 So, first of all, good morning, and I hope you had a good day yesterday, so you got a chance to see how our days will go, so the structure will be, I think, very similar throughout the two weeks. We have three lectures in the morning, and then we'll have these exercises, and it's really, really, for us, it's really key that you sit down and try to do whatever you can, because this is the best way to absorb, otherwise we keep telling you things, and you don't really learn if you don't try yourself. Okay, and we are here to help. One small comment, I think the tutorial session, I don't know, I think yesterday it was wrong at some point on the schedule, it's at 2 p.m. So I think it was written 1.50, I'm not sure, maybe it wasn't, but, okay, it starts at 2, and also I had some questions. So the room is booked from 2 to 4. On the schedule it's written, this is wrong. Why coffee break, ah, maybe, ah, it's confusing. Is coffee break at 3.30? I thought coffee break is at 4. Okay, I don't know, at some point we will have food. Okay, there are two things to correct in the schedule. So first of all, the room is booked from 2 to 4. In reality, we wanted to put on the schedule from 2 to 3.30, because we don't want to force everybody to sit here for two hours. So let's say from 2 to 3.30, we hope everybody will come and try after 3.30, if you are tired, feel free to leave, maybe some of us might leave too, but if you want to keep working and try solving your problems, you can stay until 4 to work. And I thought coffee break should be at 4 every day. And I think the other days it's at 4, so I don't know if today I will check. At some point it will be coffee break, I think it should be at 4. And today in the afternoon, we will have a socializing networking event, specifically for women and for gender minorities, so whoever self-identifies with women. And again, this will be also a type, it's not on the Adriatico, I already asked for this to be corrected, but probably it will be corrected only later. And it's also in the terrace, the same place where we had the socializing yesterday with the ice breakers, we will have some small groups discussion. And again, it's like a networking socializing opportunity. And we also have some questions about gender and mathematics and some discussions together. So hope to see all the women and gender minorities, people who later. Okay, so time to start with mathematics. And first I will give you the solution. I think most of you also did it, but let me first complete the theorem. We were proving yesterday. So we saw that if you have a rotation and the angle is rational, every orbit is periodic. But if the angle is irrational, then every orbit of the rotation is dense. So again, dense means for every, I pick any other point on the circle. And I pick an epsilon, and an epsilon neighborhood is like an arc with size epsilon. And dense just means that at some point my orbit will fill, will enter this arc. And here I wrote, when I write distance, again, you can think of this distance as measuring the smallest arc length between the two points, okay? Maybe divided by two pi, so that the total distance, total circle is one. Okay, and we did two steps. We showed, first of all, if the rotation number is irrational, then the orbit never close up. So you never get two same points. And then we used pigeonhole principle to say that if I chop my arc, my circle into capital N, sorry, this is one of them. If I chop my circle into capital N equal arcs, then, and I look at N plus one points, all distinct, automatically distinct, then I will find two points which fit in the same arc. And again, here I need to divide, my arc length has to be divided by two pi. So I want to say that the arc length between these two points falling in the same arc will be, and this is RL. And this distance is less than one over N. And as many of you did, what do I have to do next? So the key observation is to remark that R alpha is an isometry, which means that if the distance between R alpha of Z and R alpha of omega is the same than the distance between omega, Z and omega, this is obvious on the picture. So if I have two points and I rotate them, the distance is the same. So this implies plus the fact that it's an isometry, implies that I can apply R alpha to the minus k, if you want. Which is also, it's invertible and it's also an isometry. And what I get is the distance between Z and R alpha to the k minus L of Z is also less than one over N, okay? And I'm assuming, for example, that k is greater than L. Otherwise, you can do the other way around. So this tells me that R alpha to the k minus L is a rotation by a small angle, small angle. And I can choose N such that one over N is less than epsilon. So now I have my point Z. And if I move by this multiples of the rotation, if I move by the rotation of k minus L, which is a subset of the rotation that I had before, I look at these points which I get when I apply power k minus L of my rotation. These points of the orbits will be very finely spaced because I move by a small angle, are less than epsilon space, okay? So I move by a very small angle less than epsilon. So it means that every time I apply this power of the rotation, I displace myself by some quantity which is less than epsilon. And that means that if I have a target window of size two epsilon, I'm