 Yesterday, I told you about a little competition I was having for the triangle inequality that I talked a little bit about yesterday, I'll talk about it again today, to confine the nicest proof. And there was a nice proof sent to me last night that maybe I'll tell you about, because it's rather nice. Reminded me of another competition, well, it sounds lecture also. And the Americans here will know what I'm talking about. Many of you may know of a Putnam competition, one of these competitions where people have to answer clever and witty questions in a time period to show how clever they are. And it's a national competition in North America, and there's some fun questions there. And I happened to be chair of the committee setting the exam in 2002. And one of the goals in all of these everywhere in the world is you put the year into one of the questions, right? And we're really struggling as a committee to find a question with 2002 in there. And this was just soon after all these conjectures that Sam talked about about moments had been made. And I happened to have a post-doc, Nathan Inge, at the time, who was into all these moment conjectures. And I said to Nathan, you've got to help. Is there any question you could think of that leads to 2002? Nathan came back with, well, it's this divided by 12. You try and find a problem that is suitable for the partner matter. We had fun for two days, but it was always going to be a little too hard for the partner. And that's the new competition. Find a great problem involving the eighth moment that gets 2002 out of there. OK. I'll try and be a little less silly. So yesterday we got to looking at this function. I told you that the so-called pretentious R-H, I'll write it over here, says that for all s, such that the real part of s is greater than 1, for all into just k, the k-th derivative over k factorial is less than less than something depending on s 2 to the k. Or let's just give something explicit, say 1 plus t to the epsilon, where s is sigma plus i t. OK. As I said, conjecture could be any constant that depends on s, but, and we saw yesterday this implies the Riemann hypothesis, and I claim was the starting point for what we're going to do. So let's just quickly run this through Perron's formula, which we wrote up yesterday. So as is the Dirichlet series for the sum of a n over n to the s. So all I really have to do is read off the n-th coefficients to interpret this formula. So what are the n-th coefficients to sum n less than or equal to x of the n-th coefficients? Well I'll just put as approximately 1 over 2i pi, the integral just to the right of n somewhere, here I want c greater than 1, of z of s ds. And zeta prime over zeta, the n-th coefficient is minus lambda n, so let me put a minus on this side. So we've got lambda n, so minus zeta prime over zeta, the n-th coefficient. The n-th coefficient here of minus zeta, the n-th coefficient will be minus 1, and this only has one coefficient which is right at the start, so, well, we can lose the two gamma in the approximately. So well this is psi of x minus x, and this sort of tells you, well it's good to think of lambda n minus 1, you've seen the sort of balancing act in Green-Tau, you sort of take off the main term and deal with the rest in various of their works, anyway, tends to work nicely in this. I do want an x to the s over s, how did you know, okay, so we want to play the game with case derivatives, so what happens when I take a case derivative, when I've got a Dirichlet series of a sum a n over n to the s, and I take the case derivative with respect to s, then all that really happens is I just get a minus log n each time I take a derivative. So actually if I just put a k up here, case derivative, there's a plus or a minus, I'm not really concerned which, and I've got a log n to the k here. So log n is, well as we talked about last time, I'm sure you know, you can go between psi x and pi x when thinking about the counterprimes very easily by partial summation because log is a very smooth and benign function, and as long as this moment isn't incredibly high, it's not going to make a lot of difference talking about lambda n minus 1 or lambda n minus 1 times log n to the power k. So okay, so let's do what we did last time, which is just to take absolute values and see what happens. So this right hand side then is less than, less than. Remember c is 1 plus 1 over log x which allows us to say that x to the s has absolute value x to the c which is e times x. So let's see, we're going to use this formula here, we're going to assume, so this is assuming the pretentious Riemann hypothesis, and we get, well forget the constant, so we get some sort of integral between minus t and t, so we take, make the change of variable c plus i t like last time. We've got this 2 to the k, k factorial, the 1 plus t to the epsilon is obviously less than t to the, let me actually, sorry, leave it as t to the epsilon, 1 plus t to the epsilon, sorry. I've got my x, I'll pull that outside. We've got this 1 plus t in the denominator, dt, okay, so that's an upper bound. Now let's, this integral, the only integrated bit is this part and that's when you integrate it, just going to be t to the epsilon, okay, because it's really t to the epsilon minus 1, and nothing bad happens from there, t equals 0. So this is less than, less than x times t to the epsilon times 2 to the k, k factorial, which is essentially 2 to the k, what, 2k over e to the k, well, there's times a root k, but that will go into the t to the epsilon, it's just using Stirling's formula. Okay, and there is, I probably need to take account of the error term to this correctly, my notes are