 Yeah, let us begin with a sort of assortment of topics, some of which were left out earlier and some of which complete what we already discussed. The first of these has to do with fractals. I gave an example of how we could find fractal dimensions for various sets. Here is another such example. Suppose I take all numbers between 0 and 1 on the unit interval and I write this these numbers in ternary rather than binary. So, I write each number say the initial number that I start with or any number at all x element of 0 to 1. I write this as 0 point a 0 a 1 a 2 dot dot dot where each of these digits is either 0 or 1 or 2. So, I am writing everything down in ternary rather than binary. So, the allowed digits for each of the a i 0 1 or 2. So, clearly this a 0 is the coefficient of 1 over 3 to the power 1, this is the coefficient of 1 over 3 squared and so on and so forth. Then the question asked is suppose I consider the set of all such numbers in whose ternary expansion the digit 1 does not appear at all. It is clearly a set of numbers and I would like to know what is the fractal dimensionality of the set. How would you go about this? So, let S be equal to the set of all numbers in 0 1 such that the digit 1 does not appear the ternary expansion of these numbers. What is the fractal dimensionality or box counting dimension of the set? How would you go about this? What would you do? So, I am not going to permit 1 to appear here. So, this should be either 0 or 2 that should be 0 or 2 0 or 2 and so on clearly I am a subset of numbers between 0 and 1 and I ask what is the fractal dimensionality? Yes. So, how would you go about it? What would you do? Yes, but I would like to know what is the fractal dimensionality of the set? How would you go about that? Suppose A naught has a 1 there then where would it be on this? So, this is a 0 or a 2 and if it is a 0 this number is less than 1 third. So, it is clear that the number would be so here 0 here is 1 this is 1 third that is 2 thirds. Yeah, it is exactly the middle third canter set and it is not hard to find this because if this digit is 0 it is here, if the digit is 2 it is there, if the digit is 1 it is in the middle and you are not going to allow A naught to be equal to 1. So, it is this is cancelled here and then if you look at A 1 then it is again going to be the middle third that is erased each time. So, it is going to be here or here and so on. You keep doing this forever you get the middle thirds canter set. Now, what is the fractal dimensionality of the middle thirds canter set? Well, at each stage you take some interval of the preceding stage you break it up into three parts and retain only the mid the end two parts and each of these is 1 third the length of the preceding stage log 2 over log 3. So, in this case d naught equal to log 2 over log 3 and it is the middle thirds canter set. So, that is the answer there is another famous regular fractal in two dimensions the many innumerable fractals. Another one is the so called Sierpinski gasket and let me explain what its properties are. You start with a unit equilateral triangle in the first stage. So, this is the and this goes over in the next stage to the following you remove the triangle in the middle. So, this portion is gone and at the next stage you repeat this process. So, this is already gone that is a hole and then there is another hole here and another hole here and another hole here and you continue this ad infinitum. You still left with a geometrical figure at the end which is riddled with holes so to speak it is called the Sierpinski gasket. What is the fractal dimensionality of this object? You start with this figure the area here is of course this is a two dimensional figure. So, in this case you would expect d naught to be equal to log divided by what? Do you think so? Do you think so? Yeah it is equally divided. So, you are getting rid of this and then you are getting rid of this etcetera and you are left with these little things which will then get little completely by holes. So, what do you think is the fractal dimensionality of this object? You start with an area whose topological dimension is 2. So, what is the fractal dimensionality here? So, these are areas mind you. Now what is the area of this figure? So, what is the fractal dimensionality finally? Will it be greater than 1 or less than 1? So, what do you get actually? You do not necessarily have to you might get rid of the entire area. So, there is no rule saying that you should have something which is greater than 1 or less than 1 or anything like that. So, what happens in this case? How many pieces are you breaking each one into? But you are removing that it is gone. So, it is log 3 on top and each of these is one quarter of the earlier. So, it is log 3 over log 4 which is a number between 0 and 1. So, you did not say log 4 over log 3. So, this is log 3 over log 4 my mistake and so on. You can construct this in higher dimensions. I leave it to you as an exercise to tell me what the fractal dimensionality is if you did this in d dimensions d spatial dimensions. The next stage for example, you take a tetrahedron which is a simplex in three dimensional Euclidean space. You take a tetrahedron and remove the middle by exactly the same construction and you are left with tetrahedron at the corners of the original tetrahedron and you can generalize this to d dimensions. So, I leave you to tell me what d 0 is in the case of a d dimensional Cierpinski gasket in d Euclidean dimensions. You start with the tetrahedron that will be the next thing. This would be the simplex four sided figure. It is a simplex and now remove the