Asymptotic Approximation of the Apostol-Tangent Polynomials Using Fourier Series

Abstract

Asymptotic approximations of the Apostol-tangent numbers and polynomials were established for non-zero complex values of the parameter $\lambda$. Fourier expansion of the Apostol-tangent polynomials was used to obtain the asymptotic approximations. The asymptotic formulas for the cases $\lambda =1$ and $\lambda =-1$ were explicitly considered to obtain asymptotic approximations of the corresponding tangent numbers and polynomials.

1. Introduction

The tangent polynomials $T_{n }\left(z\right)$ of a degree of $n$ with a complex argument $z$ are defined by the generating function (see [1,2]).

$${\displaystyle \sum }_{n=0}^{\infty }{T}_n\left(z\right)\frac{w^n}{n!} = \left(\frac{2}{e^{2w}+1}\right)e^{zw}, \left|w\right| <\frac{\pi }{2}$$

(1)

These polynomials can be expressed in polynomial form as

$$T_n\left(z\right) = {\displaystyle \sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right) T_k z^{n-k}$$

where $T_k$ denotes the tangent numbers defined by

$$\tan w = {\displaystyle \sum }_{n=0}^{\infty } {\left(-1\right)}^{n+1}{T}_{2n+1}\frac{w^{2n+1}}{\left(2n+1\right)!}$$

(2)

It is worth mentioning that tangent numbers are the odd indices of the numbers $A_n$ of alternating permutations known as the Euler zigzag numbers. The first few values of these numbers are as follows:

$$T_0 = 1, T_1=-1, T_3 = 2, T_5 = -16, T_7=272, T_9=-7936, T_{11}=353792.$$

Clearly, $T_{n }$: = $T_n\left(0\right)$ for $n$ $\in \mathbb{N}$.

Several mathematicians were attracted to work on tangent polynomials because of the significant properties that they possessed in the field of mathematics and physics (see [3,4,5,6]). Analogues, explicit identities, and symmetric properties for tangent polynomials were derived in [2,7,8]. Some interesting Apostol analogues of the classical Bernoulli, Euler, and Genocchi polynomials were investigated by Apostol [9], Corcino, Lou, Srivastava and Araci (see [10,11,12,13,14,15,16,17]). These analogues are called the Apostol–Bernoulli, Apostol–Euler, and Apostol–Genocchi polynomials of order $m$ defined by the following relations, respectively, (see [18]): For $\lambda \in ℂ\backslash \left\{0\right\}$ and $m \in Z^{+},$

$${\displaystyle \sum }_{n = 0}^{\infty }{B}_n^m\left(z;\lambda \right)\frac{w^n}{n!} = {\left(\frac{w}{\lambda e^w-1}\right)}^m{e}^{wz},\left|w\right| < 2\pi when \lambda = 1$$

(3)

$$and \left|w+\log \lambda \right| < 2\pi when \lambda \ne 1$$

$${\displaystyle \sum }_{n = 0}^{\infty }{E}_n^m\left(z;\lambda \right)\frac{w^n}{n!} = {\left(\frac{2}{\lambda e^w+1}\right)}^m{e}^{wz},\left|w\right| < \pi when \lambda = 1$$

(4)

$$and \left|w+\log \lambda \right| < \pi when \lambda \ne 1$$

$${\displaystyle \sum }_{n = 0}^{\infty }{G}_n^m\left(z;\lambda \right)\frac{w^n}{n!} = {\left(\frac{2w}{\lambda e^w+1}\right)}^m{e}^{wz},\left|w\right| < \pi when \lambda = 1$$

(5)

$$and \left|w+\log \lambda \right| < \pi when \lambda \ne 1$$

when $m = 1,$ the above Equations (3)–(5) give the generating functions for the Apostol–Bernoulli, Apostol–Euler, and Apostol–Genocchi polynomials, respectively (see [19]). We extend the tangent polynomials as follows.

The Apostol-tangent polynomials $T_n\left(x;\lambda \right)$ in $z$ are defined by means of the generating function

$${\displaystyle \sum }_{n = 0}^{\infty }{T}_n\left(x;\lambda \right)\frac{t^n}{n!} = \left(\frac{2}{\lambda e^{2t}+1}\right)e^{xt}, \left|2t+\log \lambda \right|<\pi$$

(6)

when $\lambda = 1$, the Equation reduces to the tangent polynomials $T_n\left(x\right) = T_n\left(x;1\right).$

Lopez and Temme [20] used the Fourier series to establish the asymptotic approximations of higher-order Bernoulli and Euler polynomials. C.B. Corcino and R.B. Corcino [21] derived the asymptotics of higher-order Genocchi polynomials by employing the method in [19,22,23]. In the study of Navas et al. [24], Fourier expansion is used to obtain the asymptotic estimates for Apostol–Bernoulli and Apostol–Euler polynomials. In this paper, the asymptotic expansion of Apostol-tangent polynomials is derived using the method of Navas et al. [24].

2. Asymptotic Approximations

Let ${\mathcal{T}}_{\lambda } = \left\{\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \mathsf{\lambda }\right]: k\in \mathbb{Z}\right\}$ be the set of poles of the generating function Equation (6). The Fourier series expansion of the Apostol-tangent polynomials in terms of poles in ${\mathcal{T}}_{\lambda }$ is given in the following theorem:

Theorem 1. 

Let$\lambda \in ℂ\backslash \left\{0\right\}.$For$n\ge 1, 0\le x\le 1,$

$$\begin{gathered} \frac{T_n\left(x;\lambda \right)}{n!}=\frac{2^{n+1}}{{\lambda }^{x/2}}{\displaystyle \sum }_{kϵ\mathbb{Z}}^{ }\frac{e^{\frac{1}{2}\left(2k-1\right)\pi ix}}{{\left[\left(2k-1\right)\pi i-\log \lambda \right]}^{n+1}} \\ =\frac{1}{{\lambda }^{x/2}}{\displaystyle \sum }_{kϵ\mathbb{Z}}^{ }\frac{e^{\frac{1}{2}\left(2k-1\right)\pi ix}}{{\left\{\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \lambda \right]\right\}}^{n+1}} \end{gathered}$$

(7)

where the logarithm is taken to be the principal branch.

Proof. 

Consider the integral ${{\displaystyle \int }}_{CN}{f}_n\left(z\right)dz$ where

$$f_n\left(z\right)=\frac{2e^{xz}}{\left(\lambda e^{2z}+1\right)z^{n+1}}$$

and the circle $C_N$ is a circle about the origin of radius $\left(\frac{1}{2}\left(2N-1+ϵ\right)\pi \right), {\rm N}\in {\mathbb{Z}}^{+}$ with $ϵ$ being a fixed real number such that $ϵ\pi i\pm \log \lambda \ne 0 \left(mod \pi i\right)$.

