 Hello everyone. So, we continue with our discussion of the Legendre polynomials and Legendre functions. So, last time we discussed the Legendre polynomials, its completeness, its orthogonality and some of its properties such as the derivation of the Rodricks formula. Today, we shall use the Rodricks formula to prove that the Legendre polynomial of degree n has precisely n roots in the open interval minus 1 1 and these roots are distinct. Of course, it is a polynomial of degree n, it cannot possibly have more than n roots. The fact is that none of the roots are complex, all roots are real, all roots are distinct and they all lie in the interval minus 1 1. There is a considerable information on the distribution and location of roots and these roots were used by Gauss in 1814 in his quadrature formula that bears his name and it is a very famous quadrature formula called the Gaussian quadrature. It is the best possible quadrature formula and a discussion of it can be found in the book of S Chandrasekhar that you see in the slides Radiative Heat Transfer Dover Publications New York. So, let us get to the proof of this important result. So, we need to use the Rolls theorem repeatedly. What is Rolls theorem says? Of course, it is true for any smooth function or a continuously differentiable function, but for us it is going to be a polynomial. So, we have got a polynomial that vanishes at a and b and the derivative must vanish at least once in the open interval a, b. Now, supposing that the polynomial f has a double root at a and a double root at b that means that not only we have f of a is 0, f of b is 0 in addition to that we also have f prime of a is 0 and f prime of b is 0. So, now Rolls theorem applied to f of x gives us a point c in the open interval a, b such that f prime of c equal to 0. But now look at f prime of x that is also a polynomial and f prime of x vanishes at a and c and f prime of x vanishes at c and b. So, by application of Rolls theorem f double prime must vanish once at least in the open interval a, c it must vanish at least once in the open interval c, b thus we get that f double prime has two roots in the open interval a, b. Again we could take a function f which has a triple root at a and a triple root at b and we will conclude that f triple prime vanishes thrice in the interval a, b. In particular let us take f of x equal to 1 minus x squared the whole square or x squared minus 1 the whole square does not matter. Now x squared minus 1 the whole square has a double root at 1 and a double root at minus 1. So, by what we have just discussed the second derivative must vanish at two distinct points in minus 1, 1. Remember in the previous discussion when we talked about f double prime must vanish at least one in the open interval a, c and at least one in the open interval c, b which means that these two roots of f double prime are distinct they lie in disjoint open intervals. So, we have concluded that the second Legendre polynomial p 2 of x has two distinct roots in minus 1, 1 of course is a quadratic polynomial. So, it cannot possibly have more than two roots. So, p 2 x has two and exactly two roots in the open interval minus 1, 1. Now let us go further. So, now let us continue and look at what happens with the third derivative and as I explained if f has a triple root at a and a triple root at b then the argument all shows that f triple prime must have three distinct roots in the interval minus 1, 1 and Roderick's formula now tells you that the third Legendre polynomial p 3 x what is the third Legendre polynomial it is a constant multiple of third derivative of x squared minus 1 cube x squared minus 1 cube has a triple root at 1 and a triple root at minus 1. So, the third derivative must vanish exactly at three distinct roots in other words p 3 of x must have three distinct roots in the open interval minus 1, 1 and since p 3 of x is a cubic polynomial these are all the roots. Generally for n equal to 0, 1, 2, the polynomial p n x equal to 1 upon 2 to the power n n factorial nth derivative of x squared minus 1 to the power n has n distinct roots in the open interval minus 1, 1 and that completes the proof that the Legendre polynomial of degree n has exactly n distinct real roots in the interval minus 1. So, having completed this part let us go to the solutions of some more problems. So, the three term recursion formula the three term recursion formula is an extremely important formula and the presence of the three