5.2 The Generalized Cusick Identities 181s 2 =ai+1,j+φiψj,s 3 =ai+1,j+1.The lemma follows.
LetA
∗
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
x2 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=12 n− 1
∑j=1(−1)
i+j
Pf(n)
i
Pf(n)
j
x4 n−i−j− 2=
2 n− 1
∑i=12 n− 1
∑j=1A
(2n−1)
ij
x4 n−i−j− 2=−
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
a 11 a 12 ··· a 1 , 2 n− 1 x2 n− 2a 21 a 22 ··· a 2 , 2 n− 1 x2 n− 3.........................................
a 2 n− 1 , 1 a 2 n− 1 , 2 ··· a 2 n− 1 , 2 n− 1 1x2 n− 2
x2 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 andcolumn 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− 11
• • ··· 1 •
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
2 n=+|a∗
ij|^2 n−^2=A
∗
2 n− 2.The lemma follows by taking the square root of each side.