 So here is the definition, so it is this topic that is which is which provides lot of variety and gives you some new things okay. So what is this indirect analytic continuation? So the idea is the following, so what you do is you have so in the case of direct analytic continuation what happened is you had two domains which intersected and you had functions and analytic functions on these domains which agreed on the intersection okay. Now what you do is that you do extend it from 2 to any finite number okay, you extend it from 2 to any finite number but do not insist that all the sets in your collection you do not insist that they all intersect but you only insist that you know if there is an ordering every member intersects with the next, so what you do is you do something like this. So we have domains, so we have pairs, suppose we have pairs d alpha f alpha where d alpha are domains and f alpha from d alpha to c are analytic and an ordering a total ordering of the set A is equal to set of all alpha okay. So you know you have a collection of pairs d alpha, f alpha indexed by alpha running in an indexed capital A alright and these pairs each of these pairs consists of a domain namely an open connected set and an analytic function on the domain right and on this set A you have a total order. Let A be finite, let A be finite for simplicity okay say A equal to alpha 1, alpha 2, alpha m okay, so A is given like this okay and I have put this strictly less than to say that there is a that is because of total ordering okay that is since there is a total order in fact if it is finite then total order is always there alright but the point is why I want the ordering is the following, so the point I want is let for every i d sub alpha i intersects with d sub alpha i plus 1 and f alpha restricted to d sub alpha i intersection d sub alpha i plus 1 is equal to f alpha i plus 1 restricted to d alpha i intersection d alpha i plus 1 okay. So suppose I have this situation right in fact I could have taken the set as 1, 2, 3, etc m and then I could have simply called it as d 1, d 2, etc d 1. I could have simply called these pairs as d 1, f 1, d 2, f 2, etc d m, f m you would have made it easier to write down but the reason why I am doing this is because in practice this parameterization could be even continuous not even discrete and I am going to think of it as coming out of path alright. So but I am looking at only here I am only looking at the finite case so the situation is like this so the situation is so the diagram for this will look like this so you know the diagram will look like this so we have so here is d 1 so this is d alpha 1 then I have this is d alpha 2 and they have an untrivial intersection here then well d alpha 3 may or may not intersect d alpha 1 but it has to intersect d alpha 2 so it will be something like this so here is the intersection and then you have something here which is d alpha 4 and so on and it goes on like this until I end up with d alpha m which has some intersection of the previous and what is happening is that I have a function f alpha 1 which is analytic on d alpha 1 values in C and I have this function f alpha 2 which is analytic on d alpha 2 which is analytic on d alpha 2 and it is complex value function and on this intersection f alpha 1 and f alpha 2 so here it should be alpha i and this should be alpha i plus 1 and of course I will have to take i from 1 to m minus 1 for this to make sense so I have this situation so you can see what is happening in other words what you are having is a chain of direct analytic continuations okay we get a chain of direct analytic continuations so d alpha 1 f alpha 1 d alpha m f alpha 1. This is a chain of direct analytic continuations namely this is a direct the pair d alpha 2 f alpha 2 is a direct analytic continuation of d alpha 1 f alpha 1 okay then the pair d alpha 3 f alpha 3 is a direct analytic continuation of d alpha 2 f alpha 2 namely f alpha 3 and f alpha 2 will coincide on this intersection and this goes on alright. So every pair is a direct analytic continuation of the previous one okay and it is also direct analytic continuation of the next one okay so you have a chain of direct analytic continuations and then what we say we say that the function that you get at the end f alpha m we say that this d alpha m f alpha m we say it is an indirect analytic continuation of the first pair okay and usually the word indirect is something that I am stressing but if you see a general literature if you see general literature the word indirect is not mentioned but I am stressing indirect because in general it will be like this okay of course if there are only 2 then it is there is no difference between direct and indirect okay but the reason for calling it indirect is something very strange what can happen is I can start with a pair okay I can have a chain okay such that the ending pair has the same domain as the starting pair okay but the function I get will be completely different okay so this is a strange thing that can happen what can happen is I can start with a pair here I can have a chain but the last member has the domain the same as the first member but the function is different such a thing can happen and you know what is actually happening in this case what is actually happening in this case is that these function elements the first one that you started with and the last one that you got which is again a function on the first piece okay these 2 are 2 branches of analytic function they are different branches you arrive from one branch to another branch in this way okay that is the whole point you we saw in the previous lectures that you know if you have a general analytic function if you take if it is derivative does not vanish at a point then in a neighbourhood of that point it is it has an inverse which is also analytic alright that is the inverse function theorem it is 1 to 1 and it has an inverse but on the other hand if the derivative vanishes then it is a critical point but even for the critical even at in a neighbourhood of the critical point we have seen that you can get branches of the inverse function they are functional inverses not actually inverses because in a neighbourhood of the critical point it will be a many to one function okay it will behave like z going to z power m where m minus 1 is the order of the critical point and you will get m branches for the inverse function okay and these m branches will live on a Riemann surface as a single function but the process of going from one branch to another branch if you do it locally that comes via indirect analytic continuation that is the importance of indirect analytic continuation. So this indirect analytic continuation is something that helps you to find all helps you to exhaust all possible branches of a function okay and of course you remember you must understand that