term recursion formula is significant in these orthogonal systems. You may recall when we studied the Bessel's functions in the first chapter we got a relationship x to the power n j n x prime equal to x to the power n j n minus 1 x to the power minus n j n x prime equal to minus x to the power minus n into j n plus 1 x. So, these two relations they get a three term recursion relation between j n j n plus 1 and j n minus 1. You should go back to chapter 1 and see that just as you got a three term recursion formula for the Bessel's functions you got a three term recursion formula for the Legendre polynomials. These these three term recursion formula for Legendre polynomials is going to play a very important role later. So, we want to prove this three term recursion formula. How does one prove the three term recursion formula? Let us now as a preparation for this determine the leading coefficient of these Legendre polynomials l c I have written l c in red l c of p n x is the leading coefficient of p n x. So, the leading coefficient l c of p n x is 2 n factorial upon 2 to the power n n factorial squared equation 5.13 that you see displayed. Let us prove this what is the leading coefficient l c of p n x with the definition of p n x 1 upon 2 to the power n n factorial times the nth derivative of x squared minus 1 to the power n. Now, when you expand x squared minus 1 to the power n you are going to get x to the power 2 n plus lower order terms nth derivative of that is going to be the nth derivative of x to the power 2 n plus lower order terms, but I am looking at the leading coefficient. So, I can comfortably ignore the lower order terms. So, ignoring the lower order terms we get 1 upon 2 to the power n n factorial leading coefficient of the nth derivative of x to the power 2 n. The nth derivative of x to the power 2 n is exactly 2 n into 2 n minus 1 to the n plus 1 and I multiply numerator and denominator by n factorial. So, the numerator becomes 2 n factorial the denominator becomes 2 to the power n n factorial squared and this completes this little lemma. So, now, let us proceed with the proof of the 3 term recursion formula n plus 1 p n plus 1 minus x into 2 n plus 1 p n plus n p n minus 1 equal to 0 equation 5.14 that you see displayed in the slide. Recall that for any n p naught x dada p n x is a basis for the vector space of polynomials of degree at most n. This we have discussed before. So, no need to discuss it again. So, if you look at x p n x multiplied by this constant 2 n plus 1 that is a polynomial of degree n plus 1. So, this is going to be a linear combination of p naught x dada p n plus 1 x. So, this linear combination is written as a naught p naught x plus dot dot dot a n plus 1 p n plus 1 x. You can think of this as a Fourier expansion of x p n x with respect to the usual inner product l 2 and the orthonormal system p naught x dada p n x dada etcetera. Now, we have to determine the coefficients a naught a 1 a 2 etcetera. How do you determine the coefficients? Take the inner products on both sides with p j x. In other words 2 n plus 1 inner product x p n x p j x. Right hand side the only term that will survive is a j norm p j x the whole squared. We have already determined the norm p j x the whole squared and so, we have to compute the left hand side. Now, just write the integral for this x p n x inner product with p j x is an integral minus 1 to 1 x times p n x p j x d x. So, this can be conveniently written as inner product of p n x comma x p j x. Now, remember if j runs from 0 1 2 3 up to n minus 2 then what happens to x p j x it is a polynomial of degree at most n minus 1 because x p j x is a polynomial of degree exactly j plus 1 and I am saying j is less than or equal to n minus 2. So, x p j x is a polynomial of degree at most n minus 1 because x p j x is a polynomial of degree at most n minus 1 it is a linear combination of p naught x p 1 x dada p n minus 1 x. So, in particular since these things are all orthogonal to p n x x p j x will also be orthogonal to p n x if j runs from 0 up to n minus 2. Which means that the inner product that I had displayed in the last line in the previous slide that is equation 5.15 we will get a j will be 0 a j will be 0 for what values of j for these values of j some 0 1 2 up to n minus 2 which means that a naught a 1 a 2 dada a n minus 2 are all 0. So, our Fourier expansion simply collapses to this 2 n plus 1 x p n x equal to a n minus 1 p n minus 1 x plus a n p n x plus a n plus 