product nu in z plus alpha, but only the positive numbers, strictly positive numbers in the congruence class c minus alpha, and then it's one minus e of beta q to the power nu to the power s, okay, so that's the definition, but now just as, so if this were the, if s were one, this would be what's called the infinite polymer symbol, x q infinity, which is the product from zero to infinity of one minus q to the nx, and if you think of x as being q to the nu, where nu is let's say between zero and one, then that would exactly be q to the n plus nu shifted in that way, so this is like a polymer symbol, it is no multilayer at all, but what we know from Jacobius when s is one, you have to symmetrize, and we know from this also that we have to symmetrize, but remember when I symmetrized, there was a stupid extra term, if alpha was zero, I had an extra term log of one minus e of beta, e to the minus t to the s, but t was zero, and so it's simply one minus e of beta to the power delta alpha zero, so if alpha is zero there's this extra stupid factor, but otherwise we're going to simply symmetrize, and I think I do this for every s, certainly if it's even s it'll be minus alpha, and maybe I'm just doing still minus alpha, because I haven't, it's as I said a little more complicated if s is odd, because then you can't quite apply Poisson there, so remember it was minus one to the s beta, but when s is even you just symmetrize in alpha, and then if alpha happens to be zero, you have to, because every integer is either positive, negative, or zero, you have to include one more factor, so you can see that this will simply be the product over nu in z plus alpha, again if, so that's a very nice function, I hope I did it, no I don't quite believe this, it seems to even be right, okay, and now the claim is that this thing will be equal because of the same trace of what I've done, there's something that I can't read, which is one over tau to the one over s, but I can't read that letter in my notes, so I have to look at the text file again, oh I can't even, oh that's why I couldn't read it, that just put a star, because it's some ness, and I didn't want to read it in my notes, it's an explicit expression, you know something elementary, over tau to the one over s, but then it's a product over all z and h, but now we have to remember before the plus or minus, now the two signs, which are both plus or minus one, and I'm summing over all z in the upper half plane, such that z to the s is minus not one over tau, but epsilon one over tau, and then it's lambda one, of course you can write it all out before terms, but it's easier to write it as a product, so it's what I told you that basically alpha and beta, here we have alpha and beta in that order, first alpha and beta or maybe minus alpha and beta, and here we have first beta and then alpha, but of course with various sign possibilities, and so well here you see this already symmetric when alpha goes to minus alpha and here too, because epsilon two doesn't occur here, so I could have written this with alpha and minus alpha, okay, so this is the form that there's an explicit computable constant divided by tau to the one over s, however that isn't quite the same theta series of course, because that would say the higher s is modular, it can't be modular to co-beformed just for the reason that eta wasn't, but you still get an exact form that each analog of the theta function, which is the analog of the triple products, you put together two sums over half lattice, or two products over half lattice, the other product over the full lattice, then this thing does go, and this when you see you can't write this in terms of lambda one over is symmetric, because to symmetrize it depends whether s is even or odd, but one of us is not even an integer, I mean we're in a different world, we symmetrize in some sense with respect to roots of unity, which is why you have indeed there are a whole bunch of roots of unity here. So, okay, so that's basically with stories, so what I wrote at the end is that in both cases, if s is even or odd, you can write it a little bit uniformly, and then the formula will be that lambda s is zero of alpha beta tau times lambda s zero, so the simplification of this minus one to the alpha minus one to the s beta tau, semicolon tau will be equal to some elementary factor that one can work out, which in the case when s is even is just a constant over tau to the one of s, otherwise there's some more that we had before, but that part we don't care about because we already know it to all orders from the asymptotics, we only care about the exponentially small part, so this is what we already knew from the previous application, and now it's z in h, and now it's simply z to the s is minus, I know it's the same, epsilon one and epsilon two, or plus or minus one, and it's the same thing exactly, lambda to the one over s zero of epsilon one, epsilon two to the s beta comma semicolon, minus epsilon two alpha semicolon z, so it's kind of a bit of an s, but the basic principle remains true, that this is an exact expression, and you get it from Poisson, whereas before it was asymptotic, and you got it from Orlin-McCloryn. Okay, and so the very last stage, I