# Zeros of quasi-orthogonal $q$-Laguerre polynomials

Pinaki Prasad Kara Department of Mathematics, Sambalpur University, Sambalpur,
768019, Odisha, India
aE-mail<EMAIL_ADDRESS>bE-mail<EMAIL_ADDRESS>Priyabrat
Gochhayatb Department of Mathematics, Sambalpur University, Sambalpur, 768019,
Odisha, India
aE-mail<EMAIL_ADDRESS>bE-mail<EMAIL_ADDRESS>
###### Abstract

We investigate the interlacing of zeros of polynomials of different degrees
within the sequences of $q$-Laguerre polynomials
$\left\\{\tilde{L}_{n}^{(\delta)}(z;q)\right\\}_{n=0}^{\infty}$ characterized
by $\delta\in(-2,-1).$ The interlacing of zeros of quasi-orthogonal
polynomials $\tilde{L}_{n}^{(\delta)}(z;q)$ with those of the orthogonal
polynomials $\tilde{L}_{m}^{(\delta+t)}(z;q),m,n\in\mathbb{N},t\in\\{1,2\\}$
is also considered. New bounds for the least zero of the (order $1$) quasi-
orthogonal $q$-Laguerre polynomials are derived.

MSC: 33D15; 33D45

Keywords: Zeros, Interlacing, Stieltjes interlacing, Quasi-orthogonal
polynomials, $q$-Laguerre polynomials.

##  1 Introduction

Let $\\{t_{n}(z)\\}_{n=0}^{\infty},$ with deg($t_{n})=n,n\in\mathbb{N},$ be a
sequence of orthogonal polynomials and $w(z)$ be the corresponding weight
function on a support $(c,d).$ Then a well known fact about the zeros of
$t_{n}(z)$ is that, they are real, simple and lie in $(c,d).$ Moreover, the
zeros of $t_{n}(z)$ and $t_{n-1}(z)$ interlace, if there is exactly one zero
of $t_{n-1}(z)$ in between any two consecutive zeros of $t_{n}(z).$ Similarly,
the zeros of the two equal degree polynomials $t_{n}(z)$ and $u_{n}(z)$ are
interlacing if there is exactly one zero of $u_{n}(z)$ (equivalently,
$t_{n}(z)$) in between any two consecutive zeros of $t_{n}(z)$ (equivalently,
$u_{n}(z)$). Another interlacing property given by Stieltjes (cf. [19, Theorem
3.3.3]), usually called Stieltjes interlacing, states that, for
$m,n\in\mathbb{N}$ with $m<n-1,$ if there exists $m$ open intervals with
endpoints at the consecutive zeros of $t_{n}(z)$ then there is exactly one
zero of $u_{m}(z)$ in each of the $m$ open intervals. It follows that, if
there is at least a zero of $u_{m}(z)$ lies outside the interval
$(t_{1,n},t_{n,n});$ where $t_{1,n}$ and $t_{n,n},$ respectively, are the
smallest and the largest zero of $t_{n}(z),$ the Stieltjes interlacing will
not retain between the zeros of $u_{m}(z)$ and $t_{n}(z)$ with $n-1>m.$

The departure of zeros of polynomials from the interval of orthogonality leads
to the following definition of quasi-orthogonality:

######  Definition 1.

Suppose $\\{t_{n}(z)\\}_{n=0}^{\infty}$ is a sequence of polynomials and
deg($t_{n})=n,n\in\mathbb{N},$ and $w(z)$ is the corresponding weight function
on a support $[c,d].$ Then $\\{t_{n}(z)\\}_{n=0}^{\infty}$ is called as quasi-
orthogonal sequence of order $r<n$ if

$\int_{c}^{d}z^{i}t_{n}(z)w(z)dz\begin{cases}=0;&i=0,1,\cdots,n-r-1,\\\ \neq
0;&i=n-r.\end{cases}$ (1)

If $t_{n}(z)$ is a quasi-orthogonal polynomial of order $r$ with respect to
$w(z)>0$ on $(c,d),$ then there are minimum $(n-r)$ real and distinct zeros of
$t_{n}(z)$ in $(c,d).$ In (1), letting $r=0$ we find that
$\\{t_{n}(z)\\}_{n=0}^{\infty}$ is an orthogonal polynomial sequence. Riesz
[17] was the first mathematician who studied quasi-orthogonality of order $1$
while order $2$ was derived by Fejér [11]. Following which, Shohat [18],
Chihara [3], Dickinson [4], Draux [5], etc. considered various general cases.
For more information on quasi-orthogonality, we refer the works by Brezinski
et al. [1, 2], Joulak [13] and the references therein. Recent contribution on
the interlacing of zeros of different quasi-orthogonal polynomials found in
[6, 7, 8, 9, 10, 12]. More recently, in the context of $q$-calculus, the
quasi-orthogonality of classical orthogonal polynomials on $q$-linear and
$q$-quadratic lattices is discussed by Tcheutia et al. [20] while Kar et al.
in [14] considered the polynomials that are not present in the $q$-Askey
scheme of hypergeometric polynomials to study their quasi-orthogonality.
Pertinently, very less known on quasi-orthogonality of $q$-polynomials.

In this work, we study the interlacing properties of zeros of $q$-Laguerre
polynomials. Therefore, it is worthy to mention the orthogonality and quasi-
orthogonality of these polynomials.

The (monic) $q$-Laguerre polynomials of degree $n$ is defined as

$\tilde{L}_{n}^{(\delta)}(z;q)=\displaystyle{(-1)^{n}(q^{\delta+1};q)_{n}\over
q^{n(\delta+n)}}\ _{1}\phi_{1}\left(\begin{matrix}q^{-n}\\\
q^{\delta+1}\end{matrix};q,-q^{\delta+n+1}z\right),\qquad\delta>-1.$

These are orthogonal with respect to the weight
$\displaystyle{z^{\delta}\over(-z;q)_{\infty}}$ on $(0,\infty).$ For any value
of $\delta,$ the sequence of $q$-Laguerre polynomials is defined by the three
term recurrence relation

$\tilde{L}_{n}^{(\delta)}(z;q)=(z-a_{n})\tilde{L}_{n-1}^{(\delta)}(z;q)-{(1-q^{n-1})(1-q^{\delta+n-1})\over
q^{2\delta+4n-5}}\tilde{L}_{n-2}^{(\delta)}(z;q);\ n=2,3,\cdots,$ (2)

where

$a_{n}={1-q^{n}+q(1-q^{\delta+n-1})\over q^{\delta+2n-1}},$ (3)

and $\tilde{L}_{0}^{(\delta)}(z;q)\equiv 1,$
$\tilde{L}_{1}^{(\delta)}(z;q)=z-\displaystyle{1-q^{\delta+1}\over
q^{\delta+1}}.$

Tcheutia et al. in [20, Theorem 2.17], proved that for fixed
$\delta\in\mathbb{R}$ with $\delta\in(-k-1,-k),$
$\left\\{\tilde{L}_{n}^{(\delta)}(z;q)\right\\}_{n=0}^{\infty}$ is quasi-
orthogonal of order $k$ with respect to
$\displaystyle{z^{\delta+k}\over(-z;q)_{\infty}}.$ There are minimum $n-k$
real and distinct zeros in $(0,\infty).$ Throughout this paper we denote
$\\{z_{i,n}\\}_{i=1}^{n},\\{y_{i,n}\\}_{i=1}^{n}$ and
$\\{x_{i,n}\\}_{i=1}^{n},$ respectively, the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q),$ $\tilde{L}_{n}^{(\delta+1)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ for fixed $\delta\in(-2,-1)$ and $0<q<1.$

The following result obtained by substituting $\delta$ by $\delta+1$ in [20,
Theorem 2.18] is used in our work.

