 Maybe wait for more people a couple more people come come closer. Yeah, maybe And I start or not it's already on and The people in the waiting room are invited. Ah, so everybody is is here. Ah, they're already there. Yeah, I never understand these things so Well, there are fewer live people than Last time so I hope people keep coming in person Because it's much more fun to talk to people then only only virtually This I don't need anymore also one of the participants told me that When I answer questions from the room people on this own can't hear the question So I should repeat it. So if I forget, please remind me to repeat the question out loud I knew that but I completely forgot and I'll certainly forget again. So Today will be a little I mean, I hope it will be some fun, but not one of the Most Unexpected parts of the course partly because I'm going to explain things that I explained in the published paper And some of you may have looked at it But that is the advantage that if you need a reference there actually is one. Maybe I'll mention it at the end so I'm not yet going to talk about Koshy's form and the circle method that will come because this sort of has to come first So I won't discuss the example. I mentioned last time of Stirling's formula and why he is not the square root of 2 pi although. I'm curious next time I'll ask if anybody Look at that. But what I do want to talk about is actually several related topics of which the most important is Well, I'll just say melon transform, but I'm not sure how the name should really be pronounced whether it's Melin or Melin or I'll just say melon because I'm speaking English So the melon transform is something that one uses all the time and then I'll talk about and an extension to functions that Where the usual definition doesn't seem to quite make sense So and applications I don't have to say each time because everything in this course is not theory It's about examples and applications to concrete problems But in particular, I'll talk about L values, which I mentioned last time the special values of L series I want to give in some detail, but not the complete details the example. I mentioned last time of the Casimir effect which led to this rather bizarre function, which was the sum over all triples of integers of The square root of L squared plus M squared plus some constant times n squared Obviously very divergent the terms go to infinity and one question is how to make sense of that and having made sense of that How to make effective sense you can compute it to for instance 50 or 100 digits given lambda and also the asymptotics And that's a nice example of how to use the melon transform And if I get to it today, which I think I will I'll also talk about sums of The form so I mentioned this in the list of topics yesterday So I have some nice function f we're now nice will be made a little more precise later But mostly will not be very precise But it's such that it's small at infinity and so if t is a positive number you can ask for the Infinite sum f of nt and the question is how does that behave as t goes to zero? So and all of these topics are interrelated and they will be related Probably next time with the Cauchy's form the circle method and so on with With more applications closer to the number theory or combinatorics so I start with the Melon transform So I'll start in the classical situation when everything is well-defined. I assume that I have a function It could be complex value, but I might as well assume. It's real. I could look at the real imaginary part separately So I have some function. Let's call it phi of t defined on the positive real axis And I want to assume for the moment that the function is small here We're small For the moment, let's say could mean very small like it decays faster than any negative power of t At infinity and I'll also be assuming for the moment that it's smooth here Okay, so that's my my function. It's a test function. It might be smooth It might be only actually I think if it is continuous, it's more than enough for what I'm doing I'm not going to worry about analytic Questions like how general the theorems are in all of the examples If I will also be always be some very nice to behave function and then what defines the melon transform I won't write the word again the melon transform which I'll denote with the tilde is the integral from 0 to infinity Phi of t t to the t to the s minus 1 dt So one could say why call it s minus 1 It's just t to a complex power and of course one could put t to the s But it has better properties if you do this and one reason is this is a product of a multiplicative function of t T goes t to the s is homomorphism if you multiply two t's you multiply the values and then the merit the measure For multiplication is not dt. It's dt over t because when you're multiplying numbers, you're adding their locks So it's very natural to write it that way and this plays a role certainly in all of number theory including when you do like Jackie Langland's theory you write fancier Melon transforms over some adelic space But the idea is still the same basic thing and essentially all L functions are understood when they are understood Which they're often not by trying to write them as some known multiple of some melon transform of some known function so that's the melon transform and I want to do a couple of things with it of which the main thing is to make sense of it when the function is not Necessarily smooth at zero and not necessarily very small at infinity so I want to extend it That's not a big thing theoretically. It's easy But it's not certainly universally known and it's incredibly useful to know how to do that So in the final analysis, we're going to have allow functions that might for instance be exponentially not exponentially big But big like powers of t infinity and also big like powers of t here And then the integral will never converge for any s at all so what we want to do is take this thing which originally converges somewhere and Extended to an analytic function, but here we have a function that doesn't converge anywhere There seems to be nothing to extend But it turns out there's a very wide class of functions for which you can make perfectly good sense of this and the idea Go that goes back to Riemann and his famous paper of 1859 on the zeta function. We're essentially that's exactly what he did So I'll explain that in a second before I did that so I mentioned that this is all I've a long longish paper. It's not a research paper. It's an appendix to the book I mentioned last time by Zeitler the founding director of the Max Planck Institute for mathematics in the physical sciences in Leipzig who wrote Who was planning wrote several volumes of a many volume introduction to quantum field theory for mathematicians and Asked me about this Casimir effect, which I mentioned and then in the end I wrote a 25-page appendix explaining various useful Techniques and all of the things I'm going to take today are in there So if you've read it you can fall asleep or just remember what you read Okay, so first a couple of trivial properties just so one gets used to the function, but these are all Completely obvious. I've saved my notes. I don't remember And so here we have 5t and Here at the melon transform phi tilde of s But now if I vary 5t in various kind of trivial ways like I could rescale in the t-axe I could consider five lambda t where lambda is a positive number then it's Completely trivial from the integral to see that it's phi tilde of s But all the properties I'm going to write down there aren't properties these transformations will also be true in the more General class where the integral phi is not necessarily given phi hat but these integral because the integral might not converge But these properties will all still be true Then another one is t to the Lambda, but here I don't even have to call it lambda. I think it's it can be complex I'll call it alpha if I multiply 5t by power of t then rather trivially in this I'm just shifting S by alpha so that's completely obvious. Of course. This is also obvious just by shifting t to the monover lambda t Similarly if I take phi of t to the lambda But now lambda should again be a positive number because I want the positive real axis to go to the positive real axis Then you get lambda inverse Phi tilde all of these are completely trivial to prove I'm just listing some obvious transformation so that you know if you know it for one function you know it for all related functions If they relate in this simple way, then a special case actually is 5 1 over t which when lambdas inverse is Sorry, I wrote this is s over lambda. There still has to be an s and here. This is simply phi tilde of minus lambda of minus s Now doesn't seem to quite make sense because of lambdas minus 1 there should be a minus sign If lambdas minus 1 should be minus 5 minus s So let's pretend it's a minus. I copy this from the paper from my notes and maybe There's a slight hiccup and then the last thing is if you differentiate 5 Of course, you can then differentiate as many times as you want. Just repeat this then You replace phi tilde of s by phi tilde s minus 1 But you also multiply by 1 minus s and again you you get that assuming the integral makes sense just by integration by parts so these are kind of boring properties, but just to show that Changing phi in some simple ways changes phi tilde in a simple way So now where does this thing converge? Well, let's assume first so Well, they actually be an initial situation actually is even better It's and this is very frequently what you encounter if the function is very small by very small I'll simply mean more than polynomial PK at both 0 and infinity so when t is small It's smaller than any positive power of t and when t is large It's smaller than any negative power of t well both case than any power of t but in particular those and in that case phi tilde of s is an entire function meaning that it's simply a Holographic function in the entire complex plane because since phi convert is smaller than any power of s then the integral is Rapidly and absolutely convergent near 0 and near infinity So we want to know what happens if it's not so that this this is actually this first situation and the second situation It's let's assume that the function is smooth at 0, but not extremely small then