Lecture 51

Welcome back. We were discussing applications of Fourier transform on R and one of very important example of our function which is in the Schwarz space is the Gaussian. What we know about it not only that it is in the Schwarz space. However, if I define f of x equal to e to the power minus pi x square, then we know that this is equal to minus pi xi square. So, in other words, what we are saying is that if I have a Fourier transform as an operator if I look at f of f is equal to f hat, then what we know is that this operator what we have seen takes Schwarz space to Schwarz space and also it takes the moderate decrease function to continuous function which vanishes at infinity C0. what we see is that if we are defining f naught to be this, then f of f naught is equal to f naught as an operator. So, which is in the sense that this is the eigenvector for the operator script f. Natural question is that does that exist any other eigenvector for this operator Fourier transform. So, towards that let us look at one ODE which is called the Hermite equation which is given by y double prime minus of x square of y plus lambda y this is equal to 0 for some constant lambda. And what while dealing with uncertainty principle what we have defined the Hermite operator as minus of f double prime plus x square of f. So, now this is essentially what it is saying that this equation is L y if I do this take it to this is equal to lambda of y. So, now for the solution of this Hermite equation we have defined Hermite polynomial to be h n of x this is equal to minus 1 to the power n e to the power x dn by dx n e to the power minus of x square. So, if you can compute that if you take for any k dk by dx k e to the power minus of x square then this is going to give you some polynomial in x because each derivative first derivative is going to give 2 x then the next derivative is 4 x square kind of thing. So, this is a p of pk of x e to the power minus of x square therefore, if I am hitting with e to the power x square to this derivative then I am getting only the polynomial pn x minus 1 to the power n pn x and that is why this is a polynomial of degree n. One easily one can see that this pk is odd if k is odd and pk is even if k is even. So, here pk is odd means pk as a function of this pk of minus of x is equal to minus of pk of x and pk of this one even means pk of minus of x is equal to pk of x that we can see that. So, we can rather we will write this in this form to be more precise p 2k plus 1 is equal to minus of p 2k plus 1 2k is 2k this is what one can easily observe. Then one defines so, this is a polynomial define Hermite function as Hn of x this is equal to e to the power minus of x square by 2 Hn of x this is called the Hermite function this is the polynomial heated with the e to the power minus x square by 2. Now, one can see that this satisfies if we are defining this so, summation over k is equal to 0 to infinity Hk of x t to the power k by k factorial the Hermite function is going to satisfy this generating equation e to the power minus x square by 2 minus 2 t x plus t square. Now, let us do the sketch of a proof of this important relation. So, this is equal to e to the power minus of x square minus of x square by 2 minus 2 t x plus t square which is equal to e to the power x square by 2 it takes out then we have got this is if I am taking this out then I am left with into e to the power minus of this is x square minus t x t minus of x rho whole square that is what it is going to be okay. So, now let us try to expand this rather this is independent of t so, let us try to expand this quantity. So, now if we look at dk by dxk of e to the power minus t minus of x square this is at t is equal to 0 then this is going to give me pk of minus of x e to the power minus of x square this we have seen that if k is odd then this is odd and k is even then this is even. So, what one can write this is pk e to the power minus of pk x e to the power minus of x square this is the derivative and hence what one if I want to write the Taylor series of this function then what we get is that e to the power minus t minus of x square this is equal to summation over k is equal to 0 to infinity minus 1 to the power k this is minus 1 to the power k minus 1 to the power k and then this is the derivative at 0. So, this is the pk of x e to the power minus of x square multiplied by t to the power k by k factorial. So, this is nothing but d by dx dk by so, this just by plugging in we get that e to the power x square by 2 if I were multiplying it then this is this factor is nothing but k is equal to 0 to infinity minus 1 to the power k dk by dx k of e to the power minus of x square and then this is t to the power k by k factorial. So, now if I am hitting that with e to the power x square by 2 dk by which is nothing but minus 1 to the power k hk of x t to the power k by k factorial there is no minus negative sign here because we are introducing this. So, minus 1 to the power this. So, already this is yeah. So, this 1 is k. So, now this is hk of. So, this is what the generating relation what we have got ok. So, now once we have this thing. So, now let us try to act the Hermite operator on this. So, let us write it down that summation over k from 0 to infinity L of hk t to the power k by k factorial this is equal to I can because this is a nice sum converges uniformly I can take this out. Now this by definition is e to the power minus x square by 2 minus 2 tx plus t square that is what we have got. Now you can apply the operator L which is equal to minus d square by dx square e to the power minus x square by 2 minus 2 tx plus t square plus x square into e to the power minus x square by 2 minus 2 tx plus t square. Now if you do the simple computation what you are going to get this as e to the power minus x square by 2 minus 2 tx plus t square and then this would be 1 minus 4 t square plus 4 tx ok. So, once you have this notice that observe d by dt of e to the power minus x square by 2 minus 2 tx plus t square this is equal to e to the power minus x square by 2 minus 2 tx plus t square into minus of 2 x plus 2 t. So, therefore plugging in this value in this what we have got is that hence summation over k L of hk dk by k factorial this is equal to gather e to the power minus x square by 2 minus 2 tx plus t square plus 2 t into d by dt of e to the power minus x square minus 2 tx plus t square. So, this is by the relation what we have proved this is hk of x t to the power k by k factorial plus 2 t d by dt I can pull it outside dhk x 1 by k factorial ok. So, now this is k times t to the power k minus 1. So, this is equal to k is equal to 0 to infinity hk of x into 2 k plus 1 into t to the power k by k factorial. So, equating the coefficient we get that L of hk this is equal to 2 k plus 1 hk x. So, this shows that this is the solution of y double prime minus of x square of y plus 2 k plus 1 of y this is equal to 0. So, this hk the Hermite functions are the solution of the Hermite equation when lambda is equal to 2 k plus 1 ok. So, that is what the origin of this Hermite functions. Now, what we are interested is to take the Fourier transform of this Hermite functions. Remember that I mean we know how to compute the Fourier transform of e to the power minus pi x square. So, that suggest me to define hk star of x is equal to hk of square root of 2 pi times x. So, which is going to if we write down the relation this is hk star of x t to the power k by k factorial then this is going to be e to the power minus because this is what let us write it again k is equal to 0 to infinity hk of square root of 2 pi of x t to the power k by k factorial. This is equal to e to the power minus x square by 2 that means 2 pi x square by 2. So, which is 2 pi x square by 2 minus 2 t into 2 square root of 2 pi of x plus t square. This one can write it as e to the power minus this is pi x square minus 2 square root of pi times x into square root of 2 t because there is a plus 2 t square minus of t square this e to the power t square goes out. So, what we have left with e to the power minus pi if I take common then this is x square minus 2 x into root 2 by root pi of t plus root 2 by root pi of t whole square. This is equal to e to the power t square into e to the power minus pi x minus square root of 2 by root of pi t whole square. So, now this we can actually compute the Fourier transform of this because we know how to compute the Fourier transform of the Gaussian. So, now here this is the translation of the Gaussian. So, when we take the Fourier transform of a translating thing then it is going to be heated with e to the power minus 2 pi i xi t into f hat of xi. What I am saying is that if g of x is equal to f of x minus of t then g hat of xi this is equal to e to the power minus 2 pi i xi t into f hat of xi this we have seen. So, therefore if I take the Fourier transform of both the side we get again the Fourier transform we can take the Fourier transform inside the sum this is k equal to 0 to infinity h k star hat at xi t to the power k by k factorial. We are taking the Fourier transform in the x variable itself. So, now if we take the Fourier transform of the right hand side this is independent of x will come out of it. Now, e to the power minus pi x square Fourier transform is e to the power minus pi xi square. So, therefore now this is a translation. So, what we are going to get minus 2 pi i xi into square root of 2 by square root of pi t and then e to the power minus pi xi square. This is what it is going to be the Fourier transform of this. Now, we can write it down because we have now the let us put it in this way this is equal to e to the power minus pi xi square and then minus this is 2 times this is square root of 2 pi is coming 2 time of xi into square root of 2 pi and then this is a factor of i then this is t. So, that is what it is and then plus if I am taking the minus then I am going to get over here minus this is minus this I will make it as minus of i then this is e to the power t square then minus t square I need. So, I will take minus of i ti is there i square. So, now this I know that how to write it down this already we have seen that this is going to be summation over k from 0 to infinity hk star of xi and then this series is nothing but minus of i t to the power k by k factorial which is equal to summation over k from 0 to infinity minus i to the power k hk star of xi t to the power k by k factorial. Now, if we equate the coefficient of in this power series what we are going to get that this will imply that hk star at xi this is equal to minus i to the power k of hk hk star at xi. So, this essentially this in what we have notice is minus therefore, this is this Hermite function the dilation of the Hermite function they are going to give us the eigenvector for the operator Fourier transform. So, one knows that I mean from the linear algebra another thing if we have eigenvector then one can span get eigen basis if we have a real symmetric matrix and other thing. Now, here some function space all those members of this function space can be expressed through these eigenfunctions which are behaving very nicely vis-a-vis the Fourier transform. So, it is a very powerful tool and fruitful tool ok. So, we will look into some other application. So, maybe in the next lecture. So, thank you.