######  Lemma 1.

The zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+1)}(z;q)$ are interlacing as

$z_{1,n}<0<y_{1,n}<z_{2,n}<y_{2,n}<\cdots<z_{n,n}<y_{n,n}.$ (4)

######  Remark 1.

Replacing $n$ by $n+1$ in [20, Theorem 2.18], we also obtain

$z_{1,n+1}<0<y_{1,n}<z_{2,n+1}<y_{2,n}<\cdots<z_{n,n+1}<y_{n,n}<z_{n+1,n+1}.$
(5)

In the next section, we show that the zeros of the $q$-Laguerre polynomials of
consecutive degree are not interlacing. The interlacing results between the
zeros of the quasi-orthogonal and the orthogonal $q$-Laguerre polynomials of
the same, consecutive and non-consecutive degree with different parameter
values are also discussed in section 2. In section 3, we examine the common
zeros of polynomials that are not co-prime. The last section is devoted to the
derivation of the inner and outer bound for the only negative zero of the
quasi-orthogonal $q$-Laguerre polynomial.

##  2 Zeros of $\boldsymbol{\tilde{L}_{n}^{(\delta)}(z;q)}$ and
$\boldsymbol{\tilde{L}_{m}^{(\delta+t)}(z;q)}$ for $\boldsymbol{m\leq n+1,}$
$\boldsymbol{m,n\in\mathbb{N}}$ and $\boldsymbol{t\in\\{0,1,2\\}}$

Our first theorem proves that the interlacing does not take place between the
zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n+1}^{(\delta)}(z;q).$

######  Theorem 1.

The zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n+1}^{(\delta)}(z;q)$ are placed in the following order:

$z_{1,n}<z_{1,n+1}<0<z_{2,n+1}<z_{2,n}<\cdots<z_{n,n}<z_{n+1,n+1}.$

###### Proof.

Evaluating the mixed $q$-contiguous relation [16, (4.14)]

$\tilde{L}_{n+1}^{(\delta)}(z;q)=-{1-q^{\delta+n+1}\over
q^{\delta+2n+1}}\tilde{L}_{n}^{(\delta)}(z;q)+z\tilde{L}_{n}^{(\delta+1)}(z;q)$

at the consecutive zeros $z_{i,n}$ and $z_{i+1,n},i\in\\{1,\cdots,n-1\\},$ of
$\tilde{L}_{n}^{(\delta)}(z;q),$ we obtain

$\tilde{L}_{n+1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n+1}^{(\delta)}(z_{i+1,n};q)=z_{i,n}z_{i+1,n}\tilde{L}_{n}^{(\delta+1)}(z_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(z_{i+1,n};q).$
(6)

Now from (4), $z_{1,n}z_{2,n}<0$ and $z_{i,n}z_{i+1,n}>0$ for
$i\in\\{2,\cdots,n-1\\}$ while for $i\in\\{1,\cdots,n-1\\},$
$\tilde{L}_{n}^{(\delta+1)}(z_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(z_{i+1,n};q)<0.$
We conclude from (6) that
$\tilde{L}_{n+1}^{(\delta)}(z_{1,n};q)\tilde{L}_{n+1}^{(\delta)}(z_{2,n};q)>0$
and
$\tilde{L}_{n+1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n+1}^{(\delta)}(z_{i+1,n};q)<0$
for each $i\in\\{2,\cdots,n-1\\}.$ This implies that
$\tilde{L}_{n+1}^{(\delta)}(z;q)$ has an even number of zeros in
$(z_{1,n},z_{2,n})$ and an odd number of zeros in $(z_{i,n},z_{i+1,n})$ for
each $i\in\\{2,\cdots,n-1\\}.$ Hence, $\tilde{L}_{n+1}^{(\delta)}(z;q)$ has
minimum $n-2$ zeros in $(z_{2,n},z_{n,n}),$ plus its least zero $z_{1,n+1}<0$
and from (4) and (5), its greatest zero $z_{n+1,n+1}>y_{n,n}>z_{n,n}.$ Thus,
we have located $n$ zeros of $\tilde{L}_{n+1}^{(\delta)}(z;q)$ and still we
have to find one more zero. Since there are even number of zeros of
$\tilde{L}_{n+1}^{(\delta)}(z;q)$ in $(z_{1,n},z_{2,n})$ where
$z_{1,n}<0<z_{2,n}$ for any $n\in\mathbb{N},$ we must have either no zero or
two zeros in this single interval. Because we have precisely one negative zero
of $\tilde{L}_{n+1}^{(\delta)}(z;q)$ for $n\in\mathbb{N},$ the only option is
$z_{1,n}<z_{1,n+1}<0<z_{2,n+1}<z_{2,n}.$ This proves the theorem. ∎

It is clear from Theorem 1 that, the zeros of the quasi-orthogonal polynomials
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n+1}^{(\delta)}(z;q)$ are
interlacing in the interval $(0,\infty).$ Furthermore, the zeros of
$z\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n+1}^{(\delta)}(z;q)$
interlace.

The next result shows that, the Stieltjes interlacing fails to occur between
the zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta)}(z;q).$

######  Theorem 2.

For $n\geq 3,n\in\mathbb{N},$ if $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ are co-prime, then the zeros of
$z(z-a_{n})\tilde{L}_{n-2}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta)}(z;q)$ interlace, where $a_{n}$ is given in (3).
However, the Stieltjes interlacing will not take place between the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-2}^{(\delta)}(z;q).$

###### Proof.

Assume that $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ are co-prime. Then from (2),
$\tilde{L}_{n}^{(\delta)}(a_{n};q)\neq 0.$ Evaluate (2) at the consecutive
zeros $z_{i,n}$ and $z_{i+1,n},$ $i\in\\{2,\cdots,n-1\\},$ of
$\tilde{L}_{n}^{(\delta)}(z;q)$ to get

${\tilde{L}_{n-1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{i+1,n};q)\over\tilde{L}_{n-2}^{(\delta)}(z_{i,n};q)\tilde{L}_{n-2}^{(\delta)}(z_{i+1,n};q)}={(1-q^{n-1})^{2}(1-q^{\delta+n-1})^{2}\over
q^{4\delta+8n-10}(z_{i,n}-a_{n})(z_{i+1,n}-a_{n})}.$ (7)

