Partial Hecke-type operators and their applications

  • Aykut Ahmet Aygunes1 and

    Affiliated with

    • Yilmaz Simsek1Email author

      Affiliated with

      Boundary Value Problems20132013:46

      DOI: 10.1186/1687-2770-2013-46

      Received: 14 November 2012

      Accepted: 22 January 2013

      Published: 6 March 2013

      Abstract

      The aim of this paper is to give not only the matrix representation of partial Hecke-type operators by means of Bernoulli polynomials and Euler polynomials, but also functional equations and differential equations related to partial Hecke-type operators and special polynomials. By using these functional equations and differential equations, we derive some identities associated with special polynomials and partial Hecke-type operators. Moreover, we find several useful identities and relations using the partial Hecke operators.

      MSC:05B20, 11B68, 11F25.

      Keywords

      Bernoulli polynomial Euler polynomial partial Hecke-type operator total Hecke-type operator Hurwitz zeta function partial zeta function

      1 Introduction

      Recently, there have been many applications of Bernoulli polynomials and Euler polynomials in differential equations, in analytic number theory and in engineering. High-order linear differential-difference equations have also been solved in terms of Bernoulli polynomials. These polynomials are also related to several linear operators. In this paper, we investigate and derive several new identities related to the Hecke-type operators and generating functions for special polynomials.

      Recently, many authors introduced and investigated the following generating functions which give us the Bernoulli polynomials B n ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq1_HTML.gif and the Euler polynomials E n ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq2_HTML.gif, respectively:
      t e x t e t 1 = n = 0 B n ( x ) t n n ! ( | t | < 2 π ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ1_HTML.gif
      (1)
      and
      2 e x t e t + 1 = n = 0 E n ( x ) t n n ! ( | t | < π ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ2_HTML.gif
      (2)

      For x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq3_HTML.gif, (1) and (2) are reduced to the generating functions for the Bernoulli numbers B n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq4_HTML.gif and the Euler numbers E n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq5_HTML.gif, respectively (cf. [14]), and see also the references cited in each of these earlier works.

      The multiplication formulas for the Bernoulli and Euler polynomials are given as follows:
      k = 0 m 1 B n ( x + k m ) = m 1 n B n ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ3_HTML.gif
      (3)
      and for odd m,
      k = 0 m 1 ( 1 ) k E n ( x + k m ) = m 1 n E n ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ4_HTML.gif
      (4)

      (cf. [58]), and see also the references cited in each of these earlier works.

      The Bernoulli polynomials satisfy the following well-known identity:
      1 m k = 0 m 1 f ( x + k m ) = m n f ( m x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equa_HTML.gif

      where m and n are positive integers (cf. [5, 9, 10]).

      Bayad et al. [11] introduced and systematically studied the following family of partial Hecke-type operators on C [ x ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq6_HTML.gif.

      Throughout this paper, we use the following notations: a 1 ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq7_HTML.gif. Let N = { 0 , 1 , 2 , 3 , } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq8_HTML.gif and Z + = { 1 , 2 , 3 , } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq9_HTML.gif.

      For fixed a , N Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq10_HTML.gif and 0 k a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq11_HTML.gif, we have
      T χ a , N ( P ( x ) ) = k = 0 a 1 χ a , N ( k ) P ( x + k a ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equb_HTML.gif
      where
      χ a , N ( k ) = { ξ N k = e 2 π i k N if  N 2 , 1 a if  N = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equc_HTML.gif

      Lemma 1.1 [[11], p.114, Lemma 1]

      For any a , N Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq10_HTML.gif such that a 1 ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq7_HTML.gif, we have the following properties:
      1. (i)

        T χ a , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq12_HTML.gif preserves the degree in C [ x ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq6_HTML.gif.

