 I would say that starting next time, maybe not this time, as well as asking any questions about what I've said so far, feel free if you have a nice example of an asymptotic series or a problem that came up in some asymptotic thing, which is certainly the case of several people here, then you can say, you can either come to the board if you're here or explain it verbally. And maybe it would be fun if we could have examples that arose in real life from part-distance of the course. Because as I said, many of the tricks and methods I'm going to show are applicable to problems that come up all the time in one's research work. And so it may well be that one of you finds something in some string theory calculation or some calculation. So today is probably the last day that doesn't quite or the title is gone. But I can remind you the title of this course is a little hooperish maybe is called standard and non-standard methods in asymptotics. Actually, non-standard just means nobody told it to me. For all I know, it's very standard. I'm not an expert, but I love asymptotics and have played with it all my life. And so I have various tricks and I'm sure some are known. But today is the last lecture that is definitely not non-standard because I'm talking about the Euler-McClurin summation form and its variance. And that goes back as the name says to Euler. So it's an 18th century piece of mathematics. If any of you have looked at my appendix to Zeitler's book, which I mentioned last time in the reference I'll give at the end, I'm going to follow that exposition exactly, even including the notation. Because as I was preparing, it seemed idiotic to try to change every letter to a different letter. And I don't see any other order to do it. So if you've read that, you'll be very bored. And I hope you brought a book. And anyways, I said it's relatively standard. But still, it's not quite the usual take. And it's a lot of fun to see how these things work. And one thing I want to do with this formula, as with the asymptotic method I'll explain next time, is to give the very, very nice mnemonic how you can remember what the answer has to be. So maybe I'll start with that. I'll state the general problem, which I've already mentioned, and that there are kind of two ways that you might think how to go about answering that question. And both of them are wrong. But surprisingly, so both give the wrong answer. But if you add up the two wrong answers, you get the right answer. That's the mnemonic. That's kind of fun. I think that's my observation. I mean, it's a trivial observation. And of course, it's not really mine, but the mnemonic. So I'm going to fix two notations. I've already mentioned this in the first statement. I gave an overview of some of the problems we're looking at. Let's say this is a nice function. For the moment, I'll just think that it might be in the complex right half plane, but let's say just on the positive real line. And for the moment, I could put greater than or equal to 0. But later, I want to allow singularity to 0. There might be log t or something. So for safety or greater than or equal to 0, of course, there are two notations, are greater than or equal or is an interval. And nice means it's small that infinity, and is reasonably continuity properties. I think continuous is good enough in practice. All the functions that actually come up are given by a formula. And they're analytic. You see infinity functions. And small that infinity means it's small enough, e.g. It might be 1 over power of c and power c of t, where that power is bigger than 1. So what that means is that if I take the sum, and that's also why I omitted the term f of 0, because I don't want to necessarily assume that f of 0 makes sense, the question is this. Because f is small at infinity, let's say like 1 over t squared, or maybe exponentially small, anyway small, that means that the infinite sum f of nt is convergent for all t. So this is a function which is valid for all t bigger than 0. Even if f of t itself is valid at 0, so it might be, in the simplest case, it might be an actual smooth function even at 0, then f of 0 is well-defined. But of course, if you take t equals 0, here you'll get f of 0 infinitely often. And that would never make sense. So here I definitely want to start this summation at n equals 1, and therefore I don't use f of 0, and that's why I don't really care about f of 0. OK, so that's going to be the setup for, I think, all of today. And the question is, if we know, so f is a known function, if we know, it has an extra k, if we know how f of t looks, well, at infinity there's no real question, it's just small. But at or rather near, t equals 0, can we say how this infinite sum looks? Again, near g equals 0. Because if t is large, there's no real issue. Let's say f is exponentially large, so it's like e to the minus t. Then this sum will be e to the minus t plus e to the minus 2t plus e to the minus 3t. The first term is like f, the second is yet much smaller, and the whole thing, I mean, you see immediately but it looks like, is just a sum of separate terms whose behavior you know. And even if f isn't that drastically small, let's say f is 1 over x squared. Well, then you'll still get 1 over x squared times the sum 1 over n squared, but it's just a constant. So it's not really a big issue what happens at infinity. Now, you can read off just from the original formula. But it's 0, it's not at all clear what happens. So after all, if t is 1 millionth, then even if f is like that, it's eventually small, but let's say you have to get 10 before it's smarts getting small, then here n will have to go up to 10 million before the terms even start getting small. So of course, for a sum to converge, the terms have to go to infinity to 0. They do go to 0, but it takes a long time until they even get small at all. You need millions of terms, and you obviously don't want to add them all up. So there are two ways. I think I already said this the first day, but it's fun. So if I did say it, I'll repeat it. Simply don't remember if I didn't, I'm saying for the first time. So one is the, well, I put them in from the logical order. Euler came before Riemann. Well, it should be the Euler's a to function. Euler didn't actually do what I said, but it's based on his mathematics. Well, we can pretend that he might have done this. Actually, he wouldn't. He wouldn't have given the wrong answer. And Euler, Riemann, what he did is also not wrong. I'm just saying if you very blindly followed the simplest answer, a la Euler and a la Riemann, you get two different formulas. And the truth is, in fact, the sum of them. So Euler would be like this. We just think formally. So I'll make a back so that I can come to this later. Formally, we have G of t. So let's assume. Well, I haven't get to that, but I'm assuming. And the answer is I'm going to assume many different types of things. There's quite a bit of freedom, but let's assume first that f of t simply has a power series expansion at the origin. Here, I don't want to call this n. It's a good rule that I tried now to tell my students and also colleagues not to use the same letter in the same context for different things, even if it's a dummy letter, because it quickly leads to confusion. So let me try to remember, and if I don't remember, somebody can shout, that in my infinite sum, this variable will be called n. But in the Taylor expansion, the index will be called n, because they're playing completely different roles. And so what I want is that there's an asymptotic expansion. So here, f, remember, I said it's nice. It's maybe just smooth or continuous. It's good enough, or maybe it could be c infinity, maybe. Analytic, in most of the examples I'll have, actually all the examples are that's analytic, except at 0. But at 0, sometimes it'll be analytic there, but usually it won't. Sometimes it won't. It's only c infinity. And remember, I told you in the first lecture this was about different words actually meaning the same thing. Having an asymptotic expansion to all orders at 0 is exactly the same as the familiar property of being smooth, being differentiable to all orders. And so we know that this means that f is smooth. Smooth, I'll usually use, means c infinity, not just c1. Smooth at 0, so it is derivatives of all orders. And then, of course, by Taylor's theorem, bn, we know exactly what it is. It's simply the value at 0 of the nth derivative of f divided by n factorial. So now, formally, what do I have? g of t is the sum. So it's f of t plus f of 2t plus f of 3t, et cetera. So if I just formally insert that into the sum asymptotically, then I'll get bn, the sum bn times t to the n. But then I'll get bn times 2t to the n. But of course, 2t to the n is simply 2 to the n times t to the n. Similarly, 3t to the n is 3 to the n. So formally, I have this. Now, obviously, that can't possibly be right, because apart from the fact that I have no right to interchange in this way, the nth sum simply is always divergent, no matter unless the bn's are all 0, starting at some point. Well, in fact, unless they're all 0, this sum is always divergent. But I'm pretending that Euler weren't completely formally, which is to some extent true. But he always justified why it was allowed in that case. So this, when we know, is, well, this isn't a nice one. This is kind of the definition. This is exactly the value of z of s, which is 1 plus 2 to the minus s plus 3 to the minus s and so on, at minus n. And remember that this was defined completely rigorous, though that his arguments were a little strange, and calculated by Euler. So in other words, even though there's a totally illegitimate first step, and even that that expression, as it stands, is divergent, once I write it as z of minus n, this is a perfectly well-defined number, and this expression makes sense, because bn is finite, z of minus n is finite. I'm not saying the series converges as an absolute sum. It certainly makes sense. And if we stick in the formula that Euler found for z of minus n, it's minus 1 to the n. I hope I got this sign right, bn plus 1 over n plus 1, and then bn times t to the n. So this would be Euler's answer. I'm going to remind you, as I said a little last time, how Bernoulli numbers look. I'm going to take five minutes or so, when I finish with this introductory thing, to remind you more properties of Bernoulli numbers, because they're absolutely crucial, if you want to do any kind of asymptotics. The Bernoulli numbers have to be in a second nature to you shouldn't have to think. Now the other way would be this. So this would be the Euler answer. The other way would be this. We had our function, and my t is very small. So here's t. Then 2t is also very small, but it's not quite as small. I have 3t, 4t, 100t. I mean, they go to infinity. And I'm adding up, this is f of t. f of 2t, f of 3t, f of 4t. So I'm adding up the values over these vertical lines. And the spacing of these lines is my small number t. So that, as t goes to 0, is exacted the way that Riemann defined the integral, the integral. So let me set if, convenient notation. By that, I'm going to mean the integral from 0 to infinity, f of x dx. And then by Riemann, assuming that it's a well-behaved function, which we are assuming, this integral, you