Division Problem of a Regular Form: The Case

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Volume 2011, Article ID 794593,23pages doi:10.1155/2011/794593

Research Article

Division Problem of a Regular Form:

The Case

x

2

u

λxv

M. Mejri

Department of Mathematics, Institut Supérieur des Sciences Appliquées et de Technologie, Rue Omar Ibn El Khattab, Gabès 6072, Tunisia

Correspondence should be addressed to M. Mejri,[email protected] Received 13 December 2010; Revised 25 February 2011; Accepted 17 March 2011

Academic Editor: Heinrich Begehr

Copyrightq2011 M. Mejri. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We present a systematic study of a regular linear functional v to find all regular forms u which satisfy the equationx2_u_λxv_,_λ_∈_C_{− {}₀_}_{. We also give the second-order recurrence relation of}

the orthogonal polynomial sequence with respect to u and study the semiclassical character of the found families. We conclude by treating some examples.

1. Introduction

In the present paper, we intend to study the following problem: letvbe a regular formlinear functional, andRandDnonzero polynomials. Find all regular formsusatisfying

RuDv. 1.1

This problem has been studied in some particular cases. In fact the product of a linear form by a polynomialRx 1is studied in1–3and the inverse problemDx λ,λ∈C− {0} is considered in 4–7. More generally, when R and D have nontrivial common factor the authors of8found necessary and suﬃcient conditions foruto be a regular form. The case whereR D is treated in4,9–11. The aim of this contribution is to analyze the case in whichRx x2 _and _D_x _λx_,_λ _∈ _C_{− {}₀_}_{. We remark that}_R _and _D _{have a common}

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results concerning the class ofuare given. In the last section, some examples will be treated. The regular formsufound in theses examples are semi-classical of classs ∈ {1,2,3}14. The integral representations of these regular forms and the coeﬃcients of the second-order recurrence satisfied by the MOPS with respect touare given.

2. The Problem

x

2

_u

_λxv

LetPbe the vector space of polynomials with coeﬃcients inCandPits algebraic dual. We denote byu, fthe action ofu∈ Ponf ∈ P. In particular, we designate byu_n:u, xn_,

n≥0, the moments ofu. For any formu, any polynomialg, anyc∈C,a∈C− {0}, letu,hau,

gu, andx−c−1ube the forms defined by duality:

u, p:−u, p; hau, p

:u, hap

; gu, p:u, gp;

x−c−1u, p:u, θcp

, p∈ P,

2.1

whereθcpx px−pc/x−c;hapx pax.

We define a left multiplication of a formuby a polynomialpas

upx:

u,xpx−ξpξ

x−ξ , u∈ P

_{, p}_{∈ P.} _2.2

Let us recall that a formuis called regular if there exists a monic polynomial sequence{Pn}_n≥0,

degPnn, such that

u, PnPmrnδn,m, n, m≥0, rn/0, n≥0. 2.3

We have the following result.

Lemma 2.1see15. Letu∈ P,f ∈ P, andc∈ C. The following formulas hold:

vfx vfx vfx vθ0f

x, f ∈ P. 2.4

δfx fx, f∈ P. 2.5

x−c−1x−cu u−u₀δc, 2.6

whereδc, ppc,p∈ P.

We consider the following problem: given a regular formv, find all regular formsu

satisfying

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with constraintsu₀1,v₀1. From2.6we can deduce that

xu u₁−λδλv, 2.8

uδ λ−u₁δλx−1v. 2.9

Then the formudepends on two arbitrary parametersu₁andλ.

We notice that whenu₁λ, we encounter the problem studied in13again. We suppose that the formvhas the following integral representation:

v, f

_∞

−∞ Vxfxdx, for each polynomialf, 2.10

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

u, f

1−λP

_∞

−∞

Vx x dx

f0 u₁−λf0 λP

_∞

−∞

Vxfx

x dx, 2.11

where16,17

P

_∞

−∞

Vx

x dxlim→0

−

−∞

Vx x dx

_∞

Vx x dx

. 2.12

Let{Sn}n≥0denote the sequence of monic orthogonal polynomials with respect tov; we have

S0x 1, S1x x−ξ0,

Sn2x x−ξn1Sn1x−σn1Snx, n≥0,

2.13

with

ξn

v, xS2

nx

v, S2 n

, σ_n1

v, S2_n1

v, S2 n

, n≥0. 2.14

Whenuis regular, let{Zn}_n≥0be the corresponding MOPS:

Z0x 1, Z1x x−β0.

Zn2x x−βn1Zn1x−γn1Znx, n≥0.

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From2.7, we know that the existence of the sequence{Zn}_n≥0is among all the strictly

quasi-orthogonal sequences of order two with respect toλxv w wis not necessarily a regular form 15,18–20. That is,

xZ0x S1x c0, xZ1x S2x c1S1x b0.

xZn2x Sn3x cn2Sn2x bn1Sn1x anSnx, n≥0,

2.16

withan/0,n≥0.