         
      2. (ii)
        By induction,
        T χ a , N ( x m ) = { S 0 = 1 if m = 0 , a m x m + a m v = 0 m 1 ( m v ) S m v ( χ a , N ) x v if m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equd_HTML.gif
         
      where
      S m v ( χ a , N ) = k = 0 a 1 χ a , N ( k ) k m v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Eque_HTML.gif
      1. (iii)
        For any m Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq13_HTML.gif, let β m = ( 1 , x , x 2 , , x m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq14_HTML.gif be the canonical ℂ-basis of
        C m [ x ] = { P ( x ) C [ x ] : deg P ( x ) m } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equf_HTML.gif
         
      Then the matrix M β m ( T χ a , N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq15_HTML.gif corresponding to the operator T χ a , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq12_HTML.gif (restricted to C m [ x ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq16_HTML.gif) in the basis β m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq17_HTML.gif is represented by:
      M β m ( T χ a , N ) = ( S 0 ( χ a , N ) a 1 S 1 ( χ a , N ) a 2 S 2 ( χ a , N ) a m S m ( χ a , N ) 0 a 1 S 0 ( χ a , N ) 2 a 2 S 1 ( χ a , N ) a m ( m 1 ) S m 1 ( χ a , N ) 0 0 a 2 S 0 ( χ a , N ) a m ( m 2 ) S m 2 ( χ a , N ) 0 0 0 a m ( m 3 ) S m 3 ( χ a , N ) 0 0 0 a m S 0 ( χ a , N ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ5_HTML.gif
      (5)
      for all 0 l m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq18_HTML.gif.
      1. (iv)
        Let a , b 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq19_HTML.gif such that a b 1 ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq20_HTML.gif, then
        T χ a , N T χ b , N = T χ b , N T χ a , N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equg_HTML.gif
         

      Consequently, for a given integer n, there is only one monic polynomial P n , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq21_HTML.gif with degree n in x satisfying the functional equation (6).

      The operator T χ a , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq12_HTML.gif satisfies the following equation:
      T χ a , N ( P n , N ( x ) ) = a n P n , N ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ6_HTML.gif
      (6)

      For a 1 ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq7_HTML.gif, from (6) and Lemma 1.1, we know that P n , N ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq22_HTML.gif is a monic polynomial (cf. [11]).

      Remark 1.2 Equations (3) and (4) are closely related to the functional equation of (6). For N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq23_HTML.gif and N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq24_HTML.gif, equation (6) is reduced to P n , 1 ( x ) = B n ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq25_HTML.gif and P n , 2 ( x ) = E n ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq26_HTML.gif, respectively. For fixed a , N Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq10_HTML.gif, we know that there is only one monic polynomial satisfying (6) by Lemma 1.1, and there already exist the functional equations as (3) and (4).

      The total Hecke-type operators, associated with partial Hecke-type operators, are defined by Bayad et al. [[11], p.112, Eq. (1.6)] as follows:
      T N = a 1 ( N ) T χ a , N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equh_HTML.gif

      Theorem 1.3 [11]

      Polynomials P n , N ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq22_HTML.gif are eigenfunctions for the operators T N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq27_HTML.gif with eigenvalues N n ζ ( n , 1 N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq28_HTML.gif, that is,
      T N ( P n , N ( x ) ) = N n ζ ( n , 1 N ) P n , N ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equi_HTML.gif
      where ζ ( s , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq29_HTML.gif is the Hurwitz zeta function defined by
      ζ ( s , x ) = k 0 1 ( x + k ) s ( cf . [10, 12]) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equj_HTML.gif

      2 Differential equations related to the partial Hecke-type operators and special polynomials

      In this section, we derive some ordinary and partial differential equations not only for a generating function, but also for partial Hecke-type operators. We also give a functional equation for the generating function. We set
      F N ( t , x ) = n = 0 P n , N ( x ) t n n ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equk_HTML.gif

      We now give an explicit formula of the generating function F N ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq30_HTML.gif as follows.

      Theorem 2.1 [11]

      Generating functions for the polynomials P n , N ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq22_HTML.gif are given by
      F N ( t , x ) = { t e t x e t 1 if N = 1 , ( ξ N 1 ) e t x ξ N e t 1 if N 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equl_HTML.gif

      The polynomials P n , N ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq22_HTML.gif are the so-called Bernoulli-Euler-type polynomials.

      We derive the following partial differential equation for F N ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq30_HTML.gif as follows:
      v x v F N ( t , x ) = t v F N ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ7_HTML.gif
      (7)
      Theorem 2.2 Let v N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq31_HTML.gif. Then
      d v d x v P n , N ( x ) = { ( n ) v B n v ( x ) if N = 1 , ( n ) v P n v , N ( x ) if N 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equm_HTML.gif

      where ( n ) v = n ( n 1 ) ( n 2 ) ( n v + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq32_HTML.gif.