a disaster, okay, there should be another log, but let me just write it as this for now, let's not worry too much, so what we want to do, yeah, okay, so we're eventually going to divide, you know, when we do partial summation there'll be a divide through by log x to the k, so what I want to do is, is what we will get from partial summation when we do it, is something like, and there may be a few logs that occur because of the partial summation, but we'll get something like x t to the epsilon 2k over e log x to the k, okay, so that log x to the k comes from here, and then all we have to do is make the optimal choice for k, and the optimal choice for k will be k is log x over 2, yeah, you always optimize these things like that, so when I take k equals log x over 2, so forget that, sorry, k equals log x over 2, oh, then I get less than less than x t to the epsilon over x to the half, and so x to the half t to the epsilon, so I take t to be around x or something and I'm in good shape, so we've again proved, well proved directly now as z of s, that making this assumption about this function implies the same estimate that you get from the Riemann hypothesis as we saw last time, and what I didn't say last time is that actually this is equivalent to the Riemann hypothesis, getting a bound of x to the half, so we'll just finish this off by saying less than x to the half plus epsilon, just choosing t, say as root x or something like that, okay, so that's using the pretentious Riemann hypothesis as we call it to prove a very strong estimate on psi x minus x, and now let's just see what happens with this if we plug in a different estimate, and what we're going to be able to prove instead of this is something like log t, log 1 plus t to the k, c, some constant log 1 plus t to the k, and I'll mention in a minute how to do that, so let's actually start from reminding ourselves what we did initially to prove that this shows the Riemann hypothesis, this conjecture, if we replace this now by this estimate, this is what we'll get unconditionally, and I'll sketch that in a minute, then what we have, so let's say s0, s0 feeds hypothesis, when we look at z, that we look at z of s close to s0, then what's the radius of convergence of z of s that we can get if we make the assumption that the derivatives are bounded this way? Well, again, if you're going to get terms that have s minus s0 to the k times this thing, right, so this is going to converge in something like s minus s0 is less than some constant over log 1 plus t, just, it's the same idea of a 2 to the k that we have that, and so in particular, this gives a zero free region, so this is exactly the classical zero free region of de la Vallée-Poussin that comes out of proving this estimate. A little more transparent, I didn't get to go through the proof yesterday of a zero free region, but if you go through it, it's a little opaque, how, I mean, you get used to it, but it's a weird proof. This is, I think, a lot easier to see that why this is zero free region, it's just you have a Taylor series, it obviously converges in this ball. So if we plug that estimate, well, we can go through a zero free region and then prove the usual prime number theorem, or we could just play with it in here, and then the only real change is that we change that to 2 to C log 1 plus t to the k, and so when we come down here, we have, what, k times C log 1 plus t over E log x to the k, that's, I mean, it's all, it's just replaced 2 by C log 1 plus t, and then the optimal thing will be k is log x over C log 1 plus t, and so we end up with less than, less than x to the 1 minus some constant over log 1 plus t. Okay, something like that. Well, we have to be careful, so maybe I should say this. I'm working out to be this. We have a second error term in the prime number theorem that's x over t, maybe times some powers of log. You, as normal, you're just optimized where the 1 over t is the same as the x, the C over log t, just like in the classical proof, and so that optimization occurs where log t is like root log x, and so you end up with the upper bound x over e to the C root log x. Okay, so that's a recovery of the classical one you've all done. They'll have a little proof with a little number 10 thrown in as I talked about last time. Yeah, so the issue to improve a prime number theorem, to improve the error term in the prime number theorem is to improve this bound, if we can. Oh, actually, I think I should prove this bound. Let me go ahead and do that. Any questions on that? I was very pleased to see that sound loss control in his lecture at the back. Okay, so how might we go about proving such a bound? So we need bounds on z of s was, remember, well, I've got it up there, zeta prime zeta plus zeta minus 2 gamma. When we take the kth derivative of z of s, then we're going to have the kth derivative of zeta prime over zeta k times plus the kth derivative of zeta. So these are things we're going to need bounds on. Well, this presumably is straightforward because this guy is just the sum of, well, we'll talk about this in a minute. I'm not suggesting it's trivial, but let's say it's doable with a plus minus. Okay, so that's the kth derivative of the root of zeta. Remember, we're always to the right of 1. This one is more tricky. What do you do with it? Well, when you differentiate zeta prime over zeta, you get, what, zeta prime prime over zeta minus zeta prime over zeta all squared. You do it again. You do it again. There's some combinatorial mess, and it just so happens that that's been sorted out for us. Here it is. And this is, I guess, due to, was it Fadi Brano, Bruno? So if you find a big mess like this and it's