 So, that is the course all about Riemann hypothesis. We will mostly be looking at one particular function. This is the Riemann's zeta function. The number s is a complex number and this is an infinite sum which has many interesting properties and the course as I said will be devoted to understanding some of those. You will also look at once we have a good grasp of this function and now its connections with many interesting objects like number of primes up to a certain range. Then we will look at generalizations of this function in different domains and see what they say about those different domains. And finally, we will also look at this. The title says some applications of the zeta function or I should say the zeta function the applications of Riemann hypothesis. I have not talked about what Riemann hypothesis is which I will do later. So, the hypothesis pertains to some property of the zeta function and the hypothesis also has many consequences in various part of mathematics and computer science. So, we will look at some of those consequences. A very quick outline some of you who read that PDF file which I attached would know this. We will start with complex analysis. Now, I am sure all of you have done some course on complex analysis. Is that a correct assumption? And many of you have done it long time ago. So, I am guessing that many of you will not remember much beyond the definition of complex numbers. Is that a right assumption? No? You know more? That is good because then I have to spend less time here. What I plan to do is to start with the definition of complex number and develop the necessary tools. Not necessarily all with the proofs, but let us give you a feeling of those tools in complex analysis that we will be using in our analysis of this function. Fortunately, it turns out that we do not really need too many fancy tools. So, it is going to be not too bad, but it is going to take up some time of this course. And that time is very contingent upon how much you know or you remember about complex analysis. Once we have the basic background, we will look at well, I have already defined zeta function, but zeta function prime numbers. So, that is a very critical connection that how the zeta function started its life when Riemann was trying to understand the distribution of prime numbers. So, we will show exactly what Riemann discovered. This was a quite a remarkable piece of work by Riemann. I will get into details of that later. So, let us just stop it here. And then, so Riemann also made a conjecture about the nature of zeta function when he did this study. And that conjecture has still unproven after more than 150 years. And there are a number of mathematicians who have tried to prove it, but failed. And it is considered to be the greatest unsolved problem in mathematics. There also happens to be a one million dollar price for anybody who proves this hypothesis. Then we will prove prime number theorem. This is essentially follows from a weaker version of Riemann hypothesis that one can prove. And from that follows the prime number theorem. Let us see how many of you know what prime number theorem is. Number of primes exactly. Number of primes this theorem says number of primes less than equal to x is asymptotically x over log x. So, this we can prove. In fact, this was proved more than 100 years ago building on works of Riemann. So, once we have done with all these studies, we will move on to the generalizations of Riemann hypothesis which really come fall out of generalization of zeta function to other domains. And just as Riemann formulated his hypothesis for zeta function. For those generalized zeta function, one can formulate a similar hypothesis which is which are called generalized Riemann hypothesis or extended Riemann hypothesis depending on which domain it is being generalized to. And we will look at some of those variants. There are huge number of variants which we will not have time to go into details of all. So, we will look at some of the important ones. And in particular, we will look at generalizations to higher characters which you do not need to understand right now. Pick curves which again you do not need to understand right now if you do not know. And the maybe one more, but I will just leave it at this point. And finally, we will look at some applications. And since this course has a C S title, I will justify that by looking at some computer science applications here. So, that is the basic outline. Any questions? There is no textbook because it turns out that of course, there are large number of books written about Riemann hypothesis zeta function is connection with prime numbers. But it turns out that mathematicians have a way of writing or expressing their ideas which I cannot quite understand. So, therefore, I will have sort of gone through many of those put everything in my own way and which is what I will present. So, therefore, if there is no book which I will be following chapter by chapter. But there are reference books which basically says that everything that I cover is more or less in these books. Now, what are those? I did give couple of names in the, let us see if I can pull that out. This is the GTM series book 206 Problems in Analytical Number Theory by Ram Murthy. Ram Murthy is one of the leading number theorists in the world. So, he has written couple of really nice books on number theory. Then there is the second book is the Riemann hypothesis. This is available on the web. So, it has fairly easily exists. There is some URL which you can just google up for this. This is by Peter Borevin, Stephen Coy, Brendan Rooney. So, this is the first time I am teaching this course. So, I really do not know how the assignments or exams will evolve because it is very easy to make questions in this subject which are very hard to solve. And if you try to reduce