in this situation Sorry, I certainly didn't say I wrote here. I drew a picture. This is not comparably supported It goes to it. I said it's small small does not mean zero And so I'm small So no no this function is small I said it decays more rapidly than any power So one over t to the 10th for instance e to the minus t, but it certainly doesn't have to be zero beyond some point It's no of course. That's a special case here also this function I mean I this was meant to be a graph of function. That's very small, but it's not it's not zero No, it's just small so then the integral converges, so I didn't repeat your question the question was Does phi have to be or does one care if I is compactly supported? That's of course a special case of being small and then phi tilde would be entire But you don't need to compact support you only need small at both sides But let's assume that I'm in a much more frequent case when flies may be small often even exponentially small at infinity But it's not incredibly small it's there, but it's really smooth and remember I emphasize that last time because although everyone should know that sometimes People don't realize that it's the same thing. They've seen to say that phi is smooth that infinity means that it has a Taylor expansion which Matches the function to order zero to order one to order two to each order, so that means that phi is equivalent Asymptotic to all orders by the way, that's another confusion mathematics Sometimes people write f tilde g as the argument goes to let's say zero or infinity and only mean that the ratio Tends to one that's the weakest, but sometimes they mean in the stronger sense that the entire Asymptotic expansion is the same and I won't be careful so sometimes I've used double equipments, but it's too much trouble I obviously wouldn't write this file. They meant the weak weakest notion because I could just put a zero Why keep the smaller terms? So if I write an infinite sum, it's kind of implicit that it's meant to be Meaningful so remember so this is a zero plus a 1t plus a 2t squared and so on There's no requirement on these numbers a n that they have any growth I mean a n might grow like either the n factorial So this series may be completely divergent, but each a n individually exists and is simply of course by Taylor's formula the nth derivative of phi at Zero divided by n factorial and that might be big, but whatever it is. It's whatever it is So we're assuming that it's smooth, which means that it has an asymptotic expansion Meaning that the difference between phi and finite sum here is o of the first term that you've omitted for no matter where you truncate the sum and so then so the first thing is in this situation if it's smooth then in particular it's o of 1 because it's finite and Then you see immediately that phi tilde of s is holomorphic for real part of s bigger than zero Because this integral the only place the integral converges near zero or near infinity near infinity It's very small and near zero. This is o of one and so if s is strictly smaller than one So it's really smaller than sorry strictly bigger than Zero then I'm integrating the power which is bigger than the minus first power and I'm allowed to do that So this function will certainly be holomorphic Well the integral converges and it converges locally uniformly of blah blah blah So it's a holomorphic function in the real part In the half plane, but the real part of s is strictly positive So here I should put a dotted line and sometimes I'll write sigma because it's standard in number theory sigma for the real part of s But I'll try not to but if I use sigma, that's what I mean Okay, so now what can you do in this situation? Well the claim is in this situation when it's smaller than infinity If it's simply bounded in zero That's already enough to get that phi tilde is holomorphic in this half plane, but I'm assuming much more than that I'm assuming it has a full asymptotic development So now what can I do? Well, it's a fairly easy idea. We're worried about what happens near zero and near infinity So let me break up the integral into two pieces and I'll just go from zero to one and one to infinity But in other applications It's important to remember that I could go from zero to capital T and from capital T to infinity and have the freedom of changing T And in examples that'll be very important. So for the moment just since it plays no role yet I'll take the simplest choice and break up the interval the interval into two pieces Now I break up the second term. I'm not going to touch So all of this is when the real part of s is very big which in this case simply means Bigger than zero but I could even allow some negative powers of t then everything I'm saying is still true if s is originally big enough So this integral even if I've t blow up like t to the minus seven So even if I had some negative powers a minus one over t and son This would certainly converge and be holomorphic for real part of s large But what do I do for the first part? Well, I use that this an asymptotic sum So this is equal to this sum Zero to one and I so tracked the beginning of the asymptotic expansion So the first n terms I sum n from zero to n minus one and t to the n And then I have to add that part of course back in Okay, so But remember my s is originally very big. Well here. It's simply bigger than zero. That's good enough So n plus s minus one is bigger than minus one I can integrate that the integral of t to the n plus s minus one dt is t to the n plus s And if I so compute that at one I get one if I compute it at zero sorry t to the n plus s over n plus s and I'm evaluated that in zero and at one but at one t to the n plus s is just one and it's zero is zero Because n is greater than equal to zero and s is has is bigger than zero real part of s bigger than zero And therefore the this integral I could do each one Very easily just in my head. It's simply a n over n plus s And then the last term is I say I'm not going to touch Now this difference by definition is bounded I assume everyone knows the big o notation It means it's less than a constant which can depend on n So I should put o sub n But the cup but for all t as t goes to zero this difference is at most some fixed multiple of t to the n That's what asymptotic Expansions mean that if you take the first n terms the error is at most the order of but smaller than the last term here It's of the order of the next term So because of that this thing is holomorphic in a bigger half plane Namely if the real part of s sigma is bigger than minus n rather than bigger than zero as it was before It's now n might be hundred of subtract a hundred terms now if extended this from zero to minus a hundred So I have a much much bigger Half plane and this intervals now makes sense there because this is o of t to the n n plus s is positive And so the interval of t to the n plus s minus one near zero is finite This was holomorphic for all s and This one is not holomorphic Because it has poles, but it's it's defined for all complex numbers It's a meromorphic function and it's a simple pole at s equals s equals zero minus one minus two and so what we see is That phi tilde of s is the sum n from zero to n minus one a n over n plus s plus Meromorphic in the plane sigma real part of s bigger than minus n, but that's true for every n I just picked an n out of the sky. So if I let n go to infinity you see it's meromorphic and therefore phi tilde is Extends meromorphic the ridge the integral doesn't converge with the function So it extends. I'll just say it is meromorphic because of course the extension is then unique is Meromorphic in C with simple poles at S equals minus n where n is zero one two, etc And the residue is exactly the number a n which was the nth coefficient of the Taylor series So that's extremely simple It's certainly very well known and most of you've probably seen that but I want to emphasize that it's a very very simple argument And you don't need complex contour integrals that people sometimes take funny loops that go around zero You just break up the integral from zero to something and something to infinity And you get this very smoothly So I can first well, I can write that down the unfortunately this doesn't erase at all. Well And if I take the wet one then I Can't write again Maybe there's other chalk that Doesn't stick as much So what we have so far is in this situation if phi is Equal to some well asymptotic to the sum a and t to the n at zero and small very small at infinity Then I'll just put phi tilde is equal to the sum n from zero to infinity a n over n to the s Plus holomorphic of course this doesn't make any sense because a and might grow like n factorial that series might converge But again, I mean if you take any finite part of it That gives you fun in many poles and the further poles are more and more to the left So in other words this function as a meromorphic continuation From the blue box this one or this one this Well I'll try For this one this box There are many books. I'm not at all sure there's going to be a difference But just remind me I'll move the green one out of harm's way. Thank you Okay, so this was the first situation and by the way here we could be more general we could have a j T to the power lambda j it doesn't have to be a power series in integral parts of t I might have t to the square root of 2 t to the square root of 3 all I really need is that the lambda j So their real part actually are going to infinity and then it'll be the same Phi tilde in the same sense is equal to the sum with exactly the same proof S plus lambda j in other words the Mellon transform which originally is only defined in a half plane Subitioning to the right specifically to the right of minus lambda one if the lambs are increasing It will now extend meromorphic to the whole complex plane And it will have simple poles at the negatives of the lambda j's One can be even more general so I'm still taking the case when it's smaller than infinity, but Something at zero but in other case that some as arises you might have for instance a term I don't have to write the sum each time because it's additive if there's a term t to the lambda log t or More generally t to the log lambda log to the end of t then I will still get The pole at lambda Sorry s plus lambda that pole now will have order n plus one and