 So, we continue with the theme of this chapter which is Fourier analysis with respect to generalized complete orthogonal systems in Hilbert spaces. We have been focusing on two such system, the systems given by J0 zeta Jx where zeta Js as J goes from 0 to infinity runs through the 0s of the Bessel's function J0 which lie in the positive real line. We saw the orthogonality of that and we also saw the sequence of Legendre polynomials in great detail and we stopped last time with the Fourier Bessel expansion theorem and the Fourier Legendre expansion theorem. So, you want to take a quote unquote fairly general function f of x. Say it should be reasonably nice reasonably well behaved and you want to write this f of x as summation n from 0 to infinity cn pn x n from 0 to infinity and of course under very mild conditions this will converge in L2 of minus 1 1, but we want point wise convergence etcetera it is little more stringent. We are we are not going to prove these point wise convergence theorems and we stopped at the point where we given example of a Fourier Legendre expansion where we write e to the power itx as a series and that series is what you see displayed as equation 5.16 in this slide. So, now let us come to the slides if this was given by Lord Rayleigh as we see theory of sound volume 2 page 273. The coefficients involve the Bessel's function jn plus half t that is because the left hand side function of t and x. So, we are looking at this expansion as a function of x with respect to the Legendre polynomials pn x and the coefficients are given by 5.17. This result appears in connection with the scattering of plane waves by spherical obstacle. Now, we want to look at the proof of this expansion of Lord Rayleigh. First of all we have to establish a lemma namely equation 5.17 that jn plus half t is given by this integral 1 upon root by n factorial t upon 2 the exponent is half plus n integral minus 1 to n 1 minus x square to the power n cos dx dx ok. So, let us proceed with the proof of this 5.17. Let us begin by recalling the definition of Jpt summation n from 0 to infinity minus 1 to the power n n factorial in the denominator gamma n plus p plus 1 t upon 2 to the power p plus 2n. When p equal to half root t upon root 2 will come out and we will get summation n from 0 to infinity minus 1 to the power n factorial gamma n plus 3 by 2. What is gamma n plus 3 by 2? Gamma n plus 3 by 2 is basically n plus half gamma n plus half etc. So, we have to simplify this and you will see that you get J half t is root 2 upon t sin t and what we have to show is that root 2 upon pi t sin t is this right hand side 1 upon root pi n factorial t upon 2 half plus n exponent integral minus 1 to 1 1 minus x square to the power n cos dx dx where n is 0. Remember we are trying to look at the case n equal to 0. First we prove 5.17 for n equal to 0 and we will resort to induction. So, what do you get when you put n equal to 0 this disappears cos tx dx will be sin tx upon t that is how the 1 t cancels out and it goes to the denominator and you can verify the result trivially for n equal to 0. Now, we want to prove by induction that the result is true for all values of n and so, let us assume the result for a certain n and we will proceed to prove it for the next n. Now, remember the basic result in chapter 1 t to the power minus p J p t the whole thing prime is minus t to the power minus p J p plus 1 t. This is a basic result that we have to use and I am going to take p equal to n plus half. I am going to take p equal to n plus half. So, I need to multiply the formula for J n plus half t by t to the power minus n minus half and I do not differentiate it. So, we are assuming the result for n remember because we are assuming the result for n we are going to assume this expansion. So, we have to multiply this equation by t to the power minus half minus n. So, this will go to the left hand side and we need to differentiate needless to say I am going to differentiate under the integral sign cos tx will become minus x sin tx and I am picked up an extra factor of x and I am going to multiply and divide by 2. I am going to write as 2x into 1 minus x squared to the power n for the obvious reason. What is the obvious reason because that is a derivative of 1 minus x squared raised to the power n plus 1. When I do this differentiation I am going to pick up a factor of n plus 1. So, I need to divide by n plus 1 also and so this n factorial has become an n plus 1 factorial. The t has gone from here to the left hand side. Of course, root pi remains it is a constant and now I want to perform an integration by parts. When you perform an integration by parts the sin tx will again become a cos tx, but this time we are differentiating with respect to x remember. So, the t will come out and that is how you get a t and it become a minus sign when you integrate by parts. Now, you must use this formula that you see here and in this formula also there is a minus half. So, the minus sign will go away and you will also get this t to the power minus p and that is exactly this expression and when you put it in the right hand side you already have a t you get t to the power n plus 1 plus half and you would approve the result for n plus 1 assuming the result for n and the inductive step is complete and the induction base is the case when n equal to 0 it is a trivial verification. Now, we are ready to establish the result of Lord Ray leg on the Fourier Legendre expansion of the plane wave e to the power itx is