Lecture 44

Welcome back, in the last lecture we were discussing about the Poisson summation formula. The underlining feature of that we were periodizing the function defined on whole R and treating that periodization function and doing the Fourier series as a one periodic function and we are able to get certain information about the function knowing only at the values at the integers. So, in that philosophy there will be lot of problem which one will encounter in various other engineering and science, more specifically in the information theory where we are getting into the problem of recovering the whole function from the samples. So, what do I mean by that? Now suppose I have a function like this and only I know the values of the function only at the integer points. So, these are all the integer on the discrete points and can we recover the function f. us assume let f has moderate decay and f hat is compactly supported and support of f hat is contained in minus half to half. That means if I have minus half to half this is on the frequency side f is 0 outside this. It is a continuous function and it is 0 outside. So, now we can actually think this f hat as a one periodic function or I will take suppose f hat is something like this. Then I will consider f hat as a one periodic with getting the periodic extension so on and so forth and so now for this f hat of xi this is equal to f hat of xi and I can hit the indicator function of minus half to half of xi. This is not going to change the value of f hat because outside this because f hat is 0 outside minus half to half. So, and on xi is in minus half to half chi of minus half to half xi is 1. So, now we are we said that we are treating f hat as a one periodic function. Therefore, so now what we can get is that f hat of xi this is equal to summation n varies over z f hat, hat at n e to the power 2 pi i n xi. Now this what we had observed is that this is going to be f of minus of n if we are taking a nice function and e to the power 2 pi i n xi. So, this is by making a change of minus n goes to n this is f of n e to the power minus 2 pi i n xi. So, therefore this is nothing but indicator function of minus one half to half xi summation over n varies over z f of n e to the power minus 2 pi i n xi. So, therefore so this is this is summation n varies over z and f of n xi. So, therefore what we get is that if f hat is this then from this what we get is that f of x this is equal to integral minus half to half f hat of xi e to the power 2 pi i xi x d xi So, therefore so then we plug in the value this is equal to summation over n varies over z f of n will come out of the integral f hat has support at. So, this is and e to the power minus 2 pi i n xi into e to the power 2 pi i x xi d xi. So, this of course we know the we have computed this integral several times this is equal to sin pi of x minus of n by pi of into x minus of n. So, this is our f of x now as you can see that we are recovering f which does not have a compact support but which has a nice decay from knowing the value only at the integer because we know the value of sin pi x minus of n for every x and every integer n that is known to us. So, now only information of f at the integer we are able to get the entire f reconstruct the entire f on the real line that is quite powerful and has been used profusely in information theory signal processing and other. So, now we will also see another nice application of towards finding an infinite sum by using Poisson's summation formula. So, what we would like to prove is so summation over n varies over z 1 by n plus alpha square this is equal to pi square by sin square by alpha for alpha belongs to R minus z for all nonzero alpha this we can compute this infinite sum remember this is nothing but when we are saying this is n equal to minus infinity to infinity. So, let us see how to prove this consider g of x is equal to 1 minus mod x if mod x is lesser equal to 1 and 0 otherwise. So, this is the function and 0 otherwise we have computed the Fourier transform of this then you take g hat of xi this we know that this is sin square pi xi by pi square xi square now let me call this as h of xi when xi is not equal to 0 this is what we have seen at calculated several times. So, by Poisson's summation formula for this h what we have n varies over z h of xi plus n this is equal to summation over n varies over z h hat of n e to the power 2 pi i xi n. So, this is equal to h hat of n h hat of n is what if h is this then this is the Fourier transform is going to be h hat is going to look nothing but g of minus of x h hat of x is equal to g of minus of x. So, that means what are the integer for which this h hat a g of n is going to survive only when n is equal to 0. So, this is nothing but h hat of 0 and that is not equal to 1. So, this implies summation n varies over z sin square pi n plus alpha divided by pi square n plus alpha square this is equal to 1. So, now this sin square pi n plus pi alpha is nothing but minus 1 to the power sin of pi n plus alpha this is equal to minus 1 to the power n sin of pi alpha. Now if I take the square then is n varies over z this is sin square pi alpha divided by pi square to n plus alpha square which is equal to 1. This implies that summation n varies over z 1 over n plus alpha square this is equal to pi square by sin square pi alpha. So, this is what we know that summation over 1 over n, n equal to 1 to infinity this diverges and now for alpha not an integer what can we say about suppose minus infinity to infinity 1 by n plus alpha. You see I am not we are not talking about n equal to 1 to infinity 1 over n plus alpha. So, now this alpha is no more it is not an integer. So, for this I can write can we find the value. So, this is minus infinity to minus 1 1 by n plus alpha plus n equal to 0 is 1 by alpha plus n equal to 1 to infinity 1 by n plus alpha. So, this I can we can write plus summation over n from 1 to infinity 1 by n plus plus 1 by minus n plus alpha. Now let us try to find what is n plus alpha as we can as one can see that let us rewrite it again what we have got that 1 over alpha plus summation over n is equal to 1 to infinity 1 over minus n plus alpha plus 1 over n plus alpha this what we have got is 1 over alpha minus integral 0 to alpha pi square by sin square pi x minus 1 by x square dx this is what we have got. So, one can easily see that although it looks like that this function under the integral this has a singularity but as we know that if it goes to if x goes to 0 this tends to a finite limit. So, the singularity is not there. So, now this is equal to 1 over alpha minus this is going to be minus pi by tan pi x and then this is plus 1 over x this is 0 to alpha. we know that the limit x goes to 0 minus pi by tan pi x plus 1 by x this is equal to 0. So, this one becomes 1 over alpha plus pi by tan pi alpha minus 1 over alpha this is equal to pi by tan pi alpha. So, we can and this is what we can prove it thus summation over n from minus infinity to infinity 1 by n plus alpha this is nothing but pi by tan pi alpha. So, this for 0 less than alpha less than 1. Now if alpha is not an integer can be it is alpha is either greater than 1 or alpha is less than 0. So, now we have only proved that this is between 0 to 1 alpha lies over here what about the alpha not integer and lying here. So, as we know that this alpha I can write this as bracket of alpha plus alpha 1 where 0 is less than alpha 1 is less than 1. So, therefore if we are writing the left hand side n equal to minus infinity to infinity 1 by n plus alpha this is going to be minus infinity to infinity 1 by n plus bracket alpha plus alpha 1 this sum with the substitution is again going to be minus infinity to infinity 1 by n plus alpha 1. Now we have the result for alpha 1 because it is less than 1. So, we can write this, this is pi divided by tan pi of alpha 1. So, and very interestingly if you look at if alpha is equal to one half then there is a symmetry. So, hence what you are going to get summation minus infinity to infinity 1 plus n plus half this is the symmetry which is going to give you 0. So, this periodization technique what we have used to get 1 by n plus alpha square and then by using that we have got 1 by n plus alpha. So, there are lot of lot more application of this periodization techniques in mathematics and in other science. So, now let us see one of the nice property of the continuous function what we know that is essentially Weierstrass approximation theorem. It says that if f is a continuous function on a, b closed interval a, b then there exists the sequence of polynomial which converges to f uniformly. So, set of polynomial is dense in c, a, b that is what is the statement. So, we will see another proof not by using the Bernstein proof we will see another proof of this result in the next lecture. Thank you.