the rest you won't be one or a a lambda if there was If there was an a lambda it will now be minus one to the n times n factorial and So, you know, it's it's so you get arbitrarily Mara more arbitrary Mara morphic function you can have poles of higher order and the easiest way to see that that's the only thing It can be just think of a single term t to the lambda and now differentiate in lambda the derivative of t to the lambda Which is e to the lambda log t is t to the lambda log t and the derivative one of rest plus lamb Just minus one of rest plus lambda squared and now you just keep doing it differentiate n types So these are slightly more general things, but they quite often occur. I mean one has functions that has Have powers that are half integral or square roots or something different anyone else has a series that are that have some log terms Okay, now you can of course have exactly the same in the opposite direction So you might have a function which is very small here But we should infinity it looks like some power. So here it looks like some B lambda Well, I can call it BJT to the J or I can call it Let me keep the notations straight It's I Don't remember if I put plus or minus. I think it's Well, first of all if it's simply oh Let me start again if the function is smaller than infinity But here it's oh of t to the minus a for some a that's including what I did then fi till this already holomorphic In the half plane sigma bigger than a Sigma bigger than a So you have to go if a is very big so you have a big pole You have to go more to the right and similarly just by inverting fi with one of the forms I wrote before if it's very small than infinity But it at zero I mean but at infinity it's like t to the minus b then fi till this holomorphic in Sigma less than B so in particular if you have a function that's reasonably small but not necessarily extremely small at both ends so you have it's over t to the minus a at one end over t to The minus b then the original integral will converge in a strip and that strip will be when the real part of this sigma Between a and b But a might be bigger than b then that strip is empty then the integral never converges And even if the integral does convert somewhere you want to extend and illiterate to both the left and the right But then you just do the same thing because of course in the situation when it's small than zero But has some let's say it has some expansion bj t to the beta j or something at At to at infinity then you do exactly the same as before you split up the integral and The only difference is that the part that you have to look at is integral in a half plane far to the left rather than to the right But anyway, you just replace t by t inverse and s by essentially minus s So it's very easy here to see that we get a function which is Holomorphic in both directions But if you have both the interesting case is when the function when a so If if I both values so it's like t to the a then I'll be fine if sigmas between a and b But first of all as I said that's not all we want We want the function to be extended the whole complex plane and on top to be a is often bigger than b and then this really the empty set So the original integral may never converge, but now of course the solution is more or less obvious So I'll introduce terminology that is normally not stand. I only thought of it while I was Right now while I was erasing. Let's call the function very nice Function is Just a sum maybe even a finite sum to start with a sum of terms You know t to the alpha and because what I just said we might even allow some powers of log t Maybe it's even a finite sum. So let's so those are things we feel we can handle and now Let me imagine that my function can be corrected by a nice function at infinity and by maybe another nice function in zero Such that it's small at both ends Now then this sum might not be finite like when I had my asymptotic expansion I could correct by finite piece of it and make it smaller and smaller if I want to make it Extremely small I had to take take the limit But if we imagine that it has such a behavior at each end then I can use this trick For the left integral to extend to to the left and for the right integral to extend to the right So now let me do it. It's a very simple idea and it's very easy to see. So now let's assume that phi is Nice at both ends So again, I want to make sure I have the Same notation that I have so let's assume that it's the sum a n t to the n as t goes to zero or Again, I could also put a n Doesn't really matter. I can even Put t to the power alpha j. I mean it doesn't have to be just integers But the alpha j's are going to infinity or the real part is going to infinity Now it doesn't matter if j starts at zero one It's just to meaning this index and let's assume that it has some other expansion bj t to the beta j as t goes to infinity We're now of course the beta j's or the real parts have to go to minus infinity because the terms have to be getting smaller and smaller So let's assume that Then what I do is first pretending that the integral does converge and let's start with the function where it converges I write phi tilde of s as the sum of phi less than or equal to t tilde of s Which is the integral from 0 to t? 5 t t to the s minus 1 dt Plus the integral phi tilde t greater than t of s which is of course the same integral from t to infinity And obviously if it converges somewhere like in that strip that we had when a happened to be smaller than b Then I can add up these two things now get the right answer and although both that these pieces depend on t There's some of course won't depend on t obviously because it's the whole integral But even if the whole integral never makes sense this one I can use exactly the argument I just used that the integral on the left if Phi of t looks like this then this thing is Meromorphic in the whole half in the whole complex plane in all of C with simple poles of Residue alpha aj at s equals minus alpha j by the same argument I just gave you and If you want to make it more gen with this no point filling up the board with notation if I determine t to the alpha times log t cubed Then this would become minus three factorial divided by s plus alpha to the fourth So I mean you can get higher-order poles if you want and similarly this one is Meromorphic in all of C I mean it starts originally this one converges in a plane way to the right and this Converts in a plane way to the left and those original planes might even be disjoint But each one separately converges on a half plane and each one separately if I make this assumption of asymptotic Expansions where the exponents don't have to be just one two three They could be any sequence of it of real numbers or complex numbers with real part going to infinity and minus infinity Then I will get that these two functions separately are each Meromorphic and so this one will have simple poles and Again, it wouldn't be simplified loud log terms simple poles, and they will be BJ I think now it's minus BJ over s plus beta j Except for a change of sign. It's the same formula So it is poles at s equals minus alpha j Where alpha j at the exponent since they're the other term is poles at s equals minus beta j and the residues are AJ and minus BJ respectively now Let's assume that maybe those those two half planes don't even intersect, but still these functions each make sense So now let me put here Meromorphic continuation and Here, let me put Meromorphic continuation And let me take that as the definition before that wasn't a definition It was a fact because the original integral convert somewhere and once it converges I can break it up the integral from zero to T the integral from T to infinity But now assume it doesn't converge anywhere. I can still this has a Meromorphic continuation to all s so does that so the sum of them is a Meromorphic function in the entire complex plane with only simple poles and the Simple poles are at the minus alphas and minus beta and the residues AJ minus the corresponding BJ so I mean it could be that some alphas is on both sides and then you have to subtract AJ minus BJ Now you might say yes, but this depends on T But of course it doesn't depend on T because if you look at the definition Phi less than or equal to T tilde in some half plane is defined simply by this integral when s is very large And simply this one is defined by the same integral With T going with s little t going from T to infinity So now I see that if I take Here's some T and here's some other T prime that's a it's bigger and I compare Phi less than or equal to T prime minus Phi less than or equal to T That will just be the integral from T to T prime of Phi of T T to the s minus 1 DT But if I look at Phi greater than or equal to T prime minus Phi greater than or equal to T That will be minus the same integral And this integral this now is compactly supported this integral makes sense And it's obviously a holomorphic function for all s in the complex plane. I'm integrating over a valid interval So therefore when I add this and this I get the same as when I add this and this I'm This thing just cancels so the sum of the two things Kind of obviously is independent of the choice of T But each one has been separately continued and so in this way we have the very useful fact This was all extremely easy, but it's not obvious if you haven't thought of it certainly before Riemann I don't think anyone had ever done this order didn't do it this way at all That this is a very nice way to take something given as a melon transform and make sense of it even when the integral is completely meaningless Okay, so now let's start looking How that looks in practice So I should have mentioned that the original melon transform when I look at the original melon transform Let's call it M for melon or Malin Phi to Phi tilde and this is let's say very small Or even compactly supported small at zero and infinity and this is holomorphic functions In the whole complex plane then this function is injective if the melon transfer of its function is Identically zero the fun it then the function with zero which means of two function at the same melon transfer and they're equal And there's actually an explicit formula due I think to Melon itself, which I'll write down I don't think I'll ever use it you pick any complex number C Any