 Okay, so today I will continue using the notions we introduced last time. So essentially we had, we had introduced the metric structure we will use for the entire course in the complex, in the extended complex plane. We introduced a notion of convergence of series and we also observed that given a sequence of complex numbers, we can also consider, put it this, it's very difficult to see. Oh, is it very difficult? Okay, so if we have a sequence of complex numbers, we have the notion of convergence of the sequence and we also consider the sequence of sums up to the nth element. This is also a sequence of complex numbers, right? So we consider this sequence as well and we can also say that this sequence is convergent or not. If this is the case, we define the limit to be this. Okay, I can also put zero here so I put zero, okay? And we call it a series. So the infinite sum is, if it exists and is a complex number, not infinite. The limit is n tends to plus infinity of Sn. When this is a finite number, okay? When this series converges. We have the possibility to check if the series convergent using, for instance, the Cauchy criterion. We know that the metric space we have is complete, so we can check it. Good. I want to leave you as an exercise that if summation of Zk exists equivalently, if the limit of the partial sums is finite, then the nth element tends to zero. It cannot increase. It actually tends to zero, right? Sn tends to infinity. The sums are made with the elements which are infinitesimal. Good. So this is an exercise. Together with the notion of convergence of series, we also have the convergence of the associated real series. That is to say, we can also consider the series of the modules of Zk. And this is, since Zk is a real number, this is a series over the reals. If this exists, then we say that the series is absolutely convergent, okay? Or in absolute value. That's the reason why it's called absolutely convergence, okay? This notion is important because we can prove the following fact. Proposition. This is lesson 3, page 2. So if Zk is absolutely convergent, then it is convergent. There's nothing special in some sense. This is the same as in the real case. And the reason, say, very sketchy proof is the following. Well, we have to consider the convergence of what? Sorry. The convergence of this series, Z naught plus Zn. Of this sequence, right? On the other hand, we know that this, sorry, this is convergent by assumption. So to prove that the series is convergent, that is to say that the sequence of Sn, the partial sum, say it's convergent, suffices to prove that it satisfies this inequality for Nm greater than a certain m. But what is this? So if we prove this, we have the assertion. But what is this? Well, this is summation, say, assume that n is smaller than m k from n to m, okay, of Sn, you correct, correct. Sorry, if n is great, n is smaller than m, so we have exactly the opposite, right? m plus 1 to n, right? So we have a, sorry, Zk, okay, and this is a finite sum. And because of the triangle inequality, we can say that this is smaller or equal to this, n plus 1. Okay, I'm sorry. n is smaller than m, right? So we are summing from n plus 1 to m. n plus 1. Sorry, sorry, sorry, sorry, yes, sure. Thank you. Yes, n plus 1 to m, correct. And this is, in fact, smaller than epsilon, okay, because of the previous exercise if you want. So we have this check. So absolute convergence guarantees convergence, as in the real case, good. Okay, so now we go back to some notion which is real and is the following. Consider a sequence of real numbers. As you all know, this is not always the case that the sequence is either convergent or convergent. There are some other possibilities. If you don't add any other hypothesis, you can have that the sequence, for instance, minus 1, plus 1, minus 1, plus 1, plus 1, plus 1, has no limit on one side. On the other, it's not divergent. So there are some criteria for this. Monotonicity guarantees, for instance, that exist another limit, finite if the set is finite or infinite. So convergence or divergence of the sequence. In any case, we can define, and it has always meaning the lim soup and lim inf of this sequence. These two numbers always exist, possibly being plus infinity and minus infinity. And if the limit exists, lim soup and lim inf coincide. Good. So lim inf is, I will use lim soup actually more, also indicating this way which I don't like for obvious reasons in some books like this. And lim inf is, I don't like this because there is some, well, can be misleading with the conjugate. We prefer to use the overline only for conjugation. In any case, I think that you have the, well, this is limit of Rn. In any case, this is the lim as n times plus infinity of the soup, right? Lim soup is the limit of the soup, yes, of Rn, Rn plus 1, that's it. And here is the lim as n times the plus infinity of the inferior. That's why this is lim inf, okay, Rn. So we always have that this number exists, these two numbers exist, and this is greater or equal to this, okay? All right. I don't want to bother you again with the standard stuff you broke. Yes, please. This is the real, okay? In real analysis, nothing. No, no, no. I'm just, I'm just recalling you some facts from real analysis, nothing special. Well, we can use, we