n¿0

What we are looking for is a relation between the matrix F = (fi; j )i; j and F(1) = (fi; j (1))i; j . We

ÿrst obtain some relations between the moments Sn j and Sn j (1).

i; i;

Lemma 4.2. The following identities hold:

n n

1; 1

Sh 1 Sn’h (1);

1; 3; 2

Sh 1 Sn’h (1);

1; 3; 1

Sn 2

1;

Sn+1 = b3 = b3

h=0 h=0

V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275“295 285

n n

Sh 1 (1)Sn’h k;

2; 1; 1 2; 2

Sh 1 Sn’h (1) + Sn+1 (1);

2; 3; 1 1; 1

Sn 1

2;

= Sn+1 = b3

h=0 h=0

n n

Sh 2 (1)Sn’h ;

2; 1; 1

Sh 1 Sn’h (1) + Sn 1 (1):

3; 3; 1

Sn 1

3;

Sn 2

3; 2;

= = b3

h=0 h=0

From these identities, some can be deduced for the functions S i; j (z) (and for the resolvent functions

fi; j ).

Lemma 4.3. The functions S i; j satisfy the identities

S 1; 1 ’ 1 = b3 S 1; 1 S 3; 2 (1); S 1; 2 = b3 zS 1; 1 S 3; 1 (1);

z

S i; 1 = zS 1; 1 S i’1; 2 (1);

j = 1; i¿2;

S 2; 2 = b3 S 2; 1 S 3; 1 (1) + S 1; 1 (1); S 3; 2 = b3 zS 3; 1 S 3; 1 (1) + S 2; 1 (1):

The functions fi; j satisfy the identities

zf1; 1 ’ 1 = bp f1; 1 fp; q (1);

f1; j = bp f1; 1 fp; j’1 (1);

i = 1; j¿2;

fi; 1 = f1; 1 fi’1; q (1);

j = 1; i¿2;

fi; j = bp fi; 1 fp; j’1 (1) + fi’1; j’1 (1):

i¿2; j¿2;

Proof. The preceding identities are used. From the classical formulae for the usual product of two

series, the results follow. For the functions fi; j the link with the functions S i; j is used and the results

follow.

5. Matrix Stieltjes continued fraction

For sake of completeness we ÿrst recall some deÿnitions and some results of the general theory

of matrix continued fractions [13].

5.1. Matrix continued fractions

Matrix continued fractions are an extension of the vector-continued fractions introduced already

by Jacobi, and studied by several authors (see for example [6,10]). The notations used here are those

of Sorokin and Van Iseghem [13], where the matrices are p — q.

We are interested in a ratio of matrices KH ’1 , K ∈ Mp; q , H ∈ Mq; q and as usual an equivalence

relation is deÿned in the set Mp+q; q which is in fact the set of the pairs (H; K),

A; A ∈ Mp+q; q ; A ∼ A ” ∃C ∈ Mq; q ; det C = 0; A = AC:

286 V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275“295

Let Gp; q be the set of the equivalent classes of matrices (Grassmann space), then operations are

deÿned in Gp; q through the canonical injection from Mp; q to Gp; q . Denote by Iq the unit matrix of

size q — q, then

Iq Iq

: Mp; q ’ Mp+q; q ’ Gp; q ; A’ ’ Cl :

A A

We now deÿne what will be used as a quotient in the space Mp; q and will be denoted by 1=Z =T (Z).

˜

Operators T and T are the same functions deÿned, respectively, on Mp; q and Mp+q; q .

˜

Let T , deÿned on Mp+q; q , be the permutation of the rows which puts the last row at the ÿrst

place. The operator T is deÿned from Mp; q to Mp; q and by a straightforward computation we obtain

the direct deÿnition of T as a transformation of Mp; q (T (B) is deÿned if bp; q = 0):

«

1 ’bp; 1 ··· ’bp; q’1

¬ ·

¬ ·

1 ¬ b1; q b1; 1 bp; q ’ b1; q bp; 1 ··· b1; q’1 bp; q ’ b1; q bp; q’1 ·

¬ ·:

T (B) = (15)

bp; q ¬ . ·

.

¬. . ·

. .

bp’1; q bp’1; 1 bp; q ’ bp’1; q bp; 1 ··· bp’1; q’1 bp; q ’ bp’1; q bp; q’1

˜ ˜ ˜

As, clearly (T )p+q = Id, T is to be considered as a ˜partial™ quotient and if p = q then (T )p (A) = A’1 .

The previous formula could be taken as a deÿnition.

