Theory and Modern Applications

# Fourier expansions for higher-order Apostol–Genocchi, Apostol–Bernoulli and Apostol–Euler polynomials

## Abstract

Fourier expansions of higher-order Apostol–Genocchi and Apostol–Bernoulli polynomials are obtained using Laurent series and residues. The Fourier expansion of higher-order Apostol–Euler polynomials is obtained as a consequence.

## Introduction

Higher-order Apostol–Genocchi, Apostol–Bernoulli, and Apostol–Euler polynomials are defined by the following relations, respectively (see ):

\begin{aligned}& \sum^{\infty}_{n=0} G^{m}_{n}(z;\lambda) \frac{w^{n}}{n!}= \biggl( \frac {2w}{\lambda e^{w}+1} \biggr)^{m} e^{wz},\quad \vert w \vert < \pi \mbox{ when } \lambda= 1 \end{aligned}
(1.1)
\begin{aligned}& \phantom{\sum^{\infty}_{n=0} G^{m}_{n}(z;\lambda) \frac{w^{n}}{n!}={}}\mbox{and } \vert w \vert < \bigl|\log(-\lambda)\bigr| \mbox{ when } \lambda\neq1; \lambda\in\mathbb{C}, \\& \sum^{\infty}_{n=0} B^{m}_{n}(z;\lambda) \frac{w^{n}}{n!}= \biggl( \frac {w}{\lambda e^{w}-1} \biggr)^{m} e^{wz},\quad \vert w \vert < \pi \mbox{ when } \lambda= 1 \end{aligned}
(1.2)
\begin{aligned}& \phantom{\sum^{\infty}_{n=0} B^{m}_{n}(z;\lambda) \frac{w^{n}}{n!}={}} \mbox{and } \vert w \vert < \bigl|\log(-\lambda)\bigr| \mbox{ when } \lambda\neq1; \lambda\in\mathbb{C}, \\& \sum^{\infty}_{n=0} E^{m}_{n}(z;\lambda) \frac{w^{n}}{n!}= \biggl( \frac {2}{\lambda e^{w}+1} \biggr)^{m} e^{wz},\quad \vert w \vert < \pi \mbox{ when } \lambda= 1 \\& \phantom{\sum^{\infty}_{n=0} E^{m}_{n}(z;\lambda) \frac{w^{n}}{n!}=} \mbox{and } \vert w \vert < \bigl|\log(-\lambda)\bigr| \mbox{ when } \lambda\neq1; \lambda\in\mathbb{C}. \end{aligned}
(1.3)

When $$m=1$$, the above equations give the generating functions for the Apostol–Genocchi, Apostol–Bernoulli, and Apostol–Euler polynomials, respectively (see ). When $$m=1$$ and $$\lambda=1$$, the equations give the generating functions for the classical Genocchi, Bernoulli, and Euler polynomials (see [4, 10]).

New formulas for the product of an arbitrary number of the Apostol–Bernoulli, Apostol–Euler, and Apostol–Genocchi polynomials were established in  where these polynomials were referred to as Apostol-type polynomials. Further, higher-order convolutions for these polynomials were established in . New identities for the Apostol–Bernoulli polynomials and Apostol–Genocchi polynomials were also presented in .

Fourier expansion, being a sum of multiple of sines and cosines, is easily differentiated and integrated, which often simplifies analysis of functions such as saw waves which are common signals in experimentation . Real world applications of Fourier series include the use for audio compression .

Fourier expansions of Genocchi polynomials and Apostol–Genocchi polynomials were obtained by Luo (see [11, 12]) using Lipschitz summation, while Bayad  obtained Fourier expansion for the Apostol–Bernoulli, Apostol–Euler, and Apostol–Genocchi polynomials using complex analysis theory of residues. Following Luo  and Bayad , the Fourier expansion of Apostol Frobenius–Euler polynomials was derived by Araci and Acikgoz . Fourier series of periodic Genocchi functions and construction of good links between Genocchi functions and zeta function were also obtained in . Fourier series of higher-order Bernoulli and Euler polynomials were used by López and Temme  to obtain asymptotic approximations of these polynomials. Using the method in , approximations for higher-order Genocchi polynomials were derived in .

In this paper, Fourier expansions for higher-order Apostol–Genocchi, Apostol–Bernoulli, and Apostol–Euler polynomials are derived as no Fourier expansions of these polynomials are available in the literature. The method of López and Temme  is used to derive the desired Fourier expansions. It is found out that the method using Lipschitz summation is not applicable to these higher-order polynomials. Moreover, it is shown that for $$m=1$$ the Fourier series obtained reduce to those obtained in  and  . Exceptional values of the parameter λ are also considered.

## Fourier expansions

In this section Fourier expansions for higher-order Apostol-type polynomials mentioned above are presented and proved.

### Theorem 2.1

For$$\lambda\in\mathbb{C}$$, $$\lambda\neq0, -1$$, $$0< z<1$$, and$$n\geq m$$,

\begin{aligned} G_{n}^{m}(z;\lambda)&= \frac{2^{m}ne^{\pi in}}{\lambda^{z}} \binom{n-1}{m-1} \\ &\quad\times \sum_{k=-\infty}^{\infty} \sum _{\nu=0}^{m-1} \binom{m-1}{\nu }(n-v-1)! B_{\nu}^{m}(z) \frac{e^{(2k+1)\pi iz}}{[\log\lambda- (2k+1)\pi i]^{n-\nu}}, \end{aligned}
(2.1)

where$$B_{\nu}^{m}(z)=B_{\nu}^{m}(z;1)$$denotes the Bernoulli polynomials of higher order defined in (1.2).