can approximate by Riemann's sum. So this is the limit, as t goes to 0, of the corresponding sum, which is the sum of the f's. So it's the corresponding, you all know the argument. Here's f of t, the width is t. So if I just take each value and make a rectangle to the right, it might be below the thing, it might be above the thing, then the sum of these rectangles is an area which is asymptotically equal to the integral. And that sum is exactly t times the sum g of t, f of t, f of 2t, f of 3t, and so on, which is what I call g of t. So this is a fact. And therefore, that tells us that g of t is asymptotic to if divided by t. And that's simply a theorem. That's because of the definition of the Riemann integral. But it doesn't give the full asymptotic expansion. It only gives, remember I mentioned that this symbol is used, unfortunately, in mathematics for two things, even more, unfortunately. I'll use it for both because it's convenient. And it's always clear from context. This means only that the ratio of that and that is 1. So this means that g of t is asymptotically equal to constant over t. That constant is 1. Actually, I'm lying because it might not be that the ratio is 1. If if happens to be 0, I never said that f is positive. Then it's actually not 0 over t. What I really mean is it's this plus little o of 1 over t, so smaller terms. Then we're not writing. Of course, if if is literally 0, that's always a nuisance. When you talk about functions being f is asymptotic to g, you have to be careful if you subtract something from both, something can be 0, the ratio might not be 1. But here I'm using tilde in this weak sense. g of t is equal to i of t plus o of smaller terms, whereas here I'm using the sense of an asymptotic series. And that asymptotic series makes sense because every formal power series is a perfectly good asymptotic expansion. And so the proposition, which I'll prove in a bit, is that, as I already said, that the actual formula for this wide class of functions is that now asymptotic in the strong sense, to all orders, g of t has a Lorentz expansion. There's one negative power of t, which is indeed the integral, as Riemann tells you it has to be. And the other terms, and that's the beautiful thing that you should remember, that's the key property of the whole thing. The constant term, sorry, the Lorentz term, the 1 over t term is global. To compute that, you have to integrate f all the way from 0 to infinity. So if I change f by adding some little hump, which is compactly supported away from 0, I'll change that constant. But all of the other terms here in this expansion, all of them, to all orders, are just the b and t to the end, and they only depend on the behavior of f near 0. And that's all true for large class of functions. The behavior of g of t to all orders and t depends only on the behavior of f to all orders near t equals 0, except for a single constant constant over t, and that constant would be the integral. So we'll give the proof of that, which is quite easy. Later, the proof will be it's just a variant of the Euler-McLoran formula. And so my next topic, certainly many if you know it, but there are many ways to see it. Also, some of you may have just learned it without ever seeing a proof or forgot the proof. And since it's a basic, basic statement about, well, of mathematics anyway, and analysis, but especially of asymptotics, I think it's worth giving. So in order to do that, I have to tell you more about Bernoulli numbers than I already said something last time. So let me remind you of several things about Bernoulli numbers. So first of all, the Bernoulli numbers are just some numbers. It's called the BR. They're rational numbers, index by index, which is 0, 1, 2, 3, 4, et cetera. The first few are 1 minus 1 half, 1 sixth, 0, minus 1 30th. And except for B1, they're always 0 for odd indices. So now I won't put all the zeros. There's now another 0. So if I put 6, 8, 10, 12, 14, 16, that should certainly be enough to get a good feeling. They alternate in sign. So it's a sixth, minus a 30th, 1 over 42. By coincidence, this has the same value for 8 as it did for 4, minus 1 30th. 10 is the first one, which is not an integer. The reciprocal of an integer, it's 5 over 66. And you might think, aha. So something changes at 10. But actually, that's wrong. I'll explain that in one second. Something changes at 12. So 12, well, it's negative because they should alternate, is 6, 91 over 27, 30. Now, the numerator is prime. The denominator, on the other hand, is highly factored. It's 2 times 5 times 273, which is 3 times 91, which is 3 times 7 times 13. So you see that this is highly factored. It's 2 times 3 times 5 times 7 times 13. This is highly factored. It's 2 times 3 times 11. The numerators are prime. They aren't in general prime. But they don't have very small prime factors. The next one, which is, again, positive, is, again, very simple, 7, 6. And the next one, there's, again, a prime. As it happens, 36, 17 over that number I didn't remember. It's gone minus sign over 510. So these are the first few. And as I think I mentioned last time, we should really call them the Seiki Bernoulli numbers because the great mathematician Taka Kazusiki, who died in 1708. Euler was born in 1706. So a whole generation earlier, he discovered the Bernoulli numbers before or at the same time. Bernoulli did, in any case, both of them discovered them before the other had published. Because in both cases, it was published posthumously after they both died. So it was certainly independent. And he found it, more or less, in the same context, which is the formula for the sum of the first n squares cubed fourth powers of the sum of the first n fixed powers. So we have this. And since I mentioned this numerology, which isn't important for this course, let me say that b are the actually interesting numbers you should divide by r. Well, if r is not 0. And this is 1 over an integer. That's why I said 10 isn't really exceptional because if I divide by r, this is 1 1⁄2. This is 1 1⁄12. It's still 1 over an integer. This is up to sign 1 over 120, 1 over 252, 1 over 240. This is 1 over 132. It's still got numerator 1 because the 10 counts to the 5. This is a 691. This is a 3617. But 14 is still integral because I divide by 14. And in fact, it's a theorem. Well, let's take r to be even. This is true if and only if r. So let's say r is bigger than 0 and even because otherwise they're all 0 and it's stupid. It's true if and only if it is one of those numbers. Now, the proof of that is completely trivial. Namely, the bn over n factorials have an easy asymptotics. And they go rapidly to infinity. So after a bit, actually starting at 16, the br over r is bigger than 1 in absolute value. So it's certainly not 1 over an integer. But there's a deep reason for this. Who can tell me? Somebody has to say. What do these numbers remind you of? There are no number theorists in the room. Nobody who loves number theory. So let me remind you from yesterday's lecture if you were there, Ramanujan's loves, my love, and therefore the love of every self-respecting number theorist are formulas, which there are many, asymptotics, and multireforms. So I'm not going to talk about multireforms in this course. I might a little bit later at some point, but very briefly. But these are here, I should really. Maybe if I'd called it k, everybody would have thought of this immediately. Because k is the standard way used for the way to multireform. And so we have the famous Eisenstein series, e2, e4, e6, e8, e10, and e14. And each of these, for instance, e2 of tau is equal to 1 minus 24 q minus 2 plus 1 is 372 q squared. And so on, I'm going to say in a minute what these are. e4 of tau is 1 plus 240 q plus 2160 q squared. And so on, e6 of tau is 1 minus 504 q. And some more stuff. And these ek of tau, if k is 468 or bigger, so even but not 2, then e6 of tau is up to a constant, which I could write down, but I won't. It's the sum m tau plus n to the k. This is the famous Eisenstein series of weight k. The sum over all pairs of integers, except of course, 0, 0, which would give you a poll. This sum converters absolutely. And the fact that it's converters absolutely means that it's a multireform of weight k, which I remind you again means that for all matrices, a, b, c, d, where a, b, c, and d are integers, and the determinant is 1. So that's the group SL2 of z. If I replace tau by a tau plus b over c tau plus d, then this whole thing is a factor, c tau plus d to the k. And what's left is n times a tau plus b plus n times c tau plus d. But that's just some other m prime tau plus n prime. And they also run over the same lattice, because this is the automorphs of the lattice. And so you get simply the same function. And such a thing is called a multireform. And as I said yesterday, I will say many times, multireforms are absolutely wonderful things and they're beautiful properties. So e k of tau is this property. And e k of tau, the way I've normalized it, always starts 1 plus something times q and so on. And I didn't say what q was, but now I will. This formula implies in particular, since certainly the matrix 1101 is an SL2 z. This implies in particular that e k of tau plus 1 is simply equal to e k of tau. Now there's no automorphy factor because c is 0 and d is 1. And therefore, e k of tau has a Fourier expansion, which means it's a power series in q, which is e to the 2 pi i tau. So that power series always starts with 1. Well, that's an empty state, because I've normalized. To make it start with 1, actually, there is a way to make it non-empty and make this actually slightly shorter if I just sum over co-prime integers. Then I've already eliminated 1. Then you can see that any pair of integers at all, if it's not 0, is a unique integer called the GCD, positive integer, times a pair of co-prime integers. So therefore, up to a scalar factor, which is the Riemann order zeta function at k, it's the same if I include this condition. And if I do that, then the only terms that contribute to the cons term are where m is 0 and then n is plus or minus 1. k is even, so I get it twice. I have to divide by 2. So if you want to know what the constant is, you can write it like that. But you could also put 1 over 2 zeta of k times the sum I wrote before. Now, each ek starts like this. But in fact, ek of tau always starts as 1 minus 2k over bk times q. And the rest is a very nice function, which we'll come back to actually later today. This is the standard calculation. I'm not going to give it, because this is not, of course, the multiple forms. But if you look at my book, my chapter of a book, but it's the chapter I'm quoting, called the 1, 2, 3 of multiple forms, which anyway is kind of a basic book to get to know what's in it if you want to do any kind of number theory, because multiple forms are everywhere. Then you'll find the proof. It's not at all difficult, but I'm not going to give it. And the expansion of this thing is that except for the constant, which is 1, there's a common factor, which is minus 2k over bk. Remember, bk is 0, if k is up, the k isn't up, because it's 4, 6 or something bigger. Actually, even 2 is allowed. And then the coefficients here are for each q, you sum all of the divisive for each power q to the n, you