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.