 power partitions and in a sense the most fun because it's really about the circle method and To my great surprise when I did these calculations. This is only four or five months ago. It's quite recent Many many things were different from the classical case I assumed once you had the basic asymptotics everything would be essentially the same but it's not at all as you'll see today So that's quite fun and then the last topic of next time Will be about this very slowly convergent sum that I mentioned already in the Posters announced in the course and then I couldn't find my notes and didn't know where they were And I still never found them, but at least I found lots of notes on the computer Including some things had even written in tech So I will be able to tell that story So next time will be completely independent at this time and there won't be any further time. So then You're off the hook So for today I want to tell the the rest of the story of these power partitions. So remember the problem was So I'm talking about power partitions So s power especially s equals 2 I'll talk about most of the time because it's the most Interesting and then I can be you know can give actual numerical things So remember that p s of n for instance p2 of n is the number of representations of decompositions, let's say Unordered as a as an unordered sum of s powers e g of squares and the generating function of that So the sum p s of n q to the n where p s of 0 is by definition 1 Has again more or less by definition of this is what Euler found in the case S equals 1 that has the expansion 1 minus q to the m to the s and so therefore by the by Cauchy's formula we know That little p s of n is 1 over 2 pi i times the integral over any path in circling 0 inside the unit disk so for instance the circle of radius r less than 1 and then p2 of q e q divided by q to the n plus 1 So the circle method consists in looking at the singularities of p2 of q and that's what we did last time Well, the first time numerically and last time proving things and we found that p2 of q has a very well defined It blows up or sometimes goes to zero It can be both Unlike the case s equals 1 that sometimes very big sound It's very small in the neighborhood of each root of unity But it has a well-defined Asymptotics and more than asymptotic even in exact form that at every root of unity and so therefore this will be equal to the sum well, it'll be dominated by the sum over kappa in q mod z of a contribution, let's say p s comma kappa of n So this is the contribution but this isn't very precise and I'll say more about that contribution from the part of the integral here and Then it's a root of unity But that root of unity will always write as e your members e to the 2 pi i of some rational number kappa So the biggest contribution will always be if cap is 0 and zeta is 1 But the loss be contributes when cap is a half say it is minus one cap is a quarter zeta's i and so on So the that's the basic circle method and it's been known ever since the famous paper of Hardy Ramanurj on 1918 and many other later famous papers Hardy Littlewood Radomacher, you know Gradoff and any other people used it for a variety of problems so I'm going to come straight as I said on s equals 2 I'll say if you one part I'll give the form is for all s but in general for s equals 2 so I want to emphasize the differences and In fact, so they'll I'll have five numbered points and in fact that will be the whole lecture to the case a s equals 1 So the first one is ps of n Grows much more slowly Which is good and bad it means in particular that if you want to have big numbers and see the estimates You have to take a much bigger n which I'll come to in a second So roughly we know from Hardy and well, it's actually quite elementary and certainly Ramanurj I knew that already in England that for usual partitions p1 of n the log I won't put the exact constant although of course it's known it goes like the square root of n But if you take p2 of n I gave the formula in the first lecture It's of the order of the cube root of n and in general for any s It's of the order of the s plus first root of n So therefore, you know, it's it still goes to infinity, but it's exponential only in a fractional power of of n so that certainly has some effect then instead of mr. McMahon, but He's not just mr. Of course major McMahon who is the famous counterlater that Hardy knew and that Hardy and Ramanurj and talked a lot with and in the in the movie by Ken owner and others Which you should definitely see if you ever get the chance He plays a not very flattering role, but anyway, he's certainly occurs in the movie. He was quite Very old-fashioned prejudiced Certainly not happy about an Indian getting involved in any of these things But instead of in instead of major McMahon who was a brilliant calculator, but of course by hand we have electronic computers So therefore even though we have to take a much bigger end to get interesting numbers because of this slower growth It's not a problem at all. So Hardy and Ramanujan Studied in particular the famous number that they analyzed and I'll give the details in a second Studied the number which they considered huge because Using a very clever form of of Hardy of Euler major McKay and was able to complete Compute the number of partitions of P is just P1 of the number 200 and they were very, you know Impressed that won't you go that far and so that number is three nine seven two Nine nine nine for the Ramanujan dad learned this number by heart, but I've already forgotten So it was this 13 digit number Which obviously if you try to count all those partitions by just writing out one plus one plus one two hundred times and one plus one 198 times plus two you wouldn't ever get there, but there's a very nice recursion Due to Euler that is only the square of n terms So you can compute P of 200 with just of the order of 20 addition since attractions But only if you have all the previous piece You have to make a table up to 200 and of course you have to make no mistakes Otherwise you it won't work. So this was quite a tricky computation, but indeed major McMain did twice, but we can do and This took probably many certainly many hours from major McMain to do this by hand and that how start yet We can compute Qp2 of n Not just for one value Well, they all said to go up to 200 I said it's recursive, but not a not just up to 200 but up to 10 to the 5 in two seconds in less than two seconds I'll actually write down the Paris program in a second so that those of you who aren't used to Paris can see once again That it's pretty easy to To do well I can do it immediately. So you start with the power series one, but you have to Decide on the precision because it's like a real number on the computer You have to say how many digits you're giving so I start with one plus o of x to the one hundred thousand and one So if I do things to that the first the code is up to x to the hundred thousand will be correct And then having to find that you put a semicolon and now you take exactly this Expansion here I'm going to take y originally to be the denominator because it's faster to multiply than divide Well, actually I could divide and then the probe now I'll multiply because it's slightly faster. So I multiply Y Star equals in many lengths like see it means that you multiply y by what's on the right hand So you replace y but the old previous value of y which is originally one times the right-hand side and here You have x to the power X would be Q So you see this is exactly this product one minus q to the m to the s where s is now two and that's the whole program And then at the very end Since I want the reciprocal you Replace y by one over y don't really need the last semicolon And then that new y will be the power series p2 of q and you can read off all the code So this takes as I said two seconds so what you find is the so the the coefficient So in Paris language that would be the hundred thousand coefficient of this is called Paul coefficient because it's not a polynomial It's a power series, but it's the same Command so the hundred thousand coefficient. I won't write out the whole number because it is 60 digits. I'll give you the first four The first six and the last six But as I say not just get this one number to get all hundred thousand numbers takes two seconds So here they're 48 digits enough emitted So we have plenty of numerical evidence And if we have a theory just as they use this number very much to test whether their asymptotic formula was working And we can do the same, you know with with much bigger numbers So that certainly changes the nature of the game and it's important when you're thinking about asymptotics That the theory of asymptotics now is different from let's say when