Page 150 - Determinants and Their Applications in Mathematical Physics

4.10 Henkelians 3 135

Now, perform the row and column operations
i−2
i − 2

R = (−1) r R i−r , i = n, n − 1,n − 2,..., 3,
r
i
r=0
j−1
j − 1

C = (−1) r C j−r , j = n − 1,n − 2,..., 2.

r
j
r=0
The result is
2 3 n−1

∆φ 0 ∆ φ 0 ∆ φ 0 ··· ∆ φ 0 1
2 3 4
n
∆ φ 0 ∆ φ 0 ∆ φ 0 ··· ∆ φ 0 ∆α 0
3 4 5 n+1 2
n
Y =(−1) ∆ φ 0 ∆ φ 0 ∆ φ 0 ··· ∆ φ 0 ∆ α 0 ,

...................................................
n ∆ n+1 ∆ n+2 ··· ∆ 2n−2 ∆ n−1
∆ φ 0 φ 0 φ 0 φ 0 α 0 n
where
φ m+1
0
∆ φ 0 = .
m
m +1
Transfer the last column to the first position, which introduces the sign
(−1) n+1 , and then remove powers of φ 0 from all rows and columns except
the first column, which becomes
2 n−1 T

∆α 0 ∆ α 0 ∆ α 0
1 2 ··· n−1 .
φ 0 φ 0 φ 0
The other (n − 1) columns are identical with the corresponding columns of
the Hilbert determinant K n . Hence, expanding the determinant by elements
from the first column,
n
n(n−1) (n) i−1
Y = −φ 0 K i1 ∆ α 0 φ n−i .
0
i=1
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 5.8
on algebraic computing.
4.10.4 A Particular Case of the Yamazaki–Hori Determinant
Let
A n = |φ m | n , 0 ≤ m ≤ 2n − 2,

where
x 2m+2 − 1
φ m = . (4.10.29)
m +1