Page 151 - Determinants and Their Applications in Mathematical Physics

136 4. Particular Determinants

Theorem.
2
2
A n = K n (x − 1) , K n = K n (0).
n
Proof.
2
φ 0 = x − 1.
Referring to Example A.3 (with c = 1) in the section on differences in
Appendix A.8,

φ m+1
0
∆ φ 0 = .
m
m +1
Hence, applying the theorem in Section 4.8.2 on Hankelians whose elements
are differences,
m
A n = |∆ φ 0 | n
m+1
φ
= 0
m +1
n
1 2 1 3 1
n
φ 0 φ 0 φ 0 ··· φ 0
2 3 n

1 φ 2 1 φ 3 1 φ 4 ··· ···
2 0 3 0 4 0
= 1 3 1 φ 4 1 φ 5 ··· ··· .
φ
3 0 4 0 5 0

..................................
1 1 2n−1
φ n ··· ··· ··· φ 0
0
n 2n−1 n
Remove the factor φ from row i,1 ≤ i ≤ n, and then remove the factor
i
0
j−1
φ
0 from column j,2 ≤ j ≤ n. The simple Hilbert determinant K n
appears and the result is
(1+2+3+···+n)(1+2+3+···+n−1)
A n = K n φ 0
2
= K n φ ,
n
0
which proves the theorem.
Exercises
1. Define a triangular matrix [a ij ], 1 ≤ i ≤ 2n − 1, 1 ≤ j ≤ 2n − i,as
follows:
2
column 1 = 1 uu ··· u 2n−2 T ,
2 2n−2
row 1 = 1 vv ··· v .
The remaining elements are defined by the rule that the difference be-
tween consecutive elements in any one diagonal parallel to the secondary
diagonal is constant. For example, one diagonal is

1 3 3 1 3 3 3
3
u (2u + v ) (u +2v ) v
3 3