Good afternoon. So we have arrived now at the last lecture of the course. So what I want to do today is a little bit to give a little bit of an overall summary of the basic ideas that we've talked about and also give just a little bit of discussion about how this subject develops. So in a nutshell we have really looked at two basic classes of dynamical systems. We have studied contractions. Basically this is situations where the derivative is less than one and what is the characteristic of contractions in general. So the basic features of contractions is that generally they tend to have a fixed point and everything contracts to the fixed point. So I mean just the characteristics of the dynamics, the kind of dynamics associated to contractions. So contractions basically have let's say simple dynamics, fixed point, attracting fixed points. And on the opposite extreme we have expansions. So maps which expand, which roughly, so this is a kind of, these are not theorems I'm putting on, okay. We've done the theorems. Now I'm just trying to say, okay, more or less what is the philosophy that we are, how we're looking at these different systems. Derivative greater than one. And what are more or less the features that we have seen, at least in the examples that we've seen. We've seen that you have many periodic orbits, transitive dense orbits and so on, very rich dynamics in some sense. So we have rich dynamics, dense set of periodic orbits, dense orbits, and so on, chaotic dynamics. In the sense of sensitive dependence on initial conditions, okay. So this should have been clear for you from the examples, but I did not emphasize this at the time. So we proved a whole bunch of more specific results, of course, in terms of conjugacies and so on. But now I just want to step back and take a very general point of view. It makes sense that this, it kind of makes sense when you think about it. So derivative less than one means that everything is contracting to some common behavior. Derivative greater than one means that points are moving away from each other. So naturally it gives rise to sensitive dependence on initial condition and chaotic dynamics. And somehow that is also why you end up having lots of periodic orbits and lots of dense orbits, okay. It is possible to make formal statements of this kind. In fact, in the case, if you remember the last theorem last time was that if you have an expanding circle map, then you have this basically, right. So in the case of a circle map, we actually proved very directly this. If you have a C1 map of the circle where the derivative is bigger than one, then it is topologically conjugate to a full branch piecewise expanding map and so it has dense orbits and chaotic dynamics and so on. So we actually did prove a statement in one case that is exactly that. So natural question therefore is, there's two natural questions. First of all, what happens in a kind of neutral situation? And second of all, what happens when you have a system that has a combination of contraction and expansion? And that's when it starts getting very interesting, okay. So this is a little bit what I want to talk about today, an outlook. Because the case in which the derivative is always less than one is simple. The case in which the derivative is always bigger than one is in some sense simple in the sense that we can more or less describe it. As we shall see, the combination of the two can give rise to some very interesting phenomena, okay. But first of all, let me talk about, let me give you one more example of a neutral case. We have already seen the kind of neutral situation before. I don't put an equality because this is all roughly kind of it, right? So roughly derivative roughly greater than one, roughly less than one, roughly equal to one. It depends on the specific situation. So do you remember one example of a system that we've already seen in which the derivative is one, a neutral system? Do you remember an example? We saw it in the first part of the course, actually, not in this part of the course. Circle rotations, that's right, okay. Ej, circle rotations. And in what ways the dynamics of circle rotations different from either of these two? What are the features of the possible dynamics that we saw for circle rotations? If you remember, we saw that there's only two possibilities that depend on the rotation, on the rotation number, on the rotation, right? But in both possibilities, every point does the same thing. So these two possibilities are either every point is periodic, okay? And this is basically almost like the identity map. It's a kind of degenerate situation. If every point is part of the same period, that means that some power of f is the identity map, okay? It's a kind of, and the other perhaps more interesting situation is where every orbit is dense. That's called the minimal dynamics. And you can see that this is a different situation from either of these two, right? This one has a combination of dense orbits and periodic orbits. This one, every orbit is certainly not dense. It goes to a fixed point. And remember, this one does not have sensitive dependence on initial conditions, right? Because in circle rotations, if you take two points that are nearby, they stay always at the same distance apart. So it does not happen. So these are really three different kinds of dynamics. And you would not be too far along if you thought that in some sense, all every dynamical system is more or less one of these three types, these general types, okay? So what I want to do for the first 10 minutes of this class is I want to give you one more example of a system that is neutral of this kind, because so far we've only seen this one example. And then I will talk about a family of maps in which you get a combination of these three in the same family. So let me start with one more example of neutral. And I will use a symbolic model of neutral dynamics in a very similar way to which we use a symbolic model for these expanding maps, right? Using the shift map. So by symbolic model, I mean that we will, I will define a map. So this map is sometimes called the adding machine. And it is defined on the same space, sigma L plus. Let me call it a map tau. So sometimes like we call this the shift space. And we have defined so far the shift map on it. But the space itself doesn't necessarily come only with the shift map. It's a set of points. You can define all kinds of maps on this, right? And now I'm going to define a different kind of map. And I'm going to define it in this way. So here we have that some sequence x0, x1, x2 and so on. Maps to tau of x0, x1, x2, and so on. And I will call the new sequence y0, y1, y2, and so on. And I want to describe how to define this map. So informally, this is add1 and carry. What does that mean? That I'm going to take this sequence and I'm going to add 1 to this first digit. And that's it. I'm going to add 1. So this is some digit between 0 and L minus 1. So if this digit is 0, I add 1, I get 1. If it's 1, I add 1, I get 2. If it's 2, I add 1, I get 3. And that's it. So the only difference between this and this is that I add 1 in the first digit. Of course, there's a problem if this digit is L minus 1, right? Then what do I do? Well, if this digit is L minus 1, I add 1 and this becomes 0 and I carry and I add 1 to this. It's a kind of natural way. Just like in the car when you count the kilometers in the car, right? You just add 1. Or just like in normal numbers, except that it backwards. Think of this as 1 number, okay? This is the first digit, instead of the first digit, so you read the number left to right. And then when you get to 9, you go to 0 and you add 1 here and so on. Okay? It's very simple. So I will give the formal definition here, but to keep this in mind, this idea, right? So how do I define it? So first of all, if... So to do any questions about that? No? Okay. So to define it formally, let's look, first of all, at the case in which all of these are L minus 1, right? So if xi equals L minus 1 for all i greater than or equal to 0. In other words, if it's the constant, for example, if L was equal to 10 and you are looking at the kind of decimal system, then if all of these were 9, then imap, then we let yi equals 0 for all i greater than or equal to 0. That's the kind of special case because this is, of course, an infinite number. So if they're all 9s, we add 1. So if this was a 9, we add 1, it goes to 0, and we carry 1, then we add 1 to this 9, it becomes 0, and so on. And we keep carrying forever and they all become 0. Okay? Otherwise, what do we do? So otherwise, we look at the first digit that is not a 9, right?  