Some symmetry identities for the Apostol-type polynomials related to multiple alternating sums

Abstract

In recent years, symmetry properties of the Bernoulli polynomials and the Euler polynomials have been studied by a large group of mathematicians (He and Wang in Discrete Dyn. Nat. Soc. 2012:927953, 2012, Kim et al. in J. Differ. Equ. Appl. 14:1267-1277, 2008; Abstr. Appl. Anal. 2008, doi:11.1155/2008/914347, Yang et al. in Discrete Math. 308:550-554, 2008; J. Math. Res. Expo. 30:457-464, 2010). Luo (Integral Transforms Spec. Funct. 20:377-391, 2009), introduced the lambda-multiple power sum and proved the multiplication formulas for the Apostol-Bernoulli and Apostol-Euler polynomials of higher order. Ozarslan (Comput. Math. Appl. 2011:2452-2462, 2011), Lu and Srivastava (Comput. Math. Appl. 2011, doi:10.1016/j.2011.09.010.2011) gave some symmetry identities relations for the Apostol-Bernoulli and Apostol-Euler polynomials.

In this work, we prove some symmetry identities for the Apostol-Bernoulli and Apostol-Euler polynomials related to multiple alternating sums.

AMS Subject Classification: 11F20, 11B68, 11M35, 11M41.

1 Introduction, definitions and notations

The generalized Bernoulli polynomials ${\mathcal{B}}_n^{(\alpha )}(x)$ of order $\alpha \in {\mathbb{N}}_0$ and the generalized Euler polynomials ${\mathcal{E}}_n^{(\alpha )}(x)$ of order $\alpha \in {\mathbb{N}}_0$, each of degree n as well as in α, are defined respectively by the following generating functions [1–3]:

(1)

(2)

The generalized Apostol-Bernoulli polynomials ${\mathcal{B}}_n^{(\alpha )}(x;\lambda )$ of order $\alpha \in {\mathbb{N}}_0$ and the generalized Apostol-Euler polynomials ${\mathcal{E}}_n^{(\alpha )}(x;\lambda )$ of order $\alpha \in {\mathbb{N}}_0$ are defined respectively by the following generating functions [3]:

(3)

(4)

Recently, Garg et al. in [4] introduced the following generalization of the Hurwitz-Lerch zeta function $\mathrm{\Phi }(z,s,a)$:

$${\mathrm{\Phi }}_{\mu ,\nu }^{(\rho ,\sigma )}(z,s,a):=\sum _{n=0}^{\mathrm{\infty }}\frac{{(\mu )}_{\rho n}}{{(\nu )}_{\sigma n}}\frac{z^n}{{(n+a)}^s}$$

($\mu \in \mathbb{C}$, a, $\upsilon \in \mathbb{C}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\rho ,\sigma \in {\mathbb{R}}^{+}$, $\rho <\sigma$ when $s,z\in \mathbb{C}(|z|<1)$; $\rho =\sigma$ and $\mathrm{\Re }(s-\mu +\nu )>0$ when $|z|=1$). It is obvious that

$${\mathrm{\Phi }}_{\mu ,1}^{(1,1)}(z,s,a)={\mathrm{\Phi }}_{\mu }^{\ast }(z,s,a)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(\mu )}_n}{n!}\frac{z^n}{{(n+a)}^s}$$

(5)

(for details on this subject, see [3–5]).

The multiple power sums and the λ-multiple alternating sums are defined by Luo [6] as follows:

(6)

(7)

From (6) and (7), we have

$${\left(\frac{1-{\lambda }^m{e}^{mt}}{1-\lambda e^t}\right)}^l={\lambda }^{(-l)}\sum _{n=0}^{\mathrm{\infty }}\{\sum _{p=0}^n\left(\genfrac{}{}{0ex}{}{n}{p}\right){(-l)}^{n-p}{S}^{(l)}(m;\lambda )\}\frac{t^n}{n!}$$

(8)

and

$${\left(\frac{1+{(-1)}^{m+1}{(\lambda e^t)}^m}{1+\lambda e^t}\right)}^l={\lambda }^{(-l)}\sum _{n=0}^{\mathrm{\infty }}\{\sum _{p=0}^n\left(\genfrac{}{}{0ex}{}{n}{p}\right){(-l)}^{n-p}{T}^{(l)}(m;\lambda )\}\frac{t^n}{n!}$$

(9)

(see [6]).