The function $f_{n }\left(z\right)$ has poles at $z=0$ of order $n+1$ and at $z_k=\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \mathsf{\lambda }\right],$ $k\in \mathbb{Z}$. The poles $z_k$ are simple poles. Using the Cauchy Residue Theorem,

$${{\displaystyle \int }}_{C_N}^{ }{f}_n\left(z\right) dz=2\pi i Res \left(f_n\left(z\right), z=0\right)+2\pi i{\displaystyle \sum }_{kϵZ, k<N}^{ }Res \left(f_n \left(z\right),z=z_k\right).$$

We observe that, using the basic property of integration,

$$\left|{{\displaystyle \int }}_{C_N}^{ }\frac{2e^{xz } dz}{\left(\lambda e^{2z}+1\right)z^{n+1}}\right| \le {{\displaystyle \int }}_{C_N}^{ }\frac{\left|2e^{xz}\right|\left|dz\right|}{\left|\lambda e^{2z}+1\right|\left|z^{n+1}\right|}.$$

For 0 $\le x\le 1$, $\left|\lambda e^{2z}+1\right|>\left|\lambda e^{2z}\right|$. Let $z=a+bi$,

$$\frac{\left|e^{xz}\right|}{\left|\lambda e^{2z}+1\right|}=\frac{\left|e^{x\left(a+bi\right)}\right|}{\left|\lambda e^{2z}+1\right|}=\frac{e^{ax}}{\left|\lambda e^{2z}+1\right|}\le \frac{e^{x\Re \left(z\right)}}{\left|\lambda e^{2z}\right|}\le \frac{1}{\left|\lambda \right|}.$$

Thus,

$$\left|{{\displaystyle \int }}_{C_N}^{ }\frac{2e^{xz}dz}{\left(\lambda e^{2z}+1\right)z^{n+1}}\right| \le \frac{2}{\left|\mathsf{\lambda }\right|}{{\displaystyle \int }}_{C_N}^{ }\frac{\left|dz\right|}{\left|z^{n+1}\right|}=\frac{2^{n+1}}{\left|\lambda \right|\left(2N-1+ϵ\right)\pi ){)}^n}.$$

As $N\to \infty ,$ the last expression goes to 0. Hence, as $N\to \infty ,$ $n\ge 1,$

$${{\displaystyle \int }}_{C_N}^{ }\frac{2e^{xz}}{\left(\lambda e^{2z}+1\right)z^{n+1}} dz\to 0.$$

This implies that

$$0=Res \left(f_n\left(z\right), z=0\right)+{\displaystyle \sum }_{kϵZ, }^{ }Res \left(f_n \left(z\right),z=z_k\right).$$

Now, the first residue $Res \left(f_n\left(z\right),z=0\right)$ is given as

$$\begin{array}{l}Res\left(f_n\left(z\right),z=0\right)=\underset{z\to 0}{\lim }\frac{1}{n!} \frac{d^n}{d{z}^n}{\left(z-0\right)}^{n+1}\frac{1}{z^{n+1}}\left(\frac{2e^{xz}}{\lambda e^{2z}+1}\right)\\ =\underset{z\to 0}{\lim }\frac{1}{n!} \frac{d^n}{d{z}^n} \left(\frac{2e^{xz}}{\lambda e^{2z}+1}\right)\\ =\underset{z\to 0}{\lim }\frac{1}{n!} \frac{d^n}{d{z}^n} \left({\displaystyle {\displaystyle \sum }_{l=0}^{\infty }}{T}_l\left(x;\lambda \right)\frac{z^l}{l!}\right)\\ =\underset{z\to 0}{\lim }\frac{1}{n!} {\displaystyle {\displaystyle \sum }_{l=n}^{\infty }}{T}_l\left(x;\lambda \right)\frac{z^{l-n}}{\left(l-n\right)!}.\end{array}$$

Note that the limit of each term of the expansion is $0$ as $z\to 0$ except the term when $l=n.$ This gives

$$Res \left(f_n\left(z\right),z=0\right)=\frac{T_n\left(x;\lambda \right)}{n!}.$$

On the other hand, the residue $Res \left(f_n\left(z\right), z=z_k\right)$ is given by

$$Res \left(f_n\left(z\right),z=z_k\right)=\underset{z\to z_k}{\lim }\left(z-z_k\right) \frac{1}{z^{n+1}}\left(\frac{2e^{xz}}{\lambda e^{2z}+1}\right)$$

$$\begin{gathered} =\frac{2e^{x{z}_k}}{z_k^{n+1}}\underset{z\to z_k}{\lim }\left(\frac{z-z_k}{\lambda e^{2z}+1}\right)=\frac{e^{\left(x-2\right)z_k}}{\lambda z_k^{n+1}}. \\ \mathrm{Since} z_k=\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \mathsf{\lambda }\right], \end{gathered}$$

$$\begin{gathered} \mathrm{Res} \left(f_n\left(z\right),z=z_k\right)=\frac{e^{\left(x-2\right)\left\{\frac{1}{2 }\left[\left(2k-1\right)\pi i-\log \lambda \right]\right\}}}{\lambda \left\{\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \lambda \right]\right\}{ }^{n+1}} \\ =\frac{-2^{n+1 } e{ }^{{}_2{}^1\left(2k-1\right)x\pi i}}{\lambda { }^{\frac{x}{2}} \left[\left(2k-1\right)\pi i-\log \lambda \right]{ }^{n+1}}. \end{gathered}$$

Combining these residues gives,

$$0=\frac{T_n\left(x ; \lambda \right)}{n!}+{\displaystyle \sum }_{\mathrm{k}∊\mathbb{Z}} \frac{-2^{n+1 } e{ }^{{}_2{}^1\left(2k-1\right)x\pi i}}{\lambda { }^{\frac{x}{2}} \left[\left(2k-1\right)\pi i-\log \lambda \right]{ }^{n+1}}$$

Hence,

$$T_n\left(x;\lambda \right)=\frac{2^{n+1} n!}{{\lambda }^{\frac{x}{2}}}{\displaystyle \sum }_{\mathrm{k}∊\mathbb{Z}}\frac{ e{ }^{{}_2{}^1\left(2k-1\right)x\pi i}}{ \left[\left(2k-1\right)\pi i-\log \lambda \right]{ }^{n+1}} .$$

□

Corollary 1. 

Let$\lambda \in ℂ\backslash \left\{0\right\}.$For$n\ge 1$, the Fourier series of the Apostol-tangent numbers is given by

$$\frac{T_n\left(0;\lambda \right)}{n!}={\displaystyle \sum }_{k\in \mathbb{Z}}^{ }\frac{1}{{\left[\frac{1}{2}\left(2k-1\right)\pi i-\log \lambda \right]}^{n+1}},$$

(8)

where the logarithm is taken to be the principal branch.

Proof. 

This follows from Theorem 1 by taking $x=0.$ □

Proceeding as in [20], ordering of the poles of the generating function Equation (6) is carried out in the following lemma.

Lemma 1. 

Let$u_k=\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \lambda \right]$with$k\in \mathbb{Z},\lambda \in ℂ\backslash \left\{0\right\}$and$\gamma =\left(\mathit{log}\lambda \right)/2\pi i$, where the logarithm is taken to be the principal branch.

(a)

If$Im \lambda >0, then 0<\Re \mathfrak{e}\gamma <\frac{1}{2}$, and for$k\ge 1,$

$$\left|u_1\right| < \left|u_0\right| < \left|u_2\right| < \left|u_{-1}\right| < \cdots < \left|u_{-k}\right| < \left|u_{k+2}\right| <$$

(9)

(b)

If$Im \lambda <0, then-\frac{1}{2}<\Re \mathfrak{e}\lambda <0,$and for$k\ge 1$,

$$\left|u_0\right| < \left|u_1\right| < \left|u_{-1}\right| < \left|u_2\right| < \left|u_{-2}\right| < \cdots < \left|u_{-k}\right| < \left|u_{k+1}\right| <$$

(10)

(c)

If$\lambda >0 \left(positive real number\right),$then$\Re \mathfrak{e} \gamma =0,$and for$k\ge 1,$

$$\begin{gathered} \left|u_0\right| = \left|u_1\right| < \left|u_{-1}\right| = \left|u_2\right| < \left|u_{-2}\right| = \left|u_3\right| < \left|u_{-3}\right| = \left|u_4\right| < \cdots < \\ \left|u_{-k}\right| = \left|u_{k+1}\right| < \left|u_{-\left(k+1\right)}\right| = \left|u_{k+2}\right| < \end{gathered}$$

(11)

(d)

If$\lambda <0$ $\left(negative real number\right), then \Re \mathfrak{e} \gamma = \frac{1}{2}$, and for$k\ge 1,$

$$\begin{gathered} \left|u_1\right| < \left|u_0\right| = \left|u_2\right| < \left|u_{-1}\right| = \left|u_3\right| < \left|u_{-2}\right| = \left|u_4\right| < \cdots < \\ \left|u_k\right| = \left|u_{-k+2}\right| =< \end{gathered}$$

(12)

Moreover, $\left|u_k\right| \ge \pi \left(\left|k\right|-1\right) if \left|k\right| \ge 1.$

Proof. 