And then basically what we'll do is we'll keep carrying until we get to the last digit that's not a 9. So let i0 of x bar, so this is x bar and this is y bar. So let i0 of x bar equals the minimum i greater than or equal to 0, such that xi is different from l minus 1. So if i0 of x is equal to 0, that means that the first digit is not l minus 1. Okay? If i0 of x bar is equal to 1, that means that the first digit is l minus 1, but the second one is not, and so on. This is the first time where we get not l minus 1. And then we just let, so if i0 of x bar equals 0, that means that the first digit is not l minus 1. Then it's very simple. Then we let y1 equals x, sorry, y0 equals x0 plus 1. And yi equals xi for all i greater than or equal to 1. So we add one to the first digit and all the others stay the same. That is if the first digit is not l minus 1. If the first digit is l minus 1, then we look at the same, the first i0 digits which are l minus 1. If this is equal, is greater than or equal to 1. That means there's at least a certain number of digits with l minus 1. Then we let yi equals 0 for all i up to time i0. Okay? And yi equals, sorry, yi0 equals x i0. And yi equals xi for all i, strictly greater than i0. So this should be, hopefully, the formalization exactly of the add one and carry that I explained to you before. This just corresponds to the fact that you take all again, if l is equal to 10, then you have a certain number of 9s. And then when you apply this map, you add one. So all the 9s become 0s. And then the first time that is not 9, you just add one. Sorry, this should be plus 1 here, yes. You just add one. So this is a well-defined map. Any sequence, you can apply this, add one and carry, and you get another sequence. Okay? So now we want to study the dynamics of this map. You apply, you iterate this map over and over again, and you see what the dynamics is. What is the dynamics of this map? So first let's say, let's prove a little lamb about the structure of this map, and tau is a homomorphism. We have a metric and a topology on sigma l plus, and tau is a homomorphism, in particular its invertible map. So proof, let's show that tau is injective. So let x be different from x prime. And then we need to show that the images are different. So if they're different, then we have x i different from x i prime for some i greater than or equal to 0. It's kind of obvious that this is going to be true. So these are different, and therefore let i 1 equals to the minimum i for which this is true. And so what can we say about the image? Well, if x i by the definition of the map, if x i is different from x i prime, but all the previous ones are the same, then when you add one and carry, y i will necessarily be different from y i prime, whichever way, whichever situation you look at it. Then definition of tau, we have y i different from y i prime, and so because the two images are different. Because up to time i 1, all of them are the same. So even if they're, whether they're l minus 1 or less than l minus 1, it doesn't matter. When you add one to both terms, you either stop at some point before i, or whatever happens when you get to this point, the images cannot become the same. So that proves injectivity. Surjectivity is also easy. We can actually explicitly define tau as surjective. We can in fact explicitly define the inverse map. So we can explicitly, explicitly inverse map tau minus 1. So I will maybe leave that as an exercise. So that proves that it's injective. That is surjective because for any point you can construct the inverse map so you can find the point that maps to it just by reversing this. It's almost the same thing, but reverse, I'll leave this as an exercise, a useful exercise for you to think about. And finally, we show that tau is continuous. And this is also very easy. So remember that there exists n delta such that if x bar, if the distance between x bar and x bar prime is less than delta, this implies that x i equals x i prime for all i less than equal to n delta. And so, of course, this implies y i equals y i prime for all i less than equal to n delta. And this implies that the distance between y i and y i prime is less than delta. It's less than delta, so we can take epsilon equals delta. So I wrote down the proofs here so you have something to go on, but really, this is, sorry? Last one, what? Well, I'm using y to denote the image of tau. So I'm going fairly fast on this. There's nothing particularly complicated, but of course I'm suddenly, you just got used to the shift map and suddenly I'm giving you on the symbolic space a completely different map that has completely different properties. So it takes a little while when you go home. You need to, what I suggest is that you just try to prove this yourself, okay? Don't look at this because actually the arguments are just using the definition of this map and the usual properties of the metric here that you know. So it's more useful to actually try to prove it yourself obviously, like for all the lemmas and results that I give you, okay? But anyway, this is the proof that we've got here. Okay, so this is the fact that it's a homomorphism. It doesn't yet give us anything about the dynamics. So what do you think the dynamics is going to be here? Sorry? Ah, this is what you were saying. Sorry, you're right. Yes, that's what I meant. Thank you. Yes. I guess that's what you meant, yes. Okay, so do we see any fixed points, periodic points, then solve it? What is the dynamics here? It's not very easy to see but you can see maybe that there's no fixed points. Can you see that there's no fixed points? How can there be a fixed point? You take a sequence and you add one. It's not the same as the sequence you had before. There's no way. Any periodic points gets a little bit trickier with periodic points. So every orbit is dense. This is what we're going to show. No fixed points, no periodic points, every orbit is dense, just like the irrational circle rotations. In fact, there is a kind of conjugacy but we are just going to give a very elementary direct proof of this fact, yes. I just was emphasizing that this is similar, the dynamics is similar to the circle rotations but there is a way in which you can, some generalization of this adding machine can be used to code the circle rotations, yes. But that is quite involved. We're not going to do that here. I'm just going to give a simple, we're just going to take an orbit and prove that it's dense directly. But one of the connections between circular rotations and this is again, if you look at the situation on a little bit more general level, you can think of the space as a compact group in some way, just like you can think of the circle as a compact group. If you think of the circle as the unicycle in Z and you think of addition mod 1, it's kind of a compact group and adding theta is a translation on this group. And in this case, adding one is a kind of translation on this group looked at in a certain way, okay. I'm not going to go into that, but I just wanted to mention this, that there are examples of what is called translation on compact groups and often you get this kind of situation where you get a translation on a group and then you get this kind of, every orbit is dense, minimal dynamics, okay, i.e. tau is minimal, this is how, this is minimal means that every orbit is dense. So how are we going to prove it? We're going to prove it not by any sophisticated techniques, but just literally by taking an orbit and showing that it's dense. So I don't need to remind you what we need to show. So do you remember? I mean the metric here is the same metric as always. So what was the, if you remember in the same way we needed to prove that there was dense orbits for the shift map, how do we show that an orbit is dense in here? Do you remember? What is the property of the sequence of the various sequences along the orbits? So in this case, the sequence changes. So in