For $n$ substituted by $n-1$ in Theorem 1, we have the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-1}^{(\delta)}(z;q)$
interlace in $(0,\infty).$ So that for every $i\in\\{2,\cdots,n-1\\},$ the
numerator of left-hand side of (7) is negative. Also the right-hand side of
(7) is positive unless $a_{n}\in(z_{i,n},z_{i+1,n})$ for each
$i\in\\{2,\cdots,n-1\\}.$ Hence we deduce from (7) that, for every
$i\in\\{2,\cdots,n-1\\},$ $\tilde{L}_{n-2}^{(\delta)}(z_{i,n};q)$ and
$\tilde{L}_{n-2}^{(\delta)}(z_{i+1,n};q)$ have different sign except may be
for one pair $z_{j,n},z_{j+1,n},$ with $z_{j,n}<a_{n}<z_{j+1,n},2\leq j\leq
n-1.$ Since $n-2$ intervals are there with endpoints at the consecutive
positive zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ has exactly $n-3$ simple positive zeros, the
zeros of $\tilde{L}_{n-2}^{(\delta)}(z;q),$ together with the point $a_{n},$
interlace with the zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ in $(0,\infty).$
The given interlacing result holds true because from Theorem 1, we have
$z_{1,n-2}<z_{1,n}<0<z_{2,n}.$ Further, from Theorem 1,
$z_{1,n-2}<z_{1,n-1}<z_{1,n}<0<z_{2,n}<z_{2,n-1}<z_{2,n-2},$ the smallest zero
$z_{1,n-2},$ of $\tilde{L}_{n-2}^{(\delta)}(z;q),$ lies outside
$(z_{1,n},z_{n,n}),$ which shows that the Stieltjes interlacing does not hold
between the zeros of the quasi-orthogonal $q$-Laguerre polynomials
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-2}^{(\delta)}(z;q).$ ∎

In Lemma 1, we have for fixed $\delta\in(-2,-1),$ the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n}^{(\delta+1)}(z;q)$ are
interlacing. Upon replacing $n$ by $n-1$ in Remark 1, we see that the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-1}^{(\delta+1)}(z;q)$
interlace. The interlacing properties between the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-2}^{(\delta+1)}(z;q)$ are
discussed in the next theorem. Note that, for fixed $\delta\in(-2,-1)$ and
$k\geq 1,$ the sequence
$\left\\{\tilde{L}_{n}^{(\delta+k)}(z;q)\right\\}_{n=0}^{\infty}$ is
orthogonal in $(0,\infty)$ because $\delta+k>-1.$

######  Theorem 3.

For $n\geq 3,n\in\mathbb{N},$ the zeros of
$\tilde{L}_{n-2}^{(\delta+1)}(z;q),$ together with the point
$\displaystyle{1-q^{\delta+n}\over q^{\delta+2n-1}}=b_{n}$ (say), interlace
with those of $\tilde{L}_{n}^{(\delta)}(z;q)$ in $(0,\infty)$ provided these
two polynomials do not have any common zeros.

###### Proof.

Using [16, (4.14) and (4.16)] and [14, (10)], we get

$\tilde{L}_{n}^{(\delta)}(z;q)=\left(z-b_{n}\right)\tilde{L}_{n-1}^{(\delta)}(z;q)-\left(1-q^{n-1}\over
q^{\delta+2n-2}\right)z\tilde{L}_{n-2}^{(\delta+1)}(z;q).$ (8)

From (8), $z_{i,n}\neq b_{n}$ for any $i\in\\{1,\cdots,n\\},$ if not, this
would contradict our supposition. Evaluating (8) at the consecutive positive
zeros $z_{i,n}$ and $z_{i+1,n},$ $i\in\\{2,\cdots,n-1\\},$ of
$\tilde{L}_{n}^{(\delta)}(z;q)$ yields

${\tilde{L}_{n-1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{i+1,n};q)\over\tilde{L}_{n-2}^{(\delta+1)}(z_{i,n};q)\tilde{L}_{n-2}^{(\delta+1)}(z_{i+1,n};q)}={(1-q^{n-1})^{2}z_{i,n}z_{i+1,n}\over
q^{2\delta+4n-4}(z_{i,n}-b_{n})(z_{i+1,n}-b_{n})}.$ (9)

Clearly, (4) shows that, both $z_{i,n}$ and $z_{i+1,n}$ are positive for each
$i\in\\{2,\cdots,n-1\\}.$ Thus, the right-hand side of (9) is positive if and
only if for each $i\in\\{2,\cdots,n-1\\},$ $b_{n}\notin(z_{i,n},z_{i+1,n}).$
Also from Theorem 1 we have for every $n\geq 2,n\in\mathbb{N}$ and
$i\in\\{2,\cdots,n-1\\},$
$\tilde{L}_{n-1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{i+1,n};q)<0.$
If $b_{n}\notin(z_{i,n},z_{i+1,n}),$ for every $i\in\\{2,\cdots,n-1\\},$ then
we deduce from (9) that the zeros of $\tilde{L}_{n-2}^{(\delta+1)}(z;q)$ and
$\tilde{L}_{n}^{(\delta)}(z;q)$ interlace in $(0,\infty).$ Additionally, since
the point $b_{n}$ is not in $(z_{2,n},z_{n,n}),$ the zeros of
$(z-b_{n})\tilde{L}_{n-2}^{(\delta+1)}(z;q)$ and the positive zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ interlace. On the other hand, if
$b_{n}\in(z_{i,n},z_{i+1,n}),$ then we have $b_{n}\in(z_{j,n},z_{j+1,n})$ for
one $j$ with $2\leq j\leq n-1.$ This implies, in $(z_{j,n},z_{j+1,n}),$
$\tilde{L}_{n-2}^{(\delta+1)}(z;q)$ has no sign change but it changes sign in
each of the remaining $n-3$ intervals with endpoints at the consecutive
positive zeros of $\tilde{L}_{n}^{(\delta)}(z;q).$ Finally, for $i=1,$ (9)
becomes

${\tilde{L}_{n-1}^{(\delta)}(z_{1,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{2,n};q)\over\tilde{L}_{n-2}^{(\delta+1)}(z_{1,n};q)\tilde{L}_{n-2}^{(\delta+1)}(z_{2,n};q)}={(1-q^{n-1})^{2}z_{1,n}z_{2,n}\over
q^{2\delta+4n-4}(z_{1,n}-b_{n})(z_{2,n}-b_{n})}.$ (10)

Since $b_{n}\in(z_{j,n},z_{j+1,n})$ for one $j,$ $j\in\\{2,\cdots,n-1\\},$
$b_{n}\notin(z_{1,n},z_{2,n}).$ We know from Theorem 1 that
$z_{1,n-1}<z_{1,n}<0<z_{2,n}<z_{2,n-1}.$ This implies,
$\tilde{L}_{n-1}^{(\delta)}(z_{1,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{2,n};q)>0.$
Hence from (10),
$\tilde{L}_{n-2}^{(\delta+1)}(z_{1,n};q)\tilde{L}_{n-2}^{(\delta+1)}(z_{2,n};q)<0.$
Consequently, $\tilde{L}_{n-2}^{(\delta+1)}(z;q)$ has one positive zero less
than $z_{2,n}.$ Thus, in each of the cases, we have the zeros of
$(z-b_{n})\tilde{L}_{n-2}^{(\delta+1)}(z;q)$ interlace with the positive zeros
of $\tilde{L}_{n}^{(\delta)}(z;q),$ if these two polynomials do not have any
common zeros. ∎

For the rest of the results in this section we study the interlacing results
for the zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-k}^{(\delta+2)}(z;q),$ where $\delta\in(-2,-1)$ and
$k\in\\{0,1,2\\}.$ We now show a necessary and sufficient condition for the
zeros of the same degree polynomials $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ to be interlacing.

