Legendre polynomials and Rodrigues formula

[00:00:03]Good morning everyone. So in this uh video we are going to learn about legendary polomials. So first of all we need to know what is legendary differential equation. So the legendary differential equation it is given as 1 - x² and the second derivative of y with respect to x - 2x and then first derivative of y with respect to x and then n + n into n + 1 y = 0. Okay, this particular equation is called legant differential equation. So why is it important for us to solve this kind of equation? So usually in physics we encounter this type of equation in the field of electrostatics and magnetism and even in quantum mechanics while solving for the scroinger equation for like hydrogen atom we usually encounter this type of equation. Okay. So this particular equation I have written in cartician coordinates and it has also different for in spherical coordinates also. So we can also encounter uh that uh particular form of legendary differential equation.

[00:01:17]So uh the most uh general method to find out the solution for this type of equation is by using the method froines method.

[00:01:32]So in this method what we do first we are going to consider a power series solution. Okay. So suppose the solution y that is equal to summation r is equal to 0 to infinity and then a r x ^ k minus r. Okay. So this is a power series solution that we have assumed. Okay. And in this method first we are going to find out what is the expression for dy by dx here. Okay. And uh we can find out the expression like this which is r =0 to infinity a r and then we are going to differentiate this uh x part. Okay. So after differentiation we got k - r and then x ^ k - r -1. So similarly we can also find out what is the expression dy by dx2.

[00:02:35]Okay. So after finding out this expression we are going to put this expression in these terms. Okay. And then we are going to solve this equation. Okay. So after solving the equation the solution that we are going to get are called legendary functions.

[00:02:54]Okay. And that legendary functions.

[00:03:06]These legendary functions are denoted by PN X and QN X. Okay. So this P and X are called legendary function of first kind and Q and X are called legendary function of second kind. Okay. So we are basically interested in what is PNX. Okay. From method is indeed a very long method and we can find out the legendary functions from that method also but it is very uh time-taking. So there is uh one more way in which we can find out what is uh the what are the values of this pnx and that is by using the generating function and this generating function it is given as 1 - 2 xh + h² to the power - 1x 2 Okay. So if we are doing the binomial expansion of this function then also we can get the legendary functions. Okay.

[00:04:22]So after expansion what we are going to do? We are going to look at the coefficients of h to the power n. So in this expansion we are going to encounter uh various powers of h. So we are going to look at the coefficients of only h to the power n and that coefficient of h to the power n will be our legendary polomials. But it is also a headache to find out the legendary functions in this method. So there is one more technique that is called uh rodrigue formula. Okay. So we can also use rodri's formula to find out the legendary polomials okay or legendary functions and this rodri's formula it is given as pn x that is equal to 1 by 2 ^ n and then n factorial and then dn by dx to dxn and x² - 1. Okay. So this p and x this is called legendary function and depending on different values of n we are going to get many polomials. Okay.

[00:05:55]like we can have p 0x is also a polomial and then p1 x is also a polomial and then p2x px and so on. Okay. So one of the most easiest ways to calculate this p 0 p1 p2 is by using the rod's formula.

[00:06:15]Okay. So let us uh calculate some of the uh values like uh first we are going to calculate what is p 0 x. Okay. So we can get p 0 x by putting simply n is equal to 0 in the above formula.

[00:06:34]Okay. So p 0 x this will be simply 1 by 2 ^0 and then 0 factorial and then d 0 by dx0 and then x² - 1. Okay. So 2 ^0 is simply 1 and then 0 factorial is also simply 1. And we are not going to do any derivative in uh in this particular part. Okay. Sorry, the formula is uh x² - 1 to the power n also. Okay. So we are going to put zero here also and this part is also 1. So we get our first polomial that is p 0x = 1.

[00:07:26]Similarly we can also calculate what is p1x.

[00:07:31]Okay. So in order to calculate p1x we are also going to use the same formula.

[00:07:37]Okay. So again it is 1 by 2 ^ 1 and then 1 factorial and then we are going to take the derivative of x² -1 to the power 1.

[00:07:55]Okay. So uh let us solve it. So here we got 1 by 2 and then the first derivative of the term x² - 1.

[00:08:08]So this will be 1 by 2 and then the derivative of x square will be 2x. Okay.

[00:08:15]And the derivative of negative of 1 it is zero. So two cancel out and then we got x. Okay. So the value of p1x that is x. Okay. So let us calculate one more value.

[00:08:34]So now we are going to calculate p2x also by using the same method. So just put the value of n as 2.

[00:08:47]So it will be 2 ^ 2 and then 2 factorial and then 2 * derivative of the x² -1 to the power 2. Okay. So we have 2 ^ 2 that is 4 and then we have 2 factorial that is simply two and then we are going to do the derivative twice and let us expand this uh particular part.

[00:09:19]So it will be um x ^ 4 + 1 - 2x².

[00:09:29]Okay.

[00:09:31]So it will be 1 by 8 and then after first derivative we are left with 4 x cq and then - 4 x.

[00:09:47]Okay.

[00:09:55]So let us take one more derivative and then we will get 4x cube will be simply 12 x² - 4. Okay. So let us divide this uh term just uh simplify it then we will get four okay first let us take it common okay it will be 1x 8 and then 12 x² - 4 okay so it can be written as 1x 8 and then take 4 common it will be 3 x² - 1 okay so 4 and then So we got 1 by 2 and then 3x² - 1. Okay. So simply by increasing the value of 3 we can also calculate what is p3x, p4x and p5x.

