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)
nl
J] t*{j) for n n0
j = nn
1 for n — n,{)
n0 1
n 0(i)
_ l for n no
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_\ sj.±k{n)zJ, c(z, n ± A:,
j = o
A. E
Cj.±A(^)^J
