Determinants and Their Applications in Mathematical Physics

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)