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# Some identities involving q-poly-tangent numbers and polynomials and distribution of their zeros

## Abstract

In this paper we introduce the q-poly-tangent polynomials and numbers. We also give some properties, explicit formulas, several identities, a connection with poly-tangent numbers and polynomials, and some integral formulas. Finally, we investigate the zeros of the q-poly-tangent polynomials by using a computer.

## Introduction

Many mathematicians have studied in the area of the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, tangent numbers and polynomials, poly-Bernoulli numbers and polynomials, poly-Euler numbers and polynomials, and poly-tangent numbers and polynomials (see ). In this paper, we define q-poly-tangent polynomials and numbers and study some properties of the q-poly-tangent polynomials and numbers. Throughout this paper, we always make use of the following notations: $$\mathbb{N}$$ denotes the set of natural numbers and $$\mathbb{Z}_{+}= \mathbb{N} \cup \{ 0 \}$$. We recall that the classical Stirling numbers of the first kind $$S_{1}(n, k)$$ and $$S_{2}(n, k)$$ are defined by the relations (see )

$$(x)_{n}= \sum_{k=0}^{n} S_{1}(n, k) x^{k}\quad \text{and}\quad x^{n}= \sum _{k=0}^{n} S_{2}(n, k) (x)_{k},$$
(1.1)

respectively. Here $$(x)_{n}=x(x-1)\cdots (x-n+1)$$ denotes the falling factorial polynomial of order n. The numbers $$S_{2}(n, m)$$ also admit a representation in terms of a generating function,

$$\bigl(e^{t}-1\bigr)^{m} = m ! \sum _{n=m}^{\infty} S_{2}(n, m) \frac{ t^{n}}{n!} .$$
(1.2)

We also have

$$m ! \sum_{n=m}^{\infty} S_{1}(n, m) \frac{ t^{n}}{n!} = \bigl( \log(1+t)\bigr)^{m} .$$
(1.3)

We also need the binomial theorem: for a variable x,

$$\frac{1}{ (1- t)^{c}} = \sum_{n=0}^{\infty}\binom{c+n-1}{n} t^{n} .$$
(1.4)

For $$0 \leq q <1$$, the q-poly-Bernoulli numbers $$B_{n}^{(k)}$$ were introduced by Mansour  by using the following generating function:

$$\frac{ \operatorname{Li}_{k, q}(1-e^{-t})}{1-e^{-t}} =\sum_{n=0}^{\infty}{B}_{n,q}^{(k)} \frac{t^{n}}{n!} \quad (k \in\mathbb{Z}),$$
(1.5)

where

$$\operatorname{Li}_{k, q}(t)=\sum_{n=1}^{\infty}\frac{t^{n}}{[n]_{q}^{k}}$$
(1.6)

is the kth q-poly-logarithm function, and $$[n]_{q}=\frac{1-q^{n}}{1-q}$$ is the q-integer (cf. ).

The q-poly-Euler polynomials $$E_{n,q}^{(k)}(x)$$ are defined by the generating function

$$\frac{ \operatorname{Li}_{k,q}(1-e^{-t})}{e^{t}+1} e^{xt} =\sum_{n=0}^{\infty}{E}_{n,q}^{(k)}(x) \frac{t^{n}}{n!}\quad (k \in\mathbb{Z}).$$
(1.7)

The familiar tangent polynomials $$\mathbf{T}_{n}(x)$$ are defined by the generating function 

$$\biggl( \frac{2}{e^{2t}+ 1} \biggr) e^{x t}=\sum _{n=0}^{\infty}\mathbf {T}_{n}(x) \frac{t^{n}}{n!} \quad \bigl( \vert 2t \vert < \pi\bigr).$$
(1.8)

When $$x=0$$, $$\mathbf{T}_{n}(0)= \mathbf{T}_{n}$$ are called the tangent numbers. The tangent polynomials $$\mathbf{ T}_{n}^{(r)}(x)$$ of order r are defined by

$$\biggl( \frac{2}{e^{2t}+ 1} \biggr)^{r} e^{x t}=\sum _{n=0}^{\infty}\mathbf{T}_{n}^{(r)} (x) \frac{t^{n}}{n!}\quad \bigl( \vert 2t \vert < \pi\bigr).$$
(1.9)

It is clear that for $$r=1$$ we recover the tangent polynomials $$\mathbf{T}_{n}(x)$$.