bound to fall inside. Any two epsilon arc will be visited. And this is what I need to do for density, okay? I think many of you got this yourself. So the main thing was to use the rotation isometry property. You can maybe see it differently if you want. Also explicitly writing the formula mode one, you can see it, okay? So you can actually say more about an orbit of a rotation. Not only it will be dense in 0, 1, but I want to also state a theorem which was originally proved by Weyl. And this tells you indeed that the orbit is also equidistributed. So for every alpha irrational, and again, irrational is key. For every alpha irrational and for every point of the circle, the orbit under the rotation of that is equidistributed. And now I will give you the definition, IE. And let me say I will tell you two properties and actually sketch the second. So I can pick an arc and try to understand on my circle how often pick an arc. And maybe I will write it on 0, 1 as an interval, AB. So for every, let's say, interval AB in 0, 1. Sometimes I switch between multiplicative notation on the circle and additive notation on 0, 1. This is also 0, 1 with opposite points identified. So if I start plotting my orbit and recording how many times I visit this arc, there is no reason why this arc should be visited more often than any other arc of the same length, because everything in the rotation is homogeneous. So indeed, if I look, I can count. So this is cardinality. If I can count how many times between 0 and n, how many k's between 0 and n are such that our alpha to the k of z belongs to AB and I divide by n. The claim is that this will actually tend to the length of the interval, okay? So in particular, it does not matter where my interval is placed. The number of visits that I have to this interval only depends on the length. And actually, the average number of visits, if I count how many times I visit and renormalize by the total number of visits, this will tend to give me the length of the interval. So somehow this point is kind of approximating the length as a measure. So this I think is very intuitive statement and this is a good way to think of this equity distribution. In general, you say that the sequence is equity distributed module one. If this statement is true, you have a sequence of points on the real line and take the mode one. If they satisfy this, I say that they're equity distributed mode one. And more in general, essentially you can have a stronger statement. So for every continuous function, and again let me write it on additive notation like this. For every continuous function from the circle to a, and for every, I will write x now, instead of z, I could have written x also there. For every continuous function, I can do the following. I can take one over n, the sum of my function evaluated along the orbit of x. So I have a function defined on the interval, on the circle. And I sample my function at the values given by my orbit, and then this will converge to the integral of the function. So a couple of remarks is that, so we will see in the, utors, lectures, you will see, if you haven't seen already, you will see Ergodistic, Birkoff Ergodic theorem, and you will see a similar result. But what is really crucial in this statement is that this is true for every point of the interval, and not almost every point. And another thing which is key is actually this, in general, as a remark, this, the systems for which two holes, two is basically equivalent. Substitute the rotation with another map, two is equivalent to unique ergodicity, which I will not define, but probably either Irene or Davide might define. So you keep this in mind if you already know it. And I will just sketch, first of all, you could try to, two is not really directly related to one. So if you could take as function the characteristic function of the interval AB, you would immediately, the statement of two would give you this statement. But in general, you cannot do directly this, but you can try to convince that actually by some approximation argument, two implies one. So it's not obvious, but try to get convinced, especially if you haven't seen these things before. And let me just give you a sketch of how to show two. And this is, again, this nice argument due to vile. And it also gives you a criterion known as vile criterion for unique distribution. And the idea is to actually look at some special functions. So exponentials. So now I will use these complex numbers. So let me call em, em of n, the exponential, e to the 2 pi i m. Sorry, I need a bigger chalk. e to the 2 pi i, em of x. So are you bothered by my em? Actually my English, British students, hate my em and em. So in Italy, this is em, and this is em. And in the UK, this is em, and this is em. So I guess my em always looks like an em, and my em looks like a strange wiggle with too many legs, but okay, don't bother. You can, I always try not to use em and em when I teach in the UK. I try to use k and l to avoid confusion. But here I think I have only em, so if you write em, it doesn't matter. Okay, so, and you want to look at this function, these exponentials. And I want to prove, so the claim is that two holes for functions which are these exponentials, for exponentials. And this is what I want to sketch. And let me assume also that em is not zero. And a