Vector orthogonal polynomials of dimension? d

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ci;j ci;j+k ... ... ci+k?1;j ci+k?1;j+k. 1 xk. ;. Laboratoire d'Analyse Num erique et ..... Since the system is triangular with a non-zero diagonal, 0;:::;k?d?1 are zero.
Vector orthogonal polynomials of dimension ?d C. Brezinski 

J. Van Iseghem y

Dedicated to Prof. Walter Gautschi in the honor of his 65th birthday

Abstract

Vector orthogonal polynomials of dimension ?d where d is a nonzero positive integer are de ned. They are proved to satisfy a recurrence relation with d + 2 terms. A Shohat{Favard type theorem and a QD like algorithm are given.

Keywords: Orthogonal polynomials, biorthogonality, recurrence relation. AMS(MOS) Classi cation numbers: 42C05.

1 Introduction and de nitions Let c be given complex numbers for i; j = 0; 1; : : :. We de ne the linear functionals L0; L1 ; : : : on the space of complex polynomials by ?  L x =c : The formal biorthogonal polynomials with respect to the family L [2] are the polynomials satisfying the biorthogonality conditions   L x P ( )(x) = 0 for p = i; : : :; i + k ? 1: They are given by c    c + . .. . P ( )(x) =  .. ;    c + ?1 + c + ?1 1  x i;j

j

i

i;j

i

i;j

j

p

k

i;j

i;j

k

i;j

k

i

k

;j

i

k

;j

k

k

 Laboratoire d'Analyse Num erique et d'Optimisation, UFR IEEA - M3, Universite des Sciences et Technologies de Lille, 59655{Villeneuve d'Ascq cedex { France, e-mail: [email protected] y U.F.R. de Math ematiques Pures et Appliquees, Universite des Sciences et Technologies de Lille, 59655{Villeneuve d'Ascq cedex { France

1

where  is an arbitrary nonzero constant which depends on k; i; j . The polynomials have the exact degree k if and only if c    c + ?1 .. D( ) = ... 6= 0: . c + ?1    c + ?1 + ?1 In the sequel, we shall assume that this condition holds for all k; i; j and we shall consider the case where the polynomials P ( ) are monic which corresponds to the choice  = 1=D( ) . Vector orthogonal polynomials of dimension d 2 lN were introduced by Van Iseghem [8]. They correspond to the case where the linear functionals L are related by ?  ?  L x +1 = L + x (1) ?  ?  that is L x + = L + x : When d = 1, the usual formal orthogonal polynomials are recovered [1]. Polynomials of dimension d = ?1 were considered in [5]. They generalize the usual orthogonal polynomials on the unit circle and have applications in LaurentPade and two-point Pade approximation [6]. They are obtained by taking d = ?1 in the relations (1) de ning the vector orthogonal polynomials, that is by assuming that the linear functionals L are related by i;j

i;j

k

i;j

k

i

k

;j

i

k

;j

k

i;j

k

i;j

k

i

j

i

j

i

n

i

d

i

nd

j

j

i

?

?



L x = L +1 x +1 i

j

i

j



or c = c +1 +1. It was proved that, in the de nite case (that is when 8k; D(0 0) 6= 0) these polynomials satisfy a three{term recurrence relationship of the form (setting for simplicity P for P (0 0)) ;

i;j

i

;j

k

;

k

k

P +1(x) = (x + B +1 ) P (x) ? C +1 x P ?1(x) with P?1(x) = 0; P0(x) = 1 and C +1 = L0 (xP ) =L0 (xP ?1) B +1 = C +1L ?1 (P ?1) =L (P ) : k

k

k

k

k

k

k

k

k

k

k

k

k

k

The non-de nite case, which occurs when some of the determinants D(0 0) vanish, was treated by Draux [7] by an indirect approach involving the whole table of the usual formal orthogonal polynomials of dimension d = 1, and by a direct one in [3]. Formal orthogonal polynomials on a curve were considered in [4]. It was proved that the orthogonal polynomials of dimension d  1 or d  ?1 correspond to orthogonality on a curve consisting in discrete points located on the unit circle. ;