With the logarithm taken to be the principal branch,$\gamma$ $\left(\mathrm{as} \mathrm{a} \mathrm{function} \mathrm{of} \lambda \right)$ maps $\lambda \in ℂ\backslash \left\{0\right\}$ to the strip $-\frac{1}{2}<\Re \mathfrak{e} \gamma \le \frac{1}{2}$ (see [20]). To see this, write

$$\gamma =\frac{\theta }{2\pi }-i\frac{ln\left|\lambda \right|}{2\pi }$$

where $\theta =Arg \lambda$, from which we have

$$\Re \mathfrak{e} \gamma =\frac{\theta }{2\pi } and \mathrm{Im} \gamma =-\frac{ln\left|\lambda \right|}{2\pi }.$$

with $-\pi <\theta \le \pi$, we have

$$\frac{-\pi }{2\pi }\le \Re \mathfrak{e} \gamma =\frac{\theta }{2\pi }\le \frac{\pi }{2\pi }⟹-\frac{1}{2}<\Re \mathfrak{e} \gamma \le \frac{1}{2}$$

where $\Re \mathfrak{e} \gamma =0$ when $\lambda >0$ and $\Re \mathfrak{e} \gamma =\frac{1}{2}$ when $\lambda <0$. If $\mathrm{Im} \lambda >0$, then $0<\theta <\pi$. Hence, $0<\Re \mathfrak{e} \gamma <\frac{1}{2}$. On the other hand, if $\mathrm{Im} \lambda <0$, then $-\pi <\theta <0$. Hence, $-\frac{1}{2}<\Re \mathfrak{e} \gamma <0$.

To verify the chains Equations (9)–(12), let $x=\Re \mathfrak{e} \gamma$ and $y=\mathrm{Im} \gamma$. Then for $k\in \mathbb{Z}$,

$$\begin{array}{ll}{u}_k& =\frac{1}{2}\left[\left(2k-1\right)\pi i-\log \left(\lambda \right)\right]\\ & =\left(k-\frac{1}{2}\right)\pi i-\frac{\log \left(\lambda \right)}{2}\\ & =\pi i\left[k-\frac{1}{2}-\frac{\log \left(\lambda \right)}{2\pi i}\right]\\ & =\pi i\left[k-\frac{1}{2}-\left(x+iy\right)\right]\\ & =\pi \left[i\left(k-\frac{1}{2}-x\right)+y\right]\\ & =\pi \sqrt{{\left(k-\frac{1}{2}-x\right)}^2+y^2}\end{array}$$

Now, we consider two cases:

Case 1. 

$Im \lambda >0.$Then$0<x<\frac{1}{2}$and

$$\begin{gathered} \left|u_0\right| = \pi \sqrt{{\left(-\frac{1}{2}-x\right)}^2+y^2} = \pi \sqrt{{\left(\frac{1}{2}+x\right)}^2+y^2} \\ \left|u_1\right| = \pi \sqrt{{\left(\frac{1}{2}-x\right)}^2+y^2} \\ \left|u_2\right| = \pi \sqrt{{\left(\frac{3}{2}-x\right)}^2+y^2} \\ \left|u_3\right| = \pi \sqrt{{\left(\frac{5}{2}-x\right)}^2+y^2} \\ \left|u_{-1}\right| = \pi \sqrt{{\left(-\frac{3}{2}-x\right)}^2+y^2} = \pi \sqrt{{\left(\frac{3}{2}+x\right)}^2+y^2} \\ \left|u_{-2}\right| = \pi \sqrt{{\left(-\frac{5}{2}-x\right)}^2+y^2} = \pi \sqrt{{\left(\frac{5}{2}+x\right)}^2+y^2} \\ \left|u_{-3}\right| = \pi \sqrt{{\left(-\frac{7}{2}-x\right)}^2+y^2} = \pi \sqrt{{\left(\frac{7}{2}+x\right)}^2+y^2} \end{gathered}$$

From this, one can see that the order of magnitude of $u_k, k\in \mathbb{Z}$ given in Equation (9) holds.

Case 2. 

$Im \lambda <0.$Thus,$-\frac{1}{2}<x<0$. The chain of values of$u_k$can be derived similarly, in which the order of magnitude of$u_k, k\in \mathbb{Z}$given in Equation (10) holds.

Case 3. 

$Im \lambda = 0.$This means that λ is a real number, which is either positive or negative but not zero. Hence, we have the following subcases:

Subcase 1. 

If$\lambda >0$, then$\Re \mathfrak{e} \gamma = 0$. For$k\ge 0$,

$$\begin{array}{l}\left|u_k\right| = \pi \sqrt{{\left(K-\frac{1}{2}\right)}^2+y^2}\\ \left|u_0\right| = \pi \sqrt{{\left(-\frac{1}{2}\right)}^2+y^2} = \pi \sqrt{{\left(\frac{1}{2}\right)}^2+y^2}=\left|u_1\right|\\ \left|u_2\right| = \pi \sqrt{{\left(\frac{3}{2}\right)}^2+y^2} = \pi \sqrt{{\left(-\frac{3}{2}\right)}^2+y^2} = \left|u_{-1}\right|\\ \left|u_3\right| = \pi \sqrt{{\left(\frac{5}{2}\right)}^2+y^2} = \pi \sqrt{{\left(-\frac{5}{2}\right)}^2+y^2} = \left|u_{-2}\right|\\ \left|u_4\right| = \pi \sqrt{{\left(\frac{7}{2}\right)}^2+y^2} = \pi \sqrt{{\left(-\frac{7}{2}\right)}^2+y^2} = \left|u_{-3}\right|\end{array}$$

and so on. Hence,

$$\begin{gathered} \left|u_0\right| = \left|u_1\right| < \left|u_{-1}\right| = \left|u_2\right| < \left|u_{-2}\right| = \left|u_3\right| < \left|u_{-3}\right| = \left|u_4\right| < \cdots < \\ \left|u_{-k}\right| = \left|u_{k+1}\right| < \left|u_{-\left(k+1\right)}\right| = \left|u_{k+2}\right| < \cdots , \end{gathered}$$

which is exactly (11).

Subcase 2. 

If$\lambda <0, \theta =\pi$, and hence,$x = \frac{1}{2}$. For$k\ge 0$,

$$\begin{gathered} \left|u_k\right| = \pi \sqrt{{\left(k-\frac{1}{2}-x\right)}^2+y^2}; \mathrm{when} x = \frac{1}{2} \\ =\pi \sqrt{{\left(k-1\right)}^2+y^2} \\ =\left|u_{-k+2}\right| \end{gathered}$$

from which it can easily be observed that

$$\begin{gathered} \left|u_1\right| < \left|u_0\right| = \left|u_2\right| < \left|u_{-1}\right| = \left|u_3\right| < \left|u_{-2}\right| = \left|u_4\right| < \cdots < \\ \left|u_k\right| = \left|u_{-k+2}\right| < \cdots \end{gathered}$$

which is exactly the chain in (12).