From (8) and (9), for $l=1$, we have respectively

(10)

(11)

Symmetry property and some recurrence relations of the Bernoulli polynomials, Euler polynomials, Apostol-Bernoulli polynomials and Apostol-Euler polynomials have been investigated by a lot of mathematicians [1–24]. Firstly, Yang [22] proved symmetry relation for Bernoulli polynomials. Wang et al. in [1, 20, 21] gave some symmetry relations for the Apostol-Bernoulli polynomials. Kim in [8, 10, 11, 14, 15] proved symmetric identities for the Bernoulli polynomials and Euler polynomials. Luo in [6, 17] gave multiplication formulas for the Apostol-Bernoulli and Apostol-Euler polynomials. Also, he defined λ-power sums. Srivastava et al. [2, 3, 5] proved some theorems and relations for these polynomials. They proved some symmetry identities for these polynomials.

In this work, we give some symmetry identities for the Apostol-type polynomials related to multiple alternating sums.

2 Symmetry identities for the Apostol-Bernoulli polynomials

We will prove the following theorem for the Apostol-Euler polynomials, which are symmetric in a and b.

Theorem 2.1 There is the following relation between Apostol-Bernoulli polynomials and the Hurwitz-Lerch zeta function ${\mathrm{\Phi }}^{\ast }(z,s,a)$:

(12)

Proof Let $f(t)=\frac{t^{\alpha -1}{e}^{abxt}(1-{\lambda }^{ab}{e}^{abt})e^{abyt}}{{(1-{\lambda }^a{e}^{at})}^{\alpha }{(1-{\lambda }^b{e}^{bt})}^{\alpha }}$. Then

$$f(t)=\frac{1}{b^{\alpha -1}}\frac{e^{abxt}}{{(1-{\lambda }^a{e}^{at})}^{\alpha }}\left(\frac{1-{\lambda }^{ab}{e}^{abt}}{1-{\lambda }^b{e}^{bt}}\right){\left(\frac{bt}{1-{\lambda }^b{e}^{bt}}\right)}^{\alpha -1}{e}^{abyt}.$$

From (3) and (10), we write

$$\begin{array}{rcl}f(t)& =& \frac{{(-1)}^{\alpha -1}}{b^{\alpha -1}}\sum _{\beta =0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{\beta +\alpha -1}{\beta }\right){\lambda }^{\alpha \beta }{e}^{at(\beta +bx)}\frac{1}{{\lambda }^b}\sum _{r=0}^{\mathrm{\infty }}\sum _{p=0}^r\left(\genfrac{}{}{0ex}{}{r}{p}\right){(-1)}^{r-p}{S}_p(a;{\lambda }^b)\frac{b^r{t}^r}{r!}\\ \times \sum _{k=0}^{\mathrm{\infty }}{\mathcal{B}}_k^{(\alpha -1)}(ay;{\lambda }^b)\frac{b^k{t}^k}{k!},\end{array}$$

where $|log\lambda +t|<min(\frac{2\pi }{a},\frac{2\pi }{b})$. After the Cauchy product, we have

$$\begin{array}{rc}=& \sum _{n=0}^{\mathrm{\infty }}\{\sum _{s=0}^n\left(\genfrac{}{}{0ex}{}{n}{s}\right)b^{s+1-\alpha }{a}^{n-s}{(-1)}^{\alpha -1}\sum _{k=0}^s\left(\genfrac{}{}{0ex}{}{s}{k}\right)\sum _{p=0}^r\left(\genfrac{}{}{0ex}{}{r}{p}\right){(-1)}^{r-p}{\lambda }^{-b}\\ \times S_p(a;{\lambda }^b){\mathcal{B}}_k^{(\alpha -1)}(ay;{\lambda }^b)\sum _{\beta =0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{\beta +\alpha -1}{\beta }\right)\frac{{\lambda }^{\alpha \beta }}{{(\beta +bx)}^{s-n}}\}\frac{t^n}{n!}.\end{array}$$

In a similar manner,

$$\begin{array}{rcl}f(t)& =& \frac{t^{\alpha -1}{e}^{abyt}(1-{\lambda }^{ab}{e}^{abt})e^{abxt}}{{(1-{\lambda }^b{e}^{bt})}^{\alpha }{(1-{\lambda }^a{e}^{at})}^{\alpha }}\\ =& \frac{1}{a^{\alpha -1}}\frac{e^{abxt}}{{(1-{\lambda }^b{e}^{bt})}^{\alpha }}\left(\frac{1-{\lambda }^{ab}{e}^{abt}}{1-{\lambda }^a{e}^{at}}\right){\left(\frac{at}{1-{\lambda }^a{e}^{at}}\right)}^{\alpha -1}{e}^{abyt}.\end{array}$$

