 Another mathematical concept that we will recapitulate briefly is the concept of Fourier transforms. Fourier transforms is a very powerful tool in signal processing in image processing in solving differential equations and in many other areas of science and technology. For us it is a essentially required in arriving at some important results. So, we study it in a simple way. First we know we have the concept of Fourier series Fourier series. So, Fourier series is a form of expressing a periodic function in terms of some basis functions and these basis functions are generally sin and cosine functions. So, a function is said to be periodic let us say for f x is a function it is said to be periodic with a period L if f x is equal to f x plus L. For example, if f x has a behavior like this etcetera then this point and this points will define one period. The functions properties should be such that that function is integrable in some sense and in such situations we have the Fourier series discovered by Fourier in the problem of solving second order differential equations where he showed that the periodic function f x can be expanded as a linear superposition of cosine functions cos 2 n 2 n pi x divided by L plus sin functions plus of course, one constant. In a nutshell this is what Fourier expansion is the expansions assumes a significance only when we find a way how to calculate the coefficients of a n and b n and a naught and that is done by using what is known as the orthogonal property of the basis functions orthogonal property of the basis functions which are the basis functions the cos and sin function. In fact, these properties the orthogonal property is expressed in the following form if you integrate minus L by 2 to L by 2 cos 2 n pi x by L with cos 2 m pi x by L. This function is half delta where delta n m is called chronicle delta. It has a property 0 if n not equal to m equal to 1 if a similar integral also exists for the sin functions and most importantly the functions cosine and sin are orthogonal to each other all the time for cos 2 n pi x by L into sin 2 n pi x by L when integral integrate over the period one period total length of the period is L minus L by 12 by 2 this will be 0 for all values of n m and sure you heard all these properties. So, we move further from this idea. So, from the concept of Fourier transforms using cosine and sin functions we can further move to expressing it using exponential function use Euler's formula. For example, what does Euler's formula say e to the power i theta equal to cos theta plus i sin theta. It is a very far reaching result which connects an exponential function to the circular functions cos x and sin x. So, correspondingly we can write cos x equal to e to the power i x plus e to the power minus i x by 2 and sin x equal to e to the power i x minus e to the power minus i x by 2 and when we substitute this combine the coefficients a n and b n into some other coefficients a c n etcetera we can approach the same Fourier series via a method of expanding through purely exponential functions with some other coefficients we denote those coefficients as f n this f n is just a coefficient it is not f x it is not a function it is a coefficient index by n and you can write it as 2 pi n i l and similarly the unknown coefficients can be obtained by the inverse of the series this is of course for for all n and the coefficients f n correspondingly turns out to be because of the orthogonality of the exponential functions it will have the value 1 by l minus l by 2 to l by 2 e to the power minus 2 pi n i x by l f x. So, there is an expansion with the coefficients f n and there is an inversion to find those coefficients in terms of the function f x this is made possible by means of the orthogonality property that integral minus l by 2 to l by 2 e to the power 2 pi n i x by l minus 2 pi m i x by l d x will again be equal to Kronecker delta n m. So, when n equal to m only it exists and when equal to m the value is l. So, using this property we can evaluate our all the coefficients f n which is what is indicated by the inverse transform the Fourier's transform is a special situation when the period of the function tends to infinity that is l tends to infinity. My periodic function is essentially now a non periodic function something like this whose period is almost as large as the real line itself. Hence it cannot be actually called it is a periodic function with an infinite period in such a situation the Fourier series can be converted into an integral form because some is a kind of representation of integral. So, in a qualitative way we can understand. So, when the function f x is non periodic the Fourier series can be converted to an integral. The coefficients so called Fourier coefficients can now be denoted by a notation f carrot of k where k is the conjugate variable and that is defined through this integral as f x e to the power i k x dx over the entire real line and correspondingly the function f x can be obtained via inverting it once again in the form because now the period is going to be 2 pi and it will be minus infinity to infinity f dot k e to the power minus i k x dx. So, this conjugate pair basically define the concept of Fourier transform and inverse Fourier transform inverse Fourier transform. There are many useful applications of the Fourier transform one of such application is the concept of characteristic function. We define the characteristic function chi k as the Fourier transform it is actually the Fourier transform of a function f from x to the conjugate variable k space where f has a special connotation it is actually a probability density function. We will come to that later, but right now it represents the probability distribution. So, definitionally it has the same approach as the Fourier transform. So, characteristic function is a Fourier transform of a probability function probability distribution. The special nature of the probability distribution as compared to an arbitrary function is that it has to be positive and it must be normalizable the function must be integrable in the interval. This characteristic function the chi k is called the characteristic function it has certain advantages and applications. We can actually evaluate some characteristic functions to get a feeling of what it means. For example, the Fourier transform of let us consider a probability distribution very often considers in probability theory a Gaussian function. For example, let f x be 1 by sigma root 2 pi e to the power x minus x naught whole square divided by 2 sigma square. And this function looks like this it has a maximum at x equal to x naught. So, if we and this is of this function is known as the Gaussian. So, we can obtain its characteristic function chi k by Fourier transform by definition it is a integral e to the power i k x and the function itself 1 by sigma root 2 pi e to the power minus x minus x naught whole square by 2 sigma square dx. We can actually write it all in terms of exponential only by taking e to the power i k x by combining e to the power i k x with e to the power minus x minus x naught whole square by 2 sigma square that is I can write like this. I can make it a perfect square it can be made into a function some coefficient some constant a and then b x minus some c whole square. You are always right like this this is a c will involve i k and a b they will all