The Bernoulli polynomials $$\mathbf{B}_{n}^{(r)}(x)$$ of order r are defined by the following generating function:

$$\biggl( \frac{t}{e^{t}- 1} \biggr)^{r} e^{x t}=\sum _{n=0}^{\infty}\mathbf {B}_{n}^{(r)} (x) \frac{t^{n}}{n!} \quad \bigl( \vert t \vert < 2 \pi\bigr).$$
(1.10)

The Frobenius-Euler polynomials of order r, denoted by $$\mathbf{H}_{n}^{(r)}(u, x)$$, are defined as

$$\biggl( \frac{1-u}{e^{t} -u } \biggr)^{r}e^{xt}=\sum _{n=0}^{\infty} \mathbf{H}_{n}^{(r)}(u,x) \frac{t^{n}}{n!}.$$
(1.11)

The values at $$x=0$$ are called Frobenius-Euler numbers of order r; when $$r=1$$, the polynomials or numbers are called ordinary Frobenius-Euler polynomials or numbers.

The poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$ are defined by the generating function

$$\frac{ \operatorname{Li}_{k}(1-e^{-t})}{e^{2t}+1} e^{xt} =\sum_{n=0}^{\infty}{T}_{n}^{(k)}(x) \frac{t^{n}}{n!} \quad (k \in\mathbb{Z}),$$
(1.12)

where $$\operatorname{Li}_{k}(t)=\sum_{n=1}^{\infty}\frac{t^{n}}{n^{k}}$$ is the kth poly-logarithm function (see ).

Many kinds of generalizations of these polynomials and numbers have been presented in the literature (see ). In the following section, we introduce the q-poly-tangent polynomials and numbers. After that we will investigate some their properties. We also give some relationships both between these polynomials and tangent polynomials and between these polynomials and q-cauchy numbers. Finally, we investigate the zeros of the q-poly-tangent polynomials by using a computer.

## q-Poly-tangent numbers and polynomials

In this section, we define q-poly-tangent numbers and polynomials and provide some of their relevant properties.

For $$0 \leq q <1$$, the q-poly-tangent polynomials $${T}_{n,q}^{(k)}(x)$$ are defined by the generating function:

$$\frac{ 2 \operatorname{Li}_{k,q}(1-e^{-t})}{e^{2t}+1} e^{xt} =\sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!}\quad (k \in\mathbb{Z}).$$
(2.1)

When $$x=0$$, $${T}_{n,q}^{(k)}(0) = {T}_{n,q}^{(k)}(x)$$ are called the q-poly-tangent numbers. Observe that $$\lim_{q \rightarrow1} T_{n,q}^{(k)}(x) = T_{n}^{(k)}(x)$$. By (2.1), we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( \frac{ 2 \operatorname{Li}_{k,q}(1-e^{-t})}{e^{2t}+1} \biggr) e^{x t} \\ & = \sum_{n=0}^{\infty}{T}_{n,q}^{(k)} \frac{t^{n}}{n!} \sum_{n=0}^{\infty}x^{n} \frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{n} \binom{n}{l} {T}_{l,q}^{(k)} x^{n-l} \Biggr) \frac{t^{n}}{n!} . \end{aligned}
(2.2)

By comparing the coefficients on both sides of (2.2), we have the following theorem.

### Theorem 2.1

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x) = \sum_{l=0}^{n} \binom{n}{l} {T}_{l,q}^{(k)} x^{n-l}.$$

The following elementary properties of the q-poly-tangent numbers $${T}_{n,q}^{(k)}$$ and polynomials $${T}_{n,q}^{(k)}(x)$$ are readily derived form (2.1). We, therefore, choose to omit the details involved.

### Theorem 2.2

For $$k \in\mathbb{Z}$$, we have

\begin{aligned} (1)&\quad {T}_{n,q}^{(k)}(x+y) = \sum _{l=0}^{n} \binom{n}{l} {T}_{l,q}^{(k)}(x) y^{n-l}, \\ (2)&\quad {T}_{n,q}^{(k)}(2-x)= \sum _{l=0}^{n} (-1)^{l} \binom{n}{l} {T}_{n-l,q}^{(k)}(2) x^{l}. \end{aligned}

### Theorem 2.3

For any positive integer n, we have

\begin{aligned} \begin{aligned} (1)&\quad {T}_{n,q}^{(k)}(mx) = \sum _{l=0}^{n} \binom{n}{l} {T}_{l,q}^{(k)}(x) (m-1)^{n-l} x^{n-l}\quad ( m \geq1), \\ (2)&\quad {T}_{n,q}^{(k)}(x+1)- T_{n,q}^{(k)}(x)= \sum_{l=0}^{n-1} \binom{n}{l} {T}_{l,q}^{(k)}(x), \\ (3)&\quad \frac{ d}{d x} {T}_{n,q}^{(k)}(x) = n {T}_{n-1,q}^{(k)}(x), \\ (4)&\quad {T}_{n,q}^{(k)}(x)= {T}_{n,q}^{(k)} + n \int_{0}^{x} {T}_{n-1,q}^{(k)}(t) \,dt. \end{aligned} \end{aligned}
(2.3)