Hardy wrote his book or when Euler yet much earlier We're not the Euler-McClurrin summation form They were thinking of asymptotics as something that if you work very hard you could get 20 digits by working for a whole day But we think in terms of you want to be able to get hundreds of digits Maybe or 50 hundred several hundred in maybe split seconds So the fact that there are computers has certainly changed the way we think about large numbers and about asymptotics It's not just Theory anymore. It really is very practical Okay, now I'll come to the theoretical part and first the sort of least interest would be point three Four will be more interesting and different from the case for s equals one and five will be the most different in the last But already three there are several differences so Let me explain a little how this works. So as I already said each The kappa in q mod z or equivalently z to which is e to the 2 pi i kappa in the You know root of unity, which is usually bit mu infinity the union of all mu n But each kappa gives a contribution That I'll call generically P s comma kappa as already did and already wrote but this is So p as cap of n is the Contribution to the part of the integral near n. So let me write again what I wrote for ps of n, but now let me write It so instead of taking the original integral over a circle q of a certain radius I'll take it in the upper half plane. So tau is often more useful remember q is Is e of e to the 2 pi tau I'll go from some tells her which will typically be very low To get the asymptotic spin principle It's true for any tells her if you simply integrate from cell zero to tell zero plus one Then you'll take this function p 2 of q which is now p 2 of e of tau Times e of minus n tau Detail so this is an exact equation For any tells there on the upper half plane, but then So then p 2 of kappa p s comma kappa is the integral near kappa So here is kappa and kappa is a rational number. So let's say it's three-fifths We imagine a small interval around it and then you take some kind of a peak up to a certain height and will choose I mean you remember you can take your path of integration do anything you want You're gonna have peaks of various sizes as the various kappas. So near kappa will take the part of the integral and All that really will matter is the part very near kappa So it doesn't terribly matter and that will play a role in a second what you do So let's say that n is hundred thousandths before and cap is three-fifths then once you go up to three-fifths of point 602 it won't really matter what you do here because you're far away from the singularities much smaller So this is kind of a local computation and as a result the Contribution is not completely well defined. I'll write that down in a second You have a little bit of choice how you split it up now if you want an exact formula Well, then you can't get it already hard in Ramanujan couldn't get it What they did is they split up the integral by the method of minor and major minor arcs So roughly they had a rule that if you take all the kappa with the nominator up to something like up to 20 Then they had a rule how they would split up the whole arc They would take each of the kappas with those phenomena sets of finite set and associate to each one of arcs of a certain length and then take something and Just cut it off and roughly estimate the the arrow when they got to the end of the arc and then the sum of all those Errors it was a very tricky analysis. I certainly wouldn't want to embark on it I certainly haven't done it for s equals two nor do I know what you would get But they were able to show that the sum of all the errors they made by cutting off each thing The sum of all of those if they took all the nominators up to age Went to zero is H went to infinity So they actually got an exact formula in the limit which they didn't expect But at the very least they could expect that they would get the leading term to all orders But much more than that that the sub leading terms coming from other kappas would be not sub leading But the other contributions of other kappa would be important So here I'm not going to go into that. It's going to be a purely formal argument I'll just describe As some thought of it all the terms coming from the neighborhood of the given kappa So roughly this is what we're doing and so we have to remember what the formula was for ps near the point e of kappa and I could just put e to the minus x But in the formless it was convenient to call the variable Minus one to the t to the s where t is going to infinity Let's say through positive real numbers, but remember we actually did it you could go at an angle like it, you know argument one five degrees or something and then you also got a perfectly good Thing so the formula that I gave last time well, I gave the first time Experimentally, I mean we're doing a lot of calculations for for instance for s equals two and Kappa with the nominator five and then last time I explained how you can actually prove things I used this I gave the following formula for ps of e of k. It's a constant Which is not terribly important, but I'll write down what it is. I mean it is important Of course, you do the calculations So if this count constant is three or if it's seven then the corresponding calculation Contribution will be seven thirds as big So of course when you add up the contributes you have to know the constant But as far as the orders of magnitude it's irrelevant And but it was was the product of a root of unity which was e to the two pi i times generalized Etiquin sum but actually I'm going to construct s equals two and that general Detiquin sum with zero whenever s was even because of some parity So you can actually ignore that but in general in the case of hard in rum nuchand for s is one and also S or three or five there would be this root of unity and then there's still a factor which is n s of the denominator of kappa So for instance if s is one this is some of the dominant kappa divided by two pi and The whole thing to the s over two, but I'm just putting that for completeness. I gave it last time It's it's not important except that you have to know that there that there is a constant that we know what it is And then here it's e to the minus One over t to the s over two so that'd be e to the minus the square root of x excuse me I think it's e to the minus T to the s over two it's simply the square root of x if I called that number x and then the important part it Grows exponentially in t and then there's a very unimportant part But I have to write it or it wouldn't be correct say it of minus s times again t to the It's very confused right one over t to the s and write t to the minus s This with the same t to the minus s but t is very big So this is small and the only effect of this Z to any way it won't have any effect when s is two because then z of minus two is zero when s was one This was minus a 24th and when you put it to the other side and put it into the integral the effect It has already down the second is just a shift n to n minus a 24th So there's always a shift at n where you shift n by a half z of minus s that said for s equals to that's not important but this one is very important this is important and And therefore you see her that if CS of cap is big this thing is exponentially big But a big exponential if CS is positive but small it's still exponentially big but much smaller And so the order in which the various terms will contribute will depend on the CS of K and CS of K There's a formula CS of cap. I gave a CS of cap Excuse me. I said it last time is one over C times the sum of all conference the residue classes l multiple of C of Li to Li I better copy the formula because something happened My notes and I can't Read it So I'll copy it from in the official Source It's in my notes from last time Yes, I gave it then sorry if this is When I started to write down was just for C equals one CS of Kappa There's a factor gamma of one plus one of rest which rest equals one is just one divided by C and then the sum l multiple of C and then the poly logarithm function with index one plus one over s of E to the two pi I and then L to the s times Kappa So if s is one Then well gamma of two is just one so it's one over C Times the sum l multiple of C Li now is just integer Argument Li two of E, but now it's just L times Kappa And now it doesn't matter what Kappa is because Kappa's to nom day C is almost the denominator of Kappa Kappa is some a over C where a and CR co-prime So here it doesn't matter what Kappa is because it's some a over Kappa a prime to Kappa's L runs over all residue class multiple C then a l also runs over all residue classes as you just get the sum over all L multiple of C So this is the same as