Page 70 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 3: Hermite Polynomials

We have Notes
2 2
2tx t
e = e 2tx . e t
s
2
(2 ) r t
tx
=
! r ! s
r 0 s 0
(2 ) r
x
= ( 1) s .t r 2s
s
r ! !
r 0,s 0
n
Coefficient of t (for fixed value of s)
[obtained by putting r + 2s = n, i.e., r = n 2s]
(2 ) n 2s
x
s
= ( 1)
s
(n 2 )! !
s
n
The total value of t is obtained by summing over all allowed values of s, and since r n 2s
n 2s 0 or s n /2
Thus if n is even s goes from 0 to n/2 and if n is odd, s goes from 0 to (n - 1)/2.

n
( /2) n 2s
x
Coefficient of t n = ( 1) s (2 )
(n 2 )! !
s
s
s 0
x
H ( )
= n
! n
2 t n
2tx t
Hence e = H n ( )
x
! n
n 0
t n
x
2 (t x ) 2 H ( ).
x
e = ! n n
n 0
Other form for the Hermite Polynomials
Prove
1 d 2
x
H ( ) = 2 n n ...(i)
n exp 2 x
4 dx
We have
1 d 2tx 2tx
e = te
2 dx

d 1 d 2tx 2 2tx
e = 2t e
dx 2 dx

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