 So, in this study of return probability to the starting point, so far we examined the ultimate return probability. It is a matter of certain interest to understand how the first time return probability depends on the number of steps taken that is basically k dependence of f k. We demonstrate it here essentially for 1 dimensions. We have seen many results for absorber problem for 1 dimension both in terms of steps and in terms of time, but here we have to visualize this as a not an absorber problem, but merely a visiting to the origin for the first time. So, let us recall the definition. The definition of f k is stands for the probability of visiting the origin which is a starting point, visiting origin for the first time in k steps. Let us stick to 1D case. So, then it will be basically a random walk along a along the real line. Then we defined a generating function phi z. So, phi z represented a generating function which is basically step sum. It is a step sum generating function. So, it was defined as f k z to the power k and k equal to 1 is minimum. It is defined there is no meaning for returning to the origin 0th step because it was already there. And correspondingly we also defined a pi naught z as the generating function of the occupancy probability at the origin. So, that would be defined as again w n 0 z to the power n equal to 0 to infinity. So, these were the two definitions. Then we developed a relationship between the two quite generally regardless of dimensions and all as 1 minus 1 by pi naught z. So, if you know the generating function of the occupancy probability at the origin, we have the generating function of the first time return probability. So, now we specifically focus on 1D. We consider 1D case. So, the logic that we follow is if if we have results of occupancy probabilities for 1D, w n 0s are known for 1D. So, we can construct pi naught z from that we can construct phi z. And once we construct phi z and if we can expand it in powers of z, we have f k's as coefficients. So, this is the logic that we will follow. So, regarding regarding pi naught z in 1D, we know that the visitation to the origin occurs only in even number of steps for symmetric random walk specifically we focus on symmetric random walk or unbiased random walk. So, we know that w n 0 is 0 if n is odd equal to nonzero if n is even. So, specifically from the very basic equation we derived for the occupancy probability, we have shown that specifically for 2 m in general that the even an even number can always be denoted by 2 m. So, that at the origin is going to be 2 m C m 1 by 2 to the power 2 m, where binomial coefficients are defined by the notation 2 m C m for all for all m equal to 0 1. Now, now you can put m equal to 0 1 2 etcetera because automatically 2 m will be even. So, with this result we can construct the generating function steps some generating function for 1D symmetric random walk as sum over all the m's will do because anyway the odd ones are going to be 0. So, it will be w 2 m 0, but we should remember it is not z to the power m it will be z to the power of 2 m because only even terms will contribute. So, this is the definition. So, explicitly written it will have the form 2 m C m then z to the power 2 m divided by 2 to the power 2 m. So, I could write it as z square by or z to the ok z by 2 to the power 2 m for all m equal to 0 to infinity. Explicitly if we express 2 m C m it takes the form sum m equal to 0 to infinity 2 m C m is nothing, but 2 m factorial and m factorial and m factorial again. So, it will be m factorial whole square and we can write this as function of z square by 4 to the power m. So, we have an expression for pi naught in terms of a series of z square to proceed further to evaluate pi naught z or phi z we need to have one can work with series, but we need the reciprocal of the series. So, it will be a very tedious thing, but some simple solutions exist. Similarly we can identify this series we identify the series with 1 minus z square to the power minus half. I will shortly tell you how we what led us to identify that is specifically we say that series m equal to 0 to infinity 2 m factorial by m factorial whole square z square by 4 to the power m is 1 minus z square to the power minus half. The proof of this is a somewhat involved, but one can verify it by several means. One way is take a first of two terms of expansion Taylor expansion of 1 minus z square to the power minus half it can be always expand a Taylor expanded around z equal to 0 and verify that the coefficients agree and then trust that the series is ok. A more rigorous way is to consider the extended binomial expansion of the function for example, any function 1 plus x to the power nu where nu is not an integer can be written formally as this is a very formal way of writing can be written formally in the form of say r equal to 0 to infinity. Now the moment nu is not an integer it becomes a series it is not just a progression of a finite terms. So, this will have nu c r x to the power r is a formally extending the binomial expansion we have done two things we have just replaced if it was an integer n would have written n c r then it would have terminated at n, but if nu is not an integer we have to