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Differential and Integral Equation




                    Notes          First few Hermite Polynomials from Rodrigue s Formula

                                                               x
                                   From Rodrigue s Formula for  H n ( )
                                                                         2 d n  x  2
                                                                      n x
                                                             x
                                                          H n ( ) = ( 1) e  .  n  e
                                                                          dx
                                   Putting n = 0, 1, 2, 3, ... we get

                                                             x
                                                           H  ( ) =  e x 2 . e  x 2  1
                                                            0
                                                                        2 d  x 2
                                                                       x
                                                             x
                                                           H 1 ( ) = ( 1)e  e    2x
                                                                         dx
                                                                         2 d  2  2  2 d      2
                                                                      2 x
                                                             x
                                                          H  2 ( ) = ( 1) e  2  e  x  e x  2xe  x
                                                                          dx          dx
                                                                     2  2  x  2  x  2
                                                                    x
                                                                 = e  4x e    2e
                                                                 =  4x 2  2 .

                                                                         2 d 3  2
                                                                      3 x
                                                           H 3 ( ) = ( 1) e  e  x
                                                             x
                                                                          dx  3
                                                                      2 d    2    x 2
                                                                     x
                                                                 =  e      4x  2 e
                                                                       dx
                                                                 =  e x  2  2x  4x  2  2 e  x 2  8xe  x 2


                                                                                             x
                                                                 =  e x  2  8x 3  12x e  x 2  8x 3  12 .

                                                             x
                                   Similarly,             H 4 ( ) = 16x  4  48x 2  12 etc.
                                   3.4 Orthogonal Properties of Hermite Polynomials

                                   Prove


                                                            x
                                                 e  x 2 H n ( )H m ( )dx =  0,  if m  n
                                                       x
                                                                        2
                                                                       2 ( )! if m  n
                                                                         n
                                   We have                e  t  2  2tx  =  H  ( ) t n
                                                                         x
                                                                       n
                                                                   n  0     ! n
                                                                         x
                                   and                    e  s 2  2sx  =  H  m ( )  s m
                                                                   m  0    m !
                                                                           t  n     s  m
                                                       2  2tx  s 2  2sx
                                                                                  x
                                                                         x
                                                      t
                                                     e    e      =    H  n ( )  H  m ( )
                                                                   n  0     ! n  m  0  m !
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