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On the inverse problem of the product of a form by a polynomial: The cubic case

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On the inverse problem of the product of a form by a polynomial: The cubic case

P. Maroni

a,

, I. Nicolau

b

aUniversité Pierre et Marie Curie-C.N.R.S., Laboratoire d’Analyse Numérique, 4 Place Jussieu, 75252 Paris cedex 05, France bDepartamento de Matemática, Universidade de Trás-os-Montes e Alto Douro, Apartado 202, 5001-911 Vila Real, Portugal

Abstract

A form (linear functional)u is called regular if there exists a sequence of polynomials{Pn}n0, degPn=n, which is orthogonal with respect to u. On certain regularity conditions, the product of a regular form by a polynomial is still a regular form. In this paper, we consider the inverse problem: given a regular formv, find all the regular formsuwhich satisfy the relationx3u= −λv,λC− {0}. We give the second-order recurrence relation of the orthogonal polynomial sequence with respect tou. Some examples are studied.

2003 IMACS. Published by Elsevier Science B.V. All rights reserved.

Keywords: Inverse problem; Forms; Orthogonal polynomials

Introduction

The product of a linear form by a polynomial, is one of construction process of linear forms. In 1858, Christoffel proved that the product of a positive definite form by a positive polynomial is still a positive definite form [3,4]. This result has been generalized in [7], where it was proved that, under certain regularity conditions, the product of a regular formu by a polynomial R is a regular form. In particular, ifuis a semi-classical form or a second degree form, thenRuis also a semi-classical form or a second degree form, respectively. It is also interesting to consider the inverse problem, which consists of the determination of all regular formsu satisfying Ru= −λv, where v is a given regular form and λ∈C− {0}. In [9], the first author has considered the caseR(x)=xc, withc∈C, and in [13] the case R(x)=x2. He gave the regularity conditions and the coefficients of the second-order recurrence relation satisfied by the monic orthogonal polynomial sequence (MOPS) with respect tou. The structure relation of the MOPS relatively tou, in the symmetric case, is given in [1] and in the non-symmetric case is given

*Corresponding author.

E-mail addresses: maroni@ann.jussieu.fr (P. Maroni), inicolau@utad.pt (I. Nicolau).

0168-9274/03/$30.002003 IMACS. Published by Elsevier Science B.V. All rights reserved.

doi:10.1016/S0168-9274(02)00250-7

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in [2]. Many examples were also treated in [13,2,1]. In this paper, we consider the caseR(x)=x3. In the first section we will give the regularity conditions and the coefficients of the second-order recurrence relation satisfied by the MOPS with respect to u. We will also prove that if v is semi-classical then uis semi-classical and some results concerned to the class of u are given. Section 2 is devoted to the particular case of(u)2=0 and Section 3 to the case (u)2=0: regularity conditions more simple and some examples are given if the formvis symmetric and ifvis symmetric definite positive, respectively.

The regular formsufounded in the examples are semi-classical of classs∈ {1,2,3}and we present there integral representations and the coefficients of the second-order recurrence satisfied by the MOPS with respect tou.

1. The problemx3u= −λv

1.1. The main problem

Let P be the vector space of polynomials with coefficients in C and P its dual. We denote by u, f the action of uP on fP. Let us recall that a form u is called regular if there exists a polynomial sequence{Pn}n0, degPn=n, such that u, PnPm =knδn,m, n, m0, kn=0, n0; the left-multiplication hw is defined by hw, p := w, hp. We consider the following problem: given a regular formv, find all regular formsuwhich satisfy the next equation:

x3u= −λv, λ∈C− {0}, (1.1)

with constraints(u)0=1, (v)0=1,where(u)n:= u, xn, n0,are the moments ofu. Equivalently, u=δ(u)1δ+1

2(u)2δλx3v, (1.2)

where δ, f =f (0), the derivative w=Dwof the formw is defined by w, f := − w, fand the formx1wby x1w, f := w, θ0f, with in general

cf )(x):= f (x)f (c)

xc , c∈C.