From (1.6), (1.8), and (2.1), we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( 2 \frac{ \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \biggr) e^{x t} = \sum_{l=0}^{\infty}\frac{(1-e^{-t})^{l+1}}{[l+1]_{q}^{k}} \frac{ 2 e^{xt}}{e^{2t}+1} \\ & = \sum_{l=0}^{\infty}\frac{1}{[l+1]_{q}^{k}} \sum _{i=0}^{l+1} \binom{l+1}{i} (-1)^{i} \frac{ 2 e^{(x-i) t}}{e^{2t}+1} \\ & = \sum_{l=0}^{\infty}\frac{1}{[l+1]_{q}^{k}} \sum _{i=0}^{l+1} \binom{l+1}{i} (-1)^{i} \sum_{n=0}^{\infty}\mathbf{T}_{n} (x-i) \frac {t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{\infty}\frac{1}{[l+1]_{q}^{k}} \sum _{i=0}^{l+1} \binom{l+1}{i} (-1)^{i} \mathbf{T}_{n} (x-i) \Biggr) \frac{t^{n}}{n!} . \end{aligned}
(2.4)

By comparing the coefficients on both sides of (2.4), we have the following theorem.

### Theorem 2.4

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x) = \sum_{l=0}^{\infty}\frac{1}{[l+1]_{q}^{k}} \sum_{j=0}^{l+1} \binom{l+1}{j} (-1)^{j} \mathbf{T}_{n} (x-j).$$

By using the definition of tangent polynomials and Theorem 2.4, we have the following corollary.

### Corollary 2.5

For any positive integer n, we have

$${T}_{n,q}^{(k)}(2-x) = (-1)^{n} \sum _{l=0}^{\infty}\frac{1}{[l+1]_{q}^{k}} \sum _{j=0}^{l+1} \binom{l+1}{j} (-1)^{j} \mathbf{T}_{n} (x+j).$$

By (2.1), we note that

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = 2 \sum_{l=0}^{\infty}(-1)^{l} e^{2lt} \sum_{l=0}^{\infty}\frac{(1-e^{-t})^{l+1}}{[l+1]_{q}^{k}} e^{xt} \\ & = \sum_{l=0}^{\infty}\sum _{i=0}^{l} \sum_{j=0}^{i+1} \frac{ 2 (-1)^{l+j-i} \binom{i+1}{j}}{[i+1]_{q}^{k}} e^{(2l-2i-j+x)t} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{\infty}\sum_{i=0}^{l} \sum_{j=0}^{i+1} \frac{ 2 (-1)^{l+j-i} \binom{i+1}{j}(2l-2i-j+x)^{n}}{[i+1]_{q}^{k} } \Biggr) \frac{t^{n}}{n!} . \end{aligned}

Comparing the coefficients on both sides, we have the following theorem.

### Theorem 2.6

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x) = 2 \sum_{l=0}^{\infty}\sum_{i=0}^{l} \sum _{j=0}^{i+1} \frac{ (-1)^{l+j-i} \binom{i+1}{j}}{[i+1]_{q}^{k} }(2l-2i-j+x)^{n}.$$

By (1.7), (1.8), and (2.1) and by using the Cauchy product, we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \frac {1}{2} \biggl( \frac{ 2 \operatorname{Li}_{k,q}(1-e^{-t})}{e^{t}+1} \biggr) \frac{2 ( e^{t}+1)}{ e^{2t}+1} e^{xt} \\ & = \Biggl( \sum_{n=0}^{\infty}E_{n, q}^{(k)}(x) \frac{t^{n}}{n!} \Biggr) \Biggl( \sum _{n=0}^{\infty}\bigl( \mathbf{T}_{n}(1)+ \mathbf{T}_{n} \bigr)\frac{t^{n}}{n!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \frac{1}{2} \sum_{l=0}^{n} \binom{n}{l} \bigl( \mathbf{T}_{n}(1)+ \mathbf{T}_{n} \bigr) E_{n-l, q}^{(k)}(x) \Biggr). \end{aligned}
(2.5)

By comparing the coefficients on both sides of (2.5), we have the following theorem related the q-poly-Euler polynomials and tangent polynomials.

