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i; j
The same investigation holds for the other n. We consider the determinant
1 1 1
Kn Kn+1 Kn+2
pn pn+1
= Hn+1 Hn1 1 1
2
Hn+1 Hn+2 :
=
n
qn qn+1
Hn2 2 2
Hn+1 Hn+2
For a polynomial, we write sgn(q) = 1 (respectively âˆ’1) if all nonzero coe cients of q are positive
(respectively negative). We will prove that sgn( n ) = 1 for nÂ¿4.
V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275â€“295 289

We Ã¿rst prove, by induction, that sgn(Hn2 ) = (âˆ’1)nâˆ’1 .
From the recurrence relation (the same for the Kn ), with positive coe cients an ; bn
xHn2 = an Hn+2 + bn Hnâˆ’3
2 2

and from the recurrence assumption, we get
sgn(Hn2 ) = (âˆ’1)nâˆ’1 ; sgn(Hnâˆ’3 ) = (âˆ’1)nâˆ’4 â‡’ sgn(Hn+2 ) = (âˆ’1)n+1 :
2 2

The initial values are checked directly.
Let us denote
1 1 1
Kn 1 Kn2 Kn3
(n1 |n2 |n3 ) = Hn11 Hn12 Hn13 :
Hn21 Hn22 Hn23
To prove sgn( n ) = 1, we have to prove that sgn(n|n + 1|n + 2) = (âˆ’1)n . By using the recurrence
relations for the Hn , we get
an (n|n + 1|n + 2) = âˆ’bn (n âˆ’ 3|n|n + 1);

anâˆ’1 (n âˆ’ 3|n|n + 1) = âˆ’x(n âˆ’ 3|n âˆ’ 1|n) âˆ’ bnâˆ’1 (n âˆ’ 4|n âˆ’ 3|n);

anâˆ’2 (n âˆ’ 3|n âˆ’ 1|n) = âˆ’x(n âˆ’ 3|n âˆ’ 2|n âˆ’ 1) âˆ’ bnâˆ’2 (n âˆ’ 5|n âˆ’ 3|n âˆ’ 1);

anâˆ’2 (n âˆ’ 4|n âˆ’ 3|n) = x(n âˆ’ 4|n âˆ’ 3|n âˆ’ 2) âˆ’ bnâˆ’2 (n âˆ’ 5|n âˆ’ 4|n âˆ’ 3);

anâˆ’3 (n âˆ’ 5|n âˆ’ 3|n âˆ’ 1) = âˆ’bnâˆ’3 (n âˆ’ 6|n âˆ’ 5|n âˆ’ 3);

anâˆ’5 (n âˆ’ 6|n âˆ’ 5|n âˆ’ 3) = âˆ’bnâˆ’5 (n âˆ’ 8|n âˆ’ 6|n âˆ’ 5):
The recurrence assumption is taken as
sgn(k|k + 1|k + 2) = (âˆ’1)k ; k Â¡ n;

sgn(k âˆ’ 3|k âˆ’ 1|k) = (âˆ’1)k ; k Â¡ n:
Now, taking the preceding equalities from the bottom to the top prove that the recurrence assumption
is in fact true for all n, once the initial conditions are satisÃ¿ed, which is checked directly (nÂ¿4).
Then sgn(n|n + 1|n + 2) = (âˆ’1)n , and sgn( n ) = 1.
We now study the zeros properties of the polynomials qn . It is enough to investigate the be-
haviour of qn on the interval [0; +âˆž[. We have proved that all coe cients of the polynomial n
are nonnegative, as it is not identically zero, then n Â¿ 0 if x Â¿ 0.
If, for Â¿ 0, we have qn ( ) = 0, then

nâˆ’1 ( ) = âˆ’pn ( )qnâˆ’1 ( ) Â¿ 0; (17)

n( ) = pn ( )qn+1 ( ) Â¿ 0;

so qnâˆ’1 ( ) and qn+1 ( ) have di erent signs.
290 V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275â€“295

