 Okay, so let us continue with our discussion of analytic continuation, okay. So I told you last time that there are two notions of analytic continuation, one is the so called direct analytic continuation and this involves, this literally involves gluing together two analytic functions on two domains which intersect, okay and such that the analytic functions they give the same function on the intersection, okay. So direct analytic continuation means that you have a diagram like this, so this is T1, f1 and you have this for D2, f2 and we say D2, f2 is a direct analytic continuation of D1, f1 or conversely if on this intersection which is supposed to be non-empty f1 and f2 coincide and so the advantage of this is that you can define these two functions glue together to give an analytic function of union, okay. So then G from D1 union D2 to C given by G restricted to DI is equal to fi is analytic, okay. Of course the pair D1 union D2, G is certainly a direct analytic continuation of both of these and of course the further the direct analytic continuation is unique because of the identity theory which says that if two analytic functions agree on an open non-empty open subset then they have to agree on the whole domain, okay. So this is the story of direct analytic continuation and we have of course seen that given a pair D, f there exists a unique maximal extension maximal direct analytic extension D1, f1 of D, f. So we have seen this, so the and we saw these are two examples of this. So the first example was D1 is the unit disc and f1 is the power series corresponding to the geometric series, okay or rather let me call that as D and f. So you take the unit disc which is the disc of convergence of the power series given by the geometric series power series centered at 0 features of convergence is 1 then the maximal extension D1, f1 is simply the complex plane minus the 0.1 and the function which is a maximal extension is 1 by 1 minus that we have seen this. This is this was pretty easy to see because this power series represents the 1 by 1 minus z to the unit disc, okay and that is defined everywhere in the complex plane analytic except at the point z equal to 1 where it has a simple point, right. This was an easy example, okay. So the other example is that of the frame and zeta function for which the domain you are going to consider is the right half plane the open right half plane to the right of the line the vertical line passing through z equal to 1. So in this case D is set of all z complex numbers z such that real part of z is greater than 1 and the function f is zeta where zeta of z is the Riemann zeta function which is sigma n equal to 1 to infinity 1 by n power z and this is defined as sigma n equal to 1 to infinity 1 by e power dz ln n where ln n is the usual real logarithm. So the fact is of course there is some work involved in checking that zeta itself is an analytic function, okay. So that involves the use of the V extra m test which we did last time and of course we also use need to use the fact we need to use V extra m test to check that zeta this series converges to a function normally, okay in this domain which means it is the convergence is uniform on compact subsets, okay. And then further since we have already seen in earlier lecture that whenever you have sequence of analytic functions which converges normally to a limit function and the limit function is also analytic by using that fact whose proof essentially uses Morera's theorem we can check that zeta is actually analytic is an analytic function on this right half plane. It is a non-trivial theorem to show that the maximal analytic extension of zeta is defined on just like the geometric series it is defined on the whole complex plane minus 1 and at 1 you get a pole of order 1 simple pole, okay. So it is non-trivial to show that the maximal extension of zeta is defined on complex plane minus the point z equal to 1 with a simple pole at z equal to 1, okay. So we will see the proof of this later but then these two examples tell you they give you the difference in degree of difficulty, okay here is an example in which the maximal analytic extension is very easy to understand and here is one in which it is very difficult it is not directly easy to check that you have a maximal analytic extension, okay. I mean what I mean is it is not easy to check what the maximal domain on which the function extends this and what the corresponding function is, okay one has to do lot of analysis to check that just like in this case the only point where this zeta function does not extend is z equal to 1 where of course it becomes a harmonic series and you know and to do that one has to do lot of analysis. So this is the problem of direct analytic continuation. Now what we do next is go to this notion of indirect analytic continuation and so what is this indirect analytic continuation? So the word indirect is something that I am stressing usually in the literature the word that is used is just analytic continuation and I am stressing the word indirect because I want you to distinguish between these two. So this indirect analytic continuation is something it is nothing but a chain of or a sequence of direct analytic continuation successively, okay. So you know so it is a chain of direct analytic continuation. Of course I instead of the word analytic continuation I also use the word analytic extension, okay in the previous lecture so that is also often used. So we instead of saying direct analytic continuation we also say direct analytic extension and indirect analytic continuation instead of that you can also use the word indirect analytic extension, okay. So what is an indirect analytic continuation? It is a chain of direct analytic continuation so you know it is something like this. So you have sequence of domains of course in the diagrams that I am drawing I am showing bounded domains, okay but they need not be bounded, okay. So you know you have D1 so you have D alpha 1, f alpha 1 so you know I used so let me write it like this D alpha 1, f alpha 1 is the first pair which consists of an analytic function f alpha 1 on this