From (3) and (10), we write

$$\begin{array}{rcl}f(t)& =& \frac{{(-1)}^{\alpha -1}}{a^{\alpha -1}}\sum _{\beta =0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{\beta +\alpha -1}{\beta }\right){\lambda }^{\alpha \beta }{e}^{bt(\beta +ax)}\frac{1}{{\lambda }^a}\sum _{r=0}^{\mathrm{\infty }}\sum _{p=0}^r\left(\genfrac{}{}{0ex}{}{r}{p}\right){(-1)}^{r-p}{S}_p(b;{\lambda }^a)\frac{a^r{t}^r}{r!}\\ \times \sum _{k=0}^{\mathrm{\infty }}{\mathcal{B}}_k^{(\alpha -1)}(by;{\lambda }^a)\frac{a^k{t}^k}{k!}.\end{array}$$

Since $|log\lambda +t|<min(\frac{2\pi }{a},\frac{2\pi }{b})$, after the Cauchy product, we have

$$\begin{array}{rc}=& \sum _{n=0}^{\mathrm{\infty }}\{\sum _{s=0}^n\left(\genfrac{}{}{0ex}{}{n}{s}\right)a^{s+1-\alpha }{b}^{n-s}{(-1)}^{\alpha -1}\sum _{k=0}^s\left(\genfrac{}{}{0ex}{}{s}{k}\right)\sum _{p=0}^r\left(\genfrac{}{}{0ex}{}{r}{p}\right){(-1)}^{r-p}{\lambda }^{-a}\\ \times S_p(b;{\lambda }^a){\mathcal{B}}_k^{(\alpha -1)}(by;{\lambda }^a)\sum _{\beta =0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{\beta +\alpha -1}{\beta }\right)\frac{{\lambda }^{b\beta }}{{(\beta +ay)}^{s-n}}\}\frac{t^n}{n!}.\end{array}$$

Compressing to coefficients $\frac{t^n}{n!}$ and by using (5), we prove the theorem. □

Theorem 2.2 For all $a,b,m\in \mathbb{N}$, $n\in {\mathbb{N}}_0$, we have the following symmetry identity:

(13)

Proof Let $h(t)=\frac{t^{2m+1}{e}^{abxt}{(1-{\lambda }^{ab}{e}^{abt})}^m{e}^{abyt}}{{(1-{\lambda }^a{e}^{at})}^{m+1}{(1-{\lambda }^b{e}^{bt})}^{m+1}}$. Then

$$h(t)=\frac{1}{a^{m+1}b}{\left(\frac{at}{1-{\lambda }^a{e}^{at}}\right)}^{m+1}{e}^{abxt}{\left(\frac{1-{\lambda }^{ab}{e}^{abt}}{1-{\lambda }^b{e}^{bt}}\right)}^m\left(\frac{bt}{1-{\lambda }^b{e}^{bt}}\right)e^{abyt}.$$

From (3) and (8), we have

$$\begin{array}{rcl}h(t)& =& \frac{{(-1)}^m}{a^{m+1}b}\sum _{n=0}^{\mathrm{\infty }}\{\sum _{r=0}^n\left(\genfrac{}{}{0ex}{}{n}{r}\right){\mathcal{B}}_{n-r}^{(m+1)}(bx;{\lambda }^a)a^{n-r}{\lambda }^{-mb}\sum _{k=0}^r\left(\genfrac{}{}{0ex}{}{r}{k}\right)\\ \times \sum _{p=0}^k\left(\genfrac{}{}{0ex}{}{k}{p}\right){(-l)}^{k-p}{S}_p^{(m)}(a;{\lambda }^b){\mathcal{B}}_{r-k}(ay;{\lambda }^b)b^r\}\frac{t^n}{n!}.\end{array}$$

In a similar manner,

$$\begin{array}{rcl}h(t)& =& \frac{{(-1)}^m}{b^{m+1}a}\sum _{n=0}^{\mathrm{\infty }}\{\sum _{r=0}^n\left(\genfrac{}{}{0ex}{}{n}{r}\right){\mathcal{B}}_{n-r}^{(m+1)}(ay;{\lambda }^b)b^{n-r}{\lambda }^{-ma}\sum _{k=0}^r\left(\genfrac{}{}{0ex}{}{r}{k}\right)\\ \times \sum _{p=0}^k\left(\genfrac{}{}{0ex}{}{k}{p}\right){(-l)}^{k-p}{S}_p^{(m)}(b;{\lambda }^a){\mathcal{B}}_{r-k}(ax;{\lambda }^a)a^r\}\frac{t^n}{n!}.\end{array}$$

