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Some symmetric identities for the generalized Bernoulli, Euler and Genocchi polynomials associated with Hermite polynomials
- Waseem A. Khan1Email author and
- Hiba Haroon1
Received: 12 July 2016
Accepted: 20 October 2016
Published: 4 November 2016
Abstract
In 2008, Liu and Wang established various symmetric identities for Bernoulli, Euler and Genocchi polynomials. In this paper, we extend these identities in a unified and generalized form to families of Hermite–Bernoulli, Euler and Genocchi polynomials. The procedure followed is that of generating functions. Some relevant connections of the general theory developed here with the results obtained earlier by Pathan and Khan are also pointed out.
Keywords
Hermite polynomialsBernoulli polynomialsEuler polynomials and Genocchi polynomialsHermite–Bernoulli polynomialsHermite–Euler polynomialsHermite–Genocchi polynomialsSymmetric identities
Mathematics Subject Classification
05A1511B6833C4533C99
Background
Let \(H_n(x,y)\) be denoted by the 2-variable Kampé de Fériet generalization of the Hermite polynomials Bell (1934) and Dattoli et al. (1999) defined asThese polynomials are usually defined by the generating function asand reduce to the ordinary Hermite polynomials \(H_n(x)\) (see Andrews 1985) when \(y=-1\) and x is replaced by 2x.
$$\begin{aligned} H_n(x,y)=n!\sum \limits _{r=0}^{[\frac{n}{2}]}\frac{y^rx^{n-2r}}{r!(n-2r)!} \end{aligned}$$
(1)
$$\begin{aligned} e^{xt+yt^2}=\sum \limits _{n=0}^{\infty }H_n(x,y)\frac{t^n}{n!} \end{aligned}$$
(2)
The generalized Bernoulli polynomials \(B_{n}^{(\alpha )}(x)\) of order \(\alpha \), \(\alpha \in \mathbb C\), the generalized Euler polynomials \(E_n^{(\alpha )}(x)\) of order \(\alpha \), \(\alpha \in \mathbb C\) and the generalized Genocchi polynomials \(G_n^{(\alpha )}(x)\) of order \(\alpha \), \(\alpha \in \mathbb C\), each of degree n as well as in \(\alpha \) are defined respectively by the following generating functions (see Dere and Simsek 2015, [Erdelyi et al. (1953), vol. 3, p. 253 et seq.], [Luke (1969), Section 2.8] and Luo et al. 2003; Pathan 2012; Pathan and Khan 2014, 2015, 2016; Simsek 2010; Srivastava et al. 2012):
andIt is easy to see thatandMoreoverFor each \(k\in \mathbb N_0\), \( S_k(n) \) defined byis called the sum of integer powers Tuenter (2001). The exponential generating function for \(S_k(n)\) is given byFor each \(k\in \mathbb N_0\), \( T_k(n)\) defined byis called the alternating sum. The exponential generating function for \(T_k(n)\) isThe following are some special valueswhere \( \delta _{(i,j)}\) is the Kronecker delta defined by \(\delta _{(i,j)}=1\) for \( i=j \) and \( \delta _{(i,j)}=0\) for \(i\ne j\).
$$\begin{aligned} \left( \frac{t}{e^t-1}\right) ^{\alpha }e^{xt}=\sum \limits _{n=0}^{\infty }B_{n}^{(\alpha )}(x)\frac{t^n}{n!},\quad (|t| < 2\pi ; 1^{\alpha }=1) \end{aligned}$$
(3)
$$\begin{aligned} \left( \frac{2}{e^t+1}\right) ^{\alpha }e^{xt}=\sum \limits _{n=0}^{\infty }E_{n}^{(\alpha )}(x)\frac{t^n}{n!},\quad (|t| < \pi ; 1^{\alpha }=1) \end{aligned}$$
(4)
$$\begin{aligned} \left( \frac{2t}{e^t+1}\right) ^{\alpha }e^{xt}=\sum \limits _{n=0}^{\infty }G_{n}^{(\alpha )}(x)\frac{t^n}{n!}, \quad (|t| < \pi ; 1^{\alpha }=1) \end{aligned}$$
(5)
$$\begin{aligned} B_n^{(\alpha )}(x)=B_{n}^{(\alpha )}(x;1), \quad E_n^{(\alpha )}(x)=E_{n}^{(\alpha )}(x;1) \end{aligned}$$
$$\begin{aligned} G_n^{(\alpha )}(x)=G_{n}^{(\alpha )}(x;1)\quad(n\in \mathbb N) \end{aligned}$$
(6)
$$\begin{aligned} B_n (1)-B_n=\delta _{(n,1)},\quad n\ge 0 \end{aligned}$$
(7)
$$\begin{aligned} S_k(n)=\sum \limits _{j=0}^{n}j^k \end{aligned}$$
(8)
$$\begin{aligned} \sum \limits _{k=0}^{\infty }S_k(n)\frac{t^k}{k!}=\frac{e^{(n+1)t}-1}{e^t-1} \end{aligned}$$
(9)
$$\begin{aligned} T_k(n)=\sum \limits _{j=0}^{n}(-1)^j {j^k} \end{aligned}$$
(10)
$$\begin{aligned} \sum \limits _{k=0}^{\infty }T_k(n)\frac{t^k}{k!}=\frac{1-{(-e^t)}^{(n+1)}}{1+e^t} \end{aligned}$$
(11)
$$\begin{aligned}&S_k (0)=T_k(0)=\delta _{(0,k)}\nonumber \\&S_k(1)=0^k+1^k=\delta _{(0,k)}+1, \quad T_k (1)=0^k-1^k=\delta _{(0,k)}-1 \end{aligned}$$
(12)
A close relation of the power sum and the Bernoulli polynomials, also the alternate sum and the Euler polynomials, can be seen in Abramowitz and Stegun [1972, Eq.(23.1.4)] as follows
where n and k are nonnegative integers.
