 look at the second one, first and third we sorted out last time. So, the second one is product and now notice that absolute value of z i lies between half and twice of absolute value of z. So, z over z i therefore, lies absolute value of z over z i is going to be between half and. So, this quantity in the exponent is at most e square and absolute value of z i is at most 2 times absolute value of z and we are left with absolute value of z minus z i in the denominator. Now, this is where we run into problem because what if z is equal to z i then this of course, is unbounded and this which may happen right z i lies between an absolute value between z by 2 and 2 z. So, it may happen that that is if I choose my z in such a such that it equals z i in a precisely then this is unbounded and therefore, I cannot bound it. But if you recall the proof of the bound or the nature of the entire function of order 1 without any zeros in that proof that the function must be of the kind e to the a z plus b what we used was that for infinitely many r's mod f z at mod z equals r is bounded by e to the order r to the 1 plus epsilon right. We did not know this property to hold we did not require this property to hold for all r's we only required it to hold for infinitely many r's and that is what we are attempting to do here when we if you go back. So, we want to show that this is an entire function of order 1 with no zeros which we know of course, and in order to show that all we need is to show that this absolute value of this whole function is bounded by e to the order mod z to the 1 plus epsilon for infinitely many mod z's not for all mod z. So, that gives us some flexibility in choosing the z here and what we will do is we will choose a z such that z the absolute value of z is not in this z. So, consider the interval of numbers which is between mod z i minus 1 over mod z i square to 1 mod z i plus 1 over mod z i square and for each z i take the union of these intervals and we want to pick a z such that mod z is not in any of these intervals is that possible can we choose infinitely many such z's yes we can why what is the range of this one single interval here, but what is the exact value between that we can compute the exact value. One interval corresponding to i what is the range of values. So, this to this so what is the difference between the ranges so that is that is the number of values which are omitted right for your z i z to be. So, the number of values omitted or the range of values omitted has 2 by z i square right. So, total values omitted actually it is less than equal to because for two different i's this intervals may overlap and this is bounded right. Therefore, only finitely many values are ruled out so you still have the whole range the infinite many values for your z to choose from. So, you can always choose a larger infinitely many z satisfying this. Now, for such a z let us go back this is the quantity we want to estimate what is the difference between z and z i at least 1 over z i square because z is not in this interval for any z i. So, z i z z is therefore, with respect to any z i it is at least 1 over z i square away from that. So, we can replace the denominator here by 1 over z i square and get an upper bound and z i is upper bounded by 2 times z. So, you get z 8 e square mod z cube and how many times are you multiplying this well the number of times you are multiplying this is number of z i's in this range. How many z i's are there in this range we know the count e to the mod z to the 1 plus delta at most right. So, therefore, this is less than equal to 8 e square mod z cube whole to the power order no not e to the power sorry. The number of z i's number of roots are bounded by order mod z to the 1 plus delta that we shown and this is of course, the thing inside here does not move the scale here at all right. You take everything in the exponent you just maybe had multiply this with 3 log n also and which does not move this delta to anything just sort of stays the same delta prime slightly bigger delta and that is it. So, this bounds this entire this product for infinitely many z's and therefore, and for all z's we know that f z is bounded by the same. So, together this whole absolute value for infinitely many z's is bounded by e to the mod z to the 1 plus epsilon prime for some epsilon prime small enough. And therefore, f z over product over i greater than equal to 1 entire function 1 with no 0's and z equals it is convinced. So, this is the. So, at this point I am going to end my divergent into the theory of entire functions and we will step back a level up and what was that level gamma function. You are trying to get a bound on gamma prime over gamma that was the starting point and then I said that we will completely analyze the gamma function and then we will derive from that analysis the bound on gamma prime over gamma. So, let us step back into gamma function and see where we are now we know that I think we have already shown that 1 by gamma z is an entire function of order 1. Does it have any 0's? Of course, it has plenty of 0's because it has it has poles at all the negative integers. So, it has plenty of 0's. So, we stick that in, but does it have 0 at z equals 0? It does gamma 0 is what is gamma 0? It is unbounded because it is gamma 0 is gamma 1 over 0 which is unbounded. So, it does have a 0 at 0 1 over gamma z. So, that therefore, by this whole previous analysis how can we write the gamma function how inverse gamma function e to the a z plus b times z the z is for the 0 at z equals 0 and it is a 0 order 1. So, we just multiply it with z and leave it at that times product n greater than equal to 1 1 plus z over n e to the minus z. Now, here we still have not fixed the value of this constants a and b, but that is not too difficult to do. What is z gamma z at z equals 0? Is gamma 1? It is gamma 1 because gamma z is gamma z plus 1 divided by z. So, z gamma z is gamma z plus 1 as z goes to 0 z gamma z is gamma 1 which is 1. So, 1 over z gamma z is 1 at z equals 0 and we plug this into