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.

## 1 Introduction

Higher-order Apostol–Genocchi, Apostol–Bernoulli, and Apostol–Euler polynomials are defined by the following relations, respectively (see [7]):

\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 [3]). 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 [6] where these polynomials were referred to as Apostol-type polynomials. Further, higher-order convolutions for these polynomials were established in [7]. New identities for the Apostol–Bernoulli polynomials and Apostol–Genocchi polynomials were also presented in [8].

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 [9]. Real world applications of Fourier series include the use for audio compression [5].

Fourier expansions of Genocchi polynomials and Apostol–Genocchi polynomials were obtained by Luo (see [11, 12]) using Lipschitz summation, while Bayad [3] obtained Fourier expansion for the Apostol–Bernoulli, Apostol–Euler, and Apostol–Genocchi polynomials using complex analysis theory of residues. Following Luo [12] and Bayad [3], the Fourier expansion of Apostol Frobenius–Euler polynomials was derived by Araci and Acikgoz [2]. Fourier series of periodic Genocchi functions and construction of good links between Genocchi functions and zeta function were also obtained in [1]. Fourier series of higher-order Bernoulli and Euler polynomials were used by López and Temme [10] to obtain asymptotic approximations of these polynomials. Using the method in [10], approximations for higher-order Genocchi polynomials were derived in [4].

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 [10] 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 [3] and [12] . Exceptional values of the parameter λ are also considered.

## 2 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 [3] and [12],

$$G_{n}(z;\lambda)=\frac{2(n!)}{\lambda^{z}}\sum _{k=-\infty}^{\infty }\frac{e^{(2k+1)\pi iz}}{[-\log\lambda+(2k+1)\pi i]^{n}},$$
(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 [12] and Bayad [3].

### 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 [3],

$$B_{n}(z;\lambda)=\frac{-(n!)}{\lambda^{z}}\sum _{k=-\infty}^{\infty }\frac{e^{2k\pi iz}}{[-\log\lambda+2k\pi i]^{n}} \quad\mbox{for } \lambda\neq1,$$
(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 [3].

### 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 [3].

## 3 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 [3].

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}

## 4 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 [10]. Following [12] and [10] it will be interesting to consider the integral representations and asymptotic approximations of these polynomials for future study.

## References

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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

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### 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.

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### Competing interests

The authors declare that they have no competing interests.

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Corcino, C.B., Corcino, R.B. Fourier expansions for higher-order Apostol–Genocchi, Apostol–Bernoulli and Apostol–Euler polynomials. Adv Differ Equ 2020, 346 (2020). https://doi.org/10.1186/s13662-020-02802-x