From2.16, we have

Z1x θ0S2x c1, 2.17

Z_n2x θ0Sn3x cn2θ0Sn2x bn1θ0Sn1x anθ0Snx, n≥0. 2.18

Lemma 2.2. Let {Zn}n≥0 be a sequence of polynomials satisfying2.16where an,bn, and cn are

complex numbers such thatan/0 for alln≥0. The sequence{Zn}n≥0is orthogonal with respect tou

if and only if

u, Zn0, n≥1,

u, xZnx0, n≥2, u, xZ1x/0.

2.19

Proof. The conditions2.19are necessary from the definition of the orthogonality of{Zn}_n≥0

with respect tou.

Fork≥2, we haveby2.7

u, xkZ_n2xx2u, xk−2Z_n2xλv, xk−1Z_n2x, n≥0, 2.20

and from2.16, we get

u, xk_Z

n2x

λv, xk−2S_n3xλc_n2v, xk−2S_n2x

λbn1

v, xk−2Sn1x

λan

v, xk−2Snx

, n≥0.

2.21

Taking into account the orthogonality of{Sn}_n≥0, we obtain

u, xkZn2x

0, 2≤k≤n1, n≥1,

u, xn2Z_n2xλan

v, S2 n

/

0, n≥0.

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By2.19, it follows that

u, Z10, u, xZ1x/0,

u, Zn2u, xZn2x0, n≥0.

2.23

Consequently, the previous relations and2.22prove that{Zn}n≥0is orthogonal with respect

tou, which proves the Lemma.

Remark 2.3. Whenuis regular, from Theorem 5.1 in21, there exist complex numbersrn2/0,

tn2andvn2/0 such that

Z_n2x r_n2Z_n1x S_n2x t_n2S_n1x v_n2Snx, n≥0. 2.24

From2.16,2.24, and2.15we obtain the following relations:

rn2−tn2cn2−ξn20, n≥0,

rn2cn1−tn2ξn1−vn2bn1−σn20, n≥0,

rn2bn−tn2σn1−vn2ξnan0, n≥0,

rn2an−1−vn2σn 0, n≥1.

2.25

Taking into account2.16,2.18and2.19, we get

0u, xZ_n2x

u, Sn3cn2u, Sn2bn1u, Sn1anu, Sn0, n≥0,

0u, Zn2

u, θ0Sn3cn2u, θ0Sn2bn1u, θ0Sn1anu, θ0Sn, n≥0,

0Sn30 cn2Sn20 bn1Sn10 anSn0, n≥0,

2.26

with the initial conditions:

0S10 c0,

0S20 c1S10 b0,

0u, Z1u,θ0S2c1,

0/u, xZ1xu, S2c1u, S1b0.

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If we denote

Δn:

Sn20 Sn10 Sn0

u, Sn2 u, Sn1 u, Sn

u, θ0Sn2 u, θ0Sn1 u, θ0Sn

, n≥0, 2.28

from the Cramer rule we have

Δnan−Δn1, n≥0, 2.29

Δnbn1

Sn20 −Sn30 Sn0

u, Sn2 −u, Sn3 u, Sn

u, θ0Sn2 −u, θ0Sn3 u, θ0Sn

, n≥0, 2.30

Δncn2

−Sn30 Sn10 Sn0

−u, Sn3 u, Sn1 u, Sn

−u, θ0Sn3 u, θ0Sn1 u, θ0Sn

, n≥0. 2.31

Lemma 2.4. The following formulas hold:

u, SnSn0 u1−λSn0 λS1n−10, n≥0, 2.32

u, xSnx u1−λSn0, n≥1, 2.33

u,θ0SnSn0

1

2u1−λS

n0 λ

S1_n−10, n≥0, 2.34

S1n 0Sn0−S1_n−10Sn10

v, S2_n, n≥0, 2.35

whereS1n x: vθ0Sn1x,n≥0, andS1−1x 0.

Proof. Equations2.32and2.33are deduced, respectively, from2.9and2.8.

We have

v, θ2₀Sn

v, θ0Sn−θ0Snv, θ0Sn

v, θ0Sn

xθ0Sn

0 vθ0Sn

0, n≥0.

2.36

Using2.4, we get

v, θ2

0Sn

vθ0Sn0

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From2.9, we obtain

u, θ0Snδ, θ0Sn u1−λ

δ,θ0Sn

λv, θ₀2Sn

, n≥0. 2.38

According to2.5and2.37, we can deduce2.34. We have

S1₀ x 1, S1₁ x x−ξ2,

S1_n2x x−ξn2S1_n1x−σn2S1n x, n≥0.