a monochromatic plane wave remember. And so, we begin with this Anzart's that this e to the power itx is expressible as a Fourier Legendre series and the coefficients are cn. We have to determine these coefficients cn and the procedure is quite routine you take the inner product of both sides with respect to pn what is the inner product it is a usual L 2 inner product L 2 of minus 1 1 with respect to the Lebesgue measure. When you deal with the Legendre polynomials the inner product is the Lebesgue measure when you deal with Shebyshev's polynomials or Hermit polynomials the inner product is a weighted Lebesgue measure what those weights are we shall see later. And the coefficients will of course, depend on t because the left hand side depends on t. Now, when you take the inner product the right hand side will give you cn norm pnx the whole square remember the last capsule we worked out the norms of these Legendre polynomials norm pnx the whole square what was that it was 2 upon 2n plus 1 remember from the last time the norm pnx squared was 2 upon 2n plus 1 we need that here and then we have those right hand side integral minus 1 to 1 e to the power itx pnx dx. Now, the next step is quite obvious we must appeal to the Roderick's formula the Roderick's formula will be 1 upon 2 to the power n n factorial to nth derivative of x squared minus 1 to the power n nth derivative x squared minus 1 to the power n and I am going to you integrate by parts repeatedly and put the derivatives on e to the power itx. So, n times I must integrate by parts each time I integrate by parts I am going to pick up a minus 1 and so I am going to pick up n minus 1s finally I will get e to the power itx into 1 minus x squared to the power n in the Legendre polynomial formula of Roderick's formula he had x squared minus 1 to the power n here it has mysteriously become 1 minus x squared to the power n and there is no mystery here because the minus 1 to the power n which we obtained by integration by parts has been absorbed here and the n differentiations on e to the power itx has given you the it to the power n factor over here and now you must write this exponential in terms of the sine the cosine and this factor 1 minus x squared to the power n is an even function and so the sine tx times 1 minus x squared to the power n integral from minus 1 to 1 will go away and you will be left with cos tx term and that completes the verification except that now you need to divide by norm p n squared using this equation and you will finally get the stated result 5.16 you will finally get this result of Lord Rayleigh so that is a very nice explicit computation of a Fourier Legendre expansion so now let us go a little further and we shall prove two integral representations for the Legendre polynomials remember we obtained an integral representations for the Bessel's functions of integer orders j nx where n is a non-negative integer it is given by what 1 upon pi integral 0 to pi the presence of the 1 upon pi means it is an average value cos of x sine theta minus n theta d theta right in the first chapter we derived this via the Schlomelschitz formula the x the generating function for the sequence of Bessel's functions the result here is similar in spirit again you see a 1 upon pi integral 0 to pi water integrating is x plus root x squared minus 1 cos theta to the power n d theta 5.18 and this is true for non-negative integers n it is true for non-negative integers n one way to do is to directly check that this satisfies the Legendre differential equation and that it is a polynomial of degree n and so it must be a multiple of p nx put x equal to 1 and see whether they agree put x equal to 1 what happens this x squared minus 1 cos theta collapses and so the integration with respect to theta is this pi the pi cancels out you can easily check that p n of 1 is 1 and so one way to do it would be to differentiate this under the integral sign and check that the Legendres differential equation is satisfied we are going to take a slightly different approach for a change we are going to give an application of the three-term recursion formula satisfied with the Legendre sequence another approach would be to try and show that these p n and p m are orthogonal if n is not equal to m and use a fundamental orthogonality lemma that could be another approach. So, I am giving you many different ways of proving this result you can take your pick first of all let us put n equal to 0 what do you get when n equal to 0 you simply get 1 so the right hand side does give you p0x when you put n equal to 1 again you can integrate and you can check that you get x now what happens is that let us call the right hand side of 5.18 as qnx let us just call it qnx and the first observation to make is that this qnx is a polynomial it is a polynomial of degree exactly n you see what happens when you expand this integrand using the binomial theorem you are going to get a whole bunch of terms if an even power of these terms will become polynomials root of x squared minus 1 to some even power will get rid of the square root it is the odd powers that are going to be problematic the only way this can cease to be a polynomial is if some odd power of this remains when you integrate from 0 to pi but that will not happen if you take an odd power of cosine and integrate from 0 to pi the integral will be 0 because the integral from 0 to pi by 2 will cancel out the integral from pi by 2 to pi so the odd powers will not survive only the even powers will