######  Theorem 4.

For $n\geq 2,n\in\mathbb{N},$ the zeros of the same degree $q$-Laguerre
polynomials $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ interlace if and only if the smallest
positive zero of $\tilde{L}_{n}^{(\delta)}(z;q),$ that is,
$z_{2,n}>\displaystyle{-{1-q^{\delta+1}\over q^{\delta+n+1}}}=c_{n}$ (say).

###### Proof.

The sequence $\left\\{\tilde{L}_{n}^{(\delta)}(z;q)\right\\}_{n=0}^{\infty}$
satisfies the mixed $q$-contiguous relation (obtained using [16, (4.12) and
(4.15)])

$z\tilde{L}_{n}^{(\delta+2)}(z;q)=\left(z-c_{n}\right)\tilde{L}_{n}^{(\delta+1)}(z;q)-{1-q^{\delta+n+1}\over
q^{\delta+2n+1}}\tilde{L}_{n}^{(\delta)}(z;q).$ (11)

We have from (4) that $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+1)}(z;q)$ are co-prime for each fixed
$\delta\in(-2,-1).$ Hence, it follows from (11) that
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n}^{(\delta+2)}(z;q)$ can have
at most one common zero at $z=c_{n}.$ If $c_{n}<z_{2,n},$ then since the zeros
of $\tilde{L}_{n}^{(\delta+2)}(z;q)$ lie in $(0,\infty)$ and $z_{2,n}>0,$ the
polynomials $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ are co-prime. Evaluating (11) at the
consecutive zeros $z_{i,n}$ and $z_{i+1,n},i\in\\{1,\cdots,n-1\\},$ of
$\tilde{L}_{n}^{(\delta)}(z;q),$ gives

${\tilde{L}_{n}^{(\delta+1)}(z_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(z_{i+1,n};q)\over\tilde{L}_{n}^{(\delta+2)}(z_{i,n};q)\tilde{L}_{n}^{(\delta+2)}(z_{i+1,n};q)}=\displaystyle{z_{i,n}z_{i+1,n}\over(z_{i,n}-c_{n})(z_{i+1,n}-c_{n})}.$
(12)

From (4), $z_{1,n}z_{2,n}<0$ while for each $i\in\\{2,\cdots,n-1\\},$
$z_{i,n}z_{i+1,n}>0.$ Further, for every $i\in\\{1,\cdots,n-1\\},$
$\tilde{L}_{n}^{(\delta+1)}(z_{i,n};q)$
$\tilde{L}_{n}^{(\delta+1)}(z_{i+1,n};q)<0.$ Since $z_{1,n}<0<c_{n}<z_{2,n},$
we conclude from (12) that $\tilde{L}_{n}^{(\delta+2)}(z;q)$ changes sign in
$(z_{i,n},z_{i+1,n})$ for each $i\in\\{1,\cdots,n-1\\}.$ Consequently,
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ has an odd number of zeros in
$(z_{i,n},z_{i+1,n})$ for any $i\in\\{1,\cdots,n-1\\}.$ We note from [16,
Theorem 3] that,

$0<y_{1,n}<x_{1,n}<y_{2,n}<x_{2,n}<\cdots<y_{n,n}<x_{n,n}.$ (13)

Now from (4) and (13), we have $z_{n,n}<x_{n,n},$ where $x_{n,n}$ is the
largest zero of $\tilde{L}_{n}^{(\delta+2)}(z;q).$ Therefore, the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n}^{(\delta+2)}(z;q)$ are
interlacing.

If $c_{n}>z_{2,n},$ then since we have $z_{1,n}<0<z_{2,n}<c_{n}$ so from (12),
we deduce that $\tilde{L}_{n}^{(\delta+2)}(z_{1,n};q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z_{2,n};q)$ have same sign. Thus,
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ has an even number of zeros in
$(z_{1,n},z_{2,n})$ which confirms that the zeros of the polynomials
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n}^{(\delta+2)}(z;q)$ are not
interlacing. This completes the proof of the theorem. ∎

In the next theorem we derive an interlacing result between the zeros of the
above considered polynomials involving the point $c_{n}.$

######  Theorem 5.

For $n\geq 2,n\in\mathbb{N},$ let $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ be co-prime and $z_{2,n}<c_{n},$ where
$c_{n}$ is defined in Theorem 4. Then the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n}^{(\delta+2)}(z;q)$
interlace, if $c_{n}\notin(x_{i,n},x_{i+1,n})$ for each
$i\in\\{1,\cdots,n-1\\}.$ Otherwise, the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q),$ together with the point $c_{n},$ interlace
with those of $z\tilde{L}_{n}^{(\delta+2)}(z;q).$

###### Proof.

Evaluating (11) at the consecutive zeros $x_{i,n}$ and
$x_{i+1,n},i\in\\{1,\cdots,n-1\\},$ of $\tilde{L}_{n}^{(\delta+2)}(z;q),$ we
obtain

$(x_{i,n}-c_{n})(x_{i+1,n}-c_{n})\tilde{L}_{n}^{(\delta+1)}(x_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(x_{i+1,n};q)\\\
=\left({1-q^{\delta+n+1}\over
q^{\delta+2n+1}}\right)^{2}\tilde{L}_{n}^{(\delta)}(x_{i,n};q)\tilde{L}_{n}^{(\delta)}(x_{i+1,n};q).$
(14)

It is clear from (13) that,
$\tilde{L}_{n}^{(\delta+1)}(x_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(x_{i+1,n};q)<0$
for every $i\in\\{1,\cdots,n-1\\}.$ If for each $i\in\\{1,\cdots,n-1\\},$
$c_{n}\notin(x_{i,n},x_{i+1,n}),$ then the left-hand side of (14) is negative,
whence
$\tilde{L}_{n}^{(\delta)}(x_{i,n};q)\tilde{L}_{n}^{(\delta)}(x_{i+1,n};q)<0.$
Since $\tilde{L}_{n}^{(\delta+2)}(z;q)$ has positive zeros and the least zero
of $\tilde{L}_{n}^{(\delta)}(z;q)$ is negative, the only possible arrangement
of zeros is

$z_{1,n}<0<x_{1,n}<z_{2,n}<x_{2,n}<\cdots<z_{n,n}<x_{n,n}<c_{n},$

which shows the first part of the theorem.

