Open Access

Partial Hecke-type operators and their applications

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 ) https://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 ) https://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 ) , https://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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ2_HTML.gif
(2)

For x = 0 https://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 https://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 https://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 ) , https://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 ) https://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 ) , https://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 ] https://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 ) https://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 , } https://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 , } https://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 + https://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 https://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 ) , https://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 . https://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 + https://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 ) https://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 https://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 ] https://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 , https://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 . https://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 + https://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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq14_HTML.gif be the canonical -basis of
    C m [ x ] = { P ( x ) C [ x ] : deg P ( x ) m } . https://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 ) https://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 https://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 ] https://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 https://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 ) ) https://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 https://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 https://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 ) https://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 . https://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 https://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 https://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 ) . https://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 ) https://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 ) https://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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq23_HTML.gif and N = 2 https://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 ) https://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 ) https://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 + https://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 . https://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 ) https://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 https://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 ) https://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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equi_HTML.gif
where ζ ( s , x ) https://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]) . https://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 ! . https://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 ) https://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 ) https://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 . https://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 ) https://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 ) https://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 ) . https://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 https://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 , https://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 ) https://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 https://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 ! . https://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 ! https://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 https://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 ) . https://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 ) . https://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 ) . https://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 ! ) . https://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 ! . https://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 ! https://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 ) https://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 https://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 ) ) https://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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq33_HTML.gif and v N https://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 ) ) , https://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 ) . https://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 ) ) https://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 ) . https://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 ) . https://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 https://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 } https://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 https://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 https://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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq40_HTML.gif and n 1 https://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 . https://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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq40_HTML.gif and n 1 https://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 . https://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 https://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 ) . https://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 ) https://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 https://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 ] https://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 https://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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equy_HTML.gif
Setting N = 2 https://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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equz_HTML.gif

If a 1 ( 2 ) https://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 ) https://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 https://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 ] https://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 https://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 ) . https://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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq46_HTML.gif and T 2 https://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 . https://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 ) https://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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equac_HTML.gif

where ( s ) > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq49_HTML.gif, n > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq50_HTML.gif and 0 < a < F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq51_HTML.gif ( F Z + https://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 ) https://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 https://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 ) https://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 ) , https://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 ) https://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 ) . https://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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equaf_HTML.gif

Substituting F = N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq54_HTML.gif, s = n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq55_HTML.gif and a = 1 https://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 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq57_HTML.gif with n > 1 https://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 . https://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 https://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 ) , https://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 ) . https://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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq59_HTML.gif and ζ ( n ) https://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 ) , https://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 ) . https://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 ) . https://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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equ15_HTML.gif
(15)
where n Z + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq57_HTML.gif with n > 1 https://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 ) ! . https://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 https://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 ) . https://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 ) . https://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 https://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 . https://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 . https://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 https://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 ) . https://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 https://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 ) https://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 ) . https://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 https://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 . https://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 https://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 ) . https://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 https://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 https://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 ) . https://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 https://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 ) . https://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 https://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 ) https://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 ) . https://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 https://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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_Equan_HTML.gif
Proof Substituting N = 1 https://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 ) https://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 ) . https://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 ) . https://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 ) . https://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 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-46/MediaObjects/13661_2012_Article_303_IEq57_HTML.gif with n > 1 https://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 . https://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 . https://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 https://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 ) . https://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 ) . https://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 https://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 . https://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 https://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 ) . https://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 . https://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 https://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 ) . https://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 ) . https://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 ) . https://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

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© Aygunes and Simsek; licensee Springer. 2013

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