      Proof By using (7), for N 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq33_HTML.gif, we obtain
      n = 0 ( d v d x v P n , N ( x ) ) t n n ! = n = v ( n ) v P n v , N ( x ) t n n ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ8_HTML.gif
      (8)

      Therefore, by comparing the coefficients of t n n ! http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq34_HTML.gif on both sides of equation (8), we have the desired result.

      For N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq23_HTML.gif, we apply the same process. So, we omit it. □

      We set the following differential equation:
      ξ N e t ( x + y ) ( ξ N e t 1 ) 2 = ( x + y 1 ) e t ( x + y 1 ) ( ξ N e t 1 ) d d t e t ( x + y 1 ) ( ξ N e t 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ9_HTML.gif
      (9)
      Theorem 2.3
      ξ N ξ N 1 k = 0 n ( n k ) P k , N ( x ) P n k , N ( y ) = ( x + y 1 ) P n , N ( x + y 1 ) n P n + 1 , N ( x + y 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equn_HTML.gif
      Proof We make some arrangement (9) and obtain
      ξ N ( ξ N 1 ) 2 [ ( ξ N 1 ξ N e t 1 ) 2 e t ( x + y ) ] = x + y 1 ξ N 1 ( ( ξ N 1 ) e t ( x + y 1 ) ( ξ N e t 1 ) ) d d t 1 ξ N 1 ( ξ N 1 ) e t ( x + y 1 ) ( ξ N e t 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equo_HTML.gif
      Therefore,
      ξ N ( ξ N 1 ) 2 ( ξ N 1 ξ N e t 1 e t x ) ( ξ N 1 ξ N e t 1 e t y ) = ξ N ( ξ N 1 ) 2 ( n = 0 P n , N ( x ) t n n ! ) ( n = 0 P n , N ( y ) t n n ! ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equp_HTML.gif
      From the above equation, we get
      ξ N ( ξ N 1 ) 2 n = 0 k = 0 n ( ( n k ) P k , N ( x ) P n k , N ( y ) ) t n n ! = x + y 1 ξ N 1 n = 0 ( P n , N ( x + y 1 ) n ξ N 1 P n + 1 , N ( x + y 1 ) ) t n n ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equq_HTML.gif

      By comparing the coefficients of t n n ! http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq34_HTML.gif on both sides of the above equation, we have the desired result. □

      Remark 2.4 In Theorem 2.3, we obtain a convolution formula for the polynomials P n , N ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq22_HTML.gif. If we substitute x = y = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq35_HTML.gif into Theorem 2.3, then we get a convolution formula for the Eulerian-type numbers (cf. [10, 13]).

      Higher-order partial differential equation for T χ a , N ( P n , N ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq36_HTML.gif is given by the following theorem.

      Theorem 2.5 Let N 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq33_HTML.gif and v N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq31_HTML.gif. Then
      v x v T χ a , N ( P n , N ( x ) ) = ( n ) v a v T χ a , N ( P n v , N ( x ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equr_HTML.gif
      where
      ( n ) v = n ( n 1 ) ( n 2 ) ( n v + 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equs_HTML.gif
      Proof Taking v th derivative of the operator T χ a , N ( P n , N ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq36_HTML.gif, with respect to x, we obtain the following higher-order partial differential equation:
      v x v T χ a , N ( P n , N ( x ) ) = k = 0 a 1 χ a , N ( k ) v x v P n , N ( x + k a ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equt_HTML.gif
      Using Theorem 2.2, we get
      v x v T χ a , N ( P n , N ( x ) ) = ( n ) v a v k = 0 a 1 χ a , N ( k ) P n v , N ( x + k a ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equu_HTML.gif

      Thus, we get the desired result. □

      3 Matrix representations of partial Hecke-type operators

      In this section, we give some numerical examples for the matrix representations of the operator T χ a , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq12_HTML.gif. For the basis β m = { 1 , x , x 2 , , x m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq37_HTML.gif, our matrix representations contain Bernoulli polynomials and Euler polynomials for the operators T χ a , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq38_HTML.gif and T χ a , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq39_HTML.gif, respectively. Therefore, we need the following lemmas.