2.39

Thenby2.39

S1n 0Sn0−S1_n−10Sn10 σnS1_n−10Sn−10 Sn0

S1n 0 ξnS1_n−10

σn

S1_n−10Sn−10−S1_n−20Sn0

.

2.40

It follows that

S1n 0Sn0−S1_n−10Sn10 n

μ0

σμ

v, S2_n, n≥0, 2.41

hence2.35.

Proposition 2.5. One has

ΔnEnλ2FnλGn, n≥0, 2.42

where

EnSn10

μn0

1 2χ

n0

S1n 0−S_n10

χn0−

v, S2_n, n≥0,

Fn−Sn10

u₁μn0 χn0

v, S2_n

−u₁Sn10χn0−2S_n10χn0 S_n10

v, S2_n, n≥0,

Gn u21

1

2Sn10χ

n0−Sn10χn0

, n≥0,

2.43

with

χnx SnxS_n1x−Sn1xSnx, n≥0,

μnx Sn1x

S1_n−1x−Snx

S1n

x, n≥0.

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Proof. Using2.13, we, respectively, obtain

Sn20 −ξn1Sn10−σn1Sn0, n≥0,

u, Sn2u, xSn1x −ξn1u, Sn1 −σn1u, Sn, n≥0,

u, θ0Sn2u, Sn1 −ξn1u, θ0Sn1 −σn1u, θ0Sn, n≥0.

2.45

Taking into account previous relations, we obtain for2.28the following:

Δn

0 S_n10 Sn0

u, xSn1x u, Sn1 u, Sn

u, Sn1 u, θ0Sn1 u, θ0Sn

, n≥0, 2.46

that is,

Δn−u, xSn1x

Sn10 Sn0

u, θ0Sn1 u, θ0Sn

u, Sn1

Sn10 Sn0

u, Sn1 u, Sn

, n≥0. 2.47

Letn≥0; based on the relations2.32–2.34, it follows that

S_n10 Sn0

u, θ0Sn1 u, θ0Sn

μn0

1 2χ

n0

λ−χn0−

1 2u1χ

n0,

S_n10 Sn0

u, Sn1 u, Sn

χn0−

v, S2 n

λ−u₁χn0.

2.48

From2.48and2.47, we obtain the desired results.

Proposition 2.6. The form uis regular if and only ifΔn/0,n ≥ 0. Then, the coeﬃcients of the

three-term recurrence relation2.15are given by

γ1 Δ0, γ2−λΔ1Δ−20 , 2.49

γn3 ΔnΔn2

Δ2 n1

σn1, n≥0, 2.50

β0 u1, β1c1−ξ0−ξ1λb0Δ−1₀ , 2.51

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Proof

Necessity. From2.27andLemma 2.4, we get

u, xZ1xu, S2u, θ0S2S10− u, S1−S20 λS10−u21, 2.53

and again with2.27and2.42, we can deduce that

Δ0 u, S2u, θ0S2S10− u, S1−S20 u, xZ1x/0. 2.54

Moreover,{Zn}n≥0is orthogonal with respect tou, therefore it is strictly quasiorthogonal of

order two with respect toxv, and then it satisfies2.16withan/0,n≥0. This impliesΔn/0,

n≥ 0. Otherwise, if there exists ann0 ≥ 1 such thatΔn0 0, from2.29,Δ0 0, which is a

contradiction.

Suﬃciency. Let

c0−S10 ξ0, 2.55

c1−u,θ0S2, 2.56

b0 Δ0− u, S2 −c1u, S1. 2.57

We get

u, xZ1xu, S2c1u, S1b0 Δ0/0. 2.58

We haveu, Z1c1u, θ0S20.

From2.56and2.57we get

S20 c1S10 b0S20− u, S2 − u, θ0S2S10− u, S1 Δ0. 2.59

On account of2.54, we can deduce thatS20 c1S10 b00.

Then we had just proved that the initial conditions2.27are satisfied.

Furthermore, the system2.26is a Cramer system whose solution is given by2.29,

2.30, and2.31; with all these numbersan,bn, andcnn≥0, define a sequence polynomials

{Zn}_n≥0by2.16. Then it follows from2.26andLemma 2.2thatuis regular and{Zn}_n≥0is

the corresponding MOPS.

Moreover, by2.22we get

u, Z2 n2

λan

v, S2 n

, n≥0. 2.60

Makingn0 in2.60, it follows that

u, Z2 2

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Based on relations2.58,2.60,2.61, and2.29, we, respectively, obtain

γ1u, xZ1x Δ0; γ2

u, Z2 2

u, xZ1x −λΔ1Δ −2

0 ,

γn3

u, Z_n32

u, Z_n22

ΔnΔn2

Δ2 n1

σn1, n≥0.