On the other hand, if $c_{n}\in(x_{j,n},x_{j+1,n})$ for one $j,$
$j\in\\{1,\cdots,n-1\\},$ then we deduce from (14) that,
$\tilde{L}_{n}^{(\delta)}(x_{j,n};q)\tilde{L}_{n}^{(\delta)}(x_{j+1,n};q)>0$
for one $j,j\in\\{1,\cdots,n-1\\}.$ This shows that
$\tilde{L}_{n}^{(\delta)}(z;q)$ has $n-2$ sign changes between the consecutive
zeros of $\tilde{L}_{n}^{(\delta+2)}(z;q)$ in $(0,\infty)$ except for the one
interval $(x_{j,n},x_{j+1,n})$ containing the point $c_{n}.$ Since
$\tilde{L}_{n}^{(\delta)}(z;q)$ has $n-1$ positive zeros and $n-2$ sign
changes in $(0,\infty),$ no zero of $\tilde{L}_{n}^{(\delta)}(z;q)$ lies in
$(x_{j,n},x_{j+1,n})$ that contains $c_{n}$ and one zero of
$\tilde{L}_{n}^{(\delta)}(z;q)$ is either $<x_{1,n}$ or $>x_{n,n}.$ From (4)
and (13), we have $z_{n,n}<y_{n,n}<x_{n,n}.$ Hence, the only option is that
the least positive zero $z_{2,n}$ of $\tilde{L}_{n}^{(\delta)}(z;q)$ is
$<x_{1,n}.$ Since exactly one zero of $\tilde{L}_{n}^{(\delta)}(z;q)$ is
negative, therefore, the zeros of $(z-c_{n})\tilde{L}_{n}^{(\delta)}(z;q)$ and
$z\tilde{L}_{n}^{(\delta+2)}(z;q)$ interlace. This proves the theorem. ∎

We analyze the interlacing properties between the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-1}^{(\delta+2)}(z;q)$ in the
next result.

######  Theorem 6.

For $n\geq 2,n\in\mathbb{N},$ the zeros of
$z\tilde{L}_{n-1}^{(\delta+2)}(z;q)$ and $\tilde{L}_{n}^{(\delta)}(z;q)$
interlace.

###### Proof.

Using [16, (4.12) and (4.14)], we obtain

$\left({1-q^{\delta+n+1}\over
1-q^{n}}\right)\tilde{L}_{n}^{(\delta)}(z;q)={q^{n}(1-q^{\delta+1})\over(1-q^{n})}\tilde{L}_{n}^{(\delta+1)}(z;q)+z\tilde{L}_{n-1}^{(\delta+2)}(z;q).$
(15)

Evaluating (15) at the consecutive zeros $z_{i,n}$ and
$z_{i+1,n},i\in\\{1,\cdots,n-1\\},$ of $\tilde{L}_{n}^{(\delta)}(z;q)$ yields

${q^{2n}(1-q^{\delta+1})^{2}\over(1-q^{n})^{2}}\tilde{L}_{n}^{(\delta+1)}(z_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(z_{i+1,n};q)=z_{i,n}z_{i+1,n}\tilde{L}_{n-1}^{(\delta+2)}(z_{i,n};q)\tilde{L}_{n-1}^{(\delta+2)}(z_{i+1,n};q).$
(16)

For any $i\in\\{1,\cdots,n-1\\},$ (4) implies
$\tilde{L}_{n}^{(\delta+1)}(z_{i,n};q)\tilde{L}_{n}^{(\delta+1)}(z_{i+1,n};q)<0.$
Also for $i=1,$ $z_{i,n}z_{i+1,n}<0$ while $z_{i,n}z_{i+1,n}>0$ for
$i=2,\cdots,n-1.$ Therefore, from (16), we deduce that
$\tilde{L}_{n-1}^{(\delta+2)}(z;q)$ does not change sign in
$(z_{1,n},z_{2,n})$ and changes sign in $(z_{i,n},z_{i+1,n})$ for every
$i\in\\{2,\cdots,n-1\\}.$ Thus, we have minimum $n-2$ zeros of
$\tilde{L}_{n-1}^{(\delta+2)}(z;q)$ in $(z_{2,n},z_{n,n})$ and in
$(z_{1,n},z_{2,n}),$ there is either no zero or two zeros of
$\tilde{L}_{n-1}^{(\delta+2)}(z;q).$ Since $\tilde{L}_{n-1}^{(\delta+2)}(z;q)$
has exactly $n-1$ zeros, it has no zero in $(z_{1,n},z_{2,n}).$ Hence, the
remaining one zero must be larger than $z_{n,n}.$ Since $z_{1,n}<0<z_{2,n},$
the given interlacing property holds true. This proves the theorem. ∎

######  Remark 2.

From (4), $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+1)}(z;q)$ are co-prime. This implies,
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-1}^{(\delta+2)}(z;q)$ cannot
be co-prime. Otherwise, any common zero of these polynomials is also a zero of
$\tilde{L}_{n}^{(\delta+1)}(z;q).$

Next theorem shows the interlacing between the zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-2}^{(\delta+2)}(z;q).$

######  Theorem 7.

For $n\geq 3,n\in\mathbb{N},$ the zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ interlace in $(0,\infty).$ Moreover, the
zeros of $z(z+c_{n-1})\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ and
$\tilde{L}_{n}^{(\delta)}(z;q)$ interlace, where $c_{n}$ is given in Theorem
4.

###### Proof.

Using (8) and [16, (4.14) and (4.15)], we get

$\left({1-q^{n-1}\over
1-q^{\delta+n}}\right)z^{2}\tilde{L}_{n-2}^{(\delta+2)}(z;q)={q^{\delta+2n}\over
1-q^{\delta+n}}c_{n}\tilde{L}_{n}^{(\delta)}(z;q)+(z+c_{n-1})\tilde{L}_{n-1}^{(\delta)}(z;q).$
(17)

Since $\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-1}^{(\delta)}(z;q)$
are co-prime in $(0,\infty),$ from (17), the only possible common zero of
$\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ in
$(0,\infty)$ is $-c_{n-1}.$ But for $\delta\in(-2,-1)$ and $0<q<1,$
$-c_{n-1}<0$ and all the zeros of $\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ are
positive. This implies, the polynomials $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ are co-prime and hence, $z_{i,n}\neq-
c_{n-1}$ for any $i\in\\{1,\cdots,n\\}.$ Evaluating (17) at the consecutive
zeros $z_{i,n}$ and $z_{i+1,n},i\in\\{2,\cdots,n-1\\},$ of
$\tilde{L}_{n}^{(\delta)}(z;q),$ we have

