Page 78 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 3: Hermite Polynomials

Solution: We have Notes

t n t 2 2tx
H n ( ) = e ...(i)
x
! n
n 0
t n d m H ( ) = d m e t 2 2tx
x
! n dx m n dx m
n 0
= (2 )t m e t 2 2tx

t n
x
= (2 )t m H n ( ) (from (i))
! n
n 0
1
x
= 2 m t n m H n ( )
! n
n 0
Putting n m r n r m , for n 0;
,
r m , for n , r ,

t n d m H ( ) = 2 m 1 t H ( ) ...(ii)
r
x
x
! n dx m n (r m )! r m
n 0 r m
Equating the coefficient of t from the two sides, we have
n
1 d n H ( ) m 1
x
x
! n dx m n = 2 (n m )! H n m ( )
m
d n 2 n !
x
H n ( ) = H ( ) Q.E.D.
x
dx m (n m )! n m
(2 )!
n
n
Example 4: Prove that H (0) ( 1) . and (ii) H (0) 0
2n 2n 1
! n
Solution: We have
t n t 2 2tx
H n ( ) = e
x
! n
n 0
Putting x = 0
t n t 2
H n (0) = e
! n
n 0
t
( ) 2 ( ) n
t
= 1 t 2 ... ( 1) n ... ...(1)
2! ! n

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