2.62

We have proved2.49and2.50.

When{Zn}_n≥0is orthogonal, we have

β0 u1. 2.63

By2.16and the orthogonality of{Zn}_n≥0, we get

u, xZ2₁xc1

u, Z₁2u, S2Z1. 2.64

By virtue of2.13and the regularity ofuwe obtain

u, S2Z1

x2u, Z1

−ξ0ξ1

u, Z₁2λv, xZ1x −ξ0ξ1

u, Z2₁

λb0−ξ0ξ1

u, Z₁2,

2.65

and consequently, we get the second result in2.51from2.58, and2.64. From2.16, and the orthogonality of{Zn}n≥0, we have

βn2

u, Z_n22 cn2

u, Z2_n2u, Sn3Zn2, n≥0. 2.66

Using2.13,2.16, and the the orthogonality of{Sn}n≥0, we have

u, Sn3Zn2λbn1

v, S2_n1−ξn1ξn2

u, Z2_n2, n≥0. 2.67

Taking into account the previous relation,2.66becomes

βn2cn2−ξn1−ξn2λbn1

v, S2_n1

u, Z_n22 , n≥0. 2.68

From2.60and2.29, we have

v, S2 n1

u, Z2 n2

−λ−1_Δ

nΔ−1_n1σn1, n≥0. 2.69

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Moreover, if the formuis regular, for2.29,2.30, and2.31, we get

an−Δ_Δn1

n , n≥0,

2.70

b_n1Dnλ2HnλIn

Δ−1

n σn2, n≥0, 2.71

c_n2−Jnλ2LnλKn

Δ−1

n ξn2, n≥0, 2.72

where

DnSn0

v, S2_n1−χn10

−ξn1Sn20

μn0 1

2χ

n0

−ξn1

S1_n10−S_n20χn0−

v, S2_n, n≥0,

Hn u1Sn0

2χ_n10−v, S2_n1ξ_n1S_n20χn0

u₁χ_n0 μn0

u₁ξn1χn0

S1_n10−S_n20

ξn1

v, S2_n−χn0Sn20 u1Sn20

, n≥0,

In−u21

Sn0χn10 1₂ξn1Sn20χn0−Sn2 0χn0

, n≥0,

JnSn20

μn0

1 2χ

n0

S1_n10−S_n20χn0−

v, S2 n

, n≥0,

Ln u1χn0

2S_n20−S1_n10−u₁Sn20μn0 χn0

−v, S2_nSn20 u1Sn20

, n≥0,

Kn u21

1

2Sn20χ

n0−χn0S_n20

, n≥0.

2.73

In the sequel, we will assume thatvis a symmetric linear form. We need the following lemmas.

Lemma 2.7. If{yn}n≥0and{bn}n≥0are sequences of complex numbers fulfilling

y_n1anynbn1, n≥0, an/0, n≥0,

y0b0,

2.74

then

yn −1na−1n ⎛

⎝n

μ0

aμ ⎞

⎠n

ν0

−1νaν ⎛

⎝ν

μ0

a−1_μ

⎞

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Lemma 2.8. When{Sn}_n≥0given by2.13is symmetric, one has

S2n0 −1 n

σ2n1 n

μ0

σ2μ1, n≥0, S2n10 0, n≥0,

S1_2n0 −1n

n

μ0

σ2μ, n≥0, S_2n11 0 0, n≥0,

S_2n10 −1n

⎛

⎝n

μ0

σ2μ ⎞

⎠_Λ_n_, _n_≥₀_, _S

2n0 0, n≥0,

S1_2n0 0, n≥0, S_2n10 0, n≥0.

2.76

Proof. Asvis symmetric, thenξn0,n≥0, and therefore from2.13we have

S00 1, S10 0, S1₀ 0 1, S1₁ 0 0,

S_n20 −σ_n1Sn0, n≥0, S1_n20 −σn2S1n 0, n≥0,

S₀0 0, S₁0 1, S_n20 −σn1Sn0 Sn10, n≥0,

S1₀ 0 0, S1_n20 −σn2

S1n

0 S1_n10, n≥0,

S₀0 0, S₁0 0, S_n20 −σn1Sn0 2Sn1 0, n≥0.

2.77

Now, it is suﬃcient to useLemma 2.7in order to obtain the desired results.

