A Numerical Investigation on the Structure of the Zeros of the Q-Tangent Polynomials

Abstract

We introduce q-tangent polynomials and their basic properties including q-derivative and q-integral. By using Mathematica, we find approximate roots of q-tangent polynomials. We also investigate relations of zeros between q-tangent polynomials and classical tangent polynomials.

1. Introduction

For a long time, studies on q-difference equations appeared in intensive works especially by F. H. Jackson [1, 2], R. D. Carmichael [3], T. E. Mason [4], and other authors [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]. An intensive and somewhat surprising interest in q-numbers appeared in many areas of mathematics and applications including q-difference equations, special functions, q-combinatorics, q-integrable systems, variational q-calculus, q-series, and so on. In this paper, we introduce some basic definitions and theorems (see [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]).

For any $n\in \mathbb{C}$ , the q-number is defined by

$${\left[n\right]}_q=\frac{1-q^n}{1-q},\mid q\mid <1.$$ E1

Definition 1.1. [1, 2, 9, 13] The q-derivative operator of any function f is defined by

$$D_{q}f\left(x\right)=\frac{f\left(x\right)-f\left(\mathit{qx}\right)}{\left(1-q\right)x},x\ne 0,$$ E2

and $D_{q}f\left(0\right)=f^{\prime }\left(0\right)$ . We can prove that f is differentiable at 0, and it is clear that $D_q{x}^n={\left[n\right]}_q{x}^{n-1}$ .

Definition 1.2. [1, 2, 9, 13, 17] We define the q-integral as

$${\int }_0^{b}f\left(x\right)d_{q}x=\left(1-q\right)b\sum _{j=0}^{\infty }{q}^{j}f\left(q^{j}b\right).$$ E3

If this function, f(x), is differentiable on the point x, the q-derivative in Definition 1.1 goes to the ordinary derivative in the classical analysis when $q\to 1$ .

Definition 1.3. [5, 17, 18, 21] The Gaussian binomial coefficients are defined by

$${\left(\begin{array}{l}m\\ r\end{array}\right)}_q={\left[\begin{array}{l}m\\ r\end{array}\right]}_q=\left\{\begin{array}{rr}0& \text{if}r>m\\ \frac{\left(1-q^m\right)\left(1-q^{m-1}\right)\cdots \left(1-q^{m-r+1}\right)}{\left(1-q\right)\left(1-q^2\right)\cdots \left(1-q^r\right)}& \text{if}r\le m\end{array}\right.,$$ E4

where m and r are non-negative integers. For $r=0$ the value is 1 since the numerator and the denominator are both empty products. Like the classical binomial coefficients, the Gaussian binomial coefficients are center-symmetric. There are analogues of the binomial formula, and this definition has a number of properties.

Theorem 1.4. Let n, k be non-negative integers. Then we get.

1. $$\prod _{k=0}^{n-1}\left(1+q^{k}t\right)=\sum _{k=0}^n{q}^{\left(\begin{array}{l}k\\ 2\end{array}\right)}{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{t}^k,$$ E5

2. $\prod _{k=0}^{n-1}\frac{1}{\left(1-q^{k}t\right)}=\sum _{k=0}^{\infty }{\left[\begin{array}{l}n+k-1\\ k\end{array}\right]}_q{t}^k.$

Definition 1.5. [5, 26] Let z be any complex number with $\mid z\mid <1$ . Two forms of q-exponential functions are defined by

$$e_q\left(z\right)=\sum _{n=0}^{\infty }\frac{z^n}{{\left[n\right]}_q!},e_{q^{-1}}\left(z\right)=\sum _{n=0}^{\infty }\frac{z^n}{{\left[n\right]}_{q^{-1}}!}=\sum _{n=0}^{\infty }{q}^{\left(\begin{array}{l}n\\ 2\end{array}\right)}\frac{z^n}{{\left[n\right]}_q!}.$$ E6

Bernoulli, Euler, and Genocchi polynomials have been studied extensively by many mathematicians(see [22, 23, 24, 25]). In 2013, C. S. Ryoo introduced tangent polynomials and he developed several properties of these polynomials (see [22, 23]). The tangent numbers are closely related to Euler numbers.

Definition 1.6. [22, 23, 24, 25] Tangent numbers T_n and tangent polynomials $T_n\left(x\right)$ are defined by means of the generating functions

$$\begin{array}{l}\sum _{n=0}^{\infty }{T}_n\frac{t^n}{n!}=\frac{2}{e^{2t}+1}=2\sum _{m=0}^{\infty }{\left(-1\right)}^m{e}^{2\mathit{mt}},\\ \sum _{n=0}^{\infty }{T}_n\frac{t^n}{n!}=\frac{2}{e^{2t}+1}{e}^{\mathit{tx}}=2\sum _{m=0}^{\infty }{\left(-1\right)}^m{e}^{\left(2m+x\right)t}.\end{array}$$ E7