the sum Over all Li two, but it ended by a trivial argument It's Li two of one over C squared so this is Z of two pi squared over six divided by C squared So in the case when s is one this thing only depends on C in an extremely simple way But in the case when C is not one then that isn't true at all And that's what's going to create all the fun So and I already mentioned that on the first lecture on the second But I gave the special cases if I take for instance a fifth or minus a fifth in that case It's the same it's about zero point four four zero, but if I take the other still with s equals to the other two Points rational numbers with denominator five. That's plus or minus two over fifths Then this is minus zero point zero two six So it's a more than ten times smaller, but it does the opposite side Which means that at these two points p true will actually be extremely small and will contribute nothing at all to the Final asymptotics, so it's it's it's very erratic risk in the case In the hard to rum remanage in case each C contributed I mean sorry each point with the nominator C contributed something of the same order of magnitude as all of the others and this numerical comp thing with the this for instance this only depends on the nominator This depends on DS, but it's a root of unity So the orders of magnitude brought the same of course when you added them up then you know things could happen So that's the reason already that that things will be a bit different But actually I'm jumping ahead of myself because I want to tell how this integral looks So remember the integral that we're looking at this PS of K is Going to be the integral the same integral that is written here But I only take the part of the integral near some kappa, so I'm gonna have some little part of the path near kappa and So let me go back to the left of the board and erase and start again So what would you actually get is that? So if we put this maybe this part I should put here right the other part That formed for CS little CS doesn't matter So if I insert this formally into that then I find that ps kappa of N so the contribution to the number of parties of n and s powers coming for the rational point C is Roughly that with this this silly constant, which is you know known, but just do some small number some simple number Then there's the 1 over 2 pi coming from the coach the integral this an s because of a change of variables It should really be dq But the q has a t to the s because of this change of variables so don't worry about such details and then remember I told you that I have to make a shift and Is n minus sorry plus a half? Z of minus s so in the Hardy-Ramanujan case z s was one z of minus one is minus the 12th And it was n minus the 24th and anybody's ever seen the famous Hardy-Ramanujan form It's the sum of terms and each term has square roots of n minus the 24th There's the shift for s equals to we don't have to worry about it So if s is 2 then as already said Those 8 of minus 2 is 0 so n tilde is just n so you can forget the tilde and then Independent of s it's to the minus 3 house and then the important part is a transcendental function that does depend on s and Then it will come there'll be this constant Which is that's remembered the important one is CS of k and then after you make a change of variables It'll be n tilde to the power minus 1 over s So this is a constant depending on your rational number and as n goes to infinity Until it also goes to infinity. It's just equal to n plus a constant then you have smaller and smaller arguments that were Interested in this Wait a second Something has gone wrong Sorry, it's not divided by its times There's been a change of variable Change of sign the way that I'd find hs. It's CS of kappa times n to the 1 over s So n is very big and so that's going to be the final formula. Okay, so now I have to but what is h? So if you look at that integral Then I can tell you what h is but it's not well defined. You have a choice so You see if you look at this then remember we were integrating near kappa along some path and so If I take a slightly different path just from that one contribution do forming a path doesn't change anything in it in a Cauchy integral, it's only a homotopy class It's the only question is what you do away from kappa. Do you go all the way to i infinity? Do you start going up you can occur your moustaches up a bit or do you go down or do you circle zero? So, you know, how do I close it up? Do I do something like this or do I do something like maybe this or? Or do I do something like? Maybe close it up all the way to a circle or maybe I close it up like this so there are various contours you could put and The original function you can't do in the lower half plane because this product which I've erased product one over one My security and diverges, but of course the contribution near tau It's a purely exponential function e to the CSF k this that doesn't have any problem crossing any lines so I can play with this contour and I did that it was several hours of work and I haven't even written the details in the In my write-up, which I'll make available very soon if anybody wants to see it Because it didn't matter there by making the various choices you get at least three versions of This function h of x Hs of x all the same and I can check this numerically very carefully up to exponentially Small terms the thing is exponentially big exponentially small terms because It's a lot show you an example in a second for s equals one and you'll see the kind of thing that you can change and it makes no Difference all for the asymptotic formula in less somebody in the future can do a full Rademacher analysis Rademacher made the right choice of h for s equals one different from the Bauddardian Ramanage And then you could make an infinite some the infinite some converge gate the exact thing But I certainly don't know how to do that here and it definitely as far as I know hasn't been done so These which choice you make would be important if you were able to do the full analysis, but we aren't anywhere anyway So there are three the choices when you do the thing and you're actually very pretty So I called them Hsa of x I'll call them just a b and c So each of them is a contour integral, but then you can move the contour integral has a pole There's a 1 over t Somewhere and remember that they're well This e to the minus n t remember tau is related to t by this when you work at all out There's a pole term you cross you pick up a rescue and that you have some exponential and you pick up You expand it as a sum is take the sum the rest use and you get a very very rapidly convergent expansion So the simplest one is the following very nice function So you take this is obviously convergent for all x and c It's an entire function if s is 1 This is r factorial times gamma of r minus a half which up to a trivial factor is gamma of 2r and so this was Essentially like e to the square root of x or cinch of x or culture of x if x is 2 Then you have r factorial times gamma of r minus a half So that's a Bessel function the k Bessel function or some sort of a Bessel function if r is 2 and if r is sorry No, it's not that would be r for 2r for 2 sorry if s is 2 it's it's a generalized Bessel function So it's some kind of a Bessel function, but as I said you have a choice and there are two other versions which are not particularly more complicated This one you take the same you take only rs here It's rs factorial and here it's gamma of r minus a half and you multiply by s And so you see what you've done to go from a to b is every s term I've counted s times and the others have emitted but since these terms for you know We'll start out very Comfortably small and then very big and you're going to be adding up zillions of them and at the high point It's very very uniform to some so if you if s is for instance 3 and then you take every third term and multiply by 3 You get the same to exponentially small orders by Poisson as you would so that these are essentially the same is clear But still they're different functions, so you actually do it you get different numbers and the third Version is again very simple. They're very similar I have to write it down because I certainly don't remember by heart, so it's the same with this s, but here It's rs plus Three halves s so there's a shift and then here. It's r factorial times gamma of rs plus 3s over 2 plus 1 And that comes out if you do the integral in some other If you deform the contour into some other position So depending how you can deform the contour But the asymptotes of these will always be the same and the reason is when you have this path What you always do is steepest descent, so you take your function of t Which is the function I wrote here