can use, if you want to know why, and I recall, well, I don't want to, as I said, I don't want to bother you with properties of some of sequences. I want to use this lim soup for the sequences of the modules of complex numbers, okay? So, in order to apply the properties of real numbers, and you will see in a few minutes, the reason why, okay? I just want you to recall, okay, this problem, and just to have something put here. So, indeed, what I want to start from is a following. We have considered sum of complex numbers. So, starting from a sequence of numbers, we have the sums of these numbers, and this gives you another sequence of complex numbers, and we consider the converges of the second sequence of complex numbers. Now, we can also think of a sequence of functions of complex value functions, all right? And then sum them, because you can sum, okay, the complex numbers. Now, the point is, if the function depends on a complex number, is it always the case, choosing z in the complex realm, that the series of the sums is convergent? This is not always true, as we'll see. So, the next task is to define, so to describe in which set z can be taken in order to have convergence of sequence of sums of complex value functions. So, before continuing, let me just recall you another inspiring, in some sense, the fundamental examples we all know, probably. So, let me recall you that from the real, then you change a little bit. We have this sequence called ak to be z to the power k, k natural numbers. This is meaningful, because we can multiply z times zk times. So, this is nothing but z times z k times. Which means that this is the case power, okay? So, associated to this sequence of ak's, we also have sn, which is the sum of 1, which is z naught, a naught, sorry, z to the power 0, plus z, plus z squared, plus zm, correct? And we wonder if we can express this, and this is one of the few cases where you can do this, this sum, this sequence of sums in terms of z. And as you probably all remember, this is 1 minus z, the power n plus 1 over 1 minus z, right? So, that it is much easier than to consider the limit of this sum of objects to consider the limit of this, which is a fraction of complex value functions. So, notice that this is a power of a complex number, there are no coefficients in front. The generalization of this is, of course, to consider something like this, summation of, so, to consider bk to be, say, ak times z minus a to the power k, with ak being complex number, a being something different from 0. So, in the previous example, we have ak, all the ak is equal to 1, a equal to 0, but in some sense, well, the fact that we are considering z minus a instead of z is not very important. We consider, for instance, this change of variables putting w z minus a, we have, well, that bk is ak times w to the power k. So, the great difference is that we put some coefficients in front of the power, okay? And this coefficient is, in general, complex, not necessarily real, okay? So, what do we know about convergence of the series Sn written here? By the way, you probably all heard of the legend or legend of story, actually, or it's not a legend, it's a story. This formula was obtained by Gauss when he was very young, right? Okay, good, good to know, okay, this. And it can be, in fact, obtained as he did by considering that, well, if you take, well, let me make this small regression, if you consider Sn like here, so I copy this, and if you consider Sn plus 1, it's the same, right? Up to, then you multiply by z this, and you obtain everything from here to here, correct? So, then you make the subtract, and you obtain this formula here. This is the idea, because Gauss was asked to sum up to the first, sure, sure, of course, he was very young, he didn't have, well, but the same procedure works fine, but because he was bothering the teacher, right? Because he was very smart in calculations, and so the teacher said, okay, now sum up the numbers from one to five thousand, and he answered immediately, so how do you do this? Well, there's a general formula, say, good to know this, okay. Anyway, so here, in this expression, it is evident that this number can be very big or very small in modulus according to the fact that the modulus of z is greater than one or smaller than one, correct? Remember that we have, we're not, I'm not cheating you, okay? We're not abusing any formula up to now, because we have the Moir formula, okay? You can see this. Since the modulus of a number smaller than one is smaller than one, and the nth power of this number is the nth power, sorry, the nth power has a modulus, which is the nth power of the modulus, this decreases if the modulus of a starting number is smaller than one, correct? This is real analysis, and if it is greater than one in modulus, this number, okay, of course, this number, this real number increases, and so it cannot convert as a complex number as well. Remember that we have characterized the zero in the complex number as the complex number with zero modulus. It's the only number with zero modulus. So if we prove that the modulus tends to zero, it's enough. We don't care if the convergence is in some spiral or on a line, but well, this is important, correct? And on the other hand, if this converges, so this number tends to zero, as