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)
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