 So, let us continue with this discussion of some of the problems on Legendre polynomials and here the slide you see a whole bunch of them and I explained to you how to go about doing some of these things using orthogonality, normalization, Roderick's formula, three term recursion formula these four things are at a disposal. Let us take the first one Pn prime of 1 equal to 1 half of n into n plus 1. Let us use the Roderick's formula to compute this. What is the Roderick's formula again? It is 1 upon 2 to the power n n factorial times nth derivative of x squared minus 1 to the power n and I am differentiating again. So, it is n plus first derivative of x minus 1 to the power n into x plus 1 to the power n as you see displayed in the slide. I had to use the Leibniz rule for the n plus first derivative of a product. So, what is the kth derivative of a product u v? You have to apply the Leibniz rule it is k choose j, j derivative is falling on the first factor k minus j derivative is falling on the second factor sum over j. So, here what happens to this expression when you put x equal to 1 if fewer than n derivatives fall on this then one factor of x minus 1 will survive and that will disappear when you put x equal to 1. If more than n derivatives fall on this factor it will become 0. So, the only way you can get a non-zero term is exactly n derivatives fall on the first factor and exactly one derivative falls on the second factor and the corresponding coefficient is n plus 1 choose 1. If n derivatives fall x minus 1 to the power n is going to become n factorial. One derivative falls on this will be n times x plus 1 to the power n minus 1 you put x equal to 1. What are you going to get? You are going to get 2 to the power n minus 1 and the denominator you got a 2 to the power n and n factorial the n factorial cancels out you get one half of n into n plus 1 and that is how the formula has to be proved. So, here is an application of the Roderick's formula. Now, let us prove an important theorem concerning the interlacing of the zeros of the Legendre functions and this is a very important result. Now, you must compare it with the following result. Suppose I take y double prime plus y equal to 0 you know the solutions are sin x and cos x. The zeros of the sin function and the zeros of the cosine function interlace between two zeros of sin x there is a zero of cos x and vice versa. So, this result is of a similar character a similar genre if you like. So, theorem 59 suppose 0 less than or equal to p less than q and y of x and z of x are respectively solutions of the Legendre equations with parameter p and parameter q. I am excluding the uninteresting solution namely the identically zero solutions. So, I am taking non trivial solutions of the Legendre differential equations with parameter p and parameter q. Now, I am looking at two successive zeros of y x there is at least one zero of z x. How do you think about this in general? Now, you see you could compare y double prime plus y equal to 0 and take the solution sin x or you could look at y double prime plus 9 y equal to 0 sin 3 x which solution oscillates faster. Obviously, sin 3 x oscillates faster than sin x. So, this differential equation has been written in self adjoint form. So, think of this as the analog of the y double prime term in the sin x example and think of this p into p plus 1 as some kind of a frequency. So, I am saying p is less than q. So, the solutions of this differential equation oscillate with a higher frequency. So, you can think of it in this way intuitively if you wish, but the precise mathematical description is stated here between two successive zeros of y x there is a zero of z x. So, let a and b be successive zeros of y x. Remember that we have already shown that the zeros are isolated. The zeros cannot accumulate. It is a second order differential equation and the zeros cannot accumulate. We will prove the more general version later. Here you can see it because y of x and z of x are solutions of an analytic differential equation and they are power series solutions and so the zeros cannot accumulate any way. So, a and b are successive zeros and I am going to assume that minus 1 less than a less than b less than 1. We shall imitate the proof of orthogonality of the Legendre polynomials, but this time we shall work on the closed interval a b. Multiply the first equation by z x, the second equation by y x, integrate over a to b and integration by parts. One integration by parts and we must subtract. So, lots of things are going to happen. When you integrate by parts you are going to pick up boundary terms. When you integrate by parts the derivative will shift from one term to the other term. What is that going to be when you multiply the first equation by z x and do a parts then the derivative will shift from this piece to the z x piece. Over here you are multiplying the differential equation by y and when you integrate by part the derivative will shift from here to the y. In both these cases you are going to get the term minus integral from a to b 1 minus x squared y prime x z prime x dx and when I subtract the two results this term will cancel out. That is one thing that is going to happen. Another thing that is going to happen is that I am going to get p into p plus 1 integral from a to b y x z x dx. Here you are going to get integral from a to b q into q plus 1 y x z x dx. When you subtract you are going to get p into p plus 1 