it is clear that the condition modulus of this more than one is sufficient to conclude this, well, the limit is one over one minus z, which is meaningful for any z but not for z equal to one. So in some sense, this example shows you the following. So we have found that we can take an open set, a disk, in this case, the disk has radius one, and for any z inside this disk, you have convergence, and you have an explicit expression of the limit of the series. When z as modulus greater than one, the series divergent, nothing can be said about the convergence of the series when the modulus of taking z is equal to one, okay? And in fact, for z equal to one itself, this has no meaning. Let me point out that, of course, this is nothing but, I have to put here three third lesson number four, this is nothing new for you, I guess. And in fact, this was known also for the real numbers, right? Same considerations, and you conclude in the same way. The only point is that in the real numbers, you can see that there is an abstraction to consider, all right, so let me put this way. There is an abstraction when considering complex numbers of modules one in the complex case, and there is an abstraction of taking real numbers of modules one, there are only two, minus one and one, you see this on the Gauss plane, okay? So, and when you restrict to the real case, the same series is convergent when z is real, when z is in here, which is the intersection of the real axis with the unit open disk. And it is also reasonable to say that nothing can be concluded when we take a real number of modules one, because when it is one, in fact, the limit does not exist. With some manipulation, we can also consider as limit this number here, say, assume that when z is real, right, is x instead of z, right? So, we take instead of one minus x to the denominator one plus x, and this can be written as the series where x, okay, is taken. So, take minus z and the previous expression, and you will see that the limit is one over one plus z, when this limit exists, so when modules of x is smaller than one. And, well, up to here, nothing changed. So, instead of having a singularity here, you have a singularity here. Say, well, what's the difference? Well, with another slight modification, you can consider this limit of the series, I'm not mistaken. So, I consider x squared here, so here everything is 2k as a power exponent, sorry. And here, in the real case, you don't see the reason why modules of x has to be smaller than one, because there is no, well, this number never vanishes, never vanishes in the real. Well, it vanishes, of course, in the complex case. It vanishes here and here. So, this is one of the explanations why, considering the real case, you have only a partial view of this phenomenon. Well, just to, okay, anyway, let's go back to the example. And, well, first, let us give this definition, and the definition is the following. So, let us consider the generic case. I want to write it down at least once. So, we said that the example we consider so far is just a simple example of a more general case of series we will consider. And it is the following. This is called power series, and A is called the center of the power series. So, we also say the power is centered complex numbers, okay, where all the elements which appear in this expression are complex, okay, in general. So, the task for our purposes, it would be important to have this information, is to describe when and how and under which condition this has a meaning. This is a number, okay. Given a z, you can obtain, so this is a function. And as I said last time, functions were not rules in the past, but essentially were extension of polynomials, so series, okay. So, the complex case in the complex case, in particular, function were considered as power series. Good. So, there is another interpretation of this series, and I will give it here. So, assume that we have a family of complex valued functions. Nothing is assumed about A, A is a set, okay, and K varies in the natural numbers, okay. Well, we can always consider this series, this, sorry, this, pardon, this sequence summation take an element A in A, so K varies from 0 to N, this is a sum of complex numbers, right. How can we do this? Well, we can do this because C has a structure of group, so we have a sum, you see, and we can sum it. And we consider now this infinite series. So, there are several definitions of convergence for a series of functions. Why are we considering functions? Well, because in the previous, let me go back to the previous notion. We can also interpret this, each of this addendum here, summand, as a function. It is a function, it is complex valued function, definitely, and just by chance it is also a function of a complex variable, okay. So, forget this, and this is the FK in this specific example, right. So, let me give it in the more generic way I can now. So, we have this series. What does it mean that it converges at A? Well, at A it means that the sum, sorry, the sequence Sn has a limit when the sum is evaluated at A. So, it is, okay, A is fixed, we consider FK of A is a complex number, then we consider the associated series. What do we mean when we say point-wise convergence in A? We repeat the same property for any A in A, correct. Now, what do we