solid object in the middle and you are left with four tetrahedron at the ends. The shape is self similar. That is at each stage, you start with the triangle and it becomes smaller triangles. So, you start with the tetrahedron and it becomes smaller tetrahedron. So, you chop off the middle such that you have a tetrahedron at the corner here and the corner there and the corner there. What is the equation to a simplex in n dimensions? What is the equation to this tetrahedron? What would you say is the equation to this? It is a convex figure. It is called a simplex in d dimensions and now you have to generalize the idea of this object. It is a rigid object here. Two d dimensions as you go higher. I leave this as an exercise and we will give the solution subsequently. So, these are all constructs, artificial constructs made in order to illustrate various properties of fractal sets. With the middle third cantor set itself, you can generate a multi fractal. If you calculated here, if you said well, I just go through this construction at each stage and I associate a uniform measure with all these pieces, then all the generalized dimensions of this object would turn out to be just log 2 over log 3. Nothing would happen. But if I start associating different weights with these objects, different probability measures, then in principle I can end up with a multi fractal in which dq is not equal to d0 for all values of q. So, this is often used in modeling. We will not get back to this right now. Let me now introduce you to a two dimensional map which has standard properties in the sense that it is hyperbolic, it displays chaos and it gives you a little inkling of how this chaos arises. We already saw a two dimensional, the example of a two dimensional map. We saw the Baker transformation which you recall was Baker's transformation or the Baker map. We started with xn and xn plus 1 was equal to twice xn modulo 1. So, you stretched in this direction and the yn plus 1 was equal to one half yn if xn was less than equal to a half and it was half plus a half yn if xn was greater than a half in the unit square. And we saw this is an invertible map. So, the number of free images for any point is equal to 1. We also saw that this map is chaotic and that the unit square in the xy plane gets completely scrambled up as you iterate this map and that it has two Lyapunov exponents, one of which is log 2 and the other is a log half and that the sum of the Lyapunov exponents was equal to 0 as it should be because this is a measure preserving map and it mimics in that sense a Hamiltonian system in a very very crude sense. Now the other famous map which does that is called the Arnold cat map and it has got a fancy name, a more rigorous name but let me just call it Arnold's and the map is as follows says xn plus 1 is equal to xn plus yn and yn plus 1 equal to xn plus twice yn and both are modulo 1. Now let's see what this map does in pictures what it would do is the following it would start with this thing here x0 and y0. So, it start with the unit square in each of these variables and then x1 is x0 plus y0 modulo 1 and y1 is of course x0 plus 2 y0 again modulo 1. If you didn't have the modulo 1 then what it would do is to do this it would double in the x direction and it would triple in the y direction or not drawn it too well so let me slightly smaller scale. So, here is 1, here is 2, here is 1, here is 2, 3. So, what the map does is to take this guy here and distort it in this fashion such that the unit square becomes this. So, this is what the unit square goes into this stage here but you are supposed to do this modulo 1. So, what you do is to cut out all these pieces and put them back into this square what's the total area of this distorted square that just depends on the determinant of this map. So, if I call this map let's call this map A. So, what I have is xn plus 1 yn plus 1 is equal to this matrix 1112 times xn yn the whole thing is modulo 1 of course the determinant of this matrix is unity therefore it's measure preserving we guarantee that the area of this object here is exactly the same as the unit area and if I cut it and put it back here then this map looks like it's on several pieces but I could also pretend this map is on a unit torus right after all once I say it's periodic I could put this on a torus I could simply say this this this etc. all copies of this bring it back and put it on a torus and then on a torus in which the circumference in this direction is 1 and the circumference in this direction is also 1 this object whatever object you put here remains continuous you are not tearing anything but the moment you say modulo 1 and you cut and snip and so on it becomes disjoint pieces. So, what will happen is that if you took some object here some area here that area would go over and get scrambled up some of it would be here some would be here some would be here etc and since Arnold first drew a cat's picture said this is the way the cat gets deformed gets scrambled up. So, what he did was to start with something which looks like this and then that gets scrambled up as you go along it became known as a cat map however there is serious purpose to this and let's examine the properties of this map and see what happens. So, my claim is that this map produces chaos it actually scrambles it up completely there's mixing and there's exponential sensitivity to initial conditions the phase space is bounded as we see this measure preserving. So, nothing is lost the unit square remains a unit square but nearby points under iteration become go arbitrarily far away they become as big as the system size itself and