 We now proceed to prove the Polius theorem that is theorem on the probability of return to the origin as a function of the dimensionality of the space. So, in a nutshell there is an origin from which the random walker has started and he undergoes a random walk and we are asking a question what is the probability that he will return to the origin after some long number of steps. So, for that first we postulate a quantity called fn which is the probability of return. Now, when you say return it always means return to the origin the starting point for the first time in n steps. This allows us to define a ultimate probability of return. We also define another quantity called r ultimate probability of return. So, fn is first time probability of return in the first step and r is the sorry fn is the probability of return for the first time in nth step or in n steps and r is the probability of return ultimately. So, obviously the ultimate probability of return should have been a sum of various probabilities of returns at the finite steps. So, his ultimate probability of return could have happened say after hypothetically speaking at the first step or a first time in the second step or for the first time in the third step or for the first time in the fourth step and so on. So, in other words since the ultimate return could happen for the first time in any of the n steps for the first time any of the steps. We could write r therefore, equal to n equal to 1 to infinity fn. We may note that f0 is 0 since at the 0th step he was at the origin already and it is not a question of return it is a question of occupancy. The return occurs in the subsequent steps. If waiting times are allowed then even the first step could perhaps be called a return event, but right now we are not allowing for any waiting times neither there is a bias. So, we are talking of symmetric random walk and in such a case it is always the second the fourth the sixth etcetera the even steps only contribute to the ultimate probability of return. So, our aim to evaluate r as a function of dimensionality of the walker's space. Now, let us see what is how do we relate the occupancy probability to the return probabilities defined by the notation fn. If for example, at the nth step if we say that the occupancy probability is wn0 of being at the origin that could have happened because first of all at the 0th step he would be at the origin. So, if n equal to 0 is included then that is one way he could be at the origin. He could have been at origin initially or he could have been origin at some other n minus kth step and again return for the first time in the next k steps that also would contribute to the probability of finding him at the nth step. So, probability of finding him at the nth step is a superposition of the various occupancy probabilities at the previous steps waited with the probability of coming for the first time at the nth step that is mathematically we say that wn being at 0 is delta n0 at the 0th step of course, he was there with the probability unity plus the probability k equal to 1 to n wn minus k0 fk. So, stated in words it means the probability of finding at nth step equal to either walker was walker started either walker started that is n equal to 0 from origin he was walker was at origin at some other n minus kth step returned for the first time x k step return to the origin first time in the next k steps for all possible case. So, it is a basically a series of our statements he was in n minus 1th step he was at the origin and returned in the very first step normally we are talking of evenness. So, it n minus 2th step he was there and returned back in 2 steps or n minus 4th step he was there and again returned back all that will contribute is being found there at the nth step. So, this is a as many words we can say explain this mathematical equation. We will start numbering this equation. So, r equation we call as 1 and this recurrence or chain relationship. So, this is a chain relationship between. So, equation 2 is a chain relationship that links the current occupancy to the past occupancies the concept of that is first time occupancy probabilities. So, to that extent we have now linked the 2 concepts how exactly they come into picture in a random walk problem. To proceed further we define now we define a generating function for the occupancy probabilities for W n 0. This is a basically a step sum type generating function we can denote it as just to distinguish I would like to call it as a step term sum g f in instead of side sum g f which we discuss very often. So, step sum g f basically means the generating function we denote it as pi 0 z is a sum over the steps n 0 to infinity of W n 0 z to the power n. So, for example, for 1 d W n 0 is n c n by 2 2 to the power divided by 2 to the power n etcetera those formula we had seen and this is basically a sum multiply that by z to the power n and sum over all n. So, that should give you a generating function for the occupancy probability. So, again we to strengthen our connections or memory we say that pi 0 is the generating function with respect to W n 0. So, we say pi 0 z is the g f for W n 0 for the occupancy probability. Similarly, we can define a g f for the return probability f k through the notation we call it as phi z equal to k equal to w n 0. So, we define it as f k z to the power k. So, in other words phi z is g f for f k. So, call them as equation 3 and equation 4. Now, let us examine the equation 3 in the light of the chain equation 2. So, we