${\tilde{L}_{n-2}^{(\delta+2)}(z_{i,n};q)\tilde{L}_{n-2}^{(\delta+2)}(z_{i+1,n};q)\over\tilde{L}_{n-1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{i+1,n};q)}={(1-q^{\delta+n})^{2}(z_{i,n}+c_{n-1})(z_{i+1,n}+c_{n-1})\over(1-q^{n-1})^{2}z_{i,n}^{2}z_{i+1,n}^{2}}.$
(18)

We know from Theorem 1 that,
$\tilde{L}_{n-1}^{(\delta)}(z_{i,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{i+1,n};q)<0$
for $i\in\\{2,\cdots,n-1\\}.$ For $\delta\in(-2,-1),$ $-c_{n-1}<0,$ so
$-c_{n-1}<z_{2,n}$ and hence for any $i\in\\{2,\cdots,n-1\\},$ the right-hand
side of (18) is positive. Therefore, we deduce from (18) that, for each
$i\in\\{2,\cdots,n-1\\},$ $\tilde{L}_{n-2}^{(\delta+2)}(z_{i,n};q)$ and
$\tilde{L}_{n-2}^{(\delta+2)}(z_{i+1,n};q)$ have different sign. This means,
the zeros of $\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ and
$\tilde{L}_{n}^{(\delta)}(z;q)$ interlace in $(0,\infty).$

Now evaluating (17) at the zeros $z_{1,n}$ and $z_{2,n},$ of
$\tilde{L}_{n}^{(\delta)}(z;q),$ we obtain

${\tilde{L}_{n-2}^{(\delta+2)}(z_{1,n};q)\tilde{L}_{n-2}^{(\delta+2)}(z_{2,n};q)\over\tilde{L}_{n-1}^{(\delta)}(z_{1,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{2,n};q)}={(1-q^{\delta+n})^{2}(z_{1,n}+c_{n-1})(z_{2,n}+c_{n-1})\over(1-q^{n-1})^{2}z_{1,n}^{2}z_{2,n}^{2}}.$
(19)

Since the zeros of $\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ lie in
$(z_{2,n},z_{n,n}),$ no zero of $\tilde{L}_{n-2}^{(\delta+2)}(z;q)$ are in
$(z_{1,n},z_{2,n})$ which means it does not change sign in
$(z_{1,n},z_{2,n}).$ From Theorem 1,
$\tilde{L}_{n-1}^{(\delta)}(z_{1,n};q)\tilde{L}_{n-1}^{(\delta)}(z_{2,n};q)$
$>0.$ Therefore, from (19) we conclude that $-c_{n-1}\notin(z_{1,n},z_{2,n})$
and since $-c_{n-1}<0,$ it follows that $-c_{n-1}<z_{1,n}.$ Thus, the only
possibility of configuration of zeros is as

$-c_{n-1}<z_{1,n}<0<z_{2,n}<x_{1,n-2}<\cdots<x_{n-2,n-2}<z_{n,n}.$

This follows the result. ∎

##  3 Common zeros of $\boldsymbol{\tilde{L}_{n}^{(\delta)}(z;q)}$

In this section, we deal with the $q$-Laguerre polynomials that are not co-
prime. In particular, we discuss the interlacing properties between the zeros
of polynomials considered in Theorem 2 and Theorem 5.

######  Theorem 8.

For $n\geq 3,n\in\mathbb{N},$ if $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ are not co-prime, then $a_{n}$ is a common
zero of these polynomials and the zeros of $\tilde{L}_{n-2}^{(\delta)}(z;q)$
interlace with the non-common zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ in
$(0,\infty),$ where $a_{n}$ is given in (3).

###### Proof.

If $\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n-2}^{(\delta)}(z;q)$ have
common zeros in $(0,\infty),$ then from (2), $a_{n}$ is the only common zero
of these polynomials since $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-1}^{(\delta)}(z;q)$ are co-prime in $(0,\infty)$ by Theorem (1).
Suppose $z_{1,n-2}^{\prime}<z_{2,n-2}^{\prime}<\cdots<z_{n-2,n-2}^{\prime}$
are the $n-2$ non-common zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ in
$(0,\infty).$ Then $a_{n}$ belongs to one of the intervals
$\left(z_{i,n-2}^{\prime},z_{i+1,n-2}^{\prime}\right),$
$i\in\\{1,\cdots,n-3\\}.$ Evaluating (2) at the consecutive non-common zeros
$z_{i,n-2}^{\prime}$ and $z_{i+1,n-2}^{\prime},$ $i\in\\{1,\cdots,n-3\\},$ of
$\tilde{L}_{n}^{(\delta)}(z;q),$ such that
$a_{n}\notin\left(z_{i,n-2}^{\prime},z_{i+1,n-2}^{\prime}\right),$ we obtain

${\left(1-q^{n-1}\right)^{2}\left(1-q^{\delta+n-1}\right)^{2}\over
q^{4\delta+8n-10}}\tilde{L}_{n-2}^{(\delta)}(z_{i,n-2}^{\prime};q)\tilde{L}_{n-2}^{(\delta)}(z_{i+1,n-2}^{\prime};q)\\\
=\left(z_{i,n-2}^{\prime}-a_{n}\right)\left(z_{i+1,n-2}^{\prime}-a_{n}\right)\tilde{L}_{n-1}^{(\delta)}(z_{i,n-2}^{\prime};q)\tilde{L}_{n-1}^{(\delta)}(z_{i+1,n-2}^{\prime};q).$
(20)

Since the positive zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n-1}^{(\delta)}(z;q)$ are interlacing, we deduce from (20) that
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ has an odd number of zeros in each of the
intervals $\left(z_{i,n-2}^{\prime},z_{i+1,n-2}^{\prime}\right),$
$i\in\\{1,\cdots,n-3\\},$ that does not contain $a_{n}.$ Since
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ has $n-4$ non-common zeros in $(0,\infty),$
there are at most $n-4$ such intervals. Therefore, $a_{n}=z_{j,n}$ where
$j\in\\{3,\cdots,n-1\\}.$ Hence, the $n-3$ zeros of
$\tilde{L}_{n-2}^{(\delta)}(z;q)$ (including $a_{n}$) interlace with the $n-2$
non-common zeros of $\tilde{L}_{n}^{(\delta)}(z;q)$ in $(0,\infty).$ This
proves the theorem. ∎

######  Theorem 9.

Suppose $\tilde{L}_{n}^{(\delta)}(z;q)$ and $\tilde{L}_{n}^{(\delta+2)}(z;q)$
are not co-prime and the smallest positive zero of
$\tilde{L}_{n}^{(\delta)}(z;q),$ that is, $z_{2,n}<c_{n}$ for $n\geq
2,n\in\mathbb{N},$ where $c_{n}$ is given in Theorem 4. Then the above
polynomials have a common zero at $z=c_{n}$ and the positive zeros of
$\tilde{L}_{n}^{(\delta)}(z;q)$ interlace with the non-common zeros of
$\tilde{L}_{n}^{(\delta+2)}(z;q).$