$$\begin{aligned} S_k(n)=\sum \limits _{i=0}^{n}{i^k}=\frac{{B_{k+1}(n+1)}-{B_{k+1}}}{k+1} \end{aligned}$$
(13)
$$\begin{aligned} (-1)^nT_k(n)=\sum \limits _{i=0}^{n}(-1)^{n-i}i^k=\frac{E_k{(n+1)}+(-1)^nE_k{(0)}}{2} \end{aligned}$$
(14)
The hyperbolic cotangent and the hyperbolic tangent (Weisstein http://mathworld.wolfram.com) respectively are defined byandThereforewhere Pathan and Khan (2015) introduced the generalized Hermite–Bernoulli polynomials for two variables \({}_HB_n^{(\alpha )}(x,y)\) defined bywhich is essentially a generalization of Bernoulli numbers, Bernoulli polynomials, Hermite polynomials and Hermite–Bernoulli polynomials \({_HB}_n(x,y)\) introduced by Dattoli et al. [1999, p. 386 (1.6)] in the formThe purpose of this paper is to give some general symmetry identities for generalized Hermite–Euler, Hermite–Genocchi and mixed type polynomials by using different analytical means on their respective generating functions. These results extend some known identities of Hermite, Bernoulli, Euler, Genocchi and mixed type polynomials studied by Dattoli et al. (1999), Liu and Wang (2009), Pathan (2012) and Pathan and Khan (2014, 2015, 2016).
$$\begin{aligned} cothz=\frac{e^{2z}+1}{e^{2z}-1}=\sum \limits _{n=-1}^{\infty }\frac{2^n(B_{n+1}+B_{n+1}(1))}{(n+1)!}z^n=\sum \limits _{n=0}^{\infty }\frac{2^{n-1}(B_n+B_n(1))}{n!}z^{n-1} \end{aligned}$$
(15)
$$\begin{aligned} tanhz=\frac{e^{2z}-1}{e^{2z}+1}=\sum \limits _{n=1}^{\infty }\frac{2^{n+1}(2^{n+1}-1)B_{n+1}}{(n+1)!}z^n \end{aligned}$$
(16)
$$\begin{aligned} tanhz=-\sum \limits _{n=1}^{\infty }E_n (0)\frac{{(2z)}^n}{n!} \end{aligned}$$
(17)
$$\begin{aligned} E_n(0)=\frac{2(1-2^{n+1})}{(n+1)}B_{n+1}, ~n\ge 0, \quad ({\text{see}}\,[1,\,{\text{Eq.}}\,(23.1.20)]).\end{aligned}$$
$$\begin{aligned} \left( \frac{t}{e^t-1}\right) ^{\alpha }e^{xt+yt^2}=\sum \limits _{n=0}^{\infty }{}_HB_n^{(\alpha )}(x,y)\frac{t^n}{n!} \end{aligned}$$
(18)
$$\begin{aligned} \left( \frac{t}{e^t-1}\right) e^{xt+yt^2}=\sum \limits _{n=0}^{\infty }{}_HB_n(x,y)\frac{t^n}{n!} \end{aligned}$$
(19)
Symmetry identities for generalized Hermite–Euler polynomials
In this section, we establish general symmetry identities for the generalized Hermite–Euler polynomials \({}_HE_{n}^{(\alpha )}(x,y)\), of order\(\alpha \) and alternate power sum. Throughout this section \(\alpha \) will be taken as an arbitrary real or complex parameter.
Theorem 1
For integers
\(n\ge 0\), \(a\ge 1\)
and
\(b\ge 1,\) if a
and
b
have
the same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}{}_HE_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{k}b^{n-k} {}_HE_{n-k}^{(\alpha )}(ax,a^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(20)
Proof
Let \(G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }}\)
Replacing n by \(n-k\) in the R.H.S. of above equation, we getUsing a similar plan, we getBy comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S of above Eqs. (21) and (22), we arrive at the desired result.
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt} \\& ={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(a-1)}\frac{(bt)^i}{i!}\right) } \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(ay)\frac{(bt)^k}{k!}}\right) \\&={\left( {\sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}}\right) }{\left( \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{\infty } {b^{i+k}}T_i{(a-1)}E_k^{(\alpha -1)}(ay)\frac{t^{i+k}}{i!k!}\right) } \\&={\sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}}{\sum \limits _{k=0}^{\infty }{b^k}\sum \limits _{i=0} ^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{k!}} \\ G(t)&={\sum \limits _{n=0}^{\infty }\sum \limits _{k=0}^{\infty }{a^n}{b^k}{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k} \left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\frac{t^{n+k}}{k!n!}} \end{aligned}$$
$$\begin{aligned} G(t)={\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}{}_HE_{n-k} ^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\right] }\frac{t^{n}}{n!} \end{aligned}$$
(21)
$$\begin{aligned} G(t)={\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^{n-k}}{a^k}{}_HE_{n-k} ^{(\alpha )}(ax,a^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(b-1)}E_{k-i}^{(\alpha -1)}(by)\right] }\frac{t^{n}}{n!} \end{aligned}$$
(22)
Remark 1
For \(z=0\) in Theorem (1), the result reduces to known result of Liu and Wang [2009, p. 3348, (2.1)].
Corollary 1
For integers
\( n\ge 0\), \(a\ge 1\)
and
\( b\ge 1 \) , if
a
and
b
have the same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}E_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay) \nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{k}b^{n-k} E_{n-k}^{(\alpha )}(ax)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(23)
Theorem 2
For integers
\( n\ge 0\), \(a\ge 1\)
and
\( b\ge 1 \) , if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}{}_HE_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}{n+1}\\ ~l~\end{array}\right) \frac{B_{n+1-l}}{n+1} a^{n-l}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k}a^{k}{}_HE_{l-k}^{(\alpha )}(ax,a^{2}z)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(24)
Proof
Let \(G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }}\)
Replacing n by \(n-k\) in the R.H.S. of above equation, we getSince G(t) is not symmetric in a and b, thus G(t) can also be expanded asUsing identity (7) and comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (25) and (26), we get the desired result.
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\\ & ={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(a-1)}\frac{(bt)^i}{i!}\right) } \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(ay)\frac{(bt)^k}{k!}}\right) \\& ={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{k=0}^{\infty }{b^k}\sum \limits _{i=0} ^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{k!}\right) } \end{aligned}$$
$$\begin{aligned} G(t)={\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}{}_HE_{n-k} ^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\right] }\frac{t^{n}}{n!} \end{aligned}$$
(25)
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{at})}^b}{1+e^{at}}\right) }{\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b} \right) }{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& ={\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(b-1)}\frac{(at)^i}{i!}\right) } \left( -\sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) \nonumber \\&\qquad \times \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(by)\frac{(at)^k}{k!}}\right) \nonumber \\G(t)&=\left( -\sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) {\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!} \right) }\nonumber \\&\qquad \times {\left( \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{k}a^k\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(b-1)}E_{k-i}^{(\alpha -1)}(by)\frac{t^k}{k!} \right) } \end{aligned}$$
(26)
Remark 2
For \( z=0\) in Theorem (2), the result reduces to known result of Liu and Wang [2009, p. 3349, Theorem (4)].