mean when we say that this series is absolutely convergent at the point A? Well, again, we remember that associated to this series of complex numbers, when FK is evaluated at A, the associated series of moduli is a series of real numbers, study the convergence of this series of real numbers, and then we conclude that if the series of the moduli converges, then the convergence of the series said to be absolute, okay, or in absolute value. There is another convergence which is also very important to know, notionally, the uniform convergence, right? Do you have a notion of uniform convergence? Well, I think that in, also in real analysis fundamental, you say that you probably know that if you have a sequence of continuous functions and you consider the uniform convergence set, of course, a sequence of continuous function on a set, so we have topology and so on, okay, now we are not assuming a day as a topology, but we have to want to have a notion of continuity, okay? So the uniform limit, or the uniform converges guarantees that the limit is, the limit of a sequence of continuous functions is continuous. How do we write uniform convergence in this special case? Well, we can say that the sequence is this uniformly convergent a if for any epsilon positive, and there exists a capital N such that for any N and M greater than N, we have that S N A minus S M A is smaller than epsilon for any A in A, correct? So uniformly means without any dependence from the point A chosen in A. All right. There is another notion, well maybe this terminology can be standard. This is not standard. I think that normal convergence is somehow, or normality property is, because normal comes out in several fields in mathematics. So normally, well in this special definition, it means that you can control each F N of Z as Z varies in a set B with, sorry, F K, with a constant number M K, okay? And M K is of course real number, positive real number, M K has to be positive real number. And we say that, well the sequence, sorry, that the series converges normally if the series associated of M K is convergent in the real sense. And I, well I probably will use it sometimes. And probably we don't need a special definition for this because it is just an application of other comparison theorems to be applied just to have this. In any case, in some books you find it, this is the bias stress M test. The use of letter M is not by chance because normally we call it M test or bias stress M test. So you test that the series of function is convergent using the fact that you know that you can control the moduli of the summands using a constant depending on the index and not depending on Z. When the Z varies in a subset. Last thing I had to say just at the very beginning of this, never use the index I in complex analysis for summations, okay? This is just, it's not compulsory, it's not, but it's warmly recommended, okay? Not to use I. That's why I use K. We use other letters. Of course it's up to you, but don't use I. And it's obvious why, okay? Finally, of course, if the series is not, is not convergent, well it is divergent. And so going back to what we had obtained, let me prove this first important result, okay? So first, consider a power series expansion, okay? Z minus I to the power K, okay? And then consider the number modules of ak, which is a real number, and take the kth root of this number, okay? And define, actually calculate the lim soup. That's why I had to recall you this definition. Lim soup always exists. Lim is not guaranteed, but lim soup exists and it can be plus infinity. Because this number here is non-negative, right? So we have three possibilities. Either this is plus infinity, or it is zero, or it is, or it is a number. Yes, sure. Call it l, right? In any case, we have one of these, one of these results. So the limit lim soup exists. And the first case, we say that r is zero, whatever r will mean, we'll say. Pardon me. When r, when this lim soup is zero, r is plus infinity. And when the lim soup of the case power of ak is l is a finite number, real number, we call that r is one over l, okay? In some sense, this is the definition with the extension for these two cases in the obvious way, okay? So that we can say that one over r is lim soup of ak one over k, as k tends to plus infinity, with the convention that r is zero when this is plus infinity and r is plus infinity, okay? When this limit is zero, correct? So this is an associative number to the power series considered. So the theorem goes back, this is a result due to Hadamard, it's the following. Now, assume you have the power expansion as before and define r to be one over lim soup with the convention with the convention we adopted just so far, right? So r is zero when this number is plus infinity, so the denominator diverges, then r is zero when r is plus infinity means that this number tends to zero. Then this number is in fact a measure of the ray of the of the of the circle center that a for which z has to be considered to have a convergent power expansion. This is the meaning of the symbol r. So if so if z and c is such that modules of z minus a is greater than r, then the series diverges and if z and c is such that z minus a smaller than r, then the series converges and something more is added. If you take, assume that