minus q into q plus 1 integral from a to b y x z x dx that term is survived and then there will be the boundary terms when I integrate by parts. Unlike the previous case these boundary terms will not all vanish because we are not integrating from minus 1 to 1 we are integrating from a to b. Earlier when we did orthogonality of Legendre polynomials integration was from minus 1 to 1 and this 1 minus x squared was responsible for the vanishing of the boundary term. So, here the boundary terms will survive and these boundary terms have been written out when you integrate by parts. One of these boundary terms has been written in red. Why is it written in red? Because this is the one that is going to cancel out. Why is going to vanish at a and b because a and b are successive zeros of y x. So, this red thing is going to drop out and what is going to remain will be 1 minus b squared z b y prime b minus 1 minus a squared z a y prime a. In the other term that I talked about I have not done anything to it I just written it as it is. Now remember that so far I have not put any conditions on y and z except that a and b are successive zeros of y of x. Now since a and b are successive zeros y does not vanish in between a and b. So, I may assume without loss of generality that y is positive on the open interval a b. In the other case if y is everywhere negative simply work with minus y in place of y nothing will change. So, replacing y by minus y if necessary I may assume that y of x is positive. Now z of x does not vanish in a b that we make an assumption that is we prove it by contradiction and we will try to arrive at a contradiction. We shall assume that z of x is non vanishing in the interval a b. Again replacing z by minus z if necessary I may assume that z of x is always positive on a b and we will arrive at a contradiction. With this z of a is positive z of b is positive y of a is 0 y of b is 0 and y is positive which means the function y is increasing at a because it from 0 it is becoming positive and it is decreasing at b because from positive it has become 0. So, y prime of a must be greater than or equal to 0 and y prime of b must be less than or equal to 0. So, y prime of b is less than or equal to 0 z of b is positive 1 minus b squared is positive y prime of a is non negative. So, this piece is also negative. So, this is negative or 0 this piece is negative or 0 y of x and z of x are both positive in the interval a b. So, this integral is positive and this number p into p plus 1 minus q into q plus 1 is negative because p is strictly less than q. So, 0 or negative 0 or negative strictly negative and the sum is 0 and that is a contradiction. So, we have proved the theorem. So, this is what I said has been summarized in this slide and here is a little exercise for you. Can it happen that y of x and z of x have a common 0? Please read this proof very carefully and try to rule out a case that y of x and z of x have a common 0. Another important exercise try to prove that if 0 less than or equal to p less than q then between two successive zeros of J p x there is a 0 of J q x. Try to prove a similar result for the Bessel's functions zeros of the Bessel's function. Given a sequence a n of real or complex numbers the generating function is by definition the power series summation n from 0 to infinity a n t to the power n. Now the theorem 60 says summation n from 0 to infinity t to the power n p n x equal to 1 upon root of 1 minus 2 x t plus t squared. That is a generating function for the sequence of Legendre polynomial can be written in closed form. Now this result is extremely important in classical potential theory. For a connection between classical potential theory see for example this book of Ramsey Newtonian attraction Cambridge University Press page 131 and following or page 121 to 134 are the more comprehensive and classic treaties of Oliver Diamond Kellogg foundations of potential theory dover 1953. Both these references are being given the Kellogg's reference is very classical and a very important reference in the potential theory. So we want to prove this theorem 60. So let us complete the proof of this. Remember this must be compared to the generating function for the sequence of Bessel's functions. There are several proofs of this important theorem and we select the one from Courant and Hilbert's monumental treaties methods of mathematical physics volume 1 only a sketch of proof will be provided the details can be worked out easily. Some details will be slightly cumbersome and integral computation for instance. But you can do it exercise 11 in the previous slide tells you that if you take this function 1 minus 2 x t plus t squared. So it is 1 plus some something t squared minus 2 x t that something is small in absolute value. So I can apply the binomial series 1 plus z to the power a can be written as a binomial series. So when you write this as a binomial series that series will converge absolutely when the mod t squared minus 2 x t is less than 1. So the series converges absolutely. You can rearrange the terms in such a way that I can collect the powers of t and I can write this some function v x ts summation n from 0 to infinity r n x t to the power n where r n x is obviously going to be a polynomial of degree exactly n. When you write this expansion using the binomial series you will have to tell me why is this a polynomial of degree exactly n and the series is valid for x between minus 1 and 