whenever you are trying to talk about branches of a function that function is certainly not it does have a singular point and it has a branch cut and so on and so forth okay. So that is the importance of indirect analytic continuation okay that is the motivation for study of indirect analytic continuation but this needs to be formalized using a proper theory. So this is the first step you define what is meant by a chain of direct analytic continuation okay we say that d alpha m f alpha m is an indirect analytic continuation continuation of the first one d alpha 1 f alpha m and the point with indirect analytic continuation let me again repeat and stress is that if you start with a particular function and you do an indirect analytic continuation and assume you come back to the same domain you started with you will end up you could end up and if you try hard enough you will always end up with all possible branches of that function so you can get all the branches of the function okay by this process okay that is the that is the importance of indirect analytic continuation. So the word indirect is usually permitted in standard literature but I am stressing it okay so let us formalize this let us formalize this one would like to treat this in two possible ways that are the two possible ways of proceeding now and I will do both of them the first one is trying to do it using power series okay so let me let us think of indirect analytic continuation using power series so the idea is the following. So let me write that down indirect analytic continuation using power series so basically what you are doing is this means that you are trying to do indirect analytic continuation where your pairs d come d alpha f alpha they consist of a domain and a power series which is convergent and represents an analytic function on that domain okay so you are trying to get a chain of direct analytic continuation of functions that define the power series. Let f be analytic on a domain let f be analytic on a domain right and so I let me let me draw picture so here is my complex plane and well here is my domain d and here is my function f okay and I can do the following thing if you give me a point z1 in the domain okay then the function f is analytic at z1 certainly because it is analytic on the whole domain since it is analytic at z1 I can write out its power series centered at z1 okay. So the power series centered at z1 as you know will be power series which has certain radius of convergence okay and that radius of convergence the function the power series will represent the function f within that disc of convergence which is the disc centered at z1 with radius equal to radius of convergence z1 corresponding to this power series expansion. So take the Taylor expansion t f z1 at z1 of f with radius of convergence r of z1 okay so you know you know what this Taylor expansion is of course p f z1 is of z is just you know sigma n equal to 0 to infinity p z minus p z1 to the power of n by factorial n into f n of z1 where f n is the nth derivative of f this is the Taylor expansion the Taylor expansion at the point Taylor expansion centered at z1 is given by this formula okay it is just expanding f in terms of a power series centered at z1 which means that you are expanding f in terms of powers of z minus z1 positive integral powers of z minus z1 okay that is the Taylor series of f okay and of course the Taylor coefficients are given by nth derivative the nth Taylor coefficient is just the nth derivative evaluated at the at that point divided by factorial n right and r of z1 r of z1 is the is the radius of convergence of convergence of this of this power series it is a radius of convergence okay. Of course you know I want to look at the case I want to eliminate a silly case the silly case is when the radius of convergence is infinite okay if the radius of convergence is infinite it means that f extends to an entire function okay because if the radius of convergence is infinite it means the disc of convergence is the whole complex it is a it is a disc centered at z1 infinite radius so the whole complex plane is contained and if you take that that analytic function that will be a maximal extension of this pair d, f and that maximal extension is entire so actually this f is the restriction of an entire function and there is nothing special about it okay. Trying to extend an analytic function is serious only for analytic functions which have singularities why do you try to extend an analytic function you try to extend an analytic function just to discover where it could have singularities where it could have branch points and so on. But if the function is entire there is nothing to check is it extends to a unique analytic function on the whole plane and the story is over we are not interested in that okay so I am going to only consider the situation where the radius of convergence is finite okay assume f does not extend to an entire function so the radius of convergence is always finite for every z1 in D so you have that means you are considering a function which has only finite radius of convergence okay so the picture is like this see if you give me z1 then there is some finite disc surrounding z1 with radius equal to radius of convergence or of z1 okay. Now the first key point that is very important to this formulation of indirect analytic continuation with respect to power series using power series is the fact that you know if I now vary this z1 okay I vary this z1 after all z1 could have been any point in D if I vary this z1 then of course these discs of convergence will vary and the radius that the corresponding radii of convergence will also vary but the key point is this variation in the radius of convergence is continuous okay so here is a fact so here is a lemma Rz Rz1 is continuous in z1 belonging to D okay mind you I could have I could have used the variable point as z but if I use a variable point of z I have to put a different variable for the variable of this power series okay mind you that this power series will coincide with f of z on D intersection this radius of convergence I mean D intersection this disc of convergence so you must understand that so let me write that note P of fz1 of z is equal to f of z on D intersection mod z-z1 strictly less than Rz1 this is of course going to be true this if you take the analytic function defined by this power series and you inter and look at the intersection of this disc of convergence with the original domain you started with there it will give back the function because after all you know that the Taylor series of a function converges to a function to that function at every point in the disc of at every point in the domain of the function okay where it is analytic okay so this Taylor series this function P fz1 as an analytic function is just f when you take the intersection of the