involve sigma square also this can be easily done. We will skip the details of derivation and when we do that and then use the standard Gaussian integral result we can easily show that the characteristic function of a Gaussian distribution is also a Gaussian distribution that is the main result it has a form e to the power i x naught k minus k square sigma square by 2. If we go back to the original distribution and compare original distribution f x it had a its variance width was measured by the standard deviation sigma. If you compare its characteristic function it is also a Gaussian but its measured here of the width the measure of the width is 1 by sigma. So, there is some reciprocity because now sigma square is in the numerator it multiplies k square it does not divide. So, which means that the standard deviation is 1 by sigma. So, this is an important important result that follows by the method of characteristic functions. We can obtain lot of information about the Gaussian distribution via the characteristic function. One of the important applications of characteristic function and therefore of Fourier transforms is in obtaining moments of the moments of the distribution. See as we go along we we will be formulating either a difference equations or a differential equations for various probability distributions. Sometimes those equations will give you the characteristic function directly and not the distribution. So, to obtain the characteristic function to obtain the distribution from the characteristic function one has to do the inversion. Some of those inversions are numerically very difficult to do. In such situations one is then interested to know at least the various measures such as the mean of the distribution, the standard deviation, the skewness of the distribution etcetera from the various moments of the unknown probability distribution. The knowledge of characteristic function enables us to obtain these moments and we show that here. Supposing we define the kth moment of a probability distribution n k as an integral over the function f x and x to the power k dx and supposing this moment exists for all integer values 0, 1, 2, 3 etcetera. Then it is possible to obtain various metrics associated with the distribution function f x. For example, what is m 0? If you put k equal to 0 we get m 0 equal to minus infinity to infinity f x dx and by normalization this will be 1. If probability density is normalized m naught will be 1. If you put k equal to 1 the moment m 1 it is going to be minus infinity to infinity x f x dx. And we know that when we weight the distribution function with the variable x itself and integrate over the space of x we get the mean value of x. We can call it as x bar. Several notations use in some sometimes we may use this notation and in statistics often the notation mu is used for the mean. So, m 1 is the mean itself. Similarly, if you put k equal to 2 we will get moment m 2 which is the integral of x square with respect to f x dx and from the m 2 we can call it as x square bar. We can define the variance as x 2 square bar minus x bar square which is m 2 minus m 1 square. One can proceed further to obtain higher moments and get skewness etcetera. We now show that these moments can be easily obtained even without knowing f x by just the knowledge of the characteristic function. We proceed as follows. From the definition chi of k as a Fourier transform of the distribution function f x. If you set k equal to 0 we see that chi 0 will be an integral of f x dx which will be 1 if f x is a normalized function. Now let us see what happens when we differentiate chi with respect to k. When we differentiate you can we assume that we can take the differential operator inside the integral then it will have the form when I differentiate with respect to k e to the power i k x will become e to the power i x e to the power i k x there is no k dependence in f x. So, it remains as such. So, chi prime k will have this form. Now if you set chi prime k at k equal to 0 which is simply called chi prime 0 that will be i can be taken out 0 to infinity x f x dx and that is by definition i into x bar or mu. So, we can find the mu the mean of the distribution by dividing by i. So, this will be minus of i chi prime 0 this is a very important result. Similarly, we can carry out a higher order differentiation of the characteristic function. So, if we do similarly chi double prime 0 you will get an i square x square i square is minus 1. So, you will get it as minus of m 2. So, the variance of the distribution sigma square as I mentioned is m 2 minus m 1 square and if you go to equation 1 here we have seen that the mean is equal to minus i chi prime 0 and if you substitute all that we are going to get sigma square is equal to chi prime 0 whole square minus chi double prime 0. So, since the differential process is easy to execute other than integration processes very often we find it convenient to solve for characteristic function establish moments and in fact, one can go beyond sigma square also you can obtain the skewness of the distribution and some parameters of this distribution give us an idea of what kind of a probability distribution it is going to be. So, this is one very very useful example of characteristic function just to illustrate the usefulness let us give an example. Supposing in the course of studying some problem we have formulated a differential equation it was possible to solve this differential equation in the Fourier conjugate space by taking Fourier transform where it became an algebraic equation and then we obtained a solution which is equivalent to getting a characteristic function of that and let us say we obtained a characteristic function let us say of this form chi k equal to 1 minus some i b k to the power minus n minus 1 where let us say n was integer could be 1 2 etcetera and b a positive constant. We can now perform as before apparently it is not easy to invert it and obtain the distribution although it is possible, but more difficult whereas the differentiating it is very easy and we can see that chi 0 is 1 it automatically satisfies the normalization condition. Then if you do the first differentiation and then put k equal to 0 we will show that it gives you a mean value mu for the distribution b into n plus 1. When you differentiate this n plus 1 will come outside with n minus 2 and then you put k equal to 0 it will be just 1 and you are going to get this. Similarly we can show by the second differentiation that the second moment m 2 is going to be minus chi double prime 0 and it will have b square n plus 1 into n plus 2. So, when you second differentiate it will be n plus 1 into n plus 2. So, when you combine these two we get the expression for the variance sigma square equal to m 2 minus m 1 square going to be can show that it will be b square into n plus 1. So, without actually inverting we have got some important characteristics of the resulting distribution says that the mean is going to be b into n plus 1 and the variance will be b square into n plus 1. So, mean and the variance has similar proportionality with respect to the index. There is a very important representation of a function which we are going to soon see via the Fourier integral formulation and that is called the Dirac delta function. We study more about the delta function in the subsequent classes. Thank you for listening.