sum the divisive n to the power k minus 1. But now, the statement that I had before, which that this number is the reciprocal of in it, well, this is the reciprocal. So this will be in z of q, if and only if, by what I already claimed, k has one of these values. But if you know multiple forms even a little bit, you know that that means that the space of cuss forms is empty. So I remind you very briefly, mk, or more precisely, mk of sl2z is the set of multiple forms of weight k. I won't write down the definition, but it's essentially, well, it is just this. It's functions f, such that f of a tau plus b over c tau plus d. Is c tau plus d to the k f of tau plus some modest growth condition? And this space, if k is, if I exclude 2, which is a little special, this space always contains the Eisenstein series, because as I said, that is a multiple form. So these are multiple forms of weight k. For the reason I said, that's not true if k is 2. If k is 2, this sum still converges. If you do it in some order, and we actually first sum the internal sum over n, and then over m, and then we'll get this formula. But it's not absolutely convergent. So the argument is not true. And it's not quite multiple. There's a slight hiccup called quasi-multi-redual. Come back to that when we do an asymptotic calculation in a few minutes. But e4, e6, and so on are multiple forms. So we certainly have e k of tau. And then mk of tau is the sum of sk, or sk of sl2z at the multiples of the Eisenstein series. And these are the cos forms which, in the case of sl2z, simply means there are multiple forms which have no constant terms. So the constant term is 0. So now, these, so if you look at the ring of all multiple forms, then it's a famous theorem that it's generated by e4 and e6. And it's a free, freely generated. There are no relations. And so a basis, if here's k and here's a basis, mk, then for 2, there's nothing. For 4, there's e4. For 6, there's e6. For 8, there's e4 squared. For 10, there's e4, e6. For 14, there's only one possibility, e4 squared, e6. So it's a one-dimensional space. So there are, again, no cos forms. But for 12, there are two different forms. e4 cubed has weight 12 and e6 squared. Let me write that in a place where you can see it. So e4 cubed, which starts 1 plus 720q, et cetera, is a multiple form of weight 12. And so is e6 squared, which starts 1 minus 2 times 504. And so when I take the difference, e4 cubed minus e6 squared, it starts 1728. And in fact, 1728 divides all of the terms. So now it starts with q minus 24q squared. And this is the famous dedicated a function, which will also, well, the 24th power. It will also come back today and many times. And you played a role in the Ramanujan talk yesterday. So delta has this famous product expansion, which I'm not going to explain now, but many of you have seen. And that's a cos form, because it starts with 1. This function is called by everybody delta. And it's a cos form. And in higher weight, SK in general, every cos form is divisible by this delta and the remaining weight is k minus 12. So the only times that you have no cos forms is when either the weight is less than 12 or the weight is bigger than 12, but there are no more forms of that weight. There were no more forms of weight too. And so m14 still works. So this is an equivalence, but that's not just a random coincidence. That's a theorem. There's a beautiful, well, conjecture in the most general situation. Certainly a theorem here, which is that if you have any two multiple forms, maybe different weights, different groups, even different types, they might be illiterate multiple forms. I don't want to go into that. Let's say for the moment, two multiple forms on SL2z. Then there's always, they're always linked by a chain of congruences. So if you say two multiple forms are congruent, if their difference is divisible by some number, say some prime number P, then we make an equivalence relation and the meta-statement is that the whole set of all nice multiple forms, heck, I can value so I'm not going to go into it, is a connected graph. You can get from any form to any other. But if you stay in the given weight, it should literally be true that there's, but it's true. Okay, so what does that mean? We have ek and we have delta. But now there has to be a congruence between them because any two multiple forms have to be related by congruence. That congruence, if I rescale ek, so it's actually called gk, I'll use it later today, if I let gk of tau, which is a multiple of ek of tau, and I define the multiple so that all of the horrible coefficients that I had here in this formula, remember all of them had a common huge factor, which was this Bernoulli number, I take it out. So now that becomes simply the constant term, bk over 2k plus the sum n from one to infinity, and here I'll use the standard notation, sigma k minus one of n cubed of n. So for instance, sigma k minus one of four is one plus two to the k minus one plus four, k minus one, as I said in general, sigma k minus one of n is the sum of the device of n to the power k minus one. So now this is all integral. These coefficients are all integral. But here I have an extra one, but if I have delta, then there has to be a congruence between them. But delta, I mean not delta, any cusp form, but a cusp form is coefficient zero. So therefore, that prime has to divide the difference between the Bernoulli number and zero, which means that p divides the numerator of bk over 2k. As soon as there's a cusp form, that's got to happen. And that's why the strange coincidence. I had not prepared any of this. It just struck me as I was telling that it's such a nice piece of mathematics to know that everybody should have seen. And if you don't know multiple forms, I hope this is already enough to begin to wet your appetite, whatever multiple forms are, even if you've never seen them. So they're defined by this very simple form, but if I define the general multiple form of weight k, it's simply this. And they have such strong properties that for instance, if you have any multiple form of weight 12 on SL2z, it has to be a combination of e4 cubed and e6 squared. So for instance, to prove the very non-trivial identity, if you try to do it directly, that e4 squared is e8. The proof is automatic if you know this, because they both have weight eight and they start with one, and it's a one dimensional space. And similarly to prove this identity, which is very non-trivial, you prove that this product is a multiple form, which is not completely easy, but then e4 cubed, e6 squared of that have to be linear dependent, and you just look at two or three coefficients and you find the coefficients. So multiple forms are an extremely strong and beautiful thing, but in particular, they have a direct arithmetic connection with the Bernoulli numbers and Zeta values, specifically as the constant term of the renormalized Eisenstein series, which I say I'll come back to. So that was a long digression, and I'm gonna remove all of this now, except the little table, and go back to, and also this is the thing about the, I put here the primes. I can mention a couple of theorems. So I talked about the numerator, that it's never one, but P divides the denominator of Bn, in fact, even a Bn over n, this is a famous theorem von Start-Kleusen, if normally if P minus one divides n. So here I just erased it, here we have the primes two, three, five, seven, and 13. If you subtract one from two, three, five, seven, and 13, you get one, two, four, six, and 12. One, two, four, six, and one, two, four, eight, and 16, and they all divide 16, and similarly here for 12, you get one, two, three, six, and 12. So you get, anyway, it always works, it's a theorem. The numerators are also very interesting. This isn't really numerated because it's really Bn over n that counts. These numerators are primes, such primes are called irregular, that was Cumber's great discovery, that you can divide primes into regular and irregular, and for any prime that occurs in the numerator of a Bn over n, so I admit the trivial factor, is by definition, that's the definition, that is an irregular prime, and the first one in the sense of the first one you see in a table of renewing numbers is 691, the second one is 3,617. Later the numerators have several prime factors often, so you get many. But the actual first irregular prime is 37, but it only occurs in B18, and the next one is I think 67, I've forgotten. There are only three under 100, and Cumber proved both cases, the first and the second case of Fermat's last theorem for all regular primes. So the primes that don't occur in the numerator of any renewing number are much, much easier to get at in deep number theory. That's not at all elementary number theory, what Cumber did, and in particular Fermat's last theorem, which has now been proved for all primes using, my while said company, using multiple forms of incredibly deep proof, but the proof of Cumber was very deep at the time, and it's still difficult, but it's much more elementary, and that's exactly for the primes that don't occur in any. So just from general background, renewing numbers are fascinating arithmetically. The denominators are very simple, we know which primes and we even know which powers. Okay, but namely P cannot divide the denominator B and more than once. So if it occurs, the only higher powers are P minus one divides N, and the power of P divides N. So we know exactly what the denominator is, but the numerator had these mysterious irregular primes that are much deeper than the regular primes, much harder to deal with from the point of view of number theory. That was again a digression, but this is not of course a number theory at all, asymptotics is kind of disjoint from number theory, but nevertheless I wanted to say some of these very, very beautiful facts. Okay, so that was just a list. And so I remind you that last time I said the standard definition of the renewing numbers, but it's kind of a little silly because it's perfectly good definition. Of course everyone uses it, including certainly me all the time, it's very convenient, but as a definition it's not satisfying. Sorry, just let me finish the sentence so answer, a mathematical definition should be of course correct and meaningful, but it should be motivated. One should not as people do all the time their papers say let F of t be the following function then formula that takes up half a page, and they don't tell anybody how they found them. Three pages later there's a theorem about F of t. A definition of mathematics should be motivated. You say we do something and we see here's an interesting function, let's call it F of t. So this is not very well motivated why the heck should we care about, forget the factorial, but okay, but why this particular function? And as I said last time there are many reasons, but you had a question, so. Yes. Oh, you even have a mic. I hope this is working, yeah. Yeah, so you said it was a digression so maybe it's not so, that's your relevant question, but I just had two questions that came to mind. So you wrote this expansion in powers of Q and it just seemed that these coefficients were growing really quickly in the beginning. No, not at all quick, but they're not. Well, polynomially. No, but I