4.9 Hankelians 2 121Proof. The sum formula forTcan be expressed in the form
n
∑j=1ψr+i+j− 1 T(n,r)
ij
=−δinT(n+1,r)
n+1,n, (4.9.10)
C
′
j=[
(r+j)ψr+j(r+j+1)ψr+j+1···(r+j+n−1)ψr+j+n− 1]T
n.(4.9.11)
Let
C
∗
j=C′
j−(r+j)Cj+1=
[
0 ψr+j+1 2 ψr+j+2···(n−1)ψr+j+n− 1]T
n. (4.9.12)
Differentiating the columns ofT,
T
′
=n
∑j=1Uj,where
Uj=∣
∣C
1 C 2 ···C
′
j
Cj+1···Cn∣
∣
n, 1 ≤j≤n.Let
Vj=∣
∣C
1 C 2 ···C
∗
j
Cj+1···Cn∣
∣
n, 1 ≤j≤n=
n
∑i=2(i−1)ψr+i+j− 1 Tij. (4.9.13)Then, performing an elementary column operation onUj,
Uj=Vj, 1 ≤j≤n− 1Un=∣
∣C
1 C 2 ···Cn− 1 C′
n∣
∣
=
∣
∣C
1 C 2 ···Cn− 1 C∗
n∣
∣+(r+n)∣
∣C
1 C 2 ···Cn− 1 Cn+1∣
∣
=Vn−(r+n)T(n+1,r)
n+1,n. (4.9.14)
Hence,
T
′
+(r+n)T(n+1,r)
n+1,n=
n
∑j=1Vj=
n
∑j=1(i−1)n
∑j=1ψr+i+j− 1 Tij=−T
(n+1,r)
n+1,nn
∑i=2(i−1)δin=−(n−1)T(n+1,r)
n+1,n.The theorem follows.