4.11 Hankelians 4 143b.
∣
∣
∣
∣
1
(i+j−2)!∣
∣
∣
∣
=
(−1)
n(n−1)/ 2
1! 2! 3!···(n−2)!n!(n+ 1)!···(2n−2)!.
The second determinant is Hankelian.
Proof. Denote the first determinant byAn. Every element in the last
row ofAnis equal to 1. Perform the column operations
C
′
j=Cj−Cj−^1 ,j=n, n−^1 ,n−^2 ,...,^2 , (4.11.30)which remove all the elements in the last row except the one in position
(n,1). After applying the binomial identity
(
nr)
−
(
n− 1r)
=
(
n− 1r− 1)
,
the result is
An=(−1)n+1∣
∣
∣
∣
(
n+j− 2n−i− 1)∣
∣
∣
∣
n− 1. (4.11.31)
Once again, every element in the last row is equal to 1. Repeat the column
operations withj=n− 1 ,n− 2 ,...,2 and apply the binomial identity
again. The result is
An=−∣
∣
∣
∣
(
n+j− 2n−i− 2)∣
∣
∣
∣
n− 2. (4.11.32)
Continuing in this way,
An=+∣
∣
∣
∣
(
n+j− 2n−i− 4)∣
∣
∣
∣
n− 4=−
∣
∣
∣
∣
(
n+j− 2n−i− 6)∣
∣
∣
∣
n− 6=+
∣
∣
∣
∣
(
n+j− 2n−i− 8)∣
∣
∣
∣
n− 8=±
∣
∣
∣
∣
(
n+j− 22 −i)∣
∣
∣
∣
2=± 1 , (4.11.33)sign(An)={
1 whenn=4m, 4 m+1−1 whenn=4m− 2 , 4 m−1,(4.11.34)
which proves (a).
Denote the second determinant byBn. DivideRiby (n−i)!, 1≤i≤n−1,and multiplyCjby (n+j−2)!, 1≤j≤n. The result is
(n−1)!n!(n+ 1)!···(2n−2)!(n−1)! (n−2)! (n−3)!···1!Bn=∣
∣
∣
∣
(n+j−2)!(n−i)! (i+j−2)!∣
∣
∣
∣
n