Comparing the coefficients of $\frac{t^n}{n!}$, we proved the theorem. □

Corollary 2.3 We put $a=b=\lambda =1$ in (13). We have

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathcal{B}}_k^{(m+1)}(y){\mathcal{B}}_{n-k}(x)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathcal{B}}_k^{(m+1)}(x){\mathcal{B}}_{n-k}(y).$$

3 Some symmetry identities for the Apostol-Euler polynomials

Theorem 3.1 Let a and b be positive integers with the same parity. Then

(14)

Proof Let $h(t)=\frac{2e^{abxt}}{{\lambda }^a{e}^{at}+1}\frac{1+{(-1)}^{a+1}{({\lambda }^b{e}^{bt})}^a}{{\lambda }^b{e}^{bt}+1}$. From (4) and (9) for $l=1$, we have

$$\begin{array}{rcl}h(t)& =& \sum _{k=0}^{\mathrm{\infty }}{\mathcal{E}}_k(bx;{\lambda }^a)\frac{a^k{t}^k}{k!}\frac{1}{{\lambda }^a}\sum _{l=0}^{\mathrm{\infty }}\sum _{p=0}^l\left(\genfrac{}{}{0ex}{}{l}{p}\right){(-1)}^{l-p}{T}_p(a;{\lambda }^b)\frac{b^l{t}^l}{l!}\\ =& \sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathcal{E}}_k(bx;{\lambda }^a){\lambda }^{-b}{a}^k{b}^{n-k}\sum _{p=0}^{n-k}\left(\genfrac{}{}{0ex}{}{n-k}{p}\right){(-1)}^{n-k-p}{T}_{n-k}(a;{\lambda }^b)\right)\frac{t^n}{n!}.\end{array}$$

Since ${(-1)}^{a+1}={(-1)}^{b+1}$, the expression for $h(t)=\frac{2e^{abxt}}{{\lambda }^b{e}^{bt}+1}\frac{1+{(-1)}^{b+1}{({\lambda }^a{e}^{at})}^b}{{\lambda }^a{e}^{at}+1}$ is symmetric in a and b. Therefore, we obtain the following power series for $h(t)$ by symmetry:

$$h(t)=\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathcal{E}}_k(ax;{\lambda }^b){\lambda }^{-a}{b}^k{a}^{n-k}\sum _{p=0}^{n-k}\left(\genfrac{}{}{0ex}{}{n-k}{p}\right){(-1)}^{n-k-p}{T}_{n-k}(b;{\lambda }^a)\right)\frac{t^n}{n!}.$$

Equating the coefficient of $\frac{t^n}{n!}$ in the two expressions for $h(t)$ gives us the desired result. □

Theorem 3.2 Let a and b be positive integers with the same parity. Then

(15)

Proof Let $k(t)=\frac{2^{\alpha +2}{e}^{abxt}{(1+{(-1)}^{a+1}({\lambda }^{ab}{e}^{abt}))}^{\alpha }}{{({\lambda }^a{e}^{at}+1)}^{\alpha +1}{({\lambda }^b{e}^{bt}+1)}^{\alpha +1}}{e}^{abyt}$. From (4) and (9), we write

$$\begin{array}{rcl}k(t)& =& {\left(\frac{2}{{\lambda }^a{e}^{at}+1}\right)}^{(\alpha +1)}{e}^{abxt}{\left(\frac{1+{(-1)}^{a+1}({\lambda }^{ab}{e}^{abt})}{{\lambda }^b{e}^{bt}+1}\right)}^{\alpha }\left(\frac{2}{{\lambda }^b{e}^{bt}+1}\right)e^{abyt}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_n^{(\alpha +1)}(bx;{\lambda }^a)\frac{a^n{t}^n}{n!}\frac{1}{{\lambda }^{b\alpha }}\sum _{n=0}^{\mathrm{\infty }}\sum _{n=0}^p\left(\genfrac{}{}{0ex}{}{p}{n}\right){(-\alpha )}^{n-p}{T}_p^{(\alpha )}(a;{\lambda }^b)\frac{b^n{t}^n}{n!}\\ \times \sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_n(ay;{\lambda }^b)\frac{b^n{t}^n}{n!}\\ =& \sum _{n=0}^{\mathrm{\infty }}\{\sum _{s=0}^n\left(\genfrac{}{}{0ex}{}{n}{s}\right){\mathcal{E}}_{n-s}^{(\alpha +1)}(bx;{\lambda }^a){\lambda }^{(-b\alpha )}{a}^{n-s}\\ \times \sum _{k=0}^s\left(\genfrac{}{}{0ex}{}{s}{k}\right)\sum _{p=0}^k\left(\genfrac{}{}{0ex}{}{k}{p}\right){(-\alpha )}^{k-p}{T}_k^{(\alpha )}(a;{\lambda }^b){\mathcal{E}}_{s-k}(ay;{\lambda }^b)b^s\}\frac{t^n}{n!}.\end{array}$$

