1 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 [1–11]). 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 [11])

$$ (x)_{n}= \sum_{k=0}^{n} S_{1}(n, k) x^{k}\quad \text{and}\quad x^{n}= \sum _{k=0}^{n} S_{2}(n, k) (x)_{k}, $$
(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 [6] 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. [6]).

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 [7–9]

$$ \biggl( \frac{2}{e^{2t}+ 1} \biggr) e^{x t}=\sum _{n=0}^{\infty}\mathbf {T}_{n}(x) \frac{t^{n}}{n!} \quad \bigl( \vert 2t \vert < \pi\bigr). $$
(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 [9]).

Many kinds of generalizations of these polynomials and numbers have been presented in the literature (see [1–11]). 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.

2 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

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

3 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 [1]) 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 [5]).

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 [5]).

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 [1]).

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

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

4 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

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

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 \(k= -3\).

Figure 1
figure 1

Zeros of \(\pmb{T_{n,q}^{(k)}(x)}\) .

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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).

Figure 2
figure 2

Stacks of zeros of \(\pmb{T_{n,q}^{(k)}(x)}\) for \(\pmb{2\leq n \leq40}\) .

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Table 1 Numbers of real and complex zeros of \(\pmb{T_{n,q}^{(k)}(x) }\)
Full size table
Table 2 Approximate solutions of \(\pmb{T_{n,q}^{(k)}(x)=0}\) , \(\pmb{k=3}\) , \(\pmb{q=1/2}\)
Full size table

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 \(k=3\).

Figure 3
figure 3

Real zeros of \(\pmb{T_{n,q}^{(k)}(x)}\) for \(\pmb{2 \leq n \leq40}\) .

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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.

Table 3 Approximate solutions of \(\pmb{T_{n,q}^{(k)}(x) =0}\) , \(\pmb{k=-3}\) , \(\pmb{q=-1/2}\)
Full size table

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.