 a very critical role later on. Now, let us move forward. So, this is the L function for different characters which are analogous to zeta function. And like in zeta function, it has a functional equation. That also played a very critical role. So, let us just check out what are the properties of zeta function. Firstly, it has a pole at 1 that played a crucial role. We will call the, when we looked at this poles of zeta prime over zeta, it had a pole at 1. So, that we did a certain residue. That residue was x and then there were other residues. Similarly, we will do the same exercise for L function. So, anyway that is the first thing about zeta function. Second one is this connection between the psi and the zeta function. Integral psi being written as integral of zeta prime over zeta also. Third was this functional equation for zeta function which allowed us to analytically extend it over the entire plane. And what are the properties of zeta function? Let us say that then we looked at actually this functional equation is allowed us to also write this here. zeta function is a product and that was very useful in analyzing the growth rate of zeta function over zeta prime over zeta along the lines vector to be the growth for the order log r. So, these are the basic things that one did for zeta function. We will do the same thing for the other thing. The first thing we have already seen is that we are at z equals 1. Now, let us come to the functional equation which details. All I will say is that you will just mostly follows what we did for zeta function. Now, if you remember how did we arrive at the functional equation? We started with the gamma function. You start with the gamma function. Gamma of integral of integral of integral of integral of zeta. Then we said let us replace t by e to the power of zeta and we said replace t by pi n square u and then gamma z become 0 to infinity. I am just thinking things that do not integral to zeta function. So, we looked at gamma z by 2. So, there we get pi to the power of zeta to be 0 to infinity. Then therefore, we get and now we say sum over all x. That gives a zeta function on the left hand side. So, integral on the right hand side and then we do the analysis on the right hand side. But now, we want to do the same thing for L function. So, up to this point it is the same no difference. But now instead of summing over all x what we do instead do is we will call the definition of the L function. We first multiply it with chi of n and then sum over all x. And then we look at this summation and earlier we played around with the just the summation without chi of n multiplied. But now we have chi of n multiplied. But the analysis of this summation is exactly the same as we did it before. And the fact that chi of n in multiplicative makes everything go through very small. There is one minor difference that we will encounter as we do the analysis. So, to cater for that we will have to do slightly differently for our aspects. So, let me split the characters into two types. What character? These are the only two possibilities for this character, now minus 1 you could either map it to minus 1 or plus 1 with that all is for the multiplicative. The chi of minus 1 whole square is 1. So, chi of minus 1 is 1. So, depending on whether it is minus 1 or plus 1 you can divide it into these two. Now, for one of these, now I think for Aalympic Pudip the question to you. In this do you see a problem for one of these two types? I think for event character this should be fine this is almost a minus 1 for one of these two types this does not quite work out once you start doing the analysis of this sum you will realize yourself. So, I will leave that this is equal to that same 1 minus z ok. So, let us define. So, first define this in the same fashion this one is zai zai zai zai zai zai zai zai zai zai zai zai zai zai exactly as in zeta function we are getting this. So, I here you say zai is equal to. So, this is the function you know they are some new things which are coming here one is these tau of chi which I will define in a moment what this is divided by square cube times of 1 minus a and then this character changes to the chi bar, chi bar is the inverse of kai remember the characters form a group. So, chi bar is simply the inverse of kai. So, the only unexplained thing here is tau, tau phi this is a famous quantity a well known quantity it is called gauss sum for a character. This is equal to I am going from 1 to q of so these gauss sums play a very, very important role when in character analysis in general, but I will not dwell into this just say that the gauss sum occurs and multiplied and all of this you can just here I just follow almost blindly what we did. So, these things will just fall out of course, you will have to use those properties of character I listed last time I will do various sum properties that is all you do not have to do that. So, that is for even characters for odd characters it changes a little bit what is u in the last? q is that character the whole character is defined as a map of z q star to complex space. So, q is that implicitly everywhere is actually I should be writing chi of q, but that is just painful to write chi of q everywhere. So, q is again one universal number fixed once and for all it is an odd character when we one needs to change the definition of chi a bit this is there is this plus 1 speaking out. So, physically one has to start with gamma of z over 2 plus 1 here and then sum over because this sum does not quite work out very well. So, again this funny things just pop out of the analysis. So, this defines the functional equation for the L function and using the functional equation now you can see that the all the L functions can be extended over the entire complex space. Once you extended over the entire complex space then again you can talk about the zeros everywhere. So, what would be the zeros of this? This L function again these are governed by the gamma. So, zeros of gamma functions are negative in T s sorry the poles are negative and they are all about one. So, the L function the trivial zeros of the L function will be at negative T s. So, those are trivial zeros just as in case of zero function and then there are non trivial zeros which are again between the strip of zero and beyond one there are no zeros. Now, what about zeros here of this L functions for odd characters? These are all negative odd in T s those are trivial zeros and again the non trivial zeros will be between zero and one. So, everything works out in the same way you can can do the rest of the analysis in the exactly the same way that is this and should talk about the rest of the analysis. So, this completely describes the L functions now I need to relate the side with the L function and the method used is again exactly the same, but there is one intermediate step which is which are this. So, again this is the cousin of psi for psi of x is just sum over n less than equal to x of lambda n and psi of x chi is summation n less than equal to x chi of n times. So, this is a quantity which we will use which we will relate to the L functions and finally, we call what is our target our target is estimate this. So, this is