So the formudepends on three arbitrary parameters(u)1, (u)2andλ= −(u)3. If we suppose that the formvpossesses the following integral representation:

v, f = +∞

−∞

V (x)f (x)dx, for each polynomialf,

whereV is a locally integrable function with rapid decay, continuous at the origin and with derivative continuous at the origin, then the formuis represented by

u, f = f (0)

1+λPf +∞

−∞

V (x) x3 dx

+f(0)

(u)1+λPf +∞

−∞

V (x) x2 dx

+1 2f(0)

(u)2+λP +∞

−∞

V (x) x dx

λPf +∞

−∞

V (x)f (x)

x3 dx, (1.3)

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where, [5,8]

P +∞

−∞

V (x)

x dx= lim

ε0+

ε

−∞

V (x) x dx+

+∞

ε

V (x) x dx

,

Pf +∞

−∞

V (x)

x2 dx= lim

ε0+

ε

−∞

V (x) x2 dx+

+∞

ε

V (x)

x2 dx−2 εV (0)

,

and Pf

+∞

−∞

V (x)

x3 dx= lim

ε0+

ε

−∞

V (x) x3 dx+

+∞

ε

V (x)

x3 dx−2 εV(0)

.

Let{Sn}n0denote the sequence of orthogonal polynomials with respect tov S0(x)=1, S1(x)=xζ0,

(1.4) Sn+2(x)=(xζn+1)Sn+1(x)σn+1Sn(x), n0.

Whenuis regular, let{Zn}n0be the corresponding orthogonal sequence Z0(x)=1, Z1(x)=xβ0,

(1.5) Zn+2(x)=(xβn+1)Zn+1(x)γn+1Zn(x), n0.

From (1.1) we know that the sequence{Zn}n0, when it exists, is among the strictly quasi-orthogonal sequences of order three with respect tov[6,10], see also [12, pp. 127, 128], this is

Z0(x)=1, Z1(x)=S1(x)+c0, Z2(x)=S2(x)+c1S1(x)+b0,

(1.6) Zn+3(x)=Sn+3(x)+cn+2Sn+2(x)+bn+1Sn+1(x)+anSn(x), n0,

withan=0, n0.Moreover, the sequence{Zn}n0is orthogonal with respect touif and only if u, Zn =0, n1;

u, xZn(x)

=0, n2,

u, xZ1(x)

=0; u, x2Zn(x)

=0, n3,

u, x2Z2(x)

=0;

(1.7) since the other orthogonality conditions have been used in (1.6). From (1.6) and (1.7) we have

0 = u, Zn+3

= u, Sn+3 +cn+2 u, Sn+2 +bn+1 u, Sn+1 +an u, Sn, n0, 0 =

u, xZn+3(x)

(1.8)

=

u, xSn+3(x) +cn+2

u, xSn+2(x) +bn+1

u, xSn+1(x) +an

u, xSn(x)

, n0, 0 =

u, x2Zn+3(x)

=

u, x2Sn+3(x) +cn+2

u, x2Sn+2(x) +bn+1

u, x2Sn+1(x) +an

u, x2Sn(x)

, n0,

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with the initial conditions:

0= u, Z1 = u, S1 +c0,

0= u, Z2 = u, S2 +c1 u, S1 +b0, 0=

u, xZ2(x)

=

u, xS2(x) +c1

u, xS1(x)

+b0(u)1, 0=

u, xZ1(x)

=

u, xS1(x)

+c0(u)1, 0=

u, x2Z2(x)

=

u, x2S2(x) +c1

u, x2S1(x)

+b0(u)2.

(1.9)

If we denote

n:=

u, Sn+2 u, Sn+1 u, Sn u, xSn+2(x) u, xSn+1(x) u, xSn(x) u, x2Sn+2(x) u, x2Sn+1(x) u, x2Sn(x)

, n0, (1.10)

then the system (1.8) is equivalent to

nan= −n+1, n0, (1.11)

nbn+1=

u, Sn+2u, Sn+3 u, Sn u, xSn+2(x)u, xSn+3(x) u, xSn(x) u, x2Sn+2(x) − u, x2Sn+3(x) u, x2Sn(x)

, n0, (1.12)

ncn+2=

− u, Sn+3 u, Sn+1 u, Sn

u, xSn+3(x) u, xSn+1(x) u, xSn(x)

− u, x2Sn+3(x) u, x2Sn+1(x) u, x2Sn(x)

, n0. (1.13)

Proposition 1.1. The formuis regular if and only if∆n=0, n0 and(u)2(u)21=0.In this case, the coefficients of the second-order recurrence relation of{Zn}n0are given by

β0=(u)1, βn+1=ζn+1+cncn+1, n0, (1.14)

γ1=(u)2(u)21, γ2= −∆0 (u)2(u)212

, (1.15)

γ3= −λ∆1[(u)2(u)21]

20 , γn+4=n+2n

2n+1 σn+1, n0.