a combinatorics book and they claim it as a combinatorialist, it's usually from like 1730. If you find this in a computer science book and they claim it for one of theirs, it's usually from 1987. So we believe a combinatorist that Bruno was probably the first. Anyway, so Which is Phi. Phi to Bruno. Phi. What language does that last thing belong to? I'm supposed to know. You're the one who speaks all the languages. Okay, so zeta prime over zeta is a big old combinatorial mass times something involving zeta differentiated a certain number of times over zeta. That's the way to look at this. And let me just note that this combinatorial sum, you can have fun with it and you can find out that this whole sum actually is less than an absolute value, 2 to the k minus 1. So actually this is, for us, this is all that we really care about, 2 to the k minus 1 times the maximum of this last thing. Okay, taking absolute values. So you're just sort of taking the worst term and assuming every term's bad. Okay, so we're going to have to work with a lot of terms like zeta ks or zeta js over zeta s. So what do we need? We need upper bounds for these guys, for the zeta j's and we need a lower bound for, or an upper bound for 1 over zeta s which is a lower bound for zeta s. As you might imagine these are fairly standard things in the literature. This is particularly to the right of 1, to the right of 1 convergences on your side. But let's just take a brief look at what might be difficult. So if I want to just sum this up, and now I'm thinking about s being sigma plus i t. And again, it's not very convergent. So sigma's just to the right of 1. I want to sum this guy up, over all integers n. Then the obvious way to do it is to say, well, this is roughly the integral. Right? I mean we approximate sums by integrals. Even though this is complex, so we've, and we have to be, you know, if it's, if it's real this is an easy approximation, easy to get good error terms. If it's complex, that's a little harder. And essentially you want something, if this is going to be a correct approximation, you want things to go smoothly. Now let's just forget the sigma for a minute and look at the end to the i t. What happens, say, is you go between 2 to the i t and 3 to the i t, things are going to be smooth if there's roughly the same angle involved. Yeah? If the, the, the gradation by angle as you move from one into the next isn't much. If it whips around the circle many, many times, then who knows where it ends up. You can't really approximate by the integral. So if t is, say, 10, 2 to the i t in angle is nowhere near 3 to the i t. Right? But a thousand to the i t is very close to a thousand and one to the i t if t is 10. If you think about it, well, it just as I go from n to n plus one, let's just compare the angles, n plus one to the i t over n to the i t looks something like one plus what, i t over n if you look at the Taylor series. Right? So where things start to get okay is when n is bigger than t. And if n is less than t, things are not okay. This is not a good approximation. So essentially we have to break this up into two parts. We have to say this is like the sum from n equals one to n of one over n to the s plus the sum n greater than n of one over n to the s where n, say, is t squared or something. To be on the safe side, something like t squared. And then this guy without a doubt looks something like the integral du over u to the s. Okay, that's easy. And this one is a mess. And the best we can do really is take absolute values so you get one over n to the sigma, so big O of a sum. So this is a problem. And there are different ways of coping with it. So let me just say when we put the log, this looks fairly benign with sigma greater than one, but now when I put on here a log n to the k, I mean the same arguments apply when I put on log n to the k, log u to the k. But now this term is going to be something somewhat significant. And if you think about sigma very close to one, the sum of log n to the k over n makes this look like a log n to the k plus one. Okay? And the other terms actually are more cooperative, they're smaller. You know, there's terms in the approximation, you've got to figure out all this, but everything else behaves, this doesn't behave. So there are different ways of coping with this. So okay, I guess the first question is, is this a good thing to put into your formula? Well, you're bounding the kth derivative, it's actually log n to the k would be more convenient. It's not actually quite true, but that's the sort of thing you're aiming for. So if we put in log n to the k plus one into our formula, we get something here like log one plus t to the 4k, maybe? Something like that, do you know? 2k, okay. He can get 2k, I can get 4k. So, 2 is better. So you can get something, but it's not as good as the classical zero free region. So in Demetrius' paper, he has the idea that one way to deal with his term is to get rid of it. It's always good. So what's his thought is, that's why last time I said I'm going to regret this notation. I'm going to now try, zeta y is going to be the exact opposite of what it was last time. Oh, I put brackets around it last time, so now I'm not putting a bracket around it. There is a significant set. So n greater than or equal to one, p divides n implies p is greater than y. And here, we don't need to take y very big, it's just like a power of log. Is that right? No, it's going to be like t, because t squared, so it's like n over there. So, we're going to work instead of zeta, we're going to work with zeta, remove the small primes. I guess in Sounds Lecture, he talked about the pain of the small primes. And