Page 45 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION
P. 45

Differential and Integral Equation




                    Notes          Self Assessment

                                   1.  Find
                                               x
                                                    x
                                          x
                                        P 1 ( ), P 2 ( ), P 3 ( )
                                       from Rodrigue formula
                                   2.3 Generating Function for Legendre Polynomials

                                   In the following we will show that  P n ( )  is the coefficient of h  in the expansion of
                                                                                     n
                                                                  x
                                                                    1
                                                        (1 2xh h 2 )  2

                                   for                  | | |,| | |
                                                              h
                                                         x
                                                              1
                                                                       n
                                                                           x
                                   i.e.            (1 2xh  h 2  )  2  =  h P n ( )                         ...(A)
                                                                   n  0
                                                              1               1
                                   Now              (1 2hx h  2 )  2  = [1 h (2x h )]  2
                                                           1           1 3 1  2      2
                                                      = 1   ( h )(2x h )  .  h  (2x h )  ...
                                                           2           2 2 2
                                                             1.3...(2n  3)  n  1  n  1
                                                          ...         h   (2x h )
                                                             2.4.6(2n  2)
                                                          1.3. ...(2n  1)  n  n
                                                                    h  (2x h )  ...
                                                                  n
                                                           2.4.6....(2 )
                                                           n
                                   Therefore the coefficients of h  are
                                          1.3.5...(2n  1)  2  1.3.5. ...(2n  3)  n  2  1.3.5...(2n  5)  n  4
                                       =             (2 )              (2 )  n  1 C           n  2 (2 )    ...(B)
                                                                                                     x
                                                       x
                                                                         x
                                                                                                  C
                                                  n
                                            2.4.6...(2 )   2.4.6 ...(2n  2)      1  2.4.6...(2n  4)  2
                                                                          n
                                          1.3.5...(2n  1)  n  2n    x n  2  (2 )(2n  2)(n  2)(n  3)
                                       =              x       (n  1)  2                  4   ...
                                               n          2n  1     2     (2n  1)(2n  3) 2 2
                                          1.3.5...(2n  1)  n  ( n n  1)  n  2  ( n n  1)(n  2)(n  3)  n  4
                                       =              x          x                   x    ...
                                               n          2(2n  1)    2.4.(2n  1)(2n  2)
                                       = P n ( )                                                           ...(C)
                                             x
                                   Thus
                                                              1       h P  ( )
                                                                       n
                                                                          x
                                                    (1 2xh h  2 )  2  =  n
                                                                   n  0
                                   Where  P n ( )  is given by
                                           x
                                                 1.3.5...(2n  1)  n  ( n n  1)  n  2  (n  2)(n  1) n  (n  3)  n  4
                                         P n ( ) =     n     x   2(2n  1)  x  2.4.(2n  1)(2n  2)  x  ...   ...(D)
                                           x




          38                                LOVELY PROFESSIONAL UNIVERSITY
   40   41   42   43   44   45   46   47   48   49   50