### Proof

Applying the Cauchy integral formula to (1.1),

$$\frac{G_{n}^{m}(z;\lambda)}{n!}=\frac{1}{2\pi i} \int_{C} \biggl(\frac {2w}{\lambda e^{w}+1} \biggr)^{m} e^{wz} \frac{dw}{w^{n+1}},$$
(2.2)

where C is a circle about zero with $$\text{radius} <|i\pi- \log\lambda|$$. Let

$$f(w)= \biggl(\frac{2w}{\lambda e^{w}+1} \biggr)^{m} \frac{e^{wz}}{w^{n+1}}.$$
(2.3)

Note that 0 is a pole of order $$n-m+1$$, while the values $$w_{k}$$ such that $$\lambda e^{w_{k}}+1=0$$ are poles of order m. For $$k\in\mathbb{Z}$$,

$$w_{k}=-\log\lambda+(2k+1)\pi i.$$
(2.4)

Let $$C_{k}$$ be a circle about 0 with $$\mathrm{radius} <|w_{k}|$$. Letting $$k\rightarrow\infty$$ and using the residue theorem,

$$\lim_{k\rightarrow\infty} \frac{1}{2\pi i} \int_{C_{k}} \biggl(\frac {2w}{\lambda e^{w}+1} \biggr)^{m} \frac{e^{wz}}{w^{n+1}} \,dz = \operatorname{Res}\bigl(f(w),0\bigr)+\sum _{k=-\infty}^{\infty} R_{k},$$
(2.5)

where $$R_{k}=\operatorname{Res}(f(w),w_{k})$$.

For $$0< z<1$$, the limit on the left-hand side of (2.5) is 0. For $$k=0$$,

\begin{aligned} R_{0}=\operatorname{Res}\bigl(f(w),0\bigr)&=\frac{1}{2\pi i} \int_{C} f(w) \,dw =\frac {G_{n}^{m}(z;\lambda)}{n!}. \end{aligned}

Then (2.5) becomes

\begin{aligned}& 0=\frac{G_{n}^{m}(z;\lambda)}{n!}+\sum_{k=-\infty}^{\infty} R_{k} \\& \quad\Leftrightarrow\quad G_{n}^{m}(z;\lambda)=-(n!)\sum _{k=-\infty}^{\infty} R_{k}. \end{aligned}
(2.6)

To compute the residues $$R_{k}$$, $$k\ge1$$, the Laurent series of $$f(w)$$ about $$w_{k}$$ will be used. Since $$w_{k}$$ is a pole of order m, its Laurent series is

$$f(w)=\sum_{r=0}^{\infty} a_{r}(w-w_{k})^{r} + \sum _{r=-1}^{-m} a_{r}(w-w_{k})^{r},$$
(2.7)

where $$a_{-1}=\operatorname{Res}(f(w),w_{k})$$.

Multiplying both sides of (2.7) by $$(w-w_{k})^{m}$$, we have

\begin{aligned} (w-w_{k})^{m}f(w)&=\sum_{r=0}^{\infty} a_{r}(w-w_{k})^{m+r}+a_{-1}(w-w_{k})^{m-1} \\ &\quad+a_{-2}(w-w_{k})^{m-2}+\cdots+a_{-m}, \end{aligned}

where $$a_{-1}$$ is now the coefficient of $$(w-w_{k})^{m-1}$$. That is, $$a_{-1}=a_{m-1}$$ in the expansion

$$(w-w_{k})^{m}f(w)=\sum _{r=0}^{\infty} a_{r}(w-w_{k})^{r}.$$
(2.8)

Let

$$G_{n}^{m}(z;\lambda)= \frac{2(n!)}{\lambda^{z}}\sum_{k=-\infty}^{\infty} \beta_{k}^{m}(n,z) \frac{e^{(2k+1)\pi iz}}{[-\log\lambda+(2k+1)\pi i]^{n}},$$
(2.9)

where $$\beta_{k}^{m}(n,z)$$ are to be determined. From  and ,

$$G_{n}(z;\lambda)=\frac{2(n!)}{\lambda^{z}}\sum _{k=-\infty}^{\infty }\frac{e^{(2k+1)\pi iz}}{[-\log\lambda+(2k+1)\pi i]^{n}},$$
(2.10)

it is seen that $$\beta_{k}^{1}(n,z)=1$$, k.