Proof. Necessity. If {Zn}n0 is orthogonal, it is strictly quasi-orthogonal with respect to v and then an=0, n0. This impliesn=0, n0. Assuming the contrary, that there exists ann01 such that

n0=0, then from (1.11),0=0= − u, x2Z2(x) u, xZ1(x) =0, which is a contradiction. Moreover (u)2(u)21= u, xZ1(x) =0.

Sufficiency. Using (1.2), the conditions u, Z1 =0, u, Z2 =0 and u, xZ2(x) =0 are satisfied for c0=ζ0(u)1,

c1= {ζ1+ζ0} +λ+(u)1(u)2

(u)2(u)12, (1.16)

b0=σ1+ζ02(u)2+

ζ0(u)1

λ+(u)1(u)2

(u)2(u)12.

Taking account of (1.16), we also have u, xZ1(x) =(u)2(u)12=0 and u, x2Z2(x)

=−λ2−2(u)1(u)2λζ0{(u)2(u)12}λ(u)23

(u)2(u)12 = −∆0 (u)2(u)211

=0.

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We had just proved that the initial conditions (1.9) are satisfied. Further, the system (1.8) is a Cramer system whose solution is given by (1.11), (1.12) and (1.13). Moreover, from (1.5) fornn+2 and (1.6) fornn+1 we can write

Sn+4(x)+cn+3Sn+3(x)+bn+2Sn+2(x)+an+1Sn+1(x)

=(xβn+3)Zn+3(x)γn+3Zn+2(x), n0.

Multiplying the above equation byZn+3(x)and applying the formu, we obtain u, Sn+4Zn+3 +cn+3 u, Sn+3Zn+3 =

u, xZn2+3(x)

βn+3

u, Z2n+3

, n0,

because{Zn}n0is orthogonal with respect tou. Now, we will replaceSn+4(x)for its expression given by (1.4) fornn+2, and then

u, xSn+3(x)Zn+3(x)

ζn+3 u, Sn+3Zn+3 +cn+3 u, Sn+3Zn+3

=

u, xZn2+3(x)

βn+3

u, Zn2+3

, n0.

Finally, replacingSn+3by its expression given by (1.6) u, xZn2+3(x)

cn+2

u, xSn+2(x)Zn+3(x)

ζn+3

u, Zn2+3 +cn+3

u, Zn2+3

=

u, xZn2+3(x)

βn+3

u, Zn2+3

, n0, that is

−cn+2

u, Zn2+3(x)

ζn+3

u, Z2n+3 +cn+3

u, Z2n+3

= −βn+3

u, Zn2+3

, n0.

As, u, Zn2+3 =0, we have

βn+3=ζn+3+cn+2cn+3, n0.

Using the same proceeding, we easily prove that βn+1=ζn+1+cncn+1, n=0,1,

that is, we have proved (1.14). Let us see now that u, Zn2+3

=

u, xn+3Zn+3(x)

=

x3u, xnZn+3(x)

= −λ

v, xnZn+3(x)

, n0, from (1.1). Using (1.6) this becomes

u, Zn2+3

= −λan

v, xnSn(x)

= −λan

v, Sn2

, n0.

Then

γn+4= u, Z2n+4

u, Z2n+3=an+1 v, Sn2+1

an v, Sn2 =an+1

an

σn+1, n0.

By virtue of (1.11), we successively obtain γn+4=n+2n

2n+1 σn+1, n0, γ1=

u, xZ1(x)

=(u)2(u)21, γ2= u, x2Z2(x)

u, xZ1(x) = − 0

[(u)2(u)21]2, γ3= u, x3Z3(x)

u, x2Z2(x)= −λa0

−∆0[(u)2(u)21]1= −λ∆1

20 (u)2(u)21 .

