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Unit 3: Hermite Polynomials




                                                                                                Notes
                                           ( /2)     (2n  2 )!x n  2
                                           n
                                                          r
                                        =     ( 1) r  n             P n ( )
                                                                       x
                                                           r
                                                  2 ( )!(n  2 )!(n r )!
                                                     r
                                           n  0
          Hence,
                                           2       2
                                                 n
                                                         dt
                                     P n ( )    t e  t  H n ( ) .
                                        x
                                                       xt
                                             ! n  0
          Self Assessment
                                                         x
          12.  From recurrence relation II Obtain the value of  H 3 ( ). Given that
                                                     x
                                         x
                                      H  2 ( )  4x  2  2; H  1 ( ) 2x
          13.  Prove that
                                      x
                                                 x
                                   H n  ( ) 4nx H  n  1 ( ) 2n H  n ( ) 0
                                                          x
          14.  Prove that
                                          dH  3 ( )  6H  ( )
                                              x
                                                     x
                                            dx      2
          3.6 Summary
              Hermite  differential equation has no  finite singular  points except  x  .   Therefore
               Frobenius method involving a power series solution is obtained.
              There are two independent solutions corresponding to two different values of indicial
               power.
              For    n  a polynomial solution called Hermite polynomial is obtained.

              Hermite polynomials are seen to be generated by a generating function.
              Orthogonal properties of Hermite polynomials are obtained. It helps in expressing any
                                      x
               polynomial in terms of  H n ( ) .
              Recurrence  relations established help in  expressing every  polynomial as  well  as  its
               derivatives in terms of two or three Hermite polynomials.

          3.7 Keywords

          Boundary Conditions are the behaviour of the solution  of the differential equations in  the
          initial value of the independent variable as well as at the final value of independent variable.

          Frobenius Method: At an ordinary point as well as at regular singular point, helps in evaluating
          the solution as a power series.
          Orthogonality relations  of  Hermite  polynomials  are relations  involving  integrals  of  two
          Hermite polynomials. These relations help us to see that  H n ( )  form a complete set.
                                                            x
          Recurrence Relations are relations between two or three polynomials for all values of n and x.
          Rodrigue Formula Expresses  H n ( )  in an  alternative way than that of finding a solution of
                                      x
          differential  equations.



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