###### Proof.

We have by (4), $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+1)}(z;q)$ are co-prime. Therefore, from (11), $c_{n}$
is the only common zero of $\tilde{L}_{n}^{(\delta)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ in $(0,\infty).$ If
$x_{1,n-1}^{\prime}<x_{2,n-1}^{\prime}<\cdots<x_{n-1,n-1}^{\prime}$ are the
$n-1$ non-common zeros of $\tilde{L}_{n}^{(\delta+2)}(z;q),$ then $c_{n}$ lies
in at most one interval
$\left(x_{i,n-1}^{\prime},x_{i+1,n-1}^{\prime}\right),$
$i\in\\{1,\cdots,n-2\\}.$ Evaluating (11) at the consecutive non-common zeros
$x_{i,n-1}^{\prime}$ and $x_{i+1,n-1}^{\prime},$ $i\in\\{1,\cdots,n-2\\},$ of
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ such that
$c_{n}\notin\left(x_{i,n-1}^{\prime},x_{i+1,n-1}^{\prime}\right),$ yields

$\left(x_{i,n-1}^{\prime}-c_{n}\right)\left(x_{i+1,n-1}^{\prime}-c_{n}\right)\tilde{L}_{n}^{(\delta+1)}\left(x_{i,n-1}^{\prime};q\right)\tilde{L}_{n}^{(\delta+1)}\left(x_{i+1,n-1}^{\prime};q\right)\\\
=\left({1-q^{\delta+n+1}\over
q^{\delta+2n+1}}\right)^{2}\tilde{L}_{n}^{(\delta)}\left(x_{i,n-1}^{\prime};q\right)\tilde{L}_{n}^{(\delta)}\left(x_{i+1,n-1}^{\prime};q\right).$
(21)

We have from (13) that, the zeros of $\tilde{L}_{n}^{(\delta+1)}(z;q)$ and
$\tilde{L}_{n}^{(\delta+2)}(z;q)$ interlace. Thus, we deduce from (21) that
$\tilde{L}_{n}^{(\delta)}(z;q)$ has an odd number of zeros in
$\left(x_{i,n-1}^{\prime},x_{i+1,n-1}^{\prime}\right)$ for each
$i\in\\{1,\cdots,n-2\\},$ that does not contain the point $c_{n}.$ From
Theorem 5, we have $z_{2,n}<x_{1,n}$ when
$c_{n}\in\left(x_{1,n},x_{n,n}\right).$ Since $\tilde{L}_{n}^{(\delta)}(z;q)$
has $n-3$ non-common zeros in $(x_{1,n},x_{n,n}),$ there are at most $n-3$
such intervals. Hence, $c_{n}=x_{j,n}$ where $j\in\\{2,\cdots,n-1\\}.$ Thus,
the $n-2$ non-common positive zeros of $\tilde{L}_{n}^{(\delta)}(z;q),$
together with the point $c_{n},$ interlace with the $n-1$ non-common zeros of
$\tilde{L}_{n}^{(\delta+2)}(z;q).$ This completes the proof. ∎

We illustrate Theorem 8 and Theorem 9 by the following numeric examples using
Mathematica 12 :

######  Example 1.

If we take $q=0.997,n=26$ and $\delta=-1.121695,$ the polynomials
$\tilde{L}_{24}^{(-1.121695)}(z;0.997)$ and
$\tilde{L}_{26}^{(-1.121695)}(z;0.997)$ have a common zero in $(0,\infty),$
namely, $a_{26}=0.167473.$ Further, the zeros of
$\tilde{L}_{24}^{(-1.121695)}(z;0.997)$ interlace with the non-common zeros of
$\tilde{L}_{26}^{(-1.121695)}(z;0.997)$ in $(0,\infty).$

######  Example 2.

When $q=0.94,n=26$ and $\delta=-1.92598,$ then $c_{26}=0.278236$ is a common
zero of $\tilde{L}_{26}^{(-1.92598)}(z;0.94)$ and
$\tilde{L}_{26}^{(-1.92598+2)}(z;0.94).$ Moreover, the positive zeros of
$\tilde{L}_{26}^{(-1.92598)}(z;0.94)$ interlace with the non-common zeros of
$\tilde{L}_{26}^{(-1.92598+2)}(z;0.94).$

##  4 Bounds for the smallest zero of
$\boldsymbol{\tilde{L}_{n}^{(\delta)}(z;q)}$

We derive the inner and outer bounds for the negative zero of
$\tilde{L}_{n}^{(\delta)}(z;q).$ To find the bounds, we make use of the
following contiguous relations:

$z^{3}\tilde{L}_{n-2}^{(\delta+3)}(z;q)={(1-q^{\delta+n+1})\over(1-q^{n-1})}b_{n}\left(z-A_{n}\right)\tilde{L}_{n-1}^{(\delta)}(z;q)+\left(z-A_{n-\delta-2}\right)\tilde{L}_{n}^{(\delta)}(z;q)$
(22)

and

$z^{4}\tilde{L}_{n-2}^{(\delta+4)}(z;q)={A_{n-\delta-2}\
b_{n}((1-q^{n})+q(1-q^{\delta+n+1}))\over
q^{n}(1-q^{\delta+1})}\left(z-B_{n}\right)\tilde{L}_{n-1}^{(\delta)}(z;q)\\\
+\left(z^{2}+{(1+q)(1-q^{\delta+2})\over
q^{\delta+n+1}}z-{A_{n-\delta-2}(1-q^{\delta+3})\over
q^{\delta+n+1}}\right)\tilde{L}_{n}^{(\delta)}(z;q),$ (23)

where $A_{n}=\displaystyle{(1-q^{\delta+1})(1-q^{\delta+2})\over
q^{\delta+1}(1-q^{\delta+n+1})},$
$B_{n}=\displaystyle{(1-q^{\delta+1})(1-q^{\delta+3})\over
q^{\delta+1}((1-q^{n})+q(1-q^{\delta+n+1}))}$ and $b_{n}$ is given in Theorem
3.

Clearly, (22) and (23) are the application of [16, (4.14) and (4.15)] in (17).

######  Theorem 10.

For $n\geq 2,n\in\mathbb{N},$ the negative zero $z_{1,n}$ of
$\tilde{L}_{n}^{(\delta)}(z;q)$ satisfies

$-c_{n-1}<B_{n}<z_{1,n}<A_{n}<0,$ (24)

where $A_{n},B_{n}$ and $c_{n}$ are respectively given in (22), (23) and in
Theorem 4.