      Lemma 3.1 Let m , n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq40_HTML.gif and n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq41_HTML.gif. Then
      k = 0 n 1 k m = B m + 1 ( n ) B m + 1 ( 0 ) m + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equv_HTML.gif
      Lemma 3.2 Let m , n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq40_HTML.gif and n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq41_HTML.gif. Then
      k = 0 n 1 ( 1 ) k k m = E m ( 1 ) n E m ( n ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equw_HTML.gif

      Proofs of Lemma 3.1 and Lemma 3.2 have been given by many authors (among others) (cf. [2, 4, 8, 10]).

      In a special case, substituting N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq23_HTML.gif into (iii) in Lemma 1.1 and using Lemma 3.1, we get
      S l , 1 ( a ) = k = 0 a 1 χ a , 1 ( k ) k l = B l + 1 ( a ) B l + 1 ( 0 ) a ( l + 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equx_HTML.gif

      According to the above equation, we are ready to give the main result of this section by the following theorem.

      Theorem 3.3 The matrix M β m ( T χ a , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq42_HTML.gif corresponding to the operator T χ a , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq38_HTML.gif (restricted to C m [ x ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq16_HTML.gif) in the basis β m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq43_HTML.gif is represented by Bernoulli polynomials as follows:
      M β m ( T χ a , 1 ) = ( B 1 ( a ) B 1 ( 0 ) a B 2 ( a ) B 2 ( 0 ) 2 a 2 B 3 ( a ) B 3 ( 0 ) 3 a 3 B m + 1 ( a ) B m + 1 ( 0 ) a m + 1 ( m + 1 ) 0 B 1 ( a ) B 1 ( 0 ) a 2 B 2 ( a ) B 2 ( 0 ) a 3 ( m 1 ) B m ( a ) B m ( 0 ) a m + 1 m 0 0 B 1 ( a ) B 1 ( 0 ) a 3 ( m 2 ) B m 1 ( a ) B m 1 ( 0 ) a m + 1 ( m 1 ) 0 0 0 ( m 3 ) B m 2 ( a ) B m 2 ( 0 ) a m + 1 ( m 2 ) 0 0 0 B 1 ( a ) B 1 ( 0 ) a m + 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equy_HTML.gif
      Setting N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq24_HTML.gif (iii) in Lemma 1.1 and using Lemma 3.2, we obtain
      S l , 2 ( a ) = k = 0 a 1 χ a , 2 ( k ) k l = E l ( 0 ) ( 1 ) a E l ( a ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equz_HTML.gif

      If a 1 ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq44_HTML.gif, then we obtain another main result by the following theorem.

      Theorem 3.4 Let a be an odd number. The matrix M β m ( T χ a , 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq45_HTML.gif corresponding to the operator T χ a , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq39_HTML.gif (restricted to C m [ x ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq16_HTML.gif) in the basis β m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq17_HTML.gif is represented by Euler polynomials as follows:
      M β m ( T χ a , 2 ) = ( E 0 ( 0 ) + E 0 ( a ) 2 E 1 ( 0 ) + E 1 ( a ) 2 a E 2 ( 0 ) + E 2 ( a ) 2 a 2 E m ( 0 ) + E m ( a ) 2 a m 0 E 0 ( 0 ) + E 0 ( a ) 2 a E 1 ( 0 ) + E 1 ( a ) a 2 ( m 1 ) E m 1 ( 0 ) + E m 1 ( a ) 2 a m 0 0 E 0 ( 0 ) + E 0 ( a ) 2 a 2 ( m 2 ) E m 2 ( 0 ) + E m 2 ( a ) 2 a m 0 0 0 ( m 3 ) E m 3 ( 0 ) + E m 3 ( a ) 2 a m 0 0 0 E 0 ( 0 ) + E 0 ( a ) 2 a m ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equaa_HTML.gif

      4 Some applications of total Hecke-type operators

      In this section, we give some applications related to eigenvalues for the total Hecke-type operators of T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq46_HTML.gif and T 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq47_HTML.gif. We derive many new identities which are related not only to the total Hecke-type operators, but also to the Riemann zeta function, the Hurwitz zeta function, Bernoulli and Euler numbers, Euler identities and the convolution of Bernoulli and Euler numbers and polynomials.

