 Now, let us continue with the Polier's theorem regarding the problem of return to the origin. So, we introduced two different concepts one of the return probability f n which is first time return probability of first time return to origin in n steps. And the second one of course, was the occupancy probabilities with which we have been dealing throughout this study. So, occupancy probability specifically at the origin is occupancy at the nth step occupancy probability at the nth step. And all these we introduced by generalizing random walk to more than one dimension. In fact, we considered the case of a d dimensional system d equal to 1 to 3 are our real Euclidean dimensions higher dimensions are introduced here for pursuing the result or consequence of a theoretical analysis. So, with this we related the two why are they are generating functions. So, we create we develop the two generating functions one is phi of z with respect to f n that is phi of z as defined as sigma f n z to the power n as a step sum generating function the step sum not the site sum. Similarly, we defined another one called phi naught z a generating function which was n equal to 0 to infinity w n 0 z to the power n. And then we connected these two related via the relationship phi z equal to 1 minus 1 by phi naught z. So, if you know the generating function of the occupancy probability at the origin one can in principle find out this generating function for the one can of course, find out the generating function for the first time visit probability. And if possible to invert one can find out the generating function one can find out the probabilities themselves. All that we are doing because our aim or aim of this study is to determine the value of determine R the ultimate probability of return return return to the origin where R was sigma f k where k equal to 1 to in n steps. So, in infinite steps in the final analysis. So, this is the way we postulated the problem. So, what is the final probability of return function is our end point. So, that led us to simple relationship finally, for R which said since R therefore, by definition since R turns out to be just phi of unity because I mean z equal to 1 the generating function is the same as the definition of the ultimate return probability R. Hence, we obtained the expression this equal to 1 minus 1 by phi naught the generating function of the occupancy probability at the origin evaluated for z equal to 1. So, this was our aim. So, now it remains to evaluate phi naught z and then phi naught 1 hence it remains to evaluate phi naught 1 let us say phi naught z. So, here we from its definition phi naught z which is basically w n 0 z to the power n. So, this now necessitates that we formally evaluate w n 0 for that we now release it to the old result where we had obtained a solution that the Fourier transform of the function at 0 that is Fourier transform of the occupancy probability was already derived as 1 by d as if you look at it had many components k 1 to d dimensional space that was defined as 1 by d sigma cos k i i equal to 1 to n all whole to the power n and from the inversion logic we had found out that w n 0 then will be corresponding to 1 by 2 pi to the power d the d fold integral minus pi to pi minus pi to pi with respect to all k variables evaluated at e to the power i k dot r, but at the origin r is 0. So, it will be just unity and we called it y to the power n where we define y was defined as 1 by d sigma cos k i i equal to 1 to d not n 1 to d that would be d distinct parameters. So, with this we can easily. So, sigma w n 0 z to the power n that will become 1 by 2 pi to the power d we will take the sum inside this integrals it will be d k 1 d k 2 etcetera d k d sigma here it will be n equal to 0 to infinity step sum. So, n equal to 0 to infinity z to the power n y to the power n. So, this basically finally implies hence phi z not phi it is we have defined it as pi naught z pi naught z is going to be 1 by 2 pi to the power d minus pi to pi minus pi to pi d k 1 up to d k d y z to the power n it will be sum inside this is how it is going to be n equal to 0 to infinity y z to the power n. So, this is a geometric series whose sum we know since sigma n equal to 0 to infinity y z to the power n is going to be y z and since well we can write it down is pi naught z will become 1 by 2 pi to the power d minus pi to pi minus pi to pi d k 1 d k d into it will just be 1 by 1 minus z y reverting back to y basically k i since y equal to 1 by d sigma k i cos k i i equal to 1 to d we will have pi naught z 1 by 2 pi to the power d minus pi to pi minus pi to pi d k 1 d k d 1 divided by 1 minus z into 1 by d sigma cos k i sigma there will be d terms this is expression we can just know. So, it is basically now the generating function for the occupancy probability at the origin is expressed in terms of an integral a d dimensional integral of various cosine of the conjugate vectors or conjugate elements with respect to each of the coordinates in the d dimensional space. So, one has to basically perform an integration. So, now the problem is reduced to performing it somewhat complicated looking integration and the quantity of interest we must recall that it is a pi naught 1 is the quantity of interest that is that pi naught 1 therefore is going to be same 1 by 2 pi to the power d same integrals d k d fold integrals, but here now z is going to be 1. So, it is just 1 by d i equal to 1 to d cos k i the entire problem of return rests on the value of this integral one is going to get. So, you can easily see that if you fix a value of d this is going to be just a number. So, to understand the evaluation of this kind of integrals we do it systematically let us start with the case of d equal to 1 then one case we obtain results, but let us understand what does it mean. So, to understand this integral start with d equal to 1 which is 1 d random walk. So, what is the problem now in 1 d random walk? So, if the there is no absorber or anything it is a infinite it is a entire real line with lattice points discrete lattice points. So, random walker starts from the origin and we are asking a question what is the probability of ultimately him returning to the origin does he or does he totally escape? Of course, we have found answers in various ways to this problem, but here specifically we see this result in the in respect of the formalism we developed and in the light of the results that we are going to get in general d dimensions. So, it will add to a new perspective. So, if you put d equal to 1 this formalism tells that pi naught 1 will be now it is only 1 integral. So, d equal to 1 means it will be 1 by 2 pi minus pi 2 pi it will be just the d k 1 only 1 variable will be there and here it is going to be d is 1 and i equal to 1 2 d will be only 1 term. So, it will be cos k 1. So, we are now looking at a function called 1 minus cos