the difficulty, it very quickly becomes very easy to solve. And for examinations one needs questions which are sort of of intermediate difficulty. So, I will make an attempt in form millerating those questions, but it may turn out that they are either too easy or too difficult. So, you will have to just suffer. There is really no other way out of this. So, two exams, Midsum and Sem, one of them will be take home. We will discuss with you which one do you want to be take home. Assignments again, it is assignments can be done very easily here because there are many things which I will not going to details which I will just leave out as assignment problems. So, assignments will be more of a dynamic in nature that during the lectures I will say assignment problem number one, assignment problem number two as we go along. Weightages, assignment, how much weightage do you want to assignments? 20 percent? 40 percentage. Is that ok? That finishes off standard beginning of every course. Any questions on this? So, let us now begin our study of zeta function. I did say that I will first talk about complex analysis, but before jumping into complex analysis let us spend some time in trying to see what this function is about. So, this let us write it down again. Now, this function of course has a parameter S and it is an infinite sum. Quite clearly, if there are some values of parameter for which it does not even converge. For example, what is zeta 1? What? ln 2 this diverges. In fact, you sum up you take S to be 1 and then you sum up for n up to x, you will get about log of x. So, as x goes to infinity this diverges. What about zeta 2? That actually does converge and we will see later on that it converges to as you said pi square by 6. Can anybody prove that it converges to pi square by 6? You know the proof? Yeah, Euler was the first mathematician to prove it. It has Euler proved it actually without complex analysis and when we look at complex analysis we will find some really nice proof of this equation. But anyway it does converge that is the whole point and this anybody can see that it converges or can is that simple to see convergence of this series? Why does it converge? We are going even before complex analysis to math 101, infinite series convergences that is one simple way of doing it. So, basically this is approximately this integral. In fact, by setting it up properly you can say show that this is less than equal to this and this integral is easy to handle this is just minus 1 over x 1 to infinity which is 1. So, zeta 1 diverges whereas, zeta 2 converges. In fact, for all the same analysis we will show that zeta s converges for all s greater than 1 where s is real number. We are not talking about complex number right now the same proof works. So, that gives you a some idea of what this function looks like and of course, for s less than equal to 1 this is going to diverge again talking about on the real line. So, on the real line this nature is the nature is quite clear at a top level that for s greater than 1 it converges for s less than equal to 1 it diverges fine. But the fundamental question is fine this is an infinite series of some form why should it be interesting why should we bother studying this series and as I have been saying already that this has some intimate connection with prime numbers. So, this connection at least one of those connections was discovered by Euler and it is a very simple observation let us call it a theorem just proved by Euler. So, this says that zeta function zeta s is equal to product over all prime numbers p of this quantity 1 minus 1 by p to the s to the minus 1 that is 1 over this quantity proof is fairly straight forward all you need to do is to observe look at start from there and let us write it in terms of prime numbers and to do that let us notice that this sum runs over all integers and by the fundamental theorem arithmetic this goes even before that right it is now I am going to high school I am sure all of you remember fundamental theorem arithmetic that is every number can be uniquely written as product of prime numbers. And so if you write that we get n greater than equal to 1 product prime p e to the i going from 1 to let us say k sub n 1 by p i to the where n is. So, n is written as a product of these primes and just rewriting this. And now let us exchange the product and sum by observing that in this sequence pick a prime particular prime say p the p 1 what all in what all ways does it occur p 1 occurs in every n which is a multiple of p 1 and more over if p 1 to the k divides n and p 1 to the k plus 1 does not divide n then it will occur as p 1 to the k s. So, basically for every prime p and every power k of the prime p p to the k occurs in all multiples of exact multiples of p to the k that is p to the k divides n, but p to the k plus 1 does not divide. So, we can write this as a product over all prime p and then sum over all k greater than equal to 1 or 0. So, notice what I have done this is product over all primes p sum over all exponents k of p and 1 over p to the k s. So, you take this sum and multiply it out over all primes p another way of seeing the same thing is that you take any num pick any set of primes here any finite set of primes in this product pick any for each prime pick any exponent k and you multiply out those primes with those exponents. So, those chosen finitely many primes with the corresponding exponent you get a number and that is a unique number that is by fundamental theorem arithmetic that this number you get because you picked up a unique set of primes and a unique set of a corresponding exponents you get a unique number and that number will be present here. And every such choice will create a unique number and every number n will be created by one such choice. So, that is the equivalence to this there is all one