      Throughout this section, we use the following notation:
      ζ ( a ) = d d s ζ ( s ) | s = a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equab_HTML.gif
      The partial zeta function H ( s , a , F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq48_HTML.gif is defined by
      H ( s , a , F ) = n a ( F ) 1 n s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equac_HTML.gif

      where ( s ) > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq49_HTML.gif, n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq50_HTML.gif and 0 < a < F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq51_HTML.gif ( F Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq52_HTML.gif) (cf. [4, 8, 10, 12, 14]).

      Theorem 4.1 The polynomials P n , N ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq22_HTML.gif are eigenfunctions for the operators T N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq27_HTML.gif with eigenvalues H ( n , 1 , N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq53_HTML.gif, that is,
      T N ( P n , N ( x ) ) = H ( n , 1 , N ) P n , N ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equad_HTML.gif

      where H ( s , a , F ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq48_HTML.gif is a partial zeta function.

      Proof
      n a ( F ) 1 n s = 1 F s ζ ( s , a F ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equae_HTML.gif
      Therefore,
      ζ ( s , a F ) = F s H ( s , a , F ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equaf_HTML.gif

      Substituting F = N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq54_HTML.gif, s = n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq55_HTML.gif and a = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq56_HTML.gif into the above equation, after using Theorem 1.3, we arrive at the desired result. □

      Theorem 4.2 Let n Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq57_HTML.gif with n > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq58_HTML.gif. Then we have
      T 2 ( E 2 n ( x ) ) = ( 1 ) n ( 1 2 2 n ) ( 2 π ) 2 n ( 4 n + 2 ) ( 2 n ) ! E 2 n ( x ) k = 1 n 1 ( 2 n 2 k ) B 2 k B 2 n 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ10_HTML.gif
      (10)
      Proof Putting N = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq24_HTML.gif in Theorem 1.3 and using
      P n , 2 ( x ) = E n ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equag_HTML.gif
      we have
      T 2 ( E n ( x ) ) = 2 n ζ ( n , 1 2 ) E n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ11_HTML.gif
      (11)
      We recall from the definition of ζ ( n , 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq59_HTML.gif and ζ ( n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq60_HTML.gif that we have
      2 n ζ ( n , 1 2 ) = ( 1 2 n ) ζ ( n ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ12_HTML.gif
      (12)
      (cf. [[10], p.96]). Combining (11) and (12), we get
      T 2 ( E n ( x ) ) = ζ ( n ) ( 1 2 n ) E n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ13_HTML.gif
      (13)
      If we replace n by 2n in the above equation, we obtain
      T 2 ( E 2 n ( x ) ) = ζ ( 2 n ) ( 1 2 2 n ) E 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ14_HTML.gif
      (14)
      From the work of Srivastava and Choi [[4], p.98], we recall that
      ζ ( 2 n ) = 2 2 n + 1 k = 1 n 1 ζ ( 2 k ) ζ ( 2 n 2 k ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ15_HTML.gif
      (15)
      where n Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq57_HTML.gif with n > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq58_HTML.gif and
      ζ ( 2 n ) = ( 1 ) n + 1 ( 2 π ) 2 n B 2 n 2 ( 2 n ) ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ16_HTML.gif
      (16)

      By substituting (15) and (16) into (14), after some elementary calculations, we arrive at the desired result. □

      Theorem 4.3 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then
      T 2 ( E 2 n ( x ) ) = ( 1 ) n + 1 ( 2 2 n 1 ) π 2 n 2 ( 2 n ) ! B 2 n E 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ17_HTML.gif
      (17)
      Proof Combining (14) and (16), we easily complete the proof of the theorem, that is,
      T 2 ( E 2 n ( x ) ) = ζ ( 2 n ) ( 1 2 2 n ) E 2 n ( x ) = ( 1 ) n + 1 ( 2 2 n 1 ) π 2 n 2 ( 2 n ) ! B 2 n E 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equah_HTML.gif

       □

      By using (10) and (17), we obtain a convolution formula (Euler identity) for Bernoulli numbers.

