 So, we have looked at convolutions in some detail so far in this course and in particular we have looked at discrete time convolutions. So, convolutions where the independent variable is a discrete variable. So, here is another way to look at discrete convolutions. I mean let us define two sequences let us say a of n and b of n where n goes from where n is takes integral values or let us say whole number values just to keep things simple n takes these values. So, to further simplify it let us assume that a of n and b of n are both nonzero up to some point after which they are 0. So, n does not go up to infinity, but it goes up to some finite value and after that the both of these sequences are 0. So, we need to find the convolution of these two sequences. So, let us define these polynomials let us define p 1 of x equal to a naught a of 0 plus a 1 x plus a 2 times x square and so on and let us define p 2 of x as b 0 plus plus. So, now let us look at what happens when you multiply these two polynomials. So, a thing to note here is that a subtle point here is that p 1 of x even though the series that we are concerned with our a n and b n p 1 of x is not really a function of n. So, the independent variable here is if you really want to look at it is x. So, these are just some functions that we have defined just as a tool to simplify this convolution. So, x is not really the variable that we are concerned about, but it is the independent variable in these two definitions. So, now let us multiply these two polynomials let q x equal to p 1 x into p 2 x. So, this is a simple multiplication of two functions. So, this is not a convolution, but let us see what happens when you do it. So, this we can expand it as now let us look at the coefficient of the term let us say x raise to 5. So, we are looking at the coefficient of x raise to 5 term. So, what are the different ways to get x raise to 5? So, to get x raise to 5 you can take the constant term from here and the x raise to 5 term from here or you can take the linear term from here and the x raise to 4 x to the power 4th term here. So, there are 6 possible combinations in which you can end up with x raise to 5. So, let us look at the c. So, what is that going to look like? So, let us call the coefficient c of 5 times x raise to 5 and that is going to look like this. So, this is a of 0 times b of 5 into x raise to 5 plus a of 1 times x times b of 4 times x raise to 4 plus a of 2 x square times b of 3 x cube plus a of 3 x cube times b of 2 x square plus these are the only ways in which you can end up with x raise to 5. If you write it in a more readable format. So, these are these are 6 terms. So, these can be represented by this summation like this. So, you have i equal to 0 to 5. So, i is essentially the parameter passed to a. So, I can expand I can write the same thing as a of i times b of 5 minus i and the whole thing multiplied by x raise to 5. So, essentially what we have done is. So, we have defined c of 5 to be the summation going from i equal to 0 to 5 a of i into b of 5 minus i. So, in general if you look at the coefficient of x raise to n it looks like something like this. So, if you if you look at this expression carefully you will realize that this is precisely the convolution of the two sequences a n and b n. So, if you are worried about this i the index going from 0 to n and not minus infinity to infinity you can check that if this i is negative a i which we defined only for positive terms. You can redefine it to be 0 for when n is negative and if you do if you do that you will you will see that this is just the convolution and the other terms the terms which are not here they are just ignored. Very good yes that is indeed very correct. In fact, you know you will now you will realize that is a very good point that Siddhant has brought out because what he shown is that what we do so frequently in high school math multiplying two polynomials is essentially an instance of discrete convolution. This is a very beautiful observation I am sure all of you must have enjoyed this. In fact, you know take it further when you multiply two numbers written to a certain base you see for example suppose I wrote two numbers say 461.52 let us say and 37.6 what is this theory this is 4 into 10 squared plus 6 into 10 plus 1 into 10 to the power 1 here 10 to the power 0 plus 5 into 10 raised to the minus 1 plus 2 10 raised to the power minus 2 and this is 3 10 raised to the 1 and so on. And now when you are multiplying these this is an instance of the polynomial multiplication which Siddhant talked about. So, in effect you know now to get the so-called coefficients of the different powers of 10 you would actually be carrying out a convolution of these digits the only problem is now in addition to convolution you would then have to carry over you know so this is an instance where you multiply two numbers to a given base you first need to do a convolution of the digits and then you need to carry over because some of those products would give you a quantity more than 10 or more than 10 and then you need to carry you know so that is a very interesting observation. In fact Siddhant has in some sense already given you an exposure to what we are going to do in module 4 there we are going to talk about the Z transform and the Z transform we are actually you will see now remember you know if some of you do module 4 if you do the second course you should try and keep this observation of Siddhant in mind and try and relate it to what we are doing in the Z transform there. So, it is very interesting, but for now even the observation in its own right that polynomial multiplication is essentially an instance of convolution is very useful to understand convolution and we must definitely appreciate Siddhant for having done a very good job in bringing this up. Thank you Siddhant. Are there any other points that you want to discuss now? So, for those of you who are interested in computer science. So, there is an algorithm which can do convolution in less time than it takes to actually find evaluate these points. So, there are algorithms for so if you want to multiply it to n two sequences of length n. So, there are algorithms that can do it in time n log n instead of the usual way which would take n square. So, this is really interesting and so you would expect that if you have to multiply two numbers each of length n each having n digits you would take n square time right. So, you have to multiply each digit by each digit in one number by every pair of digits. So, you do not actually have to do this. So, there are algorithms to do that and we would not study those algorithms, but we will I mean you will get a flavor of how the algorithm perhaps you know if some of you look at something called the fast Fourier transform it will give you a flavor of this and you know you can do the first of course this is a little beyond the scope of this discussion, but I am just mentioning it for those of you who might have some interest very good. That is a very good observation. So, you could have efficient algorithms for convolution that is a well studied theme very good. So, I think we will conclude the discussion here. We look forward to your reactions on this the whole idea of these discussions is to evoke reactions from you as well. Thank you so much.