the other case, remember, we had to say that this sequence had to contain all possible finite blocks, right, because at some point that finite block would come to the beginning and then it would be close to the point that you're trying to approximate, right? So to show that it's dense, we take an arbitrary point here and we show that the orbit comes arbitrarily close to this if we wait long enough. So in this case, the sequence changes, right, because every time you're adding one and you're carrying. But we still need to show that given any finite block, there will be some moment in the iterates of this point where that final block will occur as the initial part of that sequence. That was the essence of what we had shown before, right? So we need to show that any finite block of digits can occur, will occur as the initial digits of some iterate n over x bar. So if you do not remember why this proves density, go back and look at the argument that we used for the shift map, okay? In that case, because it was just the shift map, what we wanted is a specific sequence that contained every finite block. But in this case, that would not help us at all, because the sequence changes. But what we really use in the proof is this, is the fact that every finite block occurs as initial digits for some iterate. So how are we going to do it? Well, let's prove it first for a specific point. So we show this first for the sequence zero. So let us show that the orbit of zero is dense. So of course, for the shift map, again, just emphasizing the difference, the shift map, this is just a fixed point. But for this map, it's not a fixed point because the sequence changes. So what is the next iterate? That's right. You add one and then all the way until you get to l minus one, zero, zero, zero, zero. Okay? What's the next image? Zero one. What's the next? What's the next iterate? All the way up to l minus one, one, zero, zero, zero. Right? What's the next iterate? Zero, two. That's right. All the way up to l minus one, two. Okay? And then we continue like this until we get to zero, until we get l minus one here. And we have zero, l minus one, zero, zero, zero. All the way to l minus one, l minus one, zero, zero, zero. Okay? And what's the next iterate? Okay. Now let's take a pause and reflect and see what this tells us what is happening here. What is going to happen when we continue? So remember when we constructed for the shift map, when we constructed the sequence that we wanted, there was going to be a dense sequence, we said, okay, take all finite blocks. Right? So you start with writing zero, one, two, three up to l minus one. And then we take all the blocks of order two, zero, zero, zero, one, zero, two, then one, zero, one, one, and so on. Okay? So what do we have here? Well, we have as the finite blocks, this is what we need to show. We have of length one, we have zero, one, two, and so on up to l minus one. So all the finite blocks of length one occur as the initial digits of some iterate, zero, one, two, up to l minus one. Okay? What about the finite blocks of length two? Well, here we have zero, one, one, one, two, one, all the way to l minus one, one. Right? And then we have zero, two, one, two, and so on all the way up to l minus one, two.  Here we have, in fact, we should include these. These are one, zero, two, zero. If we look at these as blocks of length two. So this gives us, if you think about this, all possible blocks of length two. From zero, zero to l minus one, l minus one. Okay? If this was l was equal to ten, this would give us everything from zero, zero to ninety nine. Every possible combination. And you can see now that this is going to continue like this, right? So this is now zero, zero, one. And we look at all the possible blocks of length three starting from here. So these are the ones that contain some zero in the third. Here we have all the possible blocks. Up to here we have all the possible blocks of length three that have a zero in the third position. Okay? Then now we're going to start all the possible blocks of length three that have one in the third position. And now as we add one here, and then we add and carry, we will end up with all possible blocks of length three, which have a one in the third position. And then when we get l minus one, so this goes all the way to l minus one, l minus one, one. Okay? And then this will go to zero, zero, two. Okay? And then this will go all the way to l minus one, l minus one, two. Okay?  And at this point we have, if we look from the beginning, we have all possible blocks of length three. We'll occur as the initial digits. Okay? And this continues like that. So now the next step is going to be zero, zero, zero, one, zero, zero, zero.  So if you look far enough, if you give me any block of length 200, if I wait long enough, at some point I will see that specific block of length 200 occurring as the initial digits of some iterate of the point zero. So this shows that the orbit of zero is dense. So what about an arbitrary orbit? So now let x be arbitrary. How do we do that? That zero belongs to the orbit. Yeah, but in general, zero will not belong to the orbit, right? Because we will never necessarily get to zero of zeros. Very good. We need to obtain some initial finite block of zeros. And then we will be in the situation where we were before. And we can apply, this is the typical mathematician strategy. You know that, right? So I will tell a joke now, because at least once in the course, the last lecture I can tell a joke about mathematicians. So you know how mathematicians boil a pot of water to make pasta, at least Italian mathematicians, right? So you ask a mathematician, how do you boil water? I said, oh, easy. I go to the cupboard. I open the cupboard, and I take out the pot, and I go to the sink, and I fill it with water, and I put it on the stove, and I turn on the fire, and I wait for the water to boil. Okay, very good. And then the next day, and you say, okay, but what about if the pot is already on the table? He said, oh, it's very easy. I take the pot, I go to the cupboard, I put it back in the cupboard, and that's where I was before, and I can know how to do it from there. And so we do the same thing here. And like you said, we know that as long as we have enough zeros, okay, so it's enough to show, then it is enough to show that we, that some point, that some literate of x starts with an arbitrarily long sequence of zeros. Because like you said, so what I need to show is that at some literate, I get some finite block 200, say. So to show that it's dense, I fix an epsilon and I fix some z, and I need to show that it comes sufficiently close to z. So that means fixing a certain number of digits. I need to show that some literate has that number of digits. If I can show that after some time I will get enough zeros here, okay, then for a long time after that, the dynamics will look like it's just the zero sequence. It will be exactly the same thing. It will be exactly what I have here for the initial terms, okay. And this, if you think about it, is obvious that at some point you get zeros or just sketch, okay. It's obvious that you get enough zeros because you start with x0, x1, x2, x3. Let's see if I find, if I had found a clever way of saying that here. So after that most, at most l iterations, a zero will appear in the zero position, right, because we're adding one to here all the time and sooner or later we will get to l-1 and the next literate this will become zero. So within at most l iterations there will be a zero here, right. And then after that zero we keep going, we add, and then every time that we cycle through l more, every time that we add l more terms here, this will jump by one, okay. And so if we cycle around l more sums enough times, at some point this will get to l-1 and then when we cycle once more this will be zero and this will be zero. They both will be zero. And so you keep going like this, right. So after at most l squared iterations the digit zero will appear in both, both the zero, the first position and so on. So for some literate, for some literate x prime equals tau n of x bar with n less than equal to l to the n epsilon, we have x i prime equals zero for all i up to n epsilon, okay. And then after at most an additional l n