survive so this when you expand this you are going to get a polynomial and a polynomial of degree exactly n secondly when n is 0 you get 1 q0 of x is 1 which is the same as p0 of x and you put n equal to 1 you see that q1 of x and p1 of x are both equal to x now we shall show that qnx satisfies the same three term recursion formula as pnx namely we will prove that n plus 1 qn plus 1 minus x times 2n plus 1 qn plus n qn minus 1 is 0 and if you replace qn by pn we know it is true because that is the three term recursion formula for the Legendre polynomials that were established earlier so how do we prove this three term recursion formula now once you prove this three term recursion formula the first line q0 and p0 agreeing q1 and p1 agreeing will imply that qn and pn must agree for all n's in other words qnx is exactly pnx that is what we wanted to show so let us proceed to prove the three term recursion formula that you see here so call this complicated expression x plus root of x squared minus 1 cos phi call this a what do we have in the integrand qn plus 1 right integrand will have a to the power n plus 1 a to the power n plus 1 can be written as a into a to the power n what is a x plus this so x into a to the power n plus root of 1 minus x squared the cos phi has been written as d d phi of sin phi and so this is what we have now you ought to integrate from 0 to pi I know to divide by 1 upon pi of course what do you get you get integral of this thing and you need to integrate by parts this d d phi we will have to shift to the other term right so when you integrate by parts you are going to get exactly qn plus 1 will be x times qn plus n upon pi integral 0 to pi a to the n minus 1 x squared minus 1 sin squared phi d phi you have to perform the integration by parts and I am not done it here and that is an easy exercise okay so now you simply write x squared minus 1 sin squared phi as x squared minus 1 minus root of x squared minus 1 cos phi the whole squared sin squared phi is 1 minus cos squared phi nothing fancy and that can be written as x squared minus 1 minus a minus x the whole square that is 2 ax minus a squared minus 1 so qn plus 1 turns out to be x qn minus n upon pi integral 0 to pi you get this expression a to the power n plus 1 plus a to the power n minus 1 minus 2 x a to the power n d phi and this simply translates to the three-term recursion formula for the qn's as I stated earlier a very different proof can be found in the book by Byerley on Fourier series the book of Byerley is a very old book but a really remarkable book for the kind of things that are being covered I believe Byerley's book is freely available on the internet because it is a very old book I think it must be in one of these public domains now there is a second integral representation which is very remarkable and this is also due to Laplace this is theorem 62 for a non-negative integer n remember these integral representations are for the n integers again I repeat for the Bessel's functions also the integral representations were for integer values of n so pnx is 1 upon pi integral 0 to pi d theta upon x plus root of x squared minus 1 cos theta the whole thing to the power n plus 1 display equation 5.19 now there are several things we need to understand unfortunately the proof of this is not going to be routine like the previous one it is not going to be easy to simply call the right hand side qn and verify it satisfies the three-term recursion formula it is not even clear the right hand side is a polynomial even that part is not clear and validity of this I am proving only for x greater than or equal to 1 number of comments are in order first of all if x is bigger than or equal to 1 this denominator x plus root of x squared minus 1 cos theta is going to stay away from 0 in fact this term will never be 0 the denominator and since for a fixed x this is continuous and so if it is never going to be taking the value 0 its infimum as theta varies from 0 to pi will be strictly positive so as far as the integral goes it makes perfect sense in fact when I put x equal to 1 what do I get when I put x equal to 1 trivially I get 1 on the right hand side which is expected pn of 1 must be 1 and so formula is certainly correct when you put x equal to 1 now the problem arises when you put x equal to minus 1 when x equal to minus 1 the right hand side turns out to be minus 1 to the power n plus 1 as you can easily see but we know that pn of minus 1 is minus 1 to the power n remember that p2 of x is an even function p3 of x is an odd function so for negative values of x this formula doesn't appear to be correct whereas for positive values of x specifically for x equal to 1 at least the formula appears to be correct so something must be going wrong when you cross from the positive real axis to the negative real axis what goes wrong what happens when I put x equal to 0 suppose I try to put x equal to 0 here I get i I get i to the power n plus 1 i is an innocent constant but what else do I get I get a cosine theta to the power n plus 1 and n is 1 2 3 etc and what happens when theta is close to pi by 2 this integral diverges so this is a problem you need to be very careful in interpreting this the condition x greater than or equal to 1 is going to be significant in this theorem of course what happens when x is a complex number again you can check if x is a purely imaginary number suppose if x equal to i y then what happens is again you will see that the denominator will become 0 somewhere when theta