 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory, and will be about the prime number theorem. So this is the fundamental theorem of analytic number theory. And what it says is the number of primes less than x, which is usually done by pi of x, is asymptotic to x over log of x. So this is the number of primes less than x. And asymptotic means the ratio tends to one as x tends to infinity. So if x is very large, say about a billion or so, it means the number of primes less than a billion is going to be very close to a billion over log of a billion, which means if you pick a random number between one and a billion, the chance of it being prime is about one over log of a billion. The prime number theorem was first proved in about 1896 by Ademar and de la Ballet-Poussa. They gave independent proofs of it, and they both used ideas originated by Bernard Riemann. And what I'm going to do in this lecture and the next one is give a sort of rough sketch of the main ideas behind that proof. There are several technical complications involving complex analysis that I'm not going to go through in detail. Requires more complex analysis than this course really covers. In particular, in the more than a century since that proof was found, there've been several improvements in particular. There were some improvements found by Neumann and Don Zagie, which I'm going to incorporate. There are several other proofs of this. So Ademar and de la Ballet-Poussa's theorem used complex analysis. And if you're a number theorist, this is kind of annoying because results in number theory really should be proved just using number theory, not using complex numbers. And for many years, it was an open question whether you could find an elementary proof. And in 1948, Selberg actually found an elementary proof of the prime number theorem. His proof uses Selberg's identity, which is the sort of variation of what I call Selberg's identity in the previous lecture. Unfortunately, Selberg's proof was marred by a rather unfortunate priority dispute with Erdisch. What seems to have happened is that Selberg was giving some lectures on his work on finding an elementary proof just before he had, I mean, after he'd found Selberg's identity, but just before he'd figured out how to use this to prove the prime number theorem. And someone in the audience passed notes of Selberg's lecture onto Erdisch, who an Erdisch and Selberg both noticed that you could then use Selberg's identity to prove the prime number theorem. So my feeling about this is that Selberg found a proof of the prime number theorem and Erdisch found a rather minor variation of it using Selberg's ideas. So my own feeling is that Selberg is that person who should be credited with the proof. Anyway, so in the first two lectures of this course, I mentioned you could find some easy opera and lower bounds for Pi of X. So we have an upper bound. You can find an upper bound for Pi of X using the fact that the product over the primes between N and 2N is at most the binomial coefficient, which is at most 2 to the 2N. So this gives you an upper bound for Pi of X, which if you work out the details, it looks like something like Pi of X is at most about two times X over log of X. On the other hand, you can find a lower bound. Again, here use the product over all prime powers that are less than or equal to 2N or P is at least this binomial coefficient. And using this, you can find a lower bound on the number of primes and you get a lower bound of the one Pi X is greater than or equal to about a half times X over log of X. So if you look at these, it's quite easy to get the prime number theorem up to a factor of about two. And with a bit more effort, you can get this two down to a number slightly bigger than one and you can get this half up to a number slightly less than one. So however, it turns out to be very difficult to get this, to show that you can make this constant as close to one as you like and the same for this constant, which is what the prime number theorem says. So what I'll do is I'll recall the proof of the prime number theorem, sketch the proof of the prime number theorem. So this uses the Riemann-Zeta function, a zeta of S, which is one over one to the S plus one over two to the S and so on. And it was Riemann who discovered you could use the Riemann-Zeta function to estimate the number of primes less than X. Actually, you don't really use the Riemann-Zeta function. What you do is you use the log, you take the log of the Riemann-Zeta function and then take the derivative of that, which is called the logarithmic derivative of the Riemann-Zeta function and it's this function here. And as I mentioned in the previous lecture, this can be written as lambda one over one to the S plus lambda two over two to the S and so on, where lambda of P to the N is equal to the logarithm of P for all prime powers and lambda N is zero otherwise. So you can see that this lambda function is very nearly counting primes. It's counting primes except that you are not quite counting primes, you're counting prime powers and you're also weighting things by a factor of log of P. And the idea is you can estimate the following function. You define psi of N to be lambda one plus lambda two and so on all the way up to plus lambda of N. And so for example, if we take psi of 10, this is equal to log of two plus log of three plus log of two, again this comes because four is two squared plus log of five plus log of seven plus log of two again coming from two cubed plus log of three coming from three squared. And if I worked this out right, this number is about 7.83 and if you look at psi of 20, then you have to take this and you have to add log of 11 plus log of 13 plus log of two again, that comes from two to the four plus log of 17 plus log of 19. And if I work this out, this is about 19. And the idea is to show that psi of N is roughly equal to N more precisely, the ratio of psi of N and N should tend to one as N tends to infinity. And this would turn out to be more or less equivalent to the prime number theorem. So here are the steps of the proof of the prime number theorem. So step one, zeta of S, first you show that zeta of S has no zeros with the real part of S greater than or equal to one. So this is the key step. So you remember I mentioned every proof as a sort of key step plus a lot of routine details. And this is the key step. And once you've done that, well, calling the details routine is a, they're rather more difficult than being completely routine, but this is really the central step of the prime number theorem. So why should this be important? Well, Riemann showed that you can write pi of X as a sum over zeros of the Riemann zeta function. Actually, it didn't really sum