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Differential and Integral Equation




                    Notes          Self Assessment

                                                                                                     x
                                   9.  Using Rodrigue s Formula derive the Hermite s polynomials  H 2 ( ) and H 3 ( )
                                                                                             x
                                   10.  Evaluate
                                                                      2
                                                                     x
                                                                               x
                                                                          x
                                                                  x e  H 2 ( ) H  1 ( )dx
                                   11.  Evaluate
                                                                            x
                                                                     x e  x  2 H  2 ( )dx

                                   3.5 Recurrence Formula for Hermite Polynomials


                                   (I)  Prove
                                                         d
                                                          H  n ( ) = 2n H n  1 ( )   for n  1
                                                             x
                                                                          x
                                                        dx
                                       We have from generating function
                                                         H n ( )t n  2xt t  2
                                                            x
                                                            ! n  = e                                       ...(i)
                                                      n  0
                                       Differentiating both sides with respect to x, we have

                                                        n
                                                             x
                                                       t dH  n ( )  2t e  2xt t  2
                                                        n  dx    =
                                                     n  0
                                                                           x
                                                                        H n ( )t n
                                                                 =  2t     ! n
                                                                     n  0
                                                                       H x    1
                                                                         ( ) n
                                                                        n
                                                                 =  2     ! n  t                    Let n  1 n
                                                                    n  0
                                                                       H  n  1 ( )t n
                                                                            x
                                                                 =  2    n  1 !
                                                                    n  1
                                   or

                                                        n
                                                       t dH  ( )       H   ( )t  n
                                                                           x
                                                             x
                                                            n      2    n  1
                                                        n  dx    =      (n  1)!                            ...(ii)
                                                     n  0           n  1
                                                  n
                                       Comparing  t  on both sides we have
                                                                         x
                                                                                                     x
                                                             x
                                                          H n ( )  =  2  H n  1 ( )         Here   dH n ( )  H  ( )
                                                                                                            x
                                                             ! n     (n  1)!                       dx     n
                                   or


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