this expression what do you get when z is 0 this whole product is 1 everything here is 1 z equals 0 this is e to the a z vanishes. So, you get e to the b equals 1 this implies that b is. So, that takes your one constant what about the other constant 1 over z gamma z is 1. So, you take this z here down here and then take the limit now what about gamma z at z equals 1 of course that is 1. So, let us plug that in also for z equals 1 what do we get e to the a times product n greater than equal to 1 1 plus 1 by n e to the minus 1 by n this is equal to 1, but this is not. So, straightforward to work out, but we can do a little bit of trickery here to work this product what is the value to calculate the value of this product or does anyone know an expression for this product yeah that is a good idea take the log and see what works out. But, before taking the log what I am going to do is bound this up to some capital n and then eventually we will take n capital n to infinity to get the value now take the log. So, what is the log value here log of this product this equals let us simplify this let us not take the log first let us just try to simplify this this is equal to product 1 less than equal to n less than equal to capital n n plus 1 divided by n e to the minus 1 by n. So, this is e to the summation 1 less than equal to n less than capital n minus 1 by n times product of 1 going to n going from 1 to capital n n plus 1 over n. So, what is that product? So, it is like 2 by 1 times 3 by 2 times 4 by 3 times or dot up to capital n plus 1 divided by this just leaves out capital n plus 1. In fact, I do not even need to take now is the expression in the exponent come here to you what is this one the sum this is harmonic series truncated at capital n and this is log of n plus 1. In fact, let me write this as. So, look at this expression log n minus these harmonic series truncated at capital n as n tends to infinity what happens to this that is a very familiar quantity math student with our way Euler's gamma and you remember log n is very well approximated by this summation up to n 1 plus 1 half plus 1 third plus 1 quarter up to 1 by n. And the difference between log n and this sum as n tends to infinity tends towards a fixed constant gamma which is like 0.40 something that is that constant was first described by Euler it was called Euler's constant. And as n tends to infinity what happens to log n plus 1 over n it goes away 0. So, therefore, we get that this product I think I should put a minus sign here because this sum is always bigger than the log say when capital n is 3 then log 3 is very close to 1 because to the base E that is very close to 1 whereas this is 1 plus half plus 1 third to as bigger and it stays bigger. So, the difference between these two are this minus this is gamma this minus this. So, the difference is minus gamma. So, that is that is what you get and therefore, what is e to the a we just were described that this is e to the gamma. So, here is gamma now this completely describes the gamma there is nothing left out here you know each and every single constant appearing how does it factor how does it add up everything is there. And this is the expression for gamma function that now we will use to give a bound on gamma prime over gamma. And the way to do it is very simple take the log of both sides and differentiate. So, if we take the log of both sides what do we get minus log of gamma z equals small gamma z plus log z plus summation n greater than equal to 1 log of 1 plus z over n minus z over n. Now, because we have taken the log again we have to say something about the validity of this expression. So, the simplest way of describing the validity of this expression is to say that just take cut out one line that for whatever log of whatever thing we are taking that quantity we for that quantity we cut that line from 0 to negative infinity that real line. And say that on the rest of the domain it is well defined, but we are taking log of on the right hand side log of 1 plus z over n of course, we are taking log z here also. So, if you think about what we need to avoid it turns out that we need to avoid essentially the real line actually not the whole real line I think let us greater than 1 I think negative real line is what we need to avoid from 0 to all the way up to minus infinity that negative real line if you can avoid this is well defined because if it is that is right if z is minus n then this becomes 0 and then log becomes undefined. So, that is the problem we do not want to run into. So, we just avoid the negative real line for z value. So, let me note it here this is and that is what we need to keep in mind and that also takes care of all this duplicacy also once you cut out make this cut then the log function is analytic on the rest of the domain. So, everything works out fine now differentiate now we can differentiate because analytic now log of this we can differentiate and what do we get on the left hand side we get minus gamma prime over gamma equals small gamma plus 1 by z plus n greater than equal to 1 1 by n divided by 1 plus z by n minus 1 by n. So, what comes out on the right is a very simple expression and this is well defined whenever z does not take values in the negative real line. In fact, one can go one step further and say that is this part at least is well defined whenever n does not take negative integer values, but let us just stay with the negative real not z not taking negative real values. So, now in order to estimate this quantity we need to estimate the right hand side which seems like a not too difficult task and indeed it is not too difficult task, but now we have to really dig back and go one level up and recall where did we require to bound gamma prime over gamma or for what z do you remember we were looking at you are looking at this rectangle and we wanted to show this is this integral is what we are interested in. So, you want to get rid of all these three integrals and I said that we will first start with this integral which is the simplest to handle and