over zeros of the Riemann zeta function. What he summed over was poles of zeta prime S over zeta of S. But obviously a zero zeta of S is a pole of zeta prime of S over S. So if you look at his proof, he was actually using the logarithmic derivative of the zeta function. And there's a pole at S equals one, which gives you the main term, which is roughly X over log of X. Actually, it's really something more complicated called the logarithm, the integral of X, but we won't worry about that X over log of X is good enough. So, and then you get plus other terms coming from the other poles of zeta prime over zeta. And suppose, so these come from numbers S such that the zeta of S is equal to zero. And suppose S is equal to sigma plus IT, where this is real and this is imaginary. This is sort of traditional notation. Why people use sigma plus IT? I don't know, but all number theorists do so. So this is the real part and the size of the error depends on the size of sigma. The T sort of gives you a sort of oscillating contribution, which doesn't really matter too much. Now, if sigma was greater than one, this, these terms would be much bigger than X over log of X. So things would break down completely. Fortunately, it's easy to show the Riemann zeta function as no zeros bigger than S equals one. If sigma is equal to one, the terms are about equal to X over log of X in size very roughly. And the trouble is if you are having other error terms that were the same size as X over log of X, this would kind of mess up the idea that pi of X is approximately equal to X over log of X. So zeros with real part one, if there were any, then the error terms would be as large as the main estimate for pi of X and you wouldn't be able to give an asymptotic formula in the prime number theorem. And it turns out that zeros with sigma less than one are giving you smaller error terms and there's a bit of a complication because there might be an infinite number of terms with sigma slightly less than one, but that's a complication that we're going to kind of brush over. So this obviously is the question, where exactly are the zeros? And the most notorious open question in mathematics is the Riemann hypothesis, which says that all the non-trivial zeros should have real part sigma actually equal to a half, which is in some sense the best possible. And this would give us the best possible estimates on the number of primes less than X. So that's the key step. The second step is something called Neumann's Talberian theorem. Now, this is a theorem from complex analysis, which says that if you've got a reasonably nice function and the integral from one to infinity of f of X, X to the minus S, the X converges for the real part of S greater than one, and if you can extend it through a function that is holomorphic for the real part of S equal to one, then the integral converges for S equals one. So Talberian theorems are kind of theorems that tell you that something actually does converge when you hope it will converge. This theorem I'm going to skip the proof of because it involves some slight, well, involves some ideas from complex analysis. I'll try and put a link in the video description to a short proof of this by Zagia, if you want to see the details. And then step three is we use Neumann's theorem to show the integral from one to infinity of psi of X minus X over X squared, dx, we show this converges. And we're going to show it converges by using Neumann's Talberian theorem and in order to use Newton's Talberian theorem, we need to know that something is actually holomorphic for the real part of S equal to one. And this is where we use the fact that the Riemann Zeta functions know zeros with real part one because if it had zeros then the function Zeta prime over Zeta would not be holomorphic for this condition. So this is where we use step one. So once we've proved this, the remaining steps are rather easy. So step four, we deduce from step three that psi of X is asymptotic to X. So, and going from step four to step five, so step five shows that pi of X is then asymptotic to X over log of X. So going from step three to step four and step four to step five are really rather easy. And the hard parts of this proof are steps one and steps two. So I just recall some facts about Dirichlet series. If we've got a Dirichlet series A one over one to the S plus A two over two to the S and so on then we can ask, where does it converge? Well, if it converges for S equals S naught, some real number, then it also converges for S is greater than S zero. And more generally, it converges whenever the real part of S is greater than the real part of S zero. So what this means is that in the complex plane, we can find some vertical line here such that the series converges in this region here and it does not converge to the left of this line. We might also get extreme cases where it converges for all complex numbers or doesn't converge for all complex numbers. But except in those cases, there's always some line such that the series converges whenever the real part is for the right of this line. In other words, larger than this number, the real part of S zero diverges to the left of it. You notice this is actually sort of similar to power series. You remember that if you've got power series say centered at the origin, then there is, you can find some circle such that the power series converges inside the circle and diverges outside the circle. And when you're on the circle, maybe it converges and maybe it doesn't, it's kind of hard to tell. And again, the circle may be degenerate. It may be a single point in the power series might converge nowhere or it might be sort of an infinite circle and the power series converges everywhere. So convergence of Dirichlet series is rather like convergence of power series except that instead of converging inside a circle, it converges in a half plane. By the way, if a power series, if a power, the place where a power series converges in a circle is the circle of convergence is much the same as the circle of absolute convergence. In other words, if a power series is absolutely convergence in an open disk, then it's actually convergent in the unit disk. For Dirichlet series, things are slightly more complicated. For instance, if you take the series one over one to the S minus one over two to the S plus one over three to the S and so on, it converges provided the real part of S is greater than zero. But if you take the absolute value of its coefficients, you get one over one to the S plus one over two to the S and so on. And this converges whenever the real part of S is greater than one. So the half plane of convergence is no longer the same as the half plane of absolute convergence. The proof of these