Moreover,

$$\begin{array}{ll}\left|u_k\right|& =\pi \left|k-\frac{1}{2}-\gamma \right|\\ & =\pi \sqrt{{\left(k-\frac{1}{2}-x\right)}^2+y^2}\\ & \ge \pi \sqrt{{\left(k-\frac{1}{2}-x\right)}^2}\\ & =\pi \left|k-\frac{1}{2}-x\right| \mathrm{with}-\frac{1}{2}\le x\le \frac{1}{2}\\ & =\pi \left|k-\left(x+\frac{1}{2}\right)\right|\\ & \ge \pi \left(\left|k\right|-\left|x+\frac{1}{2}\right|\right) \\ & \ge \pi \left(\left|k\right|-1\right). \end{array}$$

□

The asymptotic expansion of the Apostol-tangent numbers $T_n\left(0;\lambda \right)$ is given in the next theorem.

Theorem 2 

Given$\lambda \in ℂ\backslash \left\{0\right\},$let H be a finite subset of${\mathcal{T}}_{\lambda }$satisfying

$$\mathit{max}\left\{\left|u\right|:u\in H\right\} < \mathit{min}\left\{\left|u\right|:u\in {\mathcal{T}}_{\lambda }\backslash H\right\}: = v$$

for all integers$n\ge 2,$

$$\frac{T_n\left(0;\lambda \right)}{n!}={\displaystyle \sum }_{u\in H}\frac{1}{u^{n+1}}+O\left(v^{-\left(n+1\right)}\right)$$

Proof. 

Write the series (8) as ${\displaystyle \sum }_{k\in \mathbb{Z}} \frac{1}{{\left({\mu }_k\right)}^{n+1}}$. By Lemma 1, we can relabel the set of poles by increasing order of magnitude as

$$\left|{\mu }_0\right| \le \left|{\mu }_1\right| \le \left|{\mu }_2\right| \le \cdots \le \left|{\mu }_M\right| \le \cdots$$

Since $\left|{\mu }_k\right|\ge \pi \left(\left|k\right|-1\right)$, for $k\ge 2$, the series is absolutely convergent for $n\ge 2$. For any $M>2$, the tail of the series is

$${\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left|{\mu }_k\right|}^{n+1}}=\frac{1}{{\left|{\mu }_{M+1}\right|}^{n+1}}{\displaystyle \sum }_{k=M+1}^{\infty }\frac{{\left|{\mu }_{M+1}\right|}^{n+1}}{{\left|{\mu }_k\right|}^{n+1}}$$

Since $k>M+1, {\left|\frac{{\mu }_{M+1}}{{\mu }_k}\right|}^{n+1}\le 1,$ we have ${\left|\frac{{\mu }_{M+1}}{{\mu }_k}\right|}^{n+1}\le {\left|\frac{{\mu }_{M+1}}{k}\right|}^2$ for $n\ge$ 2. Hence,

$${\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left|{\mu }_k\right|}^{n+1}}\le \frac{1}{{\left|{\mu }_{M+1}\right|}^{n+1}}{\displaystyle \sum }_{k=M+1}^{\infty }{\left|\frac{{\mu }_{M+1}}{{\mu }_k}\right|}^2$$

Let

$$C_{M,\lambda }={\displaystyle \sum }_{k=M+1}^{\infty }{\left|\frac{{\mu }_{M+1}}{{\mu }_k}\right|}^2$$

Then,

$${\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left|{\mu }_k\right|}^{n+1}}\le \frac{1}{{\left|{\mu }_{M+1}\right|}^{n+1}}\left[C_{M,\lambda }\right]=\frac{C_{M,\lambda }}{{\left|{\mu }_{M+1}\right|}^{n+1}}$$

Now, consider $C_{M,\lambda }:$

$$\begin{gathered} C_{M,\lambda }={\displaystyle \sum }_{k=M+1}^{\infty }{\left|\frac{{\mu }_{M+1}}{{\mu }_k}\right|}^2 \\ ={\displaystyle \sum }_{k=M+1}^{\infty }\frac{{\left|{\mu }_{M+1}\right|}^2}{{\left|{\mu }_k\right|}^2} \\ ={\left|{\mu }_{M+1}\right|}^2{\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left|{\mu }_k\right|}^2} \end{gathered}$$

Since

$$\begin{gathered} \left|{\mu }_k\right|=\pi \left|k-\frac{1}{2}-\gamma \right|\ge \pi \left(\left|k\right|-1\right) \\ \left|{\mu }_{M+1}\right|=\pi \left|M+1-\frac{1}{2}-\gamma \right|\ge \pi \left(\left|M+1\right|-1\right) \\ =\pi \left|M+\frac{1}{2}-\gamma \right| \end{gathered}$$

Then

$$\begin{gathered} C_{M,\lambda }={\left|M+\frac{1}{2}-\gamma \right|}^2 {\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left|k-\frac{1}{2}-\gamma \right|}^2} \\ \le 2{\left|M+\frac{1}{2}-\gamma \right|}^2{\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left(\left|k\right|-1\right)}^2} \\ \le 2{\left|M+\frac{1}{2}-\gamma \right|}^2\left(\frac{1}{M^2}+{\displaystyle \sum }_{\mathcal{l}=0}^{\infty }\frac{1}{{\left(M+\mathcal{l}\right)}^2}\right) \end{gathered}$$

With

$${\displaystyle \sum }_{\mathcal{l}=0}^{\infty }\frac{1}{{\left(M+\mathcal{l}\right)}^2}\le \underset{1}{\overset{\infty }{{\displaystyle \int }}}\frac{1}{{\left(M+x\right)}^2}dx=\frac{1}{M+1}$$

So,

$$\begin{gathered} C_{M,\lambda } \le 2{\left|M+\frac{1}{2}-\gamma \right|}^2\left(\frac{1}{M^2}+\frac{1}{M+1}\right) \\ =\frac{2{\left|M+\frac{1}{2}-\gamma \right|}^2}{M^2}+\frac{2{\left|M+\frac{1}{2}-\gamma \right|}^2}{M+1} \end{gathered}$$

Let

$${\xi }_1=\frac{{\left|M+\frac{1}{2}-\gamma \right|}^2}{M^2}\le {\left|\frac{3}{2}-\gamma \right|}^2$$

And

$${\xi }_2=\frac{{\left|M+\frac{1}{2}-\gamma \right|}^{ }}{M+1}\le \frac{\left|M+1\right|+\left|-\frac{1}{2}-\gamma \right|}{\left|M+1\right|}\le 1+\left|\frac{1}{2}+\gamma \right|.$$

Consequently,

$$\begin{gathered} C_{M,\lambda }=2{\xi }_1+{\xi }_2\left|M+\frac{1}{2}-\gamma \right| \\ \frac{C_{M,\lambda }}{{\left|{\mu }_{M+1}\right|}^{n+1}}\le \frac{2{\xi }_1}{{\left|{\mu }_{M+1}\right|}^{n+1}}+\frac{{\xi }_2\left|M+\frac{1}{2}-\gamma \right|}{{\left|{\mu }_{M+1}\right|}^{n+1}} \end{gathered}$$

where

$$\left|{\mu }_{M+1}\right|=\pi \left|M+\frac{1}{2}-\gamma \right|=\sqrt{{\left(M+\frac{1}{2}-Re \gamma \right)}^2+{\left(Im \gamma \right)}^2 }\ge M$$

