# On the generalized Apostol-type Frobenius-Euler polynomials

## Abstract

The aim of this paper is to derive some new identities related to the Frobenius-Euler polynomials. We also give relation between the generalized Frobenius-Euler polynomials and the generalized Hurwitz-Lerch zeta function at negative integers. Furthermore, our results give generalized Carliz’s results which are associated with Frobenius-Euler polynomials.

MSC:05A10, 11B65, 28B99, 11B68.

## 1 Introduction, definitions and notations

Throughout this presentation, we use the following standard notions: $N={1,2,…}$, $N 0 ={0,1,2,…}=N∪{0}$, $Z − ={−1,−2,…}$. Also, as usual denotes the set of integers, denotes the set of real numbers and denotes the set of complex numbers. Furthermore, $( λ ) 0 =1$ and

$( λ ) k =λ(λ+1)(λ+2)⋯(λ+k−1),$

where $k∈N$, $λ∈C$.

The classical Frobenius-Euler polynomial $H n ( α ) (x;u)$ of order α is defined by means of the following generating function:

$( 1 − u e t − u ) α e x t = ∑ n = 0 ∞ H n ( α ) (x;u) t n n ! ,$
(1)

where u is an algebraic number and $α∈Z$.

Observe that $H n ( 1 ) (x;u)= H n (x;u)$, which denotes the Frobenius-Euler polynomials and $H n ( α ) (0;u)= H n ( α ) (u)$, which denotes the Frobenius-Euler numbers of order α. $H n (x;−1)= E n (x)$, which denotes the Euler polynomials (cf. ).

Definition 1.1 (for details, see [16, 17])

Let $a,b,c∈ R +$, $a≠b$, $x∈R$. The generalized Apostol-type Frobenius-Euler polynomials are defined by means of the following generating function:

$( a t − u λ b t − u ) α c x t = ∑ n = 0 ∞ H n ( α ) (x;u;a,b,c;λ) t n n ! .$
(2)

Remark 1.2 If we set $x=0$ and $α=1$ in (2), we get

$a t − u λ b t − u = ∑ n = 0 ∞ H n (u;a,b,c;λ) t n n ! ,$
(3)

where $H n (u;λ;a,b,c)$ denotes the generalized Apostol-type Frobenius-Euler numbers (cf. ).

## 2 New identities

In this section, we derive many new identities related to the generalized Apostol-type Frobenius-Euler numbers and polynomials of order α.

Theorem 2.1 Let $α,β∈Z$. Each of the following relationships holds true: (4) (5) (6)

and

$H n ( − α ) ( x ; u 2 ; a 2 , b 2 , c 2 ; λ 2 ) = ∑ k = 0 n ( n k ) H k ( − α ) (x;u;a,b,c;λ) H n − k ( − α ) (x;−u;a,b,c;λ).$
(7)

Proof of (6) From (2),

$∑ n = 0 ∞ H n ( − α ) (x;u;a,b,c;λ) t n n ! ∑ n = 0 ∞ H n ( α ) (y;u;a,b,c;λ) t n n ! = c ( x + y ) t .$
(8)

Therefore,

$∑ n = 0 ∞ ( ∑ k = 0 n ( n k ) H n − k ( α ) ( y ; u ; a , b , c ; λ ) H k ( − α ) ( x ; u ; a , b , c ; λ ) ) t n n ! = ∑ n = 0 ∞ ( x ln c ) n t n n ! .$

Thus, by using the Cauchy product in (8) and then equating the coefficients of $t n n !$ on both sides of the resulting equation, we obtain the desired result.

The proofs of (4), (5) and (7) are the same as that of (2), thus we omit them. □

Observe that in (6) we have

$( ( x + y ) ln c ) n = ( H ( α ) ( y ; u ; a , b , c ; λ ) + H ( − α ) ( x ; u ; a , b , c ; λ ) ) n ,$

where $( H ( α ) ( y ; u ; a , b , c ; λ ) ) n$ is replaced by $H n ( α ) (y;u;a,b,c;λ)$.