and it's a completely elementary function this part you can forget It's very small so this is essentially x of a constant times t Minus a power of t so it's cs of kappa times t minus ts over 2 and so you take the places where the with the real part of that is Constant that stationary phase that the oscillations there cancel out It's what you're always doing this kind of thing and that tells you the part of the integral that Contributes the most exact like you do in quantum field theory when you have an integral a path integral And you look at the place where the oscillations stops you look near the maximum and so and then you follow That steepest descent for a long time so to get the asymptotes to all orders you followed that path for a while But eventually that path, you know it won't it might cross the line You can do all kinds of things and then you have say some choice how you close it up And that leads to this different formulas. I don't want to go into that Also that I thought about this several months ago I don't even remember the details But these were three of the function gate As I said it made no difference but now for comparison so the formula that you get is always the same one It's the one that I wrote here that the contribution to ps of n of a single rational number is up to some pre fact of this transcendal function of Let's say n is so I'll just take the case n equals 2. This would be this c2 of kappa Time is squared of n So it's roughly some exponential it turns out in the cube root of n But the constant in the exponential depends on c2 of kappa. So that's the important thing and the exact function will be one of these functions So just for complete this let me tell you what the asymptotics are in the case when If s is 2 I'm actually my choice I make a choice as I said and it will be the function h2 of x the c Version although it doesn't really matter if x is positive and zero if x is negative Actually, if real part of x is positive and negative and the reason is you're doing Cauchy integral in one sign The the place of steepest descent would be above the line and you can go off to infinity You get zero so it's only and it's only for one side that you get a contribution again That's a detail because these terms where x is negative would be even if you took h2c would be extremely small So it it doesn't really matter how you how you cut it off, but that's the actual functions I took and so just for curiosity I can tell you the exact asymptotics for large arguments But the form is ugly if you call h2 of x so I'll call h2 of 2x to the three halves Because then it looks nicer and this you can prove in several ways either there's an integral representation or from the ODE Or there are various ways So it's x e to the 3x of the square to 12 pi times 1 minus 17 over 36x minus 35 over 2592 x squared and so on And if you took not h2c but h2a or h2b all terms at this expansion would be the same The difference is something exponentially small that would be with e to the minus 3x So that's why it doesn't really matter And here by the way you see that this x is x but this x is x to the 3f So this x is roughly the two-thirds power of the argument And the argument here was n to the one half And so the two-thirds power of n to the one half is n to the one third which is what I told you at the beginning So I would like to say a word more about this all because it's I think this is very amusing already in the s equals one case Which is the one we know Hardy and Ramanujan as I said used one h and Hardy and rather not Hardy just rather micro known used another h And so you can say well, which one was it was it a b or c and so the answer is quite amusing So for for s equals one Hardy and Ramanujan Used So I can call the h1, but it's the Hardy-Ramanujan function And it's very elementary It's the square root of x over constant which is two square root of pi But then if you don't shift by half you don't get exponential accuracy at all e to the two square root of x But now this you see is very big if x is big But as you sum the x has a one over c in it and if x gets very small this function is not particularly small It is a constant, but Ramanujan changed it By using a slightly different contour and he used a different version of h1 Which is the same to all orders because well here you can see the order. It's just an x natural factor. So his function was Square root of x minus a half over the square root of pi without the two times the cinch hyperbolic sine Of two square root of x so cinch of x when x is big is one half e to the x Well for every axis it's one half e to the x minus one half e to the minus x That means it differs by something exponentially small from this So the big term said it no effect at all But in the infinite sum x goes to zero and then this thing is is going slowly to To zero because of the square root of x for is this one isn't and so his sum became convergent and normally converged We gave the right answer But now you could say okay, so that's the ones that historically the two important functions And i've given you three So which of my three functions h a h b and h c are these and the only reasonable thing is one of them is Hardly revolution one is rather much and the other is something else But no it's not like that at all in fact h1 a In this case is equal to the square root of x times cinch of two square root of x minus a half Times cosh Of two square root of x again divided by square root of pi And h1 b Is not different. It's the same and h1c is the square root of x times cosh of two square root of x minus a half cinch. I mean you just Outlate these series. It's completely elementary And so they're all different from both h1 hardy revolution and h1 Rather macher So we have three we had two old functions two new ones you would assume the three new ones You'd assume the three new ones is the two old ones are a new one But the three new ones are actually only two new ones because two of them coincide but not for other s But for s equals one and none of them are the same as the ones that were used before But you can see that they're all the same cinch and cosh are both e to the x to orbit multiple of something exponentially small So all of these are equal to both of those Multiple of the same e to the minus two square root of x and they're all the new combinations basically of each other I mean any two of them span the holes span the holes Same space and certainly any three for s equals two and so on. I mean this kind of a nice differential equation I think anyway here. It's certainly true So that's some discussion of the nature of the transcendental functions But now as I said comes the really surprising Thing except it's no longer such a surprise because I've mentioned it several times Which when you now make these contributions and you start adding them up They're very very erratic and the reason already explained That in the case s equals one which is the classical case c one of cap if cap is a over c the nominator c Is simply z of two over the denominator square only depends on the denominator But for already for s equals two and for all other s It depends very much on the actual number. What's important? Well, I just erased it all but I didn't really erase it because there was an hs What's important is the size of this and really the real part because it's an exponential And but here these two numbers happen to be real one was positive and one was negative So near a fifth and four fifths meaning the other standard two fifths of Unity, you know E to the two pi per fifth in here it blew up like a certain exponential And near the other two fifths of unity It actually went exponentially down like a different exponential quite a bit smaller, but now with the negative exponent so now you get You know wildly different contributions and that means that the story is much more complicated So let me again give you the numerics because this whole course is about asymptotes But very much also about numerics and the fund of asymptotics is that you actually get get numbers So in the so the point four is the dependence Remember i'm giving all the places where the story is different for as bigger than one that it was for s equals one The dependence on capital which is some number where the a doesn't it's just notation but the c is the denominator So before that when s was one it only depended on c and it went down as c grow grew in a very uniform way So if s equals one I already said that c two c one Of kappa is simply pi squared over six over c squared And so here if I take the number P of 200 free remember they were using and I take the total contribution Of all points of denominator c to this big number then if I take Uh, I'm not sure my I'll have to write quite small because I wanted to