So,

$$\begin{gathered} \frac{C_{M,\lambda }}{{\left|{\mu }_{M+1}\right|}^{n+1}}\le \frac{2{\left|\frac{3}{2}-\gamma \right|}^2}{{\pi }^{n+1}{\left|M+\frac{1}{2}-\gamma \right|}^{n+1}}+\frac{2\left(1+\left|\frac{1}{2}+\gamma \right|\right)\left|M+\frac{1}{2}-\gamma \right|}{{\pi }^{n+1}{\left|M+\frac{1}{2}-\gamma \right|}^{n+1}} \\ \le \frac{2{\left|\frac{3}{2}-\gamma \right|}^2}{{\pi }^{n+1}{\left|M+\frac{1}{2}-\gamma \right|}^{n+1}}+\frac{2\left(1+\left|\frac{1}{2}+\gamma \right|\right)}{{\pi }^{n+1}{\left|M+\frac{1}{2}-\gamma \right|}^n} \\ \le \frac{2{\left|\frac{3}{2}-\gamma \right|}^2}{{\pi }^{n+1}{\left|M\right|}^{n+1}}+\frac{2\left(1+\left|\frac{1}{2}+\gamma \right|\right)}{{\pi }^{n+1}{\left|M\right|}^n} \\ \le \frac{2{\left|\frac{3}{2}-\gamma \right|}^2}{{\pi }^{n+1}{\left|M\right|}^{n+1}}+\frac{2\left(1+\left|\frac{1}{2}+\gamma \right|\right)}{{\pi }^{n+1}{\left|M\right|}^{n+1}} \\ \le \frac{2{\left|\frac{3}{2}-\gamma \right|}^2}{{\pi }^{n+1}}+\frac{2\left(1+\left|\frac{1}{2}+\gamma \right|\right)}{{\pi }^{n+1}} \end{gathered}$$

We can see that $C_{M,\lambda }\to 0$ as $n\to \infty$ for $\left|M\right| > 2.$ Thus, the tail of the series is

$${\displaystyle \sum }_{k=M+1}^{\infty }\frac{1}{{\left|{\mu }_k\right|}^{n+1}}\to 0 \mathrm{as} n\to \infty .$$

Moreover, for fixed $M>2$ and $n\gg 0$, $C_{M,\lambda }$ is bounded and independent of M. Hence, we can replace $C_{M, \lambda }$ with $C_{\lambda }$. This completes the proof of the theorem. □

When $\lambda =1,$ $\log \lambda =0,$ and $u_k=\frac{1}{2}\left(2k-1\right)\pi i, k\in \mathbb{Z}$. Take $H=\left\{\frac{\pi i}{2}, \frac{-\pi i}{2}\right\}$. Then $v=\frac{3\pi }{2}$, and the ordinary tangent numbers $T_n=T_n\left(0;1\right)$ satisfy

$$\begin{array}{l}\frac{T_n}{n!}=\frac{T_{n }\left(0;1\right)}{n!}\\ =\frac{1}{{\left(\frac{\pi i}{2}\right)}^{n+1}}+\frac{1}{{\left(\frac{-\pi i}{2}\right)}^{n+1}}+O\left({\left(\frac{3\pi }{2}\right)}^{-\left(n+1\right)}\right)\end{array}$$

(13)

An approximation of $T_n\left(0;1\right)$ is given by

$$\frac{T_n}{n!}\approx \frac{2^{n+1}}{{\left(\pi i\right)}^{n+1}}+\frac{2^{n+1}}{{\left(-\pi i\right)}^{n+1}}$$

(14)

For even $n, n\ge 2$, it is known that $T_n=0$, which is also true when we use Equation (14). Then, we have

$$\frac{T_{2n}}{\left(2n!\right)}\approx \frac{2^{2n+1}}{{\left(\pi i\right)}^{2n+1}}+\frac{2^{2n+1}}{{\left(-\pi i\right)}^{2n+1}}=0.$$

For odd indices,

$$\begin{gathered} \frac{T_{2n-1}}{\left(2n-1\right)!}\approx \frac{2^{2n}}{{\left(\pi i\right)}^{2n}}+\frac{2^{2n}}{{\left(-\pi i\right)}^{2n}} \\ \approx \frac{2^{2n+1}}{{\left(\pi i\right)}^{2n}} \\ {T}_{2n-1}\approx \frac{{\left(-1\right)}^n{2}^{2n+1}\left(2n-1\right)!}{{\pi }^{2n}} , n\ge 1. \end{gathered}$$

(15)

$$\begin{gathered} T_{2\left(3\right)-1}\approx \frac{{\left(-1\right)}^3{2}^{2\left(3\right)+1}\left(2\left(3\right)-1\right)!}{{\pi }^{2\left(3\right)}}\approx \frac{-2^{7}5!}{\pi 6} \\ {T}_5\approx -15.97688023. \end{gathered}$$

This value is very close to the exact value of $T_5$ which is $-16.$

It is proved in the next theorem that an asymptotic approximation of the Apostol-tangent polynomials can be obtained from its Fourier series (Theorem 1) by choosing an appropriate subset of ${\mathcal{T}}_{\lambda }.$

Theorem 3. 

Given$\lambda \in ℂ\backslash \left\{0\right\},$let$H$be a finite subset of${\mathcal{T}}_{\lambda }$satisfying

$$\mathit{max}\left\{\left|u\right|:u\in H\right\}<\mathit{min}\left\{\left|u\right|:u\in {\mathcal{T}}_{\lambda } \backslash H\right\}≔v.$$

For all integers$n\ge 2$, we have uniformly for$x$in a compact subset$K$of$ℂ$,

$$\frac{T_n\left(x;\lambda \right)}{n!}={\displaystyle \sum }_{u \in H}\frac{e^{ux}}{u^{n+1}}+O\left(\frac{e^{v\left|x\right|}}{v^{n+1}}\right),$$

where the constant implicit in the order term depends on$\lambda , H$and$K.$Moreover, for$n\gg 0,$this constant can be made independent of$K$, equal to the constant for the Apostol-tangent numbers, corresponding to the case$x=0.$

Proof. 

From the generating function in Equation (6), we have

$$\frac{2e^{\left(x+y\right)z}}{\lambda e^{2z}+1}={\displaystyle \sum }_{n=0}^{\infty }{T}_n\left(x+y;\lambda \right)\frac{z^n}{n!}.$$

The left-hand side of the equation can be written as

$$\begin{gathered} \frac{2e^{\left(x+y\right)z}}{\lambda e^{2z}+1}=\frac{2e^{xz}{e}^{yz}}{\lambda e^{2z}+1}={\displaystyle \sum }_{n=0}^{\infty }{T}_n\left(x;\lambda \right)\frac{z^n}{n!}{e}^{yz} \\ =\left({\displaystyle \sum }_{n=0}^{\infty }{T}_n\left(x;\lambda \right)\frac{z^n}{n!} \right)\left({\displaystyle \sum }_{n=0}^{\infty }\frac{{\left(yz\right)}^n}{n!}\right) \\ ={\displaystyle \sum }_{n=0}^{\infty }{\displaystyle \sum }_{k=0}^n{T}_{n-k}\left(x;\lambda \right)\frac{z^{n-k}}{\left(n-k\right)!}·\frac{{\left(yz\right)}^k}{k!}·\frac{n!}{n!} \\ ={\displaystyle \sum }_{n=0}^{\infty }\left({\displaystyle \sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)T_{n-k}\left(x;\lambda \right)y^k\right)\frac{z^n}{n!}. \end{gathered}$$

Hence,

$$T_n\left(x+y;\lambda \right)={\displaystyle \sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)T_{n-k}\left(x;\lambda \right)y^k.$$

