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 a in:^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 5a 2/5 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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