there is exponential sensitivity like the Baker map because you put this because it's hyperbolic and let's see what's meant by that what are the Eigen values of this matrix. So, I have this matrix A which is equal to 1 1 1 2 what are the Eigen values of this matrix well it's clear lambda minus 1 times lambda minus 2 that's lambda squared minus 3 lambda plus 2 minus 1 plus 1 equal to 0. So, lambda plus or minus is equal to 3 plus or minus square root of so that's a 9 minus 4 is a 5 divided by 2 those are the Eigen values of this matrix we check what the Eigen vectors are as well and the Eigen vectors would satisfy if u 1 u 2 is an Eigen vector then I apply this and I insist that this be equal to u 1 u 2. So, this implies that u 1 plus u 2 be equal to lambda times u 1 the other equation doesn't give anything new. So, it's clear that the Eigen vectors for lambda plus the Eigen vector is if I choose u 1 to be equal to 1 for instance u 2 is lambda minus 1 times u 1. So, this is equal to 1 and lambda minus 1 that's 3 plus root 5 minus 2 that's 1 plus root 5 that's the Eigen direction and for lambda minus the Eigen vector is 1 and then this is a minus sign here 1 minus root 5, but since root 5 is bigger than 1 let me write this as minus root 5 minus 1 over 2 what's the significance of this number 1 plus root 5 over 2 it's a famous number it's the golden mean it's the golden mean this number is the reciprocal of that number you can see. So, we can plot what the Eigen directions are now and since this is a constant matrix this is a linear map if you forget the modulo 1 since it's constant the acrobene is the same everywhere at all points and therefore this is what controls the expansion or contraction at every point it's clear that there is an expansion in one direction and a contraction in another direction if you go back to the original picture that I drew it's quite evident that if this gets shifted out and then you end up with the sort of stretching in one direction in this fashion the part of that the parallelogram was cut and put here so there is a sort of stretching in this direction there is a contraction in the opposite direction. So, if you quantify that these are the Eigen directions on which you have the stretching and the contraction and we can easily check out what it looks like. So, here is the unit square and let's look at this Eigen direction in what direction is this relative to this what's this direction this is the x direction and that's the y direction this corresponds to x and that corresponds to y. So, vector unit vector with these as components vector with these as the x and y components would make an angle with x axis which is the tan inverse of 1 plus root 5 over 2 divided by 1 this is about 1.618 or something like that 1 plus root 5 over 2 we want the tan inverse of that. So, that's some angle is greater than 45 degrees some crazy angle it's 58 degrees or something like that this is the direction and the other guy it's easy to check that these two slopes the product of the two slopes is minus 1. So, there are right angles to each other this is transverse this direction and that is going to be something like this not hard to check that this corresponds to an expanding direction and the other corresponds to the contracting direction how do you come to that conclusion how do you do this after all if I iterated this map n times I get a to the power n acting on this x naught y naught modulo 1 right. Now, how do I conclude that there is chaos in this system and that things actually exponentially separate out in at least one direction it's the lamb does that control it as you can see and just like in the case of the one dimensional example of the Bernoulli shift where I know that xn was 2 to the n x naught times modulo 1 and the Lyapunov exponent became log 2 because that factor 2 to the power n if I take the log of it and divide by the initial separation I end up with log 2 as the answer if I divide by n and take the limit as n goes to infinity I get the Lyapunov exponent. So, that's exactly what we have to do here except there are two Lyapunov exponents what are these two what are the Lyapunov exponents for this map no no the log of these things the log of this because you see if I solve this map it's evident that I am going to get something like xn yn equal to a to the power n x naught y naught modulo 1 that's putting it back in the unit square but this guy here if I took this matrix and diagonalize this matrix and write the elements down the diagonal elements down it would be controlled by the behavior of lambda 1 to the n and lambda 2 to the power n multiplied by two guys which are the which die do the diagonalization of this matrix and then eventually the Lyapunov exponent will be governed will be given by log lambda 1 and log lambda 2 I call them lambda plus and minus so let me so these things could be written as e to the n log lambda plus e to the n log lambda minus and then this number here is a Lyapunov exponent I am sorry I use the same symbol lambda for the eigen value I should have probably used mu here and then lambda for the Lyapunov exponent. So, that's the reason for the confusion now what's log lambda plus so the Lyapunov exponent log 1 plus 3 sorry 3 plus root 5 over 2 is this positive or negative it's positive exactly and the other one is log 3 minus root 5 over 2 and what's that it's negative because this is the number less than 1 therefore it's the contracting direction and we also found the eigen directions so this is negative this direction contracts in this direction expands this point here is like a homoclinic point because that's exactly the point