1 p n plus 1 x. Now, we have to knock off this a n how do we knock off this a n remember p n x is an odd function if n is odd it is an even function if n is even. So, x p n x is an odd function if n is even and an even function if n is odd in other words the parity of x p n x agrees with the parity of p n minus 1 and the parity of p n plus 1, but disagrees with the parity of p n. Now remember that a function can be uniquely written as a sum of an odd function and an even function and therefore, a n must be 0. So, we get further shortening we can knock off this middle term and we simply get 2 n plus 1 x p n x equal to a n minus 1 p n minus 1 x plus a n plus 1 p n plus 1 x. Now, we have to put x equal to 1 when you put x equal to 1 we immediately get 2 n plus 1 equal to a n minus 1 plus a n plus 1. Remember that p k of 1 is 1 the normalization that we are talking about. So, we got an equation between a n minus 1 and a n plus 1 we need one more equation connecting a n minus 1 and a n plus 1 the way to do that would be to take the leading coefficient of both sides of this equation that you see in the very first line namely 2 n plus 1 times a leading coefficient of x p n x, but the leading coefficient of x p n x the same as a leading coefficient of p n x. The right hand side we will get a n minus 1 times a leading coefficient of p n minus 1 x plus a n plus 1 times a leading coefficient of p n plus 1 x. We already computed the leading coefficients of these Legendre polynomials explicitly you must put it in and you must get the second relationship between a n minus 1 and a n plus 1. So, now all in all they got 2 equations between a n minus 1 and a n plus 1 and that will completely determine a n minus 1 and a n plus 1 uniquely you put it in the formula and you will get the 3 term recursion formula. Now, let us go to the next item normalization constants. Now, remember that p m x and p n x in a product is 0 if m is not equal to n what happens when m is equal to n you are going to get norm p n squared. It is important to compute norm p n squared because that is a normalization constant and it will appear when you want to compute the Fourier coefficients. So, let us calculate this. Now, what you can do is you can apply Roderick's formula. Roderick's formula has a 1 upon 2 to the power n n factorial. So, to clear the denominators I get 4 to the power n n factorial squared times integral minus 1 to n p n x the whole squared dx is integral minus 1 to 1 nth derivative of x squared minus 1 to the power n multiplied by the same thing dx. We must now repeatedly integrate by parts you have to throw all the derivatives on one of the factors. All the derivatives must be removed from here and it must go to the second factor. We must perform an integration by parts n times and we already done it earlier and every time we integrate by parts there will be boundary terms and those boundary terms will drop out and the same thing is going to happen here also. When you integrate by parts the very first time you are going to become a minus sign minus 1 to 1. The derivative shifts from one of the terms to the other term you will get d n plus 1 x squared minus 1 to the power n d n minus 1 x squared minus 1 to the power n plus b 1 and b 1 is a collection of boundary terms that appear. The boundary terms will drop out. Let us examine why the boundary terms will drop out. What are the boundary terms? The boundary terms are going to be given by nth derivative of x squared minus 1 to the power n, n minus first derivatives of x squared minus 1 to the power n, x squared minus 1 to the power n has plus minus 1 has roots of multiplicity n. So, all derivatives of f up to and including order n minus 1 must vanish at plus minus 1 implying that the boundary terms b 1 must vanish. Do the integration by parts again. One more derivative will shift and we will get 4 to the power n n factorial squared integral from minus 1 to 1 p n x squared d x equal to integral minus 1 to 1 d n plus 2 x squared minus 1 to the power n times d of n minus 2 x squared minus 1 to the power n d x plus b 2 where d 2 the boundary terms again will vanish and so on. Finally, all the derivatives will shift to one of the terms n fold integration by parts integration by parts n times ultimately leads us to 4 to the power n n factorial squared integral minus 1 to 1 p n x squared d x will be minus 1 to the power n. Every time we integrate by parts you will pick up a minus sign and so all in all you have picked