go up to eight Uh, the biggest term when you compute it to high accuracy. It's very easy. We use that h It doesn't matter now if I use rather much or or um Or hardly rum legion because the number is huge So the contribution here is the number very close to the true answer And if I changed whether I used h one hardly rum nusion or ab or c I would change this by you know three to the minus 12 here. You wouldn't see it at all So the here it doesn't really that's why I said it's not really important It looks important, but it's not important exactly how you define that function Then for c equals two it's much smaller. So this is three billion The next term is only 36 000 So they decrease extremely rapidly The next term c equals three is already only 90 I'm not going to put the dot dot dot each time the next term is only five The next term is 1.424 The next is 0.071. The next is exactly zero. So some this is a cancellation And the next term is zero. So for seven there are several terms But the exponential sum of the pre coefficient that although there are several terms in a certain order Their sum is actually identically zero So you get these numbers and the sum here of these I wrote out So the sum of these up to denominator seven eight is three nine seven two nine nine nine zero two nine three eight seven Point nine seven five dot dot dot and remember the true answer is three eighty eight Well, it's an integer so it stops So you see we're within point zero two five of the answer for this absolutely gigantic number And that's what hardy and ramanujan found and where they were so happy that they'd never expected At all that their approximation could conceivably give that kind of accuracy Already that the first term is so incredibly good It gives you know mostly you know more than half of the terms the error is only 36 000 But the first three are down to an error of six the first five you're already at the nearest integer So it's an incredibly accurate thing But now as I said, it's not like that at all when s is two because of this change and when we instead do Now if s is two it's completely different remember there my test case Was a hundred thousand so p two of ten to the fifth and I gave the first well, I could give it again I gave the first six in the last Six digits so that you could sort of see the number I didn't write it all out So it was roughly three point nine times ten to the fifty nine So now again make a similar table. I'll just do it up to five And then I'll give the total contribution of all points All rational points cap of that or not. So here it'll be zero here It'll be a half here. It'll be a third and minus a third here a quarter and minus a quarter And here are four values one two three or four over five. So the total contribution the sum of the P two of ten to the five and I'll just give it two two significant digits. So There's a main term which justice for Hardy and Raman Dugan Was is already extremely near to the right answer So here this was correct the error was only thirty six thousand Here the first term is three point nine times ten to the fifty nine The second term is well the one point eight doesn't matter. It's the it's the power of ten that matters It's ten to the twenty six So then you think wow ten to the thirty three weeks more than half of the previous exponent So we've more than halved the thing and then the next term is even a lot better. So this is you know very very So this is already much nearer now So in other words if you include these two terms you're already You'll already be you know very good. So sorry much smaller. I'm this contribution is much smaller Then the next one is only minus two thousand. So it's very small compared. We're talking about ten to the sixty The two thousand is nothing but now it comes to surprise if I hadn't already kind of given it away When you get to c equals four Then it's much bigger quite a bit bigger Than the c equals two term And then when you go to five again, it's much smaller But still much bigger than the c equals three term is ten to the fifteen rather than ten to the three So it's you know wildly varying And that's completely different from what you saw here where the terms are going uniformly in size to zero I mean one happens to be zero in the next because you're adding up positive and negative terms with the size The number of digits goes, you know very very smoothly along a smooth curve and here It's not like that at all So in fact the biggest contribution Here i'll give the details in a second the biggest contribution if you continue on to get the first four is From c equals four, which means kappa is plus or minus A quarter denominator four and zeta the corresponding word if unity is i or minus i could just put plus or minus i okay So that's that's the biggest contribution and that means if that were the only contribution remember the contribution always has I think i raised it now, but the contribution i did erase it it had i to the two pi e to the two pi i times Kappa times n or actually times n tilde that that's just a shift of n by constant Times this hs and then there was a a pre-const this little c cap But it depends in an oscillatory way on kappa n So you have to add them. So if kappa is a quarter Then that means that each the two pi times kappa n is period four But since kappa is not just some interval four, but an odd interval four Exactly for it means even more not only this period four it is anti period two when you change n by two You change the sign. So let me give you a little table. So it's roughly I mean asymptotically It's anti invariant When you shift your n i mean n is a large number when you shift by two So here if you take the hardy ramanujan if you take this this was n equals 200 But if you made a table of this and next to it i the correspond table for 201 202 203 Then the first row the next row would be very close to this. It's a slowly vanishing Varying function, but the next row here the next room would be minus 36,000 Because it's just this is slowly varying function, which is roughly 36,000 times minus one to the n It depends on n mod two and the next turn there'd be three values depending on n mod three here depending n mod four But now the most important contribution is this fourth one. So it depends on n mod four So to make that come to life. I hope let me make a little table So i'll give the total error so I take p two of Why did I write delta two of n Is the value of p two of n minus the main contribution? From cap equals zero And so here i'll just put the number delta of 10 to the Of 100,000 And since they're all of the same order 10 to the 32 and I don't want to write it nine times 12 times I think I'll just put I'll divide by 10 to the 32. So let me take you the numbers Sorry, excuse me. This is going to be delta n But I am dividing them all for convenience by 32 And so i'm going to take 10 to the 5 plus and I'll take zero one two three four five six seven eight nine 10 11 So i'm giving you a table of 12 successive values But i'm grouping them four and then the next four because then you can see what's happening And so when you do this What you find Is that The terms are very very similar and growing in a smooth way here. You add 10 here. You add nine will subtract That's for zero four and eight But then if you take one five and nine you get 1.371 minus 1.375 minus 1.379 But now here you get numbers that again are Positive and they're almost the negative of this And simply the last column is then again 1.373 Remember this 1.373 is really times 10 to the 32. I didn't want to write it a hundred times about 12 times So that's the table and you can see that if you fix n is zero mode for one mode for two mode for three mode For then it's just a very slowly varying function That's the h2 term but the pre-factor depends on at one four, but it changes the sign This minus 8.3 becomes plus 8.3. This minus 1.37 becomes plus 1.37 so that's So now you get this rather strange phenomenon And so this is what I just showed you that the contributions in this case 10 to the 5 From denominator one two three four and five they vary wildly in their size And the biggest contribution is always c equals 1 but the next biggest is c equals 4 so in fact for Uh So the reason is again If I look at kappa and c2 of kappa Then if I make a table now i'm going to order these by well i'm ordering the kappa but by the real part of c2 So this c2 you can calculate to any number of digits It's just I gave the form before it's a sum of dialogs And so I'll give a little piece of the table and zero. It's 2.315 A plus or minus a quarter. That's the biggest one. It's 1.038 plus or minus 0.383 I will I prepare the table all the way up to This is the first eight. I'm not going to copy up the whole table anybody once can ask