domain which is D alpha 1 and that has a direct analytic continuation to D alpha 2, f alpha 2 which is defined here and then that further has a direct analytic continuation to D alpha 3, f alpha 3 and so on and finally I end up with D alpha n, f sub alpha n, okay. So the point about this is that every successive pair is a direct analytic continuation of the previous pair, okay and the point is that every successive pair is a direct analytic continuation of the previous and the next, okay but there is no relationship between one pair and another pair which is not its predecessor or successor, okay. So you know D alpha 1, f alpha 1 if I try to look at D alpha 1, f alpha 1 and D alpha 3, f alpha 3 there is D alpha 3, f alpha 3 need not be a direct analytic continuation of D alpha 1, f alpha 1 because to begin with they may not even intersect as I have drawn it in the picture, okay. Of course the fact is that if they intersect and if the intersection is has some intersection common with the second pair then of course they will all glue up to give a single analytic function and D alpha 3, f alpha 3 will become a direct analytic continuation of D alpha 1, f alpha 1, okay. But the problem is that first of all these two need not intersect the second problem is even if they intersect the intersection need not have anything in common with their intersections with the intersection of these two and the intersection of these two, okay. So you know you could have something like this, you could have something like this, you could have a situation like this, okay. So here is D alpha 1, f alpha 1 and here that is here and that has a direct analytic continuation D alpha 2, f alpha 2 and then you could have direct analytic continuation D alpha 3, f alpha 3 and this is a direct analytic continuation of this which means f alpha 1, f alpha 2 coincide here, this is a direct analytic continuation of this that means f alpha 2 and f alpha 3 coincide here and this is a direct analytic continuation well I mean this need not be direct analytic continuation of that. These two need not be direct analytic continuation, okay. Such a thing can happen, such a thing can happen. So you start with the function you go back you probably come back to the you come back to you analytically continue it once then you again further analytically continue it and then end up with the function which has which on the which has some region in common with the starting domain but the function may be different f alpha 3 and f alpha 1 need not coincide on this intersection, okay. So the question is this is a strange thing that happens. So what is actually happening is the following what is actually happening is that if you have a chain of so in this case you are having a chain of direct analytic continuation so you have an indirect analytic continuation D alpha 3, f alpha 3 is an indirect analytic continuation of D alpha 1, f alpha 1. Then the question is of course what is how is f alpha 3 related to f alpha 1 that is our question, okay. I told you f alpha 3 and f alpha 1 need not agree on this intersection, okay. So more generally the question is I have D alpha 1, f alpha 1 I start with a particular pair here and then I do an indirect analytic continuation and I finally end up with final pair D alpha n, f alpha n my question is how is f alpha 1, n related to f alpha 1 that is the question, okay. How is it related? So the fact is the answer to that is the following the answer to that is well this and this are not totally unrelated they are branches of a multi-valued analytic function, okay. So like for example this could be a logarithm a branch of a logarithm but then this could be some other branch of the logarithm, okay. So for example in this case you know this could be a branch of the logarithm and then when I come back this f alpha 3 could be a different branch of the logarithm, okay this can happen. If you think of the Riemann surface for log z this is what is happening as you move along the sheets you move from one branch to the next branch and if you take its image on the complex plane you will see that as you go around once your branch of the starting branch of the logarithm after you go around once around the origin the new branch that you get the new function you get which is an indirect analytic continuation of the old branch the first one that you started with is a new branch of the logarithm. So the reason why we study indirect analytic continuation is that it allows you to move from one branch of a function to another branch of a function, okay that is the importance of studying this. So well I told you that there is one way of looking at this a way of formulating this and that is in fact there are two ways of formulating indirect analytic continuation one is this as a chain the other one is to use power series, okay. So that is what we were trying to look at last class so power series formulation of indirect analytic continuation. So what is this power series formulation it is done like this so basically the idea is that you know you have you start with the point z0 you go along a path to a point z1 so this path is gamma alright. So gamma so gamma is a path you have a close interval AB on the real line which if you want you can take it as the unit interval 01, okay and you have continuous function from this to the complex numbers, okay. The image of gamma is a path with z0 equal to gamma of a and z1 is equal to gamma of b this is the starting point this is the ending point this is the path alright and the idea is that you know if you give me a general point here it is given by gamma of t where t is a point in the interval real interval a, b, okay and what we want is that at this point you know we want to look at a function which is analytic at this point and given by a power series, okay. So what is happening is that you are looking at for each t1 is given an analytic a power series ft of z which is given by so sigma n equal to 0 to infinity an of t z-gamma of t to the power of n with radius of convergence or with radius of convergence r of t and disc of convergence mod z-gamma t is less than r of t. So I have so for