Corollary 2
For integers
\( n\ge 0\), \(a\ge 1\)
and
\( b\ge 1 \) , if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}E_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}{n+1}\\ l\end{array}\right) \frac{B_{n+1-l}}{n+1} a^{n-l}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k}a^{k}E_{l-k}^{(\alpha )}(ax)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(27)
Theorem 3
For integers
\( n\ge 1\) , \(a\ge 1\)
and
\( b\ge 1\), if
a
is even and
b
is odd, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k} b^k{}_HE_{n-k}^{(\alpha )}(bx,b^2 z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0}^{n-1}\left( \begin{array}{lll}n\\ l\end{array}\right) E_{n-l}(0)a^{n-l} \sum \limits _{k=0}^{l}\left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k} a^k {}_HE_{l-k}^{(\alpha )} (ax,a^2 z)\nonumber \\&\qquad \times \sum \limits _{i=0}^k\left( \begin{array}{lll}k\\ i\end{array}\right) T_i (b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(28)
Proof
LetAnother expansion of G(t) is as follows :Now using Eq. (17) and identity (7) and comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (29) and (30), we arrive at the desired result.
$$\begin{aligned}G(t)&=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }}\nonumber \\G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\&={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(a-1)}\frac{(bt)^i}{i!}\right) } \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(ay)\frac{(bt)^k}{k!}}\right) \nonumber \\ \end{aligned}$$
(29)
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{at})}^b}{1+e^{at}}\right) } {\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b}\right) }{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\ &=\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) \left( -\sum \limits _{n=0}^{\infty } \frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) \left( \sum \limits _{i=0}^{\infty }T_i{(b-1)}\frac{(at)^i}{i!}\right) \nonumber \\&\quad \times \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)} {(by)\frac{(at)^k}{k!}}\right) \nonumber \\G(t)&=\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) \left( -\sum \limits _{n=0}^{\infty } \frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) \nonumber \\&\quad \times \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) a^kT_i{(b-1)}E_{k-i}^{(\alpha -1)}(by)\frac{t^k}{k!} \end{aligned}$$
(30)
Remark 3
For \( z=0 \) in Theorem (3), the result reduces to known result of Liu and Wang [2009, p. 3350, (2.11)].
Corollary 3
For integers
\( n\ge 1,\)
\(a\ge 1\)
and
\( b\ge 1\), if
a
is even and
b
is odd, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k} b^kE_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0}^{n-1}\left( \begin{array}{lll}n\\ l\end{array}\right) E_{n-l}(0)a^{n-l} \sum \limits _{k=0}^{l}\left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k} a^k E_{l-k}^{(\alpha )}(ax)\nonumber \\&\qquad \times \sum \limits _{i=0}^k\left( \begin{array}{lll}k\\ i\end{array}\right) T_i (b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(31)
Theorem 4
For integers
\(n\ge 0\), \(a\ge 1 \)
and
\( b\ge 1\), if
a
and
b
have same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}{}_HE_k^{(\alpha )}(bx+{\frac{b}{a}}i,b^2z) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}\sum \limits _{i=0}^{b-1}{(-1)^i}{}_HE_k^{(\alpha )} (ax+{\frac{a}{b}}i,a^2z)E_{n-k}^{(\alpha -1)}(by) \end{aligned}$$
(32)
Proof
LetWe expand G(t) as follows :Replacing n by \(n-k\) in the R.H.S. of above equation, we getUsing a similar plan, we obtainBy comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of last two Eqs. (33) and (34), we arrive at the desired result.
$$\begin{aligned} G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }} \end{aligned}$$
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\\& ={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}(-1)^ie^{bti}\right) }{\left( \sum \limits _{n=0}^{\infty }E_n^ {(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }\\&={\left( \sum \limits _{i=0}^{a-1}(-1)^i{\left( \frac{2}{e^{at}+1}\right) }^{\alpha }e^{(bx+\frac{b}{a}i)at+a^2b^2zt^2}\right) }{\left( \sum \limits _{n=0}^{\infty } E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }\\G(t)&={\left( \sum \limits _{i=0}^{a-1}(-1)^i\sum \limits _{k=0}^{\infty }{}_HE_k^{(\alpha )}(bx+\frac{b}{a}i;b^2z)\frac{(at)^k}{k!}\right) }{\left( \sum \limits _{n=0} ^{\infty }E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) } \end{aligned}$$
$$\begin{aligned} G(t)=\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^kb^{n-k}\sum \limits _{i=0}^ {a-1}(-1)^i{}_HE_k^{(\alpha )}(bx+\frac{b}{a}i;b^2z)E_{n-k}^{(\alpha -1)}(ay)\right] \frac{t^n}{n!} \end{aligned}$$
(33)
$$\begin{aligned} G(t)=\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) b^ka^{n-k}\sum \limits _{i=0}^ {b-1}(-1)^i{}_HE_k^{(\alpha )}(ax+\frac{a}{b}i;a^2z)E_{n-k}^{(\alpha -1)}(by)\right] \frac{t^n}{n!} \end{aligned}$$
(34)
Remark 4
For \(z=0\) in Theorem (4), the result reduces to known result of Liu and Wang [2009, Theorem 2.10].
Corollary 4
For integers
\(n\ge 0\), \(a\ge 1 \)
and
\( b\ge 1\), if
a
and
b
have same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}E_k^{(\alpha )}(bx+{\frac{b}{a}}i) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}\sum \limits _{i=0}^{b-1}{(-1)^i}E_k^{(\alpha )} (ax+{\frac{a}{b}}i)E_{n-k}^{(\alpha -1)}(by) \end{aligned}$$
(35)
Theorem 5
For integers
\(n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1\), if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}{}_HE_k^{(\alpha )}(bx+{\frac{b}{a}}i,b^2z) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left({n+1}\right) \frac{B_{n+1-l}}{n+1} b^{n-l} \sum \limits _{k=0}^{l}{}\left( \begin{array}{lll}l\\ k\end{array}\right) a^{n-k}b^{k}\nonumber \\&\qquad \times \sum \limits _{i=0}^{b-1}(-1)^i{}_HE_k^{(\alpha )}(ax+\frac{a}{b}i,a^{2}z) E_{l-k}^{(\alpha -1)}(by) \end{aligned}$$
(36)
Proof
LetIn view of definition (1.15), G(t) has the following expansion:Using identity (7) and comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (33) and (37), we get the result (36).
$$\begin{aligned} G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }} \end{aligned}$$
$$\begin{aligned}&G(t)={\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{at})}^b}{1+e^{at}}\right) }{\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b} \right) }{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\&\quad ={\left( -\sum \limits _{n=0}^{\infty }\frac{B_{n}+B_{n}(1)}{n!}abt^{n-1}\right) }{\left( \sum \limits _{i=0}^{b-1}{(-1)^i}\sum \limits _{n=0}^{\infty } {}_HE_n^{(\alpha )}(ax+{\frac{a}{b}}i,a^2z)\frac{(bt)^n}{n!}\right) }\nonumber \\&\qquad \times {\left( \sum \limits _{n=0}^{\infty }E_{n}^{(\alpha -1)}(by)\frac{(bt)^n}{n!}\right) } \end{aligned}$$
(37)
Remark 5
For \(z =0\) in Theorem (5), the result reduces to known result of Liu and Wang [2009, Theorem (2.13)].