this number here r is positive, so assume that r exists and is positive and take a positive real number between zero and r, you can always do this, then in particular, if r is smaller than capital R and the z minus a is smaller than that end, then the series converges uniformly in this disk. This is the content of Ademar theorem. As we will see, the proof is quite easy and keep in mind the example, the fundamental example and the fundamental example, all ak's, all the coefficients were real, well this is not very important because we are considering the modularity. But we somehow were considered the modules of all this part here and it had to be smaller. So if we had a power, each the base of this power had to be smaller than one to guarantee convergence. In this case, we don't have a power because this is the number times something which is some power. But when you consider, you see here the, sorry, you consider the kth root and you substitute, imagine you substitute this fact inside here, you are considering the kth power of this number times z minus a to the power k. When you consider the modularity, you are very close to the example given before. So in some sense, we are trying to find a balance of the way this number is increasing with this number, but not this number, this number. This is not the kth power of something, but the modularity of this is the kth power of this. This is the leading idea. Just another small remark, nothing is said about the convergence of the convergence of the series when the modulus of z minus a is equal to r. So when we are like in the case of the fundamental example, precisely on the boundary of the sequence and notice that because of the geometric property we are considering, the modulus does not depend on the argument. We always have disks, something which does not depend on the argument, depends only on the modules. So that's why this r associated to the series, the power series expansion is called radius of convergence. It can be 0, it can be infinite. So in this case, the entire plane is the disk. Otherwise, it is a disk centered at a. And for this z with the property here, this series defines a function. So we are in some sense describing the set where analytic or power, say, complex power expansion exists as functions. When the radius is 0, it means that, well, nothing can be said about convergence except one point. And there are examples. The example, the fundamental example shows you that, well, we have a series where the convergence is guaranteed only in a small radius. And we will show you also examples of functions which are defined on an entire complex plane, which with this terminology have radius of convergence infinite. So now, let us see the proof. And then I give you the examples. Now we are page 3. So first of all, for the sake of simplicity, as I said previously, we can always assume that a is 0. This is not, without loss of generality, we can do this because instead of considering, we can always transform this into akw to the power k where z minus a is w. Geometrically, we are just shifting everything from a to the origin, but the consideration for convergence is independent from shifting. We are taking this or something different, but it is not affected by any translation. So this is just for the sake of simplicity. Good. Now, assume that r is given and take modules of z smaller than r. R positive. We are considering convergence in a disk. So a is 0. So this is exactly the condition we have to prove to be sufficient for convergence of the series. Correct? Good. Since these are two real numbers, we can always take a raw real number such that modules of z is smaller than raw and smaller than r. And we observe that with this choice of r, when 1 over r is smaller than 1 over raw. Remember that 1 over r is the limb soup of the nth root of the ak's. That is to say, there exists a certain n such that for any n greater than n, we have a certain k greater than n that ak has the kth root of its modules smaller than 1 over raw. Now, what do we have? We have, as I said, keep in mind that we are balancing the growth of w, c minus a to the power k, and the growth of the modules of ak, the kth root of 1 over k. So we have a summation of ak zk. Take the modules of the summands. And this is smaller or equal of what? This is smaller or equal of 1 raw to the power n times z to the power k. From this, in fact, we have that modules of ak is smaller than 1 over raw. What do you mean? Yes, definitely. Yes, sure. Good point. Let me remark that, well, of course it's correct. When you consider a series, you are considered an infinite sum. So what is important to control is the infinite part. So call it the tail or whatever. So if you prove that something is converging from a certain n, it's enough because the other summands have a finite sum by definition, correct? So you have only some extra elements to sum, but they are finite. Their contribution can be only finite. So the difficult part to prove is that infinite. Okay, so correct. This is not true in general, let's say k greater than n, but this is enough to guarantee that entire series converges, okay? Why is it converging? We haven't proved that the series converges. Well, so from n to infinity, these numbers are controlled by these other numbers. And this number