Stability of Third Degree Linear Functionals and Rational Spectral Transformations

Abstract

In this paper, we consider linear forms of the third degree class. This means that the Stieltjes function associated with the corresponding moment sequence satisfies a cubic equation with polynomial coefficients. We introduce the notion of a primitive triple of a strict third degree form. A simplification criterion of the corresponding cubic algebraic equation is given. Moreover, we show that the class of third degree linear forms is closed under rational spectral transformations. Several consequences of this fact are deduced. In particular, we illustrate with several examples the set of third degree linear forms is stable for the most standard algebraic operations in the linear space of linear forms.

1 Introduction

The third degree regular linear forms (TDRFs) are the sole creation of Maroni and Ben Salha [6]. They arise as a natural extension of the well known second degree regular forms [23, 24]. These forms are characterized by the fact that their formal Stieltjes function $S(u)(z):=-\sum _{n\ge 0}{(u)_{n}/z^{n+1}}$ satisfies a cubic equation with polynomial coefficients

$$\begin{aligned} A(z)S^3(u)(z) + B(z)S^2(u)(z) + C(z)S(u)(z) + D(z)=0. \end{aligned}$$

Here $(u)_{n}$ denotes the n-th moment of the linear form u.

A regular linear form u is called a strict third degree form (STDRF) if it is a (TDRF) and it is not a second degree regular form, i. e. its Stieltjes function satisfies a quadratic equation with polynomial coefficients. Some (TDRF) properties are given in [5, 6]. Namely, a third degree form belongs to the Laguerre–Hahn class [6], but the converse is not true in general. The authors of [3, 5] determined all classical forms which are (TDRFs). It is worthy to mention here that Hermite, Laguerre and Bessel forms are not (TDRFs), but only Jacobi forms for some choices of the parameters possess this property. Some examples of (TDRF) arising from 2-orthogonality are studied in [6]. The resolution of the third degree equation verified by S(u) using Cardan’s formulas gives some information about u and its corresponding integral representation that is obtained due to the Hardy space theory for a half-plane. In other words, when $S(u)\in {\mathcal H}^p, 1\le p< \infty ,$ where ${\mathcal H}^p$ is the vector space of holomorphic functions g in $\mathrm{Im} z >0$ such that

$$\begin{aligned} \sup \limits _{y>0}\int _{-\infty }^{+\infty }\left| g(x+iy)\right| ^p\mathrm{d}x< +\infty , \end{aligned}$$

then the function satisfies [13]

$$\begin{aligned} g(z)=\frac{1}{\pi }\int _{-\infty }^{+\infty }\frac{\mathrm{Im} g(t+i0)}{t-z}\mathrm{d}t, \; \mathrm{Im} z> 0. \end{aligned}$$

For some examples, see [6].

On the other hand, a rational spectral transformation [29] of the Stieltjes function S(u) is a new Stieltjes function defined by

$$\begin{aligned} S (\widetilde{u})(z)=\left( \frac{a S(u)+b}{c S(u)+d} \right) (z), \quad a(z)d(z)-b(z)c(z) \ne 0, \end{aligned}$$

where a, b, c,  and d are coprime polynomials. When $c(z)=0$, the spectral transformation is said to be linear. Linear spectral transformations appear in the framework of discrete Darboux transformations based on the LU and LU factorizations of the Jacobi matrix J of the representation of the multiplication operator in terms of a orthogonal polynomial basis with respect to a linear form u (see [8, 28]). This is the matrix counterpart of the three term recurrence relation the sequence of orthogonal polynomials satisfies. Linear spectral transformations are generated by Christoffel and Geronimus transformations which yield linear forms $\widetilde{u}, \hat{u},$ defined by $\widetilde{u}= p(x) u,$ and $p(x)\hat{u}= u,$ respectively, where p(x) is a polynomial. On the other hand, rational spectral transformations appear in a natural way in the study of Laguerre–Hahn linear forms taking into account they preserve the Laguerre–Hahn character (see [12, 20]). They are generated by linear spectral transformations and associated and anti-associated transformations (see [29]). For more information concerning these last two transformations, see [26, 27].

The goal of our contribution is to search for a generator system of the set of third degree forms. More precisely, we show that the class of third degree forms is closed under rational spectral transformations of the form. Several consequences of this fact are deduced. For instance, we prove that for the most usual transformations of a linear form (associated form, Christoffel and Geronimus transformations of a form, co-recursive form, inverse form, etc...) the set of third degree forms is stable. In this way, we present a constructive way to generate third degree forms, emphasizing the algebraic approach based on the study of the vector space of linear forms and the corresponding set of Stieltjes functions associated with them.

The structure of this paper is as follows. In Sect. 2, we present the basic background concerning algebraic (topological) aspects in the theory of linear forms and Orthogonal Polynomials (OP) which will be useful in the forthcoming sections. In Sect. 3, we recall the definitions as well as the main properties of (TDRF). In Sect. 4, we introduce the notion of a primitive triple of a strict third degree form and we obtain a simplification criterion of the corresponding cubic algebraic equation verified by its formal Stieltjes function. In Sect. 5, we start by giving the functional link between two regular forms u and v assuming that the Stieltjes function of one of them is obtained by the action of a rational spectral transformation of the Stieltjes function of the other. As a consequence, we state our main result. We deal with a stability problem, i. e. we show that u is a third degree form if and only if v is also a third degree form. This leads us to state in Sect. 6 some interesting applications concerning the stability of the class of forms of the third degree by the corresponding transformations: associated form, Christoffel and Geronimus form, co-recursive form, inverse form as well as the addition of a delta Dirac linear form and its derivative. These last two transformation appear in the literature in the framework of the spectral analysis of higher order linear differential operators with sequences of orthogonal polynomials as eigenfunctions (see [14, 15]).

2 Notation and Preliminaries

Let $\mathcal {P}$ be the vector space of polynomials with coefficients in $\mathbb {C}$ (the field of complex numbers) and let $\mathcal {P}^{\prime }$ be its topological dual space, whose elements are called linear forms (or linear functionals). We denote by $\langle u, p\rangle$ the image of $p \in \mathcal {P}$ by $u \in \mathcal {P}^{\prime }$. In particular, we denote by $\left\langle u, x^{n}\right\rangle :=(u)_{n}, n \ge 0$, the moments of u. Giving $u\in \mathcal {P}^{\prime }$, you can introduce the formal series $F(u)(z):=\sum _{n\ge 0}(u)_nz^n$ or the so-called formal Stieltjes function once you know the sequence of moments $\{(u)_n\}_{n\ge 0}$ [21]

$$\begin{aligned} S(u)(z):=-z^{-1}F(u)(z^{-1})=-\sum _{n\ge 0}\frac{(u)_n}{z^{n+1}}. \end{aligned}$$

(2.1)

Remark 2.1

For any $f \in \mathcal {P}$ and $u\in \mathcal {P'}$, $S(u)(z)=f(z)$ if and only if $u=0$ and $f=0$.

Let $f, p\in \mathcal {P}$ and $u, v \in \mathcal {P}'$. By transposition of linear operators in $\mathcal {P}$ we can introduce the following transformations in $\mathcal {P}'$ [21]: The left product of u by a polynomial f,  denoted by fu,  is defined as $\langle fu, p \rangle :=\langle u, fp \rangle$; the derivative of a form $u'=Du$ is defined as $\langle u', p \rangle :=-\langle u, p' \rangle$; the division of a form by a first degree polynomial $(x-c)^{-1}u,~ c\in \mathbb {C},$ is given by $\langle (x-c)^{-1}u,p \rangle :=\langle u,\theta _c p \rangle ,$ where $\theta _c$ is the divided difference operator $(\theta _cp)(x):=\frac{p(x)-p(c)}{x-c}$; the Cauchy product of two forms uv, $\langle uv, p \rangle :=\langle u, vp \rangle ,$ where $(vp)(x):=\langle v,\frac{xp(x)-\xi p(\xi )}{x-\xi } \rangle$ is the right-multiplication of a form by a polynomial.

For any $f \in \mathcal {P}$ and $u, v \in \mathcal {P'}$, we have (see [21])

$$\begin{aligned} S(u v)(z)&=-zS(u)(z)S(v)(z), \end{aligned}$$

(2.2)

$$\begin{aligned} S(f u)(z)&=f(z)S(u)(z)+(u\theta _0 f)(z). \end{aligned}$$

(2.3)

Let us consider a polynomial sequence $\{ P_n\}_{n\ge 0}$ such that $\deg {P_n}=n,~n\ge 0.$ Then, there exists a unique sequence $\{ u_n\}_{n\ge 0}$, $u_n\in \mathcal {P}',~ n\ge 0$, called the dual sequence of $\{ P_n\}_{n\ge 0}$ such that $\langle u_n,P_m \rangle =\delta _{n,m},~ n, m \ge 0$.