for a copy of the paper It'll be on my web page in a few days. I haven't yet put it on because I'm still changing one or two formulas that they've misprints So if you take the first eight come in a rather strange order Well, I can tell you what the next ones are Actually, I was going to do at least the numbers So the next one after a quarter is a half So it's worse than a quarter but bigger than everybody else The next three are the ones with the nominator nine but with numerator one four seven not two five or eight They give a very small contribution and then the next ones So maybe I'll take this number away and just give you the the actual numbers the next one is plus or minus A fifth an eighth Then here it's plus or minus one or nine over sixteen And then the next one is plus or minus One four six nine or eleven over 25 and the next one already wrote is plus or minus a third They come in a kind of a weird order and I just wrote that one. So they're This is the order of the contributions in principle for a given n Now when you do it for a particular n like 200 It can change slightly because for instance these three the three confuses are of the same order But the c2 are exactly the same number Okay, and when you add up the confuse they cancel exactly they give zero just like we had one for Adiramanutian so the actual size of the contributions for 200,000 is again the weird order And I'm going to give a little more because it actually shows something So the biggest 17 contributions So for each c I add up all the contributors for that c For n equals for p2 of 10 to the fifth which are standard example are in order One as I said is always the biggest four was the next biggest then two we add here Then you would think it's nine and then eight, but I told you nine actually cancels completely Contribute zero the main term so eight then 16 three 25 12 then comes five 24 32 48 40 64 20 96 and 17 the actual numbers play no role This is you know a lot of computation defined in what order they contribute And this is what you find so those are the first 17 and they're And their total contribution Is what it should be which is p2 of 10 to the fifth which remember was around 3.9 times 10 to the 15 minus Approximately 8.3 times 10 to the 13 So what you're missing if you take these first four 17 contributions We're not at all down to the nearest integer. It's how did ramanution where we just five terms But we're down to 10 to the 13 which sounds like a big number Sorry, this was 59 So we're complying to comparing with almost 10 to the 60 It's you know almost 10 to the 50 times smaller. So the first roughly 50 digits 47 digits are correct However, so then you'd say great. So I just continue But the next few c the next c in order That's why I wanted to actually give the numbers because this is the actual cutoff point Where it no longer helps you might think okay I just take some more c and I'll get closer and closer as happened not at all the next c In order at least or I'll just give the next few are ag 56 There's a tendency that highly factored numbers contribute a bit more. You see here 32 48 40 64 They're not quite random, but I also don't know they're not quite systematic either I can't read my handwriting. So I'm not sure if that's a 13. I think it is but but all give much smaller contributions Like you know 10 to the 8 10 to the 9 And they're only a handful of them. I mean 5 and then after that they're yet smaller So 5 times 10 to the 9 doesn't begin to eat up the 10 to the 13 So roughly when you add this 17th, it's still better than the 17 It's better than if you stopped here. This error is a bit smaller than it was it was 10 to the 14 Now it's 10 to the 13 But if you continue and take the next 5 it remains 8.3 about the next three digits are the same 8.1 or 2 9 5 or whatever times 10 to the 13 it doesn't go down anymore. We're missing something And so that's the the last point Wait, I started at 2 and I go to 3 30 and it's 1 minute 3 so I'm okay. I can't subtract Some I'm still okay on time. I don't think I'll use up the full have now, but at least I'm not good to go over so You know, you might think Well, what's going what's going wrong now? And of course, there are several possibilities One is that I made a mistake you make mistakes and I sure you I made plenty as I did these calculations along the way But I ended up checking everything six, you know against each other. So I hope there aren't any but I can't guarantee It could be that the theory is just wrong and that what is called for the experts the minor arc So you break up the circling to the major arts Which are open intervals of the appropriately chosen size around all rational numbers of small denominator Of course every denominator is small everything is relatively ends you go up to Some size for hardy and ramanujan the denominator went up to the fourth root of n And then you take the intervals around that and then the minor arts is what happens so to speak in between And so it could be that they're having a big contribution and for all I know that still is true But in fact, it's much more interesting So now comes the the fifth point and to me by far the most fun of the whole Thing because it really means that the circle method In terms you would say See how it's in z it really has some extra pizzazz that the one for s equals one doesn't have So now I come to the as I say the last Point especially if I can read my notes So the the most subtle point So remember what the big difference is between s equals one and all other s The the circle method and the asymptotics Euler-Mclaurin kind of worked the same way But what was very different is that if s is one Then the function a to s of tau which was essentially one upper p s which was just the Dedekind a to function is a modular function So it transforms in a well-defined weight under the multiple groups. So remember that a to s Well a to of tau for instance Is q to the 124th divided by p one of q and in general a to s is q to the Minus a half say it of minus s divided by p s of q So the size of ps of q is essentially the same So that means a to of tau itself Well a to of tau inverse is what we want starts q to the minus a 24th Plus q to the 23 over 24 Plus q plus 2 q to the 47 over 24 Etc the coefficients here just the partition function one one two three five Seven etc But the exponents are minus a 24th minus 24th plus one minus a 24th plus two All of this is exponentially small as tau goes to infinity So that but that would be the same for a to s a to s of tau would also be In this case, there's no power of q. So d1. I can't do it in my head plus q plus I mean it also would have an expansion. That's not what changes, but the changes when I take gamma of tau So tau is going to some kappa Which is and gamma is some elements such that gamma of tau is going to infinity So if you write and we did this before I'm not going to write down the formula. So this is So up to a stupid factor Which is something like the square root of tau minus kappa as kappa goes to infinity when you transform tau an element of s l 2z to go to infinity Here because of the multilayer you get a factor square root of tau minus kappa or one over square root of tau minus kappa And then what you get is exactly the same expansion So the the nature of the expansion And here q is no longer e to the 2 pi i tau, but it's e of 2 pi times gamma of tau But gamma of tau is now going to infinity and so again all of these terms are exponentially small And so the function itself had only one exponentially big term And it's that term that when we multiplied by the remaining thing In the circle formula gave the best of like function because it was a product of two exponentials And one of them was this very big exponential And then the later terms won't contribute But now let's look what happens when s is not one Well, we gave the formula in I gave it last time Well, I gave the first time numerically for instance near fifth and then last time proving And so the formula is not like that at all now Instead what we'll find now For s uh equals two for instance the same for any other We find that p to of s at a cap at a root of unity which is e to the 2 pi kappa Times e to the one over t squared You see When you transfer the function I gave the exact formula another time it turned into a sum of functions Where you had for instance a to 2 of minus 1 over tau With the product roughly up to a simple fact like square root of tau of Eight of the square root of I think it was square root of tau and i times the square root of tau with index a half So there were two different terms, but each of them had an exponential product And when you do it at other seats, you'll always have a sum of terms So in general if I take a to s or p s of e to the kappa I'll