k

2

In this paper, we shall consider the case where d is negative or, in other words, the case ?d with d  1, which corresponds to linear functionals related by L (x +1) = L ? (x ) (2) that is L (x ) = L + (x +1) = L + (x + ): In that case, we shall speak of vector orthogonal polynomials of dimension ?d. We shall make use of new notations which are simpli ed with respect to the general case. Because of the pseudo{periodicity of the moments we have for i = sd, c = c0 ? . So, the determinants in the expression de ning P ( ) only depend on the di erence j ? s. Thus, after replacing j ? s by ?s with s 2 ZZ, we shall make use of the notation P instead of P ( 0) and D instead of D( 0) . Thus we have L (P ) = 0; p = sd; : : :; sd + k ? 1; c0 ?    c0 ? + . .. . P (x) = P ( 0)(x) =  .. ; c    c ?1 ? + ?1 ? 1  x c0 ?    c0 ? + ?1 . .. 1= = D = .. : . c ?1 ?    c ?1 ? + ?1 Denoting by D (x) the numerator of P , we see that ?1 D P (x) = DD(x) ; P (0) = DD ; L + (P ) = D+1 : The functionals L can be de ned for negative value of the index p, since, by the periodicity for all p and k, L (x ) = c = c + + . We shall assume that the polynomials are orthogonal of the exact dimension ?d that is that the relation (2) does not hold with ?d + 1 or ?d ? 1. These polynomials could possibly have applications in the simultaneous Pade approximation of several Laurent series and in simultaneous two-point Pade approximation. Such applications remain to be studied. j

i

j

i

i

d

i

d

j

j

i

j

nd

n

sd;j

sd;j

;j

s

k

sd;

s r

sd;

s k

r

k

s k

p

;

s

;

s

k

sd;

s k

k

k

;

s

k

;

s

k

k

;

s

;

s

k

s k

k

;

s

k

;

s

k

s k

s k

s k

s k

s k

s k

s k

s k

k

sd

s k

s k

s k

p

p

i

p;i

p

kd;i

k

2 The recurrence relationship We shall now prove that, for s xed, the orthogonal polynomials (P ) 0, of dimension ?d satisfy a recurrence relationship with d +2 terms as in the case of the dimension d [8]. For sake of simplicity, everything is written for s = 0, and P is denoted by P . We have: s k k

s k

k

3

Theorem 1 If 8k  0, P (0) 6= 0 and L (P ) 6= 0, then the monic vector orthogonal polynomials P

P

k +1

k

k +1

k

k

satisfy a relation of the form

(x) = (x + b )P (x) ? x k

d X

a P ? (x); k = 0; 1; : : : i

k

k

i

i=1

with P0(x) = 1, and P (x) = 0 for i < 0. The coecients a0; : : :; a and b (which depend on k) are solutions of the system i

d X

d

k

a L (xP ? ) = L (xP ) i = 0; : : :; d ? 1 j

i

k

j

i

k

j =1

b L (P ) = a L ? (P ? ): k

k

k

d

k

d

k

d

If k < d, the same recurrence relationship holds. The coecients a +1 ; : : :; a are zero. The coecients a0;    ; a and b (which depend on k) are solutions of the system k

k

k X

d

k

a L (xP ? ) = L (xP ) i = 0; : : :; k ? 1 j

i

k

j

i

k

j =1

b L (P ) = k

k

k X

a L (xP ? ) ? L (xP ): j

k

k

k

j

k

k

j =1

Proof : As P (0) has been assumed to be di erent from zero, it is possible to consider the polynomial P normalized by the condition P (0) = 1. So P = A P , P ? P is zero for x = 0 and has the exact degree k + 1. Let us write it as k

k

k +1

k

k

k

k

k

(P +1 ? P )(x) = x k

k X

 P (x) i

k

i

i=0 k X

8p; L (P +1 ? P ) = L ? p

k

k

p

!

 P : i

d

i

i=0

Let us suppose rst that k  d. Since L (P +1 ? P ) = 0 for p = d;    ; k ? 1, we obtain a triangular system with respect to 0 ;    ;  ? ?1, whose diagonal terms are L ? (P ? ); p = d; : : :; k ? 1. Since these quantities are di erent from zero by assumption, the existence of the following relation is proved: p

k

k

k

p

d

p

d

d

P +1(x) = ( x + 1)P (x) + x k

k

?1 X

k

k

i=k

?