take the upper limit as infinity and write nu c r and interpret the factorials in terms of gamma function. Specifically in this case let us take specifically we choose this is by hand side we know this result we have to only just verify it in a way. So, choose nu equal to minus half and x equal to minus z square. Once we choose nu equal to minus half I am only going to indicate the steps it is quite a tedious, but straight forward procedure then you will have gamma of negative numbers. So, long as these negative numbers are not integers that is gamma of minus 1 does not exist, but gamma of minus 3 by 2 exists and it is a very defined number. We can convert this gammas of negative argument into those of positive argument by what is called as a reflection formula for gamma function. For example, the reflection formula for gamma function can be used that is gamma of z and gamma of 1 minus z is equal to pi by sin pi z is in general true. So, if a z is large and positive then 1 minus z will be negative number. So, you will convert it that negative number in terms of the positive gamma function via this relationship. So, upon doing that and then simplifying we will end up this particular series with the nu equal to minus half and x is minus z square will come over to this series. So, this is the way we can arrive at this generating function solution. So, for the moment we take this result. So, if we take that result our specific expression therefore, for the generating function pi naught 1 d symmetric random walk turns out to be we can write it as 1 by square root of 1 minus z square that minus half can be rewritten like this. So, once pi naught z is known it is actually a simple function now. So, we can determine phi z the generating function as equal to 1 minus 1 by 1 minus. So, 1 minus 1 by pi naught z that is 1 by square root of 1 minus z square which will turn out to be 1 minus square root of 1 minus z square. So, we have another simple expression for phi z. So, the now the next step is we have to re express this equation 1 minus square root of 1 minus z square as a power series in z. So, as to identify the coefficients of some z to the power k to get f k. The same procedure we can follow as the extended binomial expansion of square root of 1 minus z square you can even call it as a Taylor expansion. So, upon re expanding or let us say upon expanding square root of 1 minus z square as a power series we do not do it explicitly, but the method is now almost understandable you have to develop a general expansion or you have to look up at some handbook of series of various functions. So, we can obtain the following expansion for 1 minus. So, we gets 1 minus square root of 1 minus z square equal to we take the expression it will be k equal to 1 to infinity 2 k minus 2 c k minus 1 z to the power 2 k divided by k into 2 to the power 2 k minus 1. Again you can verify it by putting the calculating the coefficients for k equal to 1, 2 etcetera and confirm that it is true after expressing the binomial in terms of the factorial functions. Here you do not need to do any negative gamma functions because there will be all integer factorials can verify it. Or you can follow the same root as I mentioned in the previous slide that is by using the extended binomial expansion one can do either that or verify it either way we have a detailed expansion. So, upon comparing this expression with the definition upon comparing this expression with the generating function with the expansion or with the definition with the definition of the definition phi z equal to sigma f k z to the power k 1 to infinity we note that f of odd number let us say k equal to we rewrite f 2 k plus 1 will be 0 because it is a expansion of z square. So, you do not have odd powers of z since z to the power 2 k plus 1 coefficients of z to the power 2 k plus 1 since coefficients z to the power odd powers of z are 0 there are not powers of z and f 2 k will be 2 k minus 2 c k minus 1 1 by k into 2 to the power 2 k minus 1 for all k equal to 1 2 etcetera that is the very lowest value will be f 2 f 0 has no meaning f 1 does not exist. So, lowest value will be f 2 and this is the expression we can write for completeness in explicit form it will be then 2 k minus 2 factorial divided by k minus 1 factorial and then 2 k minus 2 minus k minus 1 which is again k minus 1 factorial. So, it will be square of this and then here it will be we can say it is a 2 k into 2 to the power 2 k minus 2 I just took out 1 k and put it here for all k equal to 1 2 etcetera. So, now we have an explicit. So, finally, this is the exact expression then exact expression specifically f of 2 for example, if we evaluate f of 2 will be half if you put k equal to 1 you will just get this 1 by 2 term. So, it will be half there is a 50 percent chance of first time return in the very second step. Similarly, k equal to 2 if you put f 4 you will have this will be for example, it will become 2 factorial here and it will be 1 factorial square here and here you will have 4 into another 4. So, it is going to be 1 by 8 likewise you can show that f 2 k that is k equal to 3 you put in the sixth step probability will turn out to be something like 21 divided by 1024 etcetera. So, so on and so