have a sum Of terms e to the something exponentially big That's just like here here. This is e to the 124th of something then minus 23 over 24 But here the expansion only one of those exponents is positive all the others are next negative And so the whole thing can be replaced by a single exponential But here it became it was an infinite sum But the exponents eventually went to minus infinity. So it was convergent But that means that finally many of them are negative or positive So you'll find that the many terms but not just one sometimes there are none remember I told you that two fifths it starts e to a negative number negative Times t so the whole thing is exponentially small at plus or fifth. It's a positive number So it's e to a positive number times t it's exponentially big But then there's a second term that second term Happens at a fifth to be also exponentially small so we can ignore it So there's only one term that contributes here and none that contribute here But when you get to a ninth, then you have the one term that's exponentially big But the next term is still exponentially big and then the other terms are have negative exponents So you get a finite number of contributions from each one So I'll actually write the form the details don't matter, but just you see it You'll get a sum so we have this number n two of c which is the smallest interest squares This by c divided by two pi n to the three halves. All of this is just stupid factors then e to the two pi of minus n kappa and now comes a complicated mess And I'll write it out just for completeness, but the point is Only the structure of it. So I'm going to go from this set So I'm going to go I'll have the function x kappa And x I'll call kappa again a over c as I've been doing Is the set of pairs l m Where l is a residue class modular c m is an integer different from zero And they're linked by l squared a is congruent to m mod C Okay, so that's some infinite set So um, and then I'll take all functions u from this into Z graded three to zero But they're only fine at the well there are many of these but only find a demandable contributor I'll show you why in a second and then here we have e to the two pi Times some number which depends on kappa and and on this function u Over c so that's a root of unity and then the exponential that we're used to But this what I called c to a kappa before was just the leading term when u was the zero function But now we have different exponents But we only care about the ones we're c to So only find that they many terms But there exist only finitely many Of these functions u Such that this c two which I'll write down in a second just so you've seen that Such that the real part of this is is bigger than zero bigger than equal to zero So it's going to be a finite sum of positive exponentials But it's a mess because it's a whole even the indexing set of that sum It's not just something simple it's pairs l and m But they're only fine at the many believe me with this thing as positive and then multiplied by some Root of unity, so I'll give the form is just for fun Or not for fun. It's not very Not really fun at all, but just see you for completeness So we have this function u and I sum over all l m in this set But it'll be a finite sum Okay, some u of l ml and then c two of kappa mu Will actually be c two of kappa because I've I've shifted things Somewhere don't ask me too many details because I copied this all From the paper. It's a mess two pi of the c to the three halves Times the sum again over this set x kappa which The sums are called l m and then the actual it's the same mu And then it's multiplied by m minus i times the squared of m so m can be positive or negative. So Uh, that's whatever it is over the squared of two m The actual form doesn't matter at all. This is a real number. This is a rational. Well, this is an integer So this is c through divinity And this is some real number or maybe it's a complex number. It's a complex number But as I said, they're only finite the many that are going to contribute And so now when I put this into the formula Then I'll find that the total contribution to kappa is not the one that I told you before I was always just taking the leading term which was the one with the c to of kappa I don't know quite I don't quite believe the c to of kappa Minus, maybe it is that let's say it's correct But so the contribution will be that the full C to of kappa of n will have a part p to kappa of n Zero so this corresponds to the the term that we had before the leading term And then I'll just I called everything else, you know, plus Everything else so That's this means that u is that function u which is takes on non-negative values to zero or here it's Not the zero function So the zero term will be kind of usually the main term And this thing will be the sum So do I have to write the whole thing again? I think I do so it's got the same Prefactor of course that doesn't change But I'm doing something wrong This can't have been right because There is no n here. So I don't know what I did here. There's no n here. There's an n It's completely unimportant the only important so here I have the same l of kappa mu over c But now I have my same transcendental function that we remember with something like cinch of x and the hardy ramanusian case cinch of square root of x It's this function h2 of c2 of k and Mu Then times the square root of n so before I just written c2 of kappa times square root of n But there are actually many kappa comma mu and I was only concentrating on mu equals zero And so these these extra terms I say the details don't matter at all And now when you do this Things get much better and there was still a very slight puzzle at the end and I can't quite explain it Could be a numerical mistake. So now I'll end with numerical example once again Well s is always 2 now n is again 10 to the fifth And the biggest contributions if kappa If you remember an earlier table, I told you if kappa was 1 4 7 or 9 Then these had the the biggest a rather big c2 of kappa But for this particular value of n the three constants cancel. So they the main contribution So the primary contributions Of these kappas These are in the notation I've been using p2 kappa for these three kappa. Well, there are six kappas. It's up to sign p2 kappa 10 to the fifth But the the the main contribution zero Is approximately minus 3.39 times 10 to the 23 minus 1.15 times 10 to the 23 And plus 4.53 Times 10 to the 23. So they're fairly big. Remember, we were missing 10 to the 13 But they add up to zero Exactly. And so that's why I'd mentioned that before they don't contribute anything The the terms were n equals 9 but the secondary contributions But the secondary contributions the p2 for the same kappa of the plus. Well, there's only one term There are many terms, but here there are only two terms that are positive the 0th one and one other and those three together give The p2 k of n for these so it's actually six values of these for these kappa. So c equals 9 Do not give 0 and they give in fact 8.30 times 10 to the 13 And what we were missing before so previously missing just using the top terms Was 8.26 times 10 to the 13 So you see the single contribution was c equals 9 but the secondary contributes the second order exponential Which is still positive but much smaller than the biggest one is giving something very much smaller than the main term Well, it would have been 10 to the 23 it actually cancelled But now these 10 to the 13 there are three contributors. They don't cancel and they give almost all that we're missing And in fact when you look here what happens if c is 1 up to 7 there's no secondary term So the c so for those kappas The p2 kappa plus is simply zero Remember we defined the h2 to be identically zero as soon as the x1 was negative So there for those k's there's only one exponent. It's positive. So there's no contribution It's almost they're all negative. But anyway, there's no secondary contribution for c equals 8. There is The secondary term. It's not zero, but they cancel Exactly just let this primary term is cancelled for c equals 9 for c equals 9 already told you the contributions were 8.30 times 10 to the 13 The next most important ones are of c is 25 or 36 and this is of the order Of 10 to the 11. So it's 100 times smaller Which means it's changing the second digit of the 8.30. That's just we're missing and the sum of these Is now 8.25 times 10 to the 13 And what we were missing was 8.26 times 10 to the 13 So I felt pretty good. I have to admit I then spent You know like two days trying to get you know more digits to see how far it would go But whatever I did the further terms are still much smaller. They no longer contributed It's the same problems before so I'm still if I'm really