remark, so what is the integral of this exponential? So remember this exponential is nothing else than cosine. No, sorry, there is no i. No, there is i, sorry. Okay, this is, no, there is no i. Okay, so what I want to, you integrate is the absolute value of this. Okay, for the absolute value. So, no, let me say something stupid. This absolute value is one. Okay, so let me, okay, maybe say that we want to prove it for cosine and sine actually for, so think separately of real part and imaginary part. So we want to prove it for real part and imaginary part. Okay, so for cosine and sine. Okay, so the integral is zero. So what we want to understand is what happens when I compute these sums. So what happens if I compute these sums? If I compute these sums, so m of rk, I think I'm making a mess. Apologies, I think I'm making a mess. What do I want to do? So I want to compute, I want to compute what is the exponential of, I want to compute the exponential of, so I want to look at the orbit of the rotations. I'm confusing myself between additive and multiplicative notation. So let me just say, sorry, I made a mess between the two notations. So I want to compute it at, yeah, okay, r alpha to the k of x. Let's say like this, which is just x plus k alpha modulo one. And this is just e to the 2 pi i k, this is 2 pi i m x plus alpha. That is just e to the 2 pi i alpha times a m of x. Okay, sorry, did I forget something? M, did I forget something? E a m alpha, is it, did I forget this one? K, sorry, e to the, k alpha, thank you, yes, okay. So mx is here and mk alpha, yeah, okay. Thank you, I'm happy that you're not paying attention. e to the 2 pi i mk alpha times e m of x, okay, is it correct? Okay, and this I want to think of as some number elevated to, it's like some lambda to the k where lambda is this complex number. e to the 2 pi i m alpha, okay. So it's this number power k. So maybe we will go, we will go here. So if I just want to say that if I compute these sums that I still have in the statement. So if I compute one over n, the sum from k from zero to n minus one of e m of r alpha to the k of x. Plugging what I just wrote over there. This actually becomes the sum from zero to n minus one. And this is lambda to the k. It's this lambda that I wrote over there times e m of x. So you see what you get here, and actually this, this I can take out of the sum. And I just have this, okay, no, maybe I saw something stupid again. So I need to take absolute value of everything. And okay, let me just take everything in absolute value. And then you want to use, so what is this sum? So this sum is just a geometric sum. So you just get one minus lambda to the n divided by one minus lambda. And the key thing is that in absolute value, everything which is in this sum is actually, because it's a geometric sum, so you know this formula for the geometric sum, this is actually bounded by some constant. So all this quantity stays bounded, but you have in front an n. So as n goes to infinity, it actually goes to zero, okay? Okay, so I may stop a little bit there, that's the sketch. And so what you get is actually for e m for these exponentials, for the real part and the imaginary part if you want to think of them as complex functions. Actually, you do have these two holes. So the conclusion is that two holes. And now for say real part, imaginary part of e m. And then what you want to say at this point is to finish, that I will not do. So what you should check, check what happens for m is equal to zero. And then once you have it for exponential, you can extend it to f continuous. And essentially you're just using that continuous functions can be approximated by trigonometric polynomials. Actually, okay, maybe extend it, okay. And maybe this I will leave to do. So this basically gives you a sketch of two. And once you have two, try to go from two to one, okay? Okay, this I think concludes the first properties of a rotation. And so we know in rational rotations have dance orbits and the orbits are equidistributed in this strong sense of vinyl. And as I said, this criterion of using exponentials is actually can be used not only for orbits of the rotation, but if you take some other sequence on the circle, something called y criterion tells you indeed that if these complex exponentials get equidistributed, then you say that the sequence is equidistributed, modular one. So for the rotation, it just comes from this small computation. In general, you have to prove it, but you only have to verify it for this special functions. Okay, so the next step I want to, as I said, I want to kind of review Gauss map and continued fractions. So we will step out of the world of rotations and we will look at a related dynamical system. And this dynamical system is actually not entropy zero. It's kind of more in the hyperbolic world of Hannah's course. But it will act for us as the zooming lens, as the renormalization machine in order to study rotation. So now and tomorrow we will have a break and jump to the Gauss map. And see how continued fractions relate to symbolic coding for the Gauss map. And then we will use the Gauss map as a renormalization machinery and relate the Gauss map back to the rotation, okay? So break for a second and I promise you that I will define the Gauss map so that she can give you an exercise on the Gauss map