### Theorem 2.7

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x)= \frac{1}{2} \sum _{l=0}^{n} \binom{n}{l} \bigl( \mathbf{T}_{n}(1)+ \mathbf{T}_{n} \bigr) E_{n-l, q}^{(k)}(x).$$

By (1.5), (1.8), and (2.1) and by using the Cauchy product, we have

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( \frac{ \operatorname{Li}_{k,q}(1-e^{-t})}{1-e^{-t}} \biggr) \frac{2 ( 1-e^{-t})}{ e^{2t}+1} e^{xt} \\ & = \Biggl( \sum_{n=0}^{\infty}B_{n, q}^{(k)} \frac{t^{n}}{n!} \Biggr) \Biggl( \sum _{n=0}^{\infty}\bigl( \mathbf{T}_{n}(x)- \mathbf{T}_{n}(x-1) \bigr)\frac{t^{n}}{n!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{n=0}^{n} \binom{n}{l} \bigl( \mathbf{T}_{n}(x)- \mathbf{T}_{n}(x-1) \bigr) B_{n-l, q}^{(k)} \Biggr). \end{aligned}
(2.6)

By comparing the coefficients on both sides of (2.6), we have the following theorem related the q-poly-Bernoulli polynomials and tangent polynomials.

### Theorem 2.8

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x)= \sum_{l=0}^{n} \binom{n}{l} \bigl( \mathbf{T}_{n}(x)- \mathbf{T}_{n}(x-1) \bigr) B_{n-l, q}^{(k)}.$$

By (1.2), (1.5), (1.8), and Theorem 2.8, we have the following corollary.

### Corollary 2.9

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x)= \sum_{l=0}^{n} \sum_{m=0}^{l} \binom{n}{l} \frac{(-1)^{m+l} m! S_{2}(l, m)}{[m+1]_{q}^{k}} \bigl( \mathbf{T}_{n-l}(x)- \mathbf{T}_{n-l}(x-1) \bigr) .$$

## Some identities involving q-poly-tangent numbers and polynomials

In this section, we give several combinatorics identities involving q-poly-tangent numbers and polynomials in terms of Stirling numbers, falling factorial functions, raising factorial functions, Beta functions, Bernoulli polynomials of higher order, and Frobenius-Euler functions of higher order.

By (2.1) and by using the Cauchy product, we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( \frac{ 2 \operatorname{Li}_{k,q}(1-e^{-t})}{e^{2t}+1} \biggr) \bigl(1- \bigl(1-e^{-t}\bigr)\bigr)^{-x} \\ & = \frac{ 2 \operatorname{Li}_{k,q}(1-e^{-t})}{e^{2t}+1} \sum_{l=0}^{\infty}\binom{x+l-1}{l}\bigl(1-e^{-t}\bigr)^{l} \\ & = \sum_{l=0}^{\infty}\langle x\rangle _{l} \frac{(e^{t}-1)^{l}}{l!} \biggl( \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} e^{-lt} \biggr) \\ & = \sum_{l=0}^{\infty}\langle x\rangle _{l} \sum_{n=0}^{\infty}S_{2}(n, l) \frac {t^{n}}{n!} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(-l) \frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{\infty}\sum_{i=l}^{n} \binom{n}{i} S_{2}(i, l) {T}_{n-i,q}^{(k)}(-l) \langle x\rangle _{l} \Biggr) \frac{t^{n}}{n!}, \end{aligned}
(3.1)

where $$\langle x\rangle_{l}=x(x+1) \cdots(x+l-1)$$ ($$l \geq1$$) with $$\langle x\rangle_{0}=1$$.

By comparing the coefficients on both sides of (3.1), we have the following theorem.

### Theorem 3.1

For $$n \in\mathbb{Z}_{+}$$, we have

$${T}_{n,q}^{(k)}(x)=\sum_{l=0}^{\infty}\sum_{i=l}^{n} \binom{n}{i} \langle x \rangle_{l} S_{2}(i, l) {T}_{n-i,q}^{(k)}(-l).$$

By using the Jackson q-integral (see ) and Theorem 2.1, we get

\begin{aligned} \int_{0}^{1} {T}_{n,q}^{(k)}(x)\, d_{q}x & = \int_{0}^{1} \sum_{l=0}^{n} \binom {n}{l} {T}_{l,q}^{(k)} x^{n-l}\, d_{q}x \\ &= \sum_{l=0}^{n} \binom{n}{l} {T}_{l,q}^{(k)} \frac{1}{[n-l+1]_{q}} . \end{aligned}
(3.2)

By (3.2) and Theorem 3.1, we have the following theorem.

### Theorem 3.2

For any positive integer n, we have

$$\sum_{l=0}^{n} \binom{n}{l} {T}_{l,q}^{(k)} \frac{1}{[n-l+1]_{q}}=\sum _{l=0}^{\infty}\sum_{i=l}^{n} \binom{n}{i} S_{2}(i, l) {T}_{n-i,q}^{(k)}(-l) (-1)^{l} \hat{c}_{l,q},$$

where $$\hat{c}_{l, q}$$ are q-Cauchy numbers of the second kind (see ).

