1.1. General properties 11

Using the coefficients Sj.±k(n) and Cj.±k{n) we can derive a neat expansion for

arbitrary difference expressions. By (1.9) it suffices to consider (S±)k.

(1.59) R = ] T (cj +

SJSATJ, CJ.SJ

G £(Z), k G N0

Lemma 1.2. Any difference expression R of order at most 2fc + 1 can be expressed

as

k

ty^ft Cj = 5j = 0 if and only if R = 0. In other words, the set

{rJ, 5+r-7}jGn0

forms

a basis for the space of all difference expressions.

We have

k

(1.60) (S*)* = J2 {CJ-±K + ^±^ + )W ,

j=o

where Sj.±k(n) and Cjm±k(n) are defined in (1.58).

Proof. We first prove (1.59) by induction on k. The case k = 0 is trivial. Since

the matrix element rk(n, n ± k) = IIj=o ain:^j ~ ?) 7^ 0 is nonzero we can choose

Sjt(rc) = i?(n,n-|-fc+

l)/rk(n—

l,n + & — 1), Cjt(n) = R(n,n-

k)/rk(n,n

— k) and

apply the induction hypothesis to i? — (cjt — SkS+)rk. This proves (1.59). The rest

is immediate from

A : fc

(1.61) ( f i ( 4 , n ) ) ( n ) = 5 » ^ ' , (ii(c(*,.,n))(n) = £ ) * .

j=o j=o

D

As a consequence of (1.61) we note

Corollary 1.3. Suppose R is a difference expression of order k. Then R = 0 if

and only if R\Ker(r-z) — 0 for k -\- 1 values of z G C. (Here R\Ker(r-z) — 0 5a2/5

that Ru = 0 /or any solution u of ru = zu.)

Next, £(z,no,ni) = $(2, n ^ n o ) - 1 provides the useful relations

c(z,n

0

,ni) s(^,n

0

,ni)

c(2,n0 + l,ni) s(2;,n

0

4-l,ni)

_ a(ni) / s(z,ni + l,n

0

) -s(z,ni,n

0

)

~" a(n0) ^ - c ( z , n i + l,n

0

) c(z,ni,n

0

)

and a straightforward calculation (using (1.27)) yields

a(n0 + l) , ,

s(z,n,n

0

+ l) = ^——c(z,7i,n0),

/ ,x / x z-b(n0) .

x

5(2, n, n0 - 1) = c(^,n,n0) H 7——s(z,n,no),

a(n0)

, . z — b(no + l) ,

x / x

c(z,n,n

0

+ l) = -,—r c(z,n,no)H-s(z,n,n0),

a(n0)

(1.63) c(z,n,n0 - 1) = ° s(z,n,n

0

).

a(n0)