4.10 Henkelians 3 135Now, perform the row and column operations
R
′
i=
i− 2
∑r=0(−1)
r(
i− 2r)
Ri−r,i=n, n− 1 ,n− 2 ,..., 3 ,C
′
j=j− 1
∑r=0(−1)
r(
j− 1r)
Cj−r,j=n− 1 ,n− 2 ,..., 2.The result is
Y=(−1)
n∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∆φ 0 ∆
2
φ 0 ∆
3
φ 0 ··· ∆
n− 1
φ 0 1∆
2
φ 0 ∆
3
φ 0 ∆
4
φ 0 ··· ∆
n
φ 0 ∆α 0∆
3
φ 0 ∆
4
φ 0 ∆
5
φ 0 ··· ∆
n+1
φ 0 ∆
2
α 0...................................................∆
n
φ 0 ∆
n+1
φ 0 ∆
n+2
φ 0 ··· ∆
2 n− 2
φ 0 ∆
n− 1
α 0∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n,
where
∆
m
φ 0 =φm+1
0m+1.
Transfer the last column to the first position, which introduces the sign
(−1)
n+1
, and then remove powers ofφ 0 from all rows and columns exceptthe first column, which becomes
[1
∆α 0φ 0∆
2
α 0φ
2
0···
∆
n− 1
α 0φn− 1
0]T
.
The other (n−1) columns are identical with the corresponding columns of
the Hilbert determinantKn. Hence, expanding the determinant by elements
from the first column,
Y=−φn(n−1)
0n
∑i=1[
K
(n)
i 1∆
i− 1
α 0]
φn−i
0.
The proof is completed with the aid of (4.10.5) and (4.10.8) and the formula
for ∆
i− 1
α 0 in Appendix A.8. Further notes on the Yamazaki–Hori determinant appear in Section 5.8on algebraic computing.
4.10.4 A Particular Case of the Yamazaki–Hori Determinant
Let
An=|φm|n, 0 ≤m≤ 2 n− 2 ,where
φm=x2 m+2
− 1m+1