 We continue from last time, remember last time what did we do? We did a beautiful application of Fayer's theorem. We used Fayer's theorem to prove that trigonometric polynomials are dense in L2 and that enabled us to complete the proof of Parseval formula and I also in the last part of the last capsule I mentioned to you that we are going to be looking at a beautiful application to number theory and here is the result. You see it on the slide, Kronecker's theorem and Weier's equidistribution theorem. So let us recall the classical result due to Leopold Kronecker which is also known as Dirichlet's theorem in some books. Suppose alpha is an irrational number, let us consider the sequence of numbers, fractional part of alpha, fractional part of 2 alpha, fractional part of 3 alpha etc. where the brace bracket theta denotes the fractional part of theta. So you simply take the multiples, the integer multiples of alpha and just take their fractional parts. In other words you go modulo 1, you subtract the integer part and you get the fractional. So these numbers in that you see listed in 3.6 they are all in the compact interval 01 and they are distinct. Where they are distinct? Suppose for example they are not distinct. Suppose there is a pair k and l distinct positive natural numbers such that the fractional part of k alpha and the fractional part of l alpha are equal. That means k alpha minus integer part of k alpha equal to l alpha minus integer part of l alpha or k minus l of alpha is an integer which means that alpha is a rational number and that is a contradiction because we have chosen alpha to be an irrational number. Remember right in the slide alpha is an irrational number, it is important. So alpha is an irrational number, these numbers listed in 3.6 are infinitely many, no two of them are equal. So by Bolzano-Weier-Strasse's theorem you have got a compact interval 01 and you have got infinitely many points there, they must have a limit point. So sequence 3.6 must have limit points in the closed interval 01. Well that is not good enough. Kroniker's theorem tells you great deal more than that. It says that not only the sequence 3.6 has limit points, it has got lots of limit points. In fact every point of the closed interval 01 is a limit point. Stated differently the points listed in 3.6, the numbers listed in 3.6 are dense in the interval 01. So Kroniker's theorem is a density theorem and this is very classical and you will see the proof of it in elementary books in number theory or analysis. Let us prove this simple fact. Let us begin with a simple observation that instead of proving that the numbers 3.6 are dense in the closed interval 01, it is enough to prove that m plus n alpha where m and n vary over all the integers. This h is dense in r because if I prove that this h is dense in r then h intersect this interval 01 would be dense in this interval 01 and our job will be done as soon as we observe that h intersect this interval 01 is precisely the numbers listed in 3.6. It is clear that each of these numbers n alpha fractional part lies in h intersect 01. Conversely if you take a point x in h intersect the interval 01 then this x is already equal to the fractional part of x because I removed one from the interval remember and so this is going to be equal to m plus n alpha because my x is in h and so this is going to be m plus square bracket n alpha plus brace bracket n alpha and so this shows that this m plus square bracket n alpha must be 0 and so this x must be brace bracket n alpha. So we have shown that in order to prove Kronecker's theorem it suffices to show that this capital H this set capital H displayed here is dense in the set of real numbers. It is pretty clear that these numbers listed in h they form a subgroup of r that is capital H is a subgroup of the group of real numbers and capital H is not cyclic this is not a cyclic group. I would like you to explain why this is not a cyclic group because if h or a cyclic group it will be generated by a single element t it will be generated by a single element t right this t is obviously not 0 and so what are the elements of h then 0 plus minus t plus minus 2t plus minus 3t etc right and clearly this is not going to be h I cannot get m plus n alpha as multiples of integer multiples of a single t then a result now follows from a more general result in the next slide. We prove the following result theorem 34 if a subgroup of r is not cyclic then it is dense in r in our case the h in the previous slide is not cyclic and hence it is dense in r I am leaving it to you to figure out why it is not cyclic. If it were cyclic it will be generated by a single element t and explain why multiples of a single element t cannot give you all possible numbers in h and hence it must be dense in h must be dense in r and that completes the proof of chronicles theorem to show that h is dense in r where h is a non-cyclic subgroup h is non-cyclic so in particular h is not a zero subgroup take an element in h other than zero then minus x is also in h which means that h contains both x and minus x x is not zero one of these two must be positive so h contains positive elements. So now let us look at the set of all positive elements in h let us look at the set of all positive elements in h and let us take mu to be the infimum of all the positive elements in h this mu is non-negative of course it is not clear that mu is strictly positive infimum is different from minimum remember always so we will show that if this infimum is strictly positive then h is cyclic well suppose if mu is in h suppose if this infimum is in h then let us look at h intersect open interval mu to mu this h intersect open interval mu to mu is empty because if y belongs to h intersect open interval mu to mu then y minus mu will also