Theorem 1.7. For any positive integer n, we have

$$T_n\left(x\right)={\left(-1\right)}^n{T}_n\left(2-x\right).$$ E8

Theorem 1.8. For any positive integer $m\left(=\mathit{odd}\right)$ , we have

$$T_n\left(x\right)=m^n\sum _{i=0}^{m-1}{\left(-1\right)}^i{T}_n\left(\frac{2i+x}{m}\right),n\in {\mathbb{Z}}_{+}.$$ E9

Theorem 1.9. For $n\in {\mathbb{Z}}_{+}$ , we have

$$T_n\left(x+y\right)=\sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right)T_k\left(x\right)y^{n-k}.$$ E10

The main aim of this paper is to extend tangent numbers and polynomials, and study some of their properties. Our paper is organized as follows: In Section 2, we define q-tangent polynomials and find some properties of these polynomials. We consider q-tangent polynomials in two parameters and establish some relations between q-tangent polynomials and q-Euler or Bernoulli polynomials. In Section 3, we observe approximate roots distributions of q-tangent polynomials and demonstrate interesting phenomenon.

2. Some properties of the q-tangent polynomials

In this section we define the q-tangent numbers and polynomials and establish some of their basic properties. we shall also study the q-tangent polynomials involving two parameters. We shall find some important relations between these polynomials and q-other polynomials.

Definition 2.1. For $x,q\in \mathbb{C}$ , we define q-tangent polynomials as

$$\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right),\mid t\mid <\frac{\pi }{2}.$$ E11

From Definition 2.1, it follows that

$$\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(0\right)\frac{t^n}{{\left[n\right]}_q!}=\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1},$$ E12

where ${\mathcal{T}}_{n,q}$ is q-tangent number. If $q\to 1$ , then it reduces to the classical tangent polynomial(see [22, 23, 24, 25]).

Theorem 2.2. Let $x,q\in \mathbb{C}$ . Then, the following hold.

$$\mathrm{i. }{\mathcal{T}}_{n,q}+\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{2}^{n-k}{\mathcal{T}}_{k,q}=\left\{\begin{array}{ll}{\left[2\right]}_q& \text{if}n=0\\ 0& \text{if}n\ne 0\end{array}\right.,$$ E13

$$\mathrm{ii. }{\mathcal{T}}_{n,q}\left(x\right)+\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{2}^{n-k}{\mathcal{T}}_{k,q}\left(x\right)={\left[2\right]}_q{x}^n.$$

Proof. From the Definition 2.1, we have

$$\begin{array}{cc}{\left[2\right]}_q& =\left(1+e_q\left(2t\right)\right)\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{n!}\\ & =\sum _{n=0}^{\infty }\left({\mathcal{T}}_{n,q}+\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{2}^{n-k}{\mathcal{T}}_{k,q}\right)\frac{t^n}{n!}.\end{array}$$ E14

Now comparing the coefficients of tⁿ we find (i). For (ii) we use the relation

$$\begin{array}{cc}{\left[2\right]}_q{e}_q\left(\mathit{tx}\right)& =\left(1+e_q\left(2t\right)\right)\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}\\ & =\sum _{n=0}^{\infty }\left({\mathcal{T}}_{n,q}\left(x\right)+\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{2}^{n-k}{\mathcal{T}}_{k,q}\left(x\right)\right)\frac{t^n}{n!},\end{array}$$ E15

and again compare the coefficients of tⁿ.☐

Theorem 2.3. Let n be a non-negative integer. Then, the following holds

$${\mathcal{T}}_{n,q}\left(x\right)=\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\mathcal{T}}_{n-k,q}{x}^k.$$ E16

Proof. From the definition of the q-exponential function, we have

$$\begin{array}{cc}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}& =\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)=\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{{\left[n\right]}_q!}\sum _{n=0}^{\infty }{x}^n\frac{t^n}{{\left[n\right]}_q!}\\ & =\sum _{n=0}^{\infty }\left(\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\mathcal{T}}_{n-k,q}\left(x\right)x^k\right)\frac{t^n}{{\left[n\right]}_q!}.\end{array}$$ E17

The required relation now follows on comparing the coefficients of tⁿ on both sides.☐

Theorem 2.4. Let n be a non-negative integer. Then, the following holds

$${\mathcal{T}}_{n,q}=\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\left(-1\right)}^{n-k}{q}^{\left(\begin{array}{l}n-k\\ 2\end{array}\right)}{\mathcal{T}}_{k,q}\left(x\right)x^{n-k}.$$ E18

Proof. From the property of q-exponential function, it follows that

$$\begin{array}{cc}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{{\left[n\right]}_q!}& =\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)e_{q^1}\left(-\mathit{tx}\right)\\ & =\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}\sum _{n=0}^{\infty }{q}^{\left(\begin{array}{l}n\\ 2\end{array}\right)}{\left(-1\right)}^n{x}^n\frac{t^n}{{\left[n\right]}_q!}\\ & =\sum _{n=0}^{\infty }\left(\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\left(-1\right)}^{n-k}{q}^{\left(\begin{array}{l}n-k\\ 2\end{array}\right)}{\mathcal{T}}_{k,q}\left(x\right)x^{n-k}\right)\frac{t^n}{{\left[n\right]}_q!}.\end{array}$$ E19