1 provided mod t is small. In fact, if mod t is less than 0.4 that will be enough for carrying out the analysis. So now let us do the following let us now use the fundamental orthogonality lemma that is going to play a role. So let us first write down the expression for the potential v of x t with two different parameters t equal to u and t equal to v and let us multiply the two series. When you multiply the two series we will get summation j from 0 to infinity summation k from 0 to infinity r j x r k x u to the power j v to the power k. Remember that the Cauchy product of two absolutely convergent series will absolutely converge. So I can assume that mod u is less than 0.4 mod v is less than 0.4 there is no harm in doing that. Now what happens I am going to integrate both sides with respect to x in the range minus 1 1 and I am going to get 1 upon root u v l n 1 plus root u v upon 1 minus root u v that is the tedious part you will have to integrate the left hand side with respect to x remember it is an integral with respect to x is an elementary integral what a tedious integral I am sure you can do it. The right hand side you are going to get a double summation u to the power j v to the power k integral minus 1 to 1 r j x r k x dx. Now you will ask how can I take the integration under the summation sign you want uniform convergence remember that term by term integration is perfectly fine if you know that the series converges uniformly my u and the v were restricted to be less than 0.4 in absolute value you can make it even smaller and you can guarantee uniform convergence for all x in the range minus 1 1 and the integration can be taken under the summation sign. So, now left hand side you got a nice function which can be written as a power series l n 1 plus z what is l n of 1 plus z it is z minus z squared upon 2 plus z cube upon 3 minus z to the power 4 upon 4 divided r and you got l n of 1 minus z what is l n of 1 minus z minus z minus z squared upon 2 minus z cube upon 3 minus z r both these expansions are valid for mod z less than 1 remember my my u and the v are going to be pretty small in absolute value. So, there is no problem and then you write the difference l n of 1 plus root u v minus l n of 1 minus root u v and then you divide by 1 upon root u v okay. So, the logarithmic series we are going to be used and we are going to get summation n from 0 to infinity 2 times u to the power n v to the power n upon 2n plus 1. So, both sides are double power series in u v. So, compare the coefficients of u to the power j v to the power k on both sides what happens you see in the left hand side the u to the power j v to the power k only appears when j and k are equal. So, we immediately get that integral minus 1 to 1 r j x r k x dx is 0 if j is not equal to k whereas, if j is equal to k integral minus 1 to 1 r j x the whole squared dx is 2 upon 2j plus 1. Lo and behold r j x is a polynomial of degree exactly j. Now, you can apply the fundamental lemma to the sequence r naught x r 1 x r 2 x dot dot dot and we get that r j x will be a multiple of p j x. How is it that I had written that this constant of proportionality is 1 directly the norms of the both these things are 1. Remember that norm p j squared is 2 upon 2j plus 1 we did the norm calculation in the last capsule. So, the constant of proportionality has to be plus 1 or minus 1 you can figure out that it will be 1 by looking at the leading coefficient or something and so, that completes the argument that the Legendre polynomials has generating function 1 upon root 1 minus 2 x t plus t squared because I just expanded this and I found out that the coefficients r and x are exactly the Legendre polynomials p and x. Okay. So, one final comment this function v of x t is a potential due to a point mass p placed at unit distance from the origin at a point x the potential at x which is at a distance of t from the origin and x is the cosine of the angle between O x and O k. So, this has a physical interpretation and I think with this that completes the section on generating functions. Couple of exercises for you and an expansion which is which involves both the Bessel's function and the Legendre polynomials. So, let us call it the Fourier Bessel expansion rather than state the theorems on Fourier Legendre series you can consult this classic books that are cited so far. Here we look at an example which is due to Lord Rayleigh theory of sound volume 2 page 273 e to the power i t x equal to summation n from 0 to infinity 2 n plus 1 i to the power n root pi by 2 t j n plus half t p n x prove an induction first before you embark upon this expansion prove that j n plus half t equal to 1 upon root pi n factorial t by 2 to the power half plus n integral minus 1 to 1 1 minus x square to the power n cos t x dx. How do you prove this? First write down the power series for j p t where p is any non-negative real number for example put p equal to half put p equal to half find out what is j half of t you will get a very familiar looking series a sine series it would not be sine x it will be slight variant thereof and then try to differentiate and check you have the relation between j p prime j p plus 1 and j p minus 1 remember and try to use induction to prove this and this and formally deduce the result of Lord Rayleigh. Now this expansion appears in scattering theory problems on scattering of plane waves by spherical obstacles. I close this capsule here thank you very much.