real number C and you integrate over the vertical line real part of Sigma C and This Phi tilde sorry Phi. This is Phi of T Phi of T it turns out that the melon transform this case will be very small on vertical lines And so when I take Phi tilde of S on a vertical line, it's Smaller than any power of T. So if I multiply here by T to the minus S rather T to the S minus 1 as before This integral converges makes sense and it's a standard theorem, but that gives you the original function. So it's in Because I'm sleepy Thank you the question for those who didn't hear is why I wrote DT instead of DS and the reason is for the mistake Thank you Okay, so I Don't care about that form the bits inject but the new one the generalized one is not at all now We have so this was the set of functions which are small if I look at functions Functions on T. Let's say reasonably smooth continuous, whatever you need on our plus with asymptotic expansions to all orders Just as powers of T or maybe powers of P times integral powers of log T Asymptotic to asymptotic series to all orders At both zero and infinity, which is the larger class then I've just defined a melon transform And it's it's well-defined was independent the choice of T and of course it extends the old one But now it goes into meromorphic functions on C and now it's not injective at all. So let's take example one Which is if T if Phi of T is simply T to the T to the alpha Just a pure power Well, obviously the original integral is the integral from zero to infinity T to the alpha plus S minus 1 DT This always diverges it never converges So this as a definition would useless But my due definition is absolutely fine at zero the asymptotic development of this thing to all orders It's just T to the alpha and so about what I told you get one of rest plus alpha But in infinity it's also T to the alpha all orders and so part of it I also told you get minus one of rest plus alpha and so that's zero and so in this case The it's not at all injective any such function as melon transform identically zero and Therefore any finite combination of such function of course also if I put T to the alpha times log T to the n where alpha is Is a complex number and n is a non-negative integer then as I said I would have here minus one to the n n factorial over S plus alpha to the n plus one minus exactly the same thing It just drops out Anyway, it kind of has to because if I've As I already said T to the alpha log T is just the derivative in alpha of T to the alpha So if it was identically zero the derivative there is still zero So all of these things go under the melon transform to zero This came up many years ago the paper of of mine I just talked about it to Pierre to linear who's a good friend and he said over this class function So this is what I called nice before kind of finite combinations of pure powers of T complex powers or complex powers of T times Non-negative integral powers of log T. It says so there's an intrinsic description that I'll just say it for fun These are the gm finite functions So gm is the fancy language for the multiplicative group which here since we're on the real line would be the multiplicative group of Positive real numbers and that acts on functions That's we had right at the beginning five lambda T But now you see if my five T. I shouldn't use the same T If I have T with T to the alpha then five lambda T is just a multiple of that It's lambda to the alpha T to the alpha so therefore it's an eigenfunction and that's true for any lambda So therefore the span under the whole multiplicative group is just multiples of that function. It's one-dimensional Similarly if I have T to the alpha log T And I replace T by lambda T I'll get lambda alpha times T to the alpha log T plus Now I probably can't do it. It'll be lambda to the alpha log And I guess times T to the alpha So therefore it's not an eigenfunction But this the two-dimensional space spanned by T to the alpha and T to the alpha log T That space is preserved by this and you get a little triangular matrix for the action and more generally if you have a function Which is a finite sum of powers of T complex powers of T times integral powers of log T then that function Spans if you take all the functions with the same powers alphabet and less than equal to the given then that's a fine That dimensional space and it's invariant under the action the multiplicative group So that's what's called if you have any group act on any vector space G acting on V The G finite vector is a vector whose orbit under G lies in a finite dimensional space It spans something finite dimensional. It doesn't matter at all for anything, but it's a nice comment that that's Kind of the intrinsic meaning so this new thing You're losing information in a sense, but it's very boring information because of course We want to understand 5 T and we certainly understand pure powers of T So I'm throwing away pure powers of T Which actually already means we could have done with a little less work because I could have taken the powers of T on one side So track them putting them on the other side And then I'd make one of my half planes bigger and one smaller and I didn't have to do are you on both sides separately Okay, so that's kind of a stupid example where the melon transform is zero Let's take one where it's a little less trivial so I Remember recall The definition given by Euler of the gamma function. He said he called it pi of s plus one So his pi interpolated n factorial and the modern gamma interpolates n minus one factorial at some point people shifted the Thing because it was more convenient the gamma function is defined At least for real part of s bigger than zero by this integral So it's simply the melon transform of e to the minus t and more generally if phi of t is a pure Exponential e to the minus lambda t with lambda positive then phi tilde of s is Simply the gamma function multiplied by lambda minus s that was well It's obvious, but it was one of the rules I wrote at the beginning when you replace t by lambda t So this is incredibly useful because when you have a Dirichlet series a Dirichlet series is an infinite sum not of terms of S to the n like a power series, but it terms something to the s So it's a sum of terms lambda to the s and what you always do with the Dirichlet series you multiply by gamma s And then you can write it as a melon transform of something and then you can apply these methods So here so here if we apply this Then you see that of course the expansion of this here is one minus lambda t plus lambda squared over two factorial t cubed and so on and So you immediately get that Gamma I didn't really need the lambda I could have taken lambda e1 you get the gamma of s Which is only defined by this integral a real part of s bigger than zero is Meromorphic in all of c and it is simple pulse at negative or non-positive integers simple poles and the pole at s plus n where n is 0 1 2 is Is minus 1 to the n factorial of course that's very easy to show anyway because an order of course knew it very well although he didn't have the notion of complex functions and poles, but of course The basic defining property of gamut that order was trying to get it was supposed to interpolate the factorial The factorials satisfies n factorial is n times n minus 1 factorial So this fact which is a trivial consequence of this integral by integration by parts if s is large And then using that you could just move to the left because if I know s Gamma for s bigger than zero then I can get it in the half plane as bigger than minus one I just divide by s then introduce a pole at zero the next time get a pole that minus one and so on so of course In this case the application is very easy and so we have this but now one can apply it as already said to any Diraclet series and this is what Riemann did first for the Riemann's data function, so let's do that case first If I take the Riemann's data function Which is really Euler as I said last time. It's the sum one over n to the s Then you see let me call it m because later. I have too many ends M to the minus s if you pull this to the other side up to a factor gamma of s. It's just the Euler transform of e to the minus Mt so therefore if I multiply the whole Riemann's data function, this is what Euler did Gamma of s times z of s then what I find is that this is simply the Mellon transform of the function e to the minus t plus e to the minus 2t plus e to the minus 3t I mean it's the same coefficients and more generally this would work for any L series if L of s is any our any Diraclet series 1 to infinity then gamma of s L of s is always Equal to phi tilde of t Assuming this should convert somewhere so the a ands are most of polynomial growth. Otherwise would never converge We wouldn't be talking about anything. This will always be the Mellon transform of Sorry phi tilde of s and the function is simply the sum of the corresponding exponential Some which is actually a power series, but not in t, but in e to the minus t So you convert your Diraclet series you use the same coefficient But you replace n to the minus s by x to the n where x is e to the minus t And then that function has a Mellon transform and you can do that But let's first do this case. So here since all of the coefficients here are simply one here The a ands may be something very deep let the number of points on a elliptic curve of Over some field or something like that. So the in number theory these a ands may be unknown But in the case of the remount zeta function a and is always one so the coefficients are just one one one one This is a geometric series and of course we all know how to solve it So it's just one over e to the t minus one But one over e to the t minus one that's the very definition I mean you can expand of course you take the power series of e to the t which starts with one But I've subtracted then t plus t squared over two plus t cubed over six and so on and Then I can just to work this out one over t minus one half Plus I think a 12th it might be minus a 12th and so on and by definition those coefficients Are called the Bernoulli numbers How should I do it? It's br over r factorial t to the r minus one So that's the definition of the Bernoulli numbers probably everyone knows the Bernoulli numbers I'll say a little more about them later because we'll need them often for now