be in h and y minus mu is positive and strictly less than mu which is the contradiction so it is very easy to see that h intersect open interval mu to mu is empty similarly one will check that h intersect j mu j plus 1 mu is empty for each j and then h will be the cyclic group generated by mu because h does not contain the open interval mu to mu h does not contain the open interval 2 mu 3 mu h does not contain the open interval 3 mu 4 mu so from the real numbers you scoop out the intervals mu to 2 mu 2 mu to 3 mu 3 mu to 4 mu etcetera so what is left over will be exactly mu 2 mu 3 mu 4 mu and negatives over that and that would mean that h is cyclic and that is a contradiction so now we have shown that if this infimum is positive and if this infimum lies in h then h is cyclic what if the infimum is positive and the infimum doesn't belong to h we are assumed that the infimum is positive we are going to arrive at a contradiction we have two cases mu belongs to h mu doesn't belong to h mu belongs to h's case is dismissed so now let us assume that mu is positive and mu doesn't belong to h so you got an infimum of a set of numbers and the infimum doesn't belong to the set so there is a sequence xn of positive element there is a sequence xn of positive elements in h which converge to mu and remember that if you have a sequence of real numbers it has a monotone subsequence so we may assume by passing to a subsequence that the xn the sequence of xn of positive elements which converge to mu is actually decreasing it is strictly decreasing so there is no loss of generality in assuming that there is a strictly decreasing sequence of elements in h converging to mu and take epsilon equal to mu by 2 because mu is positive then there is an n naught such that xn minus mu mod less than mu by 2 for all n bigger than or equal to n naught now I am going to take two things n and m both larger than n naught both n and m are larger than n naught so this inequality will be true for both n and m which means that what does this inequality mean xn and xm both lie in the interval mu by 2 3 mu by 2 and it is a strictly decreasing sequence so mu by 2 less than xn less than xm less than 3 mu by 2 so what happens to xn minus xm xn minus xm is less than mu but xn minus xm is in h and mu and xn minus xm is positive and mu is the infimum of all the positive things so that is a contradiction because that thing is less than mu by 2 so mu bigger than 0 is ruled out now let us take up the case mu equal to 0 so the only case that is left over is mu equal to 0 we show that in this case every interval open interval of positive length contains an element of h since there is a sequence of elements in h converging to 0 remember that the infimum of positive things in h is 0 so I can pick a sequence in h consisting of positive numbers going to 0 and I got this interval ab of positive length b minus a so one-third of b minus a is positive so take a yn such that 0 less than yn less than one-third of b minus a there must be a least natural number k such that ky n exceeds a Archimedean property which means that k minus 1 yn is less than or equal to a or ky n is less than or equal to a plus yn so where are we we have a less than ky n ky n is less than a plus yn a plus yn is less than a plus one-third of b minus a which is less than b so we see that ky n is contained in this open interval ab but yn was in my subgroup h and so ky n is also in my subgroup h proof is complete so we are given a very elementary argument to prove Kronecker's theorem this argument is quite elementary but now we are going to give another proof of Kronecker's theorem but this second proof will use Phaer's theorem but what we are going to do is we are not going to just prove Kronecker's theorem we are going to prove a remarkable sharpening of Kronecker's theorem so before we take up this second argument for Kronecker's theorem and its sharpening of Kronecker's theorem let us look at a few exercises and these exercises could be discussed in elementary analysis courses take the sequence sin 1 sin 2 sin 3 dot dot dot the sequence is clearly bounded above by 1 but how do you know that 1 is the least upper bound 1 2 3 etc are measured in radians remember and so sin 1 sin 2 sin 3 are quite intractable as far as computation is concerned you cannot just compute these values and approximate them very easily that is not possible it is not clear that one is the least upper bound of the sequence but it it is true one is indeed the supremum of the sequence sin n and to prove that rigorously you will need Kronecker's theorem so that is the first application of Kronecker's theorem now let us take a second example suppose you got a continuous function f from r to r and you got two periods lambda and mu what does it mean to say that the function is periodic it means that f of x plus lambda equal to f of x for all x similarly f of x plus mu equal to f of x for all x and this lambda and mu these two numbers lambda and mu are linearly independent over the rationals remember that the real numbers form a vector space over the rationals and these two real numbers mu and lambda are linearly independent over q then you have to show that f is constant this is another application the third application is discussing periodic solutions of differential equations now this is something that we should be teaching to the students who learn differential equations at the sophomore level they got a beautiful forced harmonic oscillator y double prime plus y equal to sin root 2 t what are the frequency of the forcing function root 2 what are the natural frequency of this harmonic oscillator 1 because the complementary function is c 1 sin t plus c 2 cosine t so if you forget the right hand side if you make the right hand side 0 the solution is c 1 cos