runs from 0 to pi and again the improper integral will diverge so so along the imaginary axis the the integral doesn't even make sense okay so we will now assume that x is bigger than 1 and we will use the first formula of Laplace that you see in the slide what are the first formula of Laplace again p nx equal to 1 upon pi integral 0 to pi x plus root x squared minus 1 cos theta to the power n d theta here comes the trick you have to perform a change of variables what change of variables that you will perform you simply put x plus root x squared minus 1 cos theta equal to 1 upon x plus root x squared minus 1 cos phi and if I make the change of variables in 5.18 then you will probably believe that there is some chance of proving the result 5.19 but unfortunately life is not going to be so kind we have to make sure that this change of variables is a valid is a legal change of variables let us rewrite 5.20 as in this form 5.21 so if this particular thing has any validity at all the first thing that you need to check is that the right hand side of 5.21 must be less than or equal to 1 in absolute value so you must take the mod of the right hand side and square it and see whether you get something which is less than or equal to 1 that is a simple algebra and I am going to leave it to you so now that means that the right hand side as it stands is a number between minus 1 and 1 and the denominator is never 0 we already discussed that so certainly the equation 5.21 determines theta uniquely in the interval 0 to pi so certainly 5.21 determines theta as a function of phi that is very clear secondly the right hand side of 5.1 when will it be equal to plus 1 and when will you be equal to minus 1 or take the square of this when will be equal to 1 do the algebra you have to perform the take the square and do the algebra and you will have to check that the right hand side takes the value 1 if and only phi is pi and takes the value minus 1 if and only phi is 0 and the denominator never vanishes as I said and so the right hand side as a function of phi is a differentiable function it is a differentiable function and it takes the value in the appropriate interval and so the next thing to check is that this equation must define theta as a smooth function of phi it is a smooth function how do you check whether some expression like this when you give an expression like this how do another theta is a smooth function of phi you have to apply the implicit function theorem replace the equality by minus sign and call the resulting function f of theta phi capital f is a function of two variables and I must apply the implicit function theorem take the derivative with respect to theta straight away you get this goes away root of x squared minus 1 sin theta and that is non-zero when theta runs from 0 to pi and so the implicit function certainly confirms that theta is a smooth function of phi where phi varies over 0 to pi and then you can use the implicit differentiation and you can use implicit differentiation to calculate d theta by d phi and we see that d theta by d phi is also not 0 on the open interval 0 to pi and it confirms that the derivative is never 0 you got a continuous function you got a smooth function the derivative is never 0 theta is strictly increasing strictly decreasing function excuse me of phi and the change of variables is certainly a legal change of variables now all that remains now is to carry out the change of variable the difficult thing was to check that the change of variables is a valid change of variables once you had gone past that stage making the change of variables in an integration is a routine matter and integral transforms to 1 upon pi integral 0 to pi d phi upon x plus root of x squared minus 1 cos phi raised to the power n plus 2 sin phi by sin theta now you got the unpleasant job of writing sin phi by sin theta purely in terms of phi again you are going to use this 5.21 this equation 5.21 you have to use and you are going to use it to show that sin phi by sin theta is simply x plus root x squared minus 1 cos phi so one factor goes away and the n plus 2 becomes n plus 1 thanks to this and that completes the proof of the Laplace's second integral representation and a very different proof using methods of complex analysis is available in Courant and Hilbert's monumental treaties methods of mathematical physics volume 1 pages 503 504 I have already given you this reference couple of times before this is really one of the most remarkable books written the integrals of Laplace the two the two integrals of Laplace that were derived here were given by Pierre Simon Laplace in his great work on celestial mechanics theta the mechanics list this is a phi volume massive phi volume treaties which was reprinted by Chelsea and now we look for a result due to Neumann in 1862 use the Laplace's integral representation to prove that limit as n tends to infinity pn cos x upon n is j naught of x straight away start with the Laplace's integral of the first kind pull out the cos x by n outside the integral because the integration is with respect to theta and the factor cos x by n goes to 1 and I simply take the limit inside the integral use the dominated convergence theorem and use some routine 12 standard calculus results and you will get that the limit of pn cos x upon n as n tends to infinity is 1 upon pi integral 0 to pi e to the power ix cos theta d theta which is g0 of x that is a formula of Neumann I think with this I shall close this capsule here thank you very much