Corollary 5
For integers
\(n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1\), if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}E_k^{(\alpha )}(bx+{\frac{b}{a}}i) E_{n-k}^{(\alpha -1)}(ay) \nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}{n+1}\\ l \end{array}\right) \frac{B_{n+1-l}}{n+1} b^{n-l} \sum \limits _{k=0}^{l}{}\left( \begin{array}{lll}l\\ k\end{array}\right) a^{n-k}b^{k}\nonumber \\&\qquad \times \sum \limits _{i=0}^{b-1}(-1)^iE_k^{(\alpha )}(ax+\frac{a}{b}i) E_{l-k}^{(\alpha -1)}(by) \end{aligned}$$
(38)
Symmetric identities for Hermite–Genocchi polynomials
In this section, we derive some symmetry identities for Hermite–Genocchi polynomials \({}_HG_{n}(x,y)\). We now begin the following theorems.
Theorem 6
For integers
\(n\ge 0\), \(a\ge 1\)
and
\(b\ge 1\), if
a
and
b
have the same parity, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k+1}}{}_HG_k(bx,b^2y)T_{n-k}(a-1)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array} {lll}n\\ k\end{array}\right) {b^k}{a^{n-k+1}}{}_HG_k(ax,a^2y)T_{n-k}(b-1) \end{aligned}$$
(39)
$$\begin{aligned} b\sum \limits _{i=0}^{a-1}{(-1)^i}a^n{}_HG_n\left(bx+{\frac{b}{a}}i,b^2y\right)=a\sum \limits _{i=0}^{b-1}{(-1)^i}b^n{}_HG_n\left(ax+{\frac{a}{b}}i,a^2y\right) \end{aligned}$$
(40)
Proof
LetWe can expand P(t) as followsOn the other handComparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of above Eqs. (42) and (43), yields identity (39).
$$\begin{aligned} P(t)=: \frac{2abt(1-(-e^{bt})^a)e^{abxt+a^2b^2yt^2}}{(e^{at}+1)(e^{bt}+1)} \end{aligned}$$
(41)
$$\begin{aligned}P(t)&=b{\left( \frac{2at}{e^{at}+1}\right) }e^{abxt+a^2b^2yt^2}{\left( \frac{1-(-e^{bt})^a}{e^{bt}+1}\right) }\nonumber \\&=b{\left( \sum \limits _{k=0}^{\infty }{}_HG_{k}(bx,b^2y)\frac{(at)^k}{k!}\right) }{\left( \sum \limits _{n=0}^{\infty }T_{n}(a-1)\frac{(bt)^n}{n!}\right) }\nonumber \\& =b\sum \limits _{n=0}^{\infty }\sum \limits _{k=0}^{\infty }a^kb^n{}_HG_{k}(bx,b^2y)T_{n}(a-1)\frac{t^{n+k}}{n!k!}\nonumber \\P(t)&=\sum \limits _{n=0}^{\infty }{\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^kb^{n-k+1}{}_HG_{k}(bx,b^2y)T_{n-k}(a-1) \right] }\frac{t^n}{n!} \end{aligned}$$
(42)
$$\begin{aligned} P(t)=\sum \limits _{n=0}^{\infty }{\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) b^ka^{n-k+1}{}_HG_{k}(ax,a^2y)T_{n-k}(b-1) \right] }\frac{t^n}{n!} \end{aligned}$$
(43)
Now following (41) we expand P(t) as follows:On the similar lines, we can show thatComparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of above Eqs. (44) and (45), yields identity (40).
$$\begin{aligned}P(t)&=b{\left( \frac{2at}{e^{at}+1}\right) }e^{abxt+a^2b^2yt^2}{\left( \frac{1-(-e^{bt})^a}{e^{bt}+1}\right) } \nonumber \\&=b{\left( \frac{2at}{e^{at}+1}\right) }e^{abxt+a^2b^2yt^2}\sum \limits _{i=0}^{a-1}(-1)^ie^{bti} \nonumber \\& =b{\left( \frac{2at}{e^{at}+1}\right) }e^{a^2b^2yt^2}\sum \limits _{i=0}^{a-1}(-1)^ie^{(bx+\frac{b}{a}i)at}\nonumber \\P(t)&=b\sum \limits _{n=0}^{\infty }{\left[ \sum \limits _{i=0}^{a-1}a^n(-1)^i{}_HG_{n}\left(bx+\frac{b}{a}i,b^2y\right)\right] }\frac{t^n}{n!} \end{aligned}$$
(44)
$$\begin{aligned} P(t)=a\sum \limits _{n=0}^{\infty }{\left[ \sum \limits _{i=0}^{b-1}b^n(-1)^i{}_HG_{n}\left(ax+\frac{a}{b}i,a^2y \right)\right] }\frac{t^n}{n!} \end{aligned}$$
(45)
Remark 6
For \(y=0\) in Theorem (6) and (7), the result reduces to known result of Liu and Wang [2009, p. 3358, (4.2) and (4.3)].