k 1 1 by 1 minus cos k 1 integrated over a domain which includes 0 the domain is from minus pi to pi. So, the integrand the domain of integration minus pi to pi and integrand 1 by 1 minus cos k 1. So, what happens to 1 by 1 minus cos k 1 in this domain? So, to understand that we have to see that the domain includes 0. At k 1 equal to 0 cos k 1 becomes a unity. So, 1 minus cos k becomes 0. Hence you have a 0 in the denominator and then the integration integrate becomes doubtful. However, of course having 0 in the denominator does not necessarily make the integral divergent. So, we must first examine the power of this 0 the order of this 0. So, to do that we note that 1 minus cos k 1 we can systematically understand we know cos k 1 for example, it if you plot cos k 1 minus pi to pi if you see pi by 2 minus pi by 2 0. So, cos cosine function has this form it should be symmetric here this is unit cos 0 is 1 cos pi is minus 1. So, this is the way we see this. So, 1 minus cos pi is going to be complementary of this. So, this is cos k 1 as a function of k 1. So, 1 minus cos k 1 will be complementary of that. So, we can plot it wherever 1 is there it will become 0 and at pi by 2 it was 0. So, it will be 1 and here it is minus 1. So, it will be 1 minus of minus 1 it will be 2. So, it will have some value like this. So, at pi by 2 it will be 1 become 0 here and then again go up like this symmetric function. So, this is at k 1 equal to 0 point. So, it will have a point. So, if you now consider 1 by 1 minus cos k 1 that will have a behavior wherever 0 it will blow up and elsewhere it will be finite. So, there is a singularity. So, to know that singularity. So, this is singular k 1 equal to 0 to know the degree of or to know the order of singularity we use Taylor expansion cos k 1 that is cos k 1 equal to 1 minus k 1 square by 2 plus of the order of k 1 to the power 4. So, very close to k 1 equal to 0 the 2 terms is sufficient. Hence, 1 minus cos k 1 is going to be of the order of k 1 square by 2 thus pi 1 pi naught 1 should be of the order of 1 by 2 pi minus pi 2 pi d k 1 divided by 1 minus cos k 1 and k 1 square this value 1 minus cos k 1 as being k 1 square by 2 that value shows that singularity. So, we will write it in a close to a very close approximation d k 1 by k 1 square by 2. So, 2 I will take up. Now, as you can see this integral will diverge because the order of singularity is quadratic at the denominator and its integral is going to at it has to pass through the origin. So, the function does not exist this integral does not exist this is it is dangerous to integrate this function brute force like then you will get minus 1 by k 1 then put the values pi and minus pi you will get a nice finite answer. Although you will get a negative answer of a positive function that is a different thing, but it would give you a kind of a confusing result that there is some finite value for this integral. But the point is you are not supposed to perform this integration knowingly that there is a singularity at k 1 equal to 0 because this integral is divergent. Hence pi naught 1 will diverge hence pi naught 1 equal to infinity because it is basically an integral of a quantity which has a strong singularity at the origin and the value will be infinity. If you do it by parts minus pi to 0 and 0 to pi the 2 will add up and it will give you a singular behavior. So, as a result of this we will have R in 1 d the ultimate return probability is 1 minus 1 by pi naught 1 which is 1 minus 1 by infinity which is 1 minus 0 which is 0. So, what do we make out of this? This formalism now says that the probability of return of a 1 dimensional random walker to the origin from where he started is a certain. So, at some point in time. So, 1 d walker definitely re walker definitely returns to the origin. So, the walker starts on the origin let us say and somebody has put a trap after he left hoping that sometime he will come and he will be trapped definitely he will be trapped there is no escape for him. So, long as he executes symmetric unbiased random walk. So, that is the sort of take away from this study. In fact, if the intuitively the result can be extended to revisitation of any site even though he starts from origin and the you can ask the question what is the probability that he will visit a site? The probability is unity a 1 d random walker will visit every site at least once. In fact, if he visits once you can start the process and again make sure that he repeats again. So, every site therefore, gets visited infinitely many times in infinite time. So, that is the message in this analysis. So, this means basically this walk is recurrent that is walker visits many times over walker revisits the site infinitely many times. Now, let us come to 2 dimensional case. So, case 2 d equal to 2. So, in 2 dimensions the result specifically becomes pi naught 1 will superscript it as 2 d that is going to be now there will be 2 integrals 1 by 2 pi square minus pi to pi minus pi to pi d k 1 d k 2 of 1 by 1 minus 1 by d becomes half cos k 1 plus cos k 2 it is a bit more difficult integral. Now, we can note the following if you see our integration domain this can be treated as x and y coordinates and each of them is minus pi to pi. So, it is basically integration over a square that is we have a domain the integration domain is a square let us do it here we have a square. So, it varies from minus pi to pi along the x. So, this is origin here and here also this is pi and let us say minus pi along y. So, one is carrying out integration over this square a b c d let us say a b c d integration domain the square a b c d of side length 2 pi the function a positive function when we are integrating over the square the value that you will get in a circle I am I am I am I am constructing now a circle this is supposed to be a circle this circle will in the asymptotically will represent in some way the integration over the square. So, in other words the integration over the square the value will be always more than that we will get in the circle since my integrand is always positive cos k 1 plus cos k 2 sum will be always less than 2. So, 1 minus half of that is going to be always a positive quantity this much we know. Therefore, the true integration on the square hence integration in this integration over square will be greater than that over circle. So, if I get some value for that on the circle x we know that the true integral is more than x that is what the point I want to say. Now, integration over the circle can be performed fairly easily by transforming integration over circle may be carried out may be done or may be performed with r theta coordinates polar coordinates. So, that small transformation helps us to see the symmetry in the problem and leads us to obtain some tangible results for otherwise a fairly complicated integral. So, we do it in our next lecture. Thank you.