By (2.1) and by using the Cauchy product, we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \biggr) \bigl( \bigl(e^{t}-1\bigr)+1\bigr)^{x} \\ & = \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \sum_{l=0}^{\infty}\binom{x}{l}\bigl(e^{t}-1\bigr)^{l} \\ & = \sum_{l=0}^{\infty}(x)_{l} \frac{(e^{t}-1)^{l}}{l!} \biggl( \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \biggr) \\ & = \sum_{l=0}^{\infty}(x)_{l} \sum _{n=0}^{\infty}S_{2}(n, l) \frac {t^{n}}{n!} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)} \frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{\infty}\sum_{i=l}^{n} \binom{n}{i} (x)_{l} S_{2}(i, l) {T}_{n-i,q}^{(k)} \Biggr) \frac{t^{n}}{n!} . \end{aligned}
(3.3)

By comparing the coefficients on both sides of (3.3), we have the following theorem.

### Theorem 3.3

For $$n \in\mathbb{Z}_{+}$$ and $$0 \leq q <1$$, we have

$${T}_{n,q}^{(k)}(x)= \sum_{l=0}^{\infty}\sum_{i=l}^{n} \binom{n}{i} (x)_{l} S_{2}(i, l) {T}_{n-i,q}^{(k)} .$$

By (3.2) and Theorem 3.3, we have the following theorem.

### Theorem 3.4

For any positive integer n, we have

$$\sum_{l=0}^{n} \binom{n}{l} \frac{{T}_{n-l,q}^{(k)}}{[l+1]_{q}}=\sum_{l=0}^{\infty}\sum _{i=l}^{n} \binom{n}{i} S_{2}(i, l) {T}_{n-i,q}^{(k)}{c}_{l,q},$$

where $${c}_{l,q}$$ are q-Cauchy numbers of the first kind (see ).

By Theorem 2.2, we note that

\begin{aligned} \int_{0}^{1} y^{n} T_{n,q}^{(k)}(x+y) \,dy &= \int_{0}^{1} y^{n} \sum _{l=0}^{n} \binom{n}{l} T_{n-l,q}^{(k)}(x) y^{l} \,dy \\ &= \sum_{l=0}^{n} \binom{n}{l} T_{n-l,q}^{(k)}(x) \int_{0}^{1} y^{n+l} \,dy \\ &= \sum_{l=0}^{n} \binom{n}{l} T_{n-l,q}^{(k)}(x) \frac{1}{n+l+1} . \end{aligned}
(3.4)

From (2.1) and Theorem 2.2, we note that

\begin{aligned} \int_{0}^{1} y^{n} T_{n,q}^{(k)}(x+y) \,dy &= \frac{ y^{n} T_{n+1,q}^{(k)}(x+y)}{n+1} \bigg\vert _{0}^{1} - \int_{0}^{1} ny^{n-1} \frac{T_{n+1,q}^{(k)}(x+y)}{n+1} \,dy \\ & = \frac{T_{n+1,q}^{(k)}(x+1)}{n+1} - \frac{n}{n+1} \int_{0}^{1} y^{n-1} T_{n+1,q}^{(k)}(x+y) \,dy \\ &= \frac{T_{n+1,q}^{(k)}(x+1)}{n+1} - \frac{n}{n+1} \int_{0}^{1} y^{n-1} \sum _{l=0}^{n+1} \binom{n+1}{l} T_{l,q}^{(k)}(x)y^{n+1-l} \,dy \\ &= \frac{T_{n+1,q}^{(k)}(x+1)}{n+1} - \frac{n}{n+1} \sum_{l=0}^{n+1} \binom{n+1}{l} T_{l,q}^{(k)}(x) \frac{1}{2n-l+1}. \end{aligned}
(3.5)

Therefore, by (3.4) and (3.5), we obtain the following theorem.

### Theorem 3.5

For $$n \in\mathbb{Z}_{+}$$, we have

$$T_{n+1,q}^{(k)}(x+1) = \sum_{l=0}^{n+1} \binom{n+1}{l} T_{l,q}^{(k)}(x) \frac{n}{2n-l+1}+ \sum _{l=0}^{n} \binom{n}{l} T_{n-l,q}^{(k)}(x) \frac{n+1}{n+l+1} .$$

By (1.2), (1.10), (2.1), and by using the Cauchy product, we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \biggr) e^{x t} \\ & = \frac{ (e^{t}-1)^{r}}{r!} \frac{r!}{t^{r}} \biggl( \frac{t}{e^{t}-1} \biggr)^{r} e^{xt} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)} \frac{t^{n}}{n!} \\ & = \frac{ (e^{t}-1)^{r}}{r!} \Biggl( \sum_{n=0}^{\infty}\mathbf{ B}_{n}^{(r)}(x) \frac{t^{n}}{n!} \Biggr) \Biggl( \sum_{n=0}^{\infty}{T}_{n,q}^{(k)} \frac{t^{n}}{n!} \Biggr)\frac{r!}{t^{r}} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{n} \frac{ \binom{n}{l}}{\binom{l+r}{r}} S_{2}(l+r,r) \sum_{i=0}^{n-l} \binom{n-l}{i} \mathbf{ B}_{i}^{(r)}(x) {T}_{n-l-i,q}^{(k)} \Biggr) \frac{t^{n}}{n!} . \end{aligned}

By comparing the coefficients on both sides, we have the following theorem.