if I put this on a torus it will keep coming round and round so let's pretend that we are doing that so since this is periodic modulo 1 it means that once you go out here you are back here you start here and then this is what the unstable manifold is going to look like so here you are unstable you keep going round and round so once you're there you're going to be injected back here and then parallel to this you want to go off similarly here you're going to go off like this and then you're going to come here go off like that and this will keep happening all the time so there is at every point there's a transversal intersection there's a transverse intersection so what's happening is that you are on a torus and along almost every point on the torus you end up with except periodic unstable periodic points which we haven't yet discovered you have a contracting direction and an expanding direction they are orthogonal to each other so it's as if you've got saddle points everywhere that's exactly what happened in the case of the Bernoulli shift as well it's as if you've got separators is everywhere because you've got these unstable periodic orbits at the rational things get thrown out on both sides and that's exactly what happens here now because it's on a torus and because it maps the torus to the torus it's called a toral automorphism and because it's hyperbolic no center manifold here it's called a hyperbolic toral automorphism and the name cat map is easier to handle but there's not the only one any number of such maps could be constructed to put the integers down here such that the determinant is plus one and you end up with such a automorphism it's a paradigm of chaos it is exponentially unstable almost everywhere and it's got an even dimensional phase space it's measure preserving so this was Arnold's example trying to mimic what happens when chaos sets in in a genuine Hamiltonian system and it happens because of the hype what's called the homoclinic tangle and this gives you a sort of toy or caricature of what happens in such a system that you have this expanding and contracting directions everywhere the entire measure is preserved but it's chaotic with this layer of an ovex point let's try to find what the periodic points are where are the periodic where are the fixed points of this map and where are the periodic points of this map how would we do that well I need to know when this map leaves a point unchanged and this is not so trivial to do but let's see what do about it now notice that under iteration since we have xn plus one is equal to xn plus yn before you do the modulo one if I start with these numbers between 0 and 1 and I go to xn plus one the maximum that xn can increase by is one you can't increase by more than one similarly yn plus one is xn plus twice yn what's the maximum by which yn could increase before you do the modulo one two as we saw in the picture right could you could add a two to the unit square and then you add a square of rectangle of height three and width two so it's clear that before you do the modulo one the largest that x can increase by is unity and the maximum that y can increase by is two therefore if you have fixed points if you have a fixed point and what does it imply any fixed point of this map of the n iterate of this map so I take a to the power n and I ask what about the fixed points of a to the power n when n is one I get the fixed points of the map when n is two I get the period two cycles of this map and so on and so forth so I am not trying to find all the periodic points and all the fixed points of this map so it's evident that if x x0 for example x0 y0 is a fixed point of this object then it says xn must be equal to x0 plus some integer k which goes away in doing the modulus bad notation use it like this suppose x0 is a fixed point then after any iterations it would do this where k can be 0 or 1 or 2 up to n it can't be more than n because after n iterations a point cannot increase by more than n since in one iteration it can't increase by more than one before I do the modulus before I do the modulo trick similarly yn equal to y0 plus l if y superscript 0 is a fixed point of a to the power n and so is x0 if x0 y0 is a pair is a fixed point of the map a to the power n then you must have after any iterations this must go to this plus integer where l equal to 0 1 2 up to 2n because it could go right up to n and then in the modulo 1 it goes away that integer part goes away now this will help us find where are the fixed points of all of the map and its iterates so let's look at the map let's look at the map itself so n equal to 1 implies that any fixed point must satisfy x equal to x plus k where k can be either 0 or 1 but now you are imposed the modulo 1 condition so what's the only possibility here 0 so it must satisfy this similarly y must be equal to on the other side y plus the only thing you can have here is a 0 implies under iteration what's the only fixed point now this is the only fixed point the origin is the only fixed point of this map nothing else now I leave you to do the higher iterates so do the first iterate a squared first and so on and you discover some rational points would also be allowed and then you do a cube and so on and it will turn out that all rational points are fixed points of this map or of its iterates and this is not trivial to prove not very difficult either you can show that all rational points either are fixed points of the map the only one is the origin or lie on periodic orbits like in the case of the Bernoulli shift now the rationals are dense everywhere on the unit square the pair of rationals dense