t plus c 2 sin t but now it is a forced harmonic oscillator and so the general solution of this differential equation is c 1 cos t plus c 2 sin t plus a sin root 2 t where the a has to be computed using the method of undetermined coefficients or the method of variation of parameters use your favorite method you could use Laplace transforms if you like whatever be your favorite method to solve find the particular integral but you need to show that since the frequency of the forcing function namely root 2 is not a rational multiple of the natural frequency 1 the natural frequency and the frequency of the forcing function are linearly independent toward the rationals one of them is not a rational multiple of the other this differential equations will have very few periodic solutions which of these solutions are periodic c 1 cos t plus c 2 sin t plus a sin root 2 t the a you have to find by the method of undetermined coefficients c 1 and c 2 are arbitrary constants so I could choose c 1 to be 0 and c 2 to be 0 and my solution is simply a sin root 2 t that is clearly periodic but if one of the constants c 1 and c 2 is non-zero then the solution is not going to be periodic so in fact I have done exercise 3 for you except for finding the constant a of course you need to show the non-periodicity when c 1 or c 2 is non-zero you may have heard of less serious figures in connection with coupled harmonic oscillators in physics for example you can look at this book of Resnick and Halliday for example suppose you've got a cathodary oscilloscope and in the x direction you input a signal and you input another signal in the y direction if these two signals the frequencies are not rational multiples of each other you will only see a rectangle which is dimly lit whereas if the x input and the y input are linearly dependent over the rationals then you will see some beautiful patterns and these are lesegious figures and some of these lesegious figures are depicted in elementary physics books now you must examine this in connection with chronicers theorem because when you give an x input and you get a y input you are looking at sinusoidal inputs both these inputs are supposed to be sinusoidal and with frequencies which are incommensurate you should discuss this in the light of chronicers theorem there are three exercises for you so now let us go to Weyl's equidistribution theorem Weyl's theorem sharpens chronicers theorem now suppose for example you take two numbers a and b 0 less than a less than b less than 1 we know that the open interval a b contains infinitely many points of the sequence 3.6 let k n be the number of points among the first n that lie in the interval a b what can you say about the ratio of k n by n chronicers theorem only tells you that k n is positive if n is sufficiently large it does not tell you that this k n by n has any kind of limiting value or anything Weyl's theorem says that k n by n actually converges to b minus a what are the meaning of this in heuristic terms you are looking at the number of points in the list 3.7 that land up in the open interval a b and that number you are calling k n so you can think of a game of darts it is like a dart board which is the open interval 0 1 think of the interval 0 1 as a dart board and these points fractional part of alpha fractional part of 2 alpha fractional part of 3 alpha fractional part of 4 alpha they are the darts you have taken a fixed interval a b you have taken a fixed interval a b and each iterate 2 alpha 3 alpha 4 alpha each time you take a multiple of alpha you ask the question whether the fractional part j alpha does it fall in this interval a b or not it is like a dart game game of darts where you throw darts now in n iterations that is in the first n iterates 3.7 how many times do you land up in the open interval a b that is the question and that number is k n so k n by n is the relative frequency with which these numbers enter the interval a b and so if it is a limit of k n by n as n tends to infinity that is going to be the asymptotic relative frequency with which these numbers 3.6 enter a specified interval a b and what the theorem says is that this asymptotic relative frequency is b minus a there is a length of the interval so it means that the asymptotic relative frequency with which these numbers will enter a specific interval i only depends upon the length of the interval it does not depend on the position of the interval whether the interval is on the left or the interval is on the right it does not care it only says that the asymptotic relative frequency with which these numbers 3.6 with that will enter the interval only depends on the size of the interval of course once you prove it for intervals it will be true for all measurable subsets and so you say that these numbers 3.6 are uniformly distributed modulo one so this is a very beautiful and the most basic example of uniform distribution modulo one to prove this uniform distribution modulo one we are going to use fares theorem that the trigonometric polynomials are dense the uniform convergence or the arithmetic means or the partial sums of the Fourier series while the equidistribution theorem itself is a very special case of a very general theorem called the individual ergodic theorem of burkov we shall talk about this in the next capsule but very briefly because that is a very vast area and this is a very beautiful and the very popular example in dynamical systems the theory of major theoretic dynamics discusses these kinds of problems and there are a variety of such problems that arise in dynamical systems and number theory and how Fourier series enables us to understand these kinds of problems and here is a nice illustration of this and I think it's a good idea to stop this capsule here thank you very much