similarly, one has the pain of the small primes here. But you know, the small primes, so t, in our last example, we could take it as something like e to the root log x. In fact, that's what we chose in the end. So, oh, there it is. Yeah. So, those aren't very big primes that we're removing. If we're counting the primes up to x, removing the primes up to e to the root log x, who cares? So, if we go back to Perron's formula and everything else, and we've just deleted the small primes, it's not going to really affect very much. It's just going to be error term. But here, it means that this guy's gone. Now, the problem is that once you do that, this is not so simple, because you're not summing over all integers n, you got to sieve by the primes less than or equal to n. And so, you've got to compute some sieve estimates, but it's not a big deal. You can do it. Fairly standard techniques. And the matrix ends up with this estimate. It's basically the proof. If you could do better estimates for the case derivative of zeta, then you could do better. So, let me just formulate a version, which I know won't please Jean, but this is what I'm going to write down. So, if you're trying to understand the sum of this sort of sum, as I said, it's a sigma plus IT. And the thing that's interesting, if you like, from partial summation, the log n to the k over n to the sigma, at least when it ends with reasonable size, is not important. It's the one over n to the IT that's causing the cancellation that's of interest. And the estimate from Vinogradov, which turns out to be, is used in Vinogradov-Korobov, but it also can be used here, if you like, just as a black box, is that the sum of n to the IT n up to x, the mean value, if you like, is less than, less than exponential minus log x cubed over log t squared. And this allows you to, well, you can imagine, use partial summation with this, go through the estimates. This allows you to put a two thirds in here. Okay, maybe there's some log-log floating around, but you can put a two thirds in here. Well, you can see the two thirds rearing up there from the log x cubed over log t squared. When you put the two thirds in here and you go through the calculation over here, let's see, what do we have? We're going to end up with a two thirds here, a two thirds here, and then equating log t with log x over log t to the two thirds. So this will be to the five thirds. And so you get log x to the three fifths. So once you've done this, assume this n to the IT estimate as a black box, you can just go ahead and get the strong form of the prime number theorem. And if you do it a little more precisely, you get a log log x to the one fifth in the denominator. So you get exactly vinaigrette of Karabov. So obviously the meat lies in this estimate for proving the strongest form of the prime number theorem. This is not something easy to prove, but it doesn't use analytic continuations, a zeta function to the left of one. It does achieve our goal of getting the strongest form of the prime number theorem without venturing to the left of one. Excuse me? How do you know about the zeta function? Oh, well, okay, one over log. Let me not get into that. So, I mean, it's the first step, though you need a little more. You can get your lower bound on zeta simply by zeta sigma cubed, zeta sigma plus IT to the four, zeta sigma, again, used Maten's identity. And so since we have upper bounds on zeta to the right of one, we have a lower bound on zeta as a consequence. And I think this is quite sharp enough. So you've got to, if you start doing it with removing the small prime factors, it all works out very nicely. Okay. So let me go back to the distance function. And yeah, so, I mean, in some sense, it's a little disappointing. So Demetrius found this brilliant argument. It gives you, if you like, a pretentious proof of not only the prime number theorem, but the strongest known form of the prime number theorem, lots of innovation, lots of clever ideas, and you end up with the same result. I mean, it's such a weird result, log x, the three-fifths, log log x, the one-fifth. Well, the reason why is obvious because the deep idea that's needed to get there is the same deep idea. It sort of tells you something about, I mean, I haven't really said this before, but let me now be a little negative about what the history of working on prime number theorem from the perspective of Riemann's zeta function is. What is it that we're really doing when we prove zero free regions in the classical way? So you could argue that, and especially if you look at the proofs, you could certainly argue that here's the one line. Here's the zero free region for the Riemann's zeta function that we know. Here's the half line over here. Demetrius' proof gives the same zero free region as I pointed out. It's easy to do from this proof. Here we've sort of avoided going across the one line, although we can note that you get the same zero free region. But what are the classical proofs is, in a sense, what they do is they say we know the zeta function so well to the right of one that in order to know it just to the left of one, we can surely extrapolate. And we have to really clever to get as far as we can go. That's roughly what happens. So sadly, most stuff about zero free regions, and I hope I don't get shouted at by Henrik, is about extrapolation to what happens to the right of one. So there is some logic to never going to the left of one, because most of the consequences are more elementary statements. Okay, I guess though at the end of a day, this new approach, if you like, which isn't so new, but writing it as one whole thing is perhaps new. And the classical