To find an explicit formula for $$\beta_{k}^{m}(n,z)$$, substitute $$w_{k}=-\log\lambda+(2k+1)\pi i$$ to (2.8) and use $$f(z)$$ in (2.3) to give

$$\begin{gathered}[b]\bigl(w-\bigl[-\log\lambda+(2k+1)\pi i\bigr] \bigr)^{m} \frac{2^{m} e^{wz}}{(\lambda e^{w}+1)^{m} w^{n-m+1}}\\\quad=\sum_{r=0}^{\infty} a_{r} \bigl(w-\bigl[-\log\lambda+(2k+1)\pi i\bigr] \bigr)^{r}.\end{gathered}$$
(2.11)

Let $$s=w-[-\log\lambda+(2k+1)\pi i]$$. Then $$w=s-\log\lambda +(2k+1)\pi i$$ and (2.11) becomes

$$\frac{(-2)^{m} e^{(2k+1)\pi iz} e^{-z\log\lambda}}{[s-\log\lambda +(2k+1)\pi i]^{n-m+1}} \cdot\frac{s^{m} e^{zs}}{(e^{s}-1)^{m}}=\sum _{r=0}^{\infty} a_{r} s^{r}.$$
(2.12)

Using (1.2) and writing

$$\bigl[s-\log\lambda+(2k+1)\pi i\bigr]^{m-n-1}=\sum _{\nu=0}^{\infty} \binom {m-n-1}{\nu}s^{\nu}\bigl[-\log\lambda+(2k+1)\pi i\bigr]^{m-n-1-\nu},$$

the left-hand side of (2.12) becomes

$$\begin{gathered}[b](-2)^{m}\lambda^{-z}e^{(2k+1)\pi iz} \Biggl(\sum_{\nu=0}^{\infty} \binom{m-n-1}{\nu}s^{\nu}\bigl[-\log\lambda+(2k+1)\pi i \bigr]^{m-n-1-\nu} \Biggr) \\\quad{}\times\Biggl(\sum_{\nu=0}^{\infty} \frac{B_{\nu}^{m}(z)}{\nu!}s^{\nu}\Biggr).\end{gathered}$$
(2.13)

Applying Cauchy-product on (2.13) will yield

$$\begin{gathered}[b] \frac{(-2)^{m}\lambda^{-z}e^{(2k+1)\pi iz}}{[-\log\lambda+(2k+1)\pi i]^{n+1-m}}\sum_{r=0}^{\infty} \sum_{v=0}^{r}\binom{m-n-1}{r-\nu } \bigl[-\log\lambda+(2k+1)\pi i\bigr]^{\nu-r} \frac{B_{\nu}^{m}(z)}{\nu !}s^{r}\hspace{-12pt}\\\quad= \sum_{r=0}^{\infty} a_{r} s^{r}.\end{gathered}$$
(2.14)

Thus,

\begin{aligned}[b] a_{r}&=\frac{(-2)^{m}e^{(2k+1)\pi iz}}{\lambda^{z}[-\log\lambda+(2k+1)\pi i]^{n+1-m}}\\&\quad{}\times\sum_{\nu=0}^{r} \binom{m-n-1}{r-\nu}\bigl[-\log\lambda +(2k+1)\pi i\bigr]^{\nu-r} \frac{B_{\nu}^{m}(z)}{\nu!}.\end{aligned}
(2.15)

In particular,

$$a_{m-1}=\frac{(-2)^{m}e^{(2k+1)\pi iz}}{\lambda^{z}[-\log\lambda +(2k+1)\pi i]^{n}}\sum _{\nu=0}^{m-1}\binom{m-n-1}{m-1-\nu} \frac {B_{\nu}^{m}(z)}{\nu!}\bigl[-\log\lambda+(2k+1)\pi i\bigr]^{\nu}.$$
(2.16)

Comparing (2.6) and (2.9),

$$\beta_{k}^{m}(n,z)= \frac{\lambda^{z}[-\log\lambda+(2k+1)\pi i]^{n}}{-2e^{(2k+1)\pi iz}}a_{m-1}.$$
(2.17)

Substituting (2.16) to (2.17),

$$\beta_{k}^{m}(n,z)=(-2)^{m-1}\sum _{\nu=0}^{m-1}\binom{m-n-1}{m-1-\nu } \frac{B_{\nu}^{m}(z)}{\nu!}\bigl[-\log\lambda+(2k+1)\pi i\bigr]^{\nu}.$$
(2.18)

Using the identity

$$(-1)^{m-1+\nu}\binom{n-1}{m-1}\binom{m-1}{\nu} \frac {(n-v-1)!}{(n-1)!}=\frac{1}{\nu!}\binom{m-n-1}{m-1-\nu},$$
(2.19)

(2.18) becomes

$$\beta_{k}^{m}(n,z)=2^{m-1} \binom{n-1}{m-1}\sum_{\nu=0}^{m-1} \binom {m-1}{\nu}\frac{(n-\nu-1)!}{(n-1)!}B_{\nu}^{m}(z) \bigl[\log\lambda -(2k+1)\pi i\bigr]^{\nu}.$$
(2.20)

Substituting to (2.9), the desired Fourier expansion for $$G_{n}^{m}(z;\lambda)$$ is obtained. □

### Remark 2.2

When $$m=1$$, (2.1) reduces to

$$G_{n}(z;\lambda)=\frac{2(n!)}{\lambda^{z}}\sum _{k=-\infty}^{\infty }\frac{e^{(2k+1)\pi it}}{[-\log\lambda+(2k+1)\pi i]^{n}} ,$$

which coincides with that of Luo  and Bayad .

