10 1. Jacobi operators It can be verified directly as well. Sometimes transformations can help to simplify a problem. The following two are of particular interest to us. If u fulfills (1.19) and u(n) ^ 0, then the sequence 4(n) u(n + l)/u(n) satisfies the (discrete) Riccati equation a(n (1.52) a(n)p(n) i 4(n - 1) Conversely, if p fulfills (1.52), then the sequence b(n) (1.53) u{n) = J | * t{j) n-l J] t*{j) for n n0 j = nn 1 for n n,{) n0 -1 n 0(i) _ l for n n o V j=n fulfills (1.19) and is normalized such that U(TIQ) = 1. In addition, we remark that the sequence p(n) might be written as finite continued fraction, (1.54) a(n)(p(n) b(n) a(n l) 2 (n - 1) a(n0 + if - b{na + 1) - a(n0) for n n.Q and (1.55) a(n)4(n) a(n)z •6(n + l ) a(n0 - l)2 6(n0) - a(n())(p(n0) for n ri{). If a is a sequence with ci(n) ^ 0 and u fulfills (1.19), then the sequence (1.56) u(n) = u(n) Y[* a(j), fulfills (1.57) & (11) . u(n + 1) + a(n l)a(n l)u(n 1) + b(n)u(n) = 2w(n). a(n) Especially, taking a(n) = sgn(a(n)) (resp. a(n) = —sgn(a(n))), we see that it is no restriction to assume a{n) 0 (resp. a(n) 0) (compare also Lemma 1.6 below). We conclude this section with a detailed investigation of the fundamental solu- tions c(z,n,no) and s(z,n,no). To begin with, we note (use induction) that both c(z,n ± k,n) and s(z,n ± A:,n), A 0, are polynomials of degree at most A : with respect to z. Hence we may set (1.58) . ± A:, n) 2_\ s j.±k{n)zJ, c(z, n ± A:, j = o A.- E Cj.±A-(^)^J-
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