4.11 Hankelians 4 151=Dcn
∑r=0frcr=f 1 +n
∑r=2rfrcr− 1. (4.11.75)
Hence,
f 1 =[
Dc{cn
U(x, c− 1
)}]
c=0=Dc[
cn∣
∣
∣
∣
xi+j− 1
−(−1)i+j
c− 1i+j− 1∣
∣
∣
∣
n]
c=0=
[
Dc∣
∣
∣
∣
cxi+j− 1
−(−1)i+ji+j− 1∣
∣
∣
∣
n]
c=0=
n
∑k=1Gn(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′
)′=KnP2
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
[xn− 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}′
=KnQ2
n,
{xE}′
{(1 +x)E}′
=(KnPnQn)2. (4.11.80)