Next, we recall the definition of orthogonal polynomial sequence (OPS) with respect to a linear form and some of the main facts concerning such sequences. Let ${u} \in \mathcal {P}^{\prime }$ and $\left\{ P_{n}\right\} _{n \ge 0} \subset \mathcal {P} .$ We say that $\left\{ P_{n}\right\} _{n \ge 0}$ is an (OPS) with respect to u if it is a simple set of polynomials and the orthogonality conditions

$$\begin{aligned} \left\langle {u}, P_{n} P_{m}\right\rangle = k_{n} \delta _{n, m}, \quad n, m \ge {0}, \end{aligned}$$

hold, where $\left\{ k_{n}\right\} _{n\ge 0}$ is a sequence of nonzero complex numbers. Here, $\delta _{n, m}$ is the Kronecker’s delta. In such a case, u is called regular (or quasi-definite) and $\left\{ P_{n}\right\} _{ n \ge 0}$ is the corresponding (OPS). Consequently, $\deg P_{n}=n, ~n \ge 0$, and we can choose $P_{n}$ as monic, i. e. with unit leading coefficient ($P_{n}(x)=x^{n}+$ lower degree terms). Then the sequence $\left\{ P_{n}\right\} _{n\ge 0}$ is called a monic orthogonal polynomial sequence (MOPS). Notice that $u=(u)_{0} u_{0},(u)_{0} \ne 0.$ The linear form u is said to be normalized if $(u)_{0}=1.$ In such a case $u=u_{0}$. In this work,we will consider monic polynomial sequences and normalized forms.

It is well known that any monic (MOPS) $\left\{ P_{n}\right\} _{n \ge 0}$ is characterized by a recurrence relation of order two (Favard’s theorem)

$$\begin{aligned} P_{n+2}(x)=\left( x-\beta _{n+1}\right) P_{n+1}(x)-\gamma _{n+1} P_{n}(x), \quad n \ge 0, \end{aligned}$$

(2.4)

with initial conditions $P_{0}(x):=1$ and $P_{1}(x):=x-\beta _{0}.$ Here, $\left\{ \beta _{n}\right\} _{n\ge 0}$ and $\left\{ \gamma _{n+1}\right\} _{n \ge 0}$ are sequences of complex numbers such that $\gamma _{n+1} \ne 0$ for all n.

Moreover, the numbers $\beta _n, \gamma _{n+1},$ known in the literature as the recurrence coefficients, can be expressed as

$$\begin{aligned} \beta _n=\frac{\langle u,xP_n^2(x)\rangle }{k_n}, ~~~~\gamma _{n+1}=\frac{k_{n+1}}{k_n},~~~~n\ge 0. \end{aligned}$$

3 Third Degree Forms

Definition 3.1

[5] A linear form u is called a third degree form (TDF) if there exist three polynomials A monic, B and C such that

$$\begin{aligned} A(z)S^3(u)(z)+B(z)S^2(u)(z)+C(z)S(u)(z)+D(z)=0, \end{aligned}$$

(3.1)

where D depends on A, B, C and u.

Remark 3.1

In general, the form u is not regular. When it fulfils this property, we call it a third degree regular form (TDRF).

The form u is a (TDRF) if and only if the following conditions hold

$$\begin{aligned} \begin{aligned}&A(x) u^3-x B(x) u^2+x^2 C(x) u=0,\\&\langle u^3 ,\theta _0^2 A \rangle -\langle u^2 ,\theta _0 B \rangle +\langle u , C \rangle =0,\\&\langle u^3 ,\theta _0 A \rangle -\langle u^2 , B \rangle +\langle u ,x C(x) \rangle =0. \end{aligned} \end{aligned}$$

(3.2)

As a consequence,

$$\begin{aligned} D(z)=\big (u^3\theta _0^3 A\big )(z)-\big (u^2\theta _0^2 B\big )(z)+\big (u\theta _0 C\big )(z). \end{aligned}$$

(3.3)

Remark 3.2

• 1. A regular linear form u is called a second degree form if the corresponding Stieltjes function satisfies a quadratic equation with polynomial coefficients [23]

  $$\begin{aligned} B(z)S^2(u)(z)+C(z)S(u)(z)+D(z)=0, \end{aligned}$$

  (3.4)

  where B, C, D satisfy $B\ne 0, C^2-4BD\ne 0, D\ne 0,$ according to the regularity of u.

• 2. The polynomial D is given in terms of B, C and u as : $D(z)=-(u^2\theta _0^2B)(z)+(u\theta _0C)(z)$.

• 3. The polynomials B and C, given in (3.4), are not unique because B and C can be multiplied by an arbitrary polynomial. If the polynomials B, C and D in (3.4) are coprime, then the pair (B, C) is called a primitive pair. Such a pair is unique [23].

• 4. When the form u is a third degree regular form (TDRF) and not a second degree form, we shall call it a strict third degree regular form (STDRF) [5].

Among the most well-known forms which are of strict third degree (STDRF), we can find the Jacobi form $\mathcal {V}:=\mathcal {J}(-\frac{2}{3}, -\frac{1}{3})$ [5]. Let us remind that $\mathcal {V}$ satisfies the following equations

$$\begin{aligned} \begin{aligned}&(x+1)^2(x-1)\mathcal {V}^3=0,\\&(z+1)^2(z-1)S^3(\mathcal {V})(z)+1=0.\\ \end{aligned} \end{aligned}$$

Notice that every second degree form is semiclassical (see [3]). In [5] (respectively [3]) all the classical forms which are of second degree (respectively of strict third degree) are determined. Hermite, Laguerre, and Bessel forms are not of second degree (respectively, strict third degree). Only some Jacobi forms are of second degree (respectively, strict third degree). Indeed,

Theorem 3.1

[3]. Among the classical forms, only the Jacobi forms $\mathcal {J}(k-1/2,l-1/2)$ are second degree forms, provided $k+l\ge 0,~ k, l \in \mathbb {Z}$.

Theorem 3.2

[5]. Among the classical forms, only the Jacobi forms $\mathcal {J}(k+q/3,l-q/3)$ are (STDRFs), assuming $k+l\ge -1,~ k, l \in \mathbb {Z},~q\in \{1,2\}$.

4 Primitive Triple

Now, we will deal with the problem of simplification of equation (3.1) or (3.2).

It is obvious that a third degree form satisfies an infinite number of equations like (3.1). Indeed, multiplying (3.1) by a polynomial $\chi$ (monic) we obtain

$$\begin{aligned} \chi (z)A(z)S^3(u)(z)+\chi (z)B(z)S^2(u)(z)+\chi (z)C(z)S(u)(z)+\chi (z)D(z)=0. \end{aligned}$$

Equivalently,

$$\begin{aligned}&\chi (x)A(x)u^3-x\chi (x)B(x)u^2+x^2\chi (x)C(x)u=0,\\&\langle u^3, \theta _0^2(\chi A) \rangle -\langle u^2, \theta _0(\chi B) \rangle +\langle u, \chi C \rangle =0,\\&\langle u^3, \theta _0(\chi A) \rangle -\langle u^2, \chi B \rangle +\langle u, x\chi (x) C(x) \rangle =0. \end{aligned}$$

Lemma 4.1

Let u be a (STDRF) satisfying

$$\begin{aligned}&A_1(z)S^3(u)(z)+B_1(z)S^2(u)(z)+C_1(z)S(u)(z)+D_1(z)=0, \end{aligned}$$

(4.1)

$$\begin{aligned}&A_2(z)S^3(u)(z)+B_2(z)S^2(u)(z)+C_2(z)S(u)(z)+D_2(z)=0. \end{aligned}$$

(4.2)

Let us denote by A the greatest common divisor of $A_1$ and $A_2$. Then, there exist polynomials B, C and D such that

$$\begin{aligned} A(z)S^3(u)(z)+B(z)S^2(u)(z)+C(z)S(u)(z)+D(z)=0. \end{aligned}$$

(4.3)

Proof

If $A=\gcd {(A_1, A_2)}$, then there exist two coprime polynomials $\widetilde{A}_1$ and $\widetilde{A}_2$ such that

$$\begin{aligned} A_1=A\widetilde{A}_1,~~~~A_2=A\widetilde{A}_2. \end{aligned}$$

The operation $A_2\times$(4.1)$-A_1\times$(4.2) gives

$$\begin{aligned}&\Big (B_1(z)\widetilde{A}_2(z)-B_2(z)\widetilde{A}_1(z)\Big )S^2(u)(z)+\Big (C_1(z)\widetilde{A}_2(z)-C_2(z)\widetilde{A}_1(z)\Big )S(u)(z) \\&\quad +D_1(z)\widetilde{A}_2(z)-D_2(z)\widetilde{A}_1(z)=0. \end{aligned}$$