here, not by accident, on purpose, we have taken, remember, raw to be greater than modules of z. So that this number here is in fact converging, sorry, converging smaller than one, and therefore this series is converging because of the previous considerations, right? This is a real number. So if you want, in this case I can say we can use the m test and prove it. I conclude this, okay? So thank you for the remark. I sometimes abuse the notation and say, well, this is true for any n, for any k greater than n, then for convergence consideration it's enough. So this is, because this is smaller than one for this assumption, then we concluded this is convergent and then absolutely convergent and then convergent. So we have a series of implications which guarantees that the series we started from, the power expansion is in fact convergent. Well, slight modification of the argument applied so far guarantees that, well, if we have, as I said, z smaller than r and smaller than capital R, well, we repeat the same argument putting a raw in between the two r's, same considerations as before and then instead of having the inequalities before, depending on modules of z to the power k, we have modules of r over raw. So this is greater than n and as you see, this is smaller than one because raw is greater than r. So this is convergent because it's a power expansion, sorry, the power series with base smaller than one and as you see here, independently on the choice of z. In the previous consideration, well, we had converges but depending on the modules. And finally, well, the reverse, if z is larger than r, capital R, so it's bigger than the radius of convergence, you put a raw in between, then you obtain the other inequality with something which diverges from below. So you have an estimate like this, summation of a k z k, k greater than or not, is greater than summation of summation raw over r, the power k and this number here is greater than one. So this cannot converge because by comparison of theorem, somehow it's called the lift theorem, this diverges and this diverges. So it cannot be finite, say it this way. Now, this is how to, so up to now we have a criterion to say, well, we can consider power series, complex power series and say when these are functions or not, so we have a domain of definition, it can be empty, it can be one point, it can be a disc, it can be an entire complex plane, but we have something. Now, I'll make you calculate some ragi of convergence. Before continuing, let me recall you that, well, there are some other properties, can be useful, assume that, well, you start from, as we did, such a power expansion and assume that, assume that this limit exists. Yes. No, no, no. Okay. Okay. If I, okay, I have to repeat the question, you know, for some reasons. So, the question is about the consequences of taking C greater than R, right? So, probably I did it quite quickly, but what I'm saying is that if you take Z whose modulus is greater than R, you can always have such an estimate, correct? So, the moduli of the tail of the series cannot be finite because you have an inequality here. Assume that we are considering partial sums. These partial sums here diverge to infinity. What I'm saying is that, well, this implies that the infinite part of the series from certain N to infinity has a modulus which is not finite. You are correct. You are correct, but the series cannot converge. Are you with me or not? So, okay. Now, thank you for the question. Let's go back to the fundamental example, right? The fundamental example is the leading example for, no, it is what we actually use here, right? When you have the sum, remember that this is 1 minus Z, and you want to prove convergence of this function here. We are considering Z, it has to be smaller than 1. Otherwise, the modulus of this number here is divergent. Look, look, we can reduce, we can reduce the general problem to this situation. Probably, I didn't write it explicitly, so maybe it is worth repeating it. When you have this, okay, we started with this example, okay? The basic example was the following. Correct? And for this, everything is understood. Modules of Z greater than 1, no, convergence. Modules of Z is smaller than 1, convergence. Modules of Z equal to 1, nothing can be said, okay? Now, we somehow generalize this position by taking this. Well, the fact that we are already assuming that A is equal to Z, so the center is 0, is not important for our consideration, right? Because the radius doesn't depend on the center. Good. Now, call A, K, Z, okay? Now, we have to reproduce something, okay? So, the modulus over K Z power K, okay? I want it right this way, okay? A K, 1 over K, Z. Can I write it this way? So, and this is, call it the modulus of W. So, consideration given before is independent of the argument, and let's explain you why the radius of convergence is related to the limb soup of this number. If it is smaller than 1, greater than 1, okay? So, greater than this number or smaller than this number, which guarantees, okay? Maybe if you don't mind, okay? If I'm not convincing you, it's better if we have an extra talk somewhere later, okay? Okay. So, let me continue. Yeah, it was the kind of exercise I wanted to leave you. So, assume that this limit exists, okay? And, well, show that L is