### Theorem 3.6

For $$n \in\mathbb{Z}_{+}$$ and $$r \in \mathbb{N}$$, we have

$$T_{n,q}^{(k)}(x) = \sum_{l=0}^{n} \frac{ \binom{n}{l}}{\binom{l+r}{r}} S_{2}(l+r,r) \sum_{i=0}^{n-l} \binom{n-l}{i} {T}_{n-l-i,q}^{(k)} \mathbf{ B}_{i}^{(r)}(x) .$$

From (2.1) and Theorem 2.2, we note that

\begin{aligned}& \int_{0}^{1} y^{n} T_{n,q}^{(k)}(x+y) \,dy \\& \quad = \frac{ y^{n} T_{n+1,q}^{(k)}(x+y)}{n+1} \bigg\vert _{0}^{1} - \int_{0}^{1} \frac{ny^{n-1} T_{n+1,q}^{(k)}(x+y)}{n+1} \,dy \\& \quad = \frac{T_{n+1,q}^{(k)}(x+1)}{n+1} - \frac{n}{n+1} \int_{0}^{1} \sum_{l=0}^{\infty}\frac{1}{[l+1]_{q}^{k}} \sum_{i=0}^{l+1} \binom{l+1}{i} (-1)^{i} \mathbf{T}_{n+1} (x+y-i) y^{n-1} \,dy \\& \quad = \frac{T_{n+1,q}^{(k)}(x+1)}{n+1} \\& \qquad {}- \frac{n}{n+1} \sum_{l=0}^{\infty}\sum_{i=0}^{l+1} \sum _{j=0}^{n+1}\frac{\binom{l+1}{i} \binom{n+1}{l}}{[l+1]_{q}^{k}} (-1)^{n+1+i} \mathbf{T}_{n+1-j} (1-x+i) \int_{0}^{1} y^{n-1} (1-y)^{j} \,dy \\& \quad = \frac{T_{n+1,q}^{(k)}(x+1)}{n+1} \\& \qquad {} - \frac{n}{n+1} \sum_{l=0}^{\infty}\sum_{i=0}^{l+1} \sum _{j=0}^{n+1}\frac{\binom{l+1}{i} \binom{n+1}{l}}{[l+1]_{q}^{k}} (-1)^{n+1+i} \mathbf{T}_{n+1-j} (1-x+i) B(n, j+1), \end{aligned}
(3.6)

where $$B(n, j)$$ is the beta integral (see ).

Therefore, by (3.5) and (3.6), we obtain the following theorem.

### Theorem 3.7

For $$n \in\mathbb{Z}_{+}$$, we have

$$\sum_{l=0}^{n+1} \binom{n+1}{l} \frac{T_{l,q}^{(k)}(x) }{2n-l+1} = \sum_{l=0}^{\infty}\sum _{i=0}^{l+1} \sum_{j=0}^{n+1} \frac{\binom{l+1}{i} \binom{n+1}{l}}{[l+1]_{q}^{k}} (-1)^{n+1+i} \mathbf{T}_{n+1-j} (1-x+i) B(n, j+1).$$

By (1.2), (1.11), (2.1), and by using the Cauchy product, we get

\begin{aligned} \sum_{n=0}^{\infty}{T}_{n,q}^{(k)}(x) \frac{t^{n}}{n!} & = \biggl( \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \biggr) e^{x t} \\ & = \frac{ (e^{t}-u)^{r}}{(1-u)^{r}} \biggl( \frac{1-u}{e^{t}-u} \biggr)^{r} e^{xt} \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \\ & = \sum_{n=0}^{\infty}\mathbf{H}_{n}^{(r)}(u, x) \frac{t^{n}}{n!} \sum_{i=0}^{r} \binom{r}{i} e^{it} (-u)^{r-i} \frac{ 1}{(1-u)^{r}} \frac{ 2 \operatorname{Li}_{k, q}(1-e^{-t})}{e^{2t}+1} \\ & = \frac{ 1}{(1-u)^{r}} \sum_{i=0}^{r} \binom{r}{i} (-u)^{r-i} \sum_{n=0}^{\infty}\mathbf{H}_{n}^{(r)}(u, x) \frac{t^{n}}{n!} \sum _{n=0}^{\infty}{T}_{n,q}^{(k)}(i) \frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( \frac{ 1}{(1-u)^{r}} \sum_{i=0}^{r} \binom {r}{i} (-u)^{r-i} \sum_{l=0}^{n} \binom{n}{l} \mathbf{H}_{l}^{(r)}(u, x) {T}_{n-l}^{(k)}(i) \Biggr) \frac{t^{n}}{n!} . \end{aligned}

By comparing the coefficients on both sides, we have the following theorem.