Since u is a (STDRF), we have

$$\begin{aligned} B_1(z)\widetilde{A}_2(z)-B_2(z)\widetilde{A}_1(z)&=0,\\ C_1(z)\widetilde{A}_2(z)-C_2(z)\widetilde{A}_1(z)&=0,\\ D_1(z)\widetilde{A}_2(z)-D_2(z)\widetilde{A}_1(z)&=0. \end{aligned}$$

But the polynomials $\widetilde{A}_1$ and $\widetilde{A}_2$ are coprime. Using Bézout identity, there exist three polynomials B, C and D such that

$$\begin{aligned}&B_2=\widetilde{A}_2B , ~~~~ C_2=\widetilde{A}_2C , ~~~~ D_2=\widetilde{A}_2D,\\&B_1=\widetilde{A}_1B , ~~~~ C_1=\widetilde{A}_1C , ~~~~ D_1=\widetilde{A}_1D. \end{aligned}$$

Hence, from (4.1) or (4.2) we get (4.3). $\square$

Corollary 4.1

If u is (STDRF), then the equation (3.1) cannot be simplified if and only if the polynomials A, B, C and D are coprime.

Then, it is necessary to give a criterion which allows us to simplify the cubic algebraic equation (3.1).

Proposition 4.1

If u is (STDRF), then (3.1) cannot be simplified if and only if the polynomials A, B and C are coprime or, equivalently, if for any common zero c, the following condition holds:

$$\begin{aligned} \langle u^3, \theta _0^2\theta _c A \rangle -\langle u^2, \theta _0\theta _c B \rangle +\langle u, \theta _c C \rangle \ne 0. \end{aligned}$$

(4.4)

Proof

When A, B and C have a common zero c, then we can write

$$\begin{aligned} A(x)=(x-c)A_c(x),~~~~B(x)=(x-c)B_c(x),~~~~C(x)=(x-c)C_c(x). \end{aligned}$$

This yields

$$\begin{aligned} \begin{aligned} (\theta _0C)(x)&=C_c(x)-c(\theta _0C_c)(x),\\ (u\theta _0C)(x)&=(uC_c)(x)-c(u\theta _0C_c)(x),\\ (u^2\theta _0^2B)(x)&=(u^2\theta _0B_c)(x)-c(u^2\theta _0^2B_c)(x),\\ (u^3\theta _0^3A)(x)&=(u^3\theta _0^2A_c)(x)-c(u^3\theta _0^3A_c)(x). \end{aligned} \end{aligned}$$

(4.5)

Then, from (3.3) we get

$$\begin{aligned} D(x)=(u^3\theta _0^2A_c)(x)-(u^2\theta _0B_c)(x)+(uC_c)(x)-cD_c(x), \end{aligned}$$

(4.6)

with

$$\begin{aligned} D_c(x)=(u^3\theta _0^3A_c)(x)-(u^2\theta _0^2B_c)(x)+(u\theta _0C_c)(x). \end{aligned}$$

(4.7)

But, from

$$\begin{aligned} \big (u\theta _0C\big )(x)=\big (\theta _0(uC)\big )(x), \end{aligned}$$

we deduce

$$\begin{aligned} \big (u\theta _0C_c\big )(x)&=\big (\theta _0(uC_c)\big )(x),\\ \big (u^2\theta _0^2B_c\big )(x)&=\big (\theta _0(u^2\theta _0B_c)\big )(x),\\ \big (u^3\theta _0^3A_c\big )(x)&=\big (\theta _0(u^3\theta _0^2A_c)\big )(x). \end{aligned}$$

Replacing the above results in (4.7) we have

$$\begin{aligned} D_c(x)=\Big (\big (\theta _0(u^3\theta _0^2A_c)\big )-\big (\theta _0(u^2\theta _0 B_c)\big )+\big (\theta _0(uC_c)\big )\Big )(x), \end{aligned}$$

which is equivalent to

$$\begin{aligned} D_c(x)= & {} \frac{1}{x}\Big \{(u^3\theta _0^2A_c)(x)-(u^2\theta _0B_c)(x)+(uC_c)(x) \nonumber \\&-\Big (\langle u^3, \theta _0^2 A_c \rangle -\langle u^2, \theta _0 B_c\rangle +\langle u, C_c \rangle \Big )\Big \}. \end{aligned}$$

(4.8)

Substituting (4.6) in (4.8) we obtain

$$\begin{aligned} D(x)=(x-c)D_c(x)+\Big \{ \langle u^3, \theta _0^2 A_c \rangle -\langle u^2, \theta _0 B_c \rangle +\langle u, C_c \rangle \Big \}. \end{aligned}$$

Therefore, from $A_c=\theta _cA$, $B_c=\theta _cB$ and $C_c=\theta _cC$, we get

$$\begin{aligned} D(x)=(x-c)D_c(x)+\Big \{ \langle u^3, \theta _0^2 \theta _cA \rangle -\langle u^2, \theta _0 \theta _cB \rangle +\langle u, \theta _cC \rangle \Big \}. \end{aligned}$$

Hence the desired result. $\square$

In a natural way, this allows us to give the following definition.

Definition 4.1

Let u be a (STDRF) satisfying (3.1). When the triple (A, B, C) verifies the conditions given in Proposition 4.1, it will be said a primitive triple associated with u.

Remark 4.1

From the above, there exists a unique primitive triple for the (STDRF) forms.

5 Rational Spectral Transformation of the Third Degree Forms

5.1 Rational Spectral Transformation

A rational spectral transformation [29] of the formal Stieltjes function

$$\begin{aligned} S(z)=-\sum _{n=0}^{\infty } \frac{(u)_{n}}{z^{n+1}}, \end{aligned}$$

(5.1)

is a new formal Stieltjes function defined

$$\begin{aligned} \widetilde{S}(z)=\left( \frac{a S+b}{c S+d}\right) (z), \quad a(z)d(z)-b(z)c(z) \ne 0, \end{aligned}$$

(5.2)

where a, b, c,  and d are coprime polynomials. In particular, when $c(z)=0$, the spectral transformation (5.2) is said to be linear. Among the basic linear spectral transforms you get the Christoffel and Geronimus transformations (see [8, 28]) which constitute a set of generators of the family of linear spectral transformations (see [29])

These polynomials should be chosen in such a way that the new formal Stieltjes function $\widetilde{S}$ has the same asymptotic behavior at infinity as the initial one (5.1)

$$\begin{aligned} \widetilde{S}(z)=- \sum _{n=0}^{\infty } \frac{(\widetilde{u})_{n}}{z^{n+1}}, \end{aligned}$$

where $\{(\widetilde{u})_{n}\}_{n\ge 0}$ is the sequence of transformed moments. Hence, in general, the coefficients of the polynomials a, b, c and d depend on the moments $\{(u)_{n}\}_{n\ge 0}$ of the initial form. In particular, this means that the rational spectral transformations do not constitute an algebraic group. Indeed, for a given rational spectral transformation there exist many different reciprocal rational spectral transformations. Nevertheless, it is clear that one can always compose two rational spectral transformations and, moreover, for a given rational spectral transformation there is at least one reciprocal. On the other hand, notice that rational spectral transformations are generated by linear spectral transformations, associated and anti-associated transformations (see [29]).

As a first result, we analyze the connection between two linear forms whose formal Stieltjes functions are related by a rational spectral transformation.