approaches are really equivalent in many, many ways. But there are some advantages to the new approach, which I'm going to try and point out. So let's go back to defining the distance function. I'll do it over here. This doesn't look like mine. So typically, the distance function I'm going to be interested in. I gave you a more general definition. So the f and g, f and g are multiplicative. They live inside the unit circle, and we'll define the distance between f and g with respect to a limit x is, the square of that is the sum over the primes up to x of one minus the real part of f gp over p. So in the proof of the prime number, oh, maybe I'll go a slightly different direction. So in what we talked about yesterday is, part today, is that the meter, the prime number theorem lies in integrating zeta prime s over zeta s. And understanding that as we, well, times x to the s over s and understanding that. And as we just saw, working with zeta prime s is not that difficult. I mean, it's a pretty well understood function. It's just the sum of log n over n to the s. It's not something to fear, especially to the right of one. So the element of fear here is not zeta prime s. It's not x to the s. It's not s. It's the one over zeta s. So you can actually show that getting good estimates to that is equivalent to getting good estimates to one over zeta s. There are elementary ways to do this, but let me, let me not get into that. One over zeta s x to the s over s ds. And then if we use Perron's formula again, well, for one over two i pi, which I've missed, but you end up with the sum of the Mobius function up to x. So in fact, the prime number theorem is equivalent to, it's the same statement as that the sum of the Mobius function is little o x. So square free integers are as likely to have an even number of prime factors as an odd number of prime factors. And in fact, the bound on mu of n you get is exactly the bound we've been doing with the prime number theorem. I mean, at all, analytically, much the same arguments. Now, unlike mu of n, because mu of n is the multiplicative function, and that leads us into a whole different field of things we can work with. And so a question one might ask is not just when is mu, the mean value of mu, or the sum, and one way to interpret that is the mean value of mu of n is little o one. It tends to zero. But you can say, you can ask that in a larger context, which is when is the mean value of any multiplicative function going to zero? So can you name your multiplicative function whose mean value is not zero that lives inside the unit circle? One, enter the it. Thank you, Heidi. Okay, so one is a good one. That doesn't have mean value zero, but enter the it is a great one. Okay, forget everything I said before about the difficult bit at the start because it's not relevant. This is more or less, oh, we want the mean value, so one over x. This is more over at less one over x times the integral of u to the it dt between zero and x. And this is this is just equal to one over x times x to the one plus it over one plus it. And then when I divide by x, you get this. So for those of you who are taking notes, I'll give you a second to stare at that because it's something worth thinking through. Thank you. It's also worth getting correct. I didn't heckle you very much. Okay, so this is kind of a cool mean value because look, if I take the absolute value of x, the it over one plus it, it is what one square root of one plus t squared. It's some number, right? So when I take the mean value of n to the it, it doesn't tend to zero because if I take the absolute value, it tends to some constant bigger than zero. But what's beautiful here is it doesn't tend to anything, the mean value, right? X i t as x grows just goes around and around the circle. So we're in a little bit of a strange place with mean values, multiplicative functions in the unit circle because they don't necessarily even tend to a limit. Anyway, this is another example, class of examples. This is for any real t that the mean value does not go to zero. Can anybody give me any other example? The answer is yes. So what if I took one, so one is obviously a class from that class, take t equals zero, but instead I took f of n to be, f of p to be one on every prime except 17. And on 17, I took f of p to, f of 17 to be a half. So when I take the mean value of this thing, it's going to be some constant. I don't know what the constant is, but it's clearly not going to be zero. It's going to be something, all the terms are going to be of a decent size, or almost all the terms. Mew squared. Yeah, another one, nice one. So it's not hard to find other examples where the mean value does not go to zero, but the mean value, but those examples are, take the nice, pretty examples and mess with them. That's what it boils down to. So what is the limit of how much you can mess with the pretty examples of n to the i t? And this is what, well, we'll see the distance coming in in a minute, but you can ask, are there any other examples other than the n to the i t's and the mess of them a little bit, but doesn't really change, it just changes the constant. Are there any other examples? And the great theorem of Hollass is no, there are no other examples. So Hollass, this isn't going to obey, is it? Will you stay there? It's exciting. So Hollass's theorem says, if Hollass, I guess it's the vibration set it off, Hollass's theorem states that if, so f n multiplicative, f n less than or equal to 1, if the mean value of f of n does not tend to zero, then f is pretty close to being an n to the i t. And what does pretty close mean? Then there exists t real, actually t bounded