more thing one needs to do when we are doing manipulating infinite series is convergence we have to be very careful about convergence is before we do this. So, turns out this is convergence series again for s greater than 1 as long as s is greater than 1 this each one of this is absolutely convergent you all remember absolutely convergence and therefore, we can do this exchanges. And now it is pretty straight forward this is product over all prime p and this is familiar geometric series which equals 1 over 1 minus p 1 by p to the s. So, that expression connects this zeta function with some form of the sum of the product over prime numbers. In fact, this tells us some very simple facts about prime numbers almost immediately. For example, there are infinitely many primes how do you derive it from this theorem exactly set s equals 1 we know zeta 1 is divergent and for s equal to 1 this product on the right hand side is 1 minus 1 by p to the minus 1 if there are only finitely many primes then this product will not diverge because each term is a finite number and there are finitely many products here. So, it does not diverge, but we have to be little careful this equality only held when s is greater than 1. So, we cannot use this equality for s equal to 1. So, although this seems like a simple and direct proof, but this is not quite correct proof we have to be a little more careful here the same idea works but what one needs to say realize or do here is that we know that for this equality holds for any s greater than 1. So, set s to a number bigger than 1 and send it towards 1 as it tends towards 1 the left hand side keeps getting bigger and bigger and bigger going towards infinity the right hand side does not if there are only finitely many primes. In fact, you can place an absolute upper bound on the right hand side depending on what is the largest prime number and therefore, there exist infinitely many prime. Let us see what more can we get out of this yes currently it is all realize yes we are not it is all for real. So, I should have mentioned that when Euler started the studying of this function him and there were some previous mathematicians also who always thought of this as a function over real numbers the complex numbers were not considered at all until Riemann and that was one of the great insights of Riemann to extend the domain from real to complex. And we will see during as we go along this course the remarkable effect the simple change in perception had on the nature of this function and on everything else. So, what more can we derive out of this can we say something more about prime numbers well let see zeta 1 for x equal to 1 they told us that there are infinitely many primes. So, let us just try out zeta 2. So, we know zeta 2 is bounded certainly what actually is equal to pi square by 6 which is less than 2. So, it is some small number that is all we need to do. So, this is pi square by 6 again the exact number is not important here the only fact right now we will be using is that this is bounded. So, right hand side is bounded clearly left hand side is also bounded there is no problem about equality here because we are using s greater than 1 how can we re look at this product in some way. Let me put it this way suppose how let us consider this product again greater than 1 does this converge. So, how do we decide about convergences of series what that is one way of doing it this yeah this is the first two terms of the cosine series. But that may not be that useful because cosine is always see this is infinite product and one has to do an error estimate and it is not clear how bad that will be one ways to take convert this into some take log right. So, from here go to and look at this convergence of this series. So, this convergence of this will tell us whether this is convergent or not now is this convergent. There are some simple criteria to check for checking convergence says of series one is to check the ratio of two successive terms yes this will this be less than equal to summation 1 by square. The product of infinitely many once is always one still one, but this is also yeah product of all terms greater than 1, but they all are. So, one has to see whether see this terms are also tending towards one. So, whether in the limit the product blows up or may be stays close less than equal to a boundary you think is compare it with summation 1 by n square. So, this bounded by 1 by n square what is log 1 minus x that is 1 x plus x square by 2 plus x cube by 3 and so on. So, this is bigger than 1 by n square. The sequence these are all again tending towards 0 the successive terms get closer and closer to 0 as n goes bigger goes bigger this inside guess closer and closer to 1 and log of that is close to 0. So, let us look at a ratio of two terms we can sort out this does not look very exciting. Let us try this out expand this what is log 1 minus x is x square 1 by n square there are all pluses here. So, is it 1 by x square by 2 or x square by 2 factorial x square by 2. So, this is n to the 2 n to the 4 plus 1 by 3 n to the 6 and so on and this is 1 by n plus 1 whole square plus so on. So, let us multiply out this with n square and this with n plus 1 whole square. So, that we get n plus 1 whole square by n square times 1 plus as n tends to infinity this ratio goes to 1 and this ratio also goes to 1 right and what happens if the ratio of the successive term is going to 1 does this series converge or diverge converge ratio has to be less than 1 then it converges if it is 1 then it diverges. So, this actually product this product actually diverges oh at one you cannot say oh that is that is bad. So, then that means you will have to do more work assign a problem number 1 decide the convergence of this product and hopefully by tomorrow you will have decided and we will continue with our work.