### Theorem 2.3

For$$\lambda\in\mathbb{C}$$, $$\lambda\neq0, 1$$, $$0< z<1$$, and$$n\geq m$$,

\begin{aligned}[b] B_{n}^{m}(z;\lambda)&= \frac{ne^{(n-m)\pi i}}{\lambda^{z}}\binom {n-1}{m-1}\\&\quad\times\sum_{k=-\infty}^{\infty} \sum_{v=0}^{m-1}\binom {m-1}{v}(n-v-1)! B_{v}^{m}(z)\frac{e^{2k\pi iz}}{[\log\lambda-2k\pi i]^{n-v}}.\end{aligned}
(2.21)

### Proof

The method used in proving Theorem 2.1 will be applied here. Applying the Cauchy integral formula to (1.2), we obtain

$$\frac{B_{n}^{m}(z;\lambda)}{n!}=\frac{1}{2\pi i} \int_{C} \biggl(\frac {w}{\lambda e^{w}-1} \biggr)^{m} e^{wz}\frac{dw}{w^{n+1}},$$
(2.22)

where C is a circle about zero with $$\text{radius}<|\log\lambda|$$.

Let

$$g(w)= \biggl(\frac{w}{\lambda e^{w}-1} \biggr)^{m} \frac{e^{wz}}{w^{n+1}}.$$
(2.23)

Note that zero is a pole of order $$n-m+1$$, while the values $$u_{k}$$ such that $$\lambda e^{u_{k}}-1=0$$ are poles of order m. For $$k\in\mathbb{Z}$$,

$$u_{k}=-\log\lambda+2k\pi i.$$
(2.24)

Let $$C_{k}$$ be a circle about 0 with $$\text{radius}<|w_{k}|$$. Letting $$k\to \infty$$ and using the residue theorem,

$$\lim_{k\to\infty}{\frac{1}{2\pi i} \int_{C_{k}} \biggl(\frac {w}{\lambda e^{w}-1} \biggr)^{m} \frac {e^{wz}}{w^{n+1}}\,dw}=\operatorname{Res}\bigl(g(w),0\bigr)+\sum_{k=-\infty}^{\infty}S_{k},$$
(2.25)

where $$S_{k}=\operatorname{Res}(g(w),u_{k})$$.

For $$0< z<1$$, the limit on the left-hand side of (2.25) is 0 and

\begin{aligned} Res\bigl(g(w),0\bigr)&=\frac{1}{2\pi i} \int_{C}g(w)\,dw =\frac{B_{n}^{m}(z;\lambda)}{n!}. \end{aligned}

Then (2.25) becomes

\begin{aligned}& 0=\frac{B_{n}^{m}(z;\lambda)}{n!}+\sum_{k=-\infty}^{\infty}S_{k} \\ & \quad\Leftrightarrow \quad B_{n}^{m}(z;\lambda)=-(n!)\sum_{k=-\infty }^{\infty}S_{k}. \end{aligned}
(2.26)

To compute the residues $$S_{k}$$, use the Laurent series of $$g(w)$$ about $$u_{k}$$. Since $$u_{k}$$ is a pole of order m, the Laurent series of $$g(w)$$ about $$u_{k}$$ is

$$g(w)=\sum_{r=0}^{\infty}b_{r}(w-u_{k})^{r}+ \sum_{r=-1}^{-m}b_{r}(w-u_{k})^{r},$$
(2.27)

where $$b_{-1}=\operatorname{Res}(g(w),u_{k})$$.

Multiplying both sides of (2.27) by $$(w-u_{k})^{m}$$,

$$(w-u_{k})^{m}g(w)=\sum_{r=0}^{\infty }b_{r}(w-u_{k})^{m+r}+b_{-1}(w-u_{k})^{m-1}+b_{-2}(w-u_{k})^{m-2}+ \cdots+b_{-m},$$

where $$b_{-1}$$ is now the coefficient of $$(w-u_{k})^{m-1}$$. That is, $$b_{-1}=b_{m-1}$$ in the expansion

$$(w-u_{k})^{m}g(w)=\sum _{r=0}^{\infty}b_{r}(w-u_{k})^{r}.$$
(2.28)

Let

$$B_{n}^{m}(z;\lambda)= \frac{n!}{\lambda^{z}}\sum_{k=-\infty}^{\infty } \gamma_{k}^{m}(n,z)\frac{e^{2k\pi iz}}{[2k\pi i-\log\lambda]^{n}},$$
(2.29)

where $$\gamma_{k}^{m}(n,z)$$ are to be determined. From ,

$$B_{n}(z;\lambda)=\frac{-(n!)}{\lambda^{z}}\sum _{k=-\infty}^{\infty }\frac{e^{2k\pi iz}}{[-\log\lambda+2k\pi i]^{n}} \quad\mbox{for } \lambda\neq1,$$
(2.30)

it is seen that $$\gamma_{k}^{1}(n,z)=-1$$, k.

To find an explicit formula for $$\gamma_{k}^{m}(n,z)$$, substitute $$u_{k}=-\log\lambda+ 2k \pi i$$ and the function $$g(w)$$ in (2.23) to (2.28) to obtain

$$\bigl(w-[-\log\lambda+2k \pi i]\bigr)^{m} \frac{e^{wz}}{(\lambda e^{w}-1)^{m}w^{n-m+1}}=\sum^{\infty}_{r=0}b_{r} \bigl(w-[-\log\lambda+2k\pi i]\bigr)^{r}.$$
(2.31)

Let $$t=w-[-\log\lambda+2k\pi i]$$. Then $$w=t-\log\lambda+2k\pi i$$ and (2.31) becomes

$$\frac{\lambda^{-z}e^{2k\pi iz}}{[t-\log\lambda+2k\pi i]^{n-m+1}} \biggl( \frac{t}{e^{t}-1} \biggr)^{m}e^{tz}=\sum^{\infty}_{r=0} b_{r}t^{r}.$$
(2.32)

