Page 166 - Determinants and Their Applications in Mathematical Physics

4.11 Hankelians 4 151

n

f r c r
= D c
r=0
n
r−1
= f 1 + rf r c . (4.11.75)
r=2
Hence,
−1
f 1 = D c {c U(x, c )}
n
c=0
i+j−1
x − (−1) i+j −1
c
= D c c n
i + j − 1
n c=0

cx − (−1) i+j
i+j−1

i + j − 1
= D c
n c=0
n

= G n (x, 0,k)
k=1
= G n (x, 0), (4.11.76)
where G n (x, h, k) and G n (x, h) are defined in the first line of (4.11.55) and
(4.11.56), respectively.

E = G ,

(xE) =(xG )
2
= K n P , (4.11.77)
n
where
K n = K n (0),
P n = P n (x, 0)
D [x (1 + x) n−1 ]
n
n
= . (4.11.78)
(n − 1)!
Let
D [x n−1 (1 + x) ]
n
n
Q n = . (4.11.79)
(n − 1)!
Then,
P n (−1 − x)=(−1) Q n .
n
Since
E(−1 − x)= E(x),
it follows that
2
{(1 + x)E} = K n Q ,

n
2

{xE} {(1 + x)E} =(K n P n Q n ) . (4.11.80)