The required relation now follows immediately.☐

In what follows, we consider q-derivative of $e_q\left(\mathit{tx}\right)$ . Using the Mathematical Induction, we find.

$$\mathrm{i. }k=1:D_q^{\left(1\right)}{e}_q\left(\mathit{tx}\right)=\sum _{n=1}^{\infty }{x}^{n-1}\frac{t^n}{{\left[n-1\right]}_q!}.$$ E20

$$\mathrm{ii. }k=i:D_q^{\left(i\right)}{e}_q\left(\mathit{tx}\right)=\sum _{n=i}^{\infty }{x}^{n-i}\frac{t^n}{{\left[n-i\right]}_q!}.$$

If (ii) is true, then it follows that.

$$\begin{array}{ll}\mathrm{iii. }k=i+1:D_q^{\left(i+1\right)}{e}_q\left(\mathit{tx}\right)& =D_{q;x}^{\left(1\right)}\left(\sum _{n=i}^{\infty }{x}^{n-i}\frac{t^n}{{\left[n-i\right]}_q!}\right)\\ & =\sum _{n=i+1}^{\infty }{x}^{n-\left(i+1\right)}\frac{t^n}{{\left[n-\left(i+1\right)\right]}_q!}\\ & =t^{i+1}{e}_q\left(\mathit{tx}\right).\end{array}$$ E21

We are now in the position to prove the following theorem.

Theorem 2.5. For $k\in \mathbb{N}$ , the following holds

$$D_q^{\left(k\right)}{\mathcal{T}}_{n,q}\left(x\right)=\frac{{\left[n\right]}_q!}{{\left[n-k\right]}_q!}{\mathcal{T}}_{n-k,q}\left(x\right).$$ E22

Proof. Considering q-derivative of $e_q\left(\mathit{tx}\right)$ , we find

$$\begin{array}{ll}{D}_q^{\left(i+1\right)}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}& =\sum _{n=0}^{\infty }{D}_q^{\left(i+1\right)}{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}\\ & =\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{D}_q^{\left(i+1\right)}{e}_q\left(\mathit{tx}\right)\\ & =t^{i+1}\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)\\ & =\sum _{n=0}^{\infty }{\left[n+\left(i+1\right)\right]}_q\cdots {\left[n+2\right]}_q{\left[n+1\right]}_q\\ & \times {\mathcal{T}}_{n,q}\left(x\right)\frac{t^{n+i+1}}{{\left[n+\left(i+1\right)\right]}_q!}\\ & =\sum _{n=0}^{\infty }\frac{{\left[n\right]}_q}{{\left[n+\left(i+1\right)\right]}_q!}{\mathcal{T}}_{n-\left(i+1\right),q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!},\end{array}$$ E23

which immediately gives the required result.☐

Theorem 2.6. Let a, b be any real numbers. Then, we have

$${\int }_a^b{\mathcal{T}}_{n,q}\left(x\right)d_{q}x=\sum _{k=0}^{n+1}\frac{1}{{\left[n+1\right]}_q}\left({\mathcal{T}}_{n+1,q}\left(b\right)-{\mathcal{T}}_{n+1,q}\left(a\right)\right).$$ E24

Proof. From Theorem 2.3, we find

$$\begin{array}{ll}{\int }_a^b{\mathcal{T}}_{n,q}\left(x\right)d_{q}x& ={\int }_a^b\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\mathcal{T}}_{k,q}{x}^{n-k}{d}_{q}x\\ & {\left.=\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\mathcal{T}}_{k,q}\frac{1}{{\left[n-k+1\right]}_q}{x}^{n-k+1}\right|}_a^b\\ & =\sum _{k=0}^{n+1}\frac{{\mathcal{T}}_{n+1,q}\left(b\right)-{\mathcal{T}}_{n+1,q}\left(a\right)}{{\left[n+1\right]}_q}.\end{array}$$ E25

☐

Definition 2.7. For $x,y\in \mathbb{C}$ , we define q-tangent polynomial with two parameters as

$$\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,y\right)\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)e_q\left(\mathit{ty}\right),\mid t\mid <\frac{\pi }{2}.$$ E26

From the Definition 2.7, it is clear that

$$\begin{array}{l}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,0\right)\frac{t^n}{{\left[n\right]}_q!}=\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right),\\ \sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(0,0\right)\frac{t^n}{{\left[n\right]}_q!}=\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1},\end{array}$$ E27

where ${\mathcal{T}}_{n,q}$ is q-tangent number. We also note that the original tangent number, ${\mathcal{T}}_n$ ,

$$\underset{q\to 1}{\text{lim}}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{{\left[n\right]}_q!}=\sum _{n=0}^{\infty }{T}_n\frac{t^n}{n!}=\frac{2}{e^{2t}+1},$$ E28

where $q\to 1$ .

