Page 349 - Determinants and Their Applications in Mathematical Physics

334 Appendix
n n + r
λ nr = , 1 ≤ r ≤ n,
n + r 2r
λ n0 =1, n ≥ 0;
µ nr = 2rλ nr , 1 ≤ r ≤ n,
n
µ n0 =0, n ≥ 0. (A.10.8)
Changing the sign of x in (A.10.6),
+ +
2
(x − 1+ x ) = g n − h n 1+ x . (A.10.9)
2 2n
Hence,
+ + .
1
-
g n = (x + 1+ x ) +(x − 1+ x ) ,
2 2n
2 2n
2
+ + . +
2 −1
1
-
.
h n = (x + 1+ x ) − (x − 1+ x ) ( 1+ x ) (A.10.10)
2 2n
2 2n
2
These functions satisfy the recurrence relations
2
2
g n+1 = (1+2x )g n +2x(1 + x )h n ,
2
h n+1 = (1+2x )h n +2xg n . (A.10.11)
Let
+ + .
-
f n = 1 (x + 1+ x ) +(x − 1+ x ) . (A.10.12)
2 n
2 n
2
Lemmas.
a. f 2n = g n
b. f 2n+1 = g n+1 − g n .
2x
Proof. The proof of (a) is trivial. To prove (b), note that
+ +
1
2
-
f 2n+1 = (x + 2 1+ x )
2 1+ x )(g n + h n
+ + .
2 2
+(x − 1+ x )(g n − h n 1+ x )
2
= xg n +(1+ x )h n . (A.10.13)
The result is obtained by eliminating h n from the first line of (A.10.11).
In the next theorem, ∆ is the finite-difference operator (Appendix A.8).
Theorem A.8.
a. g m+n + g m−n =2g m g n ,
b. ∆(g m+n−1 + g m−n−1 )=2g n ∆g m−1 ,
2
c. 2x (g m+n−1 − g m−n )=∆g m−1 ∆g n−1 ,
d. ∆(g m+n−1 − g m−n )=2g m ∆g n−1 ,
2
e. g m+n+1 + g m−n = 2(1 + x )(g m + xh m )(g n + xh n ),
2
f. ∆(g m−n + g m−n )=4x(1 + x )h m (g n − xh n ),
2
g. g m+n − g m−n = 2(1 + x )h m h n ,
2
h. ∆(g m+n−1 − g m−n−1 )=4x(1 + x )h n (g m − xh m ).