still have five minutes left, this I could probably even erase, but no, this I might as well erase, the one that I shouldn't erase as long ago I wrote the asymptotic formula, that's long gone, but now it'll be the same, it'll be an exact formula, but I still see the asymptotics as the main term, so the final result, so the card, the card memory is that theta s, which I haven't even told you the definition, well theta s, you can always make a theta s by, as I did here, well in fact I've done it by some elementary thing times the product, that will always be an elementary factor times an infinite sum of the exponentially small terms, so a very rapidly convergent sum of pure exponentials. So now just to do the last two minutes, I'll give the final result, and that I'll only give for, so the card memory, so one theorem, that's sort of the main theorem, is that theta s of alpha beta tau is always equal up to an elementary factor, it's a rapidly convergent, exponentially convergent series of pure exponentials, in your combination of pure exponentials in tau to the minus one of s. I don't even have to say tau is going to infinity, it converts to every time the upper half, in particular, tau is very large, then tau to the minus s is very small, and you have, sorry, if tau is, you're right away, whatever it is, it's an infinite sum, so it's a sum, sum c nu, e to the nu, well lambda, e to the sum lambda nu times tau to the minus one of s, and that's an x. But the formula for that added to be a terrible mess, just like what we did last time, it's only when you take the log that it's a reasonable formula, and that will make all of the different, difficulties next time on Monday when you applied the circle method, that you don't know which terms are going to dominate, because you're multiplying a lot of things, and that there are combinations, and two of the terms can add up something bigger than some other term, so it becomes a bit of a mess. So this is the abstract theorem, and the explicit theorem is if s is two, which is the case we care about most, then I'll actually end up and write down the actual formula, and I'll say in words what would happen. So again, cap is a rational number, the q and the denominator is the c, I think I don't care about the numerator anymore. So then we have, it will be to all the order, well it's exact, sorry, not to all the orders, it's n of two of c, which maybe you remember is the smallest n, smallest inch or n such that its square is zero mod c, so if c is square free, it's just c, but if c is the square fact, it's a bit smaller, divided by two pi t, so t, sorry, I forgot the most important thing, it's e to the minus one over t squared, excuse me. So t is going to infinity, well t doesn't have to go to infinity, t is just positive because it's exact, but you should think of t as going to infinity, and so we write a point near an arbitrary rational point as e to the two pi kappa times e to the minus one over t squared, and then I already wrote the formula earlier when I had the proposition and the coronary, I gave this exact thing to the leading orders, and so I don't really owe you anymore the formulas for nt of c, I wrote it to remind you, ct of kappa, or maybe I called it something else, ct tau comma kappa, was a constant which was a combination of zeta function and simli, but it's the same one it runs before, and now we're left with the final sum, it's all l mod c, so that's a finite product, but it's all positive numbers m, such that m is congruent and all signs, so it's a triple product, c terms into the many terms and a sign, and that what you want is m is congruent to plus or minus, I think I do need what k is, so cap is a over c, so it's a l squared mod c, so we're taking for each m only those l's, so in particular some m's are not congruent to plus or minus a times the square, they won't contribute, that's what we saw earlier, that quadratic residues and non- quadratic residues are different things, and then the final term is one minus e to the two pi i of l over c times e to the minus e of plus or minus one over eight, so you have the two roots of i of sorry, e to the two pi times plus or minus one, and then there's a term two pi c to the three halves, and then there's this term squared of m, and then there's a term t, and so that's the final theorem when s is one, and then I can take another half a minute to say what you have to do if s is not one, so if s is bigger than two, it's similar, but the justice we said at the absent-party term, there's an extra term, there's an extra term which is e to the two pi times the generalized Dedekind sum, which I defined by four, but remember that Dedekind sum, you probably don't remember, but it was l over c, and a l to the s over c, but if s is even, then this is even, and that's all, then it's zero, so you didn't see it for s equals two, but I did write it for general s, so that's why in this form, there is no Dedekind sum, so I don't know what I'm, I was gonna write in red and put errors, but there's no real point, so the first part is there's an extra e, there's also an extra term e to the one half, z to the minus s times t, that's the one you very much see for the Dedekind data function, because