4.13 Hankelians 6 167
the upper limits that
γij=
i
∑
r=1
j
∑
s=1
βir 2
r+s− 1
kr+s− 2 βjs.
Hence,
γ 2 p+1, 2 q+1=2
2 p+1
∑
r=1
2 q+1
∑
s=1
β 2 p+1,r 2
r+s− 2
kr+s− 2 β 2 q+1,s. (4.13.10)
From the first line of (4.13.6), the summand is zero whenrandsare even.
Hence, replacerby 2r+ 1, replacesby 2s+ 1 and refer to (4.13.5) and
(4.13.6),
γ 2 p+1, 2 q+1=2
p
∑
r=0
q
∑
s=0
β 2 p+1, 2 r+1β 2 q+1, 2 s+1 2
2 r+2s
k 2 r+2s
=2
p
∑
r=0
q
∑
s=0
λprλqs
N
∑
j=1
aj(2xj)
2 r+2s
=2
N
∑
j=1
aj
p
∑
r=0
λpr(2xj)
2 r
q
∑
s=0
λqs(2xj)
2 s
=2
N
∑
j=1
ajgp(xj)gq(xj)
=α 2 p+1, 2 q+1, (4.13.11)
which completes the proof of case (i). Cases (ii)–(iv) are proved in a similar
manner.
Corollary.
|αij|n=|M|n=|N|
2
n|K|n
=|βij|
2
n
| 2
i+j− 1
ki+j− 2 |n
=
(
n
∏
i=1
βii
) 2
2
n
| 2
i+j− 2
ki+j− 2 |n. (4.13.12)
But,β 11 =1andβii=
1
2
, 2 ≤i≤n. Hence, referring to Property (e) in
Section 2.3.1,
|αij|n=2
n
2
− 2 n+2
|ki+j− 2 |n. (4.13.13)
Thus,Mcan be expressed as a Hankelian.