Corollary 6
For integers
\(n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1\), if
a
and
b
have the same parity, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k+1}}G_k(bx)T_{n-k}(a-1)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array} {lll}n\\ k\end{array}\right) {b^k}{a^{n-k+1}}G_k(ax)T_{n-k}(b-1) \end{aligned}$$
(46)
$$\begin{aligned} b\sum \limits _{i=0}^{a-1}{(-1)^i}a^nG_n\left(bx+{\frac{b}{a}}i\right)=a\sum \limits _{i=0}^{b-1}{(-1)^i}b^nG_n\left(ax+{\frac{a}{b}}i\right) \end{aligned}$$
(47)
Theorem 7
For integers
\( n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1 \), if
a
is odd and
b
is even, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^k b^{n-k+1}{}_HG_k(bx,b^2y)T_{n-k}(a-1)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}n+1\\ l\end{array}\right) \frac{B_{n+1-l}}{n+1}b^{n-l}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) b^ka^{n-k+1}{}_HG_{k}(ax,a^2y)T_{l-k}(b-1) \end{aligned}$$
(48)
$$\begin{aligned}&b\sum \limits _{i=0}^{a-1}{(-1)^i}a^n{}_HG_n\left(bx+{\frac{b}{a}}i,b^2y\right)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}n+1\\ l\end{array}\right) \frac{B_{n+1-l}}{n+1}a^{n-l+1}\sum \limits _{i=0}^{b-1}(-1)^ib^n{}_HG_{l}\left(ax+\frac{a}{b}i,a^2y\right) \end{aligned}$$
(49)
Proof
LetWe can expand P(t) as follows:Another expansion of P(t) isComparing the coefficients of \(\frac{t^n}{n!}\) in Eqs. (43) and (50), we obtain the first identity. Also comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (44) and (51), we obtain the second identity.
$$\begin{aligned} P(t)=: \frac{2abt(1-(-e^{bt})^a)e^{abxt+a^2b^2yt^2}}{(e^{at}+1)(e^{bt}+1)} \end{aligned}$$
$$\begin{aligned}&P(t)=a{\left( \frac{2bt}{e^{bt}+1}\right) }e^{abxt+a^2b^2yt^2}{\left( \frac{1-(-e^{bt})^a}{1-(-e^{at})^b}\right) }{\left( \frac{1-(-e^{at})^b}{e^{at}+1}\right) }\nonumber \\&\quad =a{\left( \sum \limits _{k=0}^{\infty }{}_HG_{k}(ax,a^2y)\frac{(bt)^k}{k!}\right) }{\left( -\sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) } {\left( \sum \limits _{l=0}^{\infty }T_{l}(b-1)\frac{(at)^l}{l!}\right) }\nonumber \\&\quad =-a{\left( \sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) }{\left( \sum \limits _{l=0}^{\infty }\sum \limits _{k=0}^{l}a^{l-k}b^k{}_HG_{k}(ax,a^2y) T_{l-k}(b-1)\frac{t^l}{{l-k}!k!}\right) }\nonumber \\&P(t)=-a{\left( \sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) }\sum \limits _{l=0}^{\infty }\sum \limits _{k=0}^{l}\left( \begin{array}{lll}l\\ k \end{array}\right) a^{l-k}{}_HG_{k}(ax,a^2y)T_{l-k}(b-1)\frac{t^{l}}{l!} \end{aligned}$$
(50)
$$\begin{aligned}&P(t)=-a{\left( \sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) }{\left( \frac{2bt}{e^{bt}+1}\right) }e^{abxt+a^2b^2yt^2}{\left( \sum \limits _{i=0} ^{b-1} (-1)^ie^{ati}\right) }\nonumber \\&\quad =-a{\left( \sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) }{\left( \sum \limits _{i=0}^{b-1}(-1)^i\sum \limits _{k=0}^{\infty }b^k{}_HG_{k} \left(ax+\frac{a}{b}i;a^2y\right)\frac{t^k}{k!}\right) } \end{aligned}$$
(51)
Remark 7
For \(y=0\) in Theorem (7), the result reduces to known result of Liu and Wang [2009, p. 3358, Theorem (4.4)].
Corollary 7
For integers
\( n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1 \), if
a
is odd and
b
is even, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^k b^{n-k+1}G_k(bx)T_{n-k}(a-1) \nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}n+1\\ l\end{array}\right) \frac{B_{n+1-l}}{n+1}b^{n-l}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) b^ka^{n-k+1}G_{k}(ax)T_{l-k}(b-1) \end{aligned}$$
(52)
$$\begin{aligned}&b\sum \limits _{i=0}^{a-1}{(-1)^i}a^nG_n\left(bx+{\frac{b}{a}}i\right)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}n+1\\ l\end{array}\right) \frac{B_{n+1-l}}{n+1}a^{n-l+1}\sum \limits _{i=0}^{b-1}(-1)^ib^nG_{l}\left(ax+\frac{a}{b}i \right) \end{aligned}$$
(53)
Theorem 8
For integers
\(n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1\), if
a
is even and
b
is odd, then the following identity holds true:
and
$$\begin{aligned}\sum \limits _{k=0}^{n}\left(\begin{array}{lll}n\\ k\end{array}\right) a^kb^{n-k+1}{}_HG_k(bx,b^2y)T_{n-k}(a-1)&=\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}n\\ k\end{array}\right) b^ka^{n-k+1}{}_HG_{k}(ax,a^2y)\nonumber \\&\quad\times (E_{n-k}(0)-T_{n-k}(b-1)) \end{aligned}$$
(54)
$$\begin{aligned}b\sum \limits _{i=0}^{a-1}(-1)^ia^n{}_HG_{n}\left(bx+\frac{b}{a}i,b^2y\right)&=\sum \limits _{l=0}^{n-1}\left( \begin{array}{lll}n\\ l\end{array}\right) a^{n-l+1}E_{n-l}(0)\nonumber \\&\quad \times \sum \limits _{i=0}^{b-1}(-1)^ib^n{}_HG_l\left(ax+\frac{a}{b}i,a^2y\right) \end{aligned}$$
(55)
Proof
Letwe expand P(t) as follows:Comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (42) and (57), we haveFinally using [Liu and Wang (2009), p. 3348, (2.4)], we arrive at the desired result (54).
$$\begin{aligned} P(t)=:\frac{2abt(1-{(-e^{bt})}^a)}{(e^{at}+1)(e^{bt}+1)}e^{abxt+a^2b^2yt^2} \end{aligned}$$
(56)
$$\begin{aligned}P(t)&=a\left( \frac{2bt}{e^{bt}+1}\right) e^{abxt+a^2b^2yt^2}\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b}\right) \left( \frac{(1-{(-e^{at})}^b)}{1+e^{at}} \right) \nonumber \\& =a\left( \sum \limits _{k=0}^{\infty }{}_HG_k(ax,a^2y)\frac{(bt)^k}{k!}\right) \left( \sum \limits _{n=1}^{\infty }E_{n}(0)\frac{(abt)^n}{n!}\right) \left( \sum \limits _{l=0}^{\infty }T_l(b-1)\frac{(at)^l}{l!}\right) \end{aligned}$$
(57)
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k+1}}{}_HG_k(bx,b^2y)T_{n-k}(a-1) \nonumber \\&\quad =\sum \limits _{l=0}^{n-1} \left( \begin{array}{lll}n\\ l\end{array}\right) E_{n-l}(0)b^{n-l}\sum \limits _{k=0}^{l}{\left( \begin{array}{lll}l\\ k\end{array}\right) } a^{n-k+1}b^k{}_HG_{k}(ax,a^2y)T_{l-k}(b-1) \nonumber \\&\quad =\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k+1}b^k{}_HG_{k}(ax,a^2y)\sum \limits _{l=k}^{n-1} \left( \begin{array}{lll}n-k\\ l-k\end{array}\right) E_{n-l}(0)b^{n-l}T_{l-k}(b-1) \end{aligned}$$
(58)
Now following (56), we expand P(t) as follows:which on equating the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (44) and (59), gives the result (55).