### Theorem 3.8

For $$n \in\mathbb{Z}_{+}$$ and $$r \in \mathbb{N}$$, we have

$$T_{n,q}^{(k)}(x) = \frac{ 1}{(1-u)^{r}} \sum _{i=0}^{r} \sum_{l=0}^{n} \binom {r}{i} \binom{n}{l} (-u)^{r-i} \mathbf{H}_{l}^{(r)}(u, x) {T}_{n-l,q}^{(k)}(i) .$$

For $$n \in\mathbb{N}$$ with $$n \geq4$$, we obtain

\begin{aligned} \int_{0}^{1} y^{n} T_{n,q}^{(k)}(x+y) \,dy =& y^{n+1} \frac {T_{n,q}^{(k)}(x+y)}{n+1} \bigg\vert _{0}^{1} - \int_{0}^{1} ny^{n+1} \frac{T_{n-1,q}^{(k)}(x+y)}{n+1} \,dy \\ =& \frac{T_{n,q}^{(k)}(x+1)}{n+1} - \frac{n,q}{n+1} \int_{0}^{1} y^{n+1} T_{n-1,q}^{(k)}(x+y) \,dy \\ =& \frac{T_{n,q}^{(k)}(x+1)}{n+1} - \frac{n T_{n-1,q}^{(k)}(x+1)}{(n+1)(n+2)} \\ &{}+(-1)^{2} \frac{n}{n+1} \frac{n-1}{n+2} \int_{0}^{1} y^{n+2} T_{n-2,q}^{(k)}(x+y) \,dy \\ =& \frac{T_{n,q}^{(k)}(x+1)}{n+1}+(-1) \frac{n T_{n-1,q}^{(k)}(x+1)}{(n+1)(n+2)} \\ &{}+(-1)^{2} \frac{n}{n+1} \frac{n-1}{n+2} \frac{T_{n-2,q}^{(k)}(x+1)}{n+3} \\ &{} +(-1)^{3} \frac{n}{n+1} \frac{n-1}{n+2} \frac{n-2}{n+3}\frac {T_{n-3,q}^{(k)}(x+1)}{n+4} \\ &{} +(-1)^{4} \frac{n}{n+1} \frac{n-1}{n+2} \frac{n-2}{n+3} \frac {n-3}{n+4} \int_{0}^{1} y^{n+4}T_{n-4,q}^{(k)}(x+y) \,dy. \end{aligned}

Continuing this process, we obtain

\begin{aligned} \int_{0}^{1} y^{n} T_{n,q}^{(k)}(x+y) \,dy =& \frac{T_{n,q}(x+1)}{n+1} \\ &{} + \sum_{l=2}^{n} \frac{n(n-1) \cdots(n-l+2)(-1)^{l-1}}{(n+1)(n+2) \cdots(n+l)} T_{n-l+1,q}^{(k)}(x+1) \\ &{} +(-1)^{n} \frac{n!}{(n+1)(n+2) \cdots(2n)} \int_{0}^{1} y^{2n}T_{0,q}^{(k)}(x+y) \,dy . \end{aligned}
(3.7)

Hence, by (3.4) and (3.7), we have the following theorem.

### Theorem 3.9

For $$n \in\mathbb{N}$$ with $$n \geq2$$, we have

\begin{aligned} &\sum_{l=0}^{n} \binom{n}{l} T_{n-l,q}^{(k)}(x) \frac{1}{n+l+1} \\ &\quad = \frac{T_{n,q}^{(k)}(x+1)}{n+1} + \sum_{l=2}^{n} \frac{n(n-1) \cdots (n-l+2)(-1)^{l-1}}{(n+1)(n+2) \cdots(n+l)} T_{n-l+1,q}^{(k)}(x+1) . \end{aligned}

