Page 66 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 3: Hermite Polynomials

Notes
which implies that a 1 0 or k 0 or both are zero, since k 1 0 for any value of k given
by (iv).

Now equating to zero the coefficient of general term, i.e., x k r in (iii), we get
a
a r 2 (k r 2)(k r 1) = 2 (k r ) 0
r
2(k r )
or a r 2 = (k r 2)(k r 1) .a r

2(k r ) 2
or a r 2 = (k r 2)(k r 1) .a r ...(vi)

Now two cases arise—
Case I: when k = 0, then from (vi), we have
2r 2
a r 2 = (r 2)(r 1) a r ...(vii)

Putting r = 0, 2, 4, etc. in (vii), we have

2 2
a 2 = a 0 a 0
2.1 2!
4 2 (4 2 ).2
a 4 = a 2 a 0
4.3 4.3.2!
2
2 ( 2 ) 2 2 ( 2)
= a 0 a 0
4! 4!
and so on.

( 2) m ( 2) ( 2m 2)
a 2m = (2 )! a 0 .
m
Again putting r = 1, 3, 5, etc.
2 2 2( 1)
a 3 = a 1 a 1
3.2 3!

6 2
a 5 = 5.4 a 3

2(6 2 )( 1)
= a 1
5.4.3.2

2 ( 1)( 3)
= ( 2) a 1
5!
and so on.
m
( 2) ( 1)( 3) ( 2m 1)
a 2m 1 = (2m 1)! a 1

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