4.11 Hankelians 4 151
=Dc
n
∑
r=0
frc
r
=f 1 +
n
∑
r=2
rfrc
r− 1
. (4.11.75)
Hence,
f 1 =
[
Dc{c
n
U(x, c
− 1
)}
]
c=0
=Dc
[
c
n
∣
∣
∣
∣
x
i+j− 1
−(−1)
i+j
c
− 1
i+j− 1
∣
∣
∣
∣
n
]
c=0
=
[
Dc
∣
∣
∣
∣
cx
i+j− 1
−(−1)
i+j
i+j− 1
∣
∣
∣
∣
n
]
c=0
=
n
∑
k=1
Gn(x, 0 ,k)
=Gn(x,0), (4.11.76)
whereGn(x, h, k) andGn(x, h) are defined in the first line of (4.11.55) and
(4.11.56), respectively.
E=G
′
,
(xE)
′
=(xG
′
)
′
=KnP
2
n
, (4.11.77)
where
Kn=Kn(0),
Pn=Pn(x,0)
=
D
n
[x
n
(1 +x)
n− 1
]
(n−1)!
. (4.11.78)
Let
Qn=
D
n
[x
n− 1
(1 +x)
n
]
(n−1)!
. (4.11.79)
Then,
Pn(− 1 −x)=(−1)
n
Qn.
Since
E(− 1 −x)=E(x),
it follows that
{(1 +x)E}
′
=KnQ
2
n
,
{xE}
′
{(1 +x)E}
′
=(KnPnQn)
2
. (4.11.80)