$$\begin{aligned}P(t) &=a\left( \frac{2bt}{e^{bt}+1}\right) e^{abxt+a^2b^2yt^2}\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b}\right) \left( \frac{(1-{(-e^{at})}^b)}{1+e^{at}} \right) \nonumber \\ \quad &=a{\left( \sum \limits _{n=1}^{\infty }E_{n}(0)\frac{(abt)^n}{n!}\right) }{\left( \sum \limits _{l=0}^{\infty }\sum \limits _{i=0}^{b-1}(-1)^ib^l{}_HG_{l}\left(ax+\frac{a}{b}i, a^2y\right)\frac{t^l}{l!}\right) } \end{aligned}$$
(59)
Remark 8
For \( y=0 \) in Theorem (8), the result reduces to known result of Liu and Wang [2009, p. 3359, Theorem (4.6)].
Corollary 8
For integers
\(n\ge 0\) , \(a\ge 1\)
and
\(b\ge 1\), if
a
is even and
b
is odd, then the following identity holds true:
and
$$\begin{aligned}\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^kb^{n-k+1}G_k(bx)T_{n-k}(a-1)&=\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}n\\ k\end{array}\right) b^ka^{n-k+1}G_{k}(ax) \nonumber \\&\quad \times (E_{n-k}(0)-T_{n-k}(b-1)) \end{aligned}$$
(60)
$$\begin{aligned}b\sum \limits _{i=0}^{a-1}(-1)^ia^nG_{n}\left(bx+\frac{b}{a}i\right)&=\sum \limits _{l=0}^{n-1}\left( \begin{array}{lll}n\\ l\end{array}\right) a^{n-l+1}E_{n-l}(0)\nonumber \\&\quad \times \sum \limits _{i=0}^{b-1}(-1)^ib^nG_l\left(ax+\frac{a}{b}i\right) \end{aligned}$$
(61)
Mixed type identities
In this section, we establish some mixed type identities involving the generalized Hermite–Bernoulli and Euler polynomials of order \(\alpha \). Throughout this section \(\alpha \) will be taken as an arbitrary real or complex parameter.
Theorem 9
For integers
\(n\ge 1\), \(a\ge 1\)
and
\(b\ge 1\), if a is even, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}{}_HB_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay) \nonumber \\&\quad =-\frac{n}{2}\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}{n-1}\\ k\end{array}\right) {a^{k+1}} {b^{n-k-1}}{}_HE_{n-1-k}^{(\alpha )}(ax,a^{2}z)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) S_i(b-1)B_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(62)
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}(-1)^i{}_HB_{k}^{(\alpha )}\left(bx+\frac{b}{a}i,b^{2}z\right) E_{n-k}^{(\alpha -1)}(ay) \nonumber \\&\quad =-\frac{n}{2}\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}{n-1}\\ k\end{array}\right) {b^k}{a^{n-k}}\sum \limits _{i=0}^{b-1} {}_HE_{k}^{(\alpha )}\left(ax+\frac{a}{b}i,a^{2}z\right)B_{n-k-1}^{(\alpha -1)}(by) \end{aligned}$$
(63)
Proof
Let
Since a is even, we also haveComparing coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of above Eqs. (65) and (66), yields the first identity (62).
$$\begin{aligned}H(t)&=:\frac{2^{\alpha -1}a^{\alpha }t^{\alpha }(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}-1)^{\alpha }(e^{bt}+1)^{\alpha }}\\ & ={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& ={\left( \sum \limits _{n=0}^{\infty }{}_HB_{n}^{(\alpha )}(bx,b^2z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty } T_i(a-1)\frac{(bt)^i}{i!}\right) }{\left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}(ay)\frac{(bt)^k}{k!}\right) }\nonumber \\&\end{aligned}$$
(64)
$$\begin{aligned}H(t)={\left( \sum \limits _{n=0}^{\infty }{}_HB_{n}^{(\alpha )}(bx,b^2z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{k=0}^{\infty } \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) b^kT_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{k!}\right) } \end{aligned}$$
(65)
$$\begin{aligned}H(t)&=-\frac{ta}{2}{\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{e^{abt}-1}{e^{at}-1}\right) }{\left( \frac{at}{e^{at}-1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\&=-\frac{ta}{2}{\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(ax,a^2z)\frac{(bt)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty } S_i(b-1)\frac{(at)^i}{i!}\right) }{\left( \sum \limits _{k=0}^{\infty }B_k^{(\alpha -1)}(by)\frac{(at)^k}{k!}\right) }\nonumber \\&=-\frac{ta}{2}{\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(ax,a^2z)\frac{(bt)^n}{n!}\right) }{\left( \sum \limits _{k=0}^{\infty } \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) a^kS_i(b-1)B_{k-i}^{(\alpha -1)}(by)\frac{t^k}{k!}\right) } \end{aligned}$$
(66)
Again, we expand H(t) as follows:On similar lines, we can expand H(t) as follows:Comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S of above Eqs. (67) and (68), we get the result (63).
$$\begin{aligned}H(t)&={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& ={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}(-1)^ie^{bti}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\&={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}(-1)^ie^{(bx+\frac{b}{a}i)}at \right) }{\left( \sum \limits _{n=0}^{\infty }E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }\nonumber \\H(t)&={\left( \sum \limits _{i=0}^{a-1}(-1)^i\sum \limits _{k=0}^{\infty }{}_HB_{k}^{(\alpha )}({bx+\frac{b}{a}i;b^2z})\frac{(at)^k}{k!}\right) }{\left( \sum \limits _{n=0}^{\infty }E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) } \end{aligned}$$
(67)
$$\begin{aligned}H(t)&=-\frac{ta}{2}{\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{e^{abt}-1}{e^{at}-1}\right) }{\left( \frac{at}{e^{at}-1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& =-\frac{ta}{2}{\left( \frac{2}{e^{bt}+1}\right) }^{\alpha }e^{abxt+a^2b^2zt^2}{\left( \sum \limits _{i=0}^{b-1}e^{ati}\right) }{\left( \frac{at}{e^{at}-1} \right) }^{{\alpha }-1}e^{abyt}\nonumber \\H(t)&=-\frac{ta}{2}{\left( \sum \limits _{i=0}^{b-1}\sum \limits _{k=0}^{\infty }{}_HE_{k}^{(\alpha )}\left(ax+\frac{a}{b}i,a^{2}z\right)\frac{(bt)^k}{k!}\right) } {\left( \sum \limits _{n=0}^{\infty }B_{n}^{(\alpha -1)}(by)\frac{(at)^n}{n!}\right) } \end{aligned}$$
(68)
Remark 9
For \(z=0\) in Theorem (9), the result reduces to known result of Liu and Wang [2009, p. 3353, Theorem (3.1)].