everywhere on the unit square and those points would lie on periodic orbits but tell me what happens to this line this line here let's look at this guy here the slope of this line 1 plus root 5 over 2 1.618 or something like that some in irrational number what's special about this line what sort of points can lie on this line yeah both x and y cannot be rational right so no rational pair can lie on this line nor can it lie on all its subsequent iterates or on this so these points will not pass through any pair of points any points whose coordinates are rational and yet they would densely fill up the entire square but they'd miss all the rational points and that's what a chaotic trajectory would do typical chaotic trajectory would do but all the rational points are left out point by me the fixed points are all rational points so a line like this this is the unstable manifold and this is stable manifold this line does not pass through any pair of points both of which are rational and nor does any of its subsequent iterates so this line here comes back and starts off here and then it after it goes here it goes here and then it comes here and goes here and so on this thing will not have any point on it both coordinates of which are rational and that would be generic that's typical so such lines this line the measure would be the total measure of this area itself and the rational points form a set of measure zero they're the unstable periodic orbits and in its neighborhood you have these expanding and contracting directions so it's fairly it's a fairly complicated picture but it's like a paradigm of chaos Hamiltonian chaos in this case because it's measure preserved we saw a little bit about what irrationals do and let me give you another illustration of a map a simpler map where you have this some properties of irrationals will emerge and do that I'm not sure if I discuss this earlier we go back to a one dimensional map where you have some interesting properties did I discuss the continued fraction map at all the Gauss map let me do that now it gives you an illustration of the continued fraction map this question was actually this map was first considered I don't know the exact history of it but Gauss did very important work on this map and it's the following question in fact the question that Gauss was interested in was the following he said suppose you take a number between 0 and 1 so I take an element x element of 0 to 1 somewhere 0 and 1 I take its reciprocal that's a number bigger than 1 I throw away the integer part keep the fraction I take its reciprocal I throw away the integer part and keep its fraction and then the question was if I took a typical number and kept doing this and iterating this procedure what's the probability that the number I end up with after a very large number of iterations is less than some given x so the question was I start with a typical number between 0 and 1 take the reciprocal throw away the integer part and keep iterating this procedure and ask finally what's the probability that the answer I have is less than some prescribed threshold x which is between 0 and 1 and he found that this answer is very interesting indeed it was proportional to log 1 plus x this probability is proportional to log 1 plus x and this excited him so much that he wrote a letter to Lagrange saying that he found the solution to this problem and what he did essentially in our terminology is the following let's see what this map is so you start with the map which says that at the n plus 1 stage you take the reciprocal of xn and throw away the integer part so it's minus 1 over xn and this guy here is the integer part so I use the square packet to denote the integer part and now the question is what is this map chaotic if so what's the layup no exponent what's the invariant density of this map etc this is the question we have to answer what is the figure of this map look like what is the graph of this map look like so if I write x goes to f of x is 1 over x minus this what is this graph look like it's already going to be mod 1 so I start with x not to start with that so I start with 0 and 1 between 0 and 1 I take the reciprocal this is bigger than 1 I throw away whatever is the integer and I still have something between 0 and 1 and I keep doing this so it remains between 0 and 1 automatically same thing right same thing okay so this tells you explicitly what you are doing it will become clear why I wrote it in this fashion what does the graph look like in fact why make a mystery of it let me tell you why I wrote it in this fashion instead of writing it as a binary number a binary decimal which is what we did for the Bernoulli shift I can also write it as a continued fraction so I can also write x not as equal to 1 over a number bigger than 1 and in standard form any continued fraction can be written as an integer plus 1 over once again an integer plus 1 over forever and in a simple notation this is called 1 over a not plus 1 over a 1 this is notation for a continued fraction so that you do not have to write this crazy diagonal everywhere so these are integers positive integers what is x 1 you take its reciprocal then you have an integer plus a fraction and you throw this integer away therefore you have this right so this immediately implies that x 1 equal to 1 over 1 over a 1 plus 1 over a 2 plus so you have lost information about this is clearly not invertible do you think the map is chaotic do you think it has fixed it has fixed points once we draw this graph it will become clear it has fixed points but do you think these fixed points would be stable no they won't because the slope of this