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z n+1
nВї0

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
Гїrst 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 Snв€’h (1);
1; 3; 2
Sh 1 Snв€’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) 275вЂ“295 285

n n
Sh 1 (1)Snв€’h k;
2; 1; 1 2; 2
Sh 1 Snв€’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)Snв€’h ;
2; 1; 1
Sh 1 Snв€’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 в€’ 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 iв€’1; 2 (1);
j = 1; iВї2;

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 в€’ 1 = bp f1; 1 fp; q (1);

f1; j = bp f1; 1 fp; jв€’1 (1);
i = 1; jВї2;

fi; 1 = f1; 1 fiв€’1; q (1);
j = 1; iВї2;

fi; j = bp fi; 1 fp; jв€’1 (1) + fiв€’1; jв€’1 (1):
iВї2; jВї2;

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 Гїrst recall some deГїnitions and some results of the general theory
of matrix continued fractions .

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 , where the matrices are p Г— q.
We are interested in a ratio of matrices KH в€’1 , K в€€ Mp; q , H в€€ Mq; q and as usual an equivalence
relation is deГїned in the set Mp+q; q which is in fact the set of the pairs (H; K),
A; A в€€ Mp+q; q ; A в€ј A в‡” в€ѓC в€€ Mq; q ; det C = 0; A = AC:
286 V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275вЂ“295

Let Gp; q be the set of the equivalent classes of matrices (Grassmann space), then operations are
deГїned in Gp; q through the canonical injection from Mp; q to Gp; q . Denote by Iq the unit matrix of
size q Г— q, then
Iq Iq
: Mp; q в†’ Mp+q; q в†’ Gp; q ; Aв†’ в†’ Cl :
A A
We now deГїne what will be used as a quotient in the space Mp; q and will be denoted by 1=Z =T (Z).
Лњ
Operators T and T are the same functions deГїned, respectively, on Mp; q and Mp+q; q .
Лњ
Let T , deГїned on Mp+q; q , be the permutation of the rows which puts the last row at the Гїrst
place. The operator T is deГїned from Mp; q to Mp; q and by a straightforward computation we obtain
the direct deГїnition of T as a transformation of Mp; q (T (B) is deГїned if bp; q = 0):
пЈ« пЈ¶
1 в€’bp; 1 В·В·В· в€’bp; qв€’1
пЈ¬ пЈ·
пЈ¬ пЈ·
1 пЈ¬ b1; q b1; 1 bp; q в€’ b1; q bp; 1 В·В·В· b1; qв€’1 bp; q в€’ b1; q bp; qв€’1 пЈ·
пЈ¬ пЈ·:
T (B) = (15)
bp; q пЈ¬ . пЈ·
.
пЈ¬. . пЈ·
пЈ­. . пЈё
bpв€’1; q bpв€’1; 1 bp; q в€’ bpв€’1; q bp; 1 В·В·В· bpв€’1; qв€’1 bp; q в€’ bpв€’1; q bp; qв€’1
Лњ Лњ Лњ
As, clearly (T )p+q = Id, T is to be considered as a вЂ˜partialвЂ™ quotient and if p = q then (T )p (A) = Aв€’1 .
The previous formula could be taken as a deГїnition.

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 в€’bp fp; q (1),
and using the expression of T (B) just recalled (15)
пЈ« пЈ¶ пЈ« пЈ¶
f1; 1 f1; 2 b3 f3; 1 (1)
1
пЈ¬ 2; 1 пЈ· 1 пЈ¬ 1; 2 пЈ·
пЈ¬f f2; 2 пЈ· = пЈ¬ f (1) f1; 1 (1)D + b3 f1; 2 (1)f3; 1 (1) пЈ·
пЈ­ пЈё DпЈ­ пЈё
f3; 1 f3; 2 f2; 2 (1) f2; 1 (1)D + b3 f2; 2 (1)f3; 1 (1)
1
=пЈ« пЈ¶
1; 1
f1; 2 (1)
f (1)
пЈ¬ пЈ·
пЈ¬ пЈ·
f2; 1 (1) f2; 2 (1)
пЈ­ пЈё
в€’b3 f3; 1 (1) z в€’ b3 f3; 2 (1)
1
=пЈ« пЈ¶:
пЈ¶ пЈ« пЈ¶ пЈ« 1; 1
f1; 2 (1)
f (1)
0 0 1 0 0
пЈ¬ пЈ· пЈ¬ пЈ·пЈ¬ пЈ·
пЈ¬0 0пЈ·+пЈ¬0 0 пЈ· пЈ¬ f2; 1 (1) f2; 2 (1) пЈ·
1
пЈ­ пЈёпЈ­ пЈёпЈ­ пЈё
f3; 1 (1) f3; 2 (1)
0 z 0 0 в€’b3
From this we have the following theorem.
V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275вЂ“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 + В· В· В·
where the parameters of the continued fraction are deГїned for all n greater than 1 by
пЈ« пЈ¶
0 В·В·В· 0
пЈ¬ пЈ·
(Ipв€’1 ) 0
P=пЈ¬ . .
пЈ¬. пЈ·
.
Cn = ; 0 пЈ·; P an p Г— q matrix: (16)
пЈ­. . пЈё
0 в€’bp+nв€’1
0 В·В·В· z
In other words; we get the recurrence relation
1
F(k) = :
P + diag(diag(Ipв€’1 ); в€’bp+kв€’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 Гїnd a continued fraction given by the following recurrence
relation:
1
F(k) = :
P + diag(diag(Ipв€’1 ); в€’bp+kв€’1 )F(k + 1)diag(diag(Iqв€’1 ); ak )
In all cases, we obtain an expansion of F in a matrix continued fraction by a generalization of
the JacobiвЂ“Perron algorithm . Using this theory , we get for the convergent di erent forms
Qn Iq в€’1
(yn ; : : : ; yn+qв€’1 ) = в€ј ; = Pn Qn ;
n
Pn n

where yn are the columns of size p + q satisfying
xyn = bn ynв€’p + an yn+q
and if yn = (Hn ; Kn )t , and n deГїnes the regular multiindex (n1 ; : : : ; np ), then
t
1 1
FHn в€’ Kn = O ;:::;O :
z n1 +1 z np +1

6. Zeros, SturmвЂ“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 deГїnition we have
пЈ« пЈ¶
1 1
Kn Kn+1 в€’1
Hn1 1
Hn+1
пЈ¬2 пЈ·
пЈ¬ Kn+1 пЈ·
2
n = пЈ­ Kn :
пЈё
Hn2 2
Hn+1
3 3
Kn Kn+1
288 V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275вЂ“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 (в€’1)n .
2.
qn is Zp+q invariant; i.e.; qn в€€ Span{xk ; k = n mod(p + q)}.
3.
4. For any positive n; the polynomials qn(p+q)+k ; k = 0; : : : ; p + q в€’ 1; have a zero of order k at
the point zero.
5. if x в†’ 0; then qn (x) в€ј ГЃn (в€’x)k ; where ГЃn Вї 0; n = k mod(p + q); k = 0; : : : ; p + q в€’ 1.

Proof. From the recurrence for the Hn , all coe cients are real and from the deГїnition of the basis
hk , Hn1 is of degree [n=2] as Hn2 is of degree k в€’ 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 anв€’1 qn and x(Hnв€’1 Hn1 в€’ Hn2 Hnв€’1 ) have the
2 1

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