can write explicitly pi 0 z equal to we have to multiply by z to the power n. So, it is going to be delta n 0 plus k equal to 1 to n W n minus k 0 f k z to the power n for all n equal to 0 to infinity. Step 0 is included because that that was that time when he was definitely at the origin which we have accounted for via delta n 0. It is a very important non trivial contribution. So, the very first term gives you 1 because if n equal to 0 z to the power n is also 1. So, first term is 1 then it becomes a double sum n equal to 0 to infinity and k equal to 1 to infinity of W n minus k f k well n equal to n minus k 0 the origin f k z to the power n. Let us interchange these terms. Let us interchange the sums this is k equal to 1 to n not infinity because there is no return after n the step when the n step process. So, since the sum is the interchanging does require us to take care of the fact that the inner term depends on the outer term. So, basically as we did in earlier the inner term varies from 1 to n. So, if you draw a k equal to n line first we supposing k is this axis and n is this axis the procedure as such of summing this is you vary k from 1 to n that is k varies horizontally you hold you fix a value of n for each n you vary k from 1 to n and then you vary n of course. So, like this you do the summing that is how we would have done fix the value of n for each n run k from 1 to n horizontally and then go on varying n. So, you go line by line upwards from the horizontal summing. Now the same area can be covered that is the upper half of this k equal to n line by going vertically in which case I will fix a value of k. So, some value of k I fix then vary n then you see that it has to vary from k equal to infinity like this. So, we can also do it vertically in which case I do this n summing first now, but k will be outside you hold the value of k and some n from k to infinity. With this we can rewrite the sum in the following form it will be pi naught z pi naught z is the generating function for the occupancy probability that will be 1 plus k equal to now 1 to infinity, but n now will be summed from k to infinity and it will be w this quantities are equal just same f k z to the power n. I can regularize this sum now by redefining a new integer space let k prime equal to n minus let n prime let n prime equal to n minus k in the inner sum that means, when n equal to 0 it will vary from n prime will vary from 0 to infinity and n minus k will be n prime n will be n prime plus k implies that n equal to n prime plus k. And now the inner sum varies from as we can see then pi naught z will be 0 to infinity 1 plus k equal to 1 to infinity and in the n prime variable it will be 0 to infinity w n prime 0 f k z to the power n prime plus k. Now both have become definite integrals definite sums which can be written resolved as 1 plus k equal to 1 to infinity z to the power k f k and n prime equal to 0 to infinity w n prime 0 z to the power n prime by going back to the definition this was the definition of the generating function for the return probability and this was the definition of the generating function for the occupancy probability which you denoted by same very same pi naught z which we have been computing. So, in other words we have a very important convolution like result which says hence pi naught z will be 1 plus pi z again pi naught z. This means that is pi z will then be pi naught z will be minus 1 divided by pi naught z or 1 minus 1 by pi naught z. It is a very beautiful result which says the generating function of the first time return probabilities plus the reciprocal of the generating function of the occupancy probabilities had to unity. It is easy to remember pi z equal to 1 minus 1 by pi naught z. So, this let us say is equation number 4 back to the return ultimate return probability if you see r was sigma f k ultimately. Now by the virtue of the definition of generating function this is nothing, but pi 1 therefore r equal to phi 1 1 minus 1 by pi naught 1 ultimate return probability expressed terms of the g f w n 0 evaluated z equal to 1. So, the problem therefore, finally reduces to the evaluation of the generating function of the occupancy probabilities w n 0. Now, let us visit back our Fourier transform solution evaluation of generating function that is pi 0 1. Let us recall the definition pi 0 1 pi 0 z in general was w n 0 summed over z to the power n for all n. This is the definition we have shown by the d dimensional solution w n 0 in as a d dimensional solution. So, this is 1 by 2 pi to the power d and to be to write it explicitly it will be a default integral minus pi 2 pi d k 1 etcetera d k d sigma cos k i divided by d by 2 pi d k i divided by 2 whole to the power n. Let us carry forward how to construct a generating function for this formulae. So, define some variable y equal to 1 by d sigma cos k i here sum is always i equal to 1 to. So, in terms of this notation y we can write w n 0 is 1 by 2 pi to the power d minus pi to pi there will be default integrals minus pi to pi. So, this is d k 1 d k d y to the power n it is very simple where y is of course, a function of where i equal to 1 by d sigma i equal to 1 to d cos k i. We can work with this multiply this by z to the power n then we can quickly note that if we take the sum inside the integral it can be converted to a geometric type series which can be summed and close form expressions can be obtained for the generating function. This exercise and the progress from there on we do in the next class. Thank you.