Theorem 2.8. Let $x,y$ be any complex numbers. Then, the following hold.

$$\mathrm{i. }{\mathcal{T}}_{n,q}\left(x,y\right)=\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\mathcal{T}}_{n-k,q}\left(x\right)y^k,$$ E29

$$\mathrm{ii. }{\mathcal{T}}_{n,q}\left(x,y\right)=\sum _{l=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\mathcal{T}}_{n-l,q}\sum _{k=0}^l{\left[\begin{array}{l}l\\ k\end{array}\right]}_q{x}^{l-k}{y}^k.$$

Proof. From the Definition 2.7, we have

$$\begin{array}{ll}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,y\right)\frac{t^n}{{\left[n\right]}_q!}& =\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)e_q\left(\mathit{ty}\right)\\ & =\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}\sum _{n=0}^{\infty }{y}^n\frac{t^n}{{\left[n\right]}_q!}.\end{array}$$ E30

Using Cauchy’s product and the method of coefficient comparison in the above relation, we find (i). Next, we transform q-tangent polynomials in two parameters as

$$\begin{array}{ll}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,y\right)\frac{t^n}{{\left[n\right]}_q!}& =\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)e_q\left(\mathit{ty}\right)\\ & =\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\frac{t^n}{{\left[n\right]}_q!}\sum _{n=0}^{\infty }{x}^n\frac{t^n}{{\left[n\right]}_q!}\sum _{n=0}^{\infty }{y}^n\frac{t^n}{{\left[n\right]}_q!}.\end{array}$$ E31

Now following same procedure as in (i), we obtain (ii).☐

Theorem 2.9. Setting $y=2$ in q-tangent polynomials with two parameters, the following relation holds

$${\left[2\right]}_q{x}^n={\mathcal{T}}_{n,q}\left(x,2\right)+{\mathcal{T}}_{n,q}\left(x\right).$$ E32

Proof. Using q-tangent polynomials and its polynomials with two parameters, we have

$$\begin{array}{l}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,2\right)\frac{t^n}{{\left[n\right]}_q!}+\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q{e}_q\left(2t\right)}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)+\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)\\ ={\left[2\right]}_q{e}_q\left(\mathit{tx}\right)\end{array}$$ E33

Now from the definition of q-exponential function, the required relation follows.☐

Theorem 2.9 is interesting as it leads to the relation

$$x^n=\frac{{\mathcal{T}}_{n,q}\left(x,2\right)+{\mathcal{T}}_{n,q}\left(x\right)}{{\left[2\right]}_q}.$$ E34

Theorem 2.10. Let $\left|q\right|<1$ . Then, the following holds

$${\mathcal{T}}_{n,q}\left(x\right)=\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\left(-1\right)}^k{\mathcal{T}}_{k,\frac{1}{q}}\left(2\right)x^{n-k}.$$ E35

Proof. To prove the relation, we note that

$$e_{\frac{1}{q}}\left(-2t\right)={\mathcal{E}}_q\left(-2t\right),$$ E36

where ${\mathcal{E}}_q\left(t\right)=e_{q^{-1}}\left(t\right)$ . Using the above equation we can represent the q-tangent polynomials as

$$\begin{array}{cc}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}& =\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)\\ & =\frac{{\left[2\right]}_q}{1+{\mathcal{E}}_q\left(-2t\right)}{\mathcal{E}}_q\left(-2t\right)e_q\left(\mathit{tx}\right)\\ & =\frac{{\left[2\right]}_q}{e_{\frac{1}{q}}\left(-2t\right)+1}{e}_{\frac{1}{q}}\left(-2t\right)e_q\left(\mathit{tx}\right)\\ & =\sum _{n=0}^{\infty }{\mathcal{T}}_{n,\frac{1}{q}}\left(2\right)\frac{{\left(-t\right)}^n}{{\left[n\right]}_q!}\sum _{n=0}^{\infty }{x}^n\frac{t^n}{{\left[n\right]}_q!}\\ & =\sum _{n=0}^{\infty }\left\{\sum _{k=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\left(-1\right)}^k{T}_{k,\frac{1}{q}}\left(2\right)x^{n-k}\right\}\frac{t^n}{{\left[n\right]}_q!},\end{array}$$ E37

which leads to the required relation immediately.☐

Now we shall find relations between q-tangent polynomials and others polynomials. For this, first we introduce well known polynomials by using q-numbers.

Definition 2.11. We define q-Euler polynomials, $E_{n,q}\left(x\right)$ , and q-Bernoulli polynomials, $B_{n,q}\left(x\right)$ , as

$$\begin{array}{l}\sum _{n=0}^{\infty }{E}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}=\frac{{\left[2\right]}_q}{e_q\left(t\right)+1}{e}_q\left(\mathit{tx}\right),\mid t\mid <\pi ,\\ \sum _{n=0}^{\infty }{B}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}=\frac{t}{e_q\left(t\right)-1}{e}_q\left(\mathit{tx}\right),\mid t\mid <2\pi .\end{array}$$ E38

Theorem 2.12. For $x,y\in \mathbb{C}$ , the following relation holds

$${\mathcal{T}}_{n,q}\left(x,y\right)=\frac{1}{{\left[2\right]}_q}\sum _{l=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q\left(\frac{{\mathcal{T}}_{n-l,q}\left(x\right)}{m^l}+\sum _{k=0}^{n-l}{\left[\begin{array}{l}n-l\\ k\end{array}\right]}_q\frac{{\mathcal{T}}_{k,q}\left(x\right)}{m^{n-k}}\right)E_{l,q}\left(\mathit{my}\right).$$ E39