For $z\in ℂ$, writing $z=0+z ($ here $y=z, x=0)$,

$$\begin{gathered} T_n\left(z;\lambda \right)={\displaystyle \sum }_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)T_{n-k}\left(0, \lambda \right)z^k \\ ={\displaystyle \sum }_{k=0}^n\frac{n!}{\left(n-k\right)!k!}{T}_{n-k}\left(0, \lambda \right)z^k \\ \frac{T_n\left(z;\lambda \right)}{n!}={\displaystyle \sum }_{k=0}^n\frac{T_{n-k}\left(0, \lambda \right)z^k}{\left(n-k\right)!k!} \\ ={\displaystyle \sum }_{k=0}^n\frac{T_{n-k}\left(0,\lambda \right)}{\left(n-k\right)!} \frac{z^k}{k!} \\ ={\displaystyle \sum }_{k=0}^n\left({\displaystyle \sum }_{u\in H}\frac{1}{u^{n-k+1}}+O\left(v^{-\left(n-k+1\right)}\right)\right)\frac{z^k}{k!} \\ ={\displaystyle \sum }_{k=0}^n\left({\displaystyle \sum }_{u\in H}\frac{1}{u^{n-k+1 }} \frac{z^k}{k!}\right)+{\displaystyle \sum }_{k=0}^{n}O\left(v^{-\left(n-k+1\right)}\right)\frac{z^k}{k!}, \end{gathered}$$

where the implicit constant $c$ in the order term is that corresponding to $z=0$ and only depends on $H$ and $\lambda$. Note also that

$$\begin{gathered} \left|{\displaystyle \sum }_{k=0}^{n}O\left(v^{-\left(n-k+1\right)}\right)\frac{z^k}{k!}\right|\le {\displaystyle \sum }_{k=0}^{n}c{v}^{-\left(n-k+1\right)}\frac{\left|z^k\right|}{k!} \\ =c{v}^{-\left(n+1\right)}{\displaystyle \sum }_{k=0}^n\frac{v^k\left|z^k\right|}{k!} \\ \le c{v}^{-\left(n+1\right)}{e}_n\left(v\left|z\right|\right), \end{gathered}$$

where

$$e_n={\displaystyle \sum }_{k=0}^n\frac{w^k}{k!}$$

To complete the proof of the theorem, it remains to show that

$$\frac{e_n^{\ast }\left(uz\right)}{u^{n+1}}=\frac{e^{uz}-e_n\left(uz\right)}{u^{n+1}}$$

is bounded. Using the Mean Value Theorem (MVT) for Banach spaces (see also [20]), we have

$$\begin{gathered} e_n^{\ast }\left(w\right) = \frac{w^{n+1}}{\left(n+1\right)!}+\frac{w^{n+2}}{\left(n+2\right)!}+\cdots \\ =\frac{w^{n+1}}{\left(n+1\right)!}\left\{1+\frac{w}{n+2}+\frac{w^2}{\left(n+3\right)\left(n+2\right)}+\cdots \right\} \end{gathered}$$

from which

$$\begin{array}{ll}\left|e_n^{\ast }\left(w\right)\right|& \le \left|\frac{w^{n+1}}{\left(n+1\right)!}\right|\left|1+\frac{w}{n+2}+\frac{w^2}{\left(n+3\right)\left(n+2\right)}+\cdots \right|\\ & \le \frac{{\left|w\right|}^{n+1}}{\left(n+1\right)!}{e}^{\Re {\mathfrak{e}}^{+}\left(w\right)}\end{array}$$

where $\Re {\mathfrak{e}}^{+}\left(w\right) = \max \left\{\Re \mathfrak{e}\left(w\right),0\right\}.$ Since $\left|u\right| \le v, \mathrm{for} \mathrm{all} u\in H, \mathrm{we} \mathrm{have}$

$$\begin{gathered} \frac{\left|e^{\ast }\left(uz\right)\right|}{\left|u^{n+1}\right|}\le \frac{e^{\left|uz\right|}{\left|uz\right|}^{n+1}}{\left|u^{n+1}\right|\left(n+1\right)!} \\ =\frac{e^{\left|uz\right|}\left|z^{n+1}\right|\left|u^{n+1}\right|}{\begin{array}{c}\left|u^{n+1}\right|\left(n+1\right)!\\ \end{array}} \\ =\frac{e^{\left|uz\right|}{\left|z\right|}^{n+1}}{\left(n+1\right)!} \\ <\frac{e^{v\left|z\right|}{\left|z\right|}^{n+1}}{\left(n+1\right)!} \end{gathered}$$

so that

$$\begin{gathered} \left|{\displaystyle \sum }_{u\in H}\frac{e^{\ast }\left(uz\right)}{u^{n+1}}\right| \le {\displaystyle \sum }_{u\in H} \frac{\left|e_n^{\ast }\left(uz\right){ }^{ }\right|}{\left|u^{n+1}\right|} \\ \le \left|H\right|e^{v\left|z\right|}\frac{{\left|z\right|}^{n+1}}{\left(n+1\right)!} \end{gathered}$$

where $\left|H\right|$ denotes the number of elements in $H$. We give the argument that

$$\left|H\right|e^{v\left|z\right|}\frac{{\left|z\right|}^{n+1}}{\left(n+1\right)!}<c e_{ }^{v\left|z\right|}{v}^{-\left(n+1\right)}$$

If

$$\left|H\right|\frac{{\left(v\left|z\right|\right)}^{n+1}}{n+1}<c,$$

which certainly holds for $n\gg 0,$ uniformly for $z$ in a compact subset $K\subset ℂ$. □

Corollary 2. 

Let K be an arbitrary compact subset of$ℂ.$The tangent polynomials satisfy uniformly on K the estimates

$$\begin{gathered} \frac{T_{2n }\left(x\right)}{\left(2n\right)!}=\frac{\left(-1\right){ }^n 2^{2n+2 }\sin \left(\frac{\pi x}{2}\right)}{{\pi }^{2n+1}}+O \left(\frac{e^{\frac{3\pi }{2} \left|x\right|}}{\left(\frac{3\pi }{2}\right){ }^{2n+1}}\right) \\ \frac{T_{2n-1 }\left(x\right)}{\left(2n-1\right)!}=\frac{\left(-1\right){ }^n 2^{n+1 }\cos \left(\frac{\pi x}{2}\right)}{{\pi }^{2n}}+O\left(\frac{e^{\frac{3\pi }{2} \left|x\right|}}{\left(\frac{3\pi }{2}\right){ }^{2n}}\right) \end{gathered}$$

where the implicit constant in the order term depends on the set K. Moreover, for$n \gg 0$, this constant can be made independent of K, equal to the constant for the tangent numbers, corresponding to the case$x=0.$

Proof. 

The tangent polynomials correspond to the case $\lambda =1$ so that

$$u_k=\frac{1}{2 }\left(2k-1\right)\pi i, k\in \mathbb{Z}.$$

Thus

$${\mathcal{T}}_1=\left\{\frac{1}{2}\left(2k-1\right)\pi i:k\in \mathbb{Z}\right\}$$

Taking

$$H=\left\{\left(2k-1\right)\pi i: k=-1,0 \right\}=\left\{-\frac{\pi i}{2},\frac{\pi i}{2}\right\},$$

then $v=\frac{\left|3\pi i\right|}{2}=\frac{3\pi }{2}$. From Theorem 3,

$$\begin{gathered} \frac{T_n\left(x;1\right)}{n!}={\displaystyle \sum }_{uϵH}\frac{e^{ux}}{u^{n+1}}+O\left(\frac{e^{v\left|x\right|}}{v^{n+1}}\right) \\ =\left(\frac{e^{-\frac{\pi ix}{2}}}{{\left(-\frac{\pi i}{2}\right)}^{n+1}}+\frac{e^{\frac{\pi ix}{2}}}{{\left(\frac{\pi i}{2}\right)}^{n+1}}\right)+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{n+1}}\right). \end{gathered}$$