They're simply defined if you wish by this generating function you expand one over e to the t minus one It's called a pole that t equals zero of the coefficient of t to the r minus one divided by multiplied by our factorial It's called br so they were actually discovered independently by Bernoulli and Around the same time it's impossible to determine who was first because in both cases was published post-tubusly after both of them Had died so they certainly didn't borrow but the other was the great Japanese mathematician Seiki who was born who died only two years after Euler was born. So he was the generation before Discovered many of the things that were discovered in Europe and quite a few others So this is a sequence of rational numbers, which is very famous most number theory certainly me know them by heart up to say 15 actually half of them are easy Because all of the odd index ones are zero except b1 which is minus a half but all the others are zero But b2 is a sixth b4 is minus the 30th b6 is 1 over 42 and so on so they're whatever they are So here you have this expansion and therefore But if I just plug this expansion into the previous theorem, then I see that this thing This Phi tilde has poles Well first of all there's the term one over T So that's a pole with residue one at s equals one, but then more generally It'll be the next pole will come from minus a half That'll be minus a half over s and the next will be 112 over s plus one and so on so does poles Simple poles at s equals one zero minus one minus two etc. A few of them. There is no pole Just to be known number zero But now that's not our Riemann zeta function That's gamma times zeta and we already know that gamma also has poles and negative integers But not it not it won't so therefore at s equals one gamma fs is fine And it's just one this product has a pole and therefore zeta fs also has a pole so therefore zeta fs has a pole With rescue one a simple pole that s equals one and then you get something It's holomorphic Near s equals one but actually it's holomorphic everywhere because the only other poles you get here are Whatever it is once one 12th, maybe minus a 12th over s plus one You have simple poles when s is a negative integer, but gamma fs We just saw also as a simple pole and so when you divide a simple pole by simple pole You get something finite no no pole at all and therefore you get the theorem that I mentioned on Tuesday more generally You get that zeta fs is holomorphic everywhere and you get the value you get the zeta minus n is simply minus one to the n Times bn plus one Divided by n plus one factorial because this is this number should be n but times n factorial So it's simply n plus one So you get this for free and that's that second form is the one Euler had found 110 years before Riemann did it in this way, but actually Euler did essentially what I did Euler used this Interval he multiplied The zeta function by gamma used his own integral Representation of gamma and then studied the behavior of that function and made sense of it. Although he only looked at real numbers Okay, so this is for the Riemann's zeta function You see that in no way that I use that it was the Riemann's zeta function So I get the theorem maybe I don't have to write it again I wrote it on Tuesday if I have any L series. I'll write that part again The sum a j over lambda j to the s so a generalized Jerclet series Not necessarily doesn't matter what the index is called now Not necessarily just one to the s 2 to the s 3 to the s and so on and the denominator the lambda j's have to go to infinity Then so this has to converge, which means the AJ's shouldn't grow They should grow less than any power of lambda j which means that if I write the corresponding Power series in e to the minus t well, it's not a power series because these are fractional powers But the corresponding purely exponential series in e to the minus lambda j t that of course converges makes sense for all positive t Because if this converges then these are exponentially small they convert as much better But now if I have this then I find again that gamma of s times L of s is the Mellon transform Of phi just by doing it one term at a time We saw the Mellon transform e to the minus lambda t is lambda to the minus s times gamma of s and therefore we get Well now I'm sorry. This is true in general, but it's it's not beautiful. Let me take a true Jerclet series for this Sorry, but because I would like the poles in order to get the cancellation to make it beautiful I would like the poles to be at integers So this thing will have an expansion. I think I shouldn't have called these age I should have had some other name because I'm using a The expansion now we assume that this thing has an expansion as some topic expansion at zero Then that'll give me some a's and then by what we already know this has a pole at At s equals minus n with residue simple pole residue a n and plus s And then if I remember that gamma s also the simple pole, then I deduce That L of s is holomorphic in all s holomorphic and in all of C and the value of L of minus n is What I wrote the other time because you take this thing and divide by the rest the residue of the gamma function So that actually often happens you have many Derclet series which have an expansion at zero Which is just a power series, but sometimes like for the remunzator function You also have a pole term 1 over t and then we just saw what happens. You have an extra pole 1 over s plus S minus 1 and that one doesn't cancel everything and so in that case L of s still has a pole That's what happens for the remunzator function But it's not really important think that it's a special case once you've done it If you have just one term you could subtract the multiple of the remunzator function kill that term So the really interesting case is this and there are you know any number of examples when when you're in this situation so Maybe I'll say one last thing and then make a very brief break. It's more than three-quarters of an hour So let me give one By the different example, but it's much the same z divided as s alpha is the Horvitz zeta function It's a generalization of alphas one is the remunzator function. So this is the sum m from zero to infinity of One over m plus alpha to the s and Then you get exactly the same thing. So this of course the corresponding phi of t would be the sum e to the minus m plus alpha t And that of course is just e to the minus alpha t over 1 minus e to the minus t And this is the generating function of the so-called Bernoulli polynomials And so you get exactly the same thing this function will have a simple pole that s equals 1 because this is a pole That t equals 0 and the values that all negative integers are given by the same form as before You replace the Bernoulli number by Bernoulli polynomial The reason I mentioned this is because I want to ask the audience a question who knows Bernoulli polynomials I'm just even the definition You certainly do Many people are not answering I mean many people have seen but you've almost certainly seen the wrong definition So let me very briefly and then I'll enter the first part there and then Well, I'm a very brief post and I'll talk about the customer effect and the next thing I had in mind will be for next time So let me just remind you briefly. So the usual definition of Bernoulli polynomials is bn of x once you've defined the Bernoulli numbers and remember the Bernoulli numbers were defined Traditionally by brt to the r over r factorial is t over e to the t minus 1 I just wrote it a minute ago dividing both sides by t, but of course, it's the same. That's a completely bad definition Remove that from your mind. That's not how you should see Bernoulli numbers. Of course, it's true But it's completely artificial Why should you care about the expansion coefficient of t over e to the t minus 1 per se and this definition is even worse so the definition in every textbook is You just take n over r binomial coefficient times br times x to the n minus r So this starts x to the n minus n over 2 x to the n minus 1 plus and so on of the last coefficient is bn So the usual definition is you first define the Bernoulli numbers by completely Artificial generating function, then you'd find the Bernoulli polynomials by sort of a random It's not completely random but by some sum you can also do it in one step by combining that bn of x Br of x t to the r over r factorial Maybe it's r minus 1 again. I'll probably get it wrong because I can't do this in my head So maybe there's a sign wrong or something But essentially replace 1 over e to the t minus 1 by e to the x t and that just combines because I'm just multiplying This power series so I can put the t here again a multiplying this power series by e to the x t Which is the sum x to the k over k factorial and that gives the binomial coefficient. That's completely artificial The right definition is completely different the right definition of the Bernoulli number is The Bernoulli polynomial is much easier. So the good definition of Bernoulli numbers is bn is by definition The value of the nth Bernoulli polynomial at zero the Bernoulli polynomial is actually much simpler object and The Bernoulli polynomial is the unique polynomial that the unique function of reasonable growth Which has the following property that if you integrate bn of x dx over any interval of length 1 Well, you can see that this will be a polynomial because it's a it's I'm integrating a polynomial and that Polynomial bn starts with x to the n. So this is If a is very large x between and a plus one is always roughly a so the leading term will be a to the n That's for sure, but here. It's simply a to the n and There's only one function only one polynomial even one function of moderate growth Which satisfies that bn is the function is average when you average over an interval of length 1 gives you the pure powers So we have two canonical bases for the space of polynomials One is powers one x x squared x cubed one is one b1 of x b2 of x b3 and this map that given f of x Let's call it I It gives me the new function integral from x to x plus one f of t dt. That's a new polynomial Let's call that you know if Of x this thing is invertible because if it sends x to the n to x to the n plus lower order terms And we just changed one basis to the other so This is a