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We have proved (1.15). ✷ 1.2. The computation of∆n

As we had seen in Proposition 1.1, it is very important to have an explicit expression forn. In first place, let us note that, from (1.2), we have

u, Sn+1(x)

=Sn+1(0)+(u)1Sn+1(0)+1

2(u)2Sn+1(0)−1 2λ

Sn(1)

(0), n0, (1.17)

with

Sn(1)(x):=

v,Sn+1(x)Sn+1(ξ ) xξ

, n0, u, xSn+1(x)

=(u)1Sn+1(0)+(u)2Sn+1(0)λ S(1)n

(0), n0, (1.18)

u, x2Sn+1(x)

=(u)2Sn+1(0)λSn(1)(0), n0, (1.19)

u, x3Sn+1(x)

=0, n0. (1.20)

Using (1.4) and (1.20), we obtain for (1.10):

n=

u, xSn+1(x) u, Sn+1 u, Sn u, x2Sn+1(x) u, xSn+1(x) u, xSn(x)

0 u, x2Sn+1(x) u, x2Sn(x)

, n0, (1.21)

that is

n =

u, xSn+1(x) u, xSn+1(x) u, xSn(x) u, x2Sn+1(x) u, x2Sn(x)

u, x2Sn+1(x) u, Sn+1 u, Sn u, x2Sn+1(x) u, x2Sn(x)

, n0. (1.22)

From (1.17), (1.18) and (1.19), we have u, xSn+1(x) u, xSn(x)

u, x2Sn+1(x) u, x2Sn(x)

=X(1)n1(0)λ2+

v, Sn2(u)1+ Yn(0)Xn(0) (u)2

λ+Xn(0)(u)22, n0, (1.23) and

u, Sn+1 u, Sn u, x2Sn+1(x) u, x2Sn(x)

=1 2

Xn(1)1

(0)λ2+ v, Sn2

Xn(0)(u)1+1

2 Yn(0)Xn(0) (u)2

λ +

Xn(0)(u)1(u)2+1

2Xn(0)(u)22

, n0, with

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Xn(x)=Sn(x)Sn+1(x)Sn+1(x)Sn(x), n0, (1.24) Xn(1)(x)=Sn(1)(x)

Sn(1)+1

(x)Sn(1)+1(x) Sn(1)

(x), n0, (1.25)

Xn(x)=Sn+1(x)Sn(1)1(x)Sn(x)Sn(1)(x), n0, (1.26) Yn(x)=

Sn(1)1

(x)Sn+1(x)Sn(x) Sn(1)

(x), n0. (1.27)

Therefore,

n = Wn(1)1(0)λ3 +

Sn(1)(0) v, Sn2

+ Wn(0)Sn(1)

(0) v, Sn2

(u)1+ Tn(0)Un(0) (u)2

λ2 +

Sn+1(0) v, Sn2

(u)21(u)2

+ Vn(0)Rn(0) (u)22 + Sn+1(0)

v, Sn2

+S(1)n (0)Xn(0)+Yn(0)Sn+1(0)

(u)1(u)2

λ

Wn(0)(u)32, n0, (1.28)

with

Wn(x)=1

2Sn+1(x)Xn(x)Sn+1(x)Xn(x), n0, (1.29) Wn(1)(x)=1

2Sn(1)+1(x) X(1)n

(x)Sn(1)+1

(x)Xn(1)(x), n0, (1.30)

Wn(x)=Sn+1(x)Xn(1)1(x)Sn(1)(x)Xn(x), n0, (1.31) Un(x)=1

2Sn+1(x) X(1)n1

(x)Sn+1(x)X(1)n1(x), n0, (1.32) Vn(x)=1

2Sn(1)(x)Xn(x)Sn(1)

(x)Xn(x), n0, (1.33)

Tn(x)=1

2Sn(1)(x) Yn(x)Xn(x)

Sn(1)

(x) Yn(x)Xn(x)

, n0, (1.34)

Rn(x)=1

2Sn+1(x) Yn(x)Xn(x)

Sn+1(x) Yn(x)Xn(x)

