4.5 Centrosymmetric Determinants 89
The elementt 2 n− 1 does not appear inTnbut appears in the bottom right-
hand corner ofBn. Hence,
∂Rn
∂t 2 n− 1
=Rn− 1 ,
∂Sn
∂t 2 n− 1
=−Sn− 1. (4.5.14)
The same element appears in positions (1, 2 n) and (2n,1) inT 2 n. Hence,
referring to the second line of (4.5.12),
T
(2n)
1 , 2 n
=
1
2
∂T
(2n)
∂t 2 n− 1
=2
∂
∂t 2 n− 1
(RnSn)
=2(Rn− 1 Sn−RnSn− 1 ). (4.5.15)
Replacingnby 2nin (4.5.13),
T
2
2 n− 1
=T 2 nT 2 n− 2 +
(
T
(2n)
1 , 2 n
) 2
=4
[
4 RnSnRn− 1 Sn− 1 +(Rn− 1 Sn−RnSn− 1 )
2
]
=4(Rn− 1 Sn+RnSn− 1 )
2
.
The sign ofT 2 n− 1 is decided by puttingt 0 = 1 andtr=0,r>0. In that
case,Tn=In,Bn=On,Rn=Sn=
1
2
. Hence, the sign is positive:
T 2 n− 1 =2(Rn− 1 Sn+RnSn− 1 ). (4.5.16)
Part (a) of the theorem follows from (4.5.11).
The elementt 2 nappears in the bottom right-hand corner ofEnbut does
not appear in eitherTnorAn. Hence, referring to (4.5.10),
∂Pn
∂t 2 n
=−Pn− 1 ,
∂Qn
∂t 2 n
=Qn− 1. (4.5.17)
T
(2n+1)
1 , 2 n+1
=
1
2
∂T 2 n+1
∂t 2 n
=
∂
∂t 2 n
(PnQn+1)
=PnQn−Pn− 1 Qn+1. (4.5.18)
Return to (4.5.13), replacenby 2n+ 1, and refer to (4.5.12):
T
2
2 n=T^2 n+1T^2 n−^1 +
(
T
(2n+1)
1 , 2 n+1
) 2
=4PnQn+1Pn− 1 Qn+(PnQn−Pn− 1 Qn+1)
2
=(PnQn+Pn− 1 Qn+1)
2
. (4.5.19)