For even indices,

$$\begin{array}{l}\frac{T_{2n^{ }}^{ }\left(x\right)}{\left(2n\right)!} = \frac{T_{2n}\left(x;1\right)}{\left(2n\right)!}=\left(\frac{e^{-\frac{\pi ix}{2}}}{{\left(-\frac{\pi i}{2}\right)}^{2n+1}}+\frac{e^{\frac{\pi ix}{2}}}{{\left(\frac{\pi i}{2}\right)}^{2n+1}}\right)+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{\begin{array}{c}2n+1\\ \end{array}}}\right)\\ =\frac{2^{2n+1}{e}^{\frac{\pi ix}{2}}}{{\left(\pi i\right)}^{2n+1}}-\frac{2^{n+1}{e}^{-\frac{\pi ix}{2}}}{{\left(\pi i\right)}^{2n+1}}+O\left(\frac{e^{\frac{3\pi }{2}}{}^{{ }_{ \left|x\right|}}}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right)\\ =\frac{2^{2n+1}\left(\cos \frac{\pi x}{2}+i\sin \frac{\pi x}{2}\right)-2^{n+1}\left(\cos \frac{\pi x}{2}-i\sin \frac{\pi x}{2}\right)}{{\left(\pi i\right)}^{2n+1}}+O\left(\frac{e^{\frac{3\pi }{2}}\left|x\right|}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right)\\ =\frac{2^{n+1}\left(\cos \frac{\pi x}{2}+i\sin \frac{\pi x}{2}-\cos \frac{\pi x}{2}+i\sin \frac{\pi x}{2}\right)}{{\left(\pi i\right)}^{2n+1}}+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right)\\ =\frac{2^{n+1}2i\sin \frac{\pi x}{2}}{{\left(\pi i\right)}^{2n+1}}+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right)\\ =\frac{2^{n+2}i\sin \left(\frac{\pi x}{2}\right)}{{\pi }^{2n+1}{i}^{2n+1}}+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right)\\ =\frac{2^{2n+2}\sin \left(\frac{\pi x}{2}\right)}{{\pi }^{2n+1}{\left(-1\right)}^n}+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right)\\ =\frac{{\left(-1\right)}^n{2}^{2n+2}\sin \left(\frac{\pi x}{2}\right)}{{\pi }^{2n+1}}+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n+1}}\right).\end{array}$$

For odd indices,

$$\begin{array}{l}\frac{T_{2n-1}\left(x\right)}{\left(2n-1\right)!} = \frac{T_{2n-1}\left(x;1\right)}{\left(2n-1\right)!}=\left(\frac{e^{\frac{-\pi ix}{2}}}{{\left(\frac{-\pi i}{2}\right)}^{2n}}+\frac{e^{\frac{\pi ix}{2}}}{{\left(\frac{\pi i}{2}\right)}^{2n}}\right)+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n}}\right)\\ =\left(\frac{e^{\frac{-\pi ix}{2}}}{{\left(\frac{\pi i}{2}\right)}^{2n}}+\frac{e^{\frac{\pi ix}{2}}}{{\frac{\left(\pi i\right)}{2}}^{2n}}\right)+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n}}\right)\\ =\left(\frac{2^n{e}^{\frac{-\pi ix}{2}}}{{\left(\pi i\right)}^{2n}}+\frac{2^n{e}^{\frac{\pi ix}{2}}}{{\left(\pi i\right)}^{2n}}\right)+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{\begin{array}{c}2n\\ \end{array}}}\right)\\ =\frac{2^n\left(\cos \frac{\pi x}{2}-i\sin \frac{\pi x}{2}\right)+2^n\left(\cos \frac{\pi x}{2}+i\sin \frac{\pi x}{2}\right)}{{\left(\pi i\right)}^{2n}}+O\left(\frac{e^{\frac{3\pi }{2}\left|x\right|}}{{\left(\frac{3\pi }{2}\right)}^{2n}}\right)\\ = \frac{2^n\left(\cos \frac{\pi x}{2}-i \sin \frac{\pi x}{2}+\cos \frac{\pi x}{2}+i \sin \frac{\pi x}{2}\right)}{\left(\pi i\right){ }^{2n}}+O \left(\frac{e^{\frac{3\pi }{2} \left|x\right|}}{\left(\frac{3\pi }{2}\right){ }^{2n}{ }^{ }}\right)\\ = \frac{2^n 2 \cos \left(\frac{\pi x}{2}\right)}{\left(\pi i\right){ }^{2n}}+O \left(\frac{e^{\frac{3\pi }{2} \left|x\right|}}{\left(\frac{3\pi }{2}\right){ }^{2n}}\right)\\ = \frac{2^{n+1} \cos \left(\frac{\pi x}{2}\right)}{{\pi }^{2n} \left(i^2\right){ }^n}+O \left(\frac{e^{\frac{3\pi }{2} \left|x\right|}}{\left(\frac{3\pi }{2}\right){ }^{2n}}\right)\\ = \frac{\left(-1\right){ }^n 2^{n+1 }\cos \left(\frac{\pi x}{2}\right)}{{\pi }^{2n}}+O\left(\frac{e^{\frac{3\pi }{2} \left|x\right|}}{\left(\frac{3\pi }{2}\right){ }^{2n}}\right). \end{array}$$

□

3. The Case When $\lambda$ Is a Negative Real Number

When $\lambda$ is a negative real number, writing $\lambda =-\left|\lambda \right|$, the generating function in Equation (6) can be written as

$$\frac{2e^{xt}}{-\left|\lambda \right|e^{2t }+1}={\displaystyle \sum }_{n=0}^{\infty }{T}_{n }\left(x;-\left|\lambda \right|\right) \frac{t^n}{n!}$$

(16)

The poles of the generating function (3.1) is given by

$${\mathcal{T}}_{-\left|\lambda \right|}=\left\{\frac{1}{2}\left[\left(2k+1\right)\pi i+\ln \left|\lambda \right|\right]:k\in \mathbb{Z}\right\}.$$

The next theorem immediately follows from Theorem 3.

Theorem 4. 

Given that$\lambda$is a negative real number, let$F$be a finite subject of${\mathcal{T}}_{-\left|\lambda \right|}$satisfying

$$\mathit{max} \left\{\left|a\right|:a\in F\right\} < \mathit{min} \left\{\left|a\right|:a \in {\mathcal{T}}_{-\left|\lambda \right|}\backslash F\right\}: = \xi .$$

For all integers$n \ge 2,$we have uniformly for$x$in a compact subset$K of ℂ,$

$$\frac{T_n \left(x ; \lambda \right)}{n !}={\displaystyle \sum }_{a \in F}^{ }\frac{e^{ax}}{a^{n+1}}+O \left( \frac{e^{\xi \left|x\right|}}{{\xi }^{n+1}}\right),$$

(17)

where the constant implicit in the order term depends on$\lambda , F, and K.$

The Apostol-tangent numbers $T_{n }\left(0;-1\right)$ corresponding to the case $\lambda =-1$ have the generating function

$$\frac{2}{-e^{2t}+1}={\displaystyle \sum }_{n=0}^{\infty }{T}_n \left(0;-1\right) \frac{t^n}{n!} ,$$

(18)

The set of poles is ${\mathcal{T}}_{-1} = \left\{k\pi i :k\in \mathbb{Z}\backslash \left\{0\right\}\right\}$. An asymptotic formula for $T_n \left(0;-1\right)$ is given in the following theorem.

Theorem 5. 

For$n\ge 3,$the Apostol-tangent numbers$T_n \left(0;-1\right)$satisfying

$$\frac{T_n \left(0;-1\right)}{n!}=\left(\frac{1}{\left(-\pi i\right){ }^{n+1}}+\frac{1}{\left(\pi i\right){ }^{n+1}}\right)+O \left(\left(2\pi \right){ }^{-\left(n+1\right)}\right).$$

(19)

In particular,

$$\frac{T_{2n-1 }\left(0;-1\right)}{\left(2n-1\right)!}=\frac{\left(-1\right){ }^n 2}{\pi { }^{2n}}+O \left(\left(2\pi \right){ }^{-\left(2n\right)}\right).$$

(20)

Proof. 