even, such that the distance between f and n to the i t, going all the way out to infinity, is bounded. So less and less than 1 and less infinity are synonymous. So this is what we mean pretentiousness. For f pretends to be n to the i t, I mean maybe it is n to the i t, maybe it's close to, you know, we could have used the word is approximately, but that's used so often. So pretentiousness. There was something that was more surprising when we initially coined the phrase of two things, really had no right to be the same. Okay, so that is what happens for general multiplicative functions. And the question in the prime number theorem is for mu of n. So we said that the prime number theorem is p n t false would be the same thing as the sum of mu n does not go to zero, which by Holast's theorem implies that mu pretends to be n to the i t, which as we said last time, from the triangle inequality, this tells us that we can just take the squares here. So mu squared is 1, n to the 2 i t pretends to be 1. And that would imply, for instance, well, there's many ways to prove that's nonsense. But one way to prove it is just to prove that would say, data has a pole at 1 plus 2 i t. So Holast's theorem, if you like, fits in to the proof of the prime number theorem very nicely, the classical proof. But here's the great thing. When you're dealing with the classical proof of the prime number theorem, you're using zeta and its properties, its analytic continuation, that it's analytic everywhere, that there's a real structure going on there. With Holast, you don't have to worry about structure. Just any multiplicative function, the same notion of pretentiousness, is all that's important. So it allows you, in the same class of kind of questions, to generalize vastly from coefficients of a Dirichlet series that has Taylor series, essentially, to Dirichlet series that don't necessarily have to be well behaved. And that ends up being an enormous difference in what you can hope to prove. So let me say a little bit more about this distance function. So let's have a look at this Dirichlet series. So let's suppose that f of n, you know, I said last time that when we do a problem where we're counting things up to x, we're not really concerned about the value of the function at primes beyond x. So when we're looking at the sum of a n and up to x, we can play with the values of a n for n greater than x. So let me assume that f of p to the k is 0 if p is greater than x. And what I want to look at is f of s, so the sum of f n over n to the s. So now it's a multiplicative function, not like here where we were taking an arbitrary Dirichlet series. And we're looking at the product of 1 plus f p over p to the s plus f p squared over p to the 2s, et cetera. But this is only over the primes up to x. So that gives us something nice and finite to play with. And where we do our integration, where we're working, we're having the real s is c plus i t and c is greater than 1. So if I want to understand the size of this, just like in Sam's lecture, the terms after the first one are all like 1 over p squared. They converge. So up to a constant, they're not relevant to the size. So up to a constant, this looks like exponential of the sum p less than or equal to x of f p over p to the s. And I've written this as p to the c plus i t. So it looks like that. So that's really just f p over p to the s rewritten. And now remember c is awfully close to 1, so I'm going to cheat a bit and say it's 1, but actually it's a cheat that's allowed because it's 1 plus 1 over log x. Again, it changes things only by a constant. And now this is a bit like when we did the distance between zeta s and zeta s0. I'm going to add in, if you like, a s0 term. So what I'm going to look at is, actually I'm going to do something weird here, start off with. So I'll put this as a minus and a minus. And now, finally, what I'm going to do is I'm going to add another term, which is I'm going to add 1 minus f p over p to the p to the minus i t over p. So this extra term I've added is a minus sum of 1 over p, which is log log x. I'm going to exponentiate it. It's a log x. So I want to put a log x in there. So this is up to a constant. So whatever. You can see where I'm headed, I hope. And so now let's take absolute values of f of s. So f at c plus i t up to a constant is when we take absolute values here, what happens to absolute values? We take the real part, x of the real part of this. Real part of 1 is easy, real part of a second, but it's not so easy. But anyway, this is log x times exponential of 1 minus the real part of f p p to the i t. But that's exactly this. Okay, so this very much tells you that this distance function is got to be relevant to understanding these integrals, this Perron's formula. So when we look at Perron's formula, we're integrating up this line just to the right of 1 of f of s x the s over s. And as we said, our approach here is just to take absolute values in the most moronic way to try and understand. We'll put some bits in to make it a little less moronic, but the key is going to be understanding the size of f as we go up that line. And the size of f, well, here we have a very good guide to it. So what's the maximum the size of f could be, say, up to height t? It's going to be reflected by the minimum that the distance can be. But that's much like what happens in whole s's there, because that's talking about minimum distance, right? There exists an n. There exists a t such that, so this will be the minimum distance, if you like. So actually, I just realized, looking at the audience, that I should say about Cuclops's proof that if you've looked at the book of Van Janssen-Koelsky, since