Let

ωλ−1u₁. 2.78

Corollary 2.9. Ifvis a symmetric form, one has

Δ2n λ2−

1n1

σ_2n1

⎛

⎝n

μ0

σ2μ1 ⎞ ⎠

⎛

⎝n

μ0

σ2μ ⎞ ⎠ 2

{ω−1Λn1}2, n≥0,

Δ2n1λ−1n ⎛

⎝n

μ0

σ2μ1 ⎞ ⎠

2⎛

⎝n

μ0

σ2μ ⎞

⎠_, _n_≥₀_,

2.79

where

Λn n

ν0

1

σ_2ν1

ν

μ0

σ_2μ1 σ2μ

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Proof. FollowingLemma 2.8, for2.43we have

E2n −1 n1

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

2⎛

⎝n

μ0

σ2μ1 ⎞

⎠₁₋_Λ_n_, _n_≥_0; _E_2n1₀_, _n_≥₀_,

F2n2ωλ−1 n1

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

2⎛

⎝n

μ0

σ2μ1 ⎞

⎠₁₋_Λ_n_Λ_n1_, _n_≥₀_,

F2n1 −1n ⎛

⎝n

μ0

σ2μ ⎞ ⎠

⎛

⎝n

μ0

σ2μ1 ⎞ ⎠ 2

, n≥0,

G2nω2λ2−1 n1

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

2⎛

⎝n

μ0

σ2μ1 ⎞

⎠_Λ2

n, n≥0; G2n10, n≥0.

2.81

As a consequence, relations2.81and2.42yield2.79.

Theorem 2.10. The formuis regular if and only ifω−1Λn1/0,n≥0, whereΛnis defined in

2.80.

In this case one has

a2n

σ_2n1 λΘnω−1Λn12

, a2n1 −λσ2n22 Θnω−1Λn12, n≥0, 2.82

b2nσ2n1, n≥0, b2n1σ2n2ω_ω−₋1₁Λ_Λn11

n1

, n≥0, 2.83

c00, c1−ωλ, c2n2

1

λΘnω−1Λn12

, n≥0,

c2n3−λσ2n2Θnω−1Λn11ω−1Λn1, n≥0,

γ1−λ2ω2, γ2−

σ2 1

λ2_ω4, γ2n4

1

λ2Θ2 n1

ω−1Λn112, n≥0,

γ_2n3λ2σ_2n22 Θ2_nω−1Λn12ω−1Λn112, n≥0,

β0λω, β1−λω−

σ1

λω2,

2.84

β_2n2 1

λΘnω−1Λn12

λσ_2n2Θnω−1Λn1ω−1Λn11,

β2n3 1

λΘn1ω−1Λn112

−λσ2n2Θnω−1Λn1ω−1Λn11, n≥0,

2.85

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Proof. FromProposition 2.6andCorollary 2.9, we can deduce thatuis regular if and only if

ω−1Λn1/0,n≥0.

Moreover, from2.70we can deduce2.82.

By2.49,2.51,2.78, and2.79, for2.55,2.56, and2.57we get

c00, c1−u1−ωλ,

b0σ1.

2.86

Whenn≥0 byLemma 2.8, for2.73we get

D2n −1 n

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

⎛

⎝n

μ0

σ2μ1 ⎞ ⎠ 2

1−Λn; D2n1 0,

H2nωλ−1 n

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

⎛

⎝n

μ0

σ2μ1 ⎞ ⎠ 2

2Λn−1; H2n10,

I2nω2λ2−1 n1

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

⎛

⎝n

μ0

σ2μ1 ⎞ ⎠ 2

Λn; I2n10,

J2n0; J2n1 −1nσ2n2 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

2⎛

⎝n

μ0

σ2μ1 ⎞

⎠₁₋_Λ_n₁₋_Λ_n1_,

L2n −1 n1

σ2n1 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

⎛

⎝n

μ0 σ2μ1 ⎞ ⎠ 2 ,

L2n1ωλ−1nσ2n2 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

2⎛

⎝n

μ0

σ2μ1 ⎞

⎠_Λ_n1 ₁₋₂_Λ_n1_Λ_n_,

K2n0, n≥0; K2n1ω2λ2−1nσ2n2 ⎛

⎝n

μ0

σ2μ ⎞ ⎠

2⎛

⎝n

μ0

σ_2μ1

⎞

⎠_Λ_n_Λ_n1.

2.87

Taking into account 2.79,2.80, and 2.86-2.87, relations 2.70,2.71and 2.72give

2.82–2.84.

As a result of relations2.82–2.84andProposition 2.6we get2.85.

Corollary 2.11. 1Ifvis a symmetric positive definite form, then the formuis regular whenω ∈

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2Whenuis regular, it is positive definite form if and only if

λω2 <0, σ

2 1

ω2 >0,

1

λ2_Θ2 n1

ω−1Λn112, n >0,

λ2σ_2n22 Θ2_nω−1Λn12ω−1Λn112, n >0.

2.88

Proof. 1 If v is positive definite, then σn1 > 0, n ≥ 0, thereforeΛn > 0, n ≥ 0 and so

ω−1Λn1/0,n≥0 under the hypothesis of the corollary.

2Ifuis regular, it is positive definite if and only ifγ_n1 >0,n≥0. ByTheorem 2.10, we conclude the desired results.