map this is some number it's constant each time it's a different constant the slope of this map is 1 over x squared the minus sign in the mod of that is 1 over x squared and since x is less than 1 this is always positive so it's clear that this map has no stable fixed points and its iterates are going to get only worse so it's a candidate for chaos once again there would of course be points which lie on periodic cycles there would be points which lie which are fixed points of this map and so on what would the fixed points look like for that we should draw this graph it's clear that if I draw just 1 over x then if I draw y equal to 1 over x that's a rectangular hyperbola which passes through the point 1 1 but it actually does some crazy thing like this and goes off and I start with a number between 0 and 1 here and the graph is always bigger than 1 but I cut this and put it back right so as long as the initial point is to the right of a half this is half all I have to do is to subtract 1 to bring this piece and put it back here so this map looks like this if the number is between 1 3rd and a half then I have to subtract 2 and bring it back and put it down here so the map looks like this little sleeper steeper slope if it's between 1 4th and 1 3rd then I have to subtract 3 and therefore this map starts looking like this and it's getting steeper and steeper as you go down it has an infinite number of branches right and all of them have slopes in magnitude much bigger than 1 as you go to the left and therefore all fixed points are unstable but what are the fixed points they are these intersections these are all the fixed points what is this point correspond to so I'd like to know what the fixed point there is for this map this is the largest fixed point of the map here and on this branch the map function f of x equal to 1 over x minus 1 because that's what I subtracted so this fixed point implies that x is equal to 1 over x minus 1 or x times x plus 1 minus 1 equal to 0 x squared plus x minus 1 equal to 0 or x equal to minus 1 plus or minus square root of 1 plus 4 5 over 2 and you want the number to be between 0 and 1 so the fixed point in fact is root 5 minus 1 over 2 so that's the only acceptable solution that's the golden mean once again what's its continued fraction representation and that tells you why this number is the most irrational number of irrational numbers exactly it's just one all the way through you see what's happening here is that you want this number to satisfy 1 over x is 1 over is x plus 1 or x equal to 1 over x plus 1 or 1 plus x but I have a substitute for x here as 1 over 1 plus x etcetera and keep going so this is equal to 1 over 1 plus 1 over 1 plus 1 over 1 plus in other words all the ai's are 1 and now you see why it's so terrible to approximate this number by a rational number because if one of these ai's was very large compared to 1 you could truncate the number at that point and give a very good approximation to the number this is what I want you did when it took pi and did this in gave a continued fraction expansion of this such that the third or fourth of these guys was 293 which when compared to a fraction was much larger and therefore he got an accurate estimate of pi to a very large number of decimal places by writing a very compact continued fraction expansion so pi all the transcendental all the irrational is not one of these very rational numbers quote and quote which is in the sense that rational approximants are very difficult to find very poor rational approximants approximate this very poorly because these numbers are all comparable to each other they are all equal in fact and they are as small as they can be 1 what's this number what do you think it is yeah it's exactly it's just 1 over 2 plus 1 over 2 plus etcetera because that number satisfies this so this is the silver mean this one is the golden mean there is a silver mean what's the value of this number what's this guy here satisfies this so let's solve this so it's x into x plus 2 minus 1 is 0 and the root that we want is this is twice root 2 so the root that we want is no medals for guessing what this is what's the continued fraction expansion of this one over all the integers are 3 so much for the fixed point and it's clear that all of them are unstable the slope is greater than 1 so much for the fixed points what about period 2 cycles of this map you would want it to flip back and forth it's quite clear that when you wrote the when you write a continued fraction expansion you would want it to flip back and forth right so period 2 cycles are all numbers whose continued fraction reads like 1 over a plus 1 over b plus 1 over a plus so perhaps you would have 1 over 1 plus 1 over 2 plus 1 over 1 plus 1 over 2 plus etcetera that would be the leading the way first the largest period 2 it flips back and forth with 1 over 2 plus 1 over 1 plus etcetera so you see the continued fraction expansion of the periodic orbits of this map have a very interesting simple expansion and that's why it's also called the continued fraction map that's exactly what it is just as the Bernoulli shift was the binary shift if you like the Bernoulli map was a binary shift similarly here this is the continued fraction map now let's try and find its invariant density what was Gauss's result here and how did he get it and it's Lyapunov exponent well the invariant measure would satisfy the Frobeni-Sparow equation so that would be rho of x equal to an integral 0 to 1 dy times rho of y times a delta function of x minus f of y but what is this f of y yeah this thing here would depend on where you are right but this f of y in