 P (x):

d

Similarly, for p = k, it follows that ?L (P ) =  ? L ? (P ? ) k

k

k

4

d

k

d

k

d

i

i

which proves that  ? is nonzero, or equivalently that the relation cannot be shortened. Let us now come back to the monic polynomials P and show how to compute the coecients of the relation k

d

k

P

k +1

(x) = (x + b )P (x) ? x k

d X

a P ? (x): i

k

k

i

i=1

Now a0 ; : : :; a and b are known to exist and to be unique. They are given by imposing that L (P +1) = 0 for i = 0;    ; d ? 1 and for i = k, that is d

k

i

k

d X

a L (xP ? ) = L (xP ) i = 0; : : :; d ? 1 j

i

k

j

i

k

j =1

b L (P ) = a L ? (P ? ): When k < d, the proof is obtained by expanding P +1 ? xP on the basis fP g. Similar relations hold for the coecients. Remark 1: It must be noticed that, if the dimension is not ?d but ?d ? 1, then the orthogonality relations used in the proof of the theorem are no longer true from i = d but only from i = d + 1. On the other hand, if the polynomials are of dimension ?d + 1 instead of d, then in the last equation (p = d ? 1) of the system de ning the a 's, all the terms L ?1 (xP ) = L0 (xP ) are zero, and then a = 0, which implies that L (P ) = 0. Because P +1 (0) = b , the equation de ning b shows the link between the non{degenerate character of the family and the nonzero value of the polynomials at zero. k

k

k

d

k

d

k

d

k

i

d

d

k

k

k

i

n

i

k

k

k

3 A Shohat{Favard theorem We shall now prove the reciprocal of the recurrence relationship, namely that if a family of polynomials satis es a recurrence relationship of the form given in the preceding theorem, then it is a family of orthogonal polynomials of dimension ?d whose moments L (x ), such that L (x ) = L + (x +1), can be calculated. Let us assume that we are given a family fP g of monic polynomials such that the recurrence relationship of the Theorem 1 holds. First of all, let us remark that the family of linear functionals fL g is de ned apart from a multiplying factor. Indeed, if L (P ) = 0 for i = 0; : : :; k ? 1, then it also holds if each functional L is replaced by L =  L where  is an arbitrary nonzero constant. For k = 0, we have P1 (x) = x + b0 and L0 (P1) = L0(x) + b0 L0(1) = 0: j

i

i

j

i

d

j

k

i

i

k

i

0

i

i

i

i

5

By the remark above, L0 (1) or L0 (x) can be arbitrarily chosen (to be a nonzero value since, otherwise, we shall have 8j; L0(x ) = 0) and the preceding relation gives the other quantity. Thus L0 (1) and L0 (x) are known. For k = 1, we have P2(x) = (x + b1 )P1 (x) ? a1 xP0(x). The condition L0 (P2) = 0 allows us to obtain the value of L0 (x2) since L0 (1) and L0 (x) are already known. We have L1 (P2 ) = L1 (xP1) + b1 L1(P1 ) ? a1L1 (xP0) = 0: By the remark above, among the quantities L1 (1); L1(x) and L1 (x2 ), two of them can be arbitrarily chosen (to be a nonzero value) and the preceding relation determines the third one. Thus, these three quantities are known. And so on. For k = d ? 1, we have P (x) = (x + b ?1 )P ?1 (x) ? a1xP ?2(x) ?    ? a ?1xP0(x). The condition L0 (P ) = 0 allows us to obtain the value of L0 (x ) since L0(1); : : :; L0 (x ?1) are already known. And so on until the condition L ?2 (P ) = 0 which gives the value of L ?2 (x ) since L ?2 (1); : : :; L ?2(x ?1 ) are already known. We have j

d

d

d

d

d

d

d

d

d

d

d

d

L ?1 (P ) = L ?1 (xP ?1) + b ?1L ?1 (P ?1) ? d

d

d

d

d

d

?1 X

d

a L ?1 (xP ? ?1) = 0: i

d

d

d

d

d

d

i

i=1

By the remark above, among the quantities L ?1 (1); : : :; L ?1(x ), d of them can be arbitrarily chosen (to be nonzero values) and the preceding relation determines the (d + 1)th one. Thus, these d + 1 quantities are known. For k = d, we have P +1 (x) = (x+b )P (x)?a1 xP ?1 (x)?  ?a xP0(x). The condition L0 (P +1 ) allows us to obtain the value of L0 (x +1 ) since L0(1); : : :; L0 (x ) are known. And so on until the condition L ?1 (P +1 ) = 0 which gives the value of L ?1 (x +1 ) since L ?1 (1); : : :; L ?1(x ) are already known. We have d

