12 1. Jacobi operators

Our next task will be expansions of c(z, n, no), s{z, n, no) for large z. Let Jni,n2

be the Jacobi matrix

/ b(rn + 1) a(ni + 1)

a(m + 1) 6(m + 2)

(1.64) Jni,n2

fc(n2 - 2) a(n2 - 2)

a ( n

2

- 2 ) b(n2-l) )

Then we have the following expansion for s(z,n, no), n no,

(1.65) s(z,n,n0) =

det(z - Jn^n) ^ - E j = lPn0,n(j

k-j

n"=n0 + l 3) Uj = la(n0+j)

where k = n — no — 1 0 and

(1.66) Pno,nU)

tr(J4,n)-E;=JPn0lnWtr(J4-l)

x

.

fc

To verify the first equation, use that if z is a zero of s(.,n,no), then (s(z,no +

1, no),..., s(z, n — 1, no)) is an eigenvector of (1.64) corresponding to the eigenvalue

z. Since the converse statement is also true, the polynomials (in z) s(z,n, no) and

det(z — Jno,n) only differ by a constant which can be deduced from (1.30). The

second is a well-known property of characteristic polynomials (cf., e.g., [91]).

The first few traces are given by

n0

+ k

tr(J

no

,

no+fc+

i) = ] T 6(j),

no+fc no+fc —1

tr(J20,no+fc+1) = £ b(j)2 + 2 £ a(j)2,

j = n

0

+ l j ' = n

0

+ l

no+fc no+fc —1

tr(J^,

n o + f e + 1

) = £ 6(i)3 + 3 £ a(i)2(6(j) + b(j + 1)),

no+fc no + fc — 1

tr(J^„

0 + f e + 1

) = £ 6 W

4

- 4 j

a(j-)2(6(j)2

+ K i + 1 W )

j ' = n

0

+ l

(1.67)

2 n

0

+/c —2

+Hi+i)

2

+^)+ 4 x;

«o'+i)2-

j = n

0

+ l