meant, it's not quite quick, it's only polynomially. I'll answer that, no, that's a very good question. Let's look at e10 of n. So it's some constant, which is b10 over 12, so it's minus one over 132. If I did it right, I may have, oh, it's bk over two k, so it's over 264. And then it's one times Q, then it's two to the ninth plus one. Now of course you can say that's a big number, it's 513, and then it's three to the ninth plus one times Q cubed, but when I get to the nth coefficient, it's n to the ninth plus other things to the ninth, Q to the n, but this is also n to the ninth times the sum of the divisive n of one over d to the ninth. Certainly, right, because the divisive d of n come in pairs, d and n over d. So this sum, sorry, d to the ninth, the sum even to infinity is less, well, so here it's strictly less, and then it's eight of nine, it is less than two. So this coefficient is, oh, it's n to the ninth. So sure it's big, but it's only polynomially big, it's n to the ninth. Q is less than one, and therefore Q to the nth is exponentially small, so these series all converge extremely rapidly, there's no problem. Okay, thank you, and then I was also wondering about... Thanks, by the way, for the question. So if you have questions, please ask, I'm in there. Very elementary, because I'm not used to this thing. And also I was wondering how you see that from this Q expansion that this gives a real number when tau is real, because I think you're summing over phases. No, no, no, tau, you don't see it because it's completely false. And in fact, well, sorry, it is true that Q is real if tau is real for a very simple logical reason. Tau isn't real, and therefore everything is true if tau were real because it isn't real. What's true is if tau is on the positive imaginary axis, then Q, which was e to the two pi i tau is certainly real, and in fact, it's between zero and one. But I didn't say because I forgot. Tau should be an, I wrote because I learned these things in Germany, in German age, some people write a, this can be age, that's a German capital H, kaktur in text. If tau must be in the upper half plane, which is the set of complex numbers, I can call them z's and stuff already, such that the imaginary part of z is strictly positive. So tau, whatever it is, is not real, it's in the strict upper half plane. And that exactly means that i tau has strictly negative real part, because i times i. And therefore, Q to the i tau, so that's if it's real, but in general, Q is less than one, and of course it's also not zero. But then in the limit, you can think of what happens if Q equals zero and sort of add a point at infinity. So indeed, Q is always less than, I don't care whether Q is real or not, but Q is always less than one in absolute value. And therefore, in the sum I just had, Q to the nth is like e to the minus sum n epsilon, it's much, much smaller than n to the ninth. So although these powers are big and look forbidding, they're actually small, as I'm told that they compare to what they're multiplying. So yeah, thanks for both questions. I'll come back, this function is going to come back even today, and possibly later in the course, well actually now, at this speed maybe it won't be today, it'll be the beginning of next time. So Thursday, but I'm not going to explain this expansion. As I said, it's easy, it's written out in many textbooks, but in particular in the one, two, three of multiple forms, which is of course the best textbook. So please everybody buy it and keep a copy under your pillow and fall asleep every day reading it. So the basics of multiple forms, I think, first I think every pure mathematician should know the basic definitions of properties because they come up really in all kinds of contexts. Certainly in everywhere in physics, electrodynamics, but also in every part of mathematics, certainly not just in number theory. Okay, but I'm not talking about multiple forms, I'll erase all of this, it was a digression, and now no more questions about it unless you have one at the end, but I should get moving. So this is the standard definition, and by the way, if you, here's an exercise, and I'll use it in a few minutes, that if I take the log of one, and it's e to the minus x, well this starts with, if x is very small, this starts as log x, I'm gonna minus sign to make it positive, then it starts log of one over x, and the further coefficients, actually I should call it t, there's no reason here to change variables, it's log of one over t, and the further coefficients are exactly like they were before, namely bk over k factorial, except that there's a t to the k over k. So the proof is obviously not very hard, the right-hand side, if you differentiate it, it's essentially the same as what we had here, but you have to differentiate and multiply by t, and then you'll check that they're consistent, but there's a small question since I've differentiated, there could be a constant of integration, you have to look at the beginning of the expansion to see that it's true, but it's completely easy. So that's, if you want yet another definition, even more artificial, if you don't like the log one over t, of course I could have put log of t over one minus e to the minus t is equal, and then I'd just have that term. So that's also definition, even less natural than the first one. As I said, another definition and another function which is very close to t over tan t, if I replace t by i t, would be x over tangent x. So this one, if I add, so remember this thing starts one minus a half t plus, sorry, one minus, yeah, that's right, plus a 12th t squared, and all the other terms are even. So if I add a half t to it, it becomes an even function, and if you think about it, you'll see that's actually t over two divided by hyperbolic tangent of t over two. So if I change t over two to ix, it becomes x over tangent x, and so that's also a nice expansion. Of course it's an even function, so it's a plus or minus, depending on the sign, bk times, I think now it's two x to the k over k factorial x to the k, and the plus or minus depends on whether k over two is even or odd. Okay, so those are kind of different ways defining them. As I told you last time, bn of x, the nth Bernoulli polynomial, is usually defined in most books by the sum k from zero to n, and then binomial coefficient, bk, x to the n minus k. So this is a monic polynomial degree n, it starts x to the n, the next term is minus n over two x to the n minus one, and the last term is always bn, because bn by definition. But then as I explained last time, this Bernoulli polynomial actually has better properties. I want to keep the proposition, so but I've written it there, so I can remove the two ways here. So the Bernoulli polynomial has lots of nice properties. One, it's practically easy. Well, first of all, I gave you the good characterization last time, which is that if you take the Bernoulli polynomial and you integrate it from x to x plus one, I'm correct, I should put a different variable x prime. So I take, so here's the Bernoulli polynomial and I go, I integrate from x to x plus one. Then what I'll get is of course a polynomial, and it's obviously still modic of degree n, because if x is large, then x prime is very near to x asymptotically, and bn of x prime is x prime to the n, which is asymptotically x to the n, so I've x to the n times one. So this certainly starts, it's a polynomial which starts with x to the n, but actually it's exactly x to the n. As I said last time, that's a complete one line characterization, neither inductive, there are also inductive definitions, but many definitions, but this is the very simplest direct intrinsic definition, bn is the polynomial whose integral from x to x plus one is x to the n. Now from that or the generating function or anything else, it's impossible not to find a proof of this if you use any definition. The derivative of any bn of x is the previous one, and even more interesting property, the difference between bn of x plus one and bn of x. Well what is it? If I integrate that difference from x to x plus one, I'll get x plus one to the n minus x to the n, but that exactly means that this difference has to be nx to the n minus one, it has to be the derivative of x to the n. And you can see that almost in your head if you think carefully from this property, it implies this immediately. And this is the most interesting property, and actually this is the most important property for annoying polynomials, but you can't use it as a definition simply because it only defines bn up to a constant. You could add a constant to bn. So let me make a little table, n and bn of x. The b zero of x is of course one, it's a monopognomic degree zero, the next is x minus a half, the next is x squared minus x plus a sixth, the next is x cubed minus three halves x plus a half x squared plus a half x. There's no constant derivatives, remember the odd Bernoulli numbers are zero, and so on. So you see that if I take b three of x plus one minus b three of x, I'll get three x squared, but that'd be equally true if I had a constant here. So this doesn't fix the constant, the constant's at what we want because that is the Bernoulli number. Actually it does if you add the previous thing because if you say this also has to have the previous one is its derivative, then this term which we do know will give that term by differentiating. We have to do it correctly. But anyway, so these two properties certainly defined bn uniquely, but that's too heavy, that's overkill as a definition, but this implies in particular, and this is what both Seke and Bernoulli wanted to have and found that if you take the sum, let me call it sort of a, so I don't have two n's floating around, but I take the sum of the first a, natural numbers to sum power n, then I can compute that if n is fixed at the sum of the first fourth powers, then I'll get a formula. Well, for instance, I think everyone knows the formula, certainly everyone knows the formula one plus two up to a, is a squared a times a plus one over two. That probably most people know the formula eight to squares, you get a times a plus one times two a plus one over six. And if you take the sum of the cubes, then you get square of a times a plus one over two. But for each power, there's a polynomial, and that polynomial I started to write it here is the n plus first polynomial of a plus one minus its constant term, which is bn plus one, and then divide it by n plus one, and that just follows from this by summing it up. So in other words, if you know the Bernoulli polynomials, you know how to add up the powers, six powers of that. And that was a problem that naturally suggests itself to mathematicians, both in the West and in the East, and it was solved, as I say, independently and at the same time by Seck and Bernoulli. Okay, so much for the Bernoulli numbers and the Bernoulli polynomials. So now I want to explain the Euler-McLachlan summation formula. So now I can move to the table. So let me take a function first, which is a smooth function, let's say, on the interval zero, one. I don't want to take any rest, let's say the closed interval. So I have a function from zero to one, and the function is, as a graph, it's smooth. It's smooth for both of them. It's smooth on that