Using (1.2) and writing

$$[t-\log\lambda+2k \pi i]^{m-n-1}=\sum^{\infty}_{\nu=0} \binom {m-n-1}{\nu} t^{\nu}(-\log\lambda+2k \pi i)^{m-n-1-\nu},$$

the left-hand side of (2.32) becomes

$$\lambda^{-z}e^{2k \pi iz} \Biggl(\sum ^{\infty}_{\nu=0} \binom {m-n-1}{\nu} t^{\nu}(-\log\lambda+2k\pi i)^{m-n-1-\nu} \Biggr) \Biggl(\sum ^{\infty}_{\nu=0} \frac{B^{m}_{\nu}(z)}{\nu!}t^{\nu} \Biggr).$$
(2.33)

Applying Cauchy-product on (2.33) will yield

$$\begin{gathered}[b] \frac{\lambda^{-z}e^{2k\pi iz}}{[-\log\lambda+2k\pi i]^{n-m+1}}\sum^{\infty}_{r=0} \Biggl\{ \sum^{r}_{\nu=0} \binom{m-n-1}{r-\nu} \frac {B^{m}_{\nu}(z)}{\nu!} (-\log\lambda+2k \pi i)^{\nu-r} \Biggr\} t^{r}\\\quad=\sum^{\infty}_{r=0}b_{r}t^{r}.\end{gathered}$$
(2.34)

Thus,

$$b_{r}=\frac{e^{2k\pi iz}}{\lambda^{z}(-\log\lambda+2k\pi i)^{n-m+1}} \sum^{r}_{\nu=0} \binom{m-n-1}{r-\nu} \frac{B^{m}_{\nu}(z)}{\nu!} (-\log\lambda+2k \pi i)^{\nu-r}.$$
(2.35)

In particular,

$$b_{m-1}=\frac{e^{2k\pi iz}}{\lambda^{z}(-\log\lambda+2k\pi i)^{n}} \sum ^{m-1}_{\nu=0} \binom{m-n-1}{m-\nu-1} \frac{B^{m}_{\nu}(z)}{\nu !} (-\log\lambda+2k \pi i)^{\nu}.$$
(2.36)

Comparing (2.26) and (2.29),

$$\gamma^{m}_{k}(n,z)= \frac{-\lambda^{z}(-\log\lambda+2k \pi i)^{n}}{e^{2k \pi iz}} \cdot b_{m-1}.$$
(2.37)

Substituting (2.36) to (2.37),

$$\gamma^{m}_{k}(n,z)=-\sum^{m-1}_{\nu=0} \binom{m-n-1}{m-\nu-1}\frac {B^{m}_{\nu}(z)}{\nu!}(-\log\lambda+2k \pi i)^{\nu}.$$
(2.38)

Using the identity in (2.19), we have

$$\gamma^{m}_{k}(n,z)=(-1)^{m} \binom{n-1}{m-1} \sum^{m-1}_{\nu=0} \binom {m-1}{\nu} \frac{(n-\nu-1)!}{(n-1)!} B^{m}_{\nu}(z) (\log\lambda-2k \pi i)^{\nu}.$$
(2.39)

Substituting (2.39) to (2.29), the desired Fourier expansion of $$B^{m}_{n}(z;\lambda)$$ is obtained. □

### Remark 2.4

When $$m=1$$, (2.21) reduces to

$$B_{n}(z; \lambda)=\frac{-(n)!}{\lambda^{z}} \sum ^{\infty}_{k=-\infty} \frac{e^{2k\pi iz}}{[-\log\lambda+2k\pi i]^{n}},$$

which coincides with that in .

### Theorem 2.5

For$$\lambda\in\mathbb{C}$$, $$\lambda\neq0,-1$$, $$0< z<1$$, and$$n \geq m$$,

\begin{aligned}[b]E^{m}_{n}(z;\lambda)&= \frac{2^{m}e^{\pi i (n+m)}}{(m-1)!\lambda^{z}}\sum^{\infty}_{k=-\infty} \sum^{m-1}_{\nu=0} \binom{m-1}{\nu}(n+m- \nu -1)!B^{n+m}_{\nu}(z) \\&\quad\times\frac{e^{(2k+1)\pi iz}}{[\log\lambda-(2k+1)\pi i]^{n+m-\nu}}.\end{aligned}
(2.40)

### Proof

Multiplying both sides of (1.3) by $$w^{m}$$ yields

\begin{aligned}& \biggl( \frac{2w}{\lambda e^{w}+1} \biggr)^{m} e^{zw}=\sum ^{\infty}_{n=0} E^{m}_{n}(z; \lambda)\frac{w^{n+m}}{n!}, \end{aligned}
(2.41)
\begin{aligned}& \sum^{\infty}_{n=0} G^{m}_{n}(z; \lambda) \frac{w^{n}}{n!}=\sum ^{\infty}_{n=0} E^{m}_{n}(z; \lambda)\frac{w^{n+m}}{n!}. \end{aligned}
(2.42)

The left hand-side of (2.42) can be written

\begin{aligned} \sum^{\infty}_{n=0} G^{m}_{n}(z; \lambda) \frac{w^{n}}{n!}&=\sum ^{\infty}_{n=-m} G^{m}_{n+m}(z; \lambda) \frac{w^{n+m}}{(n+m)!} \end{aligned}
(2.43)
\begin{aligned} &=\sum^{\infty}_{n=-m} G^{m}_{n+m}(z; \lambda) \frac{n!}{(n+m)!} \cdot \frac{w^{n+m}}{(n)!}. \end{aligned}
(2.44)