Theorem 2.2 Let $α∈N$. Then we have

$∑ k = 0 α ( α k ) ( − u ) α − k ( x ln c + k ln a ) n = ∑ p = 0 n ∑ k = 0 α ( n p ) ( α k ) ( − u ) α − k ( k ln b ) p H n − p ( α ) (x;u;a,b,c;λ).$

Proof By using (2), we get

$∑ n = 0 ∞ ( ∑ k = 0 α ( α k ) ( − u ) α − k ( x ln c + k ln a ) n ) t n n ! = ∑ n = 0 ∞ ( ∑ p = 0 n ∑ k = 0 α ( n p ) ( α k ) ( − u ) α − k ( k ln b ) p H n − p ( α ) ( x ; u ; a , b , c ; λ ) ) t n n ! .$

By equating the coefficients of $t n n !$ on both sides of the resulting equation, we obtain the desired result. □

Theorem 2.3 The following relationship holds true: (9)

Proof We set

$( 2 u − 1 ) a t − u λ b t − u c x t a t − ( 1 − u ) λ b t − ( 1 − u ) c y t = ( a t − u ) ( a t − ( 1 − u ) ) c ( x + y ) t ( 1 λ b t − u − 1 λ b t − ( 1 − u ) ) .$

From the above equation, we see that

$( 2 u − 1 ) ( ∑ n = 0 ∞ H n ( x ; u ; a , b , c ; λ ) t n n ! ) ( ∑ n = 0 ∞ H n ( y ; 1 − u ; a , b , c ; λ ) t n n ! ) = ( a t − 1 + u ) ∑ n = 0 ∞ H n ( x + y ; u ; a , b , c ; λ ) t n n ! − ( a t − u ) ∑ n = 0 ∞ H n ( x + y ; 1 − u ; a , b , c ; λ ) t n n ! .$

Therefore,

$( 2 u − 1 ) ∑ n = 0 ∞ ∑ r = 0 n ( n r ) H r ( x ; u ; a , b , c ; λ ) H n − r ( y ; 1 − u ; a , b , c ; λ ) t n n ! = ( u − 1 ) ∑ n = 0 ∞ H n ( x + y ; u ; a , b , c ; λ ) t n n ! + u ∑ n = 0 ∞ H n ( x + y ; 1 − u ; a , b , c ; λ ) t n n ! + ∑ n = 0 ∞ ∑ r = 0 n ( n r ) ( ln a ) n − r H r ( x + y ; u ; a , b , c ; λ ) t n n ! − ∑ n = 0 ∞ ∑ r = 0 n ( n r ) ( ln a ) n − r H r ( x + y ; 1 − u ; a , b , c ; λ ) t n n ! .$

Comparing the coefficients of $t n n !$ on both sides of the above equation, we arrive at the desired result. □

Remark 2.4 By substituting $a=1$, $b=c=e$, $λ=1$ into Theorem 2.3, we get Carlitz’s results (for details, see [, Eq. 2.19]) as follows:

$( 2 u − 1 ) ∑ r = 0 n ( n r ) H r ( x ; u ) H n − r ( y ; 1 − u ) = ( u − 1 ) H n ( x + y ; u ) + u H n ( x + y ; 1 − u ) + H n ( x + y ; u ) − H n ( x + y ; 1 − u ) .$

We give the following generating function of the polynomials $Y n (x;λ;a)$:

$t λ a t − 1 a x t = ∑ n = 0 ∞ Y n (x;λ;a) t n n ! (a≥1)$
(10)

(cf. [16, 17]). We also note that

$Y n (0;λ;a)= Y n (λ;a).$

If we substitute $x=0$ and $a=1$ into (10), we see that

$Y n (λ;1)= 1 λ − 1 .$

Theorem 2.5 The generalized Apostol-type Frobenius-Euler polynomial holds true as follows: (11)

Proof Substituting $c=b$ for $α=1$ into (2) and taking derivative with respect to t, we obtain

$∑ n = 0 ∞ H n + 1 ( x ; u ; a , b , b ; λ ) t n n ! = a t ln a a t − u a t − u λ b t − u b x t + ln b λ b t a t − u ( a t − u λ b t − u ) 2 b x t + ln ( b x ) a t − u λ b t − u b x t .$

Using (10), we have

$∑ n = 0 ∞ H n + 1 ( x ; u ; a , b , b ; λ ) t n n ! = ln ( a 1 u ) t ∑ n = 0 ∞ ∑ k = 0 n ( n k ) Y n − k ( 1 ; 1 u ; a ) H k ( x ; u ; a , b , b ; λ ) t n n ! + ln ( b λ u ) t ∑ n = 0 ∞ ∑ k = 0 n ( n k ) Y n − k ( 1 u ; a ) H k ( 2 ) ( x ; u ; a , b , b ; λ ) t n n ! + ln ( b x ) ∑ n = 0 ∞ H n ( x ; u ; a , b , b ; λ ) t n n ! .$