Proof. Transforming q-tangent polynomials containing two parameters, we find

$$\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)e_q\left(\mathit{ty}\right)=\left(\frac{{\left[2\right]}_q}{e_q\left(\frac{t}{m}\right)+1}{e}_q\left(\mathit{ty}\right)\right)\left(\frac{e_q\left(\frac{t}{m}\right)+1}{{\left[2\right]}_q}\right)\left(\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)\right).$$ E40

Thus, for the relation between q-tangent polynomials of two parameters and q-Euler polynomials, we have

$$\begin{array}{l}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,y\right)\frac{t^n}{{\left[n\right]}_q!}\\ =\sum _{n=0}^{\infty }{E}_{n,q}\left(\mathit{my}\right)\frac{t^n}{m^n{\left[n\right]}_q!}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}\left(\sum _{n=0}^{\infty }\frac{1}{{\left[2\right]}_q}\frac{t^n}{m^n{\left[n\right]}_q!}+\frac{1}{{\left[2\right]}_q}\right)\\ =\frac{1}{{\left[2\right]}_q}\sum _{n=0}^{\infty }\sum _{l=0}^n{\left[\begin{array}{l}n\\ l\end{array}\right]}_q{E}_{l,q}\left(\mathit{my}\right)\sum _{k=0}^{n-l}{\left[\begin{array}{l}n-l\\ k\end{array}\right]}_q\frac{{\mathcal{T}}_{k,q}\left(x\right)}{m^{n-k}}\frac{t^n}{{\left[n\right]}_q!}\\ +\frac{1}{{\left[2\right]}_q}\sum _{n=0}^{\infty }\sum _{l=0}^n{\left[\begin{array}{l}n\\ l\end{array}\right]}_q{E}_{l,q}\left(\mathit{my}\right)\frac{{\mathcal{T}}_{n-l,q}\left(x\right)}{m^l}\frac{t^n}{{\left[n\right]}_q!},\end{array}$$ E41

which on comparing the coefficients immediately gives the required relation. ☐

Corollary 2.13. From Theorem 2.12, the following hold.

$$\mathrm{i. }{\mathcal{T}}_{n,q}\left(x,y\right)=\frac{1}{{\left[2\right]}_q}\sum _{l=0}^n{\left[\begin{array}{l}n\\ l\end{array}\right]}_q\left(\frac{{\mathcal{T}}_{n-l,q}\left(x\right)}{m^l}+\sum _{k=0}^{n-l}{\left[\begin{array}{l}n-l\\ k\end{array}\right]}_q\frac{{\mathcal{T}}_{k,q}\left(x\right)}{m^{n-k}}\right)E_{l,q}\left(\mathit{my}\right).$$ E42

$$\mathrm{ii. }{\mathcal{T}}_n\left(x,y\right)=\frac{1}{2}\sum _{l=0}^n\left(\begin{array}{l}n\\ l\end{array}\right)\left(\frac{{\mathcal{T}}_{n-l}\left(x\right)}{m^l}+\sum _{k=0}^{n-l}\left(\begin{array}{l}n-l\\ k\end{array}\right)\frac{{\mathcal{T}}_k\left(x\right)}{m^{n-k}}\right)E_l\left(\mathit{my}\right).$$

Theorem 2.14. For $x,y\in \mathbb{C}$ , the following relation holds

$${\mathcal{T}}_{n-1,q}\left(x,y\right)=\frac{1}{{\left[n\right]}_q}\sum _{l=0}^n{\left[\begin{array}{l}n\\ k\end{array}\right]}_q\left(\sum _{k=0}^{n-l}{\left[\begin{array}{l}n-l\\ k\end{array}\right]}_q\frac{{\mathcal{T}}_{k,q}\left(x\right)}{m^{n-k}}-\frac{{\mathcal{T}}_{n-l,q}\left(x\right)}{m^l}\right)B_{l,q}\left(\mathit{my}\right).$$ E43

Proof. We note that

$$\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)e_q\left(\mathit{ty}\right)=\left(\frac{t}{e_q\left(\frac{t}{m}\right)-1}{e}_q\left(\mathit{ty}\right)\right)\left(\frac{e_q\left(\frac{t}{m}\right)-1}{t}\right)\left(\frac{{\left[2\right]}_q}{e_q\left(2t\right)+1}{e}_q\left(\mathit{tx}\right)\right).$$ E44