[00:10:56]Okay. So just I am writing it for you.

[00:10:58]So P3X it will be 1 by 2 5XQ - 3X and then P4X it will be around 1 by 8 and then 35 X ^ 4 - 30X² + 3. Okay. So you can try it out by putting the value of n as 3 and 4. Okay.

[00:11:29]So, usually we are encountered with many type of questions here. So, the most basic question that we are going to encounter it, we will be given some uh polomial and we are we will and we will be asked to write down that polomial in terms of legendary polomials. Okay. For example, we have a polomial like um 7 x + 4. Okay. So what we are going to do? We are going to write this particular polomial in terms of legendary polomials. Okay. So here we are only having the x term and a constant. Okay. So if we look at the polomials carefully then we will see that we have x term in p1x okay and just simply p p 0 x is a constant that is 1. Okay so we can write this particular term 7 x + 4 by using p1 and p 0. Okay.

[00:12:47]So we don't need any other polomial. So we have P1X that is equal to X. Okay. So here we are having 7 X. So 7 X will be 7 P1 X. Okay.

[00:13:07]And next we are having four. And we know P 0 X = 1. So four will be 4 P 0 X. Okay. So now we are just going to put the value in this particular expression.

[00:13:27]Okay. So after putting the value it will be simply 7 P1 X + 4 P 0 X. Okay. So this is our answer. Okay. So we have uh successfully converted this uh term 7 x + 4 into an expression of legendary polomials. Okay. So this was a very basic question. So if you are moving to more higher order like we are having terms like that has the x² in it like uh for example we have 3x² + 2x + 4. Okay. So now instead of x we are also having x² here. So if we look at this expression we can see that p2x has an x² term in it. Okay. So we are going to use p2 x p1 x and then p 0 x. Okay. So by using these three terms we can convert this particular expression into an expression of legendary polomials. Okay. So first uh let us find out what will be the value of x². Okay. So from the above we have already found out the expression for p2x. Okay. That is 1 by 2 and then 3x² - 1. Okay. So let us just uh a little bit modify it. So 3x² - 1 that is 2 p2x. Okay. And then 3x² that is equal to 2 p2x + 1. Okay. So here we are having 3x². So we don't need to we don't have to find the explicit value of x². So this will work for us.

[00:15:31]And next we are having 2x. Okay. And then we know that P1 X is equal to X. So 2X will be simply 2 * P1 X. Okay. So let us put uh these uh values in the above expression. And then instead of 3x² instead of 3x² we are going to write 2 p2 + 1. Okay.

[00:16:09]And then we have 2x. So instead of 2x we are going to write 2 p1 x. Okay. And then we have four here. So let us simplify it.

[00:16:24]So it will be 2 p2 + 2 p1 + 5. Okay. And we also know that p 0 x is 1. So 5 will be 5. P 0 X. So instead of five we are going to write 5 P 0X.

[00:16:47]So the final expression will be 2 P2 + 2 P1 and then 5 P 0 X. Okay. So this is our final expression for this particular term. Okay. So now we have converted 3x² + 2x + 4 into a combination of legendary polomials.

[00:17:15]Okay. So there is uh one moreh important uh thing that uh that is usually asked in the exams that is the orthogonal properties. Okay. So in this video I'm not going to prove the orthogonal property of these legendary polomials but instead I'm just going to write these statements. So you can have a vague idea about it. So the orthogonality the orthogonality of legendary polomials it is given as simply integral of minus1 to +1 and then we are having px and then pn x and then dx that is equal to zero. if our m is not equal to n. Okay.

[00:18:14]And then 2 by 2 n + 1 if our m is equal to n. Okay. So there is also a proof for this uh particular for this orthogonality. So we can use uh the uh generating function to prove uh this uh uh expression. Okay. So we are going to do it later and we are also going to solve uh the legendary differential equation using the provinus method in further videos. So uh for this uh video we are just going to learn how to generate the legendary polomials because this is the most important uh part in legendary differential equation. Okay. So also there are some of the important relations uh that we have in legendary polomials and uh I'm uh stating these uh relations uh to you guys like uh we have u uh the first uh important relation is that PN of 1 is equal to 1.

[00:19:36]And uh second relation is if we are having instead of x if we are having negative of x then it will be -1 to the power n pn x. Okay. And if we are having the derivative then the derivative of PN 1 that is n into n + 1 by 2 and if we are taking the derivative of pn of minus1 then it will be -1 to the power n + 1 n + 1 and then divided by 2. Okay. So these are some of the important relations.

[00:20:33]So we are going to have uh further videos in which we are going to prove these relations also. Okay. So for this particular uh video it was a short videos in which you learned about the legendary functions and the uh and I have also stated the generating functions and how to generate these uh legendary polomials using the Rodriguez formula and uh this formula is actually the simplest method to generate the uh legendary polomials and we have also generated the first few polomial like P 0 X, P1X and then P2X and we have also taken uh two different expression like 7X + 4 and then we have converted it uh or we have rewritten this particular expression in terms of legendary polomials and we have also taken a quadratic term also that is 3x² + 2x + 4 and we have also successfully written it in terms of legendary polomials and I have stated about the orthogonality relations of these legendary polomials that is pmx pn x that is equal to zero if m is not equal to n and if m is equal to n that that value will be 2 by 2n + 1 and these are some of the few few important relations that you need to take a note down okay so for this uh video u this much and uh thank you. See you in the next one.