they're both here, I need to say this. And it was certainly the inspiration. If you look at it, there's an elementary proof of the prime number theorem, or sort of a bit like this, sort of elementary. I mean, it uses ideas of a zeta function. And what you'll see, there are many elements of what I talked about in Cuclops's proof. And in fact, the origin of his proof was when he was a postdoc. He's now a professor at the University of Montreal. But when he was a postdoc, we exactly talked about combining Van Janssen-Koelsky with pretentious ideas. I mean, they were pretentious, but not pretentious enough is what it really boiled down to. He was more pretentious in here. OK, so this is Hollasser's theorem, and it's central to this subject. And as I promise you, I want to give you Harper's proof of Hollasser's theorem, and or slightly edited by Sound and I. Oh, I'm so lost with these boards. Will I ever get this down? I think they need to install a button for stupid people at me so that it rotates in a convenient way. You have to figure it out. Always look so effortless when you do it. OK, so, yeah, let me be a little more precise then about this theorem. So m f x t will be the maximum with t less than or equal to t of f c plus i t. So the maximum that the Dirichlet series gets in absolute value, where we're considering the integral. And as we said, this is like exponential of minus log x times, sorry, log x times e to the minus the minimum, t less than or equal to t, the distance squared from f to n to the i t, up to a constant. So, OK, so that's the connection. It'll be useful to have that. And now let me, I've given you sort of the vague version of Hollasser's theorem. Let me give you the less vague version of Hollasser's theorem, which is the uniform upper bound on f. Now again, if you look over there, you've got, if you look up there, the sum, take a n as f of n. You've got the Dirichlet series x to the s over s. So the x to the s looks like x. OK, that's expected. The 1 over s is probably benign. So the main thing is the value of a Dirichlet series, the biggest it can get. And Hollasser's theorem confirms this in that he actually proves that the sum n less than or equal to x of f of n is less than, less than this maximum that the Dirichlet series can be times 1 plus logarithm of that. So I'll explain that in a minute. And here we have to take some like t greater than or equal to log x. There are other versions of its theorem, but let me just state it this way. So it's an explicit way of doing the same thing. What I should have said about, something's wrong with this. I'm sorry, I meant to, I'm sorry, my notes not brilliantly written, which isn't the help to you. But let me do, it's natural to divide through by log x here. OK, that's what I should have done. And then this is less than, less than 1, this maximum. So it's a maximum Dirichlet series divided by log x. Log x is the maximum it could be. Well, you can see that because that's e to the minus the square of a real number. So it's bounded this maximum. And then log, I guess I should put absolute value since log is going to be negative here. So if you like, suppose this is 1 over 1,000, because it's less than 1, this is pretty small, it's the 1 over 1,000 you care about. So the focus is on this, that you get the maximum of the Dirichlet series divided by log. There's a little bit extra, which you can concoct an example to get this little bit extra. I probably didn't have, I did plan to do it, but I possibly didn't have time to do it. Maybe I should tell you some nice examples of, of Paul-Assisterum. I guess I only have 10 minutes, so it'll be limited what I'm going to do. So, so perhaps the most, so you've sort of got these natural examples of, of multiplicative functions that have been worked out in literature like mu n and n to the it, but there's one that's particularly lovely. And it generalizes mu n, but in a surprising way. So mu n is the multiplicative function where fp is minus 1 in all the primes. I'm not worrying much about the p-squads because they tend to only affect the constant, they don't really affect the meat of the issue. So a generalization of this would be make fp be alpha on all of the primes. Okay, so nice generalization. And if you do that, well, what Dirichlet series do you want to work with? So I said that working, that what happens on the prime squared, prime cubed, doesn't really affect things very much. So you want to pick, they're easier Dirichlet series as possible. What Dirichlet series is really easy in the sense that we are familiar with it, which is, has every fp equals alpha zeta to the alpha. So there we are. If f of s is zeta s to the alpha, then every fp equals alpha, right? So this is the subject of, I think it was originally worked out by Selberg in his referees report on Satay's paper. And this gives you, which was subsequently published, I should say. So that the mean value of this f of n, where it's alpha on every prime, looks like, and I'm not going to give you every single detail, but I'll give you the important bit. c alpha over gamma of alpha plus little o one times x log x to the alpha minus one. Huh? Oh, okay. Yeah. And c alpha is benign. Benign means well behaved, doesn't worry us. So this, okay, so what? Well, let's just verify it in cases we know. Alpha equals one, okay, that's pretty good, log x to the zero, so one and times some constant. We know the constant's one, that's easy. Alpha equals zero, the multiplicative function that's zero everywhere. This term's only log x to the minus one, that's not really telling us, because we know the mean value of the multiplicative function zero is actually