d

d

d

d d

d

d

d

d

d

d

d

d

d

d

d

d

d

L (P +1 ) = L (xP ) + b L (P ) ? d

d

d

d

d

d X

a L (xP ? ) = 0: i

d

d

d

i

i=1

But L (xP ) = L0 (P ) = 0 by the conditions on the functionals. Similarly L (xP ?1) = L0 (P ?1 ) = 0; : : :; L (xP1 ) = L0 (P1) = 0 and L (xP0) = L0(P0 ) 6= 0. Thus L (P +1 ) = b L (P ) ? a L0 (P0): Setting P (x) = 0 +    + ?1 x ?1 + x , we have L (P ) = 0 L (1) + 1 L (x) +    + ?1 L (x ?1 ) + L (x ) = 0 L (1) + 1 L0 (1) +    + ?1 L0 (x ?2) + L0 (x ?1 ): Thus, in the relation L (P +1 ) = 0, all the quantities are known except L (1) which can now be computed if b and 0 are both di erent from zero. Let us d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

d

6

remark that the condition b = 0 is equivalent to P +1 (0) = 0 and that 0 = 0 is equivalent to P (0) = 0. Moreover L (x) = L0 (1); : : :; L (x +2 ) = L0(x +1 ) are also known. And so on for any greater value of k. Thus, we have proved: d

d

d

d

d

d

d

Theorem 2 Let fP g be a family of monic polynomials such that 8k; P (0) 6= 0, k

k

and which satisfy the recurrence relationship

P +1(x) = (x + b ) ? x k

d X

a P ? (x) for k = 0; 1; : : : i

k

k

i

i=1

with P0(x) = 1 and P identically zero for i < 0. These polynomials are vector orthogonal polynomials of dimension ?d with recpect to linear functionals L whose moments L (x ) can be computed. i

i

j

i

Remark 2: As we saw before, when k = 0, there is one degree of freedom in the values of the moments since L0 (1) or L0 (x) can be arbitrarily chosen. When k = 1, there are 2 degrees of freedom, and so on. When k = d ? 1, there are d degrees of freedom. So, the vector space of the solutions has the dimension (d!), a result similar to the result which holds for the vector orthogonal polynomials of dimension d. If we de ne the polynomials Pe by Pe (x) = x P (x?1 ), then these polynomials satisfy X Pe +1(x) = (b x + 1) ? a x Pe ? (x) k

k

k

k

d

k

j

k

j

k

j

j =1

which shows, by theorem 2, that the polynomials Pe are vector orthogonal polynomials of dimension ?d only in the case d = 1 [3]. k

4 Adjacent families and a QD like algorithm Let us now come back to the de nition of the biorthogonal polynomials P ( 0) = P , and study the link between two consecutive diagonals. A QD like algorithm, as in the case of vector orthogonality of dimension d [8], will be obtained. In the sequel, all the determinants D will be assumed to be nonzero, so the biorthogonal polynomials P exist, are of the exact degree k, and the monic ones are unique for any k and s. Moreover P (0) 6= 0. The polynomials P , normalized to be 1 at zero will also be considered and we shall set P = A P . sd;

k

s k

s k

s k

s k

7

s k

s k

s k

s k

The polynomials are displayed in a two-dimensional array where k is the index of a column (increasing from the left to the right) and s is the index of a diagonal (increasing from the bottom to the top). That is, we have .. .. . . . ... . . +1    P P +1     P ?1 P +1    .. .. .. . . . . . . s

s k s

k s k

k

By analogy with the classical QD algorithm, the relations of Theorem 3 will be called the relations (q). We shall indicate the position, in the preceding table, of the polynomials which are known by a  and by a  the position of the polynomial which is computed by the relation under consideration.

Theorem 3 We have 8 > >
= L (P ) :   : q +1 = ? 1 D D +1 + Proof : We rst apply the Sylvester identity to the numerator D +1 (x) of P where the row k+1 has been put as the rst row. Renumbering the rows in the classical order, the previous equality is D +1(x)D ?1 = xD ?1(x)D +1 ? D (x)D ?+11 : (q) >

s k

s k

s k

s k

s

s k

d

sd

k

k

s k

s k

s k

s k

Dividing by D ?1 D s

k

s k +1

s

s

k

k

s k

s k

s k +1

,

s

k

and taking in account that P (x) = D (x)=D , it follows s k

s k

s k

P +1(x) = xP ?1(x) ? q +1P (x) with the given expression for q +1. By analogy with the classical QD algorithm, we shall now nd the relation (e), but, as we shall see, it can be given in two di erent forms (e1) and (e). s k

s k

s k

s k

s k

Theorem 4 The relations (e1) and (e), in the table of biorthogonal polynomials of dimension (?d), are the following: 8 > > > > >
> > > > :

P

s+1

k

?1 X

k

(x) = P (x) + s k

i=k

?