a to the minus one is minus the 12, but z to the minus two is zero, so in this particular form, you don't see what I did in general, of course you have ns of c instead of n2 of c, that's clear, you have two pi over c to the power one plus one over s, so here, well since it's two pi over c, this was obviously a brief, but it's two pi over c, the three halves will become two pi c to the one plus one over s, the congruence condition will be, of course that m is congruent to plus or minus a l to the s, of the c all of that you could guess, and the last thing is that e to the plus or minus one over eight is replaced by appropriate choices of four s roots of unity, so here s was two, so it was an eighth root of unity, and there are primitive s roots, I don't remember the exact form, it's complicated, but essentially there's a formula very much of the same sort, where you're taking all four s roots of unity, and then if you unwind that for the eta function, this is essentially eta remember, it tells you the exact transformation law, so I'm sorry today had a lot of very, very nitty gritty formulas, but the point was never the nitty gritty formulas, the point is you can get them in there simple enough that one can write them down, one doesn't necessarily want to read them, but that's a formula which started out asymptotic, because it came from oil and chlorine, it's the sum over half lattices, after you symmetrize it becomes the sum over whole lattices, then it gets both simpler and more complicated, it's simpler because the two half lattices give you a whole lattice, you can apply Poisson, the entire asymptotic thing collapses only a few terms in the beginning survive and it terminates, but to make up for it that's only the leading term, now you have all the rest of Poisson, you have all the other Fourier transforms, and they have to be computed and that's quite a lot of work, and that gives these dual things where the tau has become one over the s-thruid of tau roughly, so I'm not sure, I guess it's the same t, which I moved anyway, again the details don't matter, but roughly you're going from a number x to its minus one over s powers, but of course since one over s is not integer, there are many with roots of unity, but in the end you get an exact formula because you can use Poisson. So okay, so that was the unpleasant part of this whole calculation, which is seeing what you actually get when you don't just do asymptotics, but do it on the notes, but the next time, which that should be fun again, how does this work when you actually want the coefficients, when you want to write n as a sum of squares or fumes or whatever. So I managed to go over time even though I was under time before, I'll stop, if there are any questions, quick questions you can ask, otherwise maybe, how can you not have a question? Okay, if no one wants to ask you a question, I'm still here, people are here, can ask me privately and anybody can ask next time by Zoom or of course by email, and I'll try to answer to the extent that I know. Okay, so then we'll see you on Monday, I'm gonna say Monday I'll finish this story, and then Wednesday either I've run out of gas, I can't find any of my notes, I'll try to tell something, or I'll have something fun to say about this whole conversion series. Okay.


Can Leguta do it now? The first two and he's the champion. Yep, semifinal number one, Adler-Guta has made a beauty! Adler-Guta has got them just over three laps away from winning his first world title. Adler-Guta the leader, Bartos Mastik is second. Thurn is Ty Woffen and all, Manson is coming around the outside. But what a way to respond when the pressure's on from Adler-Guta. Fantastic start out of gate three, not the best gate bite, but the absolutely textbook star. He's got that silky start out in front, super smooth. And Nigel, it is absolutely, Art of Leguta has won four of the last five rounds of the World Championship. He's recovered from a poor start tonight. It's Art of Leguta who is the new Speedway Champion of the World! But he's saying the new World Champion, the Grand Prix Champion of Speedway, what a way to do it! He's won five Grand Prix already this year, his form has been extraordinary. And it is fully deserved, it would have been a travesty in truth if he hadn't managed to do it tonight. Being congratulated by the outgoing champion in Bartos Mastik, that's great sportsmanship from there. Clearly a very popular man with everybody coming round to shake his hand, give him a kiss, give him a cuddle. But yeah, Nigel, it was looking a bit dodgy earlier on, but by Krikey's show, great character and determination. Art of Leguta is the World Champion and for so long we've talked about him on Saifutinov being the Russian World Champion. But Art of Leguta has beaten him to the punch, Saifutinov is partner in crime in the Speedway of Nations. Great moments, these, between guys and colleagues and friends who've got at it, and Nilsen in there as well, special moments. Oh, I don't know, I don't know, it's very, very happy. Yeah, thank you, my Johnny, life, Avilina, Adam, Agnia, thank you my dad, my mom, I'm very happy today. Congratulations once again, thank you so much for this amazing job. Yeah, thank you.