Corollary 9
For integers
\(n\ge 1\), \(a\ge 1\)
and
\(b\ge 1\), if a is even, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}B_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =-\frac{n}{2}\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}{n-1}\\ k\end{array}\right) {a^{k+1}} {b^{n-k-1}}E_{n-1-k}^{(\alpha )}(ax)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) S_i(b-1)B_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(69)
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}(-1)^iB_{k}^{(\alpha )}\left(bx+\frac{b}{a}i\right) E_{n-k}^{(\alpha -1)}(ay) \nonumber \\&\quad =-\frac{n}{2}\sum \limits _{k=0}^{n-1}\left( \begin{array}{lll}{n-1}\\ k\end{array}\right) {b^k}{a^{n-k}}\sum \limits _{i=0}^{b-1} E_{k}^{(\alpha )}\left(ax+\frac{a}{b}i\right)B_{n-k-1}^{(\alpha -1)}(by) \end{aligned}$$
(70)
Theorem 10
For integers
\( n\ge 1\), \(a\ge 1\)
and
\(b\ge 1 \), if
a
is odd, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}{}_HB_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay) \nonumber \\&\quad =\sum \limits _{l=0,l\ne n-1}^{n}\left( \begin{array}{lll}n\\ l\end{array}\right) B_{n-l}{a^{n-l}}\sum \limits _{k=0,}^{l}\left( \begin{array}{lll}l\\ k\end{array}\right) {b^{n-k-1}}a^k{}_HE_{l-k}^{(\alpha )}(ax,a^{2}z)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) S_i(b-1)B_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(71)
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}(-1)^i{}_HB_{k}^{(\alpha )}\left(bx+\frac{b}{a}i,b^{2}z\right) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0,l\ne n-1}^{n}\left( \begin{array}{lll}n\\ l\end{array}\right) B_{n-l}{b^{n-l-1}}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) {b^k}{a^{n-k}}\nonumber \\&\qquad \times \sum \limits _{i=0}^{b-1}{}_HE_{k}^{(\alpha )}\left(ax+\frac{a}{b}i,a^{2}z\right)B_{l-k}^{(\alpha -1)}(by) \end{aligned}$$
(72)
Proof
LetWe expand H(t) as follows:Equating the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of above Eqs. (65) and (73) and using the identity (7), we obtain the first identity (71).
$$\begin{aligned} H(t)=:\frac{t^{\alpha }(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}-1)^{\alpha }(e^{bt}+1)^{\alpha }} \end{aligned}$$
$$\begin{aligned}H(t)&=\frac{-t}{2^{\alpha }{a^{\alpha -1}}}{\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b}\right) } {\left( \frac{e^{abt}-1}{e^{at}-1}\right) }\nonumber \\&\quad \times {\left( \frac{at}{e^{at}-1}\right) ^{{\alpha }-1}}e^{abyt} \nonumber \\& =\frac{t}{2^{\alpha }{a^{\alpha -1}}}{\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) }{\left( \sum \limits _{n=0}^ {\infty }\frac{B_n+B_n(1)}{n!}{(abt)}^{n-1}\right) }\nonumber \\&\quad \times {\left( \sum \limits _{i=0}^{\infty }S_i(b-1)\frac{(at)^i}{i!}\right) }{\left( \sum \limits _{k=0}^{\infty } B_{k}^{(\alpha -1)}(by)\frac{(at)^k}{k!}\right) } \end{aligned}$$
(73)
On the other handEquating the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of above Eqs. (67) and (74), yields the result (72).
$$\begin{aligned}H(t)&=\frac{-t}{2^{\alpha }{a^{\alpha -1}}}{\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b}\right) } {\left( \frac{e^{abt}-1}{e^{at}-1}\right) }{\left( \frac{at}{e^{at}-1}\right) ^{\alpha -1}}e^{abyt}\nonumber \\ \quad &=\frac{-t}{2^{\alpha }{a^{\alpha -1}}}{\left( -\sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}{abt}^{n-1}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }} e^{abxt+a^2b^2zt^2}\nonumber \\ &\quad \times {\left( \sum \limits _{i=0}^{b-1}e^{ati}\right) }{\left( \sum \limits _{l=0}^{\infty }B_{l}^{(\alpha -1)}(by)\frac{(at)^l}{l!}\right) }\nonumber \\ H(t)&=\frac{t}{2^{\alpha }{a^{\alpha -1}}}{\left( \sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}{abt}^{n-1}\right) }{\left( \sum \limits _{i=0}^{b-1}\sum \limits _{k=0} ^{\infty }{}_HE_{k}^{(\alpha )}(ax+\frac{a}{b}i,a^{2}z)\frac{(bt)^k}{k!}\right) }\nonumber \\&\quad \times {\left( \sum \limits _{l=0}^{\infty }B_{l}^{(\alpha -1)}(by)\frac{(at)^l}{l!}\right) } \end{aligned}$$
(74)
Remark 10
For \(z=0\) in Theorem (10), the result reduces to known result of Liu and Wang [2009, p. 3354, Theorem (3.6)].