the radius of convergence. This is the ratio test for convergence, right? Okay? But if for you it's a nobule consequence, well, it's okay. Maybe for some, sure, that's what I, what I tried exactly, okay? Now, we have some, we have some time, okay? In 20 minutes, I want to give you an example of a series whose radius of convergence is plus infinity. We have an example of a series, a series which exists and is converging only for this smaller than certain value, finite value. And, well, the example I'm thinking about is, well, the complex exponential. How can we, well, it is a very important function in complex analysis. How can we define complex exponential? So, we have, of course, the notion of complex, sorry, of a real exponential function. We can introduce the exponential in several ways. But essentially, we can characterize the exponential in this way. So, it is the only function which has this property, x is equal to x. And, okay? So, this characterizes, well, this is a fact differential equation. And this implies that f of x is e to the power x. Now, from the, from this definition, we also know that e of x is real analytic, correct? And it has an expansion. What's the expansion of e of x? So, the exponential somehow, I also use this notation is 1 plus x plus x square over 2 factorial plus x n over n factorial plus so forth. So, when you don't know what to do in general, well, start to test what you have in mind from the real case and simply substitute the x with the z, okay? And try. So, I wonder if I'm doing a mistake considering, say, x is a capital E, okay, to distinguish the other of z to be an extension over the complex numbers in this way. I consider, well, this is, in fact, it's a power series, all right? Summations of xk over k factorial, correct? And, well, I just wonder if doing this, I'm doing something silly or not. This is power expansion, a complex power expansion, for sure. No, no problem. Center dot the origin, good. Coefficients are real. So, it's in particular, say, we have that ak is 1 over k factorial. So, what is the limb soup of modules of ak? Well, modules of ak is ak in this case, because ak is real and positive. What is the limb soup of modules of ak? Okay, what is that? Well, this is the limb soup of k of 1 over k-th root of k-th factorial. So, factorial is faster than the root, so, by a comparison argument, we can conclude that this number here, the denominator, diverges. So, that this limb soup is 0. And hence, the radius of convergence of the function x per z is plus infinity. So, it's entire, and as you can see, it is, well, it is well defined. So, for any z, you can define a function, sorry, for any z, it has a meaning to consider the function x z as defined here, correct? What you cannot do at this stage is to prove or to look for a similar condition, that a similar condition holds for the complex explanation, because we still don't have any notion of derivative. We see that, well, of course, this condition is true, which means that the first coefficient is 1, a naught is 1, okay? But nothing can be said about the second. So, they're the first, sorry, the other one, correct? So, before continuing, let me just use this notation for the sequel, okay? Let us start from a series, power series, general one, okay? And assume that it has radius of convergence r. We have examples, but this set is not empty. Now, we consider another series. From this, we obtain another one, which is called, which was called by the first researchers in this field, the derived power series, derived. So, it is derived from this one, okay? Derived power series. Given this power series with radius of convergence r, the derived power series is the following. Summation for k, and then we consider k, a k, z minus a, k minus 1, okay? Sorry. Yeah, we have k greater or equal to 0. And here, the derived series, power series as index is going from 1 to infinity. This is, well, this is historically the introduction of the right power series. What do we know about the radius of convergence of this derived power series? Is it the same? Yes, it is. Luckily, because I hope that everybody remembers that this series, this, sorry, this sequence tends to 1. Because of this, we conclude that the derived power series has radius convergence r. So, the same radius of convergence of the power series we started from, which was called the primitive power series, okay? So, this terminology, of course, went in use also in other, in other, but now we want to show that, well, it's not by chance that these are called derived and primitive. So, you expect that it is a relation with some analytic operation on functions, which mean, which are supposed to be the natural one. So, derivation and integration, right? So, you say primitive of a function, a function which is the property whose its derivative is the function, the given function, vice versa. The derived function is the derivative of another one. Now, and this is exactly the case in complex analysis, as we will see. Surprisingly enough, this is the case. So, first, let me observe that the derived power series is obtained by derivating whatever it means, the term by term. So, if you apply just what you know from the real case and say, I want to consider, this is a power and I apply the rule of derivation of a power without