later. So I will first talk about the Gauss map then later continue fraction and the relation. So this Gauss map, it's a map of the unit interval. From zero one to zero one. And it's the following map, g of x, okay, is equal to zero. If x is equal to zero, that we have to consider separately. And otherwise, if x is different than zero, it's equal to the fractional part of one over x. Well, this symbol that I'm using now, so maybe I will, okay, we're, okay, let me write. Another way to write this bracket is also one over x, module one. So if this notation is commonly used, you can take real number x and write it as fractional part. We saw it already, we saw it also yesterday in Hannah's course, right? So this is the fractional part, which is the same than taking module one. And this is the integer part. So the smallest, largest integer which is less or equal to x. Okay, so let's plot the Gauss map. So you may say, okay, how do I plot? I'm sure many people have already seen it, but if you haven't seen the graph of the Gauss map, maybe it takes you a second to understand what this map is, so let's do it. So implicitly, this definition gives you a piecewise defined function. So how do I know what is fractional part of one over x? So first of all, I have to understand what is the integer part of one over x. So the integer part of one over x can be any number, can be one, can be two, it depends on how large, how small x is. The smaller x is, the larger is one over x and the larger is the integer part. So for which x is the integer part equal to n? Well, exactly when one over x has to be greater than n, but strictly less than n plus one, right? Because this is exactly when the integer part will be equal to n. But this is the same, then saying that x has to be less or equal to one over n, but greater than one over n plus one. Okay, so if I am in this range, so maybe we will call pn. We will use it also later. We will call pn this interval one over n plus one, one over n. And we will write this as x belonging to pn. And I think I still cannot do my plot. So maybe I'll finish my, so what happens? So if x belongs to pn, then Gauss of x is one over x minus integer part of one over x. Will be equal to one over x minus n. And this is true for one to the, for every n. So you have basically countably many branches. It's a piece was defined map which countably many possibilities. Again, I'm sure some people have seen this plot hundreds of time, but if you haven't, then. So I can plot one, one half, one third, one fourth, one fifth, and so on. So this is my p1 interval, this is my p2 interval, this is my p3, and so on. And on the first interval, this is just one over x minus one. You can check, you can plot it's zero at one. And it's, I know the answer, so I do the plot. So it's a piece of a hyperbola, which actually goes exactly from one, one at one half. Let's check at one half, it's two minus one, which is one, and zero at one. So if I plug one is one minus one, which is zero. And similarly, if you plot, so this is one over x minus one. And on p2, between one third and one half, I have a very similar picture. I also go from one to zero with the decreasing hyperbola. And this is one over x minus two, and so on. So this is countably many branches, each of which are pieces of hyperbola. And each of which are surjective. So if I call, say sometimes I can call GN, is the branch, G restricted to PN. And this is a surjective, it's invertible and monotone. Okay, maybe now I should have done this reality check later. Who had already, who had never seen the Gauss map? Almost nobody, so I could have, okay. I think I should have done this before, I think I spent time to draw the plot. Okay, so in any case now, Irana is sure that everybody can do an exercise. So I want to connect this map with continued fractions. So many of you told me that they've seen continued fractions. Have you seen how the symbolic coding of the Gauss map is related to continuous fraction entries? Who has seen symbolic coding of the Gauss map in relation with continued fraction? Okay, so I'm still doing something worse. Okay, and remind me again, who had never seen continued fractions? Okay, so I will just write for, just quickly write them. You can read more if you want, I can give you a reference on continued fraction, but let me recall you. So let me write what is a continued fraction. Continued fraction, and sorry, I miss my chalk, continued fraction. So most of you have seen, so if I have a number x, and let's say that x is irrational, I can actually express it uniquely in an expression of this form. I can write it as a fraction 1 over a 0 plus 1 over a 1 plus 1 over a 2. Dot, dot, dot, dot, dot, dot, dot, dot. And this is unique for irrational. And you can actually also do it for rational. So if x is rational. So there is a finite expression. So for x rational, you can do the same, but at some point your continued fraction will stop, okay? And actually, if it's rational, you can try to say in this case it's not unique. And if you want, you can try to find what are the two continue, there will be two continue fraction expansions. You can try to explicitly write them. And by convention, we write an expression, sorry, x in and as I wrote it, it's only in 0, 1. Otherwise, you can put an integer in front. Sometimes people call a 0 the integer