      Theorem 4.4 Let n > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq58_HTML.gif. Then
      B 2 n = 1 2 n + 1 k = 1 n 1 ( 2 n 2 k ) B 2 k B 2 n 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ18_HTML.gif
      (18)
      Proof Since the left-hand sides of (10) and (17) are equal, the right-hand sides of (10) and (17) must be equal. Thus, we obtain
      ( 1 ) n + 1 ( 2 2 n 1 ) π 2 n 2 ( 2 n ) ! B 2 n = ( 1 ) n ( 2 2 n 1 ) π 2 n ( 4 n + 2 ) ( 2 n ) ! k = 1 n 1 ( 2 n 2 k ) B 2 k B 2 n 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equai_HTML.gif

      After some elementary calculation in the above equation, we get the desired result. □

      Observe that the proof of (18) is also given in [4].

      Theorem 4.5 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then
      T 2 ( E 2 n ( x ) ) = e π i n π 2 n 4 ( 2 n 1 ) ! E 2 n 1 ( 0 ) E 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equaj_HTML.gif
      Proof For all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif, we have
      E 2 n 1 ( 0 ) = 4 ( 1 ) n ( 2 π ) 2 n ( 2 n 1 ) ! ( 2 2 n 1 ) ζ ( 2 n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ19_HTML.gif
      (19)
      (cf. [[4], p.131]). By using (14) and (19), we obtain
      T 2 ( E 2 n ( x ) ) = ζ ( 2 n ) ( 1 2 2 n ) E 2 n ( x ) = ( ( 2 π ) 2 n E 2 n 1 ( 0 ) 4 ( 1 ) n ( 2 n 1 ) ! ( 2 2 n 1 ) ) ( 1 2 2 n ) E 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equak_HTML.gif

      Thus, the proof is completed. □

      Theorem 4.6 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then we have
      T 2 ( E 2 n + 1 ( x ) ) = ( 1 ) n + 1 ( 1 2 2 n 1 ) ( 2 π ) 2 n + 1 E 2 n + 1 ( x ) 2 ( 2 n + 1 ) ! 0 1 B 2 n + 1 ( t ) cot ( π t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ20_HTML.gif
      (20)
      Proof Consider that n is replaced by 2 n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq62_HTML.gif in (13), we have
      T 2 ( E 2 n + 1 ( x ) ) = ζ ( 2 n + 1 ) ( 1 2 2 n 1 ) E 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ21_HTML.gif
      (21)
      For all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif, one can easily get
      ζ ( 2 n + 1 ) = ( 1 ) n + 1 ( 2 π ) 2 n + 1 2 ( 2 n + 1 ) ! 0 1 B 2 n + 1 ( t ) cot ( π t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ22_HTML.gif
      (22)
      (cf. [[4], p.99, Eq. (21)]). Hence, we have
      T 2 ( E 2 n + 1 ( x ) ) = ζ ( 2 n + 1 ) ( 1 2 2 n 1 ) E 2 n + 1 ( x ) = ( ( 1 ) n + 1 ( 1 2 2 n 1 ) ( 2 π ) 2 n + 1 2 ( 2 n + 1 ) ! 0 1 B 2 n + 1 ( t ) cot ( π t ) d t ) E 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equal_HTML.gif

      Thus, the proof is completed. □

      Theorem 4.7 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then we have
      T 2 ( E 2 n + 1 ( x ) ) = 2 ( 1 ) n ( 2 π ) 2 n ( 1 2 2 n 1 ) ( 2 n ) ! ζ ( 2 n ) E 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ23_HTML.gif
      (23)
      Proof Note that, for all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif, we have
      ζ ( 2 n ) = ( 1 ) n ( 2 n ) ! 2 ( 2 π ) 2 n ζ ( 2 n + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ24_HTML.gif
      (24)
      (cf. [[4], p.99, Eq. (22)]). By using (21) and (24), we have
      T 2 ( E 2 n + 1 ( x ) ) = ζ ( 2 n + 1 ) ( 1 2 2 n 1 ) E 2 n + 1 ( x ) = ( 2 ( 1 ) n ζ ( 2 n ) ( 2 π ) 2 n ( 2 n ) ! ) ( 1 2 2 n 1 ) E 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equam_HTML.gif