###### Proof.

At $z=z_{1,n},$ (23) gives

$z_{1,n}^{4}\tilde{L}_{n-2}^{(\delta+4)}(z_{1,n};q)={A_{n-\delta-2}\
b_{n}((1-q^{n})+q(1-q^{\delta+n+1}))\over
q^{n}(1-q^{\delta+1})}\left(z_{1,n}-B_{n}\right)\tilde{L}_{n-1}^{(\delta)}(z_{1,n};q).$
(25)

For all $n$ and $\delta,$ we have

$\lim_{z\to-\infty}\tilde{L}_{n}^{(\delta)}(z;q)=\infty.$ (26)

This implies that $(z_{1,n}-B_{n})$ must be positive since all other factors
in (25) are positive, that means

$B_{n}<z_{1,n}.$ (27)

Similarly, we have from (22) at $z=z_{1,n}$ that

$z_{1,n}^{3}\tilde{L}_{n-2}^{(\delta+3)}(z_{1,n};q)={(1-q^{\delta+n+1})\over(1-q^{n-1})}b_{n}\left(z_{1,n}-A_{n}\right)\tilde{L}_{n-1}^{(\delta)}(z_{1,n};q).$
(28)

Using (4) and (26) we observe that, the factor $(z_{1,n}-A_{n})$ must be
negative since the left-hand side of (28) is negative while the first and
final factor on the right-hand side are positive, so that

$z_{1,n}<A_{n}.$ (29)

Now for $n\geq 2,$ we see that

$(1-q^{n})+q(1-q^{n-2})>0,$

which implies

$(1-q^{n})+q(1-q^{\delta+n+1})>q^{n-1}(1-q^{\delta+3}).$

Suitable interpretation of the above inequality leads to

$-c_{n-1}<B_{n}.$ (30)

Therefore, from (27), (29) and (30), we have the inequality (24). ∎

######  Remark 3.

By using the fact as in [15, (14.21.18)], when we take limit as $q\to 1$ in
(24), it recovers the bounds, obtained by Driver and Muldoon in [10], for the
negative zero of the quasi-orthogonal Laguerre polynomials of order $1.$

Using Mathematica 12, we examine the inner and outer bounds for the smallest
zero of $\tilde{L}_{n}^{(\delta)}(z;q)$ for different values of the parameters
$q,$ $n$ and $\delta$ such that $\delta\in(-2,-1)$ which are listed in Table
1.

Table 1: Inner and outer bounds of $z_{1,n}$ for different values of $q,n$ and $\delta.$ $q$ | $n$ | $\delta$ | $B_{n}$ | $z_{1,n}$ | $A_{n}$
---|---|---|---|---|---
|  | $-1.1$ | $-0.11032$ | $-0.110294$ | $-0.10681$
| $2$ | $-1.81$ | $-0.505545$ | $-0.448837$ | $-0.20526$
|  | $-1.1$ | $-0.104315$ | $-0.104286$ | $-0.100269$
$0.23$ | $7$ | $-1.81$ | $-0.46738$ | $-0.40912$ | $-0.169571$
|  | $-1.1$ | $-0.104311$ | $-0.104283$ | $-0.100265$
| $12$ | $-1.81$ | $-0.467357$ | $-0.409097$ | $-0.169552$
|  | $-1.1$ | $-0.00598206$ | $-0.00597785$ | $-0.00580805$
| $2$ | $-1.81$ | $-0.0360942$ | $-0.0294349$ | $-0.0152326$
|  | $-1.1$ | $-0.00219276$ | $-0.00219076$ | $-0.0020879$
$0.89$ | $7$ | $-1.81$ | $-0.0114906$ | $-0.00913766$ | $-0.0038382$
|  | $-1.1$ | $-0.0016198$ | $-0.00161831$ | $-0.00153785$
| $12$ | $-1.81$ | $-0.0083227$ | $-0.00661085$ | $-0.00270734$

## Acknowledgments

Present investigation is supported under OURIIP, Govt. of Odisha, India,
Sanction Number - 1040/69/OSHEC.

## References

  * [1] C. Brezinski, K. Driver, M. Redivo-Zaglia, Quasi-orthogonality with applications to some families of classical orthogonal polynomials, Appl. Numer. Math. 48 (2004) 157–168.
  * [2] C. Brezinski, K. Driver, M. Redivo-Zaglia, Zeros of quadratic quasi-orthogonal order 2 polynomials, Appl. Numer. Math. 135 (2019) 143–145.
  * [3] T. S. Chihara, On quasi-orthogonal polynomials, Proc. Amer. Math. Soc. 8 (1957) 765–767.
  * [4] D. Dickinson, On quasi-orthogonal polynomials, Proc. Amer. Math. Soc. 12 (1961) 185–194.
  * [5] A. Draux, On quasi-orthogonal polynomials, J. Aprox. Theory 62(1) (1990) 1–14.
  * [6] K. Driver, A. Jooste, Interlacing of zeros of quasi-orthogonal Meixner polynomials, Quaest. Math. 40 (2017) 477–490.
  * [7] K. Driver, K. Jordaan, Zeros of Jacobi polynomials $P_{n}^{(\alpha,\beta)},-2<\alpha,\beta<-1$, Numer. Algor. 79(4) (2018) 1075–1085.
  * [8] K. Driver, K. Jordaan, Zeros of quasi-orthogonal Jacobi polynomials, SIGMA, 12, 042, 13 pages (2016).
  * [9] K. Driver, M. E. Muldoon, Zeros of quasi-orthogonal ultraspherical polynomials, 27 (2016) 930–944.
  * [10] K. Driver, M. E. Muldoon, Interlacing properties and bounds for the quasi-orthogonal Laguerre polynomials, Comput. Methods Funct. Theorey 15 (2015) 645–-654.
  * [11] L. Fejér, Mechanische Quadraturen mit Positiven Cotesschen Zahlen, Math. Z. 37 (1933) 287–309.
  * [12] A. Jooste, K. Jordaan, F. Toókos, On the zeros of Meixner polynomials, Numer. Math. 124 (2013) 57–71.
  * [13] H. Joulak, A contribution to quasi-orthogonal polynomials and associated polynomials, Appl. Numer. Math. 54(1) (2005) 65–78.
  * [14] P.P. Kar, K. Jordaan, P. Gochhayat, M.K. Nangho, Quasi-orthogonality and zeros of some ${}_{2}\phi_{2}$ and ${}_{3}\phi_{2}$ polynomials, J. Comput. Appl. Math. (2020) 112910, https://doi.org/10.1016/j.cam.2020.112910
  * [15] R. Koekoek, P. A. Lesky, R. F. Swarttouw, Hypergeometric Orthogonal Polynomials and Their $q$-Analogues, Springer Monographs in Mathematics, Springer Verlag, Berlin (2010).
  * [16] D.S. Moak, The $q$-analogue of the Laguerre polynomials, J. Math. Anal. 81 (1981) 20–47.
  * [17] M. Riesz, Sur le probleme des moments, III, Ark. Mat. Astron. Fys. 17(16) (1923) 1–52.
  * [18] J.A. Shohat, On mechanical quadratures, in particular, with positive coefficients, Trans. Amer. Math. Soc. 42(3) (1937) 461–496.
  * [19] G. Szegö, Orthogonal Polynomials, American Mathematical Society Colloquium Publications, Vol. XXIII, (1939).
  * [20] D.D. Tcheutia, A.S. Jooste, W. Koepf, Quasi-Orthogonality of Some Hypergeometric and $q$-Hypergeometric Polynomials, SIGMA, 14, 051, 26 pages (2018).