much smoother definition and all of the standard properties by no functions Like the fact that the derivative of bn is n times bn minus one all of those follow trivially for that because the same thing is true On the right so it's actually that's it's a nice remark many people know it I'm not saying it's a secret But most people if you ask them what are bernoy polynomials would define it in terms of bernoy numbers and bernoy polynomials by the generating function But this is a direct definition and it's not even deductive. You don't have to know the previous bn's There's a unique polynomial whose integral is a to the seven and that is b seven of x Okay, so I'll make a maybe a three-minute pause just so I can erase the board and you can breathe and then I'll I want to give the Casimir example At least briefly you can also of course ask questions except I'm supposed to repeat them if you do Thanks, look how the this talk is much better It's a good hint Yeah, now I can more or less erase except the old things that I wrote before is still there. Well Another minute. Oh, I actually had one more example Here that I meant to give Maybe I'll skip it Well, maybe I should say very briefly the other example. I have the function e to the minus n t and And from one to infinity is error to infinity doesn't really matter here What if I took the function? I might as well now take n from minus infinity to infinity But I'd n squared t and it's convenient to put pi so this function is usually called theta Jacobi's theta function and Then one sees so how does this look it looks like one plus two e to the minus pi t? plus two e to the minus four pi t Plus two e to the minus nine pi t and so Now we can do something that Riemann couldn't do because he didn't he couldn't quote the theorem about this extended Melon transform and so he had to do it by hand but he did exactly what I did But now we know that this one doesn't bother me at all So theta tilde of t I can just do this and we know that the extended melon transfer of one Which is the pure power of t is simply zero and we also know that the other ones e to the minus lambda t gives me gamma of s times one over n squared t To the s hope I did that correctly, so that is to gamma of s Z of 2s Times t to the minus s so remember what Euler did and what we did before is If I just take z of s which is the sum e to the minus n t, of course I could put a pi and t if it made me happy then this was the Melon transform of a different phi which was one of her e to the t minus one So there I could give phi in closed form. It's a geometric series here I can't because the sum x to the n is a geometric series the sum x to the n squared isn't so in that sense It's worse on the other hand. I still got z of 2s here multiplied by t to the minus Something something is wrong with this formula Sorry, everything is wrong with this formula Nobody stopped me There's no t and it's theta tilde of s the same mistake I made before mixed up t and s the theta tilde Sends e to the minus lambda t to gamma of s over lambda to the s So it's here gamma of s over n squared to the s and when you sum that you get say it of 2s Sorry, oh, yeah that also. Thank you Pi n squared. Thank you. I seem to be even sleepier than I thought that I was and I actually always called this function zeta hat some people called it lambda But I can't now because I'm using had for Melon transform. So let me abbreviate that function. Let zeta star of s be defined as Pi to the minus s over 2 Gamma of s over 2 times zeta of s Then what the difference between what Euler did and what Riemann did is that Riemann wrote the zeta function Not as the Melon transform with a geometric series sum x to the n where x is e to the minus t But of a series x to the n squared, which is a theta series But now as a consequence, so how does this function look? Let's draw a picture at infinity. It's 1 plus o of e to the minus pi t So it's exactly in the situation that I was theta of t has an asymptotic expansion to all orders in powers of 1 It's 1 plus something exponentially small But by the press all summation form which I assume most of you know and I'll certainly give it later in the course sometime So I'll just write it here theta of 1 over t will be theta of T times the squared of t So therefore as t goes to 0 theta of 1 over t is 1 to all orders So here the function will be 1 over the squared of t plus o of e to the about more or less pi over t maybe there's a power of pi So it'll be exponentially close to t to the minus half on the left and to the right and therefore we know By by what I said that this function has poles simple poles of residue whatever it is 1 at s equals one half That's from this one The minus a half and it s equals 0 which then if I change s to 2s means that z to start of s It has simple poles if s is 0 minus 1 0 1 rather than 0 or half But we also have that theta t up to power of t is the same state of 1 of t So if you take the rules from the beginning of today's Talk if you replace t by 1 over t, there's a simple relation You essentially send s to 1 minus s when you replace t by you multiply by t to the half you shift by half So therefore this function is equal to its value if you shift by half and change s to minus s Which means that zeta star of s is equal to zeta star of 1 minus s This is the famous functional equation of the Riemann's zeta function But it should be called orders function equation the orders zeta function because in orders paper that I talked of last time He states this explicitly only for real values of s He proves it for all integers by computing at positive and negative integers, and then he says Now we can interpolate the factorials by my gamma function, which he had also invented So this formula is true in integers Let's look if it's true at non-interest, and then he computes s to be half and so zeta the half is divergent Sorry three halves zeta minus half is divergent zeta three halves is very slowly convergent But he can compute both the high accuracy with this Euler-McLaurin formula Which I'll talk about next time and he checks that this is true numerically to six decimals and then he conjectures it So he doesn't he didn't make wild conjectures, wouldn't it be nice if this were true? He said this is what you'd expect and I'm sure it's true I'm sure it's important somebody will prove it and it was proved by order a hundred and ten years later So this is the functional equation of Euler But I was actually asked the question yesterday and I've been asked many times when you take these sums that I was going to talk about today I'll get to next time Some f of n t where n goes from 1 to infinity Isn't that like the Poisson summation formula? The answer is no the Poisson summation formula Says that if you have any reasonably nice function say smooth and small than infinity then the sum of n is the same as the sum of f hat of n There are many variants, but the basic form where f hat is the Fourier transform But for that it's absolutely crucial that you're summing over whole lattice minus infinity infinity If you look at this of course this sum It's just the same as f of zero plus the sum f of n That's right this f of n f of n over positive numbers plus the sum f of minus n And so that that combination is easy, but the separate ones are very hard That's where you need to work much harder So this for instance you will not get from this the values at negative integers You'll just get the zero times the values at negative integer zero if you want that you have to use the other method I mean it's you you you you pay it's easier and it's much smoother But it only works for a smaller collection of functions, but I did want to have mentioned that so now let me sorry No, no, it doesn't let's discuss this later So okay, the question was yeah, but if you can't look at the poor problem So the the the question was I have some nice smooth function And what if I just take the sum f of n or f of n t well Why don't I just extend f on the left by zero, but the problem is that it's discontinuous And so f hat of two will be it will it be huge and infinity that some will diverge You cannot apply believe me it doesn't work I mean there's there's no cheap way and there's no way to guess the formula The formula is very subtle that you get with very new in numbers and they don't show up in the press all summation Everything cancels so what happens more generally when I do the thing with f of n if f happens to be an even function a Smooth even function then everything works beautifully because although you're summing from one to infinity Up to the term f of zero which we can forget you're summing over all n and everything works So that's why e to the minus n squared t works because it's an even function of n But e to the minus n t of n is negative it's exponentially big you can't do it and and you just you lose control completely Anyway, let me go on and not be too philosophical Because I wanted to talk I don't really want to run over with the cost mere effect thing till next time It's it's just a cute example. It is no importance for anything else But I gave it as an example last time it's I want to say it so let me define f of lambda lambda will be a positive number as The sum but for convenience in a second. I'll include a factor 2 pi just makes the form this come out nicer I sum over all triples of integers So if you wish lm n in z cubed and then I take l squared plus m squared Plus lambda squared n squared that's obviously horribly divergent the terms go to infinity and the question is to make sense of it First of all so that it becomes a well-defined function and then to have an algorithm that I can compute Let's say f of 3 to 100 digits So I mean something rapidly convergent and then once I've done that I can make you imagine making a graph the function How does it look as lambda goes to infinity and lambda goes to zero and with the physicists wanted it's one of the two I meant to look it up. I've forgotten that for the cost of mere effect You need the asymptotes in one of the two directions, but anyway, we're going to get them both So the claim is so the question is one makes sense So for all lambda or for any lambda make it diverse for every lambda even if lambda zero second the numerically evaluate rapidly, so here lambda's a fixed positive real number and Thirdly the the asymptotics as