      Thus, the proof is completed. □

      Theorem 4.8 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then
      T 1 ( B 2 n ( x ) ) = ( 1 ) n + 1 2 2 n 1 π 2 n ( 2 n ) ! B 2 n B 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equan_HTML.gif
      Proof Substituting N = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq23_HTML.gif into Theorem 1.3 and by P n , 1 ( x ) = B n ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq25_HTML.gif, we have
      T 1 ( B n ( x ) ) = ζ ( n , 1 ) B n ( x ) = ζ ( n ) B n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ25_HTML.gif
      (25)
      If n is replaced by 2n in the above equation, we get
      T 1 ( B 2 n ( x ) ) = ζ ( 2 n ) B 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ26_HTML.gif
      (26)
      By using (16), we have
      T 1 ( B 2 n ( x ) ) = ( 1 ) n + 1 ( 2 π ) 2 n B 2 n 2 ( 2 n ) ! B 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ27_HTML.gif
      (27)

      Thus, the proof is completed. □

      Theorem 4.9 Let n Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq57_HTML.gif with n > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq58_HTML.gif. Then we have
      T 1 ( B 2 n ( x ) ) = ( 1 ) n ( 2 π ) 2 n ( 4 n + 2 ) ( 2 n ) ! B 2 n ( x ) k = 1 n 1 ( 2 n 2 k ) B 2 k B 2 n 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equao_HTML.gif
      Proof By using (26), (15) and (16), we have
      T 1 ( B 2 n ( x ) ) = ζ ( 2 n ) B 2 n ( x ) = ( 2 2 n + 1 k = 1 n 1 ζ ( 2 k ) ζ ( 2 n 2 k ) ) B 2 n ( x ) = 2 2 n + 1 ( k = 1 n 1 ( 1 ) k + 1 ( 2 π ) 2 k B 2 k 2 ( 2 k ) ! ( 1 ) n k + 1 ( 2 π ) 2 n 2 k B 2 n 2 k 2 ( 2 n 2 k ) ! ) B 2 n ( x ) = ( 1 ) n ( 2 π ) 2 n B 2 n ( x ) ( 4 n + 2 ) ( 2 n ) ! k = 1 n 1 ( 2 n 2 k ) B 2 k B 2 n 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equap_HTML.gif

      Thus, the proof is completed. □

      Theorem 4.10 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then
      T 1 ( B 2 n ( x ) ) = ( 1 ) n ( 2 π ) 2 n 4 ( 2 n 1 ) ! ( 2 2 n 1 ) E 2 n 1 ( 0 ) B 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equaq_HTML.gif
      Proof By using (26) and (19), we have
      T 1 ( B 2 n ( x ) ) = ζ ( 2 n ) B 2 n ( x ) = ( ( 2 π ) 2 n E 2 n 1 ( 0 ) 4 ( 1 ) n ( 2 n 1 ) ! ( 2 2 n 1 ) ) B 2 n ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equar_HTML.gif

      Thus, the proof is completed. □

      Theorem 4.11 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then
      T 1 ( B 2 n + 1 ( x ) ) = ( 1 ) n + 1 ( 2 π ) 2 n + 1 2 ( 2 n + 1 ) ! B 2 n + 1 ( x ) 0 1 B 2 n + 1 ( t ) cot ( π t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ28_HTML.gif
      (28)
      Proof By replacing n by 2 n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq62_HTML.gif in (25), we have
      T 1 ( B 2 n + 1 ( x ) ) = ζ ( 2 n + 1 ) B 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ29_HTML.gif
      (29)
      By substituting (29) into (22), we get
      T 1 ( B 2 n + 1 ( x ) ) = ( 1 ) n + 1 ( 2 π ) 2 n + 1 2 ( 2 n + 1 ) ! B 2 n + 1 ( x ) 0 1 B 2 n + 1 ( t ) cot ( π t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equas_HTML.gif

      Thus, the proof is completed. □

      Theorem 4.12 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq61_HTML.gif. Then we have
      T 1 ( B 2 n + 1 ( x ) ) = 2 ( 1 ) n ( 2 π ) 2 n ( 2 n ) ! ζ ( 2 n ) B 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ30_HTML.gif
      (30)
      Proof By using (29) and (24), we have
      T 1 ( B 2 n + 1 ( x ) ) = ζ ( 2 n + 1 ) B 2 n + 1 ( x ) = ( 2 ( 1 ) n ζ ( 2 n ) ( 2 π ) 2 n ( 2 n ) ! ) B 2 n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equat_HTML.gif

      Thus, the proof is completed. □

      By comparing (20) and (23) or (28) and (30), we arrive at the following result.

