Okay, now, let's begin the chapter 3. The last chapter of this lecture. Okay, the chapter 3 is the special functions. Okay, in this chapter we just introduced only two concrete examples of special functions called the Bessel functions and Legendre polynomial, okay. Okay, there are many more different kinds of special functions, okay. So we are going to handle only two really special cases of special examples of. Okay, the large pool of the special functions. Okay, these two special functions, Bessel functions and Legendre polynomials, they pop up in many different situations from the applied sciences, okay. Including definitely the applied mathematics too, all right. And we are interested in this top of the special functions, not just because of mathematical interest but it has the quite a wide area of the applications, okay. And the naming, the special function is a kind of the historical scandal because some people, they prefer to use the useful functions instead of the special functions, okay. They claim that nothing special in those special functions. But they are just useful, okay, anyway. First the Bessel functions. Okay, consider the following differential equation. Second order polynomial coefficients, okay, extra scalar Y double prime plus xy prime plus extra scalar minus induce mu squared y is equal to 0. Okay, think about it. What's the nature of the point X is equal to 0? Can you see that X is equal to 0 is a regular singular point, right. This is a regular singular point of this differential equation. Okay, so we'd like to solve the differential equation for the positive real X's again, okay. Okay, sure, here the prominent view, okay. Mu is, okay, non negative the real constant, okay. Not negative parameter, okay. This differential equation has a name. It's called Bessel differential equation of order mu, okay. The Bessel differential equation over the mu, right? Since the X is equal to 0 is a regular singular point, we expect there is a at least one proving this type solution, right? So again we said, y is equal to sum of zero to infinity of summation of n, x to the n plus r, right, okay, and substituting this into the differential equation, this into the differential equation. And simplify those summations then you will get, okay. R squared minus mu squared. Time to C0 X to the r plus 1 plus 1 squared minus mu squared times the C1X to D r plus 1 plus sum of the 2 to the infinity. And plus r squared minus mu square times C sub n plus C sub n -2, okay, times X to the n+r, okay. This expression must be equal to 0, right? Again, from the very first coefficients, right, from this, because we are assuming C0 is a non-zero arbitrary constant, right? So that we have initial equation. R scalar minus mu squared is equal to 0, okay, from this. We have initials, r1 is equal 2, we have to indigenous roots right way, larger one mu, possibly smaller one. That is a negative mu, right, okay. We have such a to initials, okay. When mu is equal to 0 they coincide, otherwise, they are different, right? We know that for the larger initial roots r1 equal to mu, right. There is a solution of the frothiness tideway, so, okay. Set r1 equal to mu in this expression then. Okay, you will get 0 is equal to, this party will be 0 and that will be, okay. r plus 1 cubed minus the no, r1 is equal, mu plus 1 squared minus mu squared. So that you get 2 mu plus 1. Time to C1X2 mu plus 1 plus sum of the 2 to infinity of, okay, now we have, n squared, when r was equal to mu, okay, plus 2 mu time C sub n plus C sub n minus 2 times X to the n plus r, that must be equal to 0. Right, okay, then this must be equal to 0 and those coefficients must be equal to 0 right away. From this one because mu is non negative, 2 mu plus 1 is non negative, non-zero. Okay, so that's C1 must be equal to 0. Right away, okay, and we have a two term reconsideration down there. Okay, so the conclusion we have so far, okay is, okay. So we have a C1 equal to 0 and we have a two term reconsideration, okay, what is it? N squared plus 2n mu times C sub n plus C sub n-2 is equal to 0 for or n greater n equal to 2. Right, all right, look at this one or the. Equivalently, this means the Cn is equal to -7-2 over m times of n+2 mu, right? For all n greater than or equal to 2, okay? For C1=0 and apply this reconsideration, you will get immediately there, C1 and C3 and C5 and so on, they are equal to 0, right? All the other coefficients are equal to 0, okay? The even coefficients may survive, okay? So we need the reconciliation, total reconciliation when n is an even number, even integer, right? So now we have, okay? C of 2n is equal to, right? Or let me write it in the C of 2k, when n is equal 2k, okay? -C of 2k-2, 2k times 2k+2 mu, okay? That is -C sub 2k-2 over, 2 square times k times k+mu, right? Go one step further then you will get this is -1 over 2 square times k times k+mu. Another -1 times 2 square over k -1 times k+mu -1 times C of 2k -4, right? That's the pattern, right, okay? Keep going then, okay? You will get -1 over 2 square k k+mu, another -1 over 2 square k-1, k+mu-1, okay? And the so on. Finally, you should end up by when n is equal to C0, right? So you have a C0 down there and you should have 2 square there, all right? What's the pattern then, okay? Look at this one, right? This is a series of 2, okay? There is a pattern is minus C of 2 times of k -1 over 2 square times k times k+mu, right? So that you have C0 then -1 over 2 square, 1 larger 1 times this will be equal to what is it? K plus, I cannot read it here. K+mu, right? So this is equal to 1 times of k+mu, so mu+1, right, okay? Okay? So what is it then? This is equal to, how many negative ones do we have? 1, 2, 3 and so on to k, so you have -1 to the k, okay? So you have