statements about convergence are not that difficult, but I'm going to skip them on grounds they're complex analysis rather than number theory. Next, we need to know where are the zeros of the Riemann-Zeta function? So the Riemann-Zeta function is one over one to the S plus one over two to the S plus one over three to the S and so on. And you remember this can be written as an Euler product, product over all primes of one over one minus P to the minus S. And now we notice that Zeta of S is none zero if, first of all, all factors are none zero. And secondly, the product converges through a none zero value. So obviously it's rather difficult to tell if you've got a function written as a sum of things it's pretty hard to tell when it's zero. But if it's written as a product of things, this is a lot easier because this is obviously going to be something to do with where the zeros of the product are. Well, provided the product converges, so you can get a product of things like one times a half times a third and so on and all the terms of this product are none zero, but if you take the product of all these, you're obviously going to get something zero. So it's not enough to check that the product converges, you've got to check that it converges to something none zero. And an easy way to do this is to take its logarithm. So we look at the logarithm of the product over all primes of one minus P to the minus S. Since we're taking logs, they turn reciprocals into taking minus signs. So if this converges, so if this converges, well, so we can write this as a sum of logarithms of one minus P to the minus S. And if this sum converges to something, then the product is none zero because if the sum converges to something finite then the product will be X of that finite thing and X was something finite is none zero. So what we want to do is to study the convergence of this sum here. And we know that logarithm of one minus P to the minus S is approximately P to the minus S in absolute value. So you remember log of one minus X minus log of one minus X is X plus X squared over two plus X cubed over three and if X is small this will be approximately X. So we want to know, so we want to know if the sum of P to the minus S converges. And this converges if the real part of S is greater than one. You can see that fairly easily because here this is a sum over all primes but if you take a sum over all integers, sum over N of one over N to the S this converges if the real part of S is greater than one. So if the sum over all integers converges the sum over all primes converges. So what we conclude from this is that the Riemann Zeta function is none zero for S is greater than one because of the Euler product. And what we want is to know that Zeta of S is none zero for, so it should be the real part of S greater than one for the real part of S equals one. So you can see this is right on the boundary of the region where we can prove it's none zero. So what's going on is, so we've got a complex plane here. Here we've got the Riemann Zeta function and here we've got the line with the real part of S is equal to one. And in this region here, there are no zeros because the product converges. And what we want to know is, are there zeros here? So there are no zeros in this half plane, but we want to know whether or not there are zeros on the boundary. And this turns out to be sort of very tricky to prove. Anyway, the other zeros, by the way, are in this line here. So this is called the critical line. So this is the line with real part of half. And it's known that definitely are some zeros on this line. And there are also a few zeros of the Riemann Zeta function at minus two and minus four. And so they're called the trivial zeros and everybody mostly ignores them. So we would also like to know that the Riemann Zeta function is actually defined for all complex numbers because the series for it only converges in this region. So it's not even clear that it's defined for the real part being equal to one. So how do we define the Riemann Zeta function in this region here? So what we can do, there's an easy way to do this. What we can do is we can look at the series one over one to the s minus one over two to the s plus one over three to the s and so on. And now this converges for the real part of s is greater than zero. If s is real, you can see it converges because it's just an alternating series. If s is complex, it takes a little bit more work to show that it converges. But again, I'm going to skip that. And what you notice is that this is equal one over one to the s plus one over two to the s plus one over three to the s and so on, which is minus two over times one over two to the s plus one over four to the s plus one over six to the s and so on, which is equal to Zeta of s minus, minus two Zeta of s over two to the s. So we find that one minus two over two to the s times Zeta of s is given by this series here, one over one to the s minus one over two to the s and so on. So this is defined for the real part of s is greater than zero. So this almost gives us a definition of the Riemann-Zeta function for s greater than zero, provided this thing here is non-zero. Well, this does actually have zeros. It has zeros whenever two to the s minus one is equal to one. And there are some points with real part s equals one where this holds. So this doesn't quite define Zeta of s for all numbers with real part equal to one because of the problem with this. But then we can do a different trick. We can look at Zeta of s minus three over three to the s times Zeta of s and do the same trick and show that this is also defined for real part of s greater than zero. And now we just use the fact that two to the s minus one equals one and two and three to the s minus one equal one, the only common solution is s equals one. So the first definition shows that Zeta of s is defined whenever two to the s minus one is not equal to one and real part of s is greater than zero. And this variation using three instead of two gives the same value for that. And altogether this shows that Zeta of s is defined for all numbers with real part s greater than zero except for s equals one. For s equals one, Zeta of s really does become infinite. So that gives the analytic continuation of Zeta of s to real part of s greater than zero. By extending this argument, you can actually show that Zeta of s can be defined for all complex s except for s equals one, but we don't actually need that. So I'm going to omit it. So that's a summary of the basic background for the proof of the prime number theorem. So what I'm going to do next lecture is to go through steps one to five in a little bit more detail and actually sort of sketch the proof of the prime number theorem.