Thus,

$$\sum^{\infty}_{n=0} E^{m}_{n}(z;\lambda)\frac{w^{n+m}}{n!}=\sum ^{\infty}_{n=-m} G^{m}_{n+m}(z; \lambda) \frac{n!}{(n+m)!} \cdot\frac{w^{n+m}}{(n)!}.$$
(2.45)

Comparing coefficients in (2.45) gives

$$E^{m}_{n}(z;\lambda)=\frac{n!}{(n+m)!} G^{m}_{n+m}(z;\lambda).$$
(2.46)

Using (2.1),

\begin{aligned} E^{m}_{n}(z;\lambda)&= \frac{n!}{(n+m)!} \Biggl\{ \frac {2^{m}(n+m)e^{(n+m)\pi i}}{\lambda^{z}}\binom{n+m-1}{m-1}\sum^{\infty}_{k=-\infty} \sum ^{m-1}_{v=0} \binom{m-1}{\nu }(n+m-\nu-1)! \\ &\quad \times B^{n+m}_{\nu}(z)\frac{e^{(2k+1)\pi iz}}{[\log\lambda -(2k+1)\pi i]^{n+m-\nu}} \Biggr\} . \end{aligned}
(2.47)

Simplifying

$$\frac{n!}{(n+m)!}(n+m)\binom{n+m-1}{m-1}=\frac{1}{(m-1)!},$$

and substituting to (2.47), the desired result is obtained. □

### Remark 2.6

If $$m=1$$, (2.40) reduces to

$$E_{n}(z;\lambda)=\frac{2(n!)}{\lambda^{z}}\sum ^{\infty}_{k=-\infty} \frac{e^{(2k+1)\pi iz}}{[-\log\lambda+(2k+1)\pi i]^{n+1}},$$

which coincides with the corresponding result in .

## The cases $$\lambda=-1$$ and $$\lambda=1$$

Theorem 2.1 does not apply when $$\lambda=-1$$ because for $$\lambda=-1$$, $$w_{k}=0$$, k, while Theorem 2.3 does not apply for $$\lambda=1$$ for similar reason. So these cases are considered here. Using (1.2),

$$\sum_{n=0}^{\infty} B_{n}^{m}(z;1)\frac{w^{n}}{n!}= \biggl( \frac {w}{e^{w}-1} \biggr)^{m}e^{wz},\quad \vert w \vert < 2\pi.$$
(3.1)

On the other hand, using (1.1), we get

\begin{aligned} \sum_{n=0}^{\infty}G_{n}^{m}(z;-1) \frac{w^{n}}{n!} &= \biggl(\frac{2w}{-e^{w}+1} \biggr)^{m}e^{wz} \\ &=(-2)^{m} \biggl(\frac{w}{e^{w}-1} \biggr)^{m}e^{wz},\quad \vert w \vert < 2\pi \\ &=(-2)^{m}\sum_{n=0}^{\infty} B_{n}^{m}(z;1)\frac{w^{n}}{n!}. \end{aligned}

Thus,

$$G_{n}^{m}(z;-1)=(-2)^{m} B_{n}^{m}(z;1).$$
(3.2)

Also, from (2.43),

\begin{aligned} E_{n}^{m}(z;-1) &=\frac{n!}{(n+m)!}G_{n+m}^{m}(z;-1) \\ &=\frac{n!}{(n+m)!}(-2)^{m} B_{n+m}^{m}(z;1). \end{aligned}
(3.3)

We proceed to finding the Fourier expansion for $$B_{n}^{m}(z;1)$$. The method in the previous section will be applied. First consider $$m=1$$. The Fourier expansion for $$B_{n}^{1}(z;1)=B_{n}(z;1)$$ is given in the following lemma.

### Lemma 3.1

For$$0< z<1$$and$$n\geq1$$,

$$B_{n}(z;1)=-(n!)\sum_{k=-\infty,k\neq0}^{\infty} \frac{e^{2k\pi iz}}{(2k\pi i)^{n}}.$$
(3.4)

### Proof

By (1.2)

$$B_{n}(z;1)=B_{n}^{1}(z;1)= \frac{n!}{2\pi i} \int_{C}\frac{e^{wz}}{e^{w}-1}\frac {dw}{w^{n}},$$

where C is a circle about the origin with $$\text{radius}<2\pi$$. Let $$f(w)=\frac{e^{wz}}{(e^{w}-1)w^{n}}$$. Following the method in the previous section, we obtain

$$B_{n}(z;1)=-(n!)\sum_{k=-\infty,k\neq0}^{\infty}R_{k},$$

where $$R_{k}=\operatorname{Res}(f(w),2k\pi i)$$, $$k=\pm1,\pm2, \dots$$.

These residues can be computed to be

$$R_{k}=\frac{e^{2k\pi i(z-1)}}{(2k\pi i)^{n}}.$$

Thus,

$$B_{n}(z;1)=-(n!)\sum_{k=-\infty,k\neq0}^{\infty} \frac{e^{2k\pi iz}}{(2k\pi i)^{n}}.$$

□

For $$m>1$$, the Fourier series of $$B_{n}^{m}$$(z;1) is given in the following theorem.