knowing a z is a real number, complex or whatever, modify the exponent, subtract one to the exponent, right? And multiply the coefficient by k. So, it is the, I would say the term by term derivation of a power series. I want to show you that the derived power series is in fact the derivative of this function. When it exists, it is a meaning, while it is complex differentiable in the sense that the limit of the incremental ratio exists. And the two notions coincide. So, we consider, use this notation, f of z to be the starting power series, which is a function when z is such that z minus a is smaller than r. So, assume that we have power series expansion. We can calculate the radius of convergence. Then, we take z such that z minus a is smaller, modus is smaller than r. And this defines a function for any z. And this disk, you have an output, which is a complex number. Now, we define also f1 of z to be this other series expansion, the derived series expansion, which exists as a function, complex value function, when z is in the same disk as you observed before. And then, the next task is to consider this incremental ratio of f when z minus a is smaller than r and z not minus a is smaller than r. Otherwise, it doesn't have any meaning. What does it mean? Well, it means that as soon as z is different from z not, this number can be calculated, because it is the quotient of two complex numbers. This is a complex number. This is a complex number different from zero. Well, we can calculate this and it is a function. Now, f of z can be also written this way, summation of the first nth terms and then the tail. So, I consider a0 plus a1 z minus a plus a2 z minus a square plus am z minus a to the power n. And this is polynomial of the first n terms in the power expansion. And unfortunately, well, we call it sn of z. And then we remember that, of course, this is not always the case. Well, it can be that we are considering a polynomial, but in general, we have an infinite power function. So, an infinite power series, sorry. And then we consider the rest to be denoted by rn of z. What is rn of z? rn of z is f of z minus sn of z. So, just to come back to what you pointed out before, in this case, we know that this series converges when z minus a as modules more than r. But what we have to control in case we had to is this part, the modules of this, not the modules of this, because this gives you just a finite contribution to the sum. So, I don't think we can complete the proof, but just to give you an idea, let us consider this. So, when I write f of z minus f of z0 over z minus z0, I'm actually writing sn of ze plus rn of ze with the notation so far introduced. So, the sum up to the nth term plus tail, I call it tail, because normally it's the infinite part minus sn of z minus rn of z0 over z minus z0. And I can also write it in this way sn of z, sorry, z0, right? Minus sn of z0 over z minus z0 plus rn of z minus rn of z0 over z minus z0. Nothing special. Using the notation sn and rn for f, what is the relationship with the function f1 of z that the derived function? The derived function was, sorry, the derived power series is the one obtained this way. First of all, I recall that f1 of z is summation of k a k z minus a k minus 1. And since sn of z is a0 plus a1 z minus a plus a n z minus a, which is a polynomial, we define s prime n z to be a1 plus 2 a2 z minus a plus n a n z minus a n minus 1. This is another polynomial. Notice the coefficients here. I precisely the first n minus 1 coefficients of f1 of z. So, that f1 of z is the limit as n tends to plus infinity, so s prime n, bless you. So, if we start from this fact, we have that fz minus fz0 over z minus z0 minus f prime of z0. Turns out to be, remember that the first part was this, the incremental ratio of sn, then the incremental ratio of rn. And I can always add and subtract the same amount obtaining the following. So, I added and subtracted this complex number. Remember that z0 is in the radius of convergence in particular of f and this is a polynomial. So, it is completely meaningful. Now, you can notice that this is very close to be similar to what we started in calculus one. And as I said, this is a polynomial. This function here is a polynomial z. So, that in some sense, the terms you have to control is not, the terms are finite. The difficult part to control is this. So, next time we will complete the proof of this. And in fact, we will use an expression for this, for this ratio here and control the modules of this expression and show that this, this modules, this series converges as soon as z and z0 are in the radius of convergence. All right? I forgot to say that of course, all these considerations are, can be repeated without losing any generality for a equal to zero. So, center of the origin. And in case I, I will forget next time, I just say now that from now on, for our consideration, we take a equal to zero. Okay, I stop here. And thank you for your attention. And sorry for the inconvenience of the, the homework, the missing homework except for one of you. This was my fault. I will send it again. Okay? Please answer me back and say I received something. Okay? And then you can read it. Can you read it? Good. All right. So, see you on Tuesday afternoon. Is it correct? Maybe 2 p.m., right? Good.