epsilon, at most an additional l n epsilon iterations we get the required finite, right. So epsilon was the distance that we're trying to approximate the arbitrary point z and n epsilon is the number of digits, the length of the block that we need to be able to approximate to within epsilon that number. And so within two times l to the n epsilon digit we will necessarily be able to find that particular finite block in this sequence. And this proves that every, okay, so this is the other example of a neutral behavior. These are also very interesting examples and I thought it was important to include this in your repertoire for you to realize that besides the contraction and the expansion there's this class of kind of translations or neutral behavior which also has this very interesting symbolic model and it's also a very important class of systems. Okay, let's take a couple of minutes break and then we'll come back and look at the quadsatic family. Okay, so let us now, without really going into any proofs, look a little bit at how this contracting, expanding, and neutral dynamics can interact. So suppose we take a map on the real line, x goes to x squared. So what does the graph of x squared as you know looks like this. So what is the derivative of this map? In some places the derivative is small, so the map is contracting. And in some places the derivative is large, so it's expanding. So if you look at some of the dynamics, it depends on where the points are. It could be, you could have an orbit that spends some time more in the contracting region or more in the expanding region or you could have an orbit that spends some time in both. And then it's a question of which one wins. Maybe if it spends more time in the contracting region then the dynamics is kind of contracting. If it spends more time in the expanding region the dynamics is kind of expanding. Okay, so to study this it is more interesting to look at a family of maps because as it happens if we introduce a parameter f of a and call this x squared plus a then what this does is it shifts the parabola up and down and as we shall see it is the different values of the parameter a that create difference in the dynamics and different kinds of dynamics. So for example if a is bigger than one fourth then this is the picture as we shall see.  So you see the a just shifts the map up and down right these are the coordinates x and y to the diagonal so a equals one fourth is exactly the one where you get where it's tangent. So we will analyze the different behaviors as we change the parameter a for a bigger than one fourth this is the picture what is the dynamics here can you see this what happens there's no fixed point the omega limit so you take some point x what happens to this point x y the omega limit is empty why what is the point doing it goes this way right because everything is above the diagonal so everything is going to the right so x maps to this point and what is this point this point is actually here right so this is f of x and then f of x maps to here which is here this is f2 of x and this point maps even further and so on and so every point here we have that fn of x tends to infinity for all x everything just tends to infinity nothing happens now so this is simple we've described it this is in this case this is not really it's a it's not so interesting to say whether it's a contraction or an expansion because even though strictly speaking this becomes an expansion points don't come back so it's an expansion but it's just going to plus infinity so there's no interesting dynamics there's no periodic points there's no dense orbits because even though it's expanding everything is going so those those key features I described to you essentially generally tend to work when you are in a compact space so that you have to come back right on the real line things can just move away to plus infinity minus infinity and nothing interesting happens now at this point here which is a equals one fourth something very interesting happens here which is we have a fixed point and so what is the dynamics here well you can see again this is in fact this is fixed point p what happens if you do the right of the fixed point you do the same thing right so all these points they go to plus infinity what happens if you're to the left of the fixed point it goes to p right this if you just look at this bit it's like the monotone maps that we studied the interval different morphism that we studied right at the beginning of the course right if you take the point x then you will just converge to p so in the graph you can see it like this in practice what is happening is at this point is just converging to this fixed point so this is an interesting the several interesting observation here this is a point that is attracting on one side but they're pelling on the other side right because these nearby points are moving away but these points are moving towards p what happens for points here exactly so these points they map to large number here so they map here and then they just go to plus infinity so what we have is we have this point here which is the pre-image of the fixed point let's call it p prime right and then you can see that any point that is between p and p prime will converge to this fixed point because if you take this point here after one iterate it will move here which is on this side and then it will just converge to p prime so also here we have a very clear description of the dynamics we have this interval p prime p all the points in this interval converge to this fixed point and all the points outside go to minus go to plus infinity and indeed what do we have inside if we just look inside what is the derivative inside what is the value of the derivative it's less than one except at the boundary where it's equal to one because it's tangent but everywhere else is less than one so we have the typical contracting features right everything converges to a single fixed point that's the typical contraction outside we have expansion but we don't have the currents things don't come back so they just move to infinity and nothing very interesting happens notice so in this course we have focused on structural stability when things do not change under perturbations we could have had a whole nother course focusing on bifurcations which are the situations in which instead things change when you perturb them a little bit right this is precisely a bifurcating parameter why because when you perturb just a little bit if a is just even epsilon bigger than one quarter then you have this situation here so notice there's a big bifurcation so here all the points converge to this but if you perturb it just a little bit then suddenly there's a little hole here and all the points escape through that hole to infinity okay so in some sense there isn't a kind of a continuity between what the points do asymptotically because these points asymptotically go to p but if you take the same point for a equals one fourth plus epsilon tiny then it will still come very slowly towards p but it will actually manage to cross because there will be a gap and then eventually it will go to infinity okay so the situation changes the topological conjugacy class changes this is a bifurcation okay that was just to observe to emphasize the fact that you get this change in the in the dynamics at this point now what happens when we increase this we decrease this further right and we take one fourth minus epsilon so we have this picture here so what is the dynamics here so we still have this fixed point p here and we still have that points to the right of the fixed point go to infinity we still have the symmetric fixed point p prime we still have that points to the left of p prime the next it is to the right of p so they go to infinity and we still have that this interval p p prime all the points here stay inside this interval and what is the dynamics here there's now two fixed points but now one of them is attracting and the other one is repelling which one is attracting this one is attracting look the derivative here is bigger than one because it crosses the diagonal like this all right so the derivative is bigger than one so this is repelling the derivative here is less than one this is close to zero in fact I didn't draw it very well