zero. So this has to go to zero, but gamma of zero is a pole of gamma. In fact, what do we know about gamma? Gamma's meromorphic and it has its poles at alpha equals zero minus one, minus two, etc. So this also accounts for the mu, the prime number theorem, right? If we take f is minus one, f of p is minus one. We also get that this is zero and this thing's irrelevant. So you've got a sort of balance here. At the very special points where f of n is zero minus one, actually minus two, that's outside the range I've been talking about, you do get the mean value zero. At every other complex number, it's a non-zero, I mean it's something non-zero times some particular power of log. Let's see how that compares. How big can zeta be? Yeah, so if you compare this, you've got the zeta s to the alpha, okay? And that's going, so then what we want to do is know how big is this function m. So it's going to be zeta s to the alpha. How big can that be? Remember this is zeta s where we truncate it at x. So that can be as big as log x and then to the alpha over log x. So Holast's theorem gives you the log x to the alpha minus one in the upper bound and there's just some log log extra that it gives, which we won't worry about. So Holast's theorem really does well in Selberg's beautiful example. So, well, if you look at our book at some point, you will find that you can concoct an example of log log by playing with the example where alpha equals i and changing f on the big primes. Anyway, I won't get into that here. There is an interesting question which says, when is Holast in the right ballpark? When does it give you an upper bound of the right ballpark? Because we know two examples where it's nowhere close to the right answer. The mean value is much smaller. If f is always zero, the mean value is much smaller, much more interestingly. If f of p is always minus one, the case of the Mobius function, the mean value is always much smaller than the Holast prediction. So if f is in the unit circle, is there some way to predict when this kind of fails up here? And another achievement of Cuclopolis, which I call the Cuclopolis, I'm going to have to be a bit vague because it's somewhat technical, converse theorem. So inside the unit circle, so if you like alpha, obviously f's real part greater than or equal to minus one. So the size of this thing is bigger than log x to the minus two. So what it says is something like the following. If consistently, uniformly, you have something like the mean value is less than, I don't know what the latest is, is it log x to the two plus epsilon or three or two? Two, exactly two? Two plus epsilon. Fantastic. Almost as good as possible. Then, f pretends in a certain sense, it's very similar to that, to be either zero, at least on average, or mu n n to the it. So, yeah, the only way you can sort of get a mean value of multiplicative function to consistently not be close to the right answer in Halas is in these very special cases that we already know about. Okay? So, anyway, let me not go further on that. See if everything I wanted to cover is there. It's ridiculously optimistic. Okay, maybe I'll just finish with talking about the new proof of the triangle inequality for this distance function. So, may come as a great shock to you that Terry Tao emailed me last night with a nicer proof than we had. So, if these guys lived on the unit circle, f and g, then if you look at f minus g, f of p minus g of p squared in absolute value, then what are you going to get? You're going to get a one from the absolute value of f p squared, a one from the absolute value of g p squared, and then minus twice the real part of f p g conjugate. So, twice one minus real part of f g conjugate. Okay? So, when it's on the unit circle, you can convert this into some sort of regular norm. Okay? So, we could define, oh, as far as that goes, I'll move blackboard in a second, f minus g to be the sum over p up to x of squared to be one of f p minus g p. Well, it is this. Okay. Okay. So, that's exactly the same thing as the distance function. So, the norm of f minus g is the same as the distance function. And now, we know we're comfortable with such norms, but we're only playing on the unit circle. And we know with such a norm that you have the triangle inequality, g minus h. And these are just for things on the unit circle. But, of course, with norms in this sort of, you know, Euclidean type norm, you have linearity. So, Terry's nice idea, which nobody had seen before, was that if you do have a multiplicative function with a value in the unit circle at the prime p, then why not treat this as the average of, say, two multiplicative functions that have values such that that's the mean? Okay? So, you do this at each p, f p, g p and h p. You just average. You can call this a probability argument whatever you want. You average over them all. And then you do exactly this. You just add these all together. You average over them. And that ends up giving the triangle inequality for the distance function. Is that reasonably accurate description of your argument? Yeah. Yeah. Okay. Well, I'll finish with Terry's argument. And, where did you succeed? The problem is in 1972 or so. Right. There is a, I mean, I'm not really talking about, there is a rich history of, of this. You can remember. Heidi certainly has some papers involved in that. There's some beautiful articles. I guess, I'm really focusing on where it's taking us right now, which is, you know, it's always what you want something 30 years on as found its niche again. 50 years on. Yeah, I guess 1970s, 1670s. 69 far less. Is there an inversing of 67? 63. Okay. So, on the 50th anniversary of versings there.