### Theorem 3.2

For$$0< z<1$$and$$n\geq m>1$$,

$$B_{n}^{m}(z;1)=(-1)^{m}n \binom{n-1}{m-1}\sum_{k=-\infty,k\neq0}^{\infty }\sum _{\nu=0}^{m-1}\binom{m-1}{\nu}(n-v-1)!B_{\nu}^{m}(z) (-1)^{\nu}\frac{e^{2k\pi iz}}{(2k\pi i)^{n-\nu}}.$$
(3.5)

### Proof

By the Cauchy integral formula,

$$\frac{B_{n}^{m}(z;1)}{n!}=\frac{1}{2\pi i} \int_{C}\frac {e^{wz}}{(e^{w}-1)^{m}w^{n-m+1}}\,dw, \quad \vert w \vert < 2\pi,$$
(3.6)

where C is a circle about the origin with $$\text{radius}<2\pi$$.

The complex numbers $$u_{k}=2k\pi i$$, $$k=\pm1,\pm2, \ldots$$ are poles of order m of the function

$$h(w)=\frac{e^{wz}}{(e^{w}-1)^{m}w^{n-m+1}}.$$
(3.7)

Then

$$B_{n}^{m}(z;1)=-(n!)\sum _{k=-\infty,k\neq0}^{\infty}R_{k},$$
(3.8)

where $$R_{k}=\operatorname{Res}(h(w),2k\pi i)$$, $$k=\pm1,\pm2,\dots$$.

Let

$$h(w)=\sum_{r=0}^{\infty}c_{r}(w-u_{k})^{r} + \sum_{r=-1}^{-m}c_{r}(w-u_{k})^{r}$$
(3.9)

be the Laurent series of $$h(w)$$, where

$$c_{-1}=\operatorname{Res}\bigl(h(w);u_{k}\bigr).$$
(3.10)

Multiplying both sides of (3.9) by $$(w-u_{k})^{m}$$ gives

$$(w-u_{k})^{m}h(w)=\sum_{r=0}^{\infty }c_{r}(w-u_{k})^{m+r}+c_{-1}(w-u_{k})^{m-1}+ \cdots+c_{-m},$$

where $$c_{-1}$$ is now the coefficient of $$(w-u_{k})^{m-1}$$.

That is, $$c_{-1}=c_{m-1}$$ in the expansion

$$(w-u_{k})^{m}h(w)=\sum _{r=0}^{\infty}c_{r}(w-u_{k})^{r}.$$
(3.11)

Following (3.4), write

$$B_{n}^{m}(z;1)=-(n!)\sum _{k=-\infty,k\neq0}^{\infty}\gamma _{k}^{m}(n,z;1) \frac{e^{2k\pi iz}}{(2k\pi i)^{n}},$$
(3.12)

where $$\gamma_{k}^{m}(n,z;1)$$ are to be determined. Note that $$\gamma _{k}^{1}(n,z;1)=1$$ (see (3.4)). From (3.11),

$$(w-2k\pi i)^{m}\frac{e^{wz}}{(e^{w}-1)^{m}e^{n-m+1}}=\sum _{r=0}^{\infty }c_{r}(w-2k\pi i)^{r}.$$
(3.13)

Let $$t=w-2k\pi i$$. Then $$w=t+2k\pi i$$ and (3.13) becomes

$$\frac{t^{m}}{(e^{t}-1)^{m}}e^{t}z\cdot \frac{e^{2k\pi iz}}{(t+2k\pi i)^{n-m+1}}=\sum_{r=0}^{\infty}c_{r}t^{r}.$$
(3.14)

Writing

$$(t+2k\pi i)^{m-n-1}=\sum_{\nu=0}^{\infty} \binom{m-n-1}{\nu}t^{\nu}(2k\pi i)^{m-n-1-\nu}$$
(3.15)

and using (3.1), (3.14) yields

$$\Biggl(\sum_{n=0}^{\infty}B_{n}^{m}(z;1) \frac{t^{n}}{n!} \Biggr) \Biggl(\sum_{\nu=0}^{\infty} \binom{m-n-1}{\nu}t^{\nu}(2k\pi i)^{m-n-1-\nu } \Biggr)e^{2k\pi iz}=\sum_{r=0}^{\infty}c_{r}t^{r}.$$
(3.16)

Applying Cauchy-product, (3.15) becomes

$$\frac{e^{2k\pi iz}}{(2k\pi i)^{n-m+1}}\sum_{r=0}^{\infty} \Biggl\{ \sum_{\nu=0}^{r} \binom{m-n-1}{r-\nu}\frac{B_{\nu}^{m}(z)}{\nu!}(2k\pi i)^{\nu-r} \Biggr\} t^{r}=\sum_{r=0}^{\infty}c_{r}t^{r}.$$
(3.17)

Thus,

$$c_{r}=\frac{e^{2k\pi iz}}{(2k\pi i)^{n-m+1}}\sum_{\nu=0}^{r} \binom {m-n-1}{r-\nu}\frac{B_{\nu}^{m}(z)}{\nu!}(2k\pi i)^{\nu-r}.$$
(3.18)

In particular,

$$c_{m-1}=\frac{e^{2k\pi iz}}{(2k\pi i)^{n}}\sum_{\nu=0}^{m-1} \binom {m-n-1}{m-\nu-1}\frac{B_{\nu}^{m}(z)}{\nu!}(2k\pi i)^{\nu}.$$
(3.19)