of course so the the parabola has a zero the zero here of the derivative and you can easily see that if you choose a point here it will converge to this attracting fixed point every this one will converge from this side if you take a point here it will also map here and then it will converge from this side and everything will converge if you take a point here it will converge from this side but every point inside the interval will still converge to this unique fixed point and in fact you can see that the derivative here is not less than 1 everywhere because the norm of the absolute value of the derivative here is bigger than 1, but this bigger than 1 derivative is just pushing things away from this fixed point towards this one and then in the neighborhood of this point it's contacting the derivative is less than 1. So what happens as we continue this analysis? So let me start drawing the parameter space. Here we have 1 over 4 and we saw that in this region we have just fn of x tends to infinity for all x. And then we have at this point here we have a fixed point and then here we have some region which we are going to study what happens. Here we have a region where we have an attracting fixed point. Now how far does that region go? If we decrease f further we get to a situation that looks like this and we still have the point p and we still have the symmetric p prime, p prime, this is p. What do we have in this case? So let's suppose that the derivative here, the absolute value is still less than 1 for the moment how I do it. So we still have, so things have changed, you can see that as you push the parabola down this interval is becoming a little bit bigger because as you push the parabola down this intersection with the diagonal moves to the right. You can calculate all these things explicitly as an exercise. You can just easily calculate the fixed points of these and you can see that the fixed point depend on a and you can easily calculate the actual positions of these fixed points. But as you push it down this will become bigger and then this will move so when a equals zero you will have the fixed point, this fixed point exactly at zero and then as you push it further it will start moving to the left. And here you still have that this is attracting fixed point but the derivative, the absolute value is less than 1 but the derivative is negative so that affects the way that points converge to the fixed point. And because the derivative is negative they flip, remember we had those linear maps in which you need to flip, you converge to the fixed point but you flip from one side to the other. So if you choose a point, so this is a new fixed point q, you choose a point x here then the image of this point is below q which means that it maps to some point to the left okay so yeah so this is the point q so x maps to the left and then it maps to the right and then it maps to the left and it kind of converges the graph circles round and what happens here is that the map, the point flips back and forth and each time it gets a little bit closer to the point p and it converges, it still converges, it's still attracting fixed point but locally it's orientation reversing so it converges to the fixed point by flipping both cases okay so so far there's still not really any big difference between these situations now the problem happens as you push it further then you can see that the intersection of the diagonal with this graph will be at a point where at this point the slope will have absolute value bigger than one right so at some point we'll get a situation like this okay p so the question is what is happening here now we have two fixed points but both of them are repelling right because here now the slope is bigger than one in absolute value so the points here if you start with a point near here it will still flip around but it will tend to move away from this point and the points here will also move away from this point so what happens here any ideas what do the points do where do they go and notice that here already it's more it's already start to have a little bit of an interaction between the fact that you have regions contracting regions but also expanding regions so as you move down inside the square if you want inside the interval p p prime p that is more and more the expanding regions become more and more important right that's what's happening we're losing contraction initially we just have contraction all the derivative is less than one then by shifting down the graph is always the same but because we shifted down and because the area that with the interval we're interested in is defined by this fixed point as we move down this interval will contain more and more regions where it's expanding so that's why the dynamics becomes more hybrid between contracting and expanding in particular what happens in this case anyone have a guess what happens in this case where do the points go they don't know where to go anymore because before they were all converging to this then suddenly as you change it suddenly the derivative here the slope becomes minus one and then it becomes less than minus one so it becomes negative like this and suddenly they cannot converge to that point anymore they're repelled there's no fixed point these are the only two fixed points and any point that starts here moves away from this fixed point not towards the fixed point so one way to see it is by looking at the graph of the second iterate of the map so let's look at f2 and let's look at the second iterate of the map let me draw it actually bigger okay so here we have our interval that is mapped inside itself notice that outside this interval this fixed point here always defines an interval outside this interval everything is trivial everything just moves away it's not interesting so really what we just interested is in this interval here in fact this p prime is just minus p I could have written minus p okay so this is a fixed point so now what I'm going to draw is the graph of f2 so what does the graph of f2 look like I don't have time to go through this in much detail okay this is not really I'm not giving this full proof of this so I'm just going to sketch it but at home you can go through it again and convince yourself better but if you think about the image of this so the image of the fixed point of course will be just the fixed point okay and and if you look at this point here the image of zero what is the first second iterate well the first iterate is this right first iterate is up here and the second iterate is somewhere here well I do actually I'm sorry I do this much lower parameter than I should have done really I want something like I will draw you what it looks like and then you know the graph is not drawn exactly to scale but if you look at just if you calculate the parameter where the I the derivative here becomes minus one and then you look a little bit after that parameter what you will get is this picture here so you will get this fixed point will be here and then you will get that the dynamics is something like this of f2 is the second iterate of the map so how many fixed points does this have four fixed points and where are the fixed points here where are these four fixed points here this is four fixed points for f2 there's two new fixed points there's this fixed point and this fixed point and then there's the two fixed points that were originally also here which are this and this this is the point p and this is the point q and what is the dynamics of this map of the second iterate of this map in other words what are these two points here are they attracting or appelling I mean I'm drawing some approximate pictures okay it's not but if you draw the picture if you really study the picture you will see that if the derivative is close to minus one here then you will get what the way it looks like here which is that these are attracting this here you can see that it's attracting the derivative is minus one here and here also it should be like that if I don't it properly it should be something like this the slope should be small should be less than one this is a better picture okay what does that mean that means that if you start with a point here x it will converge to either this fixed point if you start on this side it will converge to