" P s k;i

d

" ? = ? L + (P+1) L + (P ? ) s k;k

sd

k

d

sd

k

s k

s k

d

8

s+1 i

(x) k  d



...



 

8 > > > > >


A +1 P A s

k

s+1

k

s k

?1 X

e P (x) k  d 

k

(x) = P (x) + x s k

i=k

s k;i

?

s i

d

...

  e ? = L L + ((PP ) )  + ? ? Proof : The proof is based on the biorthogonality property of the polynomials. P +1 is monic and of degree k, and so it can be expanded on the basis fP +1g 0 > > > > :

s k;k

d

sd

k

s k

k

d

s k

sd

d

s k

s i

P

s+1 k

(x) = P (x) + s k

?1 X

k

P i

s+1 i

i

(x):

i=0

From the orthogonality with respect to the functional L , it follows that p

L (P ) = 0; p = sd; : : :; sd + k ? 1 L (P +1 ) = 0; p = (s + 1)d; : : :; (s + 1)d + k ? 1: Writing these equations for p = (s + 1)d;    ; sd + k ? 1; k > d, we obtain 8 0 = 0 L( +1) (P0 +1) < p = (s + 1)d                             : p = sd + k ? 1 0 = 0 L + ?1 (P0 +1) +    +  ? ?1 L + ?1(P ?+1?1): Since the system is triangular with a non-zero diagonal, 0 ; : : :;  ? ?1 are zero and the result follows. Writing down the equations for the following values of p, a triangular system is obtained with unknowns  ? ; : : :;  ?1. The coecient of smallest index is  ? L + (P ?+1 ) = ?L + (P ) and so it is nonzero in the non{degenerate case, i.e. the relation cannot be shortened. Thus we obtain the rst formula (e1) s k

p

p

s

k

s

s

d

sd

s

k

k

d

sd

k

k

k

P

s+1 k

d

d

sd

s k

k

d

d

k

k

s k

sd

d

?1 X

k

(x) ? P (x) = s k

i=k

?

s k

k

" P s k;i

s+1 i

(x):

d

This relation can also be written for the diagonal s. For conveniency, let us now use the polynomials P normalised by P (0) = 1. We have, by expansion on the basis fP g s i

s k

s k

i

?1

X P +1 (x) ? P (x) = x 0 P (x): k

s

k

s k

i

i=0

9

s i

The same proof leads to the following expressions 8 > > > > >
> > > > :

a P (x) k  d s k;i

?

s i

d

 a ? = ? L L + ((PP ) ) : + ? ? Writting these expressions for the monic polynomials P , P = (D =D ?1)P = A P , we obtain the second relation (e) s k

s k

s k;k

d

sd

k

s k

k

d

s k

sd

d

s k

8 > > > > >


A +1 P A s

k

s k

s+1 k

?1 X

s k

i=k

?

s k;i

s k

s k

e P (x) k  d 

k

(x) = P (x) + x

s k

s i

d

s k

...

   e ? = L L + ((PP ) ) : + ? ? In all the cases, the relations cannot be shortened if all the determinants D are nonzero. Let us now show how the quantities " can be expressed in terms of the quantities e . P is expressed using (e1) in terms of the "'s. Then each term xP is expressed in terms of the diagonal s + 1 by using the relation (q); it nally follows that 8 A +1 ? e > > > ?1 = 1 > A > > < > > > > :

s k;k

d

sd

k

s k

k

d

s k

sd

d

s k

s k;i

s k;i

s k

s i

s

s k;k

k

s k

+1 " = e ?1 + q +1 e i = k ? d + 1; : : :; k ? 1 " ? = e ? q +1 ? +1: Now, as in the classical case, two di erent proofs of the diagonal relation will give the QD algorithm. Let us use (q) and then (e1) P +1(x) = xP ?1(x) ? q +1P (x)

> > > > > > :

s k;i

s k;k

s i

s k;i

s k;k

d

s k

d

s k



s k;i

s k

d

s k

= x P (x) ? s k

?1 X

k

i=k

= (x ? q

s k +1

s k

s

?

s i

k;i

d

)P (x) ? x s k

?1 X

k

i=k

= (x ? q

s k +1



" ?1P (x) ? q

)P (x) ? x s k

10

P (x) s k

" ?1 P (x) s k;i

?

s i

d

?1 X

k

i=k

s k +1

?