lambda goes to zero or to infinity So those are the three questions and so here is first the answer And then I'll just sketch very briefly what you do and you've seen most of it already And this is as I said actually everything I said today is in this paper, which is my appendix It's on my website, but it's also it's an appendix to a book It is a beautiful book by tidler his name I mentioned its volume one of his several volume Introduction to quantum field theory and so there's an appendix by by me and this example and more or less Everything I've said today is written there So the first statement is we're going to generate so I want to first make sense of this So I define z of lambda s so this was f of lambda I define it without the minus 2 pi as the sum l m and m But now I put a prime. I'm going to take a Dirichlet series But it's not an ordinary Dirichlet series unless lambda squared is an integer Because this thing is not an integer. It's whatever it is But it is a generalized Dirichlet series because remember a generalized Dirichlet series was one over any positive numbers to the power s That fixed positive numbers and only s is varying. So this is absolutely in the line of what I said the prime means So this means we're summing Some prime is the sum over z cubed minus the Point zero zero zero so where it would be infinite here It didn't matter because the square root of zero is zero so I could include that case Now you can see that formally if I take this thing with s is minus a half It's exactly what it's going to be and in fact that's going to be the answer So we're going to make sense of this for all s and then my f of lambda will simply be minus 2 pi z of Lambda minus a half, but that's not yet an answer because I haven't made sense of this when s this converges When real part of s is big of big enough and big enough here is bigger than three halves That's just a very quick estimate by counting the number of lattice points in some big ellipsoid Okay, so the claim is the claim is here one So I'm answering this question part one is z of lambda s has a meromorphic extension So this converges only in a half plane Meromorphic continuation to all s in C with a simple pole at Three halves where it had to have a pole because that's where it stops converging the rest do there at three halves is two pi over lambda and No other poles So it's even a little simpler than it was for the Riemann's a to function where we had two poles So there's a unique pole simple pole that s equals three halves and what's more if I define Z star of lambda s so I can't use hat because I'm using it for something else If I define this just I did for the Riemann's a to function with 2s So I put pi to the minus s gamma of s z of lambda s That's the definition this only makes sense for the moment of real part of s is bigger than three halves But if you believe that it's a meromorphic continuation, this is also meromorphic now It's got two poles the one it had at three halves and another one. You'll see in a second that's a It's zero and this satisfies the functional equation It's not the same lambda lambda goes to lambda inverse and here s goes to three halves minus s Yeah No, this is what I wrote I said this is like what we did for the Riemann's a to function, but there I'd say it of 2s and That was times pi to the minus s gamma of s That's a to 2s is the sum n squared to the minus s You see that here if I didn't have that and if lambda per zero this would be say it of 2s And that's the reason that it's one half so it's it is what I wrote is is what I meant Which is not always the case in particular today So it's really pi to the minus s gamma of s c of lambda s So I claim these two things it is a meromorphic continuation and then that at least makes sense of the function Because then f of lambda will formally as I said it's just minus 2 pi z of lambda minus a half But if I put s is minus a half This is pi to the half state of minus a half that when you work is not exactly the minus 2 pi that I included So then I simply define this to be z hat of Lambda minus a half which is by the functional equation Essentially the same one. I'm not going to do it in my head lamb to the something Up to a constant some stupid power. It's gonna be one over lambda and Three halves minus s which is two So I guess the power is lambda the one half or minus half can't do it in my head So you see that at s equals to this thing converges, but it's miserable converges It's only barely beyond the limit of it stops converging at three halves to it converters very slowly But it does converge the lamb has become one over lambda So now bits of to find the function is some very simple multiple of the new function and Now it's n squared over lambda squared But now to the to the power two rather than to the power, you know one half in numerator So these doubts convergence of if you believe me on this then I've done the first step I've made sense of it But it's now been replaced by convergence series, but extremely solely convergent So what did we actually do well, of course, you now know what we do because it's just a special case of what we've been doing We see That I have to make the five t here Will be the sum l m and n and I should take m different from zero, but I don't have to Maybe I should put a pi to the minus s so maybe I'll put a pi here just to make myself happier Here I sum over all l m and n now one of those terms is when they're all zero That was forbidden here because you'd have shared minus s But it's certainly okay here I just have a term one and we know that under the melon transform as I've generalized it that term just goes away So five tilted s is the sum of the corresponding things for s isn't zero So therefore it's pi to the minus s gam of s times what I called z of lamb s And that's exactly what I just called z star of lamb s So in other words, this thing is already a melon transform of a very very simple function But now if you look at this you see that this just splits as the fate of t that we already had that was e to the minus Pi l squared t again the same thing and here theta of lambda squared t and now you can see I could actually stop It's all over but they're shouting because we know that theta is one to all orders of infinity this theta also So this is one to all orders of infinity of course Thank you for nothing one term is one and all the other terms are Exponentially small so e to the minus something positive times t so to infinity this functions like one, but it's zero Remember theta of t was to all orders one over squared of t So this is one over squared of t This is one over squared of t again here one over lambda squared of t So it looks like one over lambda t to the three outs to all orders as that gives me a simple pole Which has residue one over lambda maybe two pi over lambda if you keep track of the constants and there are no other poles Because that's how it is But now if you remember how I define this thing It's the sum phi less than or equal t plus phi greater than or equal t and each of that is the integral from let's say t to infinity of Each the minus whatever it is which happens here to be l squared, but you could do this for any Dirichlet series You take what's called the partial l function. Let me just put here any number x t This is the function that we know if t were zero would just be gamut x x this this is it is many notations Let's call the gamut s of x. It's a partial gamma function. It can be computed extremely rapidly and It's and of course, it's exponentially small. I mean up to power of x. It's just e to the minus x t So this is a function which of x is large is extremely small whatever t is And so what you get just by breaking up the integral the way that we already saw is that this say the star of lambda t Which I just defined as this melon transform is equal on the nose and it's a three line calculation One line calculation It's equal on the nose to the sum of two terms. The first one is one over the square root of t times the sum over again all integers l m and n and Gamma minus a half of pi times t. I might get the pies wrong But it doesn't really matter and then the same number we had before So this looks complicated, but remember that gamma minus a half of x is exponentially small So if t is something reasonable, you know two or something then only a handful of terms will contribute to this because this sum of squares After a few terms will be bigger than 50 and you'll have something in an exponential of minus a hundred So this is extremely rapidly convergent. So this an exponentially convert exponentially rapidly Convergent sum as opposed what we had before exponentially convergent now But I'm not finished because this is exactly the formula for one of the two integrals I had But I to do the other one, but the other one after a simple change of variables is exactly the same story Except that the s has gone to n I could do it for any s, but so I this should be for s Here the way I'm doing it now. It's simply f f of lambda So I'm already specialized you can do it for every s and then you'd have a t to the s and t to the three halves minus s I'm not bothering so here. It's exactly the same thing But now it's the other one be in general gamma s and gamma three halves minus s now The t has become one over t because I inverted and the quadratic form is the same Except that the lambda squared has become lambda to the minus two But whatever lambda is or whatever t is this is again, you know Exponential more than exponential its Gaussian. It's e to the minus a square Well, not really because when you have three squares, there are lots of sums of three squared to the box It's but it's more or less. It's exponentially rapidly convergent And so this second sum is also exponentially rapidly convergent and it's a sum of only two terms It's not an infinite sum. There are two contributions t big and t's to the right of t into the left of t and Both converge extremely fast as I said in the paper. I give for a small list of you know Five or six values of lambda. I get the 50 value 50 digits value of lambda takes a second in Paris But what's really beautiful in the method? That's what I emphasize before remember we defined our integral originally by splitting up into two pieces But then if you change the splitting point you change both the integrals by the same thing it cancelled