honest missing about the 10 to the 11 out of 10 to the 15 If I have no idea where it came from it's hard to believe that this 8.25 Which should have been 8.26 is the coincidence So I assumed that that there are numerical small numerical mistakes somewhere that are creeping in And that it will give something to Arbitrary small orders, but there are several possibilities. I may have made a numerical mistake. The theory may be Completely wrong in some way that I don't see It could be that Remember I had a choice whether I used h2ab or c And I actually took h2 to be h2c If something was positive and zero if something was negative But it could be that those terms but I don't think so because they're exponentially small and my error is still 10 to the 11 Those should be like 10 to the minus Something so I don't think it's that so it could be a mistake. It could be that the theory is wrong It could be that the you need a different h2 and the fourth is of course it could be a contribution to the minor arcs I'm pretty sure that the true answer is the first one somewhere. There's a small numerical hitch Remember we're already in the Well over the 50th digit compared to the original one or roughly the 50th digit is a very very high precision competition We can easily go wrong. So the exact detail I don't think anybody really cares about getting the exact number of Parties of 100,000 into squares from an asymptotic formula But the fact that the circuit method has so much Intrinsic number theory coming in and such a subtle behavior when it's not multinar I thought this was interesting enough to tell in quite some detail So that's the end Now it's actually 12 minutes before the end of today. So I can stop earlier to certainly take questions But that's all I'll say about this story and as I say next song will be completely separate Probably also not a full hour and a half because I don't think it's it's a nice story, but it's not that exciting But it's one more quite different type of asymptotic question that one can ask So I'll stop for today multiple questions anybody here or on zoom About either these three lectures or any of the previous ones The manual I can always count on you for a question question So just as a curiosity, although you know this sequence of contributions as you listed in one table Like the biggest contribution is from the first route of unity one, right? And then It came to the i and minus i which were related to the denominator four and then the third one I don't recall what it was. I think that was a half It was a half and so on. So you have a chaotic thing. So the first one Got to 10 to the 23 digits and then There was a 10 to the 32 coming later on right But I suppose I mean even when with the status that you have now, of course, there is a threshold I mean of course you can for surely say that The contributions which are way ahead in the table won't beat this 10 to the 32, right? I would say that in more than that I told you that if I took the first I wrote it out in detail the first five No, it's not on the board anymore. Let me write it very quickly without the mantis. I think it's called So if 345 then the comp used just the number of digits was 59. I'll just put the number 26 3 32 and 15 So the biggest contribution after the main term is 10 to the 32 and then I gave the next ones I gave a long list. I wrote it on the board all 17 that probably people thought I was a bit nuts to give so much Maybe I was but the main contribution order for that n is where one then four then two then eight Then 16 I gave the first 17 And then this gave me an error Of age times 10 to the 13 And then I said that if I take the next five they're way smaller But I went up to about 300 and they're all very much smaller So much smaller than even if you had them up and I'm actually convinced But that's another place that he could have gone wrong that I won't suddenly find actually when I went up to a thousand In the end I let the computer run for three days All the further contributions were very very much smaller So there will it will never happen that a later thing Overcomes not just the 10 to the 32 that even the 10 to the 13 So after a while I'm really left with an irreducible Peace that I can't get rid of by more terms But as far as as far as the theoretical part goes, I mean do you have I mean You mean can I estimate yeah, can you give an upper bound for can you actually prove mathematically that no contribution That is what estimate means. I mean estimate means to prove That something is less than something absolute value I think I certainly could because if c is very large Then the size of the c2 of capital will eventually get smaller and smaller And so whether it's positive I don't know the exact arguments, but like it's a sum of dialogues the dialogue and behaves very regularly So I think I could make a rigorous proof that if I take let's say So I went up think up to c is a thousand and none of the c's except these 17 The fact that this number in any way shape or form they were much smaller like less than 10 to the 8 I think I can prove that after a thousand all the confused will be less than this just by estimated the dialogue I didn't bother to do it because It's not that interesting. We're only talking about one number n is You were only talking about the case s is 2 and n is 10 to the fifth But in principle for any given one you could say if if you take very large c Compared to n it will be negligible But the problem is that the small c are jumping up and down there There's no way that you could see prince Why is this so ridiculously much further smaller than the ones for 1 2 4 or 5 3 is really small 3 is a big denominator So after in hardly rummage the big one is 1 the next is minus 1 and the next 2 are 3rd and minus 2 But here these are 10 to the 3 as opposed to 10 to the 32 And there seems to be no reasons just when you add up certain dialogues You happen to get a complex numbers real numbers is positive but very small and there doesn't seem to be any way to predict that But I can do what you said if I if c is big enough Then I can show eventually none of the terms can beat any of these and that's what I roughly did to say It won't help and if I keep going here, I listed the next five, but they're much smaller They won't help but then these secondary terms Also then ship in but then I did the tertiary in my program if I didn't make mistakes for all See up to quite a big point. I'd all of the contributors. I mean this entire horrible sum. It's a real mess And Eventually it looked like I Included all terms that would contribute and that everything else would be less than 10 to 11 I was left with this Approximately 10 to 11. I couldn't get rid of this. I said, I don't know But my option wasn't to give very rigorous proofs because anyway The method is a little non-regress because I haven't given the estimate even near a single point kappa Nor do I so I don't know exactly to what extent well in principle I do the the leading term is a pure exponential remember the actual formula for eta Like the one I showed 8 of minus 8 to 2 of minus 1 over tau is the product between 8 is it was always a pure or for every s I showed that a to s of gamma tau Was equal or from of kappa Plus the constant over t to the s This was always up to some power of t of sum. So if you put some power of t, I don't remember t to the minus says It was a sum Of pure exponential. So as a function of t, it was a convergent sum. That was the main theorem And only find that the many of those have a positive exponent And the others I think will contribute nothing but at least something very small So in principle, you know, it is a rigorous method But for each individual contribution of what I haven't done at all is look at the overlap of various intervals If there's anything like small arcs and big arcs and it's not a piece of analysis A piece of number theory I'm trying to find out what the contributions are and how they depend on the numerics of c and n And what n looks like modulo c and so on and I did not try to prove anything rigorously My guess is to be a lot of work like if one is a a good student It would be maybe a full phd thesis like three years of work It's not something even a really good analyst could sit down and do in In an afternoon or probably even in a week. I think it's It would be really and it's probably very interesting to do But it's it's definitely a non-trivial problem to actually do this rigorously and that wasn't my There's so many places. I was throwing away things. I didn't worry about that one particularly Or anyone in outer space have a question Otherwise we can go home five minutes early Have a drink So I see you all again. I hope At least most of you on Wednesday at two and then that's the last time