## Zeros of the q-poly-tangent polynomials

This section aims to demonstrate the benefit of using a numerical investigation to support theoretical prediction and to discover new interesting pattern of the zeros of the poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$. The q-poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$ can be determined explicitly. A few of them are

\begin{aligned}& T_{0,q}^{(k)}(x)=0, \\& T_{1, q}^{(k)}(x)=1, \\& T_{2,q}^{(k)}(x)= -3 + \frac{2}{_{q}^{k}} + 2 x, \\& T_{3,q}^{(k)}(x)= 4 - \frac{12}{_{q}^{k}} + \frac{6}{ _{q}^{k}} - \biggl( 9 - \frac{6}{_{q}^{k}} \biggr) x + 3 x^{2} , \\& T_{4,q}^{(k)}(x) = 3 + \frac{38}{_{q}^{k}} - \frac{60}{ _{q}^{k}}+ \frac{24}{ _{q}^{k}}+ \biggl( 16 -\frac{48}{ _{q}^{k}} + \frac{24}{ _{q}^{k}} \biggr) x \\& \hphantom{T_{4,q}^{(k)}(x) ={}}{}- \biggl( 18 - \frac{12}{ _{q}^{k}} \biggr) x^{2} + 4 x^{3}, \\& T_{5, q}^{(k)}(x) = -14 - \frac{60}{_{q}^{k}} + \frac{ 330}{ _{q}^{k}} - \frac{ 360}{_{q}^{k}}+ \frac{120}{_{q}^{k}} + \biggl( 15 + \frac{190}{_{q}^{k}} - \frac{300}{_{q}^{k}} + \frac{120}{_{q}^{k} } \biggr) x \\& \hphantom{T_{5, q}^{(k)}(x) ={}}{}+ \biggl( 40 - \frac{120}{_{q}^{k}} + \frac{60}{_{q}^{k}} \biggr) x^{2} - \biggl( 30 + \frac{20}{_{q}^{k}} \biggr) x^{3} + 5 x^{4}. \end{aligned}

We investigate the beautiful zeros of the q-poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$ by using a computer. We plot the zeros of the q-poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$ for $$n= 20$$, $$q=1/2, -1/2$$, $$k=-3, 3$$ and $$x \in \mathbb{C}$$ (Figure 1). In Figure 1 (top-left), we choose $$n=20$$, $$q=1/2$$, and $$k= 3$$. In Figure 1 (top-right), we choose $$n=20$$, $$q=-1/2$$, and $$k= 3$$. In Figure 1 (bottom-left), we choose $$n=20$$, $$q=1/2$$, and $$k= -3$$. In Figure 1 (bottom-right), we choose $$n=20$$, $$q=-1/2$$, and $$k= -3$$.

Stacks of zeros of $$T_{n,q}^{(k)}(x)$$ for $$2 \leq n \leq40$$ from a 3-D structure are presented (Figure 2). In Figure 2, we choose $$k= 3$$, $$q=1/2$$. Our numerical results for approximate solutions of real zeros of $$T_{n,q}^{(k)}(x)$$ are displayed (Tables 1, 2).

The plot of real zeros of $$T_{n,q}^{(k)}(x)$$ for the $$2\leq n \leq40$$ structure is presented (Figure 3). In Figure 3, we choose $$k=3$$.

We observe a remarkable regular structure of the real roots of the q-poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$. We also hope to verify a remarkable regular structure of the real roots of the q-poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$ (Table 1).

Next, we calculated an approximate solution satisfying q-poly-tangent polynomials $$T_{n,q}^{(k)}(x)=0$$ for $$x \in \mathbb{R}$$. The results are given in Table 2 and Table 3.

By numerical computations, we will present a series of conjectures.

### Conjecture 4.1

Prove that $$T_{n,q}^{(k)}(x)$$, $$x \in \mathbb{C}$$, has $$\operatorname{Im}(x)=0$$ reflection symmetry analytic complex functions. However, $$T_{n,q}^{(k)}(x)$$ has no $$Re(x)= a$$ reflection symmetry for $$a \in\mathbb{R}$$.

Using computers, many more values of n have been checked. It still remains unknown if the conjecture fails or holds for any value n (see Figures 1, 2, 3).

We are able to decide if $$T_{n,q}^{(k)}(x)=0$$ has $$n-1$$ distinct solutions (see Tables 1, 2, 3).

### Conjecture 4.2

Prove that $$T_{n,q}^{(k)}(x)=0$$ has $$n-1$$ distinct solutions.

Since $$n-1$$ is the degree of the polynomial $$T_{n,q}^{(k)}(x)$$, the number of real zeros $$R_{ T_{n,q}^{(k)}(x)}$$ lying on the real plane $$\operatorname{Im}(x)=0$$ is $$R_{T_{n,q}^{(k)}(x)}=n-C_{T_{n,q}^{(k)}(x)}$$, where $$C_{T_{n,q}^{(k)}(x)}$$ denotes complex zeros. See Table 1 for tabulated values of $$R_{T_{n,q}^{(k)}(x)}$$ and $$C_{T_{n,q}^{(k)}(x)}$$. The authors have no doubt that investigations along these lines will lead to a new approach employing numerical method in the research field of the poly-tangent polynomials $$T_{n,q}^{(k)}(x)$$, which appear in mathematics and physics.

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## Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No. 2017R1A2B4006092).

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