Alrighty, finally back with some videos. I have a ton of cool ideas lined up, but let's get right to it. I have a really cool video for you discussing a major renovation that is coming to an attraction at Canada's Wonderland. Well, let's just get it straight out and say it's a roller coaster. One of the roller coasters at Canada's Wonderland is about to receive an extremely pricey renovation. With that being said, so excited to talk about it. I've known this information for about two months now, but I've been waiting for the right time to share it. Wait, hold on, I'm getting a phone call. Oh, Canada's Wonderland's calling. One second, let me take this. Hi, this is Brendan. Yes, yes, yes, I am making a video right now. No, no, no, no. It's not about the Kingswood renovation quite yet. Don't worry. No, I'm not leaking anything about that. No worries. All right, have a good one. Woo, got a little nervous there. With that being said, that was so cheesy. I feel like I should edit that out of the video. But with that being said, I'm gonna talk about Vortex. Vortex at Canada's Wonderland, for those of you that have had your eyes peeled, can pick up on some of the things that are going on that definitely pinpoint towards a renovation coming. Now, this isn't why I know that a renovation is coming to Vortex. We've known because we know people in the community. We know a lot of companies that work on attractions, especially a lot of American companies. With that being said, Vortex is supposed to be, and I'm using the word supposed to, and I'm gonna talk about this a little later on the video, be getting a extreme makeover. When I say extreme, I'm talking about making the ride brand new. The goal for the renovation, from my understanding, is to have the ride running the best it's ever run, better than when it opened back in whenever it opened. I don't know the date at recording this video. I don't write notes when I make my videos. I know that upsets a lot of core enthusiasts, but I like to just rant. So a lot of my videos are just literally me talking right now. But with that being said, they want the ride to run better than it did when it first opened. I don't know what that entirely entails, but I do know that there's gonna be a lot of retracking. Not necessarily when I use the word retracking, removing a piece of tracking, giving it a brand new piece of track. They're gonna have machines going in and literally analyzing every single little tiny crack fracture on the track. Any crack fracture on a support column and completely redo it or rework it. With that being said, they're also gonna rework some of the systems on the ride, whether that be the speakers, the station, the wheels, the pistons, the trains ordering brand new trains. A lot of that is up in the air. I don't know exactly what it's going to entail. I just know that Wonderland is spending a fortune on it. Now, the reason I said that this is up in the air now is this was a guaranteed thing, by the way, for 2023. It was already paid for, which makes me think it'll still happen. But for those of you that don't know, Leviathan had an extremely unexpected, pricey, maintenance cost added to it this season. They had to order a brand new lift motor and the estimated price tag on that was about $500,000. So I'm not sure if that's gonna push this back then because of that unexpected price tags. Member Cedar Fair does hold its parks to really strict budgets in terms of maintenance. So with that being said, something like this might be pushed back, but I do know that there is track work that needs to be done on Vortex.  But for those of you that have had your eyes peeled, you'll notice a lot of chalk markings on the track segments and the support columns of Vortex. That's your telltale sign that something is coming. Now, don't get too excited.   No, it's gonna stay the exact same. There's no track being added to it. There's no new element being added to it. It's just taking something super valuable, super important to Wonderland's history. And to me, honestly, I think it's an icon in the park and just making sure it stays a lot longer. So we now know Vortex has no plans to be removed at Canada's Wonderland and it is here to stay. So I'm super excited about that. I hope you guys are super excited about that. But because of that, I do hope it gets a new paint scheme. I would love to see a more rustic look. I really hate the silver support columns. So I'd love to see them blended in the area. Now, with knowing the amount of money that Cedar Fair is about to spend on Canada's Wonderland over the next three years, I have a feeling that they're gonna re-theme this and give it some importance. I could see the station even getting some rework done. I think the trains are gonna stay the same. I actually love the new trains that they did a couple years back. To me, it's just about color scheme and maybe some rework on the station and the queue line for the ride. So that'd be awesome to see. But that's all the information I have on Vortex right now. Just keep your eyes peeled. It's gonna take place over the winter. For those of you that want even further proof as to this is happening, you'll notice last season, they laid down a lot of gravel in the lake for machinery to come in and work on Vortex.  So another telltale sign. Anyways, thanks so much for watching our videos and supporting the channel. There are about one to two videos coming out a day now. I'm literally committed to keeping this channel going and refueled and just ready to go now. So super excited. And yeah, hopefully this was the news you guys were looking for. I know a lot of people were like RMC, RMC, no Wonderland. I said it was something Wonderland wouldn't announce and RMC is something Wonderland wouldn't announce. So definitely, yeah, super excited. Thanks so much guys, bye.