Thus as in Theorem 2.12, we have

$$\begin{array}{l}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x,y\right)\frac{t^n}{{\left[n\right]}_q!}\\ =\left(\sum _{n=0}^{\infty }\frac{t^{n-1}}{m^n{\left[n\right]}_q!}-\frac{1}{t}\right)\sum _{n=0}^{\infty }{B}_{n,q}\left(\mathit{my}\right)\frac{t^n}{m^n{\left[n\right]}_q!}\sum _{n=0}^{\infty }{\mathcal{T}}_{n,q}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}\\ =\sum _{n=0}^{\infty }\left(\sum _{l=0}^n{\left[\begin{array}{l}n\\ l\end{array}\right]}_q\sum _{k=0}^{n-l}{\left[\begin{array}{l}n-l\\ k\end{array}\right]}_q\frac{{\mathcal{T}}_{k,q}\left(x\right)}{m^{n-k}}{B}_{l,q}\left(\mathit{my}\right)\right)\frac{t^{n-1}}{{\left[n\right]}_q!}\\ -\sum _{n=0}^{\infty }\left(\sum _{l=0}^n{\left[\begin{array}{l}n\\ l\end{array}\right]}_q\frac{{\mathcal{T}}_{n-l,q}\left(x\right)}{m^l}{B}_{l,q}\left(\mathit{my}\right)\right)\frac{t^{n-1}}{{\left[n\right]}_q!}.\end{array}$$ E45

The required relation now follows on comparing the coefficients. ☐

Corollary 2.15. From the Theorem 2.14, the following relations hold.

$$\mathrm{i. }{\mathcal{T}}_{n-1,q}\left(x,y\right)=\frac{1}{{\left[n\right]}_q}\sum _{l=0}^n{\left[\begin{array}{l}n\\ l\end{array}\right]}_q\left(\sum _{k=0}^{n-l}{\left[\begin{array}{l}n-l\\ k\end{array}\right]}_q\frac{{\mathcal{T}}_{k,q}\left(x\right)}{m^{n-k}}-\frac{{\mathcal{T}}_{n-l,q}\left(x\right)}{m^l}\right)B_{l,q}\left(\mathit{my}\right).$$ E46

$$\mathrm{ii. }{\mathcal{T}}_{n-1}\left(x,y\right)=\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{l}n\\ l\end{array}\right)\left(\sum _{k=0}^{n-l}\left(\begin{array}{l}n-l\\ k\end{array}\right)\frac{{\mathcal{T}}_k\left(x\right)}{m^{n-k}}-\frac{{\mathcal{T}}_{n-l}\left(x\right)}{m^l}\right)B_l\left(\mathit{my}\right).$$

3. The observation of scattering zeros of the q-tangent polynomials

In this section, we will find the approximate structure and shape of the roots according to the changes in n and q. We will extend this to identify the fixed points and try to understand the structure of the composite function using the Newton method.

The first five q-tangent polynomials are:

$$\begin{array}{l}{\mathcal{T}}_{0,q}\left(x\right)=\frac{1+q}{2},\hfill \\ {\mathcal{T}}_{1,q}\left(x\right)=\frac{1}{2}\left(1+q\right)\left(-1+x\right),\hfill \\ {\mathcal{T}}_{2,q}\left(x\right)=\frac{1}{2}\left(1+q\right)\left(1+q\left(-1+x\right)+x-x^2\right),\hfill \\ {\mathcal{T}}_{3,q}\left(x\right)=\frac{1}{2}\left(1+q\right)\left(-1+q\left(2-\left(-2+q\right)q\right)-x+q^{3}x-\left(1+q+q^2\right)x^2+x^3\right),\hfill \\ {\mathcal{T}}_{4,q}\left(x\right)=\frac{1}{2}\left(1+q\right)(\left(-1+q\right)\left(1+q\right)\left(1+\left(-4+q\right)q\right)\left(1+q+q^2\right)\\ -{\left(1+q\right)}^2\left(1+\left(-3+q\right)q\right)\left(1+q^2\right)x\hfill \\ +\left(-1+q\right)\left(1+q^2\right)\left(1+q+q^2\right)x^2-\left(1+q\right)\left(1+q^2\right)x^3+x^4).\hfill \end{array}$$ E47

Using Mathematica, we will examine the approximate movement of the roots. In Figure 1, the x-axis means the numbers of real zeros and the y-axis means the numbers of complex zeros in the q-tangent polynomials. When it moves from left to right, it changes to n = 30, 40, 50, and when it is fixed at q = 0.1, the approximate shape of the root appears to be almost circular. The center is identified as the origin, and it has 2.0 as an approximate root, which is unusual.

Figure 1.

Zeros of ${\mathcal{T}}_{n,0.1}\left(x\right)$ for n = 30, 40, 50.

Figure 2 shows the shape of the approximate roots when n is changed to the above conditions and fixed at q = 0.5.

Figure 2.

Zeros of ${\mathcal{T}}_{n\mathrm{,0.5}}\left(x\right)$ for n = 30, 40, 50.

In Figure 2, the shape of the root changes to an ellipse, unlike the q = 0.1 condition, and the widening phenomenon appears when the real number is 0.5. In addition, like the previous Figure 1, we can see that it has a common approximate root at 2.0. In the following Figure 3, n of the far-left figure is 30, and it increases by 10 while moving to the right, and the far-right figure shows the shape of the root when n = 50 and is fixed at q = 0.9.

Figure 3.

Zeros of ${\mathcal{T}}_{n\mathrm{,0.9}}\left(x\right)$ for n = 30, 40, 50.

In Figure 3, the roots have a general tangent polynomial shape with similar properties (see [22, 23, 24, 25]). If each approximate root obtained in the previous step is piled up according to the value of n, it will appear as shown in Figure 4. The left Figure 4 is q = 0.1 with n from 1 to 50. The middle Figure 4 is q = 0.5 with n from 1 to 50. The right Figure 4 is q = 0.9 with n from 1 to 50.