, n0. (1.35)

Moreover, if the formuis regular we have for (1.11), (1.12) and (1.13) an= −n+1

n

, n0, (1.36)

bn+1=σn+2

u, Sn+2 u, xSn+2(x) u, Sn u, xSn+2(x) u, x2Sn+2(x) u, xSn(x) u, x2Sn+2(x) 0 u, x2Sn(x)

n1, n0, (1.37) cn+2=ζn+2

u, xSn+2(x) u, Sn+1 u, Sn u, x2Sn+2(x) u, xSn+1(x) u, xSn(x)

0 u, x2Sn+1(x) u, x2Sn(x)

n1, n0, (1.38) by virtue of (1.4).

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1.3. Some results on the semi-classical particular case

Let us recall that a formuis called semi-classical if there exists two polynomialsΦandψ such that usatisfies the functional equation

D(Φu)+ψu=0.

The class of the semi-classical form uis s =max(degΦ−2,degψ−1) if and only if the following condition is satisfied

c

ψ(c)+Φ(c)+u, θcψ+θc2Φ>0, (1.39)

wherec∈ {x: Φ(x)=0}[11].

In the sequel the formvwill be supposed semi-classical of classs satisfyingD(Φv)+ψv=0.

From (1.1), it is clear that the formu, when it is regular, is also semi-classical and satisfies D(Φu) + ˜ψu=0,

with

Φ(x) =x3Φ(x) and ψ(x)˜ =x3ψ(x). (1.40)

The class ofuis at mosts˜=s+3.

Proposition 1.2. The class ofudepends only on the zerox=0.

For the proof we use the following lemma:

Lemma 1.3. For all rootcofΦwe have u, θcψ˜ +θc2Φ

=

ψ(c)+Φ(c)

(u)2+c(u)1+c2

λ v,

θc2Φ+θcψ

(1.41) and

ψ(c)˜ +Φ(c)=c3

ψ(c)+Φ(c)

. (1.42)

Proof. Letcbe a root ofΦ, then we can write

Φ(x) =x3(xc)Φc(x) withΦc(x)=cΦ)(x). (1.43) So from (1.40) and (1.43) we have

u, θcψ˜ +θc2Φ

= u, θc

ξ3ψ +

u, θc

ξ3Φc

. (1.44)

Using the definition of the operatorθc, it is easy to prove that, for two polynomialsf andg, we have

θc(f g)(x)=g(x)(θcf )(x)+f (c)(θcg)(x). (1.45)

Takingg(x)=x3andf (x)=Φc(x), we obtain u, θc

ξ3Φc

=

u, x3cΦc)(x)+Φc(c) θcξ3

(x)

= u, x3

θc2Φ (x)

+

(u)2+c(u)1+c2 Φ(c),

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because

cΦc)(x)=

θccΦ) (x)=

θc2Φ

(x), Φc(c)=cΦ)(c)=Φ(c)and θcξ3

(x)=x2+cx+c2. Using (1.2), we obtain

u, θc

ξ3Φc

=

δ(u)1δ+1

2(u)2δλx3v, x3 θc2Φ

(x)

+

(u)2+c(u)1+c2 Φ(c), hence

u, θc

ξ3Φc

=

(u)2+c(u)1+c2

Φ(c)λ v, θc2Φ

. (1.46)

Now, takingg(x)=x3andf (x)=ψ(x)in (1.45), we obtain u, θc

ξ3ψ

=

u, x3cψ)(x)+ψ(c) θcξ3

(x)

=

u, x3cψ)(x) +

(u)2+c(u)1+c2 ψ(c).

Using (1.2), we obtain u, θc

ξ3ψ

=

δ(u)1δ+1

2(u)2δλx3v, x3cψ)(x)

+

(u)2+c(u)1+c2 ψ(c).

Then u, θc

ξ3ψ

=

(u)2+c(u)1+c2

ψ(c)λ v, θcψ. (1.47)

Replacing (1.46) and (1.47) in (1.44), we obtain (1.41).

From (1.40), we haveΦ(c)=c3Φ(c)andψ(c)˜ =c3ψ(c), hence (1.42).Proof of Proposition 1.2. Letcbe a root ofΦsuch thatc=0.