3. Some Results on the Semiclassical Case

Let us recall that a formvis called semiclassical when it is regular and its formal Stieltjes functionS·;vsatisfies15

φzSz;v CzSz;v Dz, 3.1

whereφmonic,C, andDare polynomials with

Dz −vθ0φ

z vθ0Cz,

Sz;v −

n≥0

v_n zn1.

3.2

The class of the semi-classical formvissmaxdegφ−2,degC−1if and only if the following condition is satisfied22:

c

|Cc||Dc|>0, 3.3

wherec∈ {x:φx 0}, that is,φ,C, andDare coprime.

In the sequel, we will suppose that the form vis semi-classical of classs satisfying

3.1.

Proposition 3.1. Whenuis regular, it is also semi-classical and satisfies

φzSz;u CzSz;u Dz, 3.4

where

φz z3φz, Cz z3Cz−z2φz,

Dz zz u₁−λCz λz2Dz u₁−λφz.

3.5

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Proof. We need the following formula:

Sz;fwfSz;w wθ0f

z, w∈ P, f∈ P. 3.6

From2.7, we haveSz;x2_u _λS_z_;_xv_{. Using}_3.6_{, we get}

z2Sz;u z u₁λzSz;v λ. 3.7

Diﬀerentiating the previous equation, we obtain

z2Sz;u 2zSz;u 1λzSz;v λSz;v. 3.8

By3.1we can deduce3.4and3.5.

Sincevis a semi-classical,Sz;vsatisfies3.1whereφ,CandDare coprime. Letcbe a zero ofφdiﬀerent from 0, which implies thatφc 0. We know that|Cc| |Dc|/0.

IfCc/0, thenCc/0. ifCc 0, thenDc λc2Dc/0. Hence|Cc||Dc|/0.

Corollary 3.2. Introducing

ϑ1 : u1−λφ0, ϑ2: u1−λ

C0 φ0,

ϑ3:C0 u1−λ

C0 φ0λD0,

3.9

1ifϑ1/0, thenss3;

2ifϑ10 andϑ2/0, thenss2;

3ifϑ1ϑ20 andφ0/0 orϑ3/0, thenss1.

Proof. 1From3.9and3.5, we obtainC0 0,D0 ϑ1/0. Therefore, it is not possible

to simplify, which means that the class ofuiss3.

2Ifϑ1 0, then from3.5we haveC0 D0 0. Consequently,3.4–3.6is

divisible byz. Thus,ufulfils3.4with

φz z2φz, Cz z2Cz−zφz,

Dz z u₁−λCz λzDz u₁−λθ0φz.

3.10

IfD0 ϑ2/0, it is not possible to simplify, which means that the class ofuiss2.

3Whenϑ1 ϑ2 0, then it is possible to simplify3.4–3.10byz. Thus,ufulfils

3.4with

φz zφz, Cz zCz−φz,

Dz u₁−λθ0Cz θ02φz

λDz Cz.

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Since we haveC0 −φ0,D0 ϑ3, then we can deduce that ifφ0/0 orϑ3/0, it is not

possible to simplify, which means that the class ofuiss1.

4. Some Examples

In the sequel the examples treated generalize some of the cases studied in13.

4.1. v

the Generalized Hermite Form

Let us describe the casev:Hτ, whereHτis the generalized Hermite form. Here is1

ξn0, n≥0, σn1 1₂n1τ1 −1n, n≥0. 4.1

From4.1, we get

n

μ0

σ_2μ1 Γnτ3/2

Γτ1/2 , n≥0,

n

μ0

σ2μ Γn1, n≥0. 4.2

We wantΛn

_n

ν01/σ2ν1νμ0σ2μ1/σ2μ,n≥0.

But from4.1and4.2, we have 1/σ2ν1νμ0σ2μ1/σ2μ 1/Γτ1/2hν, with

hn Γ

nτ1/2

Γn1 , n≥0, 4.3

fulfilling

n1hn1−nhn

τ1

2

hn, n≥0, 4.4

and so

Λn 1

τ1/2Γτ1/2

n

ν0

ν1hν1−νhν 1

Γτ3/2

Γnτ3/2

Γn1 , n≥0. 4.5

Then we getTable 1.