5.2. Continued fraction associated to the resolvent function

From the recurrence relations written for the functions fi; j , we get with D =1=f1; 1 =z ’bp fp; q (1),

and using the expression of T (B) just recalled (15)

« «

f1; 1 f1; 2 b3 f3; 1 (1)

1

¬ 2; 1 · 1 ¬ 1; 2 ·

¬f f2; 2 · = ¬ f (1) f1; 1 (1)D + b3 f1; 2 (1)f3; 1 (1) ·

D

f3; 1 f3; 2 f2; 2 (1) f2; 1 (1)D + b3 f2; 2 (1)f3; 1 (1)

1

=«

1; 1

f1; 2 (1)

f (1)

¬ ·

¬ ·

f2; 1 (1) f2; 2 (1)

’b3 f3; 1 (1) z ’ b3 f3; 2 (1)

1

=« :

« « 1; 1

f1; 2 (1)

f (1)

0 0 1 0 0

¬ · ¬ ·¬ ·

¬0 0·+¬0 0 · ¬ f2; 1 (1) f2; 2 (1) ·

1

f3; 1 (1) f3; 2 (1)

0 z 0 0 ’b3

From this we have the following theorem.

V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275“295 287

Theorem 5.1. The resolvent function associated to the operator A has an expansion as a matrix

continued fraction

Ip C1 C2

;

P+ P+ P + · · ·

where the parameters of the continued fraction are deÿned for all n greater than 1 by

«

0 ··· 0

¬ ·

(Ip’1 ) 0

P=¬ . .

¬. ·

.

Cn = ; 0 ·; P an p — q matrix: (16)

. .

0 ’bp+n’1

0 ··· z

In other words; we get the recurrence relation

1

F(k) = :

P + diag(diag(Ip’1 ); ’bp+k’1 )F(k + 1)

If the normalization does not impose an = 1, A has two diagonals an and bn ; genetic sums can also

be computed in that case and we would ÿnd a continued fraction given by the following recurrence

relation:

1

F(k) = :

P + diag(diag(Ip’1 ); ’bp+k’1 )F(k + 1)diag(diag(Iq’1 ); ak )

In all cases, we obtain an expansion of F in a matrix continued fraction by a generalization of

the Jacobi“Perron algorithm [10]. Using this theory [13], we get for the convergent di erent forms

Qn Iq ’1

(yn ; : : : ; yn+q’1 ) = ∼ ; = Pn Qn ;

n

Pn n

where yn are the columns of size p + q satisfying

xyn = bn yn’p + an yn+q

and if yn = (Hn ; Kn )t , and n deÿnes the regular multiindex (n1 ; : : : ; np ), then

t

1 1

FHn ’ Kn = O ;:::;O :

z n1 +1 z np +1

6. Zeros, Sturm“Liouville problem

From now on, the parameters an and bn are supposed to be positive, which is equivalent to say

that system S is positive.

The proofs are written in the case p = 3; q = 2. By deÿnition we have

«

1 1

Kn Kn+1 ’1

Hn1 1

Hn+1

¬2 ·

¬ Kn+1 ·

2

n = Kn :

Hn2 2

Hn+1

3 3

Kn Kn+1

288 V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275“295

Hence, the common denominator is

Hn1 1

Hn+1

qn = det :

Hn2 2

Hn+1

Lemma 6.1. The following statements hold:

1. qn has real coe cients; is of degree n.

The leading coe cient of qn has sign (’1)n .

2.

qn is Zp+q invariant; i.e.; qn ∈ Span{xk ; k = n mod(p + q)}.

3.

4. For any positive n; the polynomials qn(p+q)+k ; k = 0; : : : ; p + q ’ 1; have a zero of order k at

the point zero.

5. if x ’ 0; then qn (x) ∼ Án (’x)k ; where Án ¿ 0; n = k mod(p + q); k = 0; : : : ; p + q ’ 1.

Proof. From the recurrence for the Hn , all coe cients are real and from the deÿnition of the basis

hk , Hn1 is of degree [n=2] as Hn2 is of degree k ’ 1 if n = 2k and k if n = 2k + 1, so the degree of

qn is n.

For the sign of the leading coe cient, we get that an’1 qn and x(Hn’1 Hn1 ’ Hn2 Hn’1 ) have the

2 1

same leading coe cient, so the sign of this leading coe cient is (’1)n , once the initial property is

veriÿed. It is the same for the term of lowest degree.

Invariance of qn follows from invariance of Hn and Kn . It follows also that qn(p+q)+k has a zero

of order k at point zero.

The following lemma plays the main role for the proof of theorem B.

Lemma 6.2. The following statements hold:

1. The polynomials qn(p+q)+k ; k = 0; : : : ; p + q ’ 1 have exactly n positive zeros.

2. The positive zeros of qn and qn+1 interlace.

i; j

i; j

3. The rational functions z n have positive residues at all poles; only some residues at the

point z = 0 may be equal zero.

Proof. Let us consider

1 1

Kn Kn+1

pn

1; 1

n= ; pn = :

qn Hn2 2

Hn+1