Thus, after some elementary calculations, we arrive at (11). □

Theorem 2.6 Let $|u|<1$ and $m∈N$. Then we have

$H ( − m ) (u;a,b,c;λ)= ∑ k = 0 n ( n k ) H k ( − α ) (−x;u;a,b,c;λ) H n − k ( α − m ) (x;u;a,b,c;λ).$
(12)

Proof In (2), we replace α by −α, then we set

$( a t − u λ b t − u ) − α c ( − x ) t ∑ n = 0 ∞ H n ( α − m ) (x;u;a,b,c;λ) t n n ! = ( a t − u λ b t − u ) − m .$

By using (2), we get

$∑ n = 0 ∞ H n ( − α ) (−x;u;a,b,c;λ) t n n ! ∑ n = 0 ∞ H n ( α − m ) (x;u;a,b,c;λ) t n n ! = ∑ n = 0 ∞ H n ( − m ) (u;a,b,c;λ) t n n ! .$

Therefore,

$∑ n = 0 ∞ ∑ k = 0 n ( n k ) H k ( − α ) (−x;u;a,b,c;λ) H n − k ( α − m ) (x;u;a,b,c;λ) t n n ! = ∑ n = 0 ∞ H n ( − m ) (u;a,b,c;λ) t n n ! .$

Comparing the coefficients of $t n n !$ on both sides of the above equation, we arrive at (12). □

## 3 Interpolation function

In this section, we give a recurrence relation between the generalized Frobenius-Euler polynomials and the Hurwitz-Lerch zeta function. Recently, many authors have studied not only the Hurwitz-Lerch zeta function, but also its generalizations, for example (among others), Srivastava , Srivastava and Choi  and also Garg et al. . The generalization of the Hurwitz-Lerch zeta function $Φ(z,s,a)$ is given as follows:

$Φ μ , ν ( ρ , σ ) (z,s,a):= ∑ n = 0 ∞ ( μ ) ρ n ( ν ) σ n z n ( n + a ) s$

($μ∈C$, $a,υ∈C∖ Z 0 −$, $ρ,σ∈ R +$, $ρ<σ$ when $s,z∈C$ ($|z|<1$); $ρ=σ$ and $ℜ(s−μ+ν)>0$ when $|z|=1$). It is obvious that

$Φ μ , 1 ( 1 , 1 ) (z,s,a)= Φ μ ∗ (z,s,a)= ∑ n = 0 ∞ ( μ ) n n ! z n ( n + a ) s$
(13)

and

$Φ n ∗ (z,s,a)= ∑ n = 0 ∞ ( n ) n n ! z n ( n + a ) s =Φ(z,s,a),$

where $Φ(z,s,a)$ denotes the Lerch-Zeta function (cf. [6, 19, 21, 24]).

Relation between the generalized Apostol-type Frobenius-Euler polynomials and the Hurwitz-Lerch zeta function is given as follows.

Theorem 3.1 Let $| λ u |<1$. We have

$H n ( α ) (x;u;a,b,c;λ)= ∑ k = 0 α ( α k ) ( − u ) α − k − 1 G ( − n ; x , λ u ; a , b , c ; α , k ) ,$
(14)

where

$G(s;x,β;a,b,c;α,j)= ∑ m = 0 ∞ ( m + α − 1 m ) β m ( x ln c + j ln a + m ln b ) s ,|β|<1.$

Proof From (2), we have

$∑ n = 0 ∞ H n ( α ) (x;u;a,b,c;λ) t n n ! = ∑ j = 0 α ( α j ) ( − u ) α − j − 1 ∑ m = 0 ∞ ( m + α − 1 m ) ( λ u ) m e α ( x ln c + k ln a + m ln b ) .$

Therefore,

$∑ n = 0 ∞ H n ( α ) ( x ; u ; a , b , c ; λ ) t n n ! = ∑ n = 0 ∞ ∑ k = 0 α ( α k ) ( − u ) α − k − 1 ∑ m = 0 ∞ ( m + α − 1 m ) ( λ u ) m ( x ln c + k ln a + m ln b ) n t n n ! .$

Comparing the coefficients of $t n n !$ on both sides of the above equation, we have arrive at (14). □