Proposition 4.1. Ifv Hτis the generalized Hermite form, then the formuτ, ω, λgiven by

2.9has the following integral representation:

uτ, ω, λ, ff0 λω−1f0 λ

Γτ1/2P

_∞

−∞

|x|2τ

x e

−x2

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Table 1

Δn Δ2n −1

n1 λ2

Γτ1/2Γnτ1/2Γ

2_n₁_ω₋₁_Λ

n12, n≥0,

Δ2n1 −1n

λ

Γ2_τ₁_/₂Γ

2_n_τ₃_/₂_Γ_n₁_, _n_≥_0.

an a2n

nτ1/22 λΓτ1/2

hn

ω−1Λn12

, n≥0, a2n1−λΓτ1/2n1

hn1ω−1Λn1 2

, n≥0.

bn b2nnτ1/2, n≥0, b2n1 n1ω−1Λn11

ω−1Λn1

, n≥0.

cn c00, c1−ωλ, c2n2 1 λ

nτ1/2

Γτ1/2

hn

ω−1Λn12

, n≥0,

c2n3−λ

n1Γτ1/2

nτ1/2hn

ω−1Λn11ω−1Λn1, n≥0.

γ1−λ2ω2, γ2−τ1/2 2

λ2_ω4 ,

γn1 γ2n3−

λ2_Γ2_τ₁_/₂

h2

n1

ω−1Λn112ω−1Λn12, n≥0,

γ2n4−

1 λ2_Γ2_τ₁_/₂

nτ3/22h2

n1

ω−1Λn114

, n≥0.

β0ωλ, β1−ωλ−τ1/2

λω2 ,

βn β2n3−λn1Γτ1/2

nτ1/2hn

ω−1Λn11ω−1Λn1−nτ3/2 λΓτ1/2

hn1

ω−1Λn112

, n≥0,

β2n2

1 λ

1

Γτ1/2

nτ1/2hn

ω−1Λn12

λΓτ1/2

hn1

ω−1Λn11ω−1Λn1, n≥0.

It is a quasi-antisymmetric (uτ, ω, λ_2n2 0,n≥0) and semi-classical form of classssatisfying

the following functional equation:

τ0, ω /1, z3Sz;u0, ω, λ −z22z21Sz;u0, ω, λ

−2z3−2λωz2λω−1, s3,

τ0, ω1, zSz;u0,1, λ −2z21Sz;u0,1, λ−2z−2λ, s1,

4.7

τ /0, ω /1, z3Sz;uτ, ω, λ −z22z2−2τ1Sz;uτ, ω, λ−2z3

−2λωz22τz2τλω−1 λω−1, s3,

τ /0, ω1, z2Sz;uτ,1, λ z−2z22τ−1Sz;uτ,1, λ−2z2−2λz2τ, s2.

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Proof. It is well known that the generalized Hermite form possesses the following integral

representation1:

v, f

_∞

−∞

1

Γτ1/2|x|

2τ_e−x2

fxdx, Rτ>−1

2, ∀f ∈ P. 4.9

Following2.11, we obtain4.6. Also the formuis quasi-antisymmetric because it satisfies

u, x2n2λv, x2n10, n≥0, 4.10

sincevis symmetric by hypothesis.

Whenτ 0,vis classical and satisfies3.4with22

φx 1, Cz −2z, Dz −2. 4.11

Then,ϑ1λω−1,ϑ20.

Now, it is suﬃcient to useCorollary 3.2andProposition 3.1in order to obtain4.7. Ifτ /0, the formvis semi-classical of class one and satisfies3.4with23

φx x, Cz −2z22τ, Dz −2z. 4.12

Thereforeϑ10,ϑ2λω−12τ1,ϑ32τ.

ByProposition 3.1andCorollary 3.2we can deduce4.8.

4.2. v

the Corecursive of the Second Kind Chebychev Form

Let us describe the casev : J−1/2,1/2; it is the corecursive of the second kind Chebychev

functional. Here is1

ξ0−

1

2, ξn10, n≥0, σn1 1

4, n≥0. 4.13

In this case we have the following result.

Lemma 4.2. Forn≥0, one has

S2n0 −

1n

22n , S2n10

−1n 22n1, S

1

2n0

−1n 22n , S

1

2n10 0,

S_2n0 n−1

n1

22n−1 , S

2n10 n1−

1n 22n ,

S1_2n0 0,

S1_2n10 n1−1

n

22n , S

2n0 nn1

−1n1 22n−2 , S

2n10 nn1−

1n1 22n−1 .

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Proof. The proof is analogous for the demonstration ofLemma 2.8.

FollowingLemma 4.2, for2.44we have

χ2n0

2n1

24n , n≥0; χ2n10

n1

24n1, n≥0; χ

2n0 0, n≥0;

χ_2n10 n1

24n , n≥0; μ2n0 −

n

24n−1, n≥0; μ2n10 −

n1

24n1, n≥0.

4.15

Therefore, we get for2.42

Δ2n n2n1−

1n1 26n λ

2 ₈_n_n₁_u

1−1

−1n 26n1λ

n12n1u2₁−1

n1

26n , n≥0,

Δ2n1 n12n1−1 n1

26n3 λ

2

8n12u₁1−1

n

26n4λn12n3u

2 1

−1n1

26n3 , n≥0.