Proposition 5.1

Let u and v be two forms in $\mathcal {P}'$. The formal Stieltjes functions S(u) and S(v) associated with the forms u an v (resp.) are related by a rational spectral transformation,

$$\begin{aligned} {S}(v)(z)=\left( \frac{a S(u)+b}{c S(u)+d}\right) (z), \quad a(z)d(z)-b(z)c(z) \ne 0, \end{aligned}$$

(5.3)

where a, b, c,  and d are coprime polynomials, if and only if u and v are related by

$$\begin{aligned}&c(x)uv- xd(x)v+xa(x)u=0, \end{aligned}$$

(5.4)

$$\begin{aligned}&(uv\theta _0c)(x)- (vd)(x)+(ua)(x)- xb(x)=0. \end{aligned}$$

(5.5)

Proof

Relation (5.3) is equivalent to

$$\begin{aligned} c(z)S(u)(z)S(v)(z)+d(z)S(v)(z)=a(z)S(u)(z)+b(z). \end{aligned}$$

Multiplying both sides of the last equation by z and using (2.2), we can write

$$\begin{aligned} -c(z)S(uv)(z)+zd(z)S(v)(z)=za(z)S(u)(z)+zb(z). \end{aligned}$$

Therefore, using (2.3), the last equation becomes

$$\begin{aligned} S(c(x)uv)(z)&-S(xd(x)v)(z)+S(xa(x)u)(z)\\&\quad =(uv\theta _0c)(z)-(vd)(z)+(ua)(z)-zb(z), \end{aligned}$$

that reads as

$$\begin{aligned} S\big (c(x)uv-xd(x)v+xa(x)u\big )(z)=P(z)\in \mathcal {P}, \end{aligned}$$

with

$$\begin{aligned} P(z)=(uv\theta _0c)(z)-(vd)(z)+(ua)(z)-zb(z). \end{aligned}$$

From Remark 2.1,

$$\begin{aligned} c(x)uv-xd(x)v+xa(x)u=0,~~\text {in}~\mathcal {P}',~~~~\text {and}~~~~P(z)=0, \end{aligned}$$

which is what we wanted to prove.

Conversely, let us assume that (5.4) and (5.5) hold. When considering the Stieltjes functions of the linear forms in both sides of (5.4) and taking formulas (2.2), (2.3) and (5.5) into account, we obtain

$$\begin{aligned} c(z)S(u)(z)S(v)(z)+d(z)S(v)(z)=a(z)S(u)(z)+b(z), \end{aligned}$$

which is the desired expression. $\square$

Remark 5.1

In particular, when $c(z)=0$, i. e. the formal Stieltjes functions S(u) and S(v) associated with the forms u an v,  respectively, are related by a linear spectral transformation

$$\begin{aligned} {S}(v)(z)=\left( \frac{a S(u)+b}{d}\right) (z), \quad a(z)d(z)\ne 0, \end{aligned}$$

where a, b and d are coprime polynomials, if and only if u and v are related by

$$\begin{aligned}&d(x)v=a(x)u,\\&(v\theta _0d)(x)-(u\theta _0a)(x)+b(x)=0. \end{aligned}$$

5.2 Algebraic Properties

Proposition 5.2

Let u and v be two forms in $\mathcal {P}'$. If the formal Stieltjes functions S(u) and S(v) associated with the forms u and v, respectively, are related by a rational spectral transformation,

$$\begin{aligned} S(v)(z)=\frac{X(z)+Y(z)S(u)(z)}{U(z)+V(z)S(u)(z)}, \end{aligned}$$

(5.6)

where $VX-UY\ne 0,~X, Y, U, V \in \mathcal {P},$ then u is a (TDF) if and only if v is also a (TDF). Moreover, if u satisfies (3.1), then for the linear form v you get

$$\begin{aligned} \widetilde{A}(z)S^3(v)(z)+\widetilde{B}(z)S^2(v)(z)+\widetilde{C}(z)S(v)(z)+\widetilde{D}(z)=0, \end{aligned}$$

(5.7)

where

$$\begin{aligned} {\left\{ \begin{array}{ll} \widetilde{A}=AU^3-BVU^2+CUV^2-DV^3,\\ \widetilde{B}=-3AXU^2+B(YU^2+2XUV)-C(XV^2+2UVY)+3DYV^2,\\ \widetilde{C}=3AUX^2-B(VX^2+2XYU)+C(UY^2+2XYV)-3DVY^2,\\ \widetilde{D}=-AX^3+BYX^2-CXY^2+DY^3. \end{array}\right. } \end{aligned}$$

(5.8)

Proof

Relation (5.6) is equivalent to

$$\begin{aligned} S(u)(z)=\frac{X(z)-U(z)S(v)(z)}{-Y(z)+V(z)S(v)(z)}, \end{aligned}$$

(5.9)

so the proof of the sufficient condition follows in a straightforward way using the necessary condition.

The condition is necessary: let u be a (TDF) such that (3.1) holds, i. e.

$$\begin{aligned} A(z)S^3(u)(z)+B(z)S^2(u)(z)+C(z)S(u)(z)+D(z)=0. \end{aligned}$$

According to (5.9) we easily obtain

$$\begin{aligned} A\bigg ( \frac{X-US(v)}{-Y+VS(v)} \bigg )^3+B\bigg ( \frac{X-US(v)}{-Y+VS(v)} \bigg )^2+C\bigg ( \frac{X-US(v)}{-Y+VS(v)} \bigg )+D=0. \end{aligned}$$

In other words,

$$\begin{aligned} A\Big (X-US(v)\Big )^3&+B\Big (X-US(v)\Big )^2\Big (-Y+VS(v)\Big ) \\&+C\Big (X-US(v)\Big )\Big (-Y+VS(v)\Big )^2 \\&+D\Big (-Y+VS(v)\Big )^3=0. \end{aligned}$$

By developing $\big (X-US(v)\big )^k$ and $\big (-Y+VS(v)\big )^k, k=1, 2, 3$, after some computations we find

$$\begin{aligned} A\Omega _1+B\Omega _2+C\Omega _3+D\Omega _4=0, \end{aligned}$$

with

$$\begin{aligned} \Omega _1&=-U^3S^3(v)+3XU^2S^2(v)-3UX^2S(v)+X^3,\\ \Omega _2&=VU^2S^3(v)-(YU^2+2XUV)S^2(v)+(VX^2+2XYU)S(v)-YX^2,\\ \Omega _3&=-UV^2S^3(v)+(XV^2+2UVY)S^2(v)-(UY^2+2XYV)S(v)+XY^2,\\ \Omega _4&=V^3S^3(v)-3YV^2S^2(v)+3VY^2S(v)-Y^3, \end{aligned}$$

which completes the proof. $\square$

Corollary 5.1

Let us assume that the polynomials A, B, C and D satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} U^3A-VU^2B+UV^2C-V^3D=\widetilde{A},\\ -3XU^2A+(YU^2+2XUV)B-(XV^2+2UVY)C+3YV^2D=\widetilde{B},\\ 3UX^2A-(VX^2+2XYU)B+(UY^2+2XYV)C-3VY^2D=\widetilde{C},\\ -X^3A+YX^2B-XY^2C+Y^3D=\widetilde{D}. \end{array}\right. }\nonumber \\ \end{aligned}$$

(5.10)

If we denote by $\mathcal {D}$ the determinant of the above linear system, then we have

$$\begin{aligned} \mathcal {D}=\left( VX-UY\right) ^6. \end{aligned}$$

(5.11)

6 Applications

6.1 The rth Kind Associated form of a Third Degree Form

Let $\{ P_n^{(1)}\}_{n\ge 0}$ be the sequence of associated polynomials of first kind for the sequence of monic polynomials $\{ P_n\}_{n\ge 0}$ orthogonal with respect to a normalized linear form u. It is well known, see [10, 21, 25, 26], that

$$\begin{aligned} P_n^{(1)}(x):=(u\theta _0P_{n+1})(x)=\bigg \langle u,\frac{P_{n+1}(x)- P_{n+1}(\xi )}{x-\xi } \bigg \rangle . \end{aligned}$$

More generally, the sequence of associated polynomials of $(r+1)$th kind, $r\ge 1,$ see [10, 21], is defined by iteration

$$\begin{aligned} P_n^{(r+1)}(x)=\big ( P_n^{(r)} \big )^{(1)}(x),~~~~ u_n^{(r+1)}=\big ( u_n^{(r)} \big )^{(1)},~~~~n, r \ge 0. \end{aligned}$$

If $\{ P_n\}_{n\ge 0}$ is a MOPS with respect to the form u, then for $r\in \mathbb {N}$, the associated sequence of polynomials of rth kind $\left\{ P_n^{(r)}\right\} _{n\ge 0}$ is also orthogonal with respect to a form $u^{(r)}$ and satisfies the recurrence relations [21]

$$\begin{aligned} \begin{aligned}&P_0^{(r)}(x)=1, ~~~~P_1^{(r)}(x)=x-\beta _0^{(r)},\\&P_{n+2}^{(r)}(x)=(x-\beta _{n+1}^{(r)})P_{n+1}^{(r)}(x)-\gamma _{n+1}^{(r)}P_n^{(r)}(x),~~~~n\ge 0, \end{aligned} \end{aligned}$$

where $\beta _{n}^{(r)}=\beta _{n+r},~\gamma _{n+1}^{(r)}=\gamma _{n+1+r},~n\ge 0$. For the two sequences of associated polynomials of the rth and $(r+1)$th kind, the following relation holds [21, 26]

$$\begin{aligned} P_{n+1}^{(r+1)}(x)P_{n+1}^{(r)}(x)-P_{n+2}^{(r)}(x)P_{n}^{(r+1)}(x)=\prod _{\nu =0}^{n}{\gamma _{\nu +r+1}},~~~~n\ge 0, \end{aligned}$$

(6.1)

as well as, for $r\ge 1$,

$$\begin{aligned} P_{n}^{(r)}(x)= \bigg \langle P_{r-1}u,\frac{P_{n+r}(x)- P_{n+r}(\xi )}{x-\xi } \bigg \rangle . \end{aligned}$$

(6.2)

The formal Stieltjes function $S\left( u^{(r)}\right)$ is a rational spectral transformation of the formal Stieltjes function S(u) of $\left\{ P_{n}\right\} _{n \ge 0}$ given by (see [21, 29])

$$\begin{aligned} S\left( u^{(r)}\right) (z)=\frac{P_{r}(z) S(u)(z)-P_{r-1}^{(1)}(z)}{\gamma _{r} P_{r-1}(x) S(u)(z)-\gamma _{r} P_{r-2}^{(1)}(z)}, \end{aligned}$$

(6.3)

where $\{{P}_{n}^{(1)}\}_{n\ge 0}$ is the sequence of associated polynomials of the first kind. As usual, by convention we suppose that $P_n(x)=0$ and $P_n^{(r)}(x)=0$ for $n<0$.

