The Legendre and the associated 
                        differential equation 
          
                  This is an article from my home page: www.olewitthansen.dk 
          
          
          
          
                                      
                                                     
          
          
          
          
          
          
          
          
          
          
          
          
          
          
          
          
          
          
          
          
         Ole Witt-Hansen                                         (2020) 
                           
     Contents 
      
       1. The Legendre equation.............................................................................................................1 
       2. Series expansion solution to the equation................................................................................1 
       3. Rodrigues formula....................................................................................................................4 
       3.1 Differentiating a product n-times...........................................................................................4 
       3. Proof of Rodrigues formula.....................................................................................................5 
       4. Associated Legendre polynomials:..........................................................................................7 
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
                                     The Legendre and associated Polynomials                       1        
           1. The Legendre equation 
           The Legendre equation comes frequently up in physics, and dealing with the properties of the 
           solutions is a substantial part of most standard books in mathematical physics. 
           However the Legendre polynomials are rather complex mathematics, and most often only the 
           results are presented, but not the detailed derivation. The Legendre differential equation reads: 
            
                                            2
                                          d y      dy
                                        2
           (1.1)                    (1 x )   2x    n(n1)y0  
                                             2
                                          dx       dx
            
           There are two linear independent classes of solutions to the equation: P (x) and Q (x), we shall 
                                                                          n         n
           here only occupy ourselves with P (x). P (x) are called Legendre polynomials.  
                                          n     n
           The most compact way to state the solution to the Legendre equation is the Rodrigues formula: 
            
                                                    n
                                            1    d
                                                  
                                                       2   n
           (1.2)                     P (x)          (x 1)  
                                                  
                                     n
                                            n
                                           2 n! dx
                                                  
                                    
           Unfortunately it is not possible to verify Rodrigues formula directly by insertion. 
           One has either to use the series expansion solution to the Legendre equation together with applying 
           the binomial formula on Rodrigues equation or some mathematical tricks, as we shall see below. 
            
           Especially in connection with physics n is often replaced with l. In connection with the series 
           expansion, it is in fact necessary. 
            
                                            2
                                          d y      dy
                                        2
                                    (1 x )   2x    l(l 1)y  0 
                                             2
                                           dx      dx
            
           2. Series expansion solution to the equation 
           Assume a series expansion solution: 
            
                                                   2     3     4        n
                                    y  a  a x  a x  a x  a x ...a x ... 
                                        0   1    2     3     4        n
                                                     2      3          n1
                                    y' a 2a x3a x 4a x ...na x     ... 
                                        1    2     3      4          n
                                                        2               n2
                                    y'' 2a 6a x12a x ...n(n1)a x     ... 
                                          2    3      4               n
                                    
                                        2                     2                n2
                                    (1 x )y'' 2a 6a x12a x ...n(n1)a x    ...
                                                2    3      4                n
                                                                                      
                                          2      3       4               n
                                    (2a x 6a x 12a x ...n(n1)a x ...
                                        2      3       4               n
                                                      2      3      4            n
                                    2xy'2xa 4a x 6a x 8a x ...2na x ... 
                                               1    2      3      4            n
                                                     2           3            4              n
            n(n1)y n(n1)a n(n1)a xn(n1)a x n(n1)a x n(n1)a x ...n(n1)a x ...
                             0          1          2           3            4              n
              
           We put the three terms in the Differential equation together, collecting the coefficients of the 
           various terms having the power of x. Since the collected series is identically zero, each of the 
           coefficients to a certain power of x must necessarily be zero. It is rather messy to complete this  
                                                               The Legendre and associated Polynomials                                                                   2              
                   task, but below we have established a schema for the powers from 0 to 3 and for n. 
                    
                    
                    
                    
                                                                    2            3           n 
                                 const                 x          x            x           x
                   y'’ 
                                  2a                   6a          12a          20a         (n2)(n1)a                
                                      2                    3            4            5                           n2
                      2
                   -x y’’                               
                                                                   2a   6a   n(n1)a  
                                                                          2           3                     n
                   -2xy’          
                                                       2a   4a   6a   2na  
                                                              1           2           3             n
                                                       l(l 1)a  
                                                                   l(l 1)a              
                                                              1            2
                                                                               l(l 1)a
                   l(l+1)y 
                                  l(l 1)a                                             3    l(l 1)a  
                                              0                                                        n
                                                
                                                                   
                    
                   For the first four terms we get: 
                                                                              l(l 1)
                   const:   2a l(l 1)a  0  a                                       a  
                                   2                0                 2                    0
                                                                                   2
                                                                                           (l 1)(l  2)
                    x:         6a 2a l(l 1)a 0                      a                                 a  
                                  3        1               1                      3                            1
                                                                                                   6
                    
                                                                                             (l  2)(l  3)                l(l 1)(l  2)(l  3)
                      2
                    x :        12a 2a 4a l(l1)a                         a                                a                                     a  
                                    4        2        2                2            4                             2                                       0
                                                                                                    12                                  4!
                    
                                                                                     (l  4)(l 3)
                      3
                    x :        20a 12a l(l 1)a                    a                               a
                                     5         3                3            5                            3
                                                                                            20
                                                                                                                              
                                                                                     (l 1)(l 3)(l  2)(l  4)
                                                                                                                        a
                                                                                                                           1
                                                                                                      5!
                    
                                                               2           2
                               (n2)(n1)a                (l ln n)a 0 
                     n                               n2                             n
                   x  :                                                                                 
                               (n2)(n1)a                (l n)(l  n1)a 0
                                                     n2                              n
                    
                                                                                     (l  n)(l  n 1)
                        (2.1)                                           a                                  a  
                                                                         n2                                   n
                                                                                       (n2)(n1)
                    
                   The solution then separates in two series, one for even power of x, and one for odd power of x, 
                   corresponding to the solutions P (x). 
                                                                    n
                    
                                                        l(l 1)           l(l 1)(l  2)(l  3)
                                                                                                                
                                                                     2                                   4
                                         y  a 1                  x                                  x ... 
                                                                                                                
                                                0
                                                            2!                         4!
                                                                                                                
                                                                                                                                     
                                                         (l 1)(l  2)             (l 1)(l  2)(l 3)(l  4)
                                                                                                                                 
                                                                              3                                           5
                                               a x                         x                                          x ...
                                                                                                                                 
                                                 1
                                                                 3!                                 5!
                                                                                                                                 
                                         
                   It is important to notice that for integer l, the solutions to Lagrange’s differential equation are 
                   polynomials. This follows from the recursion relation: