In this work, we prove that the formal Stieltjes of q-Laguerre -hahn forms is a solution of many q-Ricatti equations. As a consequence , we show that the class of those forms depends on k ∈ ℕ. Some examples are highlighted.
The \(q\)-Laguerre-Hahn regular form \(u_q\) ( linear functional) its formal Stieltjes function \(S(z,u_q)=-\sum\limits_{k\geq0}\frac{(u)_k}{z^{k+1}}\), where \((u_q)_n:=\left\langle u_q,x^n\right\rangle,\,n\geq0\), fulfills the following \(q\)-Riccati equation [1– 3] \[\phi(q^{-1}z)H_{q^{-1}}(S(z,u_q))=B(z)S(z,u_q)S(q^{-1}z,u_q)+C(z)S(z,u_q)+D(z), \label{eq1.1} \tag{1}\] where \(\phi\), \(B\), \(C\), \(D\) are a polynomials (\(\phi\) monic) and \(H_q\) the \(q\)-derivative operator.
The monic orthogonal polynomials sequence associated to \(u_q\) is said to be \(H_q\)-Laguerre-Hahn orthogonal polynomials.
When the Eq. (1) can not be simplifying by any polynomial the class of \(u_q\) is defined by \(s(u_q):=\max(\deg B-2,\deg C-1,\deg D)\) [3].
In the case where \(B=0\), we deal with the \(H_q\)-semi-classical forms [4] in particular the symmetrical \(H_q\)-semi-classical forms of class one are treated in [5].
Moreover, the symmetrical \(H_q\)-Laguerre-Hahn of class zero and one are exhaustively studied in [6, 7] and some examples of non symmetrical \(H_q\)-Laguerre-Hahn of class two are given in [8] via Cristoffel an Geronimus transformations.
Notice that the first order \(q\)-difference equation for Stieltjes function is studied in [1] for the \(H_q\)-classical forms.
In this work, we prove that when the form \(u_q\) is \(q\)-Laguerre-Hahn form of class \(s(u_q)\) satisfying (1), its Stieltjes function \(S(z,u_q)\) satisfies also the \(q^{k+1}\)-Riccati equation, \(k\in\mathbb{N}\):
\[\begin{aligned} \label{eq1.2} (q^{-k-1}-1)zR_{k+1}(z)H_{q^{-k-1}}\left(S(z,u_q)\right)=&-B_{k+1}(z)S(z,u_q)S(q^{-k-1}z,u_q) -\left(R_{k+1}(z)+C_{k+1}(z)\right)S(z,u_q)\notag\\ &-D_{k+1}(z), \end{aligned} \tag{2}\] where \(B_{k+1}, R_{k+1}, C_{k+1}\) and \(D_{k+1}\) are a polynomials for \(k\in\mathbb{N}\) (see Proposition 1 below), which means that the same form \(u_q\) is also \(q^{k+1}\)-Laguerre-Hahn ( respectively \(H_{q^{k+1}}\)-semi-classical) form relatively to the operator operator \(H_{q^{k+1}}\) . Consequently, the class of the form \(u_q\) depends on the integer \(k\) with respect to the operator \(H_{q^{k+1}}\). Of course, we have \(s(u_q)=s(u_{q{k+1}})\) since \(q\) is replaced by \(q^{k+1}\).
Some examples are highlighted, when \(u_q:=\mathcal{T}_q\) (respectively \(u_q:=\mathcal{U}_q\)), where \(\mathcal{T}_q\) (respectively \(\mathcal{U}_q\)) is the \(q\)-Chebyshev form of the first kind ( respectively the second kind) which is \(H_q\)-classical form [9], we establish the \(q\)-difference Eq. (2) satisfied by \(S(z,\mathcal{T}_q)\) (respectively \(S(z,\mathcal{U}_q)\) ) and we prove that \(\mathcal{T}_q\) (respectively \(\mathcal{U}_q\)) it is a \(H_{q^{k+1}}\)-semi-classical form of class \(2k,\,k\geq0\). In the case where \(u_q:=u_q(\omega)\) where \(u_q(\omega)\) is the symmetric Brenk type form which is \(H_q\)-semi-classical form of class \(s(u_q(\omega))=1\) [8], on one hand, we determine the Stieltjes function \(S(z,u_q(\omega))\), in particular, we show that \(S(z,u_q(0))\) it is a solution of an \(q\)-difference equation involving the Jacobi triple product identity [10, 11], also the polynomials \(B_{k+1},\,R_{k+1},\,C_{k+1}\) and \(D_{k+1}\) are calculated. As a consequence, we prove that the form \(u_q(\omega)\) is also \(H_{q^{k+1}}\)-semi-classical form of class \(2k+1,\,k\geq0\). On the other hand, we establish the \(q^{k+1}\)-difference Eq. (2) satisfied by \(S(z,u_q^{(1)}(\omega))\), which allowed us to show that the first associated form \(u_q^{(1)}(\omega)\) it is \(q^{k+1}\)-Laguerre-form of class \(2k+1,\,k\geq0.\)
Let \(\mathcal{P}\) be the vector space of polynomials with coefficients in \(\mathbb{C}\) and let \(\mathcal{P} '\) be its dual. We denote by \(\left\langle w,f\right\rangle\) the effect of \(w\in\mathcal{P} '\) on \(f\in\mathcal{P}\). In particular, we denote by \((w)_n:=\left\langle w,x^n\right\rangle,\,n\geq 0\), the moments of \(w\). Let us introduce some useful operations in \(\mathcal{P}'\). For any linear form \(w\), any polynomial \(g\) and any \(a\in\mathbb{C}-\{0\},\, \in\mathbb{C}\), we let \(g w\), \(h_a w\) and \(H_q w\) be the forms (linear functionals) defined by duality \[\left\langle g w,f\right\rangle:=\left\langle w,gf\right\rangle, \, \left\langle h_a w,f\right\rangle := \left\langle w,h_a f\right\rangle,\,\, \left\langle H_q w,f\right\rangle;=-\left\langle w,H_qf\right\rangle,\,f\in\mathcal{P},\] where \[\begin{aligned} H_q(f)(x)&=\displaystyle\frac{f(qx)-f(x)}{(q-1)x}, x\neq0,\,q\in \tilde{\mathbb{C}}:=\mathbb{C}-\left(\{0\}\cup \mathop{\bigsqcup} \limits_{n\geq0}\{z\in\mathbb{C},\,z^n=1\}\right),\\ H_q(f)(0)&=f'(0). \end{aligned}\]
We also define the right-multiplication of a form by a polynomial with \[(wf)(x):=\left\langle w,\frac{xf(x)-\xi f(\xi)}{x-\xi}\right\rangle, \,w\in\mathcal{P}',\,f\in\mathcal{P}.\]
This allows to define the product of two forms: \[\left\langle uw,f\right\rangle:=\left\langle u,wf\right\rangle,\,\,u,w\in\mathcal{P}',\,f\in\mathcal{P}.\]
The formal Stieltjes function of \(w\in \mathcal{P}'\) is defined by \[S(z,w)=-\sum\limits_{n\geq0}\frac{(w)_n}{z^{n+1}},\,z\in \mathbb{C}-\{0\}.\]
Next, we define \(H_q(S(z,w))\) as following \[H_q(S(z,w))=\frac{S(qz,w)-S(z,w)}{(q-1)z},\,z\in \mathbb{C}-\{0\}.\]
We call polynomial sequence (PS), the sequence of polynomials \(\{P_n\}_{n\geq 0}\) when \(\deg P_n=n,\,n\geq 0\). Then any polynomial \(P_n\) can be supposed monic and the sequence becomes a monic polynomial sequence (MPS). The MPS \(\{P_n\}_{n\geq 0}\) is orthogonal (MOPS) with respect to \(w\in\mathcal{P} '\) when the following conditions hold \(\left\langle w,P_mP_n\right\rangle =r_n\delta _{n,m},\,n,m\geq 0, \,r_n\not= 0,\,n\geq 0.\) In this case the form \(w\) is said regular. The form \(w\) is called normalized if \((w)_0 =1\). In this paper, we suppose that the forms are normalized. Thus, \(\{P_n\}_{n\geq 0}\) satisfies the standard recurrence relation \[\left\lbrace \begin{array}{l} P_0(x)=1\hspace{0.25cm},\hspace{0.25cm}P_1(x)=x-\beta _0,\\ P_{n+2}(x)=(x-\beta _{n+1})P_{n+1}(x)-\gamma _{n+1}P_n(x) ,\,n\geq 0;\,\gamma_{n+1}\neq0,\,n\geq0. \end{array} \right.\label{eq2.1} \tag{3}\] The regular form \(w\) is said to be symmetric when \((w)_{2n+1}=0,\,n\geq0\), or equivalently \(\beta_n=0,\,n\geq0\) in (3) [12].
Given a regular form \(w\) and the corresponding MOPS \(\{P_n\}_{n\geq0}\) satisfying (3), we define the first associated sequence \(\{P_n^{(1)}\}_{n\geq0}\) by [12] \[P_n^{(1)}(x)=\left\langle w, \frac{P_{n+1}(x)-P_{n+1}(\xi)}{x-\xi}\right\rangle=(w\theta_0P_{n+1})(x).\label{eq2.2} \tag{4}\]
We know that \(\{P_n^{(1)}\}_{n\geq0}\) is a MOPS with respect to \(w^{(1)}\) satisfying (3) with [12] \[\beta_n^{(1)}=\beta_{n+1},\,\,\gamma_{n+1}^{(1)}=\gamma_{n+2},\,n\geq0.\label{eq2.3} \tag{5}\]
Definition 1. [3] A form \(w\) is called \(q\)-Laguerre-Hahn when it is regular and its formal Stieltjes function \(S(.,w)\) satisfies the follwoing \(q\)-Riccati equation \[(h_{q^{-1}}\phi)(z)H_{q^{-1}}(S(z,w))=B(z)S(z,w)S(q^{-1}z,w)+C(z)S(z,w)+D(z),\label{eq2.4} \tag{6}\] with \(\phi\), \(B\), \(C\) and \(D\) are a polynomials (\(\phi\) monic).
The class of the \(q\)-Laguerre-Hahn satisfying ((6)) is \(s(w)=\max(\deg (B)-2,\deg (C)-1,\deg (D))\) if and only if [3] \[\prod\limits_{c\in Z_\phi}\Bigl\{\mid B(cq)\mid+\mid C(cq)\mid+\mid D(cq)\mid\Bigr\}>0,\label{eq2.5} \tag{7}\] where \(Z_\phi\) is the set of roots of \(\phi\).
Remark 1. 1. When the conditions (7) are realized, Eq. (6) can not be simplified by any polynomial [3].
2. When \(B=0\), we deal with the \(H_q\)-semi-classical forms [4].
3. The case, when \(B=0\), \(\deg(\phi)\leq2 ,\deg(C)\leq1\), \(\deg(D)\leq0\), corresponds to the \(H_q\)-classical forms [13].
4. By extension, the integer \(s(w)\) is also the class of the MOPS \(\{P_n\}_{n\geq 0}\) with respect to \(w\).
In what follows, we assume that \(q\in\tilde{\mathbb{C}}\).
We are going to use the following notations and results [14].
For \(a,a_1,…,a_n\): \[(a,q)_n:=\left\lbrace \begin{array}{l} 1,\,\,\,\,n=0,\\ \prod\limits_{\nu=0}^{n-1}(1-aq^\nu),\,\,n\geq1, \end{array} \right.\label{eq2.6} \tag{8}\] \[(a,q)_{\infty}= \prod\limits_{\nu\geq0}(1-aq^\nu),\,\mid q\mid<1,\label{eq2.7} \tag{9}\] \[(a_1,…,a_n;q)_\infty=\prod\limits_{j=1}^n(a_j,q)_\infty,\,\mid q\mid<1. \label{eq2.8} \tag{10}\]
In the sequel, we suppose that \(u_q\) is \(q\)-Laguerre-Hahn form of class \(s(u_q)\) satisfying the \(q\)-Ricatti equation [3] \[\phi(q^{-1}z)H_{q^{-1}}(S(z,u_q))=B(z)S(z,u_q)S(q^{-1}z,u_q)+C(z)S(z,u_q)+D(z).\label{eq3.1} \tag{11}\]
Lemma 1. The Eq. (11) it is equivalent to \[B_1(z)S(z,u_q)S(q^{-1}z,u_q)+R_1(z)S(q^{-1}z,u_q)+C_1(z)S(z,u_q)+D_1(z)=0,\label{eq3.2} \tag{12}\] where
\[\begin{aligned} \label{eq3.3} &B_1(x)=(q^{-1}-1)xB(x),\, R_1(x)=-\phi(q^{-1}x),\notag\\ &C_1(x)=\phi(q^{-1}x)+(q^{-1}-1)xC(x),\,D_1(x)=(q^{-1}-1)xD(x). \end{aligned} \tag{13}\]
Moreover, we have \[(R_1, D_1),\,(C_1, D_1),\,(B_1, R_1),\,(B_1, C_1)\neq(0,0),\label{eq3.4} \tag{14}\] \[B_1D_1-R_1C_1\neq0.\label{eq3.5} \tag{15}\]
Proof. As, we have \(H_{q^{-1}}(S(z,u_q))=\frac{S(q^{-1}z,u_q)-S(z,u_q)}{(q^{-1}-1)z}\), Eq. (11) can be written as following \[\begin{aligned} &(q^{-1}-1)zB(z)S(z,u_q)S(q^{-1}z,u_q)-\phi(q^{-1}z)S(q^{-1}z,u_q)\\ &\qquad+\left(\phi(q^{-1}z)+(q^{-1}-1)zC(z)\right)S(z,u_q)+(q^{-1}-1)zD(z)=0.\\ \end{aligned}\] Which provides (12) with (13).
Since the \(S(z,u_q)\) is not a rational function [12], we deduce the relations (14). Suppose that \(B_1D_1-R_1C_1=0\). Multiplying both sides identities of (12) by the polynomial \(B_1\), we obtain \[\begin{aligned} &B_1^2(z)S(z,u_q)S(q^{-1}z,u_q)+B_1(z)R_1(z)S(q^{-1}z,u_q)+B_1(z)C_1(z)S(z,u_q)+B_1(z)D_1(z)=0, \end{aligned}\] but \(B_1D_1=R_1C_1\), we get \[\begin{aligned} &B_1^2(z)S(z,u_q)S(q^{-1}z,u_q)+B_1(z)R_1(z)S(q^{-1}z,u_q)+B_1(z)C_1(z)S(z,u_q)+C_1(z)R_1(z)=0. \end{aligned}\]
Which is equivalent to \[\left(B_1(z)S(q^{-1}z,u_q)+C_1(z)\right)\left(B_1(z)S(z,u_q)+R_1(z)\right)=0.\]
Therefore \(B_1(z)S(q^{-1}z,u_q)+R_1(z)=0\) or \(B_1(z)S(z,u_q)+C_1(z)=0\), thus \(S(z,u_q)\) is a rational function. Which is absurd, then \(B_1D_1-R_1C_1\neq0\). Hence (15). ◻
Remark 2. Let \(M(z)\) be the matrix defined as following \[M(z)= \left( \begin{array}{cc} -R_1(z) & B_1(z) \\ -D_1(z) & C_1(z)\\ \end{array} \right).\label{eq3.6} \tag{16}\]
Proposition 1. The Stieltjes function \(S(z,u_q)\) satisfies the following \(q^{-k-1}\)-difference equation
\[\begin{aligned} \label{eq3.7} B_{k+1}(z)S(q^{-k-1}z,u_q)S(z,u_q)+R_{k+1}(z)S(q^{-k-1}z,u_q) +C_{k+1}(z)S(z,u_q)+D_{k+1}(z)=0, \,k\in\mathbb{N},& \end{aligned} \tag{17}\] where \[\left( \begin{array}{c} B_{k+1}(z) \\ C_{k+1}(z) \\ \end{array} \right)=M(q^{-k}z)\left( \begin{array}{c} B_{k}(z) \\ C_{k}(z) \\ \end{array} \right),\,k\geq0,\] \[\left( \begin{array}{c} R_{k+1}(z) \\ D_{k+1}(z) \\ \end{array} \right)=M(q^{-k}z)\left( \begin{array}{c} R_{k}(z) \\ D_{k}(z) \\ \end{array} \right) \,k\geq 0, \label{eq3.8} \tag{18}\] with \[C_0=1, R_0=-1, B_0=D_0=0.\label{eq3.9} \tag{19}\] Moreover \[\left( \begin{array}{c} B_{k+1}(z) \\ C_{k+1}(z) \\ \end{array} \right)=M(q^{-k}z)M(q^{-(k-1)}z)…M(z) \left( \begin{array}{c} 0 \\ 1 \\ \end{array}\right),k\geq0.\label{eq3.10} \tag{20}\] \[\left( \begin{array}{c} R_{k+1}(z) \\ D_{k+1}(z) \\ \end{array} \right)=M(q^{-k}z)M(q^{-(k-1)}z)…M(z) \left( \begin{array}{c} -1\\ 0 \\ \end{array}\right),k\geq0.\label{eq3.11} \tag{21}\]
In addition \(u_q\) is also \(q^{-k-1}\)-Laguerre-Hahn form satisfying the \(q^{k+1}\)-Ricatti equation \[\begin{aligned} \label{eq3.12} \varphi_{k+1}(q^{-k-1}z) H_{q^{-k-1}}\left(S(z,u_q)\right)=&-B_{k+1}(z)S(z,u_q)S(q^{-k-1}z,u_q) \notag\\ &-\left(R_{k+1}(z)+C_{k+1}(z)\right)S(z,u_q)-D_{k+1}(z). \end{aligned} \tag{22}\] where \[\varphi_{k+1}(z)=(1-q^{k+1})zR_{k+1}(q^{k+1}z).\label{eq3.13} \tag{23}\]
Proof. By induction on \(k\in\mathbb{N}\).
For \(k=0\) it is the \(q\)-difference Eq. (12).
Assume that for an integer \(k\geq0\) we have
\[\begin{aligned} \label{eq3.14} B_{k}(z)S(q^{-k}z,u_q)S(z,u_q)+R_{k}(z)S(q^{-k}z,u_q) +C_{k}(z)S(z,u_q)+D_{k}(z)=0, \,k\in\mathbb{N}&. \end{aligned} \tag{24}\]
From the Eq. (12), we get \[S(z,u_q)=-\frac{R_{1}(z)S(q^{-1}z,u_q)+D_{1}(z)}{B_{1}(z)S(q^{-1}z,u_q)+C_{1}(z)}.\]
Therefore \[S(q^{-k}z,u_q)=-\frac{R_{1}(q^{-k}z) S(q^{-k-1}z,u_q)+D_{1}(q^{-k}z)}{B_{1}(q^{-k}z)S(q^{-k-1}z,u_q)+C_{1}(q^{-k}z)}.\]
Based on the previous relation, (24) becomes \[\begin{aligned} -B_{k}(z)S(z,u_q)&\frac{R_{1}(q^{-k}z) S(q^{-k-1}z,u_q)+D_{1}(q^{-k}z)}{B_{1}(q^{-k}z)S(q^{-k-1}z,u_q)+C_{1}(q^{-k}z)} -R_{k}(z)\frac{R_{1}(q^{-k}z) S(q^{-k-1}z,u_q)+D_{1}(q^{-k}z)}{B_{1}(q^{-k}z)S(q^{-k-1}z,u_q)+C_{1}(q^{-k}z)}\\ &+C_{k}(z)S(z,u_q)+D_{k}(z)=0. \end{aligned}\]
Equivalently \[\begin{aligned} \label{eq3.15} &B_{k+1}(z)S(z,u_q)S(q^{-k-1}z,u_q) +R_{k+1}(z)S(q^{-k-1}z,u_q)+C_{k+1}(z)S(z,u_q) +D_{k+1}(z)=0, \end{aligned} \tag{25}\] where \[\begin{aligned} \label{eq3.16} &B_{k+1}(z)=B_1(q^{-k}z)C_k(z)-R_1(q^{-k}z)B_k(z),\notag\\ &C_{k+1}(z)=C_1(q^{-k}z)C_k(z)-D_1(q^{-k}z)B_k(z),\notag\\ &R_{k+1}(z)=B_1(q^{-k}z)D_k(z)-R_1(q^{-k}z)R_k(z),\notag\\ &D_{k+1}(z)=C_1(q^{-k}z)D_k(z)-D_1(q^{-k}z)R_k(z). \end{aligned} \tag{26}\]
Thus we get (17) with (18).
Based on (26), we deduce (20) and (21).
Eq. (22) can be deduced from (17) and Lemma 1. ◻
Remark 3. Based on Eq. (22), we see that the form \(u_q\) is also \(q^{k+1}\)-Laguerre-Hahn and its class depends on the integer \(k\) with respect to the operator \(H_{q^{-k-1}}\). Of course the form \(u_{q^{k+1}}\) remains of class \(s(u_q)\).
Lemma 2. When \(B=0\) i.e \(B_1=0\), we have
\[\begin{aligned} \label{eq3.17} &M(q^{-k}z)M(q^{-(k-1}z)… M(z)= \left( \begin{array}{cc} (-1)^{k+1}\prod\limits_{n=0}^kR_1(q^{-n}z) & 0\\ A_k(z) & \prod\limits_{n=0}^kC_1(q^{-n}z) \\ \end{array} \right),\,k\geq0, \end{aligned} \tag{27}\] with \[A_k(z)=-\left(\prod\limits_{n=0}^kC_1(q^{-n}z)\right) \sum\limits_{n=0}^k(-1)^n\frac{D_1(q^{-n}z)}{R_1(q^{-n}z)} \prod\limits_{\mu=0}^n\frac{R_1(q^{-\mu}z)}{C_1(q^{-\mu}z)},\,k\geq0.\label{eq3.18} \tag{28}\]
Proof. Since \(B_1=0\), from Lemma 1, we get \(C_1 R_1\neq0\).
By induction on \(k,\,k\in \mathbb{N}\).
For \(k=0\), it is (16).
Suppose that for an integer \(k\), we have (27). We may write \[M(q^{-(k+1)}z)M(q^{-k}z)… M(z)=M(q^{-(k+1)}z)\left(M(q^{-k}z)… M(z)\right),\] using the hypothesis of the induction and (16), we get \[\begin{aligned} &M(q^{-(k+1)}z)\left(M(q^{-k}z)… M(z)\right)\\ &\qquad= \left( \begin{array}{cc} -R_1(q^{-(k+1)}z) & B_1(q^{-(k+1)}z) \\ -D_1(q^{-(k+1)}z) & C_1(q^{-(k+1)}z)\\ \end{array} \right)\left( \begin{array}{cc} (-1)^{k+1}\prod\limits_{n=0}^kR_1(q^{-n}z) & 0\\ A_k(z) & \prod\limits_{n=0}^kC_1(q^{-n}z) \\ \end{array} \right)\\ &\qquad=\left( \begin{array}{cc} (-1)^{k}\prod\limits_{n=0}^{k+1}R_1(q^{-n}z) & 0\\ x_k(z) & \prod\limits_{n=0}^{k+1}C_1(q^{-n}z) \\ \end{array} \right), \end{aligned}\] with \(x_k(z)=C_1(q^{-k-1}z)A_k(z)+(-1)^kD_1(q^{-k-1}z)\prod\limits_{n=0}^{k}C_1(q^{-n}z)\).
From (28), it follows that \[\begin{aligned} x_k(z)&=(-1)^kD_1(q^{-k-1}z)\prod\limits_{n=0}^{k}R_1(q^{-n}z)-C_1(q^{-k-1}z)\left(\prod\limits_{n=0}^kC_1(q^{-n}z)\right) \sum\limits_{n=0}^k(-1)^n\frac{D_1(q^{-n}z)}{R_1(q^{-n}z)} \prod\limits_{\mu=0}^n\frac{R_1(q^{-\mu}z)}{C_1(q^{-\mu}z)}\\ &=-\left(\prod\limits_{n=0}^{k+1}C_1(q^{-n}z)\right) \sum\limits_{n=0}^{k+1}(-1)^n\frac{D_1(q^{-n}z)}{R_1(q^{-n}z)} \prod\limits_{\mu=0}^n\frac{R_1(q^{-\mu}z)}{C_1(q^{-\mu}z)}. \end{aligned}\]
Hence the desired result. ◻
Corollary 1. When \(u_q\) is \(H_q\)-semi-classical form \((B_1=0)\), we have \[R_{k+1}(z)S(q^{-k-1}z,u_q) +C_{k+1}(z)S(z,u_q)+D_{k+1}(z)=0, \,k\in\mathbb{N}, \label{eq3.19} \tag{29}\] where \[R_{k+1}(z)=(-1)^k\prod\limits_{n=0}^kR_1(q^{-n}z),\label{eq3.20} \tag{30}\] \[C_{k+1}(z)=\prod\limits_{n=0}^kC_1(q^{-n}z),\label{eq3.21} \tag{31}\] \[D_{k+1}(z)=\left(\prod\limits_{n=0}^kC_1(q^{-n}z)\right) \sum\limits_{n=0}^k(-1)^n\frac{D_1(q^{-n}z)}{R_1(q^{-n}z)} \left(\prod\limits_{\mu=0}^n\frac{R_1(q^{-\mu}z)}{C_1(q^{-\mu}z)}\right), \,k\in\mathbb{N}. \label{eq3.22} \tag{32}\]
Moreover, the form \(u_q\) is also \(H_{q^{k+1}}\)-semi-classical, satisfying
\[\begin{aligned} \label{eq3.23} &\varphi_{k+1}(z)H_{q^{-k-1}}\left(S(z,u_q)\right)=-\left(R_{k+1}(z)+C_{k+1}(z)\right)S(z,u_q)-D_{k+1}(z), \end{aligned} \tag{33}\] with \[\varphi_{k+1}(z)=(q^{-k-1}-1)zR_{k+1}(z).\label{eq3.24} \tag{34}\]
Proof. Since \(B_1=0\), by (26) and Proposition 1, we get successively \[\left( \begin{array}{c} B_{k+1}(z) \\ C_{k+1}(z) \\ \end{array} \right)=\left( \begin{array}{cc} (-1)^{k+1}\prod\limits_{n=0}^kR_1(q^{-n}z) & 0\\ A_k(z) & \prod\limits_{n=0}^kC_1(q^{-n}z) \\ \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \\ \end{array}\right),k\geq0,\] and \[\left( \begin{array}{c} R_{k+1}(z) \\ D_{k+1}(z) \\ \end{array} \right)=\left( \begin{array}{cc} (-1)^{k+1}\prod\limits_{n=0}^kR_1(q^{-n}z) & 0\\ A_k(z) & \prod\limits_{n=0}^kC_1(q^{-n}z) \\ \end{array} \right) \left( \begin{array}{c} -1 \\ 0 \\ \end{array}\right),k\geq0.\]
Which implies (29)-(32) and \(B_{k+1}=0,\,k\geq0\).
Based on (29) and Proposition 1, we get (33). Hence the desired results. ◻
In the following we study some examples of \(H_q\)-semi-classical and \(q\)-Laguerre-Hahn forms.
Let \(u_q:=\mathcal{T}_q\), we know that \(\mathcal{T}_q\) it is of class \(s=0\) and we have [9] \[\left\lbrace \begin{array}{l} \gamma_{1}^{\mathcal{T}_q}=\frac{q}{q+1},\,\,\, \gamma_{n+1}^{\mathcal{T}_q}=\frac{q^{n+1}}{(q^{n}+1)(q^{n+1}+1)},\,n\geq1,\\ (z^2-q^2)H_{q^{-1}}(S(z,\mathcal{T}_q))+qzS(z,\mathcal{T}_q)=0. \end{array} \right. \label{eq4.1} \tag{35}\]
Proposition 2. For \(k\ge0\), the Stieltjes function \(S(.,\mathcal{T}_q)\) satisfies the following \(q^{k+1}\)-difference equation \[\varphi_{k+}(q^{-k-1}z,\mathcal{T}_q)H_{q^{-k-1}}(S(z,\mathcal{T}_q)) +X_{k+1}(z,\mathcal{T}_q)S(z,\mathcal{T}_q)=0,\label{eq4.2} \tag{36}\] where \[\begin{aligned} \label{eq4.3} &\varphi_{k+1}(z,\mathcal{T}_q)=(z^2,q^2)_{k+1},\notag\\ &X_{k+1}(z,\mathcal{T}_q)=\frac{1}{(q^{-k-1}-1)z}\Bigl\{(q^{-2}z^2,q^{-2})_{k+1}- (q^{-1}z^2,q^{-2})_{k+1}\Bigr\},\,k\geq0. \end{aligned} \tag{37}\]
Furthermore, the form \(\mathcal{T}_q\) it is \(H_{q^{k+1}}\)-semi-classical form of class \(2k\).
Proof. From (35) and Lemma 1, we get \[q^2(1-q^{-2}z^2)S(q^{-1}z,\mathcal{T}_q)-q^2(1-q^{-1}z^2)S(z,\mathcal{T}_q)=0.\]
Which implies that \(S(z,\mathcal{T}_q)\) satisfies (12) with \[R_1(z)=1-q^{-2}z^2,C_1(z)=-(1-q^{-1}z^2),\,D_1(z)=0,\,B_1(z)=0.\label{eq4.4} \tag{38}\]
Taking into account Corollary 1, we obtain the following \(q^{k+1}\)-difference equation \[R_{k+1}(z)S(q^{-k-1}z,\mathcal{T}_q)+C_{k+1}(z)S(z,\mathcal{T}_q)=0.\label{eq4.5} \tag{39}\] with \[\begin{aligned} \label{eq4.6} &R_{k+1}(z)=(-1)^k(q^{-2}z^2,q^{-2})_{k+1},\notag\\ &C_{k+1}(z)=(-1)^{k+1}(q^{-1}z^2,q^{-2})_{k+1},\,\, k\geq0. \end{aligned} \tag{40}\]
Based on (39), (40) and Corollary 1, we deduce (36) with (37).
Let \(c\) be zero of \(\varphi_{2k+2}(x,\mathcal{T}_q)\), then \(c^2=q^{-2m},\,0\leq m\leq k\).
We have \(X_{k+1}(cq^{k+1},\mathcal{T}_q)=X_{k+1}(q^{k+1-2m},\mathcal{T}_q)\), using (37) we obtain \[\begin{aligned} X_{k+1}(cq^{k+1},\mathcal{T}_q)&=\frac{1}{(1-q^{k+1})c} \left\{\prod\limits_{n=0}^k(1-q^{2(k-n-m)})- \prod\limits_{n=0}^k(1-q^{2(k-n-m)+1})\right\}\notag\\ &=-\frac{\prod\limits_{n=0}^k(1-q^{2(n-m)+1} )}{(1-q^{k+1})c}\neq0. \end{aligned}\]
From (7), we can not simplify Eq. (36). Since \(\deg(X_{k+1}(x,\mathcal{T}_q))=2k+1\), we deduce that \(\mathcal{T}_q\) is \(H_{q^{k+1}}\)-semi-classical of class \(s=\deg(X_{k+1}(x,\mathcal{T}_q))-1=2k,\,k\geq0.\) Hence the desired result. ◻
Let \(u_q:=\mathcal{U}_q\), we know that \(\mathcal{U}_q\) it is of class \(s=0\) and we have [9] \[\left\lbrace \begin{array}{l} \gamma_{n+1}^{\mathcal{U}_q}=\frac{q^{n+2}}{(q^{n+1}+1)(q^{n+2}+1)},\,n\geq0,\\ (z^2-1)H_{q^{-1}}(S(z,\mathcal{U}_q))-zS(z,\mathcal{U}_q)-q-1=0. \end{array} \right. \label{eq4.7} \tag{41}\]
Proposition 3. One has
\[\begin{aligned} \label{eq4.8} h_{q^{-k-1}}\varphi_{k+1}(z,\mathcal{U}_q)H_{q^{-k-1}}(S(z,\mathcal{U}_q)) &+X_{k+1}(z,\mathcal{U}_q)S(z,\mathcal{U}_q)+T_{k+1}(z,\mathcal{U}_q)=0, \end{aligned} \tag{42}\] where
\[\begin{aligned} \label{eq4.9} &\varphi_{k+1}(z,\mathcal{U}_q)=(qz^2,q^2)_{k+1},\notag\\ &X_{k+1}(z,\mathcal{U}_q)=\frac{1}{(q^{-k-1}-1)z}\Bigl\{(q^{-1}z^2,q^{-2})_{k+1}- (q^{-2}z^2,q^{-2})_{k+1})\Bigr\},\notag\\ &T_{k+1}(z,\mathcal{U}_q)=\frac{1-q^{-2}}{q^{-k-1}-1}(q^{-2}z^2,q^{-2})_{k+1}\times\sum\limits_{n=0}^k\frac{(-1)^nq^{-n}}{1-q^{-2n-1}z^2} \frac{(q^{-1}z^2,q^{-2})_{n+1}}{(q^{-2}z^2,q^{-2})_{n+1}},\,k\geq0. \end{aligned} \tag{43}\]
In addition, the form \(\mathcal{U}_q\) it is \(H_{q^{k+1}}\)-semi-classical form of class \(2k\), \(k\geq0\).
Proof. Based on (41) and Lemma (1), we get \[R_1(z)S(q^{-1}z,\mathcal{U}_q)+C_1(z)S(z,\mathcal{U}_q)+D_1(z)=0,\label{eq4.10} \tag{44}\] with \[R_{1}(z)=1-q^{-1}z^2,\,C_1(z)=-(1-q^{-2}z^2),\,D_1(z)=(q^{-2}-1)z.\label{eq4.11} \tag{45}\]
Indeed by virtue of Corollary 1, we get
\[\begin{aligned} \label{eq4.12} &R_{k+1}(z)=(-1)^k(q^{-1}z^2,q^{-2})_{k+1},\notag\\ &C_{k+1}(z)=(-1)^{k+1}(q^{-2}z^2,q^{-2})_{k+1},\notag\\ &D_{k+1}(z)=(-1)^{k+1}(1-q^{-2})z(q^{-2}z^2,q^{-2})_{k+1}\times\sum\limits_{n=0}^k\frac{q^{-n}}{1-q^{-2n-1}z^2} \frac{(q^{-1}z^2,q^{-2})_{n+1}}{(q^{-2}z^2,q^{-2})_{n+1}},\,k\geq0, \end{aligned} \tag{46}\] and \[R_{k+1}(z)S(q^{-k-1}z,\mathcal{U}_q)+ C_{k+1}(z)S(z,\mathcal{U}_q)+D_{k+1}(z)=0,\,k\geq0.\label{eq4.13} \tag{47}\]
Taking into account (46), (47) and Corollary 1, we obtain (42) with (43).
Let \(c\) be zero of \(\phi_{2k+2}(x,\mathcal{U}_q)\), by (43), we get \(c^2=q^{-2m-1},\,0\leq m\leq k.\)
From (43), we obtain \[\begin{aligned} X_{k+1}(q^{k+1}c,\mathcal{U}_q)&=\frac{-1}{(q^{-k-1}-1)cq^{k+1}} \Bigl\{(q^2,q^{-2m})_{k+1}-(q^2,q^{-2m+1})_{k+1}\Bigr\}\\ &=\frac{(q^2,q^{-2m+1})_{k+1}}{(q^{-k-1}-1)cq^{k+1}}\neq0,\,k\geq0. \end{aligned}\]
Based on (7), (42) cannot be simplified. Then the class of \(\mathcal{U}_q\) is \(2k,\,k\geq0\) since \(\deg(X_{k+1}(x,\mathcal{U}_q))=2k+1\) and \(\deg(T_{k+1}(x,\mathcal{U}_q))=2k,\,k\geq0\). ◻
Let \(u_q(\omega)\) be the symmetric form of Brenk type which is \(H_q\)-semi-classical form of class one [5, pp.20] [8, pp.134] \[\left\lbrace \begin{array}{l} \gamma_{2n+1}^{u_q(\omega)}=q^{-4n-3}(1-\omega q^{2n}),\,\,\gamma_{2n+2}^{u_q(\omega)}=q^{-4n-5}(1-q^{2n+2}) ,\,\,n\geq0,\\ \omega\neq q,\,\omega\neq q^{2n},\,n\geq0,\\ \phi(q^{-1}z)H_{q^{-1}}(S(z,u_q(\omega)))=C(z)S(z,u_q(\omega))+D(z),\\ \phi(z)=z(z^2+\omega q^{-3}),\,C(z)=q^{-3}(q-1)^{-1}(qz^2+\omega-q),\\ D(z)=q^{-1}(q-1)^{-1}z. \end{array} \right. \label{eq4.14} \tag{48}\]
We recall the Jacobi triple product identity [10, 11]: \[\sum\limits_{n=-\infty}^{+\infty}q^{-n^2}z^n= (q^{-2},-\frac{z}{q},-\frac{1}{qz};q^{-2})_\infty,\,\,\mid q\mid>1,\,z\in\mathbb{C}-\{0\}.\label{eq4.15} \tag{49}\]
Proposition 4. On has \[S(z,u_q(0))=-\frac{1}{z}\sum\limits_{n=0}^{+\infty}q^{-n^2} \left(\frac{1}{q^2z^2}\right)^n,\,\,\,\mid q\mid>1,\,z\in\mathbb{C}-\{0\},\label{eq4.16} \tag{50}\] \[\begin{aligned} \label{eq4.17} -zS(z,u_q(0))-\frac{1}{q^2z}S(\frac{1}{q^2z},u_q(0))-1&= (q^{-2},-\frac{1}{q^2z^2},-qz^2;q^{-2})_\infty\mid q\mid>1,\,z\in\mathbb{C}-\{0\}, \end{aligned} \tag{51}\] \[S(z,u_q(\omega))=-\frac{1}{z}\sum\limits_{n=0}^{+\infty}(-\omega q^{-3})^n \frac{(\omega^{-1},q^{-2})_n}{z^{2n}},\,\mid q\mid>1,\,\mid z\mid>\mid \omega q^{-3}\mid^{{\frac{1}{2}}},\,\omega\neq0.\label{eq4.18} \tag{52}\]
Proof. From (48), we may write \[\left\langle H_q\left(x(x^2+\omega q^{-3})u_q\right)- (q-1)^{-1}\left(x^2+(\omega-1)q^{-3}\right)u_q,x^{2n}\right\rangle=0,\,n\geq0.\]
Equivalently \[(q^{2n}-1)\left\langle u_q,x^{2n-1}(x(x^2+\omega q^{-3})\right\rangle+ \left\langle u_q,x^{2n}\left(x^2+(\omega-1)q^{-3}\right)\right\rangle=0,\,n\geq0.\]
Thus \[(u_q(\omega))_{2n+2}=(q^{-2n-3}-\omega q^{-3})(u_q(\omega))_{2n},\,n\geq0.\]
Since \((u_q(\omega))_0=1\), then \[(u_q(\omega))_{2n+2}=\prod\limits_{k=0}^n(q^{-2k-3}-\omega q^{-3}),\,n\geq0.\label{eq4.19} \tag{53}\]
When \(\omega=0\), by (53), we get \[(u_q(0))_{2n}=\prod\limits_{k=0}^{n-1}q^{-2k-3} =q^{-n^2-2n},\,n\geq0.\label{eq4.20} \tag{54}\]
From (54) and the fact that \((u_q(0))_{2n+1}=0,\,n\geq 0\), we obtain (50).
By (49), we may write \[\sum\limits_{n=-\infty}^{+\infty}\frac{q^{-n^2}}{(q^2z^2)^n}= (q^{-2},-\frac{1}{q^3z^2},-qz^2;q^{-2})_\infty,\,\,\mid q\mid>1,\,z\in\mathbb{C}-\{0\}.\]
Which is equivalent to
\[\begin{aligned} \label{eq4.21} \sum\limits_{n=0}^{+\infty}\frac{q^{-n^2}}{(q^2z^2)^n} +& \sum\limits_{n=0}^{+\infty}q^{-n^2}(q^2z^2)^n-1=(q^{-2},-\frac{1}{q^3z^2},-qz^2;q^{-2})_\infty,\,\,\mid q\mid>1,\,z\in\mathbb{C}-\{0\}, \end{aligned} \tag{55}\] and by (50), we get
\( -zS(z,u_q(0)) + \sum\limits_{n=0}^{+\infty}q^{-n^2}(q^2z^2)^n-1=(q^{-2},-\frac{1}{q^3z^2},-qz^2;q^{-2})_\infty,\,\,\mid q\mid>1,\,z\in\mathbb{C}-\{0\}.\)
Based on (50), we have \(\sum\limits_{n=0}^{+\infty}q^{-n^2}(q^2z^2)^n=-\frac{1}{q^2z} S(\frac{1}{q^2z},u_q(0))\), which provides (51).
In the case where \(\omega\neq0\), (53), implies that \[(u_q(\omega))_{2n}=( -q^{-3}\omega)^n(\omega^{-1},q^{-2})_n,\,n\geq0.\]
Then, we obtain (52).
Let \(a_n = \frac{( -q^{-3}\omega)^n(\omega^{-1},q^{-2})_n}{z^{2n}},\,n\geq0\). Then
\[\lim\limits_{n\longrightarrow+\infty}\mid\frac{a_{n+1}}{a_n}\mid= \lim\limits_{n\longrightarrow+\infty}\frac{\mid\omega q^{-3}\mid}{\mid z\mid^2}(1-\omega^{-1}q^{-2n})=\frac{\mid\omega q^{-3}\mid}{\mid z\mid^2},\,\mid q\mid>1.\]
By the D’Alembert criterion, we conclude the convergence of the series (52) when \(\mid z\mid>\mid \omega q^{-3}\mid^{{\frac{1}{2}}}\). ◻
Proposition 5. The Stieltjes function \(S(z,u_q(\omega))\) satisfies the following \(q^{-k-1}\) -difference equation for \(k\geq0\)
\[\begin{aligned} \label{eq4.22} &X_{k+1}(z,u_q(\omega))S(q^{-k-1}z,u_q(\omega))+Y_{k+1}(z,u_q(\omega))S(z,u_q(\omega)) +Z_{k+1}(z,u_q(\omega))=0, \end{aligned} \tag{56}\] where \[\begin{aligned} \label{eq4.23} X_{k+1}(z,u_q(\omega))&=\prod\limits_{n=0}^k\left(\omega+q^{-2n+1}z^2\right),\,\, Y_{k+1}(z,u_q(\omega))=-q^{k+1},\notag\\ Z_{k+1}(z,u_q(\omega))&=q^{k+1}z\sum\limits_{n=0}^k\frac{q^{-2n+1}}{\omega+q^{-2n+1}z^2} \prod\limits_{\mu=0}^n\left(\omega+q^{-2\mu+1}z^2\right). \end{aligned} \tag{57}\]
Moreover, the form \(u_q(\omega)\) it is \(H_{q^{k+1}}\)-semi-classical form of class \(2k+1,\,k\geq0\) satisfying the following equation:
\[\begin{aligned} \label{eq4.24} &\varphi_{k+1}(q^{-k-1}z,u_q(\omega))H_{q^{-k-1}}\left(S(z,u_q(\omega))\right)=-\left(X_{k+1}(z,u_q(\omega))+Y_{k+1}(z,u_q(\omega))\right) S(z,u_q(\omega))-Z_{k+1}(z,u_q(\omega)), \end{aligned} \tag{58}\] where \[\varphi_{k+1}(z,u_q(\omega))=(1-q^{k+1})zX_{k+1}(q^{k+1}z,u_q(\omega)).\label{eq4.25} \tag{59}\]
Proof. From (48) and Lemma 1, we get
\(-q^{-4}z(qz^2+\omega) S(q^{-1}z,u_q(\omega))+q^{-3}zS(z,u_q(\omega))-q^{-2}z^2=0.\)
Then \[R_1(z)S(q^{-1}z,u_q(\omega))+C_1(z)S(z,u_q(\omega))+D_1(z)=0,\label{eq4.26} \tag{60}\] where \[R_1(z)=qz^2+\omega,\,C_1(z)=-q,\,D_1(z)=q^{2}z.\label{eq4.27} \tag{61}\]
By virtue of Corollary 1, it follows that \[R_{k+1}(z)S(q^{-k-1}z,u_q(\omega)) +C_{k+1}(z)S(z,u_q(\omega))+D_{k+1}(z)=0, \,k\in\mathbb{N}, \label{eq4.28} \tag{62}\] where \[\begin{aligned} \label{eq4.29} R_{k+1}(z)&=(-1)^k\prod\limits_{n=0}^k(\omega+q^{-2n+1}z^2),\,\, C_{k+1}(z)=(-1)^{k+1}q^{k+1},\notag\\ D_{k+1}(z)&=(-1)^kq^{k+1}z \sum\limits_{n=0}^k\frac{q^{-2n+1}}{q^{-2n+1}z^2+\omega} \prod\limits_{\mu=0}^n(q^{-2\mu+1}z^2+\omega), \,k\in\mathbb{N}. \end{aligned} \tag{63}\]
Taking into account (62) and (63), we obtain (56) with (57).
From (57) and Corollary 1, we get (58).
Let \(c\) be a zero of \(\varphi_{k+1}(z,u_q(\omega))\), by (34), we have \(c=0\), or \(X_{k+1}(q^{k+1}z,u_q(\omega))=0\).
If \(c=0\), we have \(X_{k+1}(0,u_q(\omega))+Y_{k+1}(0,u_q(\omega))=\omega^{k+1}-q^{k+1}\neq0\). By (7), we can not simplify (58) by the factor \(z\).
In the case where \(X_{k+1}(q^{k+1}c,u_q(\omega))=0\), (57) implies that \(Y_{k+1}(q^{k+1}c,u_q(\omega))=-q^{k+1}\neq0\), then, (58) is not simplified by the factor \(z-cq^{k+1}\) by virtue of (7).
Moreover, we have \(\deg(X_{k+1}+Y_{k+1})=2k+2\), \(\deg(Z_{k+1})=2k+1\), then \(u_q(\omega)\) \(H_{q^{k+1}}\)-semi-classical form of class \(2k+1\). ◻
In the following, we study the first associated form \(u_q^{(1)}(\omega)\).
Proposition 6.For \(k\geq0\), the form \(u_q^{(1)}(\omega)\) it is \(q^{k+1}\)-Laguerre-Hahn form of class \(2k+1\) and its Stieltjes function satisfy the following \(q^{k+1}\)-Ricati equation \[\begin{aligned} \label{eq4.30} &\varphi_{k+1}^{(1)}(q^{-k-1}z,u_q^{(1)}(\omega))H_{q^{-k-1}}(S(z,u_q^{(1)}(\omega)))\notag\\&\qquad=B_{k+1}^{(1)}(z,u_q^{(1)}(\omega))S(z,u_q^{(1)}(\omega))S(q^{-k-1}z,u_q^{(1)}(\omega))+X_{k+1}^{(1)}(z,u_q^{(1)}(\omega))S(z,u_q^{(1)}(\omega))+Y_{k+1}^{(1)}(z,u_q^{(1)}(\omega)), \end{aligned} \tag{64}\] with
\[\begin{aligned} \label{eq4.31} \varphi_{k+1}^{(1)}(z,u_q^{(1)}(\omega))=&\gamma_1^{u_q(\omega)}\Bigl\{\varphi_{k+1}(z,u_q(\omega)) +(q^{k+1}-1)z\bigl(X_{k+1}(q^{k+1}z,u_q(\omega))\notag\\ &+Y_{k+1}(q^{k+1}z,u_q(\omega)) -q^{k+1}zZ_{k+1}(q^{k+1}z,u_q(\omega))\bigr)\Bigr\},\notag\\ B_{k+1}^{(1)}(z,u_q^{(1)}(\omega))=&-(\gamma_1^{u_q(\omega)})^2Z_{k+1}(z,u_q(\omega)),\notag\\ X_{k+1}^{(1)}(z)=&\gamma_1^{u_q(\omega)}\Bigl\{X_{k+1}(z,u_q(\omega))+Y_{k+1}(z,u_q(\omega))-(1+q^{-k-1})zZ_{k+1}(z,u_q(\omega))\Bigr\},\notag\\ Y_{k+1}^{(1)}(z,u_q^{(1)}(\omega))=&q^{-k-1}z\Bigl\{X_{k+1}(z,u_q(\omega))+Y_{k+1}(z,u_q(\omega))-z Z_{k+1}(z,u_q(\omega))\Bigr\}-\varphi_{k+1}(q^{-k-1}z,u_q(\omega)), \end{aligned} \tag{65}\] where \(X_{k+1}(z,u_q(\omega)),\,Y_{k+1}(z,u_q(\omega)),\,Z_{k+1}(z,u_q(\omega))\) and \(\varphi_{k+1}(z,u_q(\omega))\) are the polynomials defined by (57) and (59).