interval. Okay, then I do the following calculation. I want to calculate the interval of f of x dx in some other way. So the idea is this. The first step is completely artificial, but then you'll see that inductively you can keep doing it, and then it's not artificial at all. You say that this is the same as d of x minus a half, because after all, d of a half is zero. But that is the integral from zero to one of f of x, d, of the first Bernoulli polynomial, because the first Bernoulli polynomial was x minus a half. So now by integration by parts, that's the same as the product f of x, b one of x, evaluated at one and at zero and you subtract minus the integral from zero to one, b one of x, f prime of x dx. Now b one of one is a half, and b one of zero is minus a half. So you get f of one minus minus f of zero. So what you actually get is the average value, which is what you'd expect, because I have my smooth function, let me make it more interesting. Here's the value, and if I take the average value of f of zero and f of one, that's the simplest approximation, the simplest thing to guess. If you only know the values of f of the two endpoints, then the obvious guess is it's the average. So we start f of zero, and now the second term, I can do it again because we just have the property that the derivative of each bn is n times the previous one. So b one, I can again write as the derivative of b two of x, but now divide it by two, because b two prime is two times b one. But now I can do the same thing. So now I'll get f of zero plus f of one over two, and now again by integration by partial of minus f prime times b two of x over two, evaluate it at one and at zero. But another simple property that I didn't write here, if all is easy for all of this, is that once n is bigger than one, actually if n is different from one, then normally bn of zero is bn, which is the definition, but bn of one is also bn. And in fact, more generally, not that I'll need it for anything, bn is symmetric or anti-symmetric around the point of half. So in particular, if n is the values at zero and one, that the same up to sine minus one to the n. But except for b one, that sine is always one, because if n is odd, then the Bernoulli number is zero anyway. So when I do this next calculation, b two of x times f prime of x at zero and one, in both cases, I simply get b two of, b two over two. Here I'll leave f prime of one, now minus f prime of zero. And then by integration by parts again, you get the value at the end points, you have to change the sine, so the minus becomes plus. And now you get b two of x times d of f prime of x, d two of x over two, dx. And now you see what's going on, I can do it again. Now I write b two of x over two, which is two factorial, as d of b three of x over three factorial, because that's this formula. And so I keep going, and so by induction, I'm sorry that it's hard to read, that's this last time Belter said that, one of the chalks, I thought I took blue chalk, but basically none of the chalks are very good. And so if I erase the board, it's still full of all the previous formulas, and it makes it very hard to read anything, but I mean, there is probably a wet sponge, but then I have to dry the board like they do in Germany, that's also in use, and so also I don't see it. So if I keep going, just by induction, the induction is very easy, but I better take my notes so I don't get it completely wrong. What I'll get the first term is the only exception one, because all of the terms with higher derivatives will have a bn, or actually bn plus one. And remember, except for b one, the Bernoulli polynomial is the same value at one and zero. So when I have a product of a Bernoulli number and a derivative, out of simply the Bernoulli polynomial times derivative at zero and one, I'll get the Bernoulli number times always the difference of the derivatives except for the very first time where I get this sum divided by two, which it should be, it had to be the average value. So here's the form, if I do this capital N times, then I'll have n minus one terms, and they look like this, minus one to the n, bn plus one over n plus one factorial. And then as I said, all the terms except for the zero derivative, you're taking the difference of fn of zero and fn of one, the nth derivative of f at one minus the nth derivative at zero. And then the remainder term, you'll have dn of x over n factorial dx times the capital Nth derivative of f. Now let me say that f is actually, not just from between zero and one, but also between one and two, between two and three and so on. So now I will have the same formula. If I look at the interval, not from zero to one, but from one to two or from two to three. So more generally, if I just shift things by m, I'll get that the integral from m minus one to m of f of x dx. A little bit f of m minus one plus f of m over two plus the sum n from one to capital N minus one, minus one to the n, bn plus one over n plus one factorial times fn of the right endpoint minus f the nth derivative of the left endpoint. But now there's a slight thing to worry about, which is that this was minus one. I get minus one to the n, integral from zero to, from m minus one to m. Here I'll look at the nth derivative of m on this interval. But now it's not bn of x. I've shifted the interval by m minus one. It's bn shifted by x minus m plus one. But m minus one is just the integer part. So let me define that this is a very important definition that comes up in many places in mathematics. Not everyone used this notation, but it's fairly standard. The reduced Bernoulli polynomial is defined as bn of x minus its integer part, if you prefer bn of the fractional part. So now between zero and one is the Bernoulli polynomial, and then you just turn it into a periodic function. So there's a slight, so let's do it for instance, b2 of x, b2 of x bar. Well, at zero and at one, we already said it's x squared minus x plus sixth. So it starts at a sixth. Its lowest value is here at minus a 12. That's here. But then remember these two values are equal. So now when I make a periodic, it's a step function that looks like that, but it's a continuous function because bn of zero and bn of one with the same. So bn bar of x is always a continuous function. b zero bar of x is very continuous. It's even analytic. It's the only one that's analytic because of course it's simply the constant. I don't know if you make that periodic. Only b1 of x causes you a headache because remember the graph is from zero to one. It was x minus a half. So it goes from minus a half to half. And then it's a salt tooth function because I'm going to make it periodic. And so I could say I put the value minus a half here or plus a half but I don't do either one. I take the average value because that's what you do when you do periodic transforms. So b, I won't ever use it, but b1 of x, b1 bar of x is the fractional part of x minus a half of x is not an integer but it's zero if x is an integer. But anyway, let's say here my n is not to be one. It's going to be large very soon. So here, this bn, I could have equally written bn bar of x because between zero and one they're the same. And I don't care about the values it's zero and one I'm integrating. It's measured zero. So therefore, here I can put bn bar, it was really bn bar of x minus m plus one that is the integer part. And so that's the formula that I now get. And it doesn't matter at all. And the m's only play a role that the limits of integration with the function is the nth derivative and then this bn bar. So now I continue that calculation and now you see the whole point and why this only works with renewing numbers. There are lots of other sequence of polynomials that of this property that if I have any sequence of polynomials let's call them bn of x, I don't have to write. Well, I can always make a sequence of polynomials with this property because of already a b10, I know the derivative of b11 that determines it up to a constant. I pick any constant I want. So I integrate the predecessor and then I just keep doing it. So there are lots of sequences like that but only the renewing numbers will have the property that except for b1, the values at zero and one will be the same. And therefore, what we got here is a difference between m and m minus one. Anybody who works with series looks at this and the only reasonable thing to do with that is sum it over m and it telescopes. So for all natural numbers m, I now find that the n from zero to capital M of f of x dx is equal to now the first term is slightly different. It doesn't tell the scope but because of the halves you get the whole sum which is what you'd expect for integral. I mean roughly you should have the sum of the terms should be roughly the integral but the end terms you don't know whether you should put them on the left or the right. So indeed you count them with multiplicity half is what you always do if you have a sum. Let that you count the edge terms with multiplicity half because of the symmetry left and the right of the edge. So if I add up these terms f of m minus one plus f of m sum summing. This process is you sum m goes from one up to capital M. Then I'll get f of zero half a time f of one half and another half. So I'll get all of that. Then these middle terms will telescope as I just said. But here of the Bernoulli number, I don't really need minus one to the end except for one exceptional case. It's n is always odd since n plus one has to be even. So this is always minus except for one term but it's easier to put it than to write a separate term. And now this term as I said will simply telescope. So although it's a sum of many terms in the end it only cares about the end terms m and zero. And then the last term is I'm not sure if I lost a sign so it might be minus. I hope I got it right. But this would be n from zero to m, f n of x e n but now this periodic version of Bernoulli number over n factorial dx. That is the Euler-McLaurin summation formula. So maybe I should put a big red circle around it because this is one of the most important about this, certainly one of the easiest and oldest formulas of all of analysis and you use it all the time in life. So the Euler-McLaurin summation formula. Yes, I'm listening. It's just when you write bn plus one do you divide by n plus one or n plus one factorial? That was, thank you. That's you divide by n plus one factorial naturally. What's the title? In this formula that I'm going to have laid you divide by n plus one but we'll get there. Okay, thank you. So that's the Euler-McLaurin formula. Now let's assume I can now erase this because either you take notes you're not but anyway you're not going to read it again at the beginning. So the end of the formula is still there. Now we go back and assume that the function is small at infinity. But let's assume that f and small just means let's say for the moment it tends to zero, tends to zero, little wrapped, I don't really care but it should be smoothly. So I don't want the function, for instance you could take e to the minus x times sine of e to the 10x. Then if you differentiate it, it'll be huge. So I want a function that also the derivatives go to zero. So I want that each fixed derivative is small at infinity, simply meaning it goes