up n factors of minus 1 integral minus 1 to 1 d 2 n x squared minus 1 to the power n x squared minus 1 to the power n d x. Now, what are the 2 nth derivative of x squared minus 1 to the power n? It is simply 2 n factorial. So, we picked up a 2 n factorial here. So, integral minus 1 to 1 this minus 1 to the power n has been clubbed inside and written as 1 minus x squared to the power n d x and it is an even function twice the integral from 0 to 1 1 minus x squared to the power n d x. And what you do is that you put x squared equal to u you put x squared equal to u 2 x d x will be equal to d u. So, 2 d x will be d u by rho 2 u. So, this is what we get its integral from 0 to 1 1 minus u to the power n u to the power minus half. The beta function is staring at you and it can be written as a beta integral. I have recalled for you the definition of the beta function. We already did this when we derived the Ramanujam formula in the chapter on Fourier transform, but I had displayed it again for convenience here p is n plus 1 and the q is half. So, what is the formula? 4 to the power n n factorial squared integral minus 1 to 1 p n x the whole squared d x is 2 n factorial beta n plus 1 half. Use the beta gamma relation, what is the beta gamma relation? Gamma p gamma q equal to beta p q into gamma p plus q. That will give you 4 to the power n factorial integral minus 1 to 1 p n x the whole squared d x will be what happens is that you will get beta n plus 1 comma half will be equal to gamma n plus 1 which is n factorial into gamma half which is root pi divided by gamma n plus 3 by 2 that is what you get. And we have used the fact that gamma half is root pi. So, to simplify the expression further we use the relation gamma x plus 1 equal to x gamma x and you will get gamma n plus 3 by 2 will be equal to n plus half gamma n plus half. Gamma n plus half will be n minus half gamma n minus half so on. So, keep going and ultimately there will be another gamma half and the root pi root pi will cancel out and you will get this expression and this expression simplifies to 2 upon 2 n plus 1. So, for n equal to 0 we simply get that norm p naught squared is 2 for n equal to 1 we get that norm p 1 squared is 2 by 3 which checks out via direct calculations since p naught of x is 1 and p 1 of x equal to x. Always when you derive these things you can always double check with p naught x and p 1 x just to make sure that there are no errors in your computations. So, this calculates the norm of the Legendre polynomial the inner product of p n with itself with respect to the usual l 2 inner product. There are some more problems on the Legendre polynomials and here is a list of them. The first problem says p n prime 1 equal to 1 half n into n plus 1 this we will do in the next capsule. The second one says that p n plus 1 prime minus x p n prime equal to n plus 1 p n, third one x squared minus 1 p n prime minus n x p n plus n p n minus 1 equal to 0 p n plus 1 prime minus p n minus 1 prime equal to 2 n plus 1 p n. Now, this must be compared with the corresponding results for Bessel's function. You have a relationship between J p, J p minus 1 and J p plus 1 and 1 involving the derivative. Now, how do you do these things? For example, take the third one for instance x squared minus 1 p n prime that is a polynomial of degree n plus 1. So, you know that it must be a linear combination of p naught p 1 p 2 that is the p n plus 1. So, you write that linear combination and you figure out what are the coefficients or better still you can club the first two together x squared minus 1 p n prime minus n x p n they are together form a polynomial of degree n plus 1 and write it as a linear combination of p naught p 1 dot dot p n plus 1 and all the coefficients except p n plus 1 will be 0 and that coefficient will be n. You can use parity arguments to simplify your life and you should use normalization constants. You can use the Roderick's formula, you can use a three term recursion formula, you got a whole lot of things for you to use. So, the next problem says use the method of series solutions to find the power series of 1 plus x to the power a where a is any real number. Find an ODE satisfied by this function we will need this later. So, that is why I am putting this exercise here. This has nothing to do with Fourier series it is something that we will need to find the generating function for the sequence of Legendre polynomials. I think it is a good place to stop this capsule.