Proof. We know that [12] \[S(z,u_q(\omega))=\frac{-1}{\gamma_1^{u_q(\omega)}S(z,u_q^{(1)}(\omega))+z}.\label{eq4.32} \tag{66}\]
Applying the operator \(H_{q^{-k-1}}\) to (66), we get \[\begin{aligned} &H_{q^{-k-1}}(S(z,u_q(\omega)))=\frac{\gamma_1^{u_q(\omega)}H_{q^{-k-1}}(S(z,u_q^{(1)}(\omega))) +1}{\left(\gamma_1^{u_q(\omega)}S(z,u_q^{(1)}(\omega))+z\right) \left(\gamma_1^{u_q(\omega)}S(q^{-k-1}z,u_q^{(1)}(\omega))+q^{-k-1}z\right)}. \end{aligned}\]
Multiplying both sides identities by \(\phi(q^{-k-1}z)\), and using (58), we get \[\begin{aligned} &-\left(X_{k+1}(z,u_q(\omega))+Y_{k+1}(z,u_q(\omega))\right)S(z,u_q(\omega)) -Z_{k+1}(z,u_q(\omega))\\ &\qquad=\frac{\gamma_1^{u_q(\omega)}\phi(q^{-k-1}z)H_{q^{-k-1}}(S(z,u_q^{(1)}(\omega))) +\phi(q^{-k-1}z)}{\left(\gamma_1^{u_q(\omega)}S(z,u_q^{(1)}(\omega))+z\right) \left(\gamma_1^{u_q(\omega)}S(q^{-k-1}z,u_q^{(1)}(\omega))+q^{-k-1}z\right)}, \end{aligned}\] and by (66), it follows that \[\begin{aligned} &\frac{X_{k+1}(z,u_q(\omega))+Y_{k+1}(z,u_q(\omega))}{\gamma_1^{u_q(\omega)} S(z,u_q^{(1)}(\omega))+z} -Z_{k+1}(z,u_q(\omega))\\ &\qquad=\frac{\gamma_1^{u_q(\omega)}\phi(q^{-k-1}z)H_{q^{-k-1}}(S(z,u_q^{(1)}(\omega))) +\phi(q^{-k-1}z)}{\left(\gamma_1^{u_q(\omega)}S(z,u_q^{(1)}(\omega))+z\right) \left(\gamma_1^{u_q(\omega)}S(q^{-k-1}z,u_q^{(1)}(\omega))+q^{-k-1}z\right)}. \end{aligned}\]
Equivalently \[\begin{aligned} \gamma_1^{u_q(\omega)}\phi_{k+1}(q^{-k-1}z)H_{q^{-k-1}}(S(z,u_q^{(1)}(\omega)))= &\gamma_1^{u_q(\omega)}\left(X_{k+1}(z)+Y_{k+1}(z)-zZ_{k+1}(z)\right) S(q^{-k-1}z,u_q^{(1)}(\omega))\\ &-(\gamma_1^{u_q(\omega)})^2Z_{k+1}(z)S(z,u_q^{(1)}(\omega)) S(q^{-k-1}z,u_q^{(1)}(\omega))\\ &-\gamma_1^{u_q(\omega)}q^{-k-1}zZ_{k+1}(z)S(z,u_q^{(1)}(\omega))\\ &+ q^{-k-1}z\left(X_{k+1}(z)+Y_{k+1}(z)-zZ_{k+1}(z)\right)-\phi_{k+1}(q^{-k-1}z). \end{aligned}\]
Since \(S(q^{-k-1}z,u_q^{(1)}(\omega))=(q^{-k-1}-1)zH_{q^{-k-1}}(S(z,u_q^{(1)}(\omega))) +S(z,u_q^{(1)}(\omega))\). We deduce deduce (64) with (65).
Let \(c\) be a zero of \(\varphi_{k+1}^{(1)}(z,u_q^{(1)})\), and suppose that we can simplify (64) by the factor \(z-cq^{k+1}\). From (7), we get \[\begin{aligned} &\varphi_{k+1}(c,u_q(\omega)) +(q^{k+1}-1)c+ \left(X_{k+1}(q^{k+1}c,u_q(\omega))+Y_{k+1}(q^{k+1}c,u_q(\omega))-q^{k+1}cZ_{k+1} (q^{k+1}c,u_q(\omega))\right)=0,\\ &Z_{k+1}(q^{k+1}c,u_q(\omega))=0,\\ &X_{k+1}(q^{k+1}c,u_q(\omega))+Y_{k+1}(q^{k+1}c,u_q(\omega))-(1+q^{k+1})c Z_{k+1}(q^{k+1}c,u_q(\omega)),\\ &c\Bigl\{X_{k+1}(q^{k+1}c,u_q(\omega))+Y_{k+1}(q^{k+1}c,u_q(\omega))-q^{k+1}c Z_{k+1}(q^{k+1}c,u_q(\omega))\Bigr\}-\varphi_{k+1}(c,u_q(\omega))=0. \end{aligned}\]
Equivalently \[\begin{aligned} \phi_{k+1}(c)=0,\,Z_{k+1}(q^{k+1}c)=0,\,X_{k+1}(q^{k+1}c)+Y_{k+1}(q^{k+1}c)=0. \end{aligned}\]
Which means that (58) can be simplified by the factor \(z-cq^{k+1}\). Which is contradictory by virtue of(7).
From (57), we may write for \(k\geq0\) \[\begin{aligned} \label{eq4.33} &X_{k+1}(z,u_q(\omega))=q^{-k^2+1}z^{2k+2}+…,\notag\\ &Y_{k+1}(z,u_q(\omega))=-q^{k+1},\notag\\ &Z_{k+1}(z,u_q(\omega))=q^{-k^2+k+2}z^{2k+1}+…,\notag\\ &\varphi_{k+1}(z,u_q(\omega))=(1-q^{k+1})q^{k^2+4k+3}z^{2k+1}+…. \end{aligned} \tag{67}\]
By (65), we have \[\label{eq4.34} \deg(B_{k+1}^{(1)},u_q^{(1)}(\omega))=2k+1. \tag{68}\]
Always by (65) and (67) we have \(X_{k+1}(z,u_q(\omega))=-(\gamma_1^{u_q(\omega)})^2q^{-k^2+k+2}z^{2k+2}….\) Then \[\deg(X_{k+1}^{(1)})(z,u_q^{(1)}(\omega))=2k+2.\label{eq4.35} \tag{69}\] From (65) and (69), we deduce that \(\deg(Y_{k+1}^{(1)})\leq 2k+2\), but \(Y_{k+1}^{(1)}\) is an odd polynomial, then \[\deg(Y_{k+1}^{(1)})\leq2k+1.\label{eq4.36} \tag{70}\]
Finally, by virtue of (67)-(70) and (7), we deduce that the class of \(u_q(\omega)\) is \(2k+1\). Hence the desired result. ◻
Remark 4. When \(k=0\), based on (65), (68), we obtain \[\begin{aligned} &\varphi_1^{(1)}(z,u_q^{(1)},\omega))=\gamma_1^{u_q(\omega)}q^4(1-q)z(q^{-3}+z^2),\, B_1^{(1)}(z,u_q^{(1)},\omega))=-(\gamma_1^{u_q(\omega)})^2qz,\\ &X_1^{(1)}(z,u_q^{(1)},\omega))=-\gamma_1^{u_q(\omega)}(q-\omega+q^2z^2),\, Y_1^{(1)}(z,u_q^{(1)},\omega))=(\omega-1)z, \end{aligned}\] and by (64), we get \[\begin{aligned} \label{eq4.37} q^{-1}z(q^{-3}+q^{-2}z^2)H_{q^{-1}}(S(z,u_q^{(1)}(\omega))=&q^{-5}\frac{1-\omega} {q-1}S(z,u_q^{(1)}(\omega)S(q^{-1}z,u_q^{(1)}(\omega)\notag\\ &+\frac{q^{-4}}{q-1}(q-\omega+q^2z^2)S(z,u_q^{(1)}(\omega)+ \frac{z}{q(q-1)}. \end{aligned} \tag{71}\]
Thus we recover the case studied in [8, pp.134] that \(u_q^{(1)}(\omega)\) it is \(H_{q^{-1}}\)-Laguerre forms of class one. But there are some misprints in formula (4.9) given in [6] and it must be written as (71).
The authors are very grateful to the referees for the constructive and valuable comments and recommendations.
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