forth you can you can calculate the return probabilities for any value from this exact expression. Now, a matter of interest is asymptotics how does f k behave for large k this is always a matter of interest. For that we do asymptotics our very useful tool is the Sterling's approximation once again. So, asymptotics of f 2 k in let us say 1 d we just indicate it as a 1 d problem. So, then we note we need asymptotic approximations for 2 k factorial and all. So, we note that in general 2 r factorial let us see r factorial square how does it behave if we go by Sterling's approximation this is going to be 2 r to the power 2 r e to the power minus 2 r square root of 2 pi into 2 r r factorial square is going to be r to the power r then e to the power minus r root 2 pi r whole square. So, if you now lot of terms get cancelled for example, this can be written as 2 to the power 2 r r to the power 2 r e to the power minus 2 r square root of 4 pi r. So, 2 I take out square root of pi r and denominator if you square it it is r to the power 2 r e to the power minus 2 r 2 pi r square root of 2 pi r square is simply 2 pi r. So, we see these terms cancel e to the power minus 2 r also cancels and here we will get 1 by root 2 cancels. So, we will get 1 by root pi r. So, this is going to be 2 to the power 2 r divided by root of pi r. Hence, if you substitute this in our expression for f of let us say 2 r plus 2 if you see it will be same expressions 2 to the power 2 r divided by root pi r multiplied by you will have 1 by r plus 1 here and we will have 1 by 2 into 2 to the power 2 r. So, 2 to the power 2 r cancels again. So, we have an expression 1 by 2 root pi 1 by r to the power half into r plus 1. So, for large r which we are discussing as r goes to infinity we can ignore 1 and this will have the behavior 1 by 2 root pi r to the power minus 3 by 2. So, the first time return probability to the origin varies as the step number to the power minus 3 3rd 3 by 2 not 3 3rd 3 half that is r to the power minus 3 by 2. We can recall at this point in time that the occupancy probability is distinct from the first time return probability we have been emphasizing that fact and to be very specific our occupancy probability let us say if you use that notation n for the steps it will it will be the exact expression is always 2 to the power minus 2 n 2 n factorial divided by n factorial square and this same asymptotics if you do will have 1 by root pi n or it will go as 1 by root pi n to the power minus half. So, here it goes as the step to the power minus half whereas, the first time return probability goes as a step to the power minus 3 by 2. So, the return probability decays much faster than the occupancy probability. So, pictorially we can represent this by the following graph. So, let us say we now uniformly denote all steps by the notation k or we use the r for f and n for occupancy w, but you can uniformize it by the notation k then the w function w k in fact, for even k only it if it decays as k to the power half k to the power minus half then the first time visit probability will go as f k as k to the power minus 3 by 2 much faster. So, the occupancy probability decays more slowly that is what we understand from this. So, this is the one example of explicitly calculating the visit the visit to the origin probability in one dimension. So, all these results of a 1 D for 1 D and then symmetric random walk this exercise can also become completed for 2 D. In 2 dimensions the problem of course, is a little more complicated, but I can indicate the steps. The steps is we have the occupancy probability for 2 D at the origin an explicit expression which includes all those 2 m factorials etcetera, but any different form and then you can construct it is generating function. However, the generating function for the 2 D occupancy probability is slightly complicated it is not expressible in terms of an algebraic function of the type square root of 1 minus z square instead it involves. So, for 2 D the generating function pi naught now 2 D symmetric random walk it has what is called as an elliptic integral some of the first kind k function k z square where k is the elliptic integrals these are defined as integral 0 to pi by 2 D x or D theta divided by square root of 1 minus n square sin square theta this is the kind of function this is k m this is the definition. So, to complete this we can reach up to the pi naught 2 D. However, reciprocal of that is difficult to express in terms of simple functions. So, one has to literally expand invert this elliptic integral or elliptic function and then term by term we can identify the coefficients of f 2 k it is possible to do, but a general expression is not easy. However, it is possible to obtain the asymptotic form and we find that the asymptotic form here decays as with a different power law we will not explicitly attempt this exercise. So, with this we have sort of got a grasp of what this return probability problems are and from now on we move to a more general formalisms of random walk specifically we try to understand whether some steady state situations also exist in an infinite dimensional systems such as nearest neighbor random walk that we discussed in the next lecture. Thank you.