part, and then they start from a 1. Here, actually, I will restrict the numbers in 0, 1 because those are the only ones we are working with. And then I start from a 0 because the coding is slightly nicer. Okay, so this notation I will just write like this, and this brackets, a1, a2, an, and if it's infinite, then you'll just put the dots. Okay, so the claim is that the continued fraction of x is actually related to the orbit of the Gauss map. And I think Hannah will spend more time on coding later, but I just need to tell you what they mean by coding this map and what is the itinerary of this map. So I have this nice partition. I have this partition into the intervals P1, P2, Pn. And these intervals are this joint, how I define them. And so Pi intersected Pj is empty if i is different than j. And also, their union is actually 0, 1 with 0 out. They kind of accumulate towards 0. So if I add 0, this is a partition of 0, 1. So what I want to do is to iterate my Gauss map. So I look at the orbit under the Gauss map of x, of a point x, do my dynamical systems with Gauss. And I want to record at each stage in which of these intervals my iteration lands. So I take my map. If you want to do the trick that Hannah told us yesterday, I draw the diagonal, I take a point, and then I go to my map, to the diagonal, to my map. I mean, I apply the map g. And as I move around, I look in which of these small intervals I land. Look at the record, record the Pn, the intervals, the Pn, the Pn which are visited. So maybe let me write it like this. Definition, the itinerary, the itinerary. And I'm going to call it ai, not by ai from 0 to infinity. The itinerary of this orbit, this will be a sequence, is the sequence of integers, ai. By the way, sorry, I think if you have still continuous fractions, you know this. But if you haven't seen it, I should do what I said that makes no sense if I don't tell you what are these ai. So the ai are integers. So positive is non-negative. So ai greater than 0 are integers in this natural numbers. And they are called entries of this continued fraction expansion. So otherwise, this is not true. You can write it in infinitely many ways if I don't put restrictions on ai. So yeah, this is, so again, the few that haven't seen continuous fractions, I will post some lecture notes and I spend a little bit more time on them. But I already bored people going too slowly on the Gauss map, so I will not do more on continuous fractions. OK, so the itinerary. The itinerary is the sequence of integers, ai, such that, so gi of x belongs to pai. So I look at gi and I look at which pn it belongs to and that I record. So I can meditate for a second if you haven't seen it, but this is just telling you. So this means that the iciterate length in pai. So each of this ai will be a number from 1 to n. And I'm assuming that I'm not hitting 0, which actually will be the case, but I will tell you later why. OK, let me assume that x is irrational in this definition. Otherwise, this is not well-defined. And again, it's in 0, 1. And I claim that you are not going to hit 0, so this will be well-defined. And now, OK, now I can go here. And now the proposition, or the key link, is that, so I can x is irrational, OK. x is irrational, ai, OK. x is irrational with itinerary with respect to Gauss ai. Then this itinerary is really giving you the continued fraction expansion. So this is a 0, a 1, a n. So I can recover my point as a continued fraction. It's actually not especially deep. It's just in some sense. The Gauss map is really made in order to do that. But it's a good point of view to think of a continued fraction as symbolic coding. This is what I mean when I say the continued fraction is the symbolic coding of the sequence of entries in the continued fractions are giving you the sequence which codes records where my orbit lands and which branches of the Gauss map I will be using. And I think we did it in their own order because Hannah, you will do symbolic coding for doubling map in relation to binary expansion. So the same relation that there is between doubling map and binary coding, there is between Gauss map and continued fractions. OK, maybe just let's check this. So first let's remark again that by definition of itinerary, by definition of plus the definition of pn, we have the following, that if I look at Gauss to the n, 1 over Gauss to the n of x. This is equal to an. Sorry, did I say it right? Yes. So if you want again, let me say it again. By definition of itinerary, g to the n of x belongs to p an. This is the definition of itinerary. And if you remember what is pn and what I said at the beginning, that the integer part is equal to something exactly when I belong to p of that thing. So is it this clear to everybody? Or did I skip too? Is it OK? You want me to? OK, so gn belongs to pn. Maybe if you want, you can add an equivalence here. And this is just saying that no, I don't think I can. Here I'm also using the 1 over x belongs to, sorry, x belongs to pn. This is what we did a little bit before. x belongs to pn. It just means the integer part of 1 over x is equal to n. That's what I'm also using in this. OK. And then basically, you just have to do it by induction. So again, I think I will leave you the details. But so let's just do first. Let's do it for n is equal to 1. So in particular, for