Oh, yes, welcome back to the Twitch stream. We are not live streaming on Twitch. At least I hope we're not Okay, we're not like I said last week this week. We are actually in Hong Kong. We are pre-recording this before we leave I hope that you can understand sometimes when we travel It is kind of difficult to like make the best videos you've ever seen because our workload is essentially double right now So for those of you that like these videos, I hope that you have a good time and if this isn't your style I will see you next week when we get back from Hong Kong I'm excited because I know that there are a couple videos of me as a chair as horror films Which I'm very excited to watch. I also genuinely really like making videos like these because it gives me a chance to sit back and enjoy The content, you know what I'm saying? We like content too guys. I like content just as much as y'all. First We got to start streaming on Twitch on accident. Don't do that Julian's best background comments. Yes So when we film and edit videos, it's all like in one day Like I have never changed that really with the exception of right now I genuinely forget some of the stuff that happens in the videos right now Is probably the first time that I've watched this back since I posted this I'm not sitting and watching it a couple months out Like I forgot that you've made that a bunch of speed-a-man jokes. Use the worldwide web, use speed-a-man When I'm doing a video where I'm concentrating on something that I'm really trying to pull off I don't hear your jokes. Like I don't hear them until I edit. You're that far in the zone. Yeah I don't even register like what I'm editing. I'm like, oh my gosh Like literally did not even experience Julian saying that in person.  I don't remember Where I've where I've started where I've gone Marvel's not knowing where he is for one minute straight You mean every day. Oh, he's so cute. Fun fact this video got demonetized.   I know he like really gets me No, he doesn't he doesn't he doesn't know where he is. It's happening.  I think you're gonna say your face is a drug because I was gonna say Then I'm addicted to that All right, I've seen one of these I think there's a few though. All right I'm ready because people were tweeting them at me non-stop.  Okay with scary music.  Are you calling me a rat of sorts? Wow a bald chair camouflage white-headed blue striped rat Yeah, sure. That's what I'm calling you. I want to see the other one. There's the other and how they compare It's amazing to me that two separate people. Look it. They were uploaded the same day So the first one is a Harvey Rayans version, but they both had the idea to make a horror at it Oh, it just came to them.  You disappear and that appears And that was content. This one is Jenna telling Julian to stop it compilation I feel like this is it's gonna trigger me because when I tell you to stop it It's you're either doing something gross or a lake funny. No wonderful invasive sweet Frustrating lovely. No, none of those things.  I help you so much You help me by jamming your fake acrylic nails into a banana. Oh, I like I really don't tell you to stop it unless It's like I've had it already like I've passed several stages of you Just I don't know I personally think you should kind of just like let me go a little more like see what I'm up to during some of those Clips I was like in physical pain or something frustrating and you're just mocking me Oh, yeah, Jenna Marbles being a good mother for three minutes. I don't know if this is gonna fit him Okay, I don't appreciate I thought this was gonna be complimentary. What is this? Why did you just zoom in on poor Kermit's little weenus? He's self-conscious about it. What do you think about a peanut butter bottle?  Well done snatched peach being adored.   I just all of a sudden got really hungry after watching that clip. That's disgusting Why would someone think that you burping the ABC's is our funniest moments that is disgusting because they have sense of you Who can relate like your burps were yes, they were gross because they're burps But they were just the means to the end you understand right I had to burp who could relate because I think it might be The first person who's burped illogical Eric for those of you that don't have time to sit down and watch our podcast I feel like that's a great. Yeah little shortcut. It's like a high. Let's just watch the funniest moments One of the best discoveries we've had from this video series. We found J&J memes He's killing it. He's killing and he deserves it this art. So which one do you want to start? Oh He even put the letterbox like it looks this looks like a film Can I ask Jane James?  