Proposition 6.1

Let u be a (TDRF). Then, for any $n\ge 0$, the associated form of $(n+1)$th kind $u^{(n+1)}$ is also (TDRF) and satisfies

$$\begin{aligned}&A(n+1;z)S^3\left( u^{(n+1)}\right) +B(n+1;z)S^2\left( u^{(n+1)}\right) \nonumber \\&\quad +C(n+1;z)S\left( u^{(n+1)}\right) +D(n+1;z)=0, \nonumber \\ \end{aligned}$$

(6.4)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} k_{n+1}A(n+1;z)&{}=\gamma _{n+1}^3\bigg \{ \Big ( P_{n-1}^{(1)}(z)\Big )^3A(z)- \Big ( P_{n-1}^{(1)}(z)\Big )^2P_n(z)B(z)\\ &{}+ P_{n-1}^{(1)}(z)P_n^2(z)C(z)-P_n^3(z)D(z) \bigg \},\\ k_{n+1}B(n+1;z)&{}=\gamma _{n+1}^2\bigg \{3 P_{n}^{(1)}(z) \Big ( P_{n-1}^{(1)}(z)\Big )^2A(z)\\ &{}- \bigg (\Big ( P_{n-1}^{(1)}(z)\Big )^2P_{n+1}(z)+2 P_n(z) P_{n-1}^{(1)}(z)P_{n}^{(1)}(z)\bigg )B(z)\\ &{}+\bigg (P_{n}^{(1)}(z)P_{n}^2(z)+2P_{n-1}^{(1)}(z)P_{n}(z)P_{n+1}(z)\bigg )C(z)\\ &{} -3P_{n}^2(z)P_{n+1}(z)D(z) \bigg \},\\ k_{n+1}C(n+1;z)&{}=\gamma _{n+1}\bigg \{3 P_{n-1}^{(1)}(z) \Big ( P_{n}^{(1)}(z)\Big )^2A(z)\\ &{}- \bigg (\Big ( P_{n}^{(1)}(z)\Big )^2P_{n}(z)+2 P_{n-1}^{(1)}(z) P_{n}^{(1)}(z)P_{n+1}(z)\bigg )B(z)\\ &{}+\bigg (P_{n-1}^{(1)}(z)P_{n+1}^2(z)+2P_{n}^{(1)}(z)P_{n}(z)P_{n+1}(z)\bigg )C(z)\\ &{} -3P_{n}(z)P_{n+1}^2(z)D(z) \bigg \},\\ k_{n+1}D(n+1;z)&{}=\Big ( P_{n}^{(1)}(z)\Big )^3A(z)- \Big ( P_{n}^{(1)}(z)\Big )^2P_{n+1}(z)B(z)\\ &{}+ P_{n}^{(1)}(z)P_{n+1}^2(z)C(z)-P_{n+1}^3(z)D(z), \end{array}\right. }\nonumber \\ \end{aligned}$$

(6.5)

where $k_{n+1}$ is chosen to make $A(n+1;z)$ monic.

Moreover, if (A, B, C) is a primitive triple, then the triple $\left( A(n+1;z), B(n+1;z), C(n+1;z)\right)$ defined as above has the same property.

Proof

We know that the MOPS of the $(r+1)$th kind $\left\{ P_{n}^{(r+1)}\right\} _{n\ge 0}$ is orthogonal with respect to $u^{(r+1)}$ with formal Stieltjes function [21]

$$\begin{aligned} \gamma _{n+r+1}S\big ( u^{(n+r+1)} \big )(z)=-\frac{P_{n}^{(r+1)}(z)+P_{n+1}^{(r)}(z)S\left( u^{(r)}\right) (z)}{P_{n-1}^{(r+1)}(z)+P_{n}^{(r)}(z)S\left( u^{(r)}\right) (z)} , ~~~~n, r \ge 0. \nonumber \\ \end{aligned}$$

(6.6)

More generally, suppose that u is regular and $u^{(r)}$ is a (TDRF) for an index $r\ge 0$ :

$$\begin{aligned} A(r;z)S^3\left( u^{(r)}\right) (z)&+B(r;z)S^2\left( u^{(r)}\right) (z) \nonumber \\&+C(r;z)S\left( u^{(r)}\right) (z)+D(r;z)=0. \end{aligned}$$

(6.7)

Applying Proposition 5.2 for $X=-P_{n}^{(r+1)},~Y=-P_{n+1}^{(r)},~ U=\gamma _{n+r+1}P_{n-1}^{(r+1)}$ and $V=\gamma _{n+r+1}P_{n}^{(r)}$, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{k_{n+r+1}}{k_r}A(n+r+1;z)=&{}\gamma _{n+r+1}^3\bigg \{ \Big ( P_{n-1}^{(r+1)}(z)\Big )^3A(r;z)\\ &{}- \Big ( P_{n-1}^{(r+1)}(z)\Big )^2P_n^{(r)}(z)B(r;z)\\ &{}+ P_{n-1}^{(r+1)}(z)\Big (P_n^{(r)}(z)\Big )^2C(r;z)-\Big (P_n^{(r)}(z)\Big )^3D(r;z) \bigg \},\\ \frac{k_{n+r+1}}{k_r}B(n+r+1;z)=&{}\gamma _{n+r+1}^2\bigg \{3 P_{n}^{(r+1)}(z) \Big ( P_{n-1}^{(r+1)}(z)\Big )^2A(r;z)\\ &{}- \bigg (\Big ( P_{n-1}^{(r+1)}(z)\Big )^2P_{n+1}^{(r)}(z)\\ &{}+2 P_n^{(r)}(z) P_{n-1}^{(r+1)}(z)P_{n}^{(r+1)}(z)\bigg )B(r;z)\\ &{}+\bigg (P_{n}^{(r+1)}(z)\Big (P_{n}^{(r)}(z)\Big )^2\\ &{}+2P_{n-1}^{(r+1)}(z)P_{n}^{(r)}(z)P_{n+1}^{(r)}(z)\bigg )C(r;z)\\ &{}-3\Big (P_{n}^{(r)}(z)\Big )^2(z)P_{n+1}^{(r)}(z)D(r;z) \bigg \},\\ \frac{k_{n+r+1}}{k_r}C(n+r+1;z)=&{}\gamma _{n+r+1}\bigg \{3 P_{n-1}^{(r+1)}(z) \Big ( P_{n}^{(r+1)}(z)\Big )^2A(r;z)\\ &{}- \bigg (\Big ( P_{n}^{(r+1)}(z)\Big )^2P_{n}^{(r)}(z)\\ &{}+2 P_{n-1}^{(r+1)}(z) P_{n}^{(r+1)}(z)P_{n+1}^{(r)}(z)\bigg )B(r;z)\\ &{}+\bigg (P_{n-1}^{(r+1)}(z)\Big ( P_{n+1}^{(r)}(z)\Big )^2\\ &{}+2P_{n}^{(r+1)}(z)P_{n}^{(r)}(z)P_{n+1}^{(r)}(z)\bigg )C(r;z)\\ &{}-3P_{n}^{(r)}(z)\Big ( P_{n+1}^{(r)}(z)\Big )^2 D(r;z) \bigg \},\\ \frac{k_{n+r+1}}{k_r}D(n+r+1;z)=&{}\bigg \{ \Big ( P_{n}^{(r+1)}(z)\Big )^3A(r;z)\\ &{}- \Big ( P_{n}^{(r+1)}(z)\Big )^2P_{n+1}^{(r)}(z)B(r;z)\\ &{}+ P_{n}^{(r+1)}(z) \Big ( P_{n+1}^{(r)}(z)\Big )^2C(r;z)-\Big (P_{n+1}^{(r)}(z)\Big )^3D(r;z) \bigg \}. \end{array}\right. } \end{aligned}$$