(e ??1 1 + q +1 e ?1)P (x)

d+1

s k;i

s i

s k;i

s i

?xe ?1? q ? +1P ? (x): s

k;k

d

s k

s k

d

d

Similarly, using (e) and then (q), we get X A +1 +1 e +1 P (x) (x) ? x (x) = A +1 P +1 +1 = ? +1  X A +1  +1 = A +1 xP (x) ? q +1 P ( x ) ?x e +1 P (x) +1 +1 = ? +1 ! ! +1 A A +1 A +1 +1 +1 P +1(x) = A +1 ? e +1 x ? q +1 A +1 A +1 P (x) ?1 ?1 X X A +1 +1 A ?xq +1 e +1 P (x): e P ( x ) ? x +1 A +1 A +1 = ? = ? +1

P

k

s

s k +1

k

s

s k

k

s k

i

k

;i

s i

d

k

s k

s k

s k

s

s

k

k

s k

i

k

s k

k

s

s

s k

s k

s

s k

k

k

s k;i

s

k

i

k

s k

s i

i

d

Finally, identifying the two expressions, we get rst A +1 A +1 +1 +1 A = q q +1 +1 ; +1 +1 A +1 A A +1 ? e from which it follows that s k

s

k

s k

s k

s k

s k

s

s k

k

e

s k +1;k

s k

s

k

s k

k

s k

s k

+1=

s i

d

s k

k

s k

;k

;i

k +1

Y

i=1

q

s k +1;k

k

;i

s i

d

= 1;

q : s i

s+1 i

This last relation together with the remaining part of the identi cation, gives the QD algorithm 8 > > > > > > >
> > > > > > :

e

s k +1;k

+1 =

k +1

Y

i=1

e q s k;i

s k +1

+e

s k +1;i

q

q

s i

s+1 i

= e ??1 1 + q +1 e ?1; i = k ? d + 1; : : :; k ? 1 s k;i

s i

s k;i

e ? q +1 = e ?1? q ? +1 : The rst d columns (q ) ; i = 1; : : :; d are the initializations of the algorithm. They are given by the formula of Theorem 3 (namely q1 = c0 ? +1=c0 ? ), e = 0; i < 0. If the d columns q +1; : : :; q + , and also the column (E + ?1) where E = (e ) ; i = k ? d; : : :; k, are known, then e + + ?1 is computed by the rst relation, e + for i = n; : : :; n + d ? 1 is computed by the second relation, and nally q + +1 is obtained from the third relation. s k;k

d

s

s k

k;k

d

s k

d

s i s

s

n

s k

s k;i i

s n

s n

n

;

d;i

d

11

;

s n

d

s n

s

d;n

d

d

s k;i

s

s

References [1] C. Brezinski, Pade-Type Approximation and General Orthogonal Polynomials, ISNM 50, Birkhauser, Basel, 1980. [2] C. Brezinski, Biorthogonality and its Applications to Numerical Analysis, Marcel Dekker, New York, 1991. [3] C. Brezinski, A uni ed approach to various orthogonalities, Ann. Fac. Sci. Toulouse, ser.3, vol.1, fasc.3, (1992) 277{292. [4] C. Brezinski, Formal orthogonality on an algebraic curve, Annals of Numer. Math., to appear. [5] C. Brezinski, M. Redivo Zaglia, Orthogonal polynomials of dimension ?1 in the non-de nite case, Rend. Mat. Roma, ser. VII, 13 (1993), to appear. [6] A. Bultheel, Laurent Series and their Pade Approximations, Birkhauser, Basel, 1987. [7] A. Draux, Polyn^omes Orthogonaux Formels{Applications, LNM 974, Springer Verlag, Berlin, 1987. [8] J. Van Iseghem, Vector orthogonal relations. Vector qd{algorithm,J. Comput. Appl. Math., 19 (1987) 141{150.

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