to zero. Which lots and lots of functions do. Obviously anything that's exponentially small in a smooth enough way. Then I can simply take, so this is the Euler-McLaurin summation formula but now as a special case of it I can integrate all the way to infinity. And now I get a half f of zero plus the sum n from one to infinity let's call it m from one to infinity f of n because the sum now goes to infinity. And then I have the sum n from zero to n minus one, minus one to the n, b little n over n plus one factorial, this time I put it in. And then f derivative of m as m goes to infinity is zero. So there's a minus sign and it's simply fn of zero. And finally the last term, I'll just put plus or minus because I don't really care and it's too much trouble, it's the integral from zero to infinity of the nth derivative of f times bn bar of x over n factorial. But now n is fixed, I'm not letting n go to infinity as well, n is fixed. And so bn bar of x over n factorial is some function but it's periodic and continuous, so it's bounded. So this is over one. And this function, I'm assuming it goes rapidly to zero. It's sufficiently rapidly that the integral, so this is convergent. But still I can't estimate this. But now let's take t going to zero, well t is just positive, but mentally we take it going to zero. And now I'm going to integrate, I'm going to replace f of x by f of tx. Then here I'll leave mt and that was exactly my g of t that I was talking about. So now we're almost finished with the proof of this. So this is, of course this one, if I have f of tx dx, I just replace x by x over t, it's obviously one over t times the integral. So you already see the g of t is if over t, which was the first part of the theorem I wanted to get, minus a half f of zero. But if n is zero, you have b one over one, which is minus a half v zero t to the zero, so that's okay. And these terms are exactly the same because remember, I'd written it already and removed it, but bn was the nth coefficient, which by Taylor is one over n factorial times the nth derivative. So therefore, bn plus one over n plus one bn is bn plus one over n plus one factorial times the nth derivative, except that now, because of the nth derivative of the function of tx is t to the n times the old nth derivative. And then finally here, the capital nth derivative is t to the n times the value of the nth derivative of fn of x. So I think it means, I can't do this in my head, it's t to the n, and then I replace x by x over t, then it goes back to being x, and I think it's t to the n minus one, so the coefficient would be right, and here I've done something very weird. Now it's x over t, and so it's got a completely different periodicity. But who cares? bn is fixed, and bn is bounded. So whatever this is, this is o of one. And this whole integral is also, it's just a constant. So it's o of one, I mean it's fixed, it's a number. But it's bounded by a number, so it's o of one. And here I've t to the n minus one, it may be t to the n, t to the n plus one, who knows. But you see that as n gets larger, I'm not saying anything converges, but you'll get more and more terms, and so this is an asymptotic expansion, that completes the proof. I'll stop talking for a minute so you can catch your breath, and also so I can look how far I've gotten. I see, I thought I didn't have enough, and I see I've covered three pages out of seven, so it doesn't look like I'll finish what I prepared, but that's okay. I'm going to go on in a second. Well, let me just assume, so when you use the Euler-McLoran summation formula, it's very useful, it's very useful anyway, but there are two basic scenarios, and both are important for asymptotics. One is you take this m to be infinite, so then you have to assume f is nice and small than infinity. That's the infinite version, which is less familiar and less in the textbooks, and this is essentially just the infinite version of the Euler-McLoran summation formula, but I really wanted to prove the whole thing for you. That was statement one. But to find out when it's also very useful, because let's say f is something, and you don't want to sum to infinity, and there's no t anymore, you have the fixed function, remember what we did here, we had to rescale by t, because if I just ask what's the sum of f of n, well that's just a number, it doesn't really make any sense. To make any sense for a problem of this type, you have a variable. So here I put an f of nt, but I sum to infinity. But of course what you can also do is say I sum from m from one to capital M, and ask what can I say about that? For large ember I fix the function and truncate rather than going to infinity but rescaling, and then this is exactly what this formula does for you. So let me give another example. So an example of the finite use, and this is the kind of thing we'll talk about starting certainly on Thursday, is here, so an example, I'm going to take f of x to be one over x squared. But of course if I take one over x squared, the infinite sum is fine, or even the finite sum, I have a little bit of trouble with f of zero, and also the integral of one over x squared. So as I did it here, I take, that won't work, I take x plus one squared, or I use the integral from one to m, and do the same thing that I just did. Obviously I didn't have to start at zero, it was a choice. I was obviously on the intervals of length one and adding them up. So if I do this, then what I'll get, and that's exactly remember I told you that when Euler computed the, solved the Basel problem by computing the sum one over two squared, what he did was to invent the Euler-McLean summation formula and get a very accurate version for the sum of the first m reciprocal squares. And so it's some concept, which people already knew was finite, and let's call it state of two, that's the infinite sum, but then there's a remainder, and the remainder you can now, I want to copy because I'll get all the signs wrong. The remainder according to his function formula is exactly minus one over m, plus a half over m squared, minus a sixth over n cubed, plus a 30th over n to the fifth. And now it's kind of nice because these are exactly the numbers. I think I removed my table, but these are exactly the numbers that are in the table on the nose. Minus, well if I put state of two minus that, minus a half plus a sixth, minus the 30th plus a 42nd, it continues 130th minus five over 660th. They're exactly the Bernoulli numbers. So it's exactly minus bk for k, m to the minus k divided by m. So well it's the sum of that over k. So this is, of course it's not just for one over n squared. You can equally do one over n cubed and Euler did, and one over n to the fourth and so on. When you computed one over n cubed, you could get 20 digits, and he said, I'm sorry, I was not able to recognize that. For his eight of four, he recognized this by the fourth over 90, and he said it would be an interesting question to identify this number, but now 300 years have gone by, and we still don't know, 400 years. No, 300 years, 350. We still don't know what state of three is in terms of anything else, and nobody seriously believes that there is a form, it's a new number. So, but you can compute it easily to 20 or 30 digits even by hand. If you went into a little arithmetic as Euler was, just by using this summation formula. And this is a typical asymptotic thing. You have the sum, and you want to find the infinite sum. So you say the infinite sum is whatever it is. Let's just call it S for sum, but then there's a correction term, and if you can write down that correction term a priori, then you know everything. And if the function is very regular in principle, you can do it this way, but there's another way which I'll talk about next time. So, okay, so those are the two basic use of the Euler-McLean summation formula for finite sums of a fixed function, f of one plus f of two up to f of m, as it may not of course have an infinite sum. So there might be, for instance, a case that you could also do, if you take one over m, then of course these diverges like log m, and by definition of the Euler constant, the next term is gamma, because that's how you define it. And then there's something that I can't do in my head, it's undoubtedly one over two m, and then there'll be something that I certainly don't know in my head. Like it, but these numbers will again be the very nori numbers, and it's exactly the same formula. The only difference is because of the slight divergence, there's a slight fill up, and you get the Euler's constant, I'll come back to that a little later, as I hope still today. Okay, so those are the two versions, finite and infinite after you do the scheduling, so the rescaling. So let me say very briefly that there are several variants of this, so that at least I can finish today, and then I may start with a slightly non-trivial example, or I may not and do it less next time, and then I'll come to the interesting examples next time, and then there'll be a pause, and I'll come back to that later. So variant one is you take a G, let's call it alpha of T, G of T, I fix alpha bigger than zero, and G of T now I'm fixing it. F is also some fixed function, and this will be alpha T, alpha plus one T, and so on. So if the alpha is one, it's what we had before, so this is the same sum as before, let's say alpha is a third, so here's zero, here's T, here's two T, then if alpha is a third, I take a third of T, four thirds of T, seven thirds of T, and so on. So if you do exactly the same Riemann and Euler joke that we did before, then you'll see that the leading term, of course, this is still a Riemann integral, the fact that I've shifted, it doesn't change anything, it's still a Riemann integral for one over T times the integral, but the other terms would now be the sum. So again, I'm assuming, remember, I'm always assuming, I forgot to put it in the proposition because I'd written it before, I'm assuming that my N has an expansion, the closest is our BN. So now, if I do it a la Euler, I have the sum, now it's F from zero to infinity, F of M plus alpha times T, and so now by the same argument, F of T is the sum BNT to the N, but now I replace T by M plus alpha times T then T to the N is T to the N, times M plus alpha to the N, but that's simply the Horvitz data function, data alpha at minus N. So which, and this one again is up to a sign which I won't try to remember, I can look it up in my notes, is BN of alpha, BN plus one of alpha over N plus one. So if alpha is one, it's exactly the form that we had before, okay? And this form is still due to Euler and the same proof I actually gave it last time when I talked about how to compute Dirichlet series. I remind you, Z of S alpha is defined originally, this is the Horvitz data function, it's defined originally if real part of S is bigger than one and it's simply the sum M from zero to infinity, not one over M to the S as the Riemann's data function is, but shifts it, M plus alpha to the S. And then exactly the same argument that we used for the Riemann's data function said that this is a simple pole at S equals one, that simple pole is reflected in this integral term and then the other term is what I just said. So I'll write it at random hoping I get it right and then check in my