She's like created a narrative from like two years of three years apart in Kermit's life We have seen this one, but it's definitely worth watching again. Also, J&J memes.  I don't know what that is like an MTV intro What the This really is it's like a it's like the intro to our life Jen and Julian. What's a soulmate click on it?  I can't watch this.  Just stop being over here and you What's beef sticks well Thanks a lot arm. Thanks arm Like shouldn't I be the one to measure my own beef sticks Well, it could have something to do with the preservation of gluten-free products the parties at Taco Bell It could have something to do with the preservation of gluten I was like unexpectedly funny wow that type of like repeating and like cutting off people at it is like my favorite Oh shit, we did what did we do?   What is this trap remix to my ear blood?  It's two years ago. He made that it never saw it He makes like trap edits of people on YouTube. That's such that's such a cool channel. I love that Aggressively moving her eyebrows while talking Make sure you subscribe to my channel I put it in a piece of Wednesday slash Thursday Here we go A lot of you guys have been tweeting me pictures of things that you'd like me to do trying to do something in my technique okay, I'm Feeling personally attached right now ever since I started posting on YouTube people are like what are her eyebrows?  What do you want me to do? Oh, why I can't Use my eyebrows to talk Anyone's eyebrows are aggressive if you zoom in that far well overall I'd say that that was very fun. Please don't stare at my eyebrows up close You're gonna see in action stop Everyone hates my eyebrows. I hate the eyebrow leaders that are like only the shape of your brows. No, you're Oh, they're beautiful. Maybe for a video. I'll go to a brow professional so you guys can see I'm not lying They literally look at me and go. Oh, no, and then they don't know what to do. It's a thing. They're perfect. Oh Well, I hope you guys enjoyed this hopefully a little bit We are traveling right now So this is like, you know what we can handle at the moment But I know that those of you that like these videos you like them a lot So I have a lot of fun. This was like a good time for me. I did too and honestly Thank you guys even if we don't find it right away.   We will be back from Hong Kong. So hopefully we won't be too From Hong Kong I'm feeling attacked. Make sure you subscribe to my channel.  Your those shoulders are aggressive and I'm gonna make a compilation of your shoulders being aggressive Please stop making fun of my eyebrows 2k18. Goodbye


Welcome back fellow techies to our latest tutorial video, I am The Technician Project and today's tutorial is going to show you how to change your audio effects from control view to editor view. Now because we're currently in logic version 10.5.1 and we're going to show you now that we're going to upgrade our effects. On this channel here which says Frank, I want to add a EQ, I'm going to add stereo EQ. Now this message pops up and it was a bit off putting to begin with because it wants to change my preferences of a plugins to control the views. Now I pressed OK originally and we're going to go into it. Now this is the control view and sometimes it's a little bit off putting, it's a little bit odd. Now very difficult, very complex. If you're not used to using it this way it is very difficult. What you can do is two ways of changing this. There's the simple way and the hard way. We're going to do the simple way first. If you just go to view here and where it says controls you change it to editor and that will change it to the view we're all used to seeing with the EQ. The other way to do it is to go to logic pro X, preferences, display and under this setting we go to mixer and we want to in plugin windows we want to open in controller view. We want that bit on the tip because we want to open in editor view which is this one here and we apply those settings by hitting exit. Now it should open a compressor now in editor view. It's going to ask us again, we press cancel, there's our editor view. Now you can't change back the control view from the editor view, you can only change forward the control view per editor. You can't go from editor to control. Thanks for watching. Don't forget to like the video if you enjoyed it and subscribe for more and also ding dong that notification bell to get further updates. Thanks again. Goodbye.