Comparing (3.8) and (3.12),

$$\gamma_{k}^{m}(n,z;1)=\sum _{\nu=0}^{m-1}\binom{m-n-1}{m-\nu-1} \frac {B_{\nu}^{m}(z)}{\nu!}(2k\pi i)^{\nu}.$$
(3.20)

Applying (2.19),

$$\gamma_{k}^{m}(n,z;1)=(-1)^{m-1} \binom{n-1}{m-1}\sum_{\nu =0}^{m-1} \binom{m-1}{\nu}\frac{(n-\nu-1)}{(n-1)!}B_{\nu}^{m}(z) (-2k\pi i)^{\nu}.$$
(3.21)

Substituting to (3.12), the theorem follows. □

### Remark 3.3

When $$m=1$$, the formula in Lemma 3.1 and Theorem 3.2 agrees with that obtained in .

Using (3.2) and (3.3) the following corollary is a direct consequence of Theorem 3.2.

### Corollary 3.4

For$$0< z<1$$and$$n\geq m>1$$,

\begin{aligned}& G_{n}^{m}(z;-1)=2^{m}n \binom{n-1}{m-1}\sum_{k=-\infty,k\neq 0}^{\infty}\sum _{\nu=0}^{m-1}\binom{m-1}{\nu}(n- \nu-1)!B_{\nu}^{m}(z) (-1)^{\nu}\frac{e^{2k\pi iz}}{(2k\pi i)^{n-\nu}}, \\& E_{n}^{m}(z;-1)=\frac{2^{m}}{(m-1)!}\sum _{k=-\infty ,k\neq0}^{\infty}\sum_{\nu=0}^{m-1} \binom{m-1}{\nu}(n+m-\nu -1)!B_{\nu}^{n}(z) (-1)^{\nu}\frac{e^{2k\pi iz}}{(2k\pi i)^{n+m-\nu}}. \end{aligned}

## Conclusion

It is seen that the Fourier expansions for higher-order Apostol–Genocchi, Apostol–Bernoulli, and Apostol–Euler polynomials are readily obtained using the method of Lopez and Temme . Following  and  it will be interesting to consider the integral representations and asymptotic approximations of these polynomials for future study.

## References

1. Araci, S., Acikgoz, M.: Applications of Fourier series and zeta functions to Genocchi polynomials. Appl. Math. Inf. Sci. 12(5), 951–955 (2018)

2. Araci, S., Acikgoz, M.: Construction of Fourier expansion of Apostol Frobenius–Euler polynomials and its application. Adv. Differ. Equ. 2018, Article ID 67 (2018). https://doi.org/10.1186/s13662-018-1526-x

3. Bayad, A.: Fourier expansions for Apostol–Bernoulli, Apostol–Euler and Apostol–Genocchi polynomials. Math. Comput. 80(276), 2219–2221 (2011). https://doi.org/10.1090/S0025-5718-2011-02476-2

4. Corcino, C., Corcino, R.: Asymptotics of Genocchi polynomials and higher order Genocchi polynomials using residues. Afr. Math. (2020). https://doi.org/10.1007/s13370-019-00759-z

5. Fixed Point (https://math.stackechange.com/users/30261/fixed-point): Real world application of Fourier series. Nov. 24, 2013. https://math.stackexchange.com/q/579695

6. He, Y., Araci, S., Srivastava, H.M.: Some new formulas for the products of the Apostol type polynomials. Adv. Differ. Equ. 2016, Article ID 287 (2016). https://doi.org/10.1186/s13662-016-1014-0

7. He, Y., Araci, S., Srivastava, H.M., Abdel-Aty, M.: Higher-order convolutions for Apostol–Bernoulli, Apostol–Euler and Apostol–Genocchi polynomials. Mathematics 6, Article ID 329 (2019). https://doi.org/10.3390/math6120329

8. He, Y., Araci, S., Srivastava, H.M., Acikgoz, M.: Some new identities for the Apostol–Bernoulli polynomials and the Apostol–Genocchi polynomials. Appl. Math. Comput. 262, 31–41 (2015). https://doi.org/10.1016/j.amc.2015.03.132

9. Hollingsworth, M.: Applications of the Fourier series (2019)

10. López, J.L., Temme, N.M.: Large degree asymptotics of generalized Bernoulli and Euler polynomials. J. Math. Anal. Appl. 363(1), 197–208 (2010). https://doi.org/10.1016/j.jmaa.2009.08.034

11. Luo, Q.-M.: Fourier expansions and integral representations for Genocchi polynomials. J. Integer Seq. 12, Article ID 09.1.4 (2009)

12. Luo, Q.-M.: Extensions of the Genocchi polynomials and their Fourier expansions and integral representations. Osaka J. Math. 48, 291–309 (2011)

### Acknowledgements

The authors would like to thank Cebu Normal University for the financial support to this research project.

Not applicable.

## Funding

This research project is partially funded by Cebu Normal University.

## Author information

Authors

### Contributions

CC was the one who conceptualized the problem and the method to be used in solving the problem. She did the introduction and derived the Fourier expansion of higher-order Apostol–Genocchi and Apostol–Bernoulli polynomials. RC derived the Fourier expansion of higher-order Apostol–Euler polynomials, and he wrote Sect. 3. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Roberto B. Corcino.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests.

## Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and Permissions 