notes if I got the sign right. I think it's the correct sign, but if it's not it's a minus sign, it actually seems to be correct, N plus one. And now it's the same principle that I told you before in this sum, only this term is global, only this depends on the whole integral from zero to infinity. So if I change my F to another F, which is exactly the same to all, or it can even be different everywhere, but it's the same to all orders. So it's infinitely oscillatory at the origin. Then only the integral term will change one over T and the whole thing will be identically zero. In particular, if you have a function that's infinitely small that's zero, for instance function would just support it away from zero. So I might have a function which looks like that and sum that. Well then you find it's very nice that you just get the integral over T, the Riemann term. There is no correction at all. The correction only comes from the M term and it is BN plus one of alpha. Of course it depends on the alpha on the shift. So that's the first variant and it's often useful. And of course as a variant of the variant, we could also take the sum, so what if G of T is the sum, now I go back to one to infinity, but I put in for instance a direct that character any periodic function, it doesn't matter what. For instance a direct that character if you don't know what they are, it doesn't matter any periodic function. For instance minus one to the M. Well if it's a periodic function let's say this period three, then I can write, well again I want the T. Then I can replace T by T over three. I mean if I take all the multiples of three out of chi of zero times the sum F of N times three T, then all of the F of one, but it'll be shifted by a third. So from this form I get that. And then it's exactly the same principle and this will be exactly what you would get from the Euler, that the Riemann integral of the Euler argument. Except that now what you'll get is simply the value by the chi in the second place. So it's L of S chi, not L of chi S. So it's L of minus N of chi. If chi is the trivial character, L of S chi is the Riemann's A function. So that's, so you have both a shifted version and a twisted version, but they're the same because the twist, this is just a sum of finitely many shifts after rescaling. So those are frequently useful, but they're very simple variants. Then another variant, what if F had a more general expansion at zero and this is frequently useful. So for instance, what if F of T, at the term I'll give a simple example, what if there were one extra term, B half T to the one half, but everything else was like it was before. Well, you can just guess immediately what is G of T to find as before is the sum F of N T will be I F of T. It still converges, it still is integral. So I F still makes sense. And then there'll be B zero times A to the zero plus B minus, B half times A to the minus a half times T to the half and so on. Because you just do the Euler argument and this T to the half is multiplied by the sum one to the half plus two to the half plus three to the half minus A to the minus half. And it's completely legitimate by the same argument. So more generally, if F of T is some kind of a sum, lambda to the lambda's could even be complex but the real part of lambda, let's say, is bigger than minus one and goes to infinity and then I'll have some B lambda T to the lambda. So I need this in order for the function at zero to be integrable and for the thing to make sense. Then G will be exactly what you'd expect plus the sum B lambda T to the lambda, same lambda, sorry, Z of minus lambda. So you have the Riemann term and similarly, of course, you can do a shift. I'm not going to combine them all and have the orbit say the function of lambda. Now you might say, what if lambda is, let's just stick to real numbers because complex X points never happen in real life. What if lambda is less than minus one? Well, if it's strictly less, remember, the whole thing is additive. Whatever I do, if I know the answer F1 and F2, then G1 plus G2 is, I just add them. So if I have a term one over T squared, it's of course completely idiotic. Then I could just say that F of T be equal to one over T squared. Then G of T is the sum one over N squared T squared. It's simply Z of two, T to the minus two, obviously. So it's clear that if I have a term B lambda T to the lambda and lambda is less than minus one, then minus lambda is bigger than one. This is actually convergent. This part is okay. What isn't okay is, of course, IF is not divergent, so you have to subtract off the singularity, but why bother? You can just write your function as the sum of two functions, one like this and one that's straight. So there's only one interesting case. What if we have a term, T inverse? Well now, but the same principle, I don't have to write the whole inference on. I'll just say what T inverse does. Because, well, no, I do have to say it because the integral of what makes sense. So here, the variant will be as follows. If F of T looks like some B minus one over T plus the sum, well, I'll write like that, the sum from minus one infinity, E and T to the N. And G of T, well, it's not very surprising that the B minus one over T will give you a term log one over T. Because there is, after all, it's one over T and we're multiplying one, two, three. So I'm multiplying by one plus a half plus a third, which is log of something. So it's reasonable that it's log of one over T. And then there's still a constant and that should be the Riemann constant. Of course, the integral no longer converts. I have to change IF, I have to tell you what the integral is. And the other terms are, of course, not affected. Because remember, everything is local. So whatever happens, the effect of this minus one term has nothing to do with the others. They don't interact at all. So all that you need is this and the proof of this is very easy. And especially if I finish it by telling you what IF star is, it's simply the integral, the same integral we had before, but I simply subtract. And then, see, it's really easy. You just take the function E to the minus X over X. Well, that's integral at infinity, it's small. And X at zero, it has a one over X. So this difference is smooth. And now I just apply the old thing to E to the minus X. Again, it's additive. I don't have to do this once for E to the minus X. So to prove this, I just take F of T to be equal to one over T, E to the minus T. Then G of T, which is the sum N from one to infinity, it's one over MT, E to the minus MT. But of course, this is simply one over T times the log of one minus E to the minus T. And that's the one I gave an exercise in the beginning. It's still the Bernoulli numbers, so we're finished. So it's absolutely easy to prove this. And then the final variance, so this would be variant two. So variant one is shifts, shifted, and or twisted. Nice to have names. Spoiler McLaren, variant two, which is written, I didn't write the words, I think, variant two is allow terms at zero, which are powers as before, but lambda is no longer in a non-negative integer, but for instance, a negative integer or even a complex number, okay? And then the final variance, and that you should be able to see in your head how it goes, is yet more general, and all of these are extremely useful in practice, allow terms T to the lambda log T or log T to some power. Remember, I've mentioned before in an earlier lecture, those are exactly the functions. The span of all functions of the form, a complex power of T times a non-negative interval power of log T, those are the functions that are GM finite. In other words, they're scalings. They're rescalings by T goes to AT, line of finite dimensional vector space. Well, if I allow that, let's just do the first one, because if you know how to do one, you can do them all. This is, of course, nothing other than the lambda derivative of T to the lambda. And therefore, you just take the old form and you differentiate it, and so you get something, so what you get, well, since I've been written down, I might as well write it down in that case, so that means that T to the lambda log lambda, of course, I'll assume that lambda is bigger than minus one. The integral is still convergent, so the IF part is the same. I only have to get the single term that you get here, and what you'll get is A to the minus lambda log one over T plus A to the prime of minus lambda times T to the lambda. So if you have a term T to the lambda, before T to the lambda, you just multiply it by Z to the minus lambda. Now we have to differentiate, take actually the negative derivative, if I didn't get any signs wrong, wrong then this is the, that's that answer. So in particular, if T is, if it's just log T, which is actually the commonest case of that, where do I, yeah, so in particular, so the easiest case, well, is if it's just log T, which case I might as well assume that the coefficient is one, because take a multiple, then G of T will be the usual IF of T, and then multiple mistakes, the next term is minus a half log, I'm using that Z to the prime of zero, minus a half I think, log of two pi, you'll get minus a half log of two pi over T, plus and then the terms that we already love, the N Z to the minus N. So actually I have now done five pages out of seven of my notes, yeah, so I was, those are all the variants, so I've given you the basic story, which is what I said at the beginning of F is nice, but nice might be smooth, but it might be less smooth, like you could have some strange powers, T to the one half or T to the minus half, or T to something log T, and you make this infinite sum, then I've told you how it looks as a corollary if the Euler-McClore summation formula, and it's many variants, and I prepared, well they're also in the written thing, but I'll give the same examples, there are four examples, two very easy and two much trickier, and they're extremely illuminating, and I absolutely want to do, they will also shed light on the relationship, which actually Manuel asked me about, between the Euler-McClore summation formula and the Poisson summation formula, which I mentioned very briefly last time. Let me say in the word, the Poisson summation formula is more useful, it's much, much simpler. What the Poisson summation formula says is if I have a function, a nice smooth function that's on the real axis, and I sum it at regular intervals, the sum f of n, or f of nT, but after rescaling, so I take the sum, then I can remind you, the sum of f of n for any nice function, n and z, is the same as the sum of f tilde of n, where n is the correctly normalised Fourier transform, and it's incredibly useful and you use it all the time, but it only applies if you have a sum over whole lattice. What we're doing here is a sum over half a lattice, and that's much, much harder, because you've killed the symmetry, a sum over lattice, well, you've also introduced an edge term. After all, if you could do any sum from one to infinity, then by replacing f by f of n plus 100, you could do the sum from 101 to infinity, and by subtracting you'd have the sum from one to 100, but you can't expect a uniform formula to sum with any function anywhere, and there's a limit. So it's obviously much, much harder to do a thing