4.16

Then we obtain

Δ2n4−1 n1

26n1 tn−x1tn−x2, n≥0,

Δ2n14−1 n1

26n4 tn−x3tn−x4, n≥0,

4.17

where

x1

1 4

−3t−2λt2−4λt−4λ2−4λ1/2

, x2

1 4

−3t−2λ−t2−4λt−4λ2−4λ1/2

,

x3

1 4

−5t−2λt2λ24λ1/2

, x4

1 4

−5t−2λ−t2λ24λ1/2

,

u₁ tλ.

4.18

On account ofProposition 2.6, we can deduce that the formugiven by2.9is regular if and only iftn−xi/0,n≥0, 1≤i≤4.

In the sequel, we suppose that the last condition is satisfied.

By virtue of4.17and Lemma 4.2, relations2.49–2.52, and2.55–2.57,2.70–

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Table 2

an a2n− 1 8

tn−x3tn−x4

tn−x1tn−x2

, n≥0, a2n1

1 8

tn−x1tn−x2

tn−x3tn−x4

, n≥0.

bn b0−2x1x2

1 4−

t

2 tλ 1 2

2_, _b

2n1

1 4

t 8

2n1t−λ

tn−x1tn−x2

, n≥0,

b2n21

4 t 8

2n3tλ

tn−x3tn−x4

, n≥0.

cn c0−

1

2, c1− 1

2−t−λ, c2n3 1 8

2t2n1n1t−λ−λ

tn−x3tn−x4

, n≥0,

c2n2−1

8

2t2n1n1tλ−λ

tn−x1tn−x2 , n≥0.

γ1−2x1x2, γ2

λ 16

x3x4

x2

1x22

,

γn1 γ2n3−

1 4

tn−x1tn1−x1tn−x2tn1−x2

tn−x32tn−x42

, n≥0,

γ2n4−1

4

tn−x3tn1−x3tn−x4tn1−x4

tn1−x12tn1−x22

, n≥0.

β0tλ, β1−t−λ−

λ 2x1x2{−

2x1x2

1 4−

t

2 tλ 1 2

2_}_,

β2n3

1 8

2t2n1n1t−λ−λ

tn−x3tn−x4

1 2

tn−x3tn−x4

tn1−x1tn1−x2

βn

t 4

2n3tλ

tn1−x1tn1−x2

, n≥0,

β2n2−

1 8

2t2n1n1tλ−λ

tn−x1tn−x2 −

1 2

tn−x1tn−x2

tn−x3tn−x4−

t 4

2n1t−λ

tn−x3tn−x4

, n≥0.

Proposition 4.3. Ifv J_−1/2,1/2 is the corecursive of the second kind Chebychev form, then the

formut, λgiven by2.9has the following integral representation:

ut, λ, f 1−λf0 tf0 λ

πP 1 −1 1 x

1−x

1xfxdx, ∀f ∈ P. 4.19

It is a semi-classical form of classssatisfying the following functional equation:

t /0, z3z2−1Sz;ut, λ −z2z2−z−1Sz;ut, λ t−2λ1z2tz−t, s3

t0, zz2−1Sz;u0, λ −z2z1Sz;u0, λ−2λ1, s1.

(22)

Proof. It is well known thatvJ−/2,1/2possesses the following integral representation1:

v, f

₁

−1

1

π

1−x

1xfxdx, f∈ P. 4.21

From2.11we easily obtain4.19. The formvsatisfies3.4with15

φx x2−1, Cz 1, Dz −2. 4.22

Therefore,ϑ1−t,ϑ2t,φ0/0.

Now, we can simply useProposition 3.1andCorollary 3.2in order to obtain4.20.

Corollary 4.4. Whent0 andλ−1, one has

βn −1n1, n≥0, γ1−

1

2, γn2− 1

4, n≥0,

zz2−1Sz;u0,−1 −z2z1Sz;u0,−1 3, s1.

4.23

Proof. FromTable 2, we reach the desired results.

Remarks 4.5. 1 One has the formh−1u0,−1 L−3/2,1/2, whereLα, βis studied in

24.

2Let{Zn1}n≥015,19be the first associated sequence of{Zn}_n≥0 orthogonal with

respect tou0,−1andβ1n ,γ_n11 the coeﬃcients of the three-term recurrence relations; we have

β1n βn1 −1n, n≥0; γ_n11 γn2−1₄, n≥0. 4.24

The sequence{Z1n }n≥0is a second-order self-associated sequence; that is,{Zn1}n≥0is

identical to its associated orthogonal sequence of second kindsee25.

Acknowledgments

Sincere thanks are due to the referee for his valuable comments and useful suggestions and his careful reading of the manuscript. The author is indebted to the proofreader the English teacher Hajer Rebai who checked the language of this work.

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