so we were at this point when we diverged into gamma function. So, let us get back to this. So, to start with we consider z in minus u plus i r minus u minus i r and since we want to avoid hitting a pole here remember there can be poles of gamma functions here. So, we want to avoid hitting a pole we take u to be because remember we were actually looking at gamma of z by 2 and analyze or rather gamma prime of z by 2 divided by gamma z by 2. So, when u is an odd integer we are essentially looking at when you look at gamma prime z over gamma z it is a the midpoint between two integers is what we are getting. So, we do not so this line never hits a pole as far away from a pole as possible. So, that is the z we are looking at now for such is it or in general for z which are reasonably far away from poles of gamma what is the bound on gamma prime and what is this bound equal to just look at this this is constant this is 1 by z and 1 by z 1 by mod z actually 1 by mod z mod z is way out there at least u which is at most 1. So, this gets absorbed into order constant here. So, what is left out is plus summation n greater than equal to 1 mod of if we can estimate this modulus of this whenever z is far away from negative integers then we get a good estimate. So, how do we estimate this when you look this looks again familiar should look familiar. So, to estimate this we again we will use the same trick that consider the sum from 1 to capital N of this quantity we derive an expression to it and then send n to. So, if we consider n equal to capital N we want to bound this what is the second guy bounded with we know that that is like log of capital N plus the Euler's constant. So, that we know we understand it very well what about the first guy this is z sticking out in all of the denominators well it would be convenient if we can replace this sum by an integral because we can integrate things far more easily then sum them up you agree right. And now there is a standard way of replacing sums with integrals that is called in given long time Euler Maclaurin. So, it is called the Euler Maclaurin formula. So, again I will take a bit of a diversion from here take you into Euler Maclaurin formula. So, this in general talks about sums like this and what this corresponds to in terms of integrals. In fact, it does not even say that you should be 1 to N you can replace it with a and b and the formula says of course, I will I am sure I remember it incorrectly. So, let me write something and when we derive it we will come back and correct it. So, that is the main part it is like this sum is same as this integral, but then there are some problems here which are the error terms minus no I do not remember let us take a guess it is probably main t and then there is this funny expression. So, this is as I said this is only my guess of what the formula is but it is close enough. So, let us derive this is it does not require any anything fancy how we are out of time. So, let us find out an expression for this there is got to do with this one let us look at this no that is not right let us just make this integral and let us do integration by parts. Now, what do you mean by integration by there is only one quantity here well. So, what I will do is times 1 times t. So, the other part is 1 this is equal to f t and then integrate 1 what do you get when you integrate 1 you get t. So, now let me let me do something funny here I integrate from k to k plus 1 just over 1 integer and integrating this I get t, but instead of actually there will be a t plus I can bring in any constant here. So, I will integrate this to t minus k minus half I am perfectly within my rights to do this. Then there is in other part is negative f prime t and then I integrate 1 in the same fashion they these are important right. So, this becomes t minus k minus half now what is the first one of these that is f k plus 1 right and what you get there is 1 minus 1 minus half. So, half of f k plus 1 and minus of minus this becomes plus plus half of minus this integral. Now, sum it over form for k going from a to b minus 1. So, I am assuming that a and b are integral points sum over this for k going from a to b minus 1. So, what do you get half of f of a plus half of f of a plus 1 then next time I am around you get half of f of a plus 1 plus half of f of a plus 2 and then again and again and again. So, you basically get half of f a plus full f a plus 1 plus f a plus 2 plus f a plus 3 all the way up to f of b minus 1 and then you get half of f b. So, this is therefore f n minus half of f a plus f b minus a to b t minus. Now, k is same as floor of t because now I am replacing the limits from k to k plus 1 to a to b. So, in within this range floor of t is exactly k when you go to the next integral range floor of t becomes k plus 1 and so on. So, that perfectly mimics good. So, I was actually right my first instincts were right this should have been a plus. So, we just take it from the right side. So, now let us just apply this onto this summation this is equal to what integral from 1 to capital N 1 over z plus t d t plus half of f of a is 1 over 1 plus n 1 plus not 1 plus z plus 1 plus 1 over z plus n plus derivative of f prime that is equal to z plus t whole squared minus sign t minus floor of t minus half. And this is first integral is what log z plus t plus half minus now in absolute value unless I will not get into the absolute value yet. This term just let us look at this this is sort of the error term for me. If you look at the numerator here always at most half right no matter what the value of t is. So, numerator is at most half the denominator is for 1 z plus t whole squared. So, if I try to get a bound on the error term the absolute value of this. So, let us just look at the error term. This is equal to half this integral is 1 by z plus t minus 1 by z plus t actually right and going from 1 to n. So, we have an estimate on the value of the error term. Now, we have everything what we want in place. So, now it is just we will put it together next time and derive the estimate.