If ψ(c)+Φ(c)=0, using (1.41), we have u, θcψ˜ +θc2Φ = 0, since v is semi-classical and so satisfies (1.39).

Ifψ(c)+Φ(c)=0 thenψ (c)˜ +Φ(c)=0,from (1.42).

In any case, we cannot simplify byxc.

Proposition 1.4. Letvbe a semi-classical form of classs satisfyingD(Φv)+ψv=0, χ1:=(u)1Φ(0)+(u)2

Φ(0)+ψ(0)

λ

v, θ0ψ+θ02Φ

, (1.48)

and

χ2:=2Φ(0)+(u)1

ψ(0)+2Φ(0) +(u)2

Φ(0)+ψ(0)

λ

v, θ02ψ+2θ03Φ

. (1.49)

The formugiven by (1.1) is also semi-classical of classs˜satisfyingD(Φu) + ˜ψu=0. Moreover, (1) ifχ1=0 thens˜=s+3,Φ(x) =x3Φ(x)andψ(x)˜ =x3ψ(x);

(2) ifχ1=0 andχ2=0 thens˜=s+2,Φ(x) =x2Φ(x)andψ (x)˜ =x2ψ(x)+xΦ(x);

(3) ifχ1=0, χ2=0 andΦ(0)=0 thens˜=s+1,Φ(x) =xΦ(x)andψ(x)˜ =xψ(x)+2Φ(x).

Proof. (1) Indeed, ψ(0)˜ +Φ(0)=0,

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and u,

θ0ψ˜ +θ02Φ (x)

=

u, x2ψ(x)+xΦ(x)

=

δ(u)1δ+1

2(u)2δλx3v, x2ψ(x)+xΦ(x)

, using (1.2), that is,

u,

θ0ψ˜ +θ02Φ (x)

=(u)1Φ(0)+(u)2

Φ(0)+ψ(0)

λ

v, θ0ψ+θ02Φ

=χ1.

Therefore, ifχ1=0 it is not possible to simplify, which means that the class ofuiss+3 andusatisfies D(Φu) + ˜ψu=0, withΦ(x) =x3Φ(x)andψ(x)˜ =x3ψ(x).

(2) Ifχ1=0 then letΦ0(x)=x2Φ(x)andψ˜0(x)=x2ψ(x)+xΦ(x). Then ψ˜0(0)+Φ0(0)=0,

and u,

θ0ψ˜0+θ02Φ0

(x)

=

u, xψ(x)+2Φ(x)

=

δ(u)1δ+1

2(u)2δλx3v, xψ(x)+2Φ(x)

, using (1.2), that is,

u,

θ0ψ˜0+θ02Φ0

(x)

=2Φ(0)+(u)1

ψ(0)+2Φ(0) +(u)2

Φ(0)+ψ(0)

λ

v, θ02ψ+2θ03Φ

=χ2.

If χ2 =0 it is not possible to simplify, which means that the class of u is s +2 and u satisfies D(Φu) + ˜ψu=0, withΦ(x) =Φ0(x)=x2Φ(x)andψ(x)˜ = ˜ψ0(x)=x2ψ(x)+xΦ(x).

(3) Ifχ1=0 andχ2=0 then letΦ1(x)=xΦ(x)andψ˜1(x)=xψ(x)+2Φ(x). Then ψ˜1(0)+Φ1(0)=3Φ(0).

If Φ(0)=0 it is not possible to simplify, which means that the class of u is s +1 and u satisfies D(Φu) + ˜ψu=0, withΦ(x) =Φ1(x)=xΦ(x)andψ(x)˜ = ˜ψ1(x)=xψ(x)+2Φ(x). ✷

2. The formu=δ(u)1δλx3v,(u)2=0

Whenvis a symmetric form, we have the following result:

Theorem 2.1. Ifvis a symmetric form,(u)2=0,(u)1=0 and(u)1+λΛn=0, n0, then the formu is regular, where

Λn= n

ν=0

ν µ=0

σ

σ+1

, n0, σ0=1. (2.1)

For the proof we use the following lemmas:

(11)

Lemma 2.2. If{yn}n0, {an}n0and{bn}n0are sequences of complex numbers fulfilling yn+1+anyn=bn+1, n0, an=0, n0,

y0=b0, then

yn=(−1)nan1 n

µ=0

aµ

n

ν=0

(−1)νaν

ν

µ=0

aµ1

bν, n0.