general would be something like 1 over y minus some integer k and all k is from 1 to infinity would contribute corresponding to all these branches so if I did that I end up with a summation from k equal to 1 to infinity of rho of well x plus k is y so y is 1 over x plus k so 1 over x plus k that goes in here let's write this this is 1 over y plus k and there's a summation over k divided by the Jacobian of the transformation the Jacobian of the transformation when I take the modulus it's just a 1 over this guy here so what would 1 over x plus k the whole square because it's divided by 1 over 1 over x plus k the whole square so this is just x plus k so let's write it properly 1 over x plus k the whole square rho of 1 over x plus k now this doesn't look like something which is going to solve very easily it's a functional equation which is completely crazy but this is what Gauss solved in effect he solved this so since we have by hindsight we have Gauss to support us let's write the solution down and see whether it works or not so the solution that he had I pointed out that he said it's proportional the probability that the final that the iterate after long time is less than a certain number x is proportional to log 1 plus x that's the cumulative probability therefore the probability density is the derivative of log 1 plus x so it's 1 over 1 plus x but you must normalize this to unity between 0 and 1 if I integrate this I get log 1 plus x right and I put in the limits at the upper limit I get log 2 in the lower limit I get log 1 which is 0 so the normalization constant is 1 over log 2 let's put that in write our exact expression this is the normalized density and the question is whether this satisfies it or not so the question we are asking is the log 2 cancels out on both sides it's a homogeneous equation so the question we are asking is is 1 over 1 plus x question mark equal to k equal to 1 to infinity 1 over x plus k the whole squared 1 over k plus sorry 1 over 1 plus 1 over x plus k so you can see that this becomes x plus k on top and this becomes x plus k plus 1 and one of these factors cancels you get this that's a convergent sum because it's going like 1 over k squared so it converges faster than 1 over k so the question is is this true this is the J e sum what do you do next do partial fractions you do partial fractions so this guy here is indeed equal to 1 over x plus k minus 1 over x plus k plus 1 and of course terms the whole thing telescopes and only the first term would contribute and only the k equal to 1 term the left hand side survives this one survives and you get indeed 1 over x plus 1 so this is the solution and we were assured that once you find a solution it's unique so this is the solution of course please be careful this sum here this sum here is strictly equal to summation k equal to 1 to infinity that I shouldn't write it as a difference of two sums because each of them diverges so as long as I keep the summation outside and the bracket inside the sum this is fine this is completely quotient and indeed it's true so the elementary yeah I don't know how Gauss did it I am sorry I should have read the history of this I don't know how Gauss did it I'll look it up I don't know how we did it but I know you are sufficiently excited to actually write a letter and see he's found the solution I don't even remember whether he wrote the letter to Laplace or to Lagrange or whatever but certainly sent a letter to somebody in France on this so now that we have row of x we can find the Lyapunov exponent of this map now what's the Lyapunov exponent physically measuring it's in some sense measuring the average the log of the average slope of this map it's log mod f prime of x so the average stretch factor it's measuring now the slope here at this point at x equal to 1 is 1 in magnitude and everywhere else it's bigger than 1 and it's in fact tending to infinity as you come here so the question is what's the average slope but it's got to be weighted with that factor out there so when you do that you get lambda equal to 1 over log 2 an integral from 0 to 1 dx over 1 plus x times log mod f prime of x but f prime of x is minus 1 over x squared so this is equal to log 1 over x squared which is minus 2 log x this is negative but so is that so the Lyapunov exponent is positive now what do we do with that integral we can always look it up we know it exists it's not singular because the log is an integrable singularity this is very harmless the rest of it is not going to vanish at all so the integral exists there's no difficulty about it but what do you think is the value of this integral unfortunately because there's a sort of mixture there's there's a rational function here and there's a logarithm here it can't be done by elementary means it's not a trivial integral you can't do this by elementary means so little harder work is involved it's a logarithmic integral it gets related to the zeta function of 2 and the answer here it turns out to be equal to pi squared over 6 log because this integral let me write that down here integral 0 to 1 tx log x or minus times 1 plus x and do this in several ways and show that this is indeed going to happen one of which would be to expand this in a binomial series and try to collect terms and so on the other would be to try to expand the log but neither case this gets related to the following sum which is zeta of 2 and that's pi squared over 6 so that's how the pi squared over 6 arises and this is defined this is called the Riemann zeta function of argument 2 and it's finite it's not hard to show that this number is pi squared over 6 it can be done in several