(6.8)

For $r=0$, we have (6.4) and (6.5) with

$$\begin{aligned} A(z)=A(0;z),~~~~B(z)=B(0;z),~~~~C(z)=C(0;z),~~~~D(z)=D(0;z),~~~~k_0=1. \end{aligned}$$

Thus

$$\begin{aligned} \mathcal {D}_{n,r}=\gamma _{n+r+1}^6\Big (P_{n-1}^{(r+1)}P_{n+1}^{(r)}-P_{n}^{(r)}P_{n}^{(r+1)} \Big )^6. \end{aligned}$$

Using (6.1), we obtain

$$\begin{aligned} \mathcal {D}_{n,r}=\left( \prod _{\nu =0}^{n}{\gamma _{\nu +r+1}}\right) ^6,~~~~n, r \ge 0. \end{aligned}$$

(6.9)

Hence our result follows. $\square$

Remark 6.1

Taking $n=0$ in (6.8), we have

$$\begin{aligned} {\left\{ \begin{array}{ll} k_{r+1}A(r+1;z)=-k_r\gamma _{r+1}^3D(r;z),~~r\ge 0,\\ k_{r+1}B(r+1;z)=k_r\gamma _{r+1}^2\Big \{C(r;z)-3(z-\beta _r)D(r;z)\Big \},~~r\ge 0,\\ k_{r+1}C(r+1;z)=k_r\gamma _{r+1}\Big \{-B(r;z)+2(z-\beta _r)C(r;z) \\ -3(z-\beta _r)^2D(r;z)\Big \},~~r\ge 0,\\ k_{r+1}D(r+1;z)=k_r\Big \{A(r;z)-(z-\beta _r)B(r;z)+(z-\beta _r)^2C(r;z) \\ -(z-\beta _r)^3D(r;z)\Big \},~~r\ge 0.\\ \end{array}\right. } \end{aligned}$$

(6.10)

6.2 The Co-recursive of a Third Degree Form

Let $\mu$ be a complex number, u a regular form and $\left\{ P_{n}\right\} _{n\ge 0}$ be its corresponding (MOPS) satisfying (2.4). We define the co-recursive sequence of polynomials $\left\{ P_{n}^{[\mu ]}\right\} _{n \ge 0}$ of $\left\{ P_{n}\right\} _{n \ge 0}$ as the family of monic polynomials satisfying the following three-term recurrence relation [21, Definition 4.2]

$$\begin{aligned} \begin{array}{l} {P_{0}^{[\mu ]}(x)=1, \quad P_{1}^{[\mu ]}(x)=x-\beta _{0}-\mu }, \\ {P_{n+2}^{[\mu ]}(x)=\left( x-\beta _{n+1}\right) P_{n+1}^{[\mu ]}(x)-\gamma _{n+1} P_{n}^{[\mu ]}(x), \quad n \ge 0}. \end{array} \end{aligned}$$

This family of polynomials has been introduced in [9]. Denoting by $u^{[\mu ]}$ its corresponding regular form, it is well known (see [21, Equation (4.14)]) that

$$\begin{aligned} u^{[\mu ]}=u\left( \delta -\mu x^{-1} u\right) ^{-1}. \end{aligned}$$

Proposition 6.2

If u is a (TDRF), then the co-recursive form of u, $u^{[\mu ]}$ is also a (TDRF) and satisfies

$$\begin{aligned} {A}^{[\mu ]}(z)S^3\left( {u}^{[\mu ]}\right) (z)&+{B}^{[\mu ]}(z)S^2\left( {u}^{[\mu ]}\right) (z) \nonumber \\&+{C}^{[\mu ]}(z)S\left( {u}^{[\mu ]}\right) (z)+{D}^{[\mu ]}(z)=0, \end{aligned}$$

(6.11)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} k{A}^{[\mu ]}(z)=A(z)-\mu B(z)+\mu ^2C(z)-\mu ^3D(z),\\ k{B}^{[\mu ]}(z)=B(z)-2\mu C(z)+3\mu ^2D(z),\\ k{C}^{[\mu ]}(z)=C(z)-3\mu D(z),\\ k{D}^{[\mu ]}(z)=D(z), \end{array}\right. } \end{aligned}$$

(6.12)

where k is a normalization constant chosen to make ${A}^{[\mu ]}$ monic. Moreover, if (A, B, C) is a primitive triple, then the triple $\left( A^{[\mu ]}, B^{[\mu ]}, C^{[\mu ]}\right)$ has the same property.

Proof

The formal Stieltjes function $S\left( u^{[\mu ]}\right)$ is a rational spectral transformation of the formal Stieltjes function S(u) of $\left\{ P_{n}\right\} _{n \ge 0}$ given by [21, Equation (4.15)]

$$\begin{aligned} S\left( u^{[\mu ]}\right) (z)=\frac{S(u)(z)}{1+\mu S(u)(z)}. \end{aligned}$$

Applying Proposition 5.2 for $X=0, Y=1, U=1$ and $V=\mu$, we obtain (6.11) and (6.12).

The determinant $\mathcal {D}_{\mu }$ of the system (6.12) is a nonzero constant. Indeed, $\mathcal {D}_{\mu }=1$, and this completes the proof. $\square$

6.3 The r-Perturbed of a Third Degree Form

Let $\left\{ \widetilde{P}_n\right\} _{n\ge 0}$ be the r-perturbed sequence of a (MOPS) $\{ P_n\}_{n\ge 0}$ defined by [17, 21]

$$\begin{aligned} \begin{aligned}&\widetilde{P}_0(x)=1, ~~~~\widetilde{P}_1(x)=x-\widetilde{\beta }_0,\\&\widetilde{P}_{n+2}(x)=(x-\widetilde{\beta }_{n+1})\widetilde{P}_{n+1}(x) -\widetilde{\gamma }_{n+1} \widetilde{P}_n(x), ~~~~n\ge 0, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&\widetilde{\beta }_0=\beta _0+\mu _0,\\&\widetilde{\beta }_n=\beta _n+\mu _n,~~~~\mu _n\in \mathbb {C};~~~~\widetilde{\gamma }_{n}=\lambda _n\gamma _n, ~~\widetilde{\gamma }_{n}\in \mathbb {C}-\{0\},~~1\le n\le r,\\&\widetilde{\beta }_n=\beta _n,~~~~\widetilde{\gamma }_{n}=\gamma _n,~~~~n\ge r+1. \end{aligned} \end{aligned}$$

(6.13)

Either $\mu _r\ne 0$ or $\lambda _n\ne 1$. Thus, the co-recursive case corresponds to the perturbed case of order 0. With the notations $\mu :=(\mu _1,..., \mu _r),~ \lambda :=(\lambda _1,...,\lambda _r),~ r{\ge }1$, we write

$$\begin{aligned} \widetilde{P}_{n}(x)=P_n\left( \mu _0; _\lambda ^\mu ; r; x\right) , ~~n\ge 0, \end{aligned}$$

and the sequence $\{ \widetilde{P}_n\}_{n\ge 0}$ is orthogonal with respect to the perturbed form $\widetilde{u}:=u\left( \mu _0; _\lambda ^\mu ; r\right)$ given by [21]

$$\begin{aligned} S(\widetilde{u})(z)=-\frac{U_r(z)+V_r(z)S(u)(z)}{X_r(z)+Y_r(z)S(u)(z)}, \end{aligned}$$

(6.14)

with transfer polynomials

$$\begin{aligned} \begin{aligned}&U_r(z)=\gamma _r\Big \{\widetilde{P}_{r-1}^{(1)}(z){P}_{r-2}^{(1)}(z)-\lambda _r{P}_{r-1}^{(1)}(z)\widetilde{P}_{r-2}^{(1)}(z)\Big \}-\mu _r{P}_{r-1}^{(1)}(z)\widetilde{P}_{r-1}^{(1)}(z),\\&V_r(z)=\gamma _r\Big \{\widetilde{P}_{r-1}^{(1)}(z){P}_{r-1}(z)-\lambda _r{P}_{r}(z)\widetilde{P}_{r-2}^{(1)}(z)\Big \}-\mu _r{P}_{r}(z)\widetilde{P}_{r-1}^{(1)}(z),\\&X_r(z)=\gamma _r\Big \{\widetilde{P}_{r}(z){P}_{r-2}^{(1)}(z)-\lambda _r{P}_{r-1}^{(1)}(z)\widetilde{P}_{r-1}(z)\Big \}-\mu _r{P}_{r-1}^{(1)}(z)\widetilde{P}_{r}(z),\\&Y_r(z)=\gamma _r\Big \{\widetilde{P}_{r}(z){P}_{r-1}(z)-\lambda _r{P}_{r}(z)\widetilde{P}_{r-1}(z)\Big \}-\mu _r{P}_{r}(z)\widetilde{P}_{r}(z), \end{aligned}\nonumber \\ \end{aligned}$$

(6.15)

where $\{\widetilde{P}_{n}^{(1)}\}_{n\ge 0}$ is the sequence of the associated polynomials of the first kind of the sequence of perturbed polynomials $\{\widetilde{P}_{n}\}_{n\ge 0}$. As usual, we suppose that $P_n(z)=0$, $P_n^{(r)}(z)=0$, $\widetilde{P}_{n}(z)=0$ and $\widetilde{P}_{n}^{(1)}(z)=0$ for $n<0$.