this fixed point okay here also if you start with some point here it will converge to this fixed point if you start with some point here converges fixed point in fact wherever you are you will converge to one of these two fixed points because wherever you are You start from here, and then you go here, and then you go here, and then you converge to this fixed point. So wherever you are, you will converge to one of these two fixed points for the second iterate of the map. What does that mean for the first iterate? So it looks like we have two points on either side. This is Q, this is one point, this is the other point. These two fixed points for F2 constitute a periodic orbit of period 2 for the original map. So what this shows is that here there actually exists a periodic orbit of period 2, and that this periodic orbit is attracting, and every orbit converges to this periodic orbit. And of course a periodic orbit of period 2 is a fixed point for the second iterate of the map. Clearly that periodic orbit of period 2 means one goes to the other comes back, so it's a fixed point for the second iterate of the map. And that's why if you draw the graph of the second iterate, you see two more fixed points because these two are actually fixed points for the second iterate. This is called a period doubling bifurcation. Because you have a bifurcation, you've changed. Here there was no periodic orbit of period 2 because everything was converging to this point and here suddenly you have a bifurcation and you create, first you had one attracting fixed point, now this fixed point becomes repelling and it gives birth to a attracting periodic orbit that has period 2. So what you have here, for some parameter which we can call a1, we have a period doubling bifurcation. And then here for some region here we have a periodic attracting orbit of period 2. So the fact that we have an attracting periodic orbit of period 2 as opposed to an attractive fixed point is the first indication that the expansion is starting to play a role. Attracting periodic orbit of period 2 is a weak form of contraction. If it was fully contraction, the derivative is less than 1 everywhere, you just have an attracting fixed point. But the fact that the expansion starts playing a role, it's a kind of weaker form of contraction. So it's still essentially contracting because you have a periodic orbit of period 2 that attracts but you also here have an expanding fixed point so you have a little bit of a combination of attraction and expansion which gives rise to this periodic orbit of period 2. So now I won't keep drawing many pictures except that to tell you that what happens to this periodic orbit of period 2, if you now continue decreasing the parameter a, what you get is that the same thing happens to these fixed points. As you continue decreasing the parameter here, you will get at some point this will go down and at some point this point will have a derivative minus 1. This diagonal here, sorry, this graph will cross the diagonal with the derivative minus 1 and then this point itself, this periodic orbit will undergo another period doubling bifurcation and it will become repelling and it will give rise to an attracting periodic orbit of period 4. That is called another period doubling bifurcation because for this F2 you will get exactly the same thing and so this will have a period doubling bifurcation and that will be a fixed point for F to the 4 for the second iterate of this. So here you have a parameter a2 and here you have another period doubling bifurcation and as it turns out you get a sequence of period doubling bifurcations until you get to a parameter a infinity. So let me just write this down. So what we get is a sequence period bifurcations accumulating on a parameter a infinity. So that means that here you have a period orbit of period 4, then you have a attracting periodic orbit of period 8, then attracting periodic orbit of period 16 and so on. You have a sequence, an infinite sequence of period doubling bifurcations. So very close to a infinity you will have an attracting periodic orbit of period 2 to the n, very large. And this is because more and more it's not so contracting anymore because more and more it's weakly contracting. So instead of a fixed point you have a periodic orbit of very large period. Now what happens at the parameter a? Parameter a is sometimes called the Feigenbaum parameter and what's very interesting that in this case you have a attracting cantor set and f the stricter to lambda is approximately is like it's not strictly conjugated but let me say it's approximately like an adding machine that we just described before. So the sequence of attracting periodic orbits becomes in this limit point a kind of adding machine which is kind of neutral. It's where the contraction and the expansion exactly balance out. Okay and now let me jump to the parameter minus 2 and we've studied the parameter minus 2 before. So what's the parameter minus 2? You remember what it looks like? So at the parameter minus 2 we have that this point we have the graph looks exactly like this and this point p is exactly 2 and this is minus 2. This is minus 2 and 2. Remember what is the dynamics of this parameter? We studied it before. This is chaotic dynamics. This is conjugate to the tent map and it's the typical chaotic dynamics. Okay. So this is chaotic dynamics and then what happens for a less than minus 2? So this is a equals minus 2. What happens for a less than minus 2? We get this picture here and what is the dynamics in this case? So this is the point p, this is the point minus p. You recognize this picture here? What's that? What's that? Have we done this before? We've studied this before? Well, it's the same. This is this kind of picture. Remember where we took interval i0, interval i1, maps to everything. This is delta. These points here they just escape to infinity or minus infinity depending on the map. And so what you have here is that most points they actually escape to plus infinity when we studied it I do it the other way around. But as long as the derivative here is bigger than 1 and the derivative is bigger than 1 then we get a canto set here on which it's conjugate to the shift on two symbols. We've studied this case. Now it turns out that in this case you can prove that the same happens even if it's just a little bit less than 2. So what we did is that this works only if the derivative is bigger than 2. But you remember that I said that this particular map here is kind of very special because of this differentiable conjugacy with the tent map. And actually even though it has a small derivative here, it's very expanding. All the periodic orbits are expanding. So because of that you can actually show that even if you take a parameter that is only a little bit less than minus 2. So if you take a parameter that is just like this, okay? So the delta is very small, this is delta, this is i0, this is i1. Then still the same happens because with more sophisticated arguments you can show that this even though the derivative here is small you can still apply, if you look at some iterates the derivative is expanding. So you can still apply all those arguments where all the intervals shrink to 0 and you can still get the conjugacy with the symbolic dynamics and so on. Which means that every way here, less than minus 2, we have a canto set conjugate to the shift, conjugate to shift map on sigma 2+. Here we have a canto set essentially conjugate, we haven't proved it but it's essentially conjugate to adding machine, to tau on sigma 2+. So we have described, although I've done this very quickly, I appreciate that. I appreciate that. We've described all the, basically described all the dynamics all the way up to infinity and all the way less than minus 2. What is remaining is this interval here. And this interval here is the object of a lot of research in the last 10, 20 years and still ongoing research because it's very difficult to study this interval here, okay? And I will just take a couple more minutes then we will finish to say something about what is known in this interval here. And this is the interval in which you really see what I was saying, the fact that you have a mixture of contracting dynamics and expanding dynamics and in the end either one wins or the other wins depending on exactly what the points do. So if we define omega to be this interval minus 2 a infinity. This is the interval where everything interesting may happen. And we define omega minus is equal to the parameters in omega, such that there exists an attracting periodic orbit. And we define omega plus is equal to the a in omega such that f of a is chaotic, I will not describe it more. But chaotic a little bit like in the case a equals 2.  