Figure 4.

Zeros of ${\mathcal{T}}_{n,q}\left(x\right)$ for q = 0.1, 0.5, 0.9, 1 ≤ n ≤ 50.

Let $f:D\to D$ be a complex function, with D as a subset of $\mathbb{C}$ . We define the iterated maps of the complex function as the following:

$$f_r:z_0\mapsto \underset{r}{\underbrace{f\left(f\right(\cdots (f}}\left(z_0\right)\cdots \left)\right))$$ E48

The iterates of f are the functions $f,f\circ f,f\circ f\circ f,\dots ,$ which are denoted $f^1,f^2,f^3,\dots$ If $z\in \mathbb{C}$ , and then the orbit of z₀ under f is the sequence $<z_0,f\left(z_0\right),f\left(f\left(z_0\right)\right),\cdots >$ .

We consider the Newton’s dynamical system as follows [12, 15, 20]:

$$\left\{{\mathbb{C}}_{\infty }:R\left(x\right)=x-\frac{\mathcal{T}\left(x\right)}{{\mathcal{T}}^{\prime }\left(x\right)}\right\}.$$ E49

R is called the Newton iteration function of $\mathcal{T}$ . It can be considered that the fixed points of R are the zeros of $\mathcal{T}$ and all the fixed points of R are attracting. R may also have one or more attracting cycles.

For $x\in \mathbb{C}$ , we consider ${\mathcal{T}}_{4,q}\left(x\right)$ , and then this polynomial has four distinct complex numbers, $a_i\left(i=\mathrm{1,2,3,4}\right)$ such that ${\mathcal{T}}_{4,q}\left(a_i\right)=0$ . Using a computer, we obtain the approximate zeros (Table 1) as follows:

i	q = 0.1	q = 0.5	q = 0.9
1	−0.672809	−0.581881 − 0.412941i	−1.10249
2	−0.0821877 − 0.710388i	−0.581881 + 0.412941i	−0.158841
3	−0.0821877 + 0.710388i	0.907024	1.84004
4	1.94818	2.13174	2.86029

Table 1.

Approximate zeros of ${\mathcal{T}}_{4,q}\left(x\right)$ .

In Newton’s method, the generalized expectation is that a typical orbit {R(x)} will converge to one of the roots of ${\mathcal{T}}_{4,q}\left(x\right)$ for $x_0\in \mathbb{C}$ . If we choose x₀, which is sufficiently close to a_i, then this proves that

$$\underset{r\to \infty }{\text{lim}}R\left(x_0\right)=a_i,\text{for}i=\mathrm{1,2,3,4}.$$ E50

When it is given a point x₀ in the complex plane, we want to determine whether the orbit of x₀ under the action of R(x) converges to one of the roots of the equation. The orbit of x₀ under the action of R also appears by calculating until 30 iterations or the absolute difference value of the last two iterations is within 10⁻⁶.

The output in Figure 5 is the last calculated orbit value. We construct a function, which assigns one of four colors for each point according to the outcome of R in the plane. If an orbit of x₀ for q = 0.1 converges to −0.672809, −0.0821877 − 0.710388i, −0.0821877 + 0.710388i and 1.94818, then we denote the red, blue, yellow, and sky-blue, respectively(the left figure). For example, the yellow region for the left figure represents the part of the basin of attraction of a₃ = −0.0821877 + 0.710388i.

Figure 5.

Orbit of x₀ under the action of R for ${\mathcal{T}}_{4,q}\left(x\right)$ for q = 0.1, 0.5, 0.9.

If we use ${\mathcal{T}}_{\mathrm{3,0.1}}\left(x\right)$ to draw a figure using the Newton method, we can obtain Figure 6. The picture on the left shows three roots, and the colors are blue, red, and ivory in the counterclockwise direction. When we examine the area closely, we can see that it converges to an approximate value in each color area. The convergence value in the blue area is $-0.379202+0.523651i$ , that in the red area is $-0.379202-0.523651i$ , and that in the ivory area is 1.8684. We can also see that it shows self-similarity at the boundary point as divided into three areas. The figure on the right is obtained by 2-times iterated q-tangent polynomials, ${\mathcal{T}}_{\mathrm{3,0.1}}^2\left(x\right)$ , and the area is divided into nine colors “gray ( $x=2.31831$ ), scarlet ( $x=1.76736+0.216319i$ ), light brown ( $x=0.137247+0.59473i$ ), sky blue ( $x=-0.604153+1.19884i$ ), blue ( $x=-0.794606+0.378411i$ ), red ( $x=-0.794606-0.378411i$ ), ivory ( $x=-0.604153-1.19884i$ ), green ( $x=0.137247-0.59473i$ ), and navy blue ( $x=1.76736-0.216319i$ ) in the counterclockwise direction. This also shows self-similarity at the boundary.

Figure 6.

Orbit of x₀ under the action of R for ${\mathcal{T}}_{3,0.1}\left(x\right),{\mathcal{T}}_{\mathrm{3,0.1}}^2\left(x\right)$ .