Proposition 6.3

If u is a (TDRF), then the perturbed form of u, $\widetilde{u}=(\mu _0; _\lambda ^\mu ; r), r\ge 0,$ is also a (TDRF) and satisfies

$$\begin{aligned} \widetilde{A}(z)S^3(\widetilde{u})(z)+\widetilde{B}(z)S^2(\widetilde{u})(z)+\widetilde{C}(z)S(\widetilde{u})(z)+\widetilde{D}(z)=0, \end{aligned}$$

(6.16)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} k\widetilde{A}(z)=X_r^3(z)A(z)-X_r^2(z)Y_r(z)B(z)+X_r(z)Y_r^2(z)C(z)-Y_r^3(z)D(z),\\ k\widetilde{B}(z)=3U_r(z)X_r^2(z)A(z)-\big (V_r(z)X_r^2(z)+2U_r(z)X_r(z)Y_r(z)\big )B(z)\\ \qquad \qquad +\big (U_r(z)Y_r^2(z)+2V_r(z)X_r(z)Y_r(z)\big )C(z)-3V_r(z)Y_r^2(z)D(z),\\ k\widetilde{C}(z)=3U_r^2(z)X_r(z)A(z)-\big (Y_r(z)U_r^2(z)+2X_r(z)U_r(z)V_r(z)\big )B(z)\\ \qquad \qquad +\big (X_r(z)V_r^2(z)+2Y_r(z)U_r(z)V_r(z)\big )C(z)-3Y_r(z)V_r^2(z)D(z),\\ k\widetilde{D}(z)=U_r^3(z)A(z)-V_r(z)U_r^2(z)B(z)+U_r(z)V_r^2(z)C(z)-V_r^3(z)D(z), \end{array}\right. } \nonumber \\ \end{aligned}$$

(6.17)

where k is a normalization constant such that $\widetilde{A}$ is monic, and $U_r(z), V_r(z), X_r(z)$ and $Y_r(z)$ are the transfer polynomials given by (6.15).

Moreover, if (A, B, C) is a primitive triple, then the triple $\left( \widetilde{A}, \widetilde{B}, \widetilde{C}\right)$ has the same property.

For the proof we use the following

Lemma 6.1

With the notations in (6.14) and (6.15) we get

$$\begin{aligned} U_r(z)Y_r(z)-V_r(z)X_r(z)=\left( \prod _{\nu =0}^{r}{\lambda _\nu }\right) \left( \prod _{\nu =0}^{r}{\gamma _\nu }\right) ^2, \end{aligned}$$

(6.18)

with $\lambda _0=1, ~~\gamma _0=1$.

Proof

From (6.15), after some computations, we deduce

$$\begin{aligned} U_r(z)Y_r(z)-V_r(z)X_r(z)&=\gamma _r^2\lambda _r\Big \{\widetilde{P}_{r-1}^{(1)}(z) \widetilde{P}_{r-1}(z) P_{r-1}^{(1)}(z) P_{r-1}(z) \\&+\widetilde{P}_{r-2}^{(1)}(z) \widetilde{P}_{r}(z) P_{r-2}^{(1)}(z) P_{r}(z) \\&- \widetilde{P}_{r-1}^{(1)}(z) \widetilde{P}_{r-1}(z) P_{r-2}^{(1)}(z) P_{r}(z) \\&-\widetilde{P}_{r-2}^{(1)}(z) \widetilde{P}_{r}(z) P_{r-1}^{(1)}(z) P_{r-1}(z) \Big \},\\ =&\gamma _r^2\lambda _r \Big ( \widetilde{P}_{r-1}^{(1)}(z)\widetilde{P}_{r-1}(z)\\&-\widetilde{P}_{r-2}^{(1)}(z)\widetilde{P}_{r}(z) \Big )\Big ( P_{r-1}^{(1)}(z)P_{r-1}(z)\\&-P_{r-2}^{(1)}(z) P_{r}(z) \Big ). \end{aligned}$$

From (6.1), we obtain

$$\begin{aligned} P_{r-1}^{(1)}(z) P_{r-1}(z)-P_{r-2}^{(1)}(z) P_{r}(z)=\gamma _r^{-1}\prod _{\nu =0}^{r}{\gamma _\nu },~~~~r\ge 0, \end{aligned}$$

and

$$\begin{aligned} \widetilde{P}_{r-1}^{(1)}(z) \widetilde{P}_{r-1}(z)-\widetilde{P}_{r-2}^{(1)}(z) \widetilde{P}_{r}(z)&=\widetilde{\gamma }_r^{-1}\prod _{\nu =0}^{r}{\widetilde{\gamma }_\nu }\\&=\lambda _r^{-1}\gamma _r^{-1}\left( \prod _{\nu =0}^{r}{\lambda _\nu }\right) \left( \prod _{\nu =0}^{r}{\gamma _\nu }\right) , ~~r\ge 0, \end{aligned}$$

with $\lambda _0=1, ~\gamma _0=1$, from (6.13), and we deduce the required identities. $\square$

Proof

From (6.14), we have

$$\begin{aligned} S(\widetilde{u})(z)=-\frac{U_r(z)+V_r(z)S(u)(z)}{X_r(z)+Y_r(z)S(u)(z)}, ~~r\ge 0. \end{aligned}$$

Applying Proposition 5.2 for $X=-U_r, Y=-V_r, U=X_r$ and $V=Y_r$, a simple computation yields (6.16) and (6.17).

The determinant $\mathcal {D}_{r}$ of the system (6.17) is different from zero. Indeed, from (5.11) and (6.14) , we get

$$\begin{aligned} \mathcal {D}_r=\Big ( V_r(z)X_r(z)-U_r(z)Y_r(z)\Big )^6. \end{aligned}$$

From identity (6.18) we have

$$\begin{aligned} \mathcal {D}_r=\left( \prod _{\nu =0}^r{\lambda _\nu }\right) ^6\left( \prod _{\nu =0}^r{\gamma _\nu }\right) ^{12}, \end{aligned}$$

(6.19)

thus, the proof is completed. $\square$

6.4 The Inverse of a Third Degree Form

Let u be a regular form and $\left\{ P_{n}\right\} _{n \ge 0}$ its corresponding (MOPS) satisfying (2.4). The form u has an inverse linear form $u^{-1}$ if and only if $(u)_0\ne 0$ and then $u u^{-1}=\delta$. The inverse form of u satisfies [21, Equation (5.27)]

$$\begin{aligned} x^2u^{-1}=-\gamma _1u^{(1)}, \end{aligned}$$

which readily gives (see [1])

$$\begin{aligned} u^{-1}=\delta -\left( u^{-1}\right) _1\delta '-\gamma _1x^{-2}u^{(1)}. \end{aligned}$$

(6.20)

In general, the form $u^{-1}$ given by (6.20) is regular if and only if $\Delta _n\ne 0,~n\ge 0$, with

$$\begin{aligned} \Delta _n=\left\langle u^{(1)}, \left( P_{n}^{(1)}\right) ^2\right\rangle \left( \gamma _1+\sum _{\nu =0}^{n}{\frac{\left( \gamma _1P_{\nu -1}^{(2)}(0)-\left( u^{-1}\right) _1P_{\nu }^{(1)}(0) \right) ^2}{\left\langle u^{(1)}, \left( P_{\nu }^{(1)}\right) ^2\right\rangle }}\right) ,~~n\ge 0. \end{aligned}$$

Proposition 6.4

If u is a (TDRF) with $\Delta _n\ne 0,~n\ge 0$, then $u^{-1}$ is also a (TDRF) and satisfies

$$\begin{aligned}&A^{(-)}(z)S^3\left( {u^{-1}}\right) (z)+B^{(-)}(z)S^2\left( {u^{-1}}\right) (z)\nonumber \\&\quad +C^{(-)}(z)S\left( {u^{-1}}\right) (z)+D^{(-)}(z)=0, \end{aligned}$$

(6.21)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} kA^{(-)}(z)=z^6D(z),\\ kB^{(-)}(z)=z^4 C(z),\\ kC^{(-)}(z)=z^2B(z),\\ kD^{(-)}(z)=A(z), \end{array}\right. }\nonumber \\ \end{aligned}$$

(6.22)

where k is a normalization constant chosen to make $A^{(-)}$ monic.