Remark 3.2 By substituting $a=1$, $b=c=e$ into (14), we have

$H n ( α ) (x;u;λ)=− ( 1 − u ) α u G ( − n ; x , λ u ; 1 , e , e ; α , 1 ) =− ( 1 − u ) α u Φ ( λ u , − n , x ) ,$

where

$G ( − n ; x , λ u ; 1 , e , e ; α , 1 ) =Φ ( λ u , − n , x ) .$

Remark 3.3 The function $G(s;x,β;a,b,c;α,j)$ is an interpolation function of the generalized Apostol-type Frobenius-Euler polynomials of order α at negative integers, which is given by the analytic continuation of the $G(s;x,β;a,b,c;α,j)$ for $s=−n$, $n∈N$.

## 4 Relations between Array-type polynomials, Apostol-Bernoulli polynomials and generalized Apostol-type Frobenius-Euler polynomial

In , Simsek constructed the generalized λ-Stirling type numbers of the second kind $S(n,v;a,b;λ)$ by means of the following generating function:

$f S , v (t;a,b;λ)= ( λ b t − a t ) v v ! = ∑ n = 0 ∞ S(n,v;a,b;λ) t n n ! .$
(15)

The generating function for these polynomials $S v n (x;a,b;λ)$ is given by

$g v (x,t;a,b;λ)= 1 v ! ( λ b t − a t ) v b x t = ∑ n = 0 ∞ S v n (x;a,b;λ) t n n !$
(16)

(cf. ).

The generalized Apostol-Bernoulli polynomials were defined by Srivastava et al. [, p.254, Eq. (20)] as follows.

Let $a,b,c∈ R +$ with $a≠b$, $x∈R$ and $n∈ N 0$. Then the generalized Bernoulli polynomials $B n ( α ) (x;λ;a,b,c)$ of order $α∈Z$ are defined by means of the following generating functions:

$f B (x,a,b,c;λ;α)= ( t λ b t − a t ) α c x t = ∑ n = 0 ∞ B n ( α ) (x;λ;a,b,c) t n n ! ,$
(17)

where

$|tln ( a b ) +lnλ|<2π.$

We note that $B n ( 1 ) (x;λ;a,b,c)= B n (x;λ;a,b,c)$ and also $B n (x;λ;1,e,e)= B n (x;λ)$, which denotes the Apostol-Bernoulli polynomials (cf. ).

Theorem 4.1 Let v be an integer. Then we have

$H n − v ( − ν ) (x;u;a,b,c;λ)= ν ! u 2 ν ( n ) v ∑ k = 0 n ( n k ) S v n ( x , 1 , b ; λ u ) Y n − k ( ν ) ( 1 u ; a ) .$

Proof Replacing c by b in (2) and after some calculations, we have

$∑ n = 0 ∞ H n ( − v ) (x;u;a,b,b;λ) t n + v n ! = ν ! u 2 ν ∑ n = 0 ∞ S ν n ( x , 1 , b ; λ u ) t n n ! ∑ n = 0 ∞ Y n ( ν ) ( 1 u ; a ) t n n ! .$

Comparing the coefficients of $t n n !$ on both sides of the above equation, we arrive at the desired result. □

Corollary 4.2

$H n − v ( − ν ) (x;u;a,b,c;λ)= ν ! u 2 ν ( n ) α ∑ k = 0 n ( n k ) S ( k , ν , 1 , b ; λ u ) B n − k ( x , a , b ; λ u ) .$

Proof Replacing c by b in (2) and after some calculations, we have

$∑ n = 0 ∞ H n − v ( − v ) (x;u;a,b,b;λ) t n + v n ! = ν ! u 2 ν ∑ n = 0 ∞ S ( n , ν , 1 , b ; λ u ) t n n ! ∑ n = 0 ∞ B n ( x , a , b ; λ u ) t n n ! .$

Comparing the coefficients of $t n n !$ on both sides of the above equation, we arrive at the desired result. □

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## Acknowledgements

Dedicated to Professor Hari M. Srivastava.

All authors are partially supported by Research Project Offices Akdeniz Universities.

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Correspondence to Yilmaz Simsek.

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The authors declare that they have no competing interests.

### Authors’ contributions

All authors completed the paper together. All authors read and approved the final manuscript.

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Kurt, B., Simsek, Y. On the generalized Apostol-type Frobenius-Euler polynomials. Adv Differ Equ 2013, 1 (2013). https://doi.org/10.1186/1687-1847-2013-1 