each t I am given a power series of course the so I am giving for each t I am giving you a power series for each t I am giving you a power series and the power series so I have to put an f sub t of z this is a power series centered at gamma of t so it is an expansion in terms of z-gamma t, alright with some coefficients and the coefficients also depend on t, alright. Well if you put t equal to t0 that means you are choosing a point in this interval so its image will be a particular point here and then ft0 of z will be a particular power series, okay and the assumption is that the radius of convergence is r of t is positive, okay and the disc of convergence is given of course by the disc centered at the center of the power series and radius equal to radius of the convergence of the power series, okay. And of course you know in all these situations you must remember that we are really not interested in the case when the radius of convergence is infinite, okay. We are only interested in the case when the radius of convergence is finite because if the radius of convergence is infinite it means it is an entire function, okay. It means the maximal extension is on the whole complex plane and the given function is actually a restriction of an entire function and there is nothing complicated happening there, okay. Your function is just a restriction of an entire function so that is not the situation we are interested in that is a very trivial situation, okay. We are interested in functions which have finite discs of convergence namely whose radii of convergence are finite and why are such functions important because on the border on the circle of convergence there are singularities, okay. So if a function represented by a power series you take a power series if it has finite radius of convergence then on the circle of convergence there is at least one point which is a singular point for the function because if no point was a singular point of the function then I can extend the function to an analytic function to a disc larger which contains the circle of convergence and that contradicts the very definition of circle of convergence, okay. So why we are interested in functions with you know finite radii of convergence is for this reason because it gives you all the functions which have singularities, okay which you can study, right. So the situation is that I am so for every point I am giving you a power series, okay and we would like this power series we would like the power series to vary continuously with respect to t, okay. That is just like gamma is a continuous function of t, okay which means that this is a continuous path, alright. We would also like f t to vary continuously with respect to t and one way of stating that is that you know if what we can what you can say is that if you take t prime close to t on the interval you can require that gamma this power series f t and f t prime they are one and they represent the same analytic function on the intersection of the two discs, okay. So which means that you are for t prime close to t you are assuming that f t prime is a direct analytic continuation of f t, okay. So you know so the additional condition that you are going to put is the following if you take so suppose this is gamma t prime which is close to gamma t this radius of convergence is r t if I take if I do it for gamma t prime I will get another disc here and it will have radius of convergence r t prime, okay. And what I am requiring is that for t prime close to t if t prime is close to t then gamma t prime is close to gamma t because gamma is continuous, okay and I am requiring that on this disc f t is defined and on this disc f sub t prime is defined and what I want is that I want that f t and f t prime are direct analytic continuation of each other in this common region that is f t should be the same as f t prime in this intersection, okay. So further for t close to t prime which should be thought of as t prime belonging to an epsilon neighbourhood of t in the on the real line that is that is for t prime belonging to t minus epsilon t plus epsilon intersection a b this is what it means, okay t prime and epsilon neighbourhood of t epsilon for some epsilon sufficiently small small and positive, okay that is what t prime close to t means or t close to t prime means, okay we must have so let me continue here we require that f t f t prime is a direct analytic continuation of f t, okay. So this is the condition that you say that you know so what is happening is that as t moves from a to b the point gamma t starts from z0 and ends with z1 and as you move close so at every point you are giving me a power series, okay and if you take nearby points the analytic functions represented by the power series are one and the same on the intersections, okay. So what you are giving me is analytic functions which are you are giving me a sequence of analytic functions that are moving along this path and in such a way that locally they are they represent the same analytic function, okay. So we say that so the point I want to make is that well if we formulate it like this if you formulate it like this then you know the if you take f sub a which is the analytic function at z0 which is gamma of a that if you look at that and if you look at f sub b which is analytic function at gamma b or gamma of b which is z1 I claim that they are indirect analytic continuation of each other, okay. So then f then this pair which consists of mod z-z0 less than r of a z0 is mind you gamma of a, gamma fa of and the initial one and the final one z-z1 less than rb z1 is the same as gamma b and fb are indirect analytic continuation of one another. They are indirect analytic continuation of each other, okay in the sense of the earlier definition. So I am just saying that the new definition I am just trying to prove that this new definition which involves continuously varying power series that is the same as this original definition I am trying to say they are the same and in so by saying that I am trying to say that this is a more you know stricter formulation of this which involves an analytic expression, alright.