disc of convergence and the original domain right in other words I am just saying that this is a direct analytic continuation of f it is a direct analytic continuation of f okay but the point is that this Rz1 is a continuous function of z1 and so okay perhaps this looks a little too cramped is continuous in the variable z1 belonging to D in fact for z1, z2 in D modulus of Rz1-Rz2 is strictly lesser than is less than or equal to mod z1-z2 in fact this is the equation the equation is R of z1-Rz2 is less than or equal to mod z1-z2 the moment you have something like this you can see very clearly that R is a continuous function because given an epsilon positive okay if I fix z1 if I fix z1 and make z2 variable and if I give epsilon given an epsilon positive how will I make mod Rz1-Rz2 less than epsilon I just have to make mod z1-z2 less than epsilon so I will have to choose delta equal to epsilon in the epsilon delta definition of continuity of R okay so I can simply choose given epsilon I can simply choose delta equal to epsilon so this inequality tells you that trivially the function R is continuous according to the epsilon delta definition of continuity okay and how does this come about this simply comes about by properties of the disc of convergence and radius of convergence see so let me quickly indicate the proof of that proof so you know the so let me draw a diagram so I have situation like this so I have this D so here is my z1 and see I have z2 and I have I have this disc of convergence at z1 with radius of convergence Rz1 so you know I have some disc like this okay and then with z2 I have another disc and well this length is Rz1 that length is Rz2 okay and well you know if I draw this radius this radial line for the disc center at z1 then you know that this is Rz1 this is Rz2 and this distance is mod z1-z2 okay and we need to show what do you need to show to show this inequality you will have to show that R modulus of Rz1 that is Rz1 lies between Rz2 plus modulus of z1-z2 and Rz2 minus of mod z1-z2 this is what you will have to show right this is what you will have to show alright and I will just explain why this is true it is enough to prove this because if you prove this by interchanging the rules of z1 and z2 by symmetry you will also get the other inequality so you will have to just understand why this is true okay and why is this true because if you contradict it you will get a contradiction. So if this is not true if not true what you will get is you will get Rz1 is greater than Rz2 plus modulus of z1-z2 this is what you will get alright and I claim that this is a contradiction this will give a contradiction to the properties of the so called radial symmetry of the radial symmetry property of convergence so I will just have to use the fact that you know if a power series has finite radius of convergence then on the circle of convergence certainly there is at least one point where it will where the corresponding function will have a singularity okay see the proof of this claim is suppose at every point on the circle of convergence suppose you can extend directly analytically the function to an analytic function then what you are saying is that the analytic function itself extends to a disk which contains the circle of convergence okay and that is a contradiction to the property of radius of convergence the property of radius of convergence is that inside the disk of convergence the power series will converge on the boundary you cannot say anything but outside the boundary it has to diverge that is the property of radius of convergence. The radius of convergence is smallest such number such that you know I mean it is a number with the property it is a unique number with the property that inside the circle of convergence the function I mean the power series converges outside it diverges you cannot say anything on the circle of convergence but from this definition it follows that if at every point on the circle of convergence you can directly analytically extend the function then this function is analytic on the whole it will be analytic on a bigger set than the disk of convergence and that is not allowed on a bigger open set on a bigger open disk than the disk of convergence which will include the circle of convergence and that is not allowed because outside the circle of convergence the power series is supposed to diverge that is the property of the radius of convergence. So if this is the case then what will happen is that this circle centered at z1 radius rz1 will certainly contain the circle centered at z2 with radius rz2 right this is what will happen you see this distance is mod z1-z2 okay this distance is mod z1-z2 so if I draw it like this this is mod z1-z2 and this remaining distance is rz2 okay so this distance is mod z1-z2 this remaining distance is rz2 their sum is rz2 plus mod z1-z2 okay so this sum is rz2 plus mod z1-z2 and if rz1 is greater than that okay then it means that this bigger disk the disk centered at z1 radius rz1 contains this smaller disk okay but then it means that the analytic function the power series centered at z2 that analytic function extends beyond the circle of convergence which it should not okay do you understand that so let me write that down and that will give you the proof this will this would mean that the disk of convergence mod z-z2 less than rz2 is contained in of the power series of f centered at z2 is contained in the disk of convergence mod z-z1 this is an rz1 implying that the analytic function tfz2 extends to the circle of convergence mod z-z2 is equal to rz2 be a contradiction and that finishes with proof okay that finishes with proof. So if you contradict this inequality what will happen is that one of the disks lies inside the other okay and then it will tell you that the power series in the smaller disk is extendable to an analytic function even on the boundary of the smaller disk but that is not supposed to happen if an analytic function if you take the analytic function defined by a power series and if it has finite radius of convergence then on the circle of convergence there has to be at least one singularity okay that is because the circle of convergence is supposed to be defined uniquely as the circle inside which the power series will converge always and outside which the power series will always diverge okay. So the fact is that that property tells you that you get this inequality and if you interchange z1 and z2 in this inequality by symmetry you get this inequality and both put together you get this inequality and this inequality tells you that the radius of convergence at each point is a continuous function on the domain okay and this lemma is the crucial starting point to define and to treat analytic continuation using power series okay. So armed with this lemma I can make definitions and I will do that in the next lecture right.