      Corollary 4.13
      0 1 B 2 n + 1 ( t ) cot ( π t ) d t = 2 ( 2 n + 1 ) π ζ ( 2 n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equau_HTML.gif

      Declarations

      Acknowledgements

      Dedicated to Professor Hari M Srivastava.

      All authors are partially supported by Research Project Offices Akdeniz Universities. We would like to thank the referees for their valuable comments.

      Authors’ Affiliations

      (1)
      Department of Mathematics, Faculty of Science, Akdeniz University, Campus

      References

      1. Ozden H, Simsek Y, Srivastava HM: A unified presentation of the generating functions of the generalized Bernoulli, Euler and Genocchi polynomials. Comput. Math. Appl. 2010, 60: 2779-2787. 10.1016/j.camwa.2010.09.031MathSciNetView Article
      2. Kim T, Rim S-H, Simsek Y, Kim D: On the analogs of Bernoulli and Euler numbers, related identities and zeta and L -functions. J. Korean Math. Soc. 2008, 45: 435-453. 10.4134/JKMS.2008.45.2.435MATHMathSciNetView Article
      3. Simsek Y: Twisted ( h , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq63_HTML.gif -Bernoulli numbers and polynomials related to twisted ( h , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq64_HTML.gif -zeta function and L -function. J. Math. Anal. Appl. 2006, 324: 790-804. 10.1016/j.jmaa.2005.12.057MATHMathSciNetView Article
      4. Srivastava HM, Choi J: Series Associated with the Zeta and Related Functions. Kluwer Academic, Dordrecht; 2001.MATHView Article
      5. Carlitz L: A note on the multiplication formulas for the Bernoulli and Euler polynomials. Proc. Am. Math. Soc. 1953, 4: 184-188. 10.1090/S0002-9939-1953-0052569-8MATHView Article
      6. Luo Q-M, Srivastava HM: Some generalizations of the Apostol-Bernoulli and Apostol-Euler polynomials. J. Math. Anal. Appl. 2005, 308: 290-302. 10.1016/j.jmaa.2005.01.020MATHMathSciNetView Article
      7. Raabe JL: Zurückführung einiger summen and bestimmten integrale auf die Jacob Bernoullische function. J. Reine Angew. Math. 1851, 42: 348-376.MATHMathSciNetView Article
      8. Srivastava HM, Kim T, Simsek Y: q -Bernoulli numbers and polynomials associated with multiple q -zeta functions and basic L -series. Russ. J. Math. Phys. 2005, 12: 241-268.MATHMathSciNet
      9. Lehmer DH: A new approach to Bernoulli polynomials. Am. Math. Mon. 1998, 95: 905-911.MathSciNetView Article
      10. Srivastava HM, Choi J: Zeta and q-Zeta Functions and Associated Series and Integrals. Elsevier, Amsterdam; 2012.MATH
      11. Bayad A, Aygunes AA, Simsek Y: Hecke operators and generalized Bernoulli-Euler polynomials. J. Algebra Number Theory, Adv. Appl. 2010, 3: 111-122.MATH
      12. Whittaker ET, Watson GN: A Course of Modern Analysis. 4th edition. Cambridge University Press, Cambridge; 1962.MATH
      13. Chu W, Zhou RR: Convolutions of Bernoulli and Euler polynomials. Sarajevo J. Math. 2010, 6: 147-163.MathSciNet
      14. Washington LC: Introduction to Cyclotomic Fields. Springer, Berlin; 1982.MATHView Article

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      © Aygunes and Simsek; licensee Springer. 2013

      This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.