Page 182 - Determinants and Their Applications in Mathematical Physics

4.13 Hankelians 6 167

the upper limits that

j
i
r+s−1
γ ij = β ir 2 k r+s−2 β js .
r=1 s=1
Hence,
2p+1 2q+1
r+s−2
γ 2p+1,2q+1 =2 β 2p+1,r 2 k r+s−2 β 2q+1,s . (4.13.10)
r=1 s=1
From the first line of (4.13.6), the summand is zero when r and s are even.
Hence, replace r by 2r + 1, replace s by 2s + 1 and refer to (4.13.5) and
(4.13.6),

p q

γ 2p+1,2q+1 =2 β 2p+1,2r+1 β 2q+1,2s+1 2 2r+2s k 2r+2s
r=0 s=0
p q
N

=2 λ pr λ qs a j (2x j ) 2r+2s
r=0 s=0 j=1
p q
N

=2 a j λ pr (2x j ) 2r λ qs (2x j ) 2s
j=1 r=0 s=0
N

=2 a j g p (x j )g q (x j )
j=1
= α 2p+1,2q+1 , (4.13.11)
which completes the proof of case (i). Cases (ii)–(iv) are proved in a similar
manner.

Corollary.

2
|α ij | n = |M| n = |N| |K| n
n
2
= |β ij | |2 i+j−1
n k i+j−2 | n
2

n
i+j−2
= β ii 2 |2 k i+j−2 | n . (4.13.12)
n
i=1
1
But, β 11 =1 and β ii = , 2 ≤ i ≤ n. Hence, referring to Property (e) in
2
Section 2.3.1,
2
n −2n+2
|α ij | n =2 |k i+j−2 | n . (4.13.13)
Thus, M can be expressed as a Hankelian.