ways the simplest of which would involve into contour integration since we haven't come out to get into that at the moment but let me define what the zeta function is it's useful information zeta of k is defined as summation n equal to 1 to infinity 1 over n to the power k and to start with you could ask what it is for positive integer values of k please notice if I put k equal to 1 it diverges it logarithmically diverges so k is 2 3 4 etc etc and for 2 it happens to be pi squared over 6 for 3 it's fairly complicated for 4 it's pi to the 4 over 90 or something like that and so on. So simple expressions are available whenever this k is an even integer when it's an odd integer it's no such simple expression is available but in we know what it is numerically this is called the Riemann zeta function because you can actually define it with any argument here I could put an x here and it's clear that if x is positive and greater than 1 then this converges even if x is irrational in fact since it's only the positive real part of x that matters you could make it a complex variable and all you need is the real part of z is greater than 1 it's an analytic function in the sense of analytic functions of a complex variable of this argument z and in the z plane it's defined by this infinite series to the right of real z equal to 1 it stops converging when z is equal to 1 in fact that series doesn't make sense anywhere on this line or to the left of it but it turns out this function has meaning even to the left of it it has a singularity at z equal to 1 it has a simple pole with residue equal to 1 and everywhere else it's analytic so in fact you can define zeta of z not by this infinite series not by the series but by a different representation even to the left of it everywhere the most important function if you like in analysis but what we have here is zeta of 2 which incident coincidentally happens to be related to the Lyapunov exponent of the continual traction map there are deep connections here there are reasons why this is so and since I have digress so far let me also mention that the most famous the most important problem in mathematics is the following is the establishing of the Riemann hypothesis which says that this function actually has zeros some points values of z it actually vanishes it's proper definition to the left is done by what's called analytic continuation that vanishes at some points and it turns out you can trivially show that it vanishes at all the even negative integers minus 2 minus 4 etc it vanishes at all those points trivially it also vanishes at points on a line perpendicular to the x to the real axis parallel to the imaginary axis cutting the real axis at the point of half so on this line on which z is equal to half plus i y it vanishes at a large number of points and it's got some symmetry properties on this it vanishes at an infinite number of points on this line and the Riemann hypothesis one form of it says that other than these trivial zeros all the other zeros of the zeta function are on this line and this has stood the 150 years of as a challenge nobody has proved this so far what's known is that there are an infinite number of zeros of the zeta function on this line what's also known is that all non-trivial zeros with probability one lie on this line you may wonder where probability comes in here but it does but what's not known is a proof it says there are no other zeros anywhere else so people have numerically try to find all the zeros on this they found the first 13 billion zeros as of last October they are all on this line yeah now you could ask what's the measure of the zeros here what's the distribution of the zeros here that distribution is related to the prime number theorem which says that the number of primes less than the number x is of the order of x over log x so the proof of the Riemann hypothesis on the number field on the complex now the numbers would involve would immediately solve a very large number of outstanding problems so much so that it's taken to be a truth and all those proofs rest on the truth of the Riemann hypothesis but this is so far defied everybody this is again the Riemann hypothesis looks very specific to a particular function and so on and so forth but it's much much more general than that and it's got generalizations so it is in fact the most important problem in mathematics now what's even more intriguing is that the distribution of zeros here is turns out to be related to the distribution of energy levels for quantum mechanical systems which are classically chaotic we don't fully understand the connections so there are incredibly deep number theoretic connections between the zeta function analytic number theory chaotic dynamics and quantum mechanics so much of this is still not known with a lot of open problems but certainly cracking any of these problems would be a major step so I didn't want to get into this but incidentally there are all sorts of other relationships which lead to these things apart from the continued fraction map leading to this for example I give you two numbers and I ask you what's the relative probability that they're co-prime with each other with the two positive integers and I ask what's the probability that they have a common divisor or they don't have a common divisor and so on that's related to 6 over pi square again so and that's not hard to prove either elementary proof for that so there are all sorts of very very interesting and deep connections but what is intriguing is there are connections with real life with the actual dynamics dynamical systems and the full reasons are not totally clear as