Determinants and Their Applications in Mathematical Physics

334 Appendix

λnr=

n

n+r

(

n+r

2 r

)

, 1 ≤r≤n,

λn 0 =1,n≥0;

μnr=

2 rλnr

n

, 1 ≤r≤n,

μn 0 =0,n≥ 0. (A.10.8)

Changing the sign ofxin (A.10.6),

(x−

√

1+x

2

)

2 n

=gn−hn

√

1+x

2

. (A.10.9)

Hence,

gn=

1

2

{

(x+

√

1+x

2

)

2 n

+(x−

√

1+x

2

)

2 n

}

,

hn=

1

2

{

(x+

√

1+x

2

)

2 n

−(x−

√

1+x

2

)

2 n

}

(

√

1+x

2

)

− 1

(A.10.10).

These functions satisfy the recurrence relations

gn+1=(1+2x

2

)gn+2x(1 +x

2

)hn,

hn+1=(1+2x

2

)hn+2xgn. (A.10.11)

Let

fn=

1

2

{

(x+

√

1+x

2

)

n

+(x−

√

1+x

2

)

n

}

. (A.10.12)

Lemmas.

a.f 2 n=gn

b.f 2 n+1=

gn+1−gn

2 x

.

Proof. The proof of (a) is trivial. To prove (b), note that

f 2 n+1=

1

2

{

(x+

√

1+x

2

)(gn+hn

√

1+x

2

)

+(x−

√

1+x

2

)(gn−hn

√

1+x

2

)

}

=xgn+(1+x

2

)hn. (A.10.13)

The result is obtained by eliminatinghnfrom the first line of (A.10.11).

In the next theorem, ∆ is the finite-difference operator (Appendix A.8).

Theorem A.8.

a.gm+n+gm−n=2gmgn,

b.∆(gm+n− 1 +gm−n− 1 )=2gn∆gm− 1 ,

c. 2 x

2

(gm+n− 1 −gm−n)=∆gm− 1 ∆gn− 1 ,

d.∆(gm+n− 1 −gm−n)=2gm∆gn− 1 ,

e. gm+n+1+gm−n= 2(1 +x

2

)(gm+xhm)(gn+xhn),

f.∆(gm−n+gm−n)=4x(1 +x

2

)hm(gn−xhn),

g.gm+n−gm−n= 2(1 +x

2

)hmhn,

h.∆(gm+n− 1 −gm−n− 1 )=4x(1 +x

2

)hn(gm−xhm).