We consider the polynomial q5n , and take as recurrence assumption that q5nâˆ’1 and q5nâˆ’2 have
n âˆ’ 1 simple positive zeros which interlace. The beginning of the induction is checked directly. Let
us denote by 1 ; : : : ; nâˆ’1 the n âˆ’ 1 positive zeros of q5nâˆ’1 , written in increasing order. It follows
that
q5nâˆ’2 ( 1 ) Â¿ 0; q5nâˆ’2 ( 2 ) Â¡ 0; : : :
hence q5n ( 1 ) Â¡ 0; q5n ( 2 ) Â¿ 0; : : :, and q5n (0) Â¿ 0. So q5n has at least one zero on each interval
]0; 1 [; ] 1 ; 2 [; : : : ; ] nâˆ’2 ; nâˆ’1 [ :
By property (2) of the previous lemma, q5n and q5nâˆ’1 have di erent signs at inÃ¿nity, hence q5n has
a zero on the interval ] nâˆ’1 ; +âˆž[. Now q5n has at least n positive zeros, but it cannot have more
so q5n has n simple positive zeros, which interlace with those of q5nâˆ’1 .
Let us come now to the polynomials pn . Similarly they are Z5 invariant too. If âˆ— are the pos-
j
âˆ— âˆ—
itive zeros of q5n , written in increasing order, then p5n ( j ) and q5nâˆ’1 ( j ) have di erent signs. So
p5n ( âˆ— ) Â¡ 0, p5n ( âˆ— ) Â¿ 0; : : : and it follows that
1 2

q5n ( âˆ— ) Â¡ 0; q5n ( âˆ— ) Â¿ 0; : : :
1 2

hence pn ( âˆ— ) and qn ( âˆ— ) have the same sign, so it follows that the residue at point âˆ—
j, which is
j j
âˆ— âˆ—
pn ( j )=qn ( j ), is positive. The same result is obtained for the other indices.

The preceding result can be rewritten in the following form, always with deÃ¿ned in (7).

Lemma 6.3. For each pair of indices (i; j); there exists a positive discrete measure such that
d n j (x)
i;
i; j
i; j
z n (z) = : (18)
zâˆ’x
Proof. The measure has mass equal to the residual at each pole of qn . The residuals being positive,
the measure is positive. The support is the same for all (i; j), as it is the set of zeros of qn .

This result can be written in a matrix form, with some matrix d of positive measures
n

d n (x)
z n (z) = : (19)
zâˆ’x

7. Lax pair

We consider matrix A for the operator of multiple on the variable x in the basis of polynomials
Hn : AH = xH , where
A(ei ) = aiâˆ’q eiâˆ’q + bi+p ai+p :
Matrix A depends on the normalization of the sequence Hn . We will keep the notation A for the
Ëœ
monic polynomials Hn0 , i.e., an = 1, and call A the generalized Jacobi case, i.e., the an and the bn
satisfy (13)
(an : : : an+pâˆ’1 )q = (bn+p : : : bn+p+qâˆ’1 )p : (20)
V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275â€“295 291

We are looking for a Lax pair for the matrix A
A = [A; B];
which deÃ¿nes the di erential equations satisÃ¿ed by the bn . As in [1], B = Ap+q (the lower part of
âˆ’
p+q
A ) gives rise to a matrix [A; B] of the same structure as A ,
p p
p+q i i
A (ei ) = i+k(p+q) ei+k(p+q) ; B(ei ) = i+k(p+q) ei+k(p+q) :
k=âˆ’q k=1

For k Â¿ 1, the coe cient of ei+k(p+q)+p in AB(ei ) and in BA(ei ) is the coe cient of ei+k(p+q)+p in
Ap+q+1 (ei ), so
iâˆ’q
i
(AB âˆ’ BA)(ei ) = ( âˆ’ i+p )ei+p
i+p+q

and A = [A; B] is equivalent to the family of di erential equations
iâˆ’q
i
bi+p = âˆ’ i+p :
i+p+q

For q = 1, i.e., the vector case [1], it is (A(ei ) = eiâˆ’1 + bi+p ei+p ),
i+pâˆ’1 i1 +p
iâˆ’1
= b i1 b i2 ;
i+p
i1 =i i2 =i+p

ï£« ï£¶
i+2p i+pâˆ’1 i1 +p i1 +p
iâˆ’q
b i1 ï£­ bi2 ï£¸ âˆ’ bi bi+p ;
i
âˆ’ = bi+p b i2 + b i2 âˆ’
i+p
i+p+q
i2 =i+p
i2 =i+pâˆ’1 i1 =i+1 i2 =i+p+1

ï£« ï£¶
i+2p i+pâˆ’1
bi+p = bi+p ï£­ b i1 ï£¸ :
bi2 âˆ’
i1 =i
i2 =i+p+1

For p = 1, and q any integer [11] (A(ei ) = eiâˆ’q + bi+1 ei+1 ),
B(ei ) = bi+1 Â· Â· Â· bi+q+1 ei+q+1 ;

bi+1 = bi+1 (bi+2 Â· Â· Â· bi+q+2 âˆ’ bi Â· Â· Â· biâˆ’q ):
For p and q greater than 1, this gives more complicated formulae. If we write
bi+p = bi+p (Ui+p âˆ’ Uiâˆ’q );
then Ui is a part of a genetic sum, there is at the Ã¿rst level of sum p terms and the Ã¿nal monomials
are of degree q. For p = 3; q = 2 the equations are
bi+3 = bi+3 (bi+6 (bi+9 + bi+7 + bi+5 ) + bi+4 (bi+7 + bi+5 ) + bi+2 bi+5 )

