Page 72 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 3: Hermite Polynomials

Notes
x
8. H 1 ( ) 2xH 0 ( ) is
x
(a) positive
(b) zero
(c) negative
(d) none of the above

x
3.3 The Rodrigue s Formula for H ( )
n
To Prove

2 d n 2
n x
H n ( ) = ( 1) e e x ...(i)
x
dx n
Proof:

( ) n
n
We have e 2tx t 2 = H x t
! n
n 0
x
x
x
or e x 2 (t x ) 2 = H 0 ( ) t 0 H 1 ( ) t H 2 ( ) t 2
0! 1! 2!
x
H x t H n 1 ( ) n 1
( ) n
n
t
! n (n 1)!
Differentiating both sides, partially with respect to t, n times and then putting t = 0, we have
n
x
H n ( ) ! n e (t x ) 2 e x 2
! n = t n ...(ii)
t 0
Now let t x , i.e., at t 0,x

t

n 2 n
or e (t x ) = e 2
t n n
t 0
n 2
= ( 1) n e x
x n

n d n x 2
= ( 1) n e
dx
2 d n x 2
n
x
x
H ( ) = ( 1) . e e ...(A)
n n
dx

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