In Figure 7, we express the coloring for ${\mathcal{T}}_{\mathrm{3,0.1}}^2\left(x\right)$ .

Figure 7.

Palette for escaping points.

Conjecture 3.1. The q-tangent polynomials always have self-similarity at the boundary.

We know that the fixed point is divided as follows. Suppose that the complex function f is analytic in a region D of $\mathbb{C}$ , and f has a fixed point at $z_0\in D$ . Then z₀ is said to be (see [6, 16, 20]):

1. an attracting fixed point if $\mid f^{\prime }\left(z_0\right)\mid <1$ ;

2. a repelling fixed point if $\mid f^{\prime }\left(z_0\right)\mid >1$ ;

3. a neutral fixed point if $\mid f^{\prime }\left(z_0\right)\mid =1$ .

For example, ${\mathcal{T}}_{\mathrm{3,0.1}}\left(x\right)$ has three points satisfying ${\mathcal{T}}_{\mathrm{3,0.1}}\left(x\right)=x$ .

That is, $x_0=-0.967484,-\mathrm{0.33466,2.41214}$ . Since

$$\left|\frac{d}{\mathit{dt}}{\mathcal{T}}_{\mathrm{3,0.1}}\left(-0.967484\right)\right|=0<1,\left|\frac{d}{\mathit{dt}}{\mathcal{T}}_{\mathrm{3,0.1}}\left(-0.33466\right)\right|=0<1$$ E51

Theorem 3.2. ${\mathcal{T}}_{\mathrm{3,0.1}}\left(x\right)$ for q = 0.1 has two attracting fixed points.

Using Mathematica, we can separate the numerical results for fixed points of ${\mathcal{T}}_{n\mathrm{,0.1}}\left(x\right)$ . From Table 2, we know that ${\mathcal{T}}_{n\mathrm{,0.1}}\left(x\right)$ have no neutral fixed point for $1\le n\le 4$ . We can also reach Conjecture 3.3.

Degree n	Attractor	Repellor	Neutral
1	0	1	0
2	1	1	0
3	2	1	0
4	1	3	0
5	1	4	0

Table 2.

Numbers of fixed points of ${\mathcal{T}}_{n\mathrm{,0.1}}\left(x\right)$ .

Conjecture 3.3. The q-tangent polynomials for $n\ge 2$ have at least one attracting fixed point except for infinity.

In Table 3, we denote ${\mathbf{R}}_{{\mathcal{T}}_{n,q}^r\left(x\right)}$ as the numbers of real zeros for rth iteration and ${\mathbf{RF}}_{{\mathcal{T}}_{n,q}^r\left(x\right)}$ as the numbers of attracting fixed point on real number. From this table, we can know that number of real fixed points of ${\mathcal{T}}_{3,q}^r\left(x\right)$ are less than two. Here, we can suggest Conjecture 3.4.

r	${\mathbf{R}}_{{\mathcal{T}}_{\mathrm{3,0.1}}^r\left(x\right)}$	${\mathbf{RF}}_{{\mathcal{T}}_{\mathrm{3,0.1}}^r\left(x\right)}$
1	3	2
2	3	2
3	3	2
4	23	2
5	2	2
6	1	1

Table 3.

The numbers of ${\mathbf{R}}_{{\mathcal{T}}_{\mathrm{3,0.1}}^r\left(x\right)}$ and ${\mathbf{RF}}_{{\mathcal{T}}_{\mathrm{3,0.1}}^r\left(x\right)}$ for $1\le r\le 6$ .

Conjecture 3.4. The q-tangent polynomials that are iterated, ${\mathcal{T}}_{\mathrm{3,0.1}}^r\left(x\right)$ , have real fixed point, $\alpha =-0.33466$ .

In the top-left of Figure 8, we can see the forms of 3D structure related to stacks of fixed points of ${\mathcal{T}}_{\mathrm{3,0.1}}^r\left(x\right)$ for $1\le r\le 6$ . When we look at the top-left of Figure 8 in the below position, we can draw the top-right figure. The bottom-left of Figure 8 shows that image and n-axes exist but not real axis in three dimensions. In three dimensions, the bottom-right of Figure 8 is the right orthographic viewpoint for the top-left figure,-that is, there exist real and n-axes but there is no image axis (Figure 8).

Figure 8.

Stacks of fixed point of ${\mathcal{T}}_{3,0.1}^r\left(x\right)$ for 1 ≤ r ≤ 6.

4. Conclusion

We can see that when q comes closer to 0, the approximate shape of the roots become increasingly more circular. Also in this situation, we can observe scattering of zeros in q-tangent polynomials around 2 in three-dimension. When q comes closer to 1, it has properties that are more symmetrical. We can also assume that the property that appears when iterating ${\mathcal{T}}_{n,q}\left(x\right)$ has self-similarity. By iterating, we can conjecture some properties about fixed points. This property warrants further study so that we can create a new property.

Acknowledgments

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT and Future Planning (No. 2017R1E1A1A03070483).

Conflict of interests

The authors declare that there is no conflict of interests regarding the publication of this paper.