with the n-terms than something without n-terms, and the n-terms screw you up, and of course you could just say, well, why don't I take my nice function, say, Gaussian, and I just cut it off at zero, but then the Fourier transform will be huge and infinity and the sum won't converge. So it simply doesn't work. So if, and that's a nice exercise, if f, in fact, let's do that now and then I'll stop with that. If bn is zero for all odd n, that means that f is, even though it's a smooth function, it's on zero infinity, now I can extend it by evenness, and it's still smooth on both sides, and now I can apply Poisson, and so what I'll get, I'll do it next time in more detail, but you'll see that in this formula, bn is zero if n is odd, but bn plus one is zero if n is even, and every n is even or odd with one exception, the one exception is b1, which is not zero, and so you get only the term i0, then g will be if over t, and you have only the term here when n is zero, in which case I've b1, which is minus a half, b0, which is f of zero, and that's to all orders, and that's exactly with the Poisson summation, sorry, this is a half, and the reason for the half is because really we're summing from one to infinity here, but I should sum from minus infinity to infinity, that would include b0, but I would have it twice, and so you need that, and then this is the Poisson summation, so of course the Poisson summation form is much stronger than that, because it tells you what the O of t to bn is, it's not nearly asymptotic, it's exact, this is merely asymptotic, so the answer is less good, but the problem is much better because it's a much harder problem, so both, neither one replaces the other in its usefulness, but I want to have said that now, I'll do that again next time. Next time is Thursday, same time, same place, and thank those of you who came personally for coming personally because it's nice to see live people, and thank those of you who came virtually because it's nice to see dead people now because it's nice that you're coming anyway. Okay, are there any questions, virtual or live, please? Yeah, I think if you just speak, I can shout it out, but it's easier if you have a microphone, easier for me too. Is there some reason why this Euler plus Riemann argument actually gives the right answer? Basically because both Euler and Riemann did it correctly, so namely I cheated on both, Riemann was trying to define an integral for much more general class of functions, and then his definition of the integral is, it is the sum in the limit, so his thing says that asymptotically, to the first order, it is if over t, so that's true. Euler of course did not do this, he wouldn't just interchange two sums and say like everything's fine, and he would know that the answer is always wrong because of course you missed the term, of course he found the Euler-McClarne summation form, he did it quite correctly, it's just a way to remember that there is a reason that formally the coefficient of one over t has to be integral basically by Riemann's argument, and the others, it can't be anything except say it of lambda, let me make a very crude argument, this is so bad that I don't really like it, let's say that you've understood, and maybe Manuel can tell this, he's a professional analyst, I'm not, what the reason is for the thing I've emphasized several times, if f is exponentially small at the origin, then you'll simply get a constant over t. Oh I do see the argument, sir, I don't need a manual, it's okay, next time, it's more serious, if the function, you see we're not talking analytic, we're talking smooth, if the function vans to all the orders at the origin, then bn is zero for all zero n, in particular for all odd n, so it's even, so I can, I can, we're assuming it's smaller than infinity, so I can make it an even function, I can apply this, and actually b zero is also zero, so you see this special case is simply Poisson, applied to a smooth function on the whole line, and therefore that tells you why you only get a one over t term if all of the coefficients at the origin are zero, but that tells you that the coefficients of g only depend, except for the one over t term, only depend on the coefficients of f, but you could say well why couldn't the coefficient in g of t to the fifth be a combination, a linear combination of b zero, b one, why is it just a multiple of b five, but that's obvious, whatever the theorem is, apply to f and apply to f of two t, then the coefficient of t to the fifth changes by, of t to the fifth, changes by two to the fifth, but in g each term changes by a different amount, the only way it can work is if it's the same n which is coming in, that's the only one which is scaling and varying, that is indeed the argument that these functions are g, m, they're finite. So in other words, if we use this to see that whatever the argument is, it must be the integral over t plus something whose individual terms depend on the individual terms, then that means that the coefficient of t to the n has to simply be a multiple of the coefficient of t to the n in f simply because nothing else is correctly scaled, that coefficient behaves by lambda to the n when you rescale by lambda, but that wouldn't be true if I took any other coefficient, therefore you get that the coefficient t to the n here or t to the lambda here has to be something times t to the lambda in f and in g, but whatever the argument is, it should be true by analytic continuation, but if lambda is very negative, then Euler's formal argument is simply correct, if this is t to the minus 3.2, then I'll get the sum n to the minus 3.2 t to the 3.2, and of course that is z to the 3.2 t to the 3.2, so it's clear that this multiple has to be z to the minus lambda. So in fact that is kind of a pure thought explanation of why this is the only thing that it could be. I don't know if that really helps, and I gave a proof anyway, but it's not a complete mystery, although I presented it a little perversely. It's mostly for the mnemonic version, if you forget all the details like what's the sign, minus 1 to the npn plus 1, you don't need that in z to the minus n, and the reason is just to do it formally, it can't be anything else, it's gotta be z to the minus n, and so that's the formal argument and the Riemann argument. So it's more a mnemonic that tells you how to remember it, there's the Riemann way of looking in the Euler way, the formal way, and they both are giving you pieces of the answer. But of course if you do the formal thing correctly, that is the Euler-Bichard summation form, that is Euler. So I'm not trying to malign him and say he didn't know how to do this. In fact, this theorem is his entirely, and this is just the Euler-Bichard summation form in the special case when m goes to infinity. But then rescaled and thought asymptotically, and that's something people don't necessarily think of, as you usually see it as a finite form. Okay, maybe we should stop, oh sorry, there are still questions, but remember if you can come, you can also ask questions the next time at the beginning, but go ahead. Okay, so very quickly, is the Bernoulli polynomial, are they orthogonal with respect to any measure? Oh, that's an extremely good question. And I know the answer, well first of all I think all polynomials are both with respect to a measure, you define the measure in that way, but I know in some of these my email, I will forget, I forget everything, if I write it down I'll lose the piece of paper. Could somebody send an email and say remember the question about orthogonality? There's a beautiful property of Bernoulli polynomials that I found, well I found several, and I found some that were certainly well known in the literature, 50 years old as I had expected. Others weren't, and they had to do with the integral, but the ordinary integral for the usual measure of one Bernoulli polynomial against another, I completely forgot what the story was, but there is a nice story. So that's a great question, because I was going to say it is nothing with this course, it's exact and not asymptotic, but I bet that if I think about it, those statements would be very useful for asymptotics and I never thought of it, so a wonderful question. I'll try to remember between now and Thursday, otherwise you or anybody's here can send me an email, don't remember orthogonality, and I'll look at what I did and try to remember with what it was. Just for fun, who knows orthogonal polynomials in the sense that they know the basic definition of properties? Well, you obviously do, yeah. So very few people, well, you obviously do and you aren't raising your hand if you're doing something else. I was once at one of the famous Gelfand seminars, well, I was several times, I was in fact the lecturer, but before they start, he would come at eight in the evening, which is when it was supposed to start and talk to the people, there were 100 people would come, he would talk to them all, mostly putting them down and we wouldn't start till nine and then go on till one in the morning sometimes. And so here he was, he cared very much about education of young people because he'd been very precocious and he brought up many young mathematicians. And so somebody was telling me, young postdoc who was there, said, you know, Israelovich, Israelovich was his name, can't even think. He said to a professor, I'm teaching a student and he's 14, but he's very, very gifted and I'm showing him. So we've already done, and then he went through a whole list of course, multivariate integration and differential equations, simple course of differential equations and this and this and this and this and I was sitting at my highest like that, wow, this kid was maybe he was 15, but anyway, he was under 16. And so Gelfand listens to this litany of all the things they've done, this was in a period of three months. He said, you know that student I told about, we've now done that. And Gelfand just did one question, does he know orthogonal polynomials? No? He's got to learn orthogonal polynomials. Everybody has to know orthogonal polynomials. And then he turned away and bought the next guy. Not one word of praise that this young man had told some kid all of this incredible amount of mathematics in three months. Everybody's got to know orthogonal polynomials. So if I answer the question next week, I'll make another digression. Anyway, that's the point of the course. It's all digression about the basic thing about orthogonal polynomials, but just as a quick test, if there's one word which is a very familiar theme in number theory and mathematics, but it's infinitely related like this to orthogonal polynomials, what's that word? That nobody gets this, even if you know, it won't strike you. It's continued fractions. It's practically the same thing. You are nodding, but even you aren't, and you certainly know orthogonal polynomials, and me too, Laguerre, there are many famous orthogonal polynomials. It's very, very close. And maybe I'll even do something about that. It doesn't have a hold on to do with that some time, but I can pretend. Okay, I think enough questions for today and any other questions, you've got to come back next time, like Scheherazade, and then you get your questions answered or not answered next time.