Taking $\mathrm{x}=0, \mathrm{F} = \left\{-\mathsf{\pi }\mathrm{i}, \mathsf{\pi }\mathrm{i}\right\}$ in Theorem 4, then $\xi =2\pi .$

Hence,

$$\begin{gathered} \frac{T_n \left(x ; \lambda \right)}{n!}={\displaystyle \sum }_{a\in F}\frac{e^{ax}}{a^{n+1}}+O \left(\frac{e^{\xi \left|x\right|}}{{\xi }^{\begin{array}{c}n+1\\ \end{array}}}\right) \\ \frac{T_n\left(0;-1\right)}{n!}=\left(\frac{1}{\left(-\pi i\right){ }^{n+1}}+\frac{1}{\left(\pi i\right){ }^{n+1}}\right)+O \left(\left(2\pi \right){ }^{-\left(n+1\right)}\right) \end{gathered}$$

(21)

for which Equation (19) follows. For $n \ge 2, \left(21\right)$ gives $T_{2n} \left(0;-1\right) \approx 0.$ Indeed $T_{2n} \left(0;-1\right) = 0$

$\forall n \ge 1.$ For $n\ge 1,$

$$\begin{gathered} \frac{{\begin{array}{c} \\ T\end{array}}_{2n-1} \left(0;-1\right)}{\left(2n-1\right)!}=\left(\frac{1}{\left(-\pi i\right){ }^{2n}}+\frac{1}{\left(\pi i\right){ }^{2n}}\right)+O \left(\left(2\pi \right){ }^{-\left(n+1\right)}\right). \\ =\left(\frac{1}{\left(\pi i\right){ }^{2n}}+\frac{1}{\left(\pi i\right){ }^{2n}}\right)+O \left(\left(2\pi \right){ }^{-\left(n+1\right)}\right) \\ =\frac{2}{\left(\pi i\right){ }^{2n}}+O \left(\left(2\pi \right){ }^{-\left(\begin{array}{c}n+1\\ \end{array}\right)}\right) \\ =\left(\frac{2}{\left(\pi \right){ }^{2n }\left(i\right){ }^{2n}}\right)+ O \left(\left(2\pi \right){ }^{-\left(\begin{array}{c}n+1\\ \end{array}\right)}\right) \\ =\frac{\left(-1\right){ }^n 2}{\left(\pi \right){ }^{2n}}+O \left(\left(2\pi \right){ }^{-\left(\begin{array}{c}n+1\\ \end{array}\right)}\right). \end{gathered}$$

□

Taking $n=4,$

$$T_7\left(0;-1\right) = \frac{2\left(7!\right)}{\left(\pi \right){ }^8} \approx 1.06233.$$

The actual value of $T_7\left(0;-1\right) = -\frac{2^8{B}_8}{8}=\frac{16}{15} \approx 1.06667$.

The Apostol-tangent polynomials $T_{n }\left(x;-1\right)$ correspond to the case $\lambda = -1.$ These polynomials have the generating function

$$\frac{2 e^{xt}}{-e^{2t}+1}={\displaystyle \sum }_{n=0}^{\infty }{T}_n \left(x;-1\right) \frac{t^n}{n!}$$

(22)

We will prove the following theorem.

Theorem 6. 

Let$K$be a compact subset of$ℂ$. The Apostol-tangent polynomials$T_n \left(x;-1\right)$satisfy uniformly on$K$the estimates

$$\begin{gathered} \frac{T_{2n} \left(x;-1\right)}{\left(2n\right)!}= \frac{\left(-1\right){ }^n 2 \mathit{sin}\left(\pi x\right)}{{\pi }^{2n+1}}= O\left(\frac{e^{2\pi \left|x\right|}}{\left(2\pi \right){ }^{2n+1}}\right) \\ \frac{T_{2n-1 }\left(x;-1\right)}{\left(2n-1\right)!}=\frac{{\left(-1\right)}^n 2 \mathit{cos}\left(\pi x\right)}{{\pi }^{2n}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right). \end{gathered}$$

Proof. 

Taking $\mathrm{F} = \left\{-\mathsf{\pi }\mathrm{i}, \mathsf{\pi }\mathrm{i}\right\}$, then $\mathsf{\xi } = 2\mathsf{\pi }$. Hence, it follows from Theorem 4 that

$$\frac{T_{n }\left(x;-1\right)}{n!}=\left(\frac{e^{\pi ix}}{{\left(\pi i\right)}^{n+1}}+\frac{e^{-\pi ix}}{{\left(-\pi i\right)}^{n+1}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{n+1}}\right).$$

For even indices,

$$\begin{gathered} \frac{T_{2n }\left(x;-1\right)}{\left(2n\right)!}=\left(\frac{e^{\pi ix}}{{\left(\pi i\right)}^{2n+1}}+\frac{e^{-\pi ix}}{{\left(-\pi i\right)}^{2n+1}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n+1}}\right) \\ =\left(\frac{e^{\pi ix}}{{\left(\pi i\right)}^{2n+1}}-\frac{e^{-\pi ix}}{{\left(\pi i\right)}^{2n+1}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n+1}}\right) \\ =\left(\frac{\cos \pi x+i\sin \pi x-\cos \pi ix+i\sin \pi x}{{\left(\pi i\right)}^{2n+1}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n+1}}\right) \\ =\frac{2i\sin \pi x}{{\left(\pi i\right)}^{2n+1}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n+1}}\right)=\frac{2\sin \pi x}{{\pi }^{2n+1}{\left(i\right)}^{2n}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n+1}}\right) \\ =\frac{{\left(-1\right)}^n 2 sin\left(\pi x\right)}{{\pi }^{2n+1}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n+1}}\right). \end{gathered}$$

For odd indices,

$$\begin{gathered} \frac{T_{2n-1 }\left(x;-1\right)}{\left(2n-1\right)!}=\left(\frac{e^{\pi ix}}{{\left(\pi i\right)}^{2n}}+\frac{e^{-\pi ix}}{{\left(-\pi i\right)}^{2n}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right) \\ =\left(\frac{e^{\pi ix}}{{\left(\pi i\right)}^{2n}}+\frac{e^{-\pi ix}}{{\left(\pi i\right)}^{2n}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right) \\ =\left(\frac{\cos \pi x+i\sin \pi x+\cos \pi x-i\sin \pi x}{{\left(\pi i\right)}^{2n}}\right)+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right) \\ =\frac{2\cos \pi x}{{\left(\pi i\right)}^{2n}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right) \\ =\frac{2\cos \pi x}{{\pi }^{2n}{\left(i\right)}^{2n}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right) \\ =\frac{{\left(-1\right)}^n 2\cos \left(\pi x\right)}{{\pi }^{2n}}+O\left(\frac{e^{2\pi \left|x\right|}}{{\left(2\pi \right)}^{2n}}\right). \end{gathered}$$

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4. Conclusions and Recommendation

The method of Navas et al. [24] is a clever way to obtain an asymptotic approximation from the Fourier series. In this paper, the method was applied to obtain asymptotic approximations of the Apostol-tangent numbers and polynomials for nonzero complex values of the parameter $\lambda$. The case when $\lambda$ is negative was explicitly considered because the poles are simply in terms of $\frac{1}{2}\ln \left|\lambda \right|$ plus odd multiples of $\frac{\pi }{2}i$. Moreover, the cases $\lambda =1$ and $\lambda =-1$ give beautiful approximations of the corresponding Tangent polynomials in terms of the sine and cosine functions depending on whether $n$ is even or odd.

The author recommends finding Fourier expansion and asymptotic approximations of higher-order Apostol-Tangent numbers and polynomials using the method employed in this paper (see also [25]). Furthermore, one may also try to consider multiple generalized Tangent polynomials and their p-adic interpolation function [26].