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Unit 2: Legendre’s Polynomials




                                                                                                Notes
          2.   Recurrence relations derived in this unit help us in finding unknown  P n ( )  in terms of
                                                                           x
               two or three known Legendre polynomial
                                              x
                                                                                     x
                   The Legendre’s polynomials  P n ( )  have zeroes at some  x  x i ,i  1, 2, ... i.e.  P 2 ( )
                    has two zeroes,  P 3 ( )  has three and so on.
                                   x
                   Legendre polynomials are quite suited in numerical evaluations of certain integrals.
          2.1 Legendre’s Differential Equation from Laplace’s Equation

          Laplace’s equation in spherical polar coordinates is

                      2 V    1          V     1   2 V
                     r             sin        2    2  0                            ...(A)
                   r    r   sin             sin
          Let us put

                                 n
                           V = r F n ( , )                                         ...(B)
          Here  F n ( , ) is a function of   and  . So

                          V
                             = n r  n  1  F
                          r           n
                          V      n F n
                             = r


                          2        2
                          V      n  F n
                           2  = r   2
          Substituting in Laplace equation, we get

                                      1            F    r  n  2 F
                            n r  n  1 F n   r  n  sin  n       n  = 0
                          r          sin               sin 2  2

                                        r  n       F n   r  n  F n
                                   n
          or                 ( n n  1)r F n   sin         2    2  = 0
                                       sin              sin
                     n
          Dividing by r , we have
                                        1          F     1   2 F
                               ( n n  1)F n   sin   n          n  = 0             ...(C)
                                       sin             sin 2   2

          Next consider the case when  F n ( , )  is independent of  , so

                                                   1         F
                                         ( n n  1) F n  sin   n  = 0              ...(D)
                                                 sin
          Let us put the independent variable   in terms of x given by
                                      x = cos

                                   d        F n  x      F n
                                     F n =         sin
                                   d        x           x



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