a different psi and this is what we really want, but in order to understand this or to understand that we have to extract out information about this we are we are going to use L function and first we will relate L function with this psi this one psi for this particular character because that psi has no character just I will show you once I am done with it how to relate that psi with this psi. And the relationship between this and L function is exactly again the same how did we derive this relationship with that till function now with the zeta function we just go ahead do the same process this is same as n greater than equal to 1 i of n lambda of n and then delta of x by n and then stick to the integral for the delta function take everything inside now because everything everywhere this chi of n is multiplying as a multiplier is present the integral that you will get will have after the rate of 4 it will be 1 over 2 pi i integral and this goes from c minus i infinity to c plus i infinity minus L prime z prime over 2 p z this one only changes that the zeta prime or zeta is multiplied by i prime and again we do the same thing we can approximate this c minus i r to c plus i r that same expression for error and now that we know L prime over L and we do the same analysis using functional equation for L prime over L and then write this in terms of various poles of this and what we will get is eventually psi of x i 0 there will be not a difference between the trivial character and all that and that difference will come precisely because of the pole being presented z equals 1 trivial character and it being upset for z equals 1 and that is very crucial because this as a pole at z equals 1 so the residue of that comes out to be x and if you recall the x was the leading term I mean that residue is what really gives the main value to psi the rest of the residues are just errors so for the trivial character that residue is present plus there will be again the same thing the sum over all non-trivial zeroes and trivial zeroes and similar Kasani expression so let me just bring that in and psi of x i where i is not equal to i 0 now this residue is not present there it is just error error is taken to mean that whatever the expression of the order this is physically let me write I will write the error when psi done with the generalized Riemann hypothesis so now there is a version of Riemann hypothesis with the Riemann hypothesis this determines what exactly this error is now there is a version of Riemann hypothesis for L functions and which is the form exactly same statement as the Riemann hypothesis except that set of zeta function we are not in particular and this hypothesis is for no matter what the character we choose all non-trivial zeroes of the corresponding function lie on that and if this is true if this is true then psi of x of chi 0 is x plus order q square root x so the error is same except that multiplier q comes in again because of deltion psi of x is just here so this is what we get just doing everything with it for one year good but that still does not has it yet answered our original question which was about that side see this is psi of the second variable side of character we are not interested in psi of the character we are interested in counting this side which is sum n this is equal to x and n being congruent to a mod q that is a quantity so how does one relate this side to the character side so again this sum over sum n which are sort of jumping over the groups we stayed within a residue class we are not q this is messier to analyze we would like to avoid it and replace it with the sum over all n's there is a simple mechanism available for it let us just consider psi of what is this chi bar of a and sum n is equal to x now what is chi bar of a times chi of n because chi is multiplicative is chi bar of a is 1 over chi of n and again using multiplicative property chi of n times chi divided by chi of a is chi of n and now what I want to do is sum over all times so there is a group of characters sum it over all times what is this to change the sum how do we know about this sum whenever n is a this is chi of 1 or let us say not just a whenever n is a that was the extended definition so whenever n is a model of q you get chi of 1 and summing over all characters you get phi of q on the other hand when if n is not equal to a so model q then what is this sum 0 q so this is equal to n less than equal to x which is precisely the quantity how does phi of q come along phi of q how does it come along when n is a mod q and this chi of 1 and then summing over all characters in the group chi of 1 is always 1 the number of characters in the group is exactly chi of 1 over all characters chi of 1 over all chi of 1 over all characters so that you get there for phi of q so that is the expression and which is very nice because now we get this very clearly that is you know what chi of x chi is so let us put this sum into chi 0 and not chi 0 this is 1 over chi of q plus 1 over chi of q this is the error part you know this this is essentially x over phi q plus the error and this is the whole of it is there so this is essentially the error anyway and the error was square root x q log square root x 1 of course you get multiplied by this but this is absolute value 1 so if we just take absolute value of the error and get rid of you can get rid of this the sum over all of this but that sum is same as this so that is what cancels out and you get order square root x q and that that is it and that is what we started out now this is assuming the average but the same technique we used to prove the prime number theorem more or less the same technique can be used to prove the similar theorem out there that is chi of x a q is x over phi q plus times 1 over small over 1 so this error is not a small but is small over 1 without any assumption any push ups on this I look very rapidly it is really repeating everything I mean try this for the echo bit try working out some of this step by step now this of course might appear but this is a number of resting application this particular chi of x a q I want to point out couple of them to justify that this course is a computer science course couple of computer science applications of this here under the device name on I guess the fact this is used very heavily in just about every but most of the number 30 for example it was used to show that primary testing is in D weighted polynomial factorization is they all arise out of this and the way it comes out is a pretty simple so these are the bottom line to all of this is that if you take a subgroup of z q star then and you want to find an element in z q star which is not that we do not have to look we do not have to search too much for that so let me just summarize it so h is a proper subgroup of z q star and as I said we are looking for an element outside the subgroup all we need to do is start from 1 to 3 4 log q to the 4 one of these will be an element can you see it better still out of time anyway let us give it I will not give it as assignment because this is very interesting and I want to do it make sure that you understand it so I will give it as an assignment and we will do it in the next class