So remember these are the three basic kinds of dynamics that I described before. The attracting that comes from some kind of contraction, even though this is not a fixed point it's a periodic orbit. Chaotic that comes from some kind of expansion, adding machine that is where they exactly balance out. Then what we have is that theorems omega minus is open and dense. So you know what open and dense in omega, right? So this is an int, we have an interval omega. It's just an interval, parameter interval. So what does it mean that the set is open and dense in an interval? What does the complement of such a set look like? I said that it's open and dense. Like a canto set, exactly. So the complement is basically a canto set, it's not exactly a canto set. It's basically a canto set. But the nice thing, the very interesting result is that the Lebesgue measure of omega plus is positive. So you did Lebesgue measure, some of you have done Lebesgue measure before. You did Lebesgue measure anyway in the previous semester on the real line. You remember that Lebesgue measure is a generalization of the notion of length, right? This is a very nice example in which you have a set that has no length because omega plus is contained in a canto set, right? So there's no interval inside omega plus. So if you did not have this Lebesgue measure, you could not measure the size of omega plus, okay? But you can easily imagine what a canto set of positive measure looks like because this open and dense set is just a union of intervals. It's a countable union of intervals. So if this countable union of intervals, you can easily construct the countable union of intervals, where each interval has length, for example, 1 over n squared or something. And then the sum of all these intervals can have length less than epsilon, right? So you can have inside here, you can have a countable union of intervals. But each one is so small that the sum of all of them has only length epsilon, which means that the complement of them must have in some way 1 minus epsilon, or at least it must have some positive measure in some sense, right? And what this means is that these two phenomena are really intertwined in a very complicated way because it means that you have here a canto set of positive measure in which the dynamics is chaotic, which means that the expansion dominates, right? So we have our map that looks like this, okay? And what I'm saying is that if you look at the points, they will spend some time in the region where the derivative is small. They will spend some time in the region where the derivative is large. And if they spend more time in the region where the derivative is large, then the dynamics will tend to be more like expanding dynamics and you'll have chaotic. If it spent more time in the regions where the derivative is small, it will tend to have an attacking periodic orbit. And both of these phenomena occur with positive probability, one for an open and then set and one for a canto set of positive measure. And the last theorem is that the Lebesgue measure of omega minus omega minus union omega plus is equal to zero, which means in particular, so these neutral situations do occur. There's an infinite number of parameters in which you do have the situation, but it's a zero probability or zero measure situation. So what we know about this quadratic family is that essentially the dynamics is either the attacking periodic kind or the chaotic kind. And these two happen for these two kind of topologically distinct kind of sets of parameters. And there's still a lot of research to understand more the properties of this. For example, a natural question is how much is this measure? And how much is this measure? Both of these has positive measure because this one is open and then so it contains intervals. So it poses a measure, but nobody knows how much this measure is. And in fact, I worked on this problem a few years ago. I proved with a colleague of mine that the Lebesgue measure of omega plus is greater than 10 to the minus 5000. Which is a very small number, and it's not a particularly good result. But part of our motivation in the result was just to show that it's possible to give an explicit bound. Because the fact that it has positive measure was proved already now more than 20 years ago, and there's been many generalizations of this theorem. But they never gave any explicit bound at all. So even though 10 to the 5000 is small, it's not the smallest number you can have. So to prove that something is positive is not the same as proving that it is. And now I'm working with some other colleague to try to generalize this and to try to have more global bounds. And to try to get, of course, a much better bound. We don't know. So the order of this interval is something like, I can't remember what the value of A infinity is, but it's 1.5, around 1.5 or so. So this interval is of a reasonable length of order one, let's say. There was a couple of years ago, only a very recent paper, that showed that the Lebesgue measure of omega minus is greater than 10% is greater than 0.1 times the Lebesgue measure of all of omega. So we know that at least 10% of this is from omega minus. But nobody knows what the other 90% is. It could also be part of omega minus, most of it, or it could be mostly omega plus, actually. So this is just to give you one example to go from, there's quite a lot of techniques that go into arriving at understanding the proofs of these kind of results. But conceptually, I just wanted to show you that what we've been doing so far, even though what we've been doing so far has been, let's say, certainly not easy, but relatively elementary. It's not that far to go. The ideas are quickly going to some kind of result, which is really an active area of research, which is this. And on this quadratic family, actually, there's a lot of research having been done, and there's a lot still being done. Okay, so with this, I think we can close this part of the course, and I'd like to thank you all very much for your attention. Thank you.


That's why you seem to be in maximum hurry. But thank you for creating this opportunity, for creating this little competition. Competition, big or small, is always creates that energy and drive to do better. So congratulations to all those who are going to win today. And also best wishes to all those who are going to win tomorrow. Good luck. Thank you.


So in section 9.1 of our lecture series, we introduced the idea of vectors and their geometric or physical representation, these arrows. And we talked about the magnitude, the direction, and the components of a given vector arrow there. And we were able to use that to help us understand some physical ideas about direction and speed of moving objects. In section 9.2, I want to continue these ideas of vector applications, focusing on things from physics, mostly on directions and force vectors of some kind, and show you really how the parallelogram rule when it comes to adding vectors naturally leads itself to trigonometric type problems. And so in this video and the next one, I want to focus on a specific problem called static equilibrium. So when an object is stationary, we say that it's in static equilibrium. That is, it's stationary, it's not moving. Now, if it's not moving, that does not mean that there's no forces acting on it. Most likely, I mean I could be wrong, but most likely you yourself as you are watching this video are in a state of static equilibrium. You're probably sitting, your computer or device you're watching on is probably also not moving relative to the Earth, of course, which is moving. But we'll get away from that issue right there, right? It's not moving, but that means there doesn't mean there's no forces acting on it. Like if you take me for example, I am not moving at the moment. I am