Since ${(-1)}^{a+1}={(-1)}^{b+1}$, the expression for $h(t)$ is symmetric in a and b.

In a similar manner, we have

$$\begin{array}{rcl}k(t)& =& \sum _{n=0}^{\mathrm{\infty }}\{\sum _{s=0}^n\left(\genfrac{}{}{0ex}{}{n}{s}\right){\mathcal{E}}_{n-s}^{(\alpha +1)}(ay;{\lambda }^b){\lambda }^{(-a\alpha )}{b}^{n-s}\\ \times \sum _{k=0}^s\left(\genfrac{}{}{0ex}{}{s}{k}\right)\sum _{p=0}^k\left(\genfrac{}{}{0ex}{}{k}{p}\right){(-\alpha )}^{k-p}{T}_k^{(\alpha )}(b;{\lambda }^a){\mathcal{E}}_{s-k}(bx;{\lambda }^b)a^s\}\frac{t^n}{n!}.\end{array}$$

Equating the coefficient of $\frac{t^n}{n!}$ in the two expressions for $k(t)$ gives us the desired result. □

Theorem 3.3 Let p, l, a, b and n be positive integers and a, b be of the same parity. Then

(16)

Proof Let $k(t)=\frac{at{e}^{ant}}{{\lambda }^a{e}^{at}-1}\frac{1+{(-1)}^b{({\lambda }^b{e}^{bt})}^a}{{\lambda }^a{e}^{at}+1}$. From (3) and (10), we have

$$\begin{array}{rcl}k(t)& =& \frac{at{e}^{ant}}{{\lambda }^a{e}^{at}-1}\frac{1+{(-1)}^{b+1}{({\lambda }^b{e}^{bt})}^a}{{\lambda }^a{e}^{at}+1}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{B}}_n(n;{\lambda }^a)\frac{a^n{t}^n}{n!}{\lambda }^{-\alpha }\sum _{n=0}^{\mathrm{\infty }}\sum _{n=0}^p\left(\genfrac{}{}{0ex}{}{p}{n}\right){(-1)}^{n-p}{T}_p(b;{\lambda }^a)\frac{a^n{t}^n}{n!}\\ =& \sum _{n=0}^{\mathrm{\infty }}\left(\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right){\mathcal{B}}_{n-l}(n;{\lambda }^a)a^{n-l}{\lambda }^{-a}\sum _{p=0}^l\left(\genfrac{}{}{0ex}{}{l}{p}\right){(-1)}^{l-p}{T}_p(b;{\lambda }^a)a^l\right)\frac{t^n}{n!}.\end{array}$$

On the other hand, we write the function $k(t)$ as

$$\begin{array}{rcl}k(t)& =& \frac{1}{2}\frac{2a{e}^{\frac{n}{2}(2at)}}{{\lambda }^{2a}{e}^{2at}-1}+\frac{{(-1)}^{b+1}{\lambda }^{ab}2at{e}^{2at(\frac{n+b}{2})}}{2({\lambda }^{2a}{e}^{2at}-1)}\\ =& \frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{B}}_n(\frac{n}{2};{\lambda }^{2a})2^n\frac{a^n{t}^n}{n!}+\frac{{(-1)}^{b+1}{\lambda }^{ab}}{2}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{B}}_n(\frac{b+n}{2};{\lambda }^{2a})2^n\frac{a^n{t}^n}{n!}\\ =& \sum _{n=0}^{\mathrm{\infty }}\left(2^{n-l}{a}^n({\mathcal{B}}_n(\frac{n}{2};{\lambda }^{2a})+\frac{{(-1)}^{b+1}{\lambda }^{ab}}{2}{\mathcal{B}}_n(\frac{b+n}{2};{\lambda }^n))\right)\frac{t^n}{n!}.\end{array}$$

Equating the coefficient of $\frac{t^n}{n!}$, we obtain (16). □