Proof

It is well known that the formal Stieltjes function $S\left( u^{-1}\right)$ is a rational spectral transformation of the formal Stieltjes function S(u) given by

$$\begin{aligned} S\left( u^{-1}\right) (z)=\frac{1}{z^2S(u)(z)}. \end{aligned}$$

Thus, the proof follows as a direct consequence of (5.7) and (5.8) for $X=1, Y=0, U=0$ and $V=z^2$. The determinant $\mathcal {D}^{(-)}$ of the system (6.22) is different from zero. Indeed, $\mathcal {D}^{(-)}=z^{12}$, which completes the proof. $\square$

6.5 Christoffel and Geronimus Transformations of a Third Degree Form

Proposition 6.5

Let u be a third degree form. Then the Christoffel transformation $v=p(x)u$ of the form u, where $p \in \mathcal {P}$, is also a third degree form and satisfies

$$\begin{aligned} A_v(z)S^3(v)(z)+B_v(z)S^2(v)(z)+C_v(z)S(v)(z)+D_v(z)=0, \end{aligned}$$

(6.23)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} kA_v(z)=A_u(z),\\ kB_v(z)=B_u(z)p(z)-3A_u(z)(u\theta _0p)(z),\\ kC_v(z)=C_u(z)p^2(z)-2 B_u(z)p(z)(u\theta _0 p)(z)+3A_u(z)(u\theta _0 p)^2(z),\\ kD_v(z)=D_u(z)p^3(z)-C_u(z)p^2(z)(u\theta _0 p)(z)+B_u(z)p(z)(u\theta _0 p)^2(z)\\ &{} \qquad \qquad \qquad -A_u(z)(u\theta _0 p)^3(z), \end{array}\right. } \nonumber \\ \end{aligned}$$

(6.24)

where k is a normalization constant chosen to make $A_v$ monic.

Proof

From (2.3), the formal Stieltjes function S(v) is a linear spectral transformation of the formal Stieltjes function S(u), given by

$$\begin{aligned} S(v)(z)=p(z)S(u)(z)+(u\theta _0 p)(z). \end{aligned}$$

Applying Proposition 5.2 for $X=u\theta _0 p,~Y=p,~U=1$ and $V=0$, we obtain (6.23) and (6.24). $\square$

Proposition 6.6

Let u be a third degree form. Then the Geronimus transformation v of the form u, i. e. $q(x)v=u,$ where $q \in \mathcal {P},$ is also a third degree form and satisfies

$$\begin{aligned} A_v(z)S^3(v)(z)+B_v(z)S^2(v)(z)+C_v(z)S(v)(z)+D_v(z)=0,\quad \end{aligned}$$

(6.25)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} kA_v(z)=A_u(z)q^3(z),\\ kB_v(z)=q^2(z)\Big \{B_u(z)+3A_u(z)(v\theta _0q)(z)\Big \},\\ kC_v(z)=q(z)\Big \{C_u(z)+2 B_u(z)(v\theta _0 q)(z)+3A_u(z)(v\theta _0 q)^2(z)\Big \},\\ kD_v(z)=D_u(z)+C_u(z)(v\theta _0 q)(z)+B_u(z)(v\theta _0 q)^2(z)+A_u(z)(v\theta _0 q)^3(z), \end{array}\right. } \end{aligned}$$

(6.26)

where k is a normalization constant chosen to make $A_v$ monic.

Proof

From (2.3), the formal Stieltjes function S(v) is a linear spectral transformation of the formal Stieltjes function S(u), given by

$$\begin{aligned} S(v)(z)=\frac{S(u)(z)-(v\theta _0 q)(z)}{q(z)}. \end{aligned}$$

Applying Proposition 5.2 for $X=-v\theta _0 q,~Y=1,~U=q$ and $V=0$, we obtain (6.25) and (6.26). $\square$

Finally, we show that if u is a third degree form then $v=u+\mu \delta _c$ and $w=u+\mu \delta '_c$, where $\mu , c \in \mathbb {C}$, are also third degree forms. Conditions about the quasi-definite character of v and w have been given in [11, 18] and [4], respectively. We also give the cubic equation satisfied by the corresponding Stieltjes functions.

Proposition 6.7

Let u be a third degree form. Then $v=u+\mu \delta _c$, where $\mu , c \in \mathbb {C}$, is also a third degree form and satisfies

$$\begin{aligned} A_v(z)S^3(v)(z)+B_v(z)S^2(v)(z)+C_v(z)S(v)(z)+D_v(z)=0, \end{aligned}$$

(6.27)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} kA_v(z)=(z-c)^3A(z),\\ kB_v(z)=3\mu (z-c)^2A(z)+(z-c)^3B(z),\\ kC_v(z)=3\mu ^2(z-c)A(z)+2\mu (z-c)^2B(z)+(z-c)^3C(z),\\ kD_v(z)=\mu ^3A(z)+\mu ^2(z-c)B(z)+\mu (z-c)^2C(z)+(z-c)^3D(z), \end{array}\right. }\nonumber \\ \end{aligned}$$

(6.28)

where k is a normalization constant chosen to make $A_v$ monic.

Proof

From the definition of the form v we have

$$\begin{aligned} (v)_n=(u)_n+\mu c^n,~~n\ge 0. \end{aligned}$$

Thus

$$\begin{aligned} S(v)(z)=S(u)(z)-\frac{\mu }{z-c}. \end{aligned}$$

In other words

$$\begin{aligned} S(v)(z)=\frac{(z-c)S(u)(z)-\mu }{z-c}. \end{aligned}$$

If $\mu \ne 0$, applying Proposition 5.2 for $X=-\mu ,~Y=z-c,~U=z-c$ and $V=0$, we deduce (6.27) and (6.28). $\square$

Proposition 6.8

Let u be a third degree form. Then $w=u+\mu \delta '_c$, where $\mu , c \in \mathbb {C},$ is also a third degree form and satisfies

$$\begin{aligned} A_w(z)S^3(w)(z)+B_w(z)S^2(w)(z)+C_w(z)S(w)(z)+D_w(z)=0, \end{aligned}$$

(6.29)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} kA_w(z)=(z-c)^6A(z),\\ kB_w(z)=-3\mu (z-c)^4A(z)+(z-c)^6B(z),\\ kC_w(z)=3\mu ^2(z-c)^2A(z)-2\mu (z-c)^4B(z)+(z-c)^6C(z),\\ kD_w(z)=-\mu ^3A(z)+\mu ^2(z-c)^2B(z)-\mu (z-c)^4C(z)+(z-c)^6D(z), \end{array}\right. }\nonumber \\ \end{aligned}$$

(6.30)

where k is a normalization constant chosen to make $A_w$ monic.

Proof

From the definition of the form w we have

$$\begin{aligned} (w)_n=(u)_n-\mu n c^{n-1}, ~~n\ge 0. \end{aligned}$$

Then,

$$\begin{aligned} S(w)(z)=S(u)(z)+\frac{\mu }{(z-c)^2}. \end{aligned}$$

Thus, the formal Stieltjes function S(w) is a rational spectral transformation of the formal Stieltjes function S(u), given by

$$\begin{aligned} S(w)(z)=\frac{(z-c)^2S(u)(z)+\mu }{(z-c)^2}. \end{aligned}$$

If $\mu \ne 0$, applying Proposition 5.2 for $X=\mu , Y=(z-c)^2, U=(z-c)^2$ and $V=0$, we obtain (6.29) and (6.30). $\square$

7 Conclusions

In this contribution, we have studied third degree linear forms and we have stated some algebraic properties using the corresponding formal Stieltjes functions. On the other hand, we have considered rational spectral transformations of such forms and we have shown the preservation of the third degree property. In a paper in progress we are focusing our attention on the set of transformations such that the third degree linear forms are preserved, far away of rational spectral transformations. Indeed, in [7] the authors show examples of cubic decompositions of preserving third degree linear forms. For higher order polynomial decompositions associated with polynomial mappings on the real line (see [16]) the problem is open.