âˆ’(bi+1 (bi+4 + bi+2 + bi ) + biâˆ’1 (bi+2 + bi ) + biâˆ’3 bi )):
Ëœ
We now go to the generalized Jacobi matrix, A(ei ) = aiâˆ’q eiâˆ’q + bi+p ei+p , ai and bi satisfying (13). If
Hn0 are the monic polynomials, we consider Hn = un Hn0 as deÃ¿ned in Section 3 and deÃ¿ne the new
parameters
q=p+q
p=p+q
an = ; bn+p = Ã¿n+p :
n
292 V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275â€“295

Ëœ Ëœ
In the scalar case, A is symmetric and B must be skew-symmetric, here something of the same
Ëœ
kind is recovered, we obtain as a solution for B (for any matrix M+ and Mâˆ’ are the upper and
lower part of the matrix minus the diagonal),
q p
Ëœp+q Ëœp+q
Ëœ
B= (A ) + âˆ’ (A ) âˆ’ ;
p+q p+q

p p
i+p
i
aiâˆ’q = âˆ’ (biâˆ’q âˆ’ bi+p iâˆ’q ) =âˆ’ aiâˆ’q (Viâˆ’q âˆ’ Viâˆ’pâˆ’2q );
iâˆ’pâˆ’q
p+q p+q

q q
iâˆ’q
i
bi+p = (ai+p âˆ’ aiâˆ’q i+p ) = bi+p (Ui+p âˆ’ Uiâˆ’q ):
i+p+q
p+q p+q
As the an and the bn are linked by (13), both equations are equivalent up to a change of variable.
We Ã¿rst write the equations in a and b, then in and Ã¿ which are the dual forms of the same
equation. The equation in Ã¿ is exactly the same as the equation obtained in the Ã¿rst case with the
matrix A associated to the unitary polynomials.
The Ã¿nal results in the case p = 3; q = 2 are
bi+3 = 2 bi+3 (Ui+3 âˆ’ Uiâˆ’2 );
5

Ui = ai+3 (bi+6 (bi+9 ai+7 ai+5 + ai+4 (bi+7 ai+5 + ai+2 bi+5 ))

+ai+1 (bi+4 (bi+7 ai+5 + ai+2 bi+5 ) + aiâˆ’1 bi+2 bi+5 ));

aiâˆ’2 = âˆ’ 3 aiâˆ’2 (Viâˆ’2 âˆ’ Viâˆ’7 );
5

Viâˆ’2 = bi+3 (bi+6 ai+4 ai+2 ai + ai+1 (bi+4 ai+2 ai + aiâˆ’1 (bi+2 ai + aiâˆ’3 bi ))):
5=2
5=3
If these equations are written in terms of n, and Ã¿n we have an =cn cn+1 = and bn+3 =cn cn+1 cn+2 =
n
Ã¿n+3 , then we get
5 bn+3
5 an
n= Ã¿n+3 = Ã¿n+3
n
3 an 2 bn+3
= n (Vn âˆ’ Vnâˆ’5 ); =Ã¿n+3 (Un+3 âˆ’ Unâˆ’2 ):
The V and U are computed in terms of ck , then in terms of and Ã¿, and we Ã¿nally get
= n (Vn âˆ’ Vnâˆ’5 );
n

Vn = n+2 ( n+4 ( n+6 + n+3 ) + n+1 n+3 ) + nâˆ’1 n+1 n+3 ;

Ã¿n+3 = Ã¿n+3 (Un+3 âˆ’ Unâˆ’2 );

Un+3 = Ã¿n+6 (Ã¿n+9 + Ã¿n+7 + Ã¿n+5 ) + Ã¿n+4 (Ã¿n+7 + Ã¿n+5 ) + Ã¿n+2 Ã¿n+5 :
V. Sorokin, J. Van Iseghem / Journal of Computational and Applied Mathematics 122 (2000) 275â€“295 293

8. Dynamical systems

Consider the moment problem dealing with a positive sequence S. still deÃ¿ned by (7), let us
denote
p+q
2i
k
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