over 2 to the 2k, and what? K times k-1 times 1, that is k factorial. Then you have times the k+mu, k+mu-1, and so on. And finally the mu+1, right, okay? Or 1+mu instead, right? Time C of 0, right? That's the formula, right? That's something we can get from the recurrence formula, right? And then master V equal to C0 then it must be true for all k greater than or equal to 1, right, okay? Okay? To have a solution of the in a smart form, right? We need a clever choice of this under the constant C0, okay? We need the smart choice of this arbitrary constant C0, okay? It's a convention, right? You can take a C0 to be any non zero number but by convention we choose, okay? C of zero to be 1/2 to the mu gamma of 1+mu, right? Take C0 to be this one, okay? What is the gamma here? Here the gamma of alpha is equal to, this is the famous the gamma function integral from 0 to infinity of t to the alpha -1 times e to the -t dt, okay? For all positive alpha, okay? This is the so called the gamma function, okay? This is another example of this special function, okay? You can prove easily there, right? This improper integral converges for all positive values of alpha, okay? What is important? What is interesting for the gamma then? Alpha times the gamma of alpha. This is equal to gamma of alpha+1, okay? I would like to ask you to prove it, right? Prove this identity, okay? Simple exercises from calculus, okay? In particular, the gamma of n+1 is equal to for any non negative integer n, by this formula, this is n times the gamma of n. That is equal to n times n-1 times gamma of n-1, okay? Keep going then you're going to see that n(n-1) times and 1 times the gamma of 1 finally, okay? How much is gamma over 1? When alpha is equal to 1 this is the integral from 0 to infinity of e to the -t dt, that is the one way. This is 1. So what is this one? This is n factorial, right, okay? So gamma of n+1 for any non negative integer, say 0,1,2 and so on, that is just the factorial, right? So, another name for this gamma function is by this property, right? This is generalized Factorial function, okay? Okay, so using this notation, right, using this gamma function, okay? I'm choosing that C0 little to be 1 over 2 mu gamma over 1 plus mu, right? So let's plug in that one here, okay? So for the C0 I'm taking, this is 1 over 2 mu gamma of 1 plus mu, okay? Okay, this is the formula for C sub 2k, right, in general, right, for all k greater than equal to 1, right? Okay, let's erase this part, okay? Let's see what happen to the Ck, When we have such a choice, right, okay, for the C0, okay? So that this expression will be, okay, can you see that? This is the, first, -1 to the k into top, okay? This times that makes 2 to the 2k plus mu, k factorial survives, 1 plus mu times gamma 1 plus mu. That'll be gamma of 2 plus mu, right? Then you have 2 plus mu down there, multiply this one, it will gamma of 3 plus mu, right? Keep going then, finally you are going to get gamma of k plus mu plus 1, okay? Are you following me, right? That is the C sub 2k, right, okay? For all k greater than equal to 0, right? Greater than equal to 1 bar, okay? In fact, when k is equal to 0, this just coincide with our choice for the C0, right? This one, right? This formula, right? It holds for all k greater than equal to 1, but when k is equal to 0 it'll be reduced to exactly this one. So that it's true even for k is equal to 0, right, okay? Now we have the following conclusion now, right? So then, the one solution we expect from the Fibonacci method is, okay, y1 corresponding to the larger in this root r is equal to mu, right? Okay, we have it, right, here. Summation k is equal to 0 to infinity, okay, (-1) to the k, okay, C sub 2k, 2 to the 2k plus mu k factorial, gamma of k plus mu plus 1, okay, times x to the 2k, right, okay? Plus mu, okay, I forgot, okay? This and that you can combine it. So 0 to infinity -1 to the k over k factorial gamma of 1 plus mu plus k times x over 2 to the 2k plus mu, right? We have such a solution, okay? This expression, they call it as J sub mu of x, right? Okay, that's a notation, okay? And they call it, okay, call J mu of x the beta function of, Of the first kind of order mu, okay? That's the solution for the beta differential equation of order mu, right, okay? We can see that, okay, this infinity right? This infinity right? This infinity J mu x, right, is the summation 0 to infinity, blah, blah, blah. It converges for all nonnegative values of x, right? What's the second solution then, okay? For the other initial root, the small initial root r is equal to negative mu, right, okay? For r2 is equal to negative mu. To the same thing as we did for the r is equal to plus mu, okay? It's exactly the same thing, okay? Then you will get a solution, right? That is a J of minus mu, okay, simply replacing this positive mu by minus mu, okay? This -1 to the k, k factorial gamma of 1 minus mu plus k, and x over 2 to the 2k minus mu, okay, is also a solution, okay? This is another solution, okay, called the beta function of, Of first kind of all the minus mu, okay? But one thing you must be a little bit careful, okay? For mu strictly positive, okay, J minus mu of x, right, it may contain negative powers of x, okay, as you can see from this expression, right, okay? You may have, Some negative powers of x. Then what, okay? It is impossible for those function J minus mu of x converges to when x is equal to 0, right? So in that case, okay, J minus mu of x converges only on, okay, positive real axis, right? Not including 0, okay, not including 0, okay? The next time we'd like to discuss the following question, okay? J mu and the J minus mu, are they really linearly independent, okay? That'll be the things we will discuss in the next time, okay?