sitting in a chair, which we'll just think of it like this right here. Here's my chair. But there are forces acting upon it right now, right? Gravity is pulling this chair towards the center of the Earth. On the other hand, there is a force exerted by the ground pushing back against gravity and these forces are equal and opposite in their direction, which is why my chair or myself is not accelerating towards the center of the Earth. If you don't believe in normal force, ask your big brother to sit on your head sometime and then you will feel his weight, but you also feel the force of the floor pushing back as you feel you're still being crushed right there, right? And so static equilibrium means that the sum of all the forces acting on an object is the zero vector. There's no motion because the net sum is zero. And so that's a very important thing for us here because when it comes to vectors, when you add two vectors together, right, you put the tail to the head, something like this. If the force is zero, that would mean that in the end, when you connect all of these forces together, just drawing some hypothetical examples here, when you put all of the forces together, head to tail, head to tail, head to tail, static equilibrium means you form a polygon. That is, you eventually stop where you started. The net force is zero. And so when you have three vectors in play here, a polygon is a triangle. Ooh, trigonometry seems like it might be relevant for such a situation. Let's consider such an example.  So that's his weight, which is the force due to gravity.  Kids like to be suspended in the air on the swing in a suspenseful moment like, what's going to happen when I fall?  Clearly, it's not going to happen. But Stacy, she's holding Danny back on his swing in such a way that it forms a 30-degree angle and she's holding him right there. So this little diagram is to try to show you what's going on right here. So this right here represents Danny. He's at the middle of all of this. His weight is pushing down on him at 42 degrees. But like I said, Stacy's pulling him and suspending him in the air. But then there's also a chain or a string or a rope of some kind, depends on the type of swing there, right? But there's some type of rope that's suspended to the swing and it has a tension. So the tension is how much force is being pulled on our rope or chain, what have you. And it forms an angle so that with respect to the vertical here, this tension is 30 degrees like so. So we have the rope of the swing. We have the sister pulling, holding the boy up and then he has his weight. So there are three vectors. These three forces are acting on Danny all at the same time. So gravity pushes him down. Stacy's pulling him to the side, but the rope is also pulling him back towards the swing set. But he's not moving. He's in static equilibrium, which means that the combined forces of his weight, Stacy and the tension, they add up to be zero. They have static equilibrium. It adds up to be zero in that situation. Now in this format, it might not be exactly sure what we're going to do with this thing here. But what we want to do is we want to find out the tension of the ropes of the swing. So how hard is the rope pulling Danny? And also how hard is Stacy pulling Danny, right? What are the magnitudes of these vectors? Now the weight vector is pretty easy. That was given to us. Danny weighs 42 pounds. Now pounds is a unit of force. It measures how hard something is being pushed. So pounds represents how hard is gravity pulling on things. So the weight of Danny is 42 degrees and gravity always points downward. So we know it's going to go in a downward position. We were told that Stacy pulls him in a horizontal direction. So that forms a right angle here. And then you also give this mention of the 30 degrees, which we see in the diagram is representing this angle right here is 30 degrees. So because we're in static equilibrium, what we can do is we can move the arrows around thus to form a polygon, which in the case there's three vectors, here's going to be a triangle. So what happens if I were to move the force vector up like so? We have this weight vector up like so. We end up with something like the following intermediate diagram. So we have our weight vector. We have Stacy pulling like so.  And then the next thing is what if we move the tension over here? Again, what we're trying to do is we're trying to move it so that everything goes from head to tail. So tail to head, tail to head, tail to head like so. We can rearrange these vectors in the following picture like so, so that we then get this diagram for static equilibrium. You should be able to form a polygon right here. Now the angle between S and W should still be a right angle. What about this 30 degrees? How do I conclude this is 30 degrees right here? Well, notice that these lines right here are one parallel to each other. So the weight is a vertical vector pointing downwards. And the tension forms a 30 degree angle with respect to the vertical as we see illustrated right here. So these two lines as they're both vertical lines are parallel to each other. But then the tension vector, if we extend that line, it forms a transversal of two parallel lines. And so this angle right here, its corresponding angle would be this angle right here. And so by the alternate to your angle theorem, those angles will actually be congruent to each other. And so that's why we are able to conclude that this is a 30 degree angle inside of this triangle. And so now let's draw our triangle with our new light. We have this triangle, it's a right triangle. It's actually a 30, 60, 90 triangle if we want to be more precise. We know this side is 42, like so, 30, 60, 90 triangle, oops, 60 degrees. And we want to figure out the missing pieces here. So the hypotenuse is going to be the tension. And then this right here, the magnitude would be how hard Stacy's pulling this thing. Well, since it's a 30, 60, 90 triangle, we can dramatically simplify the trigonometry here. The hypotenuse will be twice the short leg, right? Which we don't know what that is yet, right? We know the leg associated to 60 degrees. So to get 30 degrees, we're going to take the 42 and divide it by the square root of 3, like so. Then to get the tension, we would double that. And so we end up with 84 over the square root of 3, like so. Which 84 over the square root of 3, that's going to be approximately 48.5 pounds. So the rope is holding what feels like about 50 pounds. So that's heavier than Danny, right? Stacy, on the other hand, she's holding him up. You take 42 over the square root of 3, you'll end up with 24.2 pounds, like so. So from Stacy's perspective, she's not holding the whole weight of Danny. Because honestly, the rope is doing a lot of the work there. It's doing more work than it was unnecessary. That's because the rope is picking up the slack from Stacy as well. And so we see that the rope, it has a tension of 48.5 pounds. And Stacy has, she's holding up Danny with a force of 24.2 pounds right there. And so this little example of static equilibrium shows you how when one works with force vectors, it naturally leads itself to trigonometric settings. Now this question was easy compared to what you could see for two reasons. One, the relationship here that Stacy's holding, her force was horizontal to the weight. You have this right angle. This forms a right triangle, which makes the trigonometry so much easier with right triangles. You just have your usual Soca Toa. That's one thing that made it easy. Plus also, since it was a 30, 60, 90 triangle, that also simplified the trigonometry dramatically. But of course, with the use of a scientific calculator, if we had something like a 20 degree angle, that wouldn't have made much of a difference. We just take sine and cosine of 20 degrees. We can handle such a thing, right? And but what happens when you lose the right angle here? Because we don't always expect these forces to be acting perpendicular to each other. So as we continue section 9.2 in the next lecture, lecture 29, that is, we'll look at examples of forces that are acting in static equilibrium, but don't necessarily form right angles. Thus, complicated trigonometry, but not beyond what we've already learned in this lecture series.