n is equal to 1, we have that n is equal to 0. Equal to 0. We have that a0 is the integer part of 1 over x. And that means that g of x is equal to 1 over x minus a0. And now you can solve when you get that solved for x. So x is 1 over a0 plus g of x. You always check my calculations. I know, so I'll check with me that I did. So I bring a0 from the other side and then take 1 over. And then I think you can kind of see how it will go. And so if you want, we can do one more step. Then if you continue for n is equal to 1, if you want, you have that integer part of 1 over g of x. Again, by the remark is actually a1 is equal to a1. And that means, again, that g square of x, which is the fractional part of 1 over gx, is actually 1 over gx minus a1. And what does this give me? Again, I have to solve for g of x. g of x is, hopefully, it's 1 over a1 plus g squared. Let's check. So I bring a1 from the other side and take 1 over. And now this bit, if you plug it inside the previous bit, I'm not doing the proof by induction. I'm doing the first two steps, which I think are more educational for the first time. If I plug g of x inside this, you see what you get after two steps? You get 1 over a0 plus, and I have to plug g of x. So 1 over a1 plus g squared of x. So you see, you kind of start seeing that basically the orbit of the Gauss map iterates of the orbit and code the reminder you have to put if you truncate this continued fraction at, finally, many stages. So I think that the easy exercise for today will also be to, I mean, I know it's not difficult, but just do the induction. Finish the proof by induction, because just to work yourself through practicing this. So maybe complete, complete the proof. Just do the inductive step in general. And I'm almost out of time, but I have just a couple of minutes to maybe tell you some properties of this link between continued fractions. And so maybe I'll tell them as facts. So the facts that I'm not going to prove, but they're nice to know, is that actually, if you look at the orbit, so if you have the orbit of the Gauss map, you hit 0. So gn of x is equal to 0 for some n. If and only if, if and only if, what? Yes, if and only if, x is rational. So irrational numbers, sorry. That's why if you assume irrational, I told you you can be sure that you don't hit 0. And another fact, what are the periodic points of the Gauss map? Ah, maybe another more important fact before I tell you that. So if x has a certain continuous fraction, a0, a1, dot, dot, dot, dot, dot, then what will be the continued fraction expansion of g of x? Yes, exactly, a1, a2. So the Gauss map acts as a shift. So g acts as a shift on the entry as a shift. So it moves the entries by, it's really a shift in the sense of symbolic dynamics, actually. And then you can find out which points are periodic for the Gauss map. So which points are, how will periodic points for the Gauss map look like? Ah, what, how about, sorry, sorry, sorry. I went too fast. How about fixed points? If I have a fixed point for the Gauss map, so if, so fixed points for the Gauss map, there are countably many fixed points because I intersect the diagonal many times. And they basically have to have the same entries. So if you have a point which is a continuous fraction entries is k, k, k, k, k. This will be fixed by the Gauss map. And for example, this is, this is exactly what the golden mean look like. So I think I hope I'd say by, so for example, the golden mean has 111111 by continuous, the simplest number from the point of view of continuous fraction. It has periodic entries given by one, and it's fixed by the Gauss map, which is immediate by this remark that the Gauss map shifts the entries. And the next one is the silver mean, which I don't remember, it's probably square root of two. And this is called silver mean and so on. And finally, if you try to see what are periodic points, I went too fast from fixed points to periodic points. But periodic points are related in general to quadratic irrational. So quadratic irrationals are numbers which have essentially this form, a square root of d plus b over c, where here a, b, and c are integers, and d, a, b, c, and d are integers. So this indeed, the golden mean was an example, square root of 5 minus 1 over 2. And quadratic irrationals, so x, there are also solutions of a degree to a quadratic equation. So basically, quadratic irrational, x is quadratic irrational, if and only if it's a pre-periodic point. So pre-periodic. So pre-periodic means that g, there exists an n such that, there exists an n such that the gn of x is, what am I saying? Periodic, yeah, it's fine. It's just, this map is not invertible, right? So no, there exists an n such that gn of x is periodic, let's say like this. And I think I run already out of time. But for example, this implication, the whole, but you can actually try to prove an easier case. So if you have a periodic point, for example, of period 2, try to prove, so try for period 2 and then try to prove that if you have a periodic point, x periodic, and if you want to try period 2 first and then do the general case, then it's quadratic irrational. So more in general, this pre-periodic just means that the entries will have some beginning, which is whatever, and then they will repeat periodically after some time. OK? OK, thanks. That's done, yes.