Corollary 10
For integers
\( n\ge 1\), \(a\ge 1\)
and
\(b\ge 1 \), if
a
is odd, then the following identity holds true:
and
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}B_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0,l\ne n-1}^{n}\left( \begin{array}{lll}n\\ l\end{array}\right) B_{n-l}{a^{n-l}}\sum \limits _{k=0,}^{l}\left( \begin{array}{lll}l\\ k\end{array}\right) {b^{n-k-1}}a^kE_{l-k}^{(\alpha )}(ax) \nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) S_i(b-1)B_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(75)
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}(-1)^iB_{k}^{(\alpha )}\left(bx+\frac{b}{a}i\right) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0,l\ne n-1}^{n}\left( \begin{array}{lll}n\\ l\end{array}\right) B_{n-l}{b^{n-l-1}}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) {b^k}{a^{n-k}} \nonumber \\&\qquad \times \sum \limits _{i=0}^{b-1}E_{k}^{(\alpha )}\left(ax+\frac{a}{b}i\right)B_{l-k}^{(\alpha -1)}(by) \end{aligned}$$
(76)
Theorem 11
For integers
\( n\ge 1\), \(a\ge 1 \)
and
\( b\ge 1\) , the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}{}_HB_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0} ^{a-1}(-1)^i{}_HB_{k}^{(\alpha )}\left(bx+\frac{b}{a}i,b^{2}z\right)E_{n-k}^{(\alpha -1)}(ay) \end{aligned}$$
(77)
Proof
LetOn the other handComparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S of above Eqs. (78) and (79), yields the desired result.
$$\begin{aligned}&H(t)=:\frac{2^{\alpha -1}a^{\alpha }t^{\alpha }(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}-1)^{\alpha }(e^{bt}+1)^{\alpha }} \nonumber \\&\quad ={\left( \sum \limits _{n=0}^{\infty }{}_HB_{n}^{(\alpha )}(bx,b^2z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty } T_i(a-1)\frac{(bt)^i}{i!}\right) }{\left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}(ay)\frac{(bt)^k}{k!}\right) }\nonumber \\&\quad ={\left( \sum \limits _{n=0}^{\infty }{}_HB_{n}^{(\alpha )}(bx,b^2z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{k=0}^{\infty } \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) b^kT_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{k!}\right) } \end{aligned}$$
(78)
$$\begin{aligned}H(t)&={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}} e^{abyt}\nonumber \\& ={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}(-1)^ie^{bti}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}} e^{abyt}\nonumber \\& ={\left( \frac{at}{e^{at}-1}\right) ^{\alpha }}e^{a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}(-1)^ie^{(bx+\frac{b}{a}i)}at \right) }{\left( \sum \limits _{n=0}^{\infty }E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) } \end{aligned}$$
(79)
Remark 11
For \(z=0\) in Theorem (11), the result reduces to known result of Liu and Wang [2009, p. 3356, Theorem (3.9)].
Corollary 11
For integers
\( n\ge 1\), \(a\ge 1 \)
and
\( b\ge 1\), the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}B_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0} ^{a-1}(-1)^iB_{k}^{(\alpha )}\left(bx+\frac{b}{a}i\right)E_{n-k}^{(\alpha -1)}(ay) \end{aligned}$$
(80)
Theorem 12
For integers
\(n\ge 1\), \(a\ge 1\)
and
\(b\ge 1\), the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}{}_HE_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) S_i(a-1)B_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0} ^{a-1}{}_HE_{k}^{(\alpha )}\left(bx+\frac{b}{a}i,b^{2}z\right)B_{n-k}^{(\alpha -1)}(ay) \end{aligned}$$
(81)
Proof
LetAnother expansion of H(t) is as follows:Equating the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of above Eqs. (82) and (83), yields the desired result.
$$\begin{aligned}H(t)&=:\frac{t^{\alpha }(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}-1)^{\alpha }}\nonumber \\H(t)&=\frac{-t}{2^{\alpha }{b^{\alpha -1}}}{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }}}e^{abxt+a^2b^2zt^2}{\left( \frac{e^{abt}-1}{e^{bt}-1}\right) } {\left( \frac{bt}{e^{bt}-1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& =\frac{-t}{2^{\alpha }{b^{\alpha -1}}}{\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0} ^{\infty }S_i(a-1)\frac{(bt)^i}{i!}\right) }{\left( \sum \limits _{k=0}^{\infty }B_{k}^{(\alpha -1)}(ay)\frac{(bt)^k}{k!}\right) }\nonumber \\& =\frac{-t}{2^{\alpha }{b^{\alpha -1}}}{\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) } {\left( \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{k}b^kS_i(a-1)B_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{{(k-i)!}i!}\right) } \end{aligned}$$
(82)
$$\begin{aligned}H(t)&=\frac{-t}{2^{\alpha }{b^{\alpha -1}}}{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }}}e^{abxt+a^2b^2zt^2}{\left( \frac{e^{abt}-1}{e^{bt}-1}\right) } {\left( \frac{bt}{e^{bt}-1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& =\frac{-t}{2^{\alpha }{b^{\alpha -1}}}{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }}}e^{abxt+a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}e^{bti}\right) } {\left( \sum \limits _{n=0}^{\infty }B_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }\nonumber \\& =\frac{-t}{2^{\alpha }{b^{\alpha -1}}}{\left( \sum \limits _{n=0}^{\infty }B_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{a-1} \sum \limits _{k=0}^{\infty }{}_HE_{k}^{(\alpha )}\left(bx+\frac{b}{a}i,b^2z\right)\frac{(at)^k}{k!}\right) } \end{aligned}$$
(83)
Remark 12
For \(z=0\) in Theorem (12), the result reduces to known result of Liu and Wang [2009, p. 3356, Theorem (3.11)].
Corollary 12
For integers
\(n\ge 1\), \(a\ge 1\)
and
\(b\ge 1\), the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^k}{a^{n-k}}E_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) S_i(a-1)B_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0} ^{a-1}E_{k}^{(\alpha )}(bx+\frac{b}{a}i)B_{n-k}^{(\alpha -1)}(ay) \end{aligned}$$
(84)
Conclusion
The definition and generating function of the generalized Hermite–Bernoulli, Euler and Hermite–Genocchi polynomials plays a major role in obtaining new expansions, identities and representations. We can introduce and study a class of related generalized polynomials by defining Gould-Hopper Bernoulli, Euler and Genocchi polynomials.The Eq. (18) may be derived from (85) for \(r = 2\).
$$\begin{aligned} \left( \frac{t}{e^t-1}\right) ^{\alpha }e^{xt+yt^r}=\sum \limits _{n=0}^{\infty }{}_HB_n^{(\alpha , r)}(x,y)\frac{t^n}{n!} \end{aligned}$$
(85)
This process can easily be extended to establish multi-variable Hermite–Bernoulli, Euler and Genocchi polynomials and Hermite Apostle type Bernoulli, Euler and Genocchi polynomials.
Declarations
Authors' contributions
The opinions of both the authors are considered in carrying out this work and they together drafted this manuscript. Both authors read and approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
Department of Mathematics, Integral University, Lucknow, India
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