Lemma 2.3. When{Sn}n0given by(1.4)is symmetric we have S2n+2(0)=(−1)n+1

n µ=0

σ+1, n0, S2n+1(0)=0, n0,

S2n(1)+2(0)=(−1)n+1 n µ=0

σ+2, n0, S2n(1)+1(0)=0, n0, S2n (0)=0, n0,

S2n(1)

(0)=0, n0, S2n(1)+1

(0)=(−1)n n µ=0

σ+1Λn, n0 and S2n(1)+1

(0)=0, n0.

Proof of Lemma 2.3. Asvis symmetric thenζn=0, n0 and so, from (1.4), we have S0(0)=1, S1(0)=0, S0(1)(0)=1, S(1)1 (0)=0, Sn+2(0)= −σn+1Sn(0), n0, Sn(1)+2(0)= −σn+2S(1)n (0), n0, S0(0)=0, S1(0)=1,

S0(1)

(0)=0, S1(1)

(0)=1, Sn+2(0)= −σn+1Sn(0)+Sn+1(0), n0,

Sn(1)+2

(0)= −σn+2

Sn(1)

(0)+Sn(1)+1(0), n0, and

S0(1)

(0)=0, S1(1)

(0)=0, Sn(1)+2

(0)= −σn+2

Sn(1)

(0)+2 Sn(1)+1

(0), n0.

Now, it is enough to use Lemma 2.2 in order to obtain the results. ✷

Proof of Theorem 2.1. Following Lemma 2.3 we have for (1.25), (1.26), (1.30) and (1.31):

Xn(0)=0, X(1)2n(0)=S2n(1)(0) S2n(1)+1

(0), (2.2)

X2n(1)+1(0)= −S2n(1)+2(0) S2n(1)+1

(0), n0, W2n+1(0)=S2n+2(0)S2n(1)(0)

S2n(1)+1

(0), W2n(0)=0, n0, W2n(1)(0)= − S2n(1)+1

(0)2

S2n(1)(0), W2n(1)+1(0)=0, n0.

Therefore we have for (1.28):

2n=S2n(1)(0) v, S2n2

λ2, n0,

(12)

or

2n=(−1)n n µ=0

σ

v, S2n2

λ2, n0, (2.3)

and

2n+1 = − S2n(1)+1 (0)2

S(1)2n(0)λ3 + S2n+2(0)S2n(1)(0)

S2n(1)+1 (0)

S2n(1)+1 (0)

v, S2n2+1 (u)1λ2 +S2n+2(0)

v, S2n2+1

(u)21λ, n0.

As

v, S2n2+1

= −S(1)2n(0)S2n+2(0), n0, (2.4)

we have

2n+1= −S2n(1)(0) S2n(1)+1

(0)λS2n+2(0)(u)1

2

λ, n0, and using Lemma 2.3, we obtain

2n+1=(−1)n+1 n µ=0

σ+1

λΛn+(u)1

2

v, S2n2+1

λ, n0. (2.5)

Under the hypothesis of the theorem it is evident thatn=0, n0 and thereforeuis regular by virtue of Proposition 1.1. ✷

Let us defineωby

(u)1=ωλ. (2.6)

Therefore we obtain for (2.5)

2n+1=(−1)n+1 n µ=0

σ+1{Λn+ω}2

v, S22n+1

λ3, n0. (2.7)

Corollary 2.4. If(u)2=0 andv is a symmetric positive definite form then the form uis regular when ω∈C− ]−∞,0].

Proof. Ifvis positive definite thenσn+1>0, n0.ThereforeΛn>0, n0 and soω+Λn=0, n0, under the hypothesis of the corollary. ✷

Moreover, ifvis symmetric positive definite,(u)2=0 andω∈C− ]−∞,0], we have from (1.16) b0=σ1+ λ

(u)1

=σ1+ 1

ω, (2.8)

c0= −(u)1= −ωλ, (2.9)

c1= − λ

(u)21= − 1

ω2λ, (2.10)

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