5.2 The Generalized Cusick Identities 181
s 2 =ai+1,j+φiψj,
s 3 =ai+1,j+1.
The lemma follows.
Let
A
∗
2 n
=|a
∗
ij
| 2 n,
Pf
∗
n=
(
A
∗
2 n
)
1 / 2
. (5.2.15)
Lemma 5.3.
2 n− 1
∑
i=1
(−1)
i+1
Pf
(n)
i
x
2 n−i− 1
=Pf
∗
n− 1
.
Proof. Denote the sum byFn. Then, referring to (5.2.13) and Section 3.7
on bordered determinants,
F
2
n=
2 n− 1
∑
i=1
2 n− 1
∑
j=1
(−1)
i+j
Pf
(n)
i
Pf
(n)
j
x
4 n−i−j− 2
=
2 n− 1
∑
i=1
2 n− 1
∑
j=1
A
(2n−1)
ij
x
4 n−i−j− 2
=−
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
a 11 a 12 ··· a 1 , 2 n− 1 x
2 n− 2
a 21 a 22 ··· a 2 , 2 n− 1 x
2 n− 3
.........................................
a 2 n− 1 , 1 a 2 n− 1 , 2 ··· a 2 n− 1 , 2 n− 1 1
x
2 n− 2
x
2 n− 3
··· 1 •
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
2 n
. (5.2.16)
(It is not necessary to putaii= 0, etc., in order to prove the lemma.)
Eliminate thex’s from the last column and row by means of the row and
column operations
R
′
i=Ri−xRi+1,^1 ≤i≤^2 n−^2 ,
C
′
j=Cj−xCj+1,^1 ≤j≤^2 n−^2. (5.2.17)
The result is
F
2
n
=−
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
a
∗
11
a
∗
12
··· a
∗
1 , 2 n− 1
•
a
∗
21
a
∗
22
··· a
∗
2 , 2 n− 1
•
.....................................
a
∗
2 n− 1 , 1
a
∗
2 n− 1 , 2
··· a
∗
2 n− 1 , 2 n− 1
1
• • ··· 1 •
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
2 n
=+|a
∗
ij|^2 n−^2
=A
∗
2 n− 2.
The lemma follows by taking the square root of each side.