And so it has to be independent of t. So you compute this numerically when t is one two You know minus minus 1.5 and each time you get a 50-digit number It's a sum of completely different terms, but the sums are equal to 50 digits So you normally get the value you get a certificate that you haven't made a mistake in your calculation or in your program When you're thinking so that's extremely nice And there are many many situations that work like that You have a free parameter and the the final thing has to be independent the parameter And that's wonderful because there's no way in numerical calculations to know if you've made a mistake You get a 50-digit number nobody knows if they're the right 50 digits It's crucial in all numerical calculations to have a way to check like a functional equation or or in this case That it depends on a parameter and it shouldn't depend so now once you have this I'll just write the last part because Well, I'm not not quite running out of time. I actually have Two-thirds, but I can even write here what gamma minus a half and gamma is gamma minus a half of x If it's just the special case of what I wrote it's e to the minus xt dt over t to the three halves Which is essentially what's called in tables and it's pre-programmed in many packages like in Paris It's called the complementary error function. So this function is just a function You can call just like you would call sine of x. It's it's a standard function. It's already on your computer And the other one gamma two of x at least of x is not zero It's simply one over x plus one over x squared e to the minus x So that's really a standard function. And so, you know, this is completely computable It's a it's a very very rapidly convergent sum of extremely simple and easily computable functions So it's absolutely computable to arbitrary precision. So that answers the second question and Finally the third question is I'll just give the answer But you can you can try it as an exercise if you do this as an exercise just using this Description theta t squared theta lambda squared t you have to think a little how it works Then remember I told you that this state of t squared at infinity is just one term one And so is that and similarly at minus at zero. It's just one term So we know very precise how this looks at both ends But then we've thrown away one term because the L and n equals zero term We're secretly removed and in our melon transfer I've remembered that I moved the term one over and so it gives me two more terms I end up with four terms one coming from the behavior infinity one behavior minus infinity and two are the one The L and n equals zero term that I emitted but here I'm including in both terms So I have to include gamma minus a half of zero on gamma two of zero after some renormalization And so when you do that what you find is very pretty That the asymptotic expansion is actually much simpler than the exact expansion Namely as lambda goes to zero There are only two terms so f of lambda is equal to 2c over 3c is about 0.91596554 But I'll say what it is exactly in a second 2c over 3 times lambda inverse So there's a pole term then there's a linear term pi over 3 lambda and then plus O of It's to all these terms. It's actually lambda e to the minus pi over lambda. Well, I can drop the lambda because it's small It's exponentially small so in this case It's very explicit and as lambda goes to infinity then there are two other terms completely different from these two one of them is pi squared over 45 and It's lambda cubed and the other is another constant C prime which is 1.4 3 7 7 4554 dot dot dot and So it's C primed here. There's zero power and here. Well, there is actually one of her lambda But it helps a little but here this is squared of lambda. It hurts a little so To be honest I have to include it, but it's still exponentially small so in other words in both directions and Then in the paper I'd already given the table with the exact values for lambda is one tenth one half One two and ten and if I took the two values each to 50 digits And then I took this thing for lamb's a tenth and this for lamb is ten then like 40 of the 50 digits for the same And you see it's really incredibly close. So it's not just an asymptotic expansion It's an almost exact because it terminates to all orders I could have just dropped this and said the asymptotic expansion to all orders is just two terms at infinity and two other terms at zero And what's needed for the actual customer effect in electro Dynamics is I think it's this one with the 45. I've actually forgotten. I'm sorry And I haven't actually told you but C is so I can end with that Let L4 it's actually L minus 4 of s be I'll write it in a fancy way that number theorists like It's the chronicle symbol minus 4 over n n to the minus s So that's a derelict series, but the chronicle simply simply one if n is one mod 4 Minus one if n is 3 mod 4 and 0 of n is odd So this just the alternating sum of 1 over n to the s for odd numbers And so Leibniz already had found that L minus 4 of 1 is pi over 4 and Just as Euler showed that the Riemann zeta function his zeta function at even integers was a rational multiple of the Powers of pi simply this one at odd numbers. It's pi cubed. I didn't look it up or work it out. It's pi cubed times something Rational Actually, it's not it happens to be rational It's really rational number times the square root of 4, but the square root of 4 happens to be rational If I'd minus 5 it would be like that So all of these numbers are known and the C And also there's a functional equation, but for the same argument and so this is simply L minus 4 of 2 It's called Catalan's constant. It's maybe the fourth most famous constant in mathematics after E and pi and gamma It's not the fourth. I don't know the order Nobody's ever I think fix it and this function rather surprisingly is also expressed in closed form in terms of the very Same form maybe not surprisingly given what we did except now It's the value of three halves Which you can also compute as many digits as you want in many different ways So we get we have an explicit formula for the expansion coefficients So this is a very nice exam because you start with something that's looks ridiculous There's some square root of L squared plus M squared plus. Well, let's just take mp1 L squared plus M squared plus n squared it looks ridiculous But you can make sense, but you can compute it to arbitrary precision You can compute its asymptotics is that coefficient changes and it's actually important for an application This was really needed in in classical physics Okay, so it's exactly 530. It's very unusual for me to end on time But so if you have questions Please do except I have to remember to repeat them The question can also be about last time or about anything else. Yeah No, I don't think so wait the first let me repeat the question for people are listening virtually The question is I gave a recipe given some function phi which had reasonably nice behavior at zero and infinity I defined a phi tilde of s and the requirements were that if I was very small in zero and infinity My new function had to be the original melon transform given by an integral So that's unique and even injective and of course it has to be linear I can't do something different for every function It has to be a linear procedure linear functional and the question was I gave a procedure that the science defy Meromorphic function phi tilde of s. What if I thought of something else? Couldn't there be other answers and the answer is basically I think no So what I showed is already what I did there was a free parameter There was a big amount of freedom, but if you did it with different values of the parameter it didn't affect anything and That almost tells you that it's unique because whatever else you do should also work for those two pieces separately But those two pieces are Meromorphic and a half plane or holomorphic in fact and half plane So any other function that agrees with them in the half plane is and it's also meromorphic is the same function by uniqueness And the sum is the same so based I think it's Inconceivable that there's any other way of course I can't claim that if somebody else did something completely different But it seems to be impossible that you could find certainly any natural procedure that would Give you a melon transform for every function in this class Which for the funds where it's convergent with the original one, but which wouldn't be this one It seems to be completely impossible and here You saw that it was and we had something to depend on this parameter t. So this formula is not a formula It's a formula for a function of lambda and s and t But in fact, it doesn't depend on t and in ever will but that's not really a proof that there can't be anything else But I'm sure there isn't So if you're asking from my opinion, my opinion is I'm sure there isn't and I'm not sure if It's a little hard to prove that there's no other Nothing one can't think of anything else But it is really very economical and also very simple once you've got used to it another answer It's of course of Riemann did it then it must be the right thing to do because Riemann was Riemann So he wouldn't have done if it wasn't the right thing woman then And no one virtual well people aren't shy They would ask questions if they had them so then we stop and we continue on Tuesday By the way for people here in the ICT P. I'm sure you've all seen the posters There's a big event in the ICT P on Monday, which is the Ramanujan day I think we've never had a Ramanujan day before for it's quite a few years The ICT P gives a yearly Ramanujan price to young brilliant mathematician It's something that somehow reminds one of Ramanujan But now they've decided to have a yearly Ramanujan day and since I now have a position here and my position is called the Ramanujan International chair I'm supposed to give kind of an inaugural lecture, but it's supposed to be very popular because there will be Hope some non mathematicians in the group So there'll be a fun lecture and I think it's streamed or the one kids here on zooms also people aren't here It's called Ramanujan and the partition function, but it's very low key I mean if you think this is low key wait for you to see that okay, so That's to see you all on Tuesday or on Monday