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Derivation of identities involving some special polynomials and numbers via generating functions with applications
Applied Mathematics and Computation 220 (2013) 518–535
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Derivation of identities involving some special polynomials and numbers via generating functions with applications q Moawwad El-Mikkawy ⇑, Faiz Atlan Mathematics Department, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
a r t i c l e
i n f o
Keywords: Bernoulli numbers Bernoulli polynomials Euler numbers Euler polynomials Genocchi numbers Genocchi polynomials Generating functions Maple
a b s t r a c t The current article focus on the ordinary Bernoulli, Euler and Genocchi numbers and polynomials. It introduces a new approach to obtain identities involving these special polynomials and numbers via generating functions. As an application of the new approach, an easy proof for the main result in [6] is given. Relationships between the Genocchi and the Bernoulli polynomials and numbers are obtained. Some interesting identities are discovered. Ó 2013 The Authors. Published by Elsevier Inc. All rights reserved.
1. Introduction and basic definitions The use of polynomials in many areas of science and engineering is quite remarkable: numerical analysis, operator theory, special functions, complex analysis, statistics, sorting and data compression, etc. Throughout this paper we shall focus on three special polynomials. These useful polynomials are the Bernoulli, Euler and Genocchi polynomials. The Bernoulli polynomials and the Bernoulli numbers, for example, are of fundamental importance in several parts of analysis and in the calculus of finite difference. These polynomials and numbers have applications in many fields. For example, in numerical analysis, statistics and combinatorics. The interested reader may refer to [1,2,4,6–9,11,15–17,19,29,31,41,44] and the references therein. The basic properties of Bernoulli numbers and polynomials are well known and are outlined, for example in [5,10,12,14,22,34,39,42]. The generating functions have an important role in many branches of mathematics, statistics and computer science, see for example [26,27,36,43]. The main object of the current paper is to show that the generating functions approach can be employed efficiently to obtain old and new identities involving the Bernoulli, Euler and Genocchi polynomials and numbers. The current paper is organized as follows: In the next section, we list some important properties, without proofs, for some exponential generating functions. The main results of this paper are given in Sections 3 and 4. Finally, a conclusion is given in Section 5. Throughout the paper, dnm is the kronecker symbol which is equal to 1 or 0 according as n ¼ m or not. Also empty summation is assumed equal to zero. Definition 1.1 [36]. The falling factorial of x; ðxÞn is defined by
q This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial-No Derivative Works License, which permits non-commercial use, distribution, and reproduction in any medium, provided the original author and source are credited. ⇑ Corresponding author. E-mail addresses: [email protected], [email protected] (M. El-Mikkawy), [email protected] (F. Atlan).
0096-3003/$ - see front matter Ó 2013 The Authors. Published by Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.06.014
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
ðxÞn ¼
1 if xðx 1Þðx 2Þ . . . ðx n þ 1Þ if
519
n ¼ 0; n P 1:
Definition 1.2 [13]. Let f is a real valued function. The forward difference operator, D of f is defined by
Df ðxÞ ¼ f ðx þ 1Þ f ðxÞ: Generating functions are one of the most surprising, useful, and clever tools in mathematics, computer science and statistics. By using generating functions, we can transform problems about sequences which they generate into problems about real valued functions. Definition 1.3 [36]. The ordinary generating function (OGF) of a sequence < a0 ; a1 ; a2 ; . . . > is defined by
f ðxÞ ¼
1 X an xn : n¼0
Definition 1.4 [26]. The exponential generating function (EGF) of a sequence < a0 ; a1 ; a2 ; . . . > is defined by
f ðxÞ ¼
1 X xn an : n! n¼0
Throughout this paper, we shall be concerned with the exponential generating functions. For these generating functions, we have 1 1 1 X xn X xn X xn an þ bn ¼ ðan þ bn Þ n! n¼0 n! n¼0 n! n¼0 1 X xn an n! n¼0
!
1 X
xn bn n! n¼0
! ¼
1 X xn cn ; n! n¼0
and
ð1Þ
where cn ¼
n X n k¼0
k
ak bnk ;
n P 0:
ð2Þ
If gðxÞ is the generating function of the sequence < a0 ; a1 ; a2 ; . . . >, then we may write the correspondence between the sequence and its generating function by using a double-sided arrow as follows
< a0 ; a1 ; a2 ; . . . > $ gðxÞ ¼ a0 þ a1 x þ a2 x2 þ
ð3Þ
Definition 1.5 [18]. The sequences < B0 ; B1 ; B2 ; . . . > of the Bernoulli numbers, Bn ; n P 0 and < B0 ðxÞ; B1 ðxÞ; B2 ðxÞ; . . . > of the Bernoulli polynomials, Bn ðxÞ; n P 0 are defined by the exponential generating functions
GðtÞ ¼
et
1 X t tn ¼ Bn 1 n¼0 n!
ð4Þ
1 X tex t tn ¼ B ðxÞ ; n et 1 n¼0 n!
ð5Þ
and
Fðx; tÞ ¼
respectively. The first 6 Bernoulli polynomials are
B0 ðxÞ ¼ 1 B1 ðxÞ ¼ x
1 2
B4 ðxÞ ¼ x4 2 x3 þ x2
B2 ðxÞ ¼ x2 x þ
1 30
B5 ðxÞ ¼ x5
1 6
B3 ðxÞ ¼ x3
5 4 5 3 1 x þ x x: 2 3 6
The first 11 Bernoulli numbers are
B0 ¼ 1 B1 ¼ B6 ¼
1 42
1 2
B2 ¼
1 6
B7 ¼ 0 B8 ¼
1 30
B3 ¼ 0 B4 ¼
1 30
B9 ¼ 0 B10 ¼
5 : 66
B5 ¼ 0
3 2 1 x þ x 2 2
520
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
2. Properties of the generating functions G (t) and F (x,t) In this section we are going to list some properties, without proofs, for the exponential generating functions GðtÞ and Fðx; tÞ given in the previous section.
Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property
1. GðtÞ ¼ Fð0; tÞ. 2. GðtÞ ¼ t þ GðtÞ. 3. GðtÞ ¼ Fð1; tÞ. 4. GðtÞ ¼ et GðtÞ. 5. ðet 1Þ GðtÞ ¼ t. t 6. GðtÞ Gð2tÞ ¼ et þ1 . t 7. ðe þ 1Þ Gð2tÞ ¼ 2 GðtÞ. 8. Fðx þ 1; tÞ Fðx; tÞ ¼ t ext . @ 9. @x Fðx; tÞ ¼ t Fðx; tÞ. 10. Fðx; tÞ ¼ ext GðtÞ. 11. Fðx þ y; tÞ ¼ eyt Fðx; tÞ. 12. ðet 1Þ Fðx; tÞ ¼ t ext . 13. Fð1 x; tÞ ¼ Fðx; tÞ. 14. ðet þ 1Þ Fðx; 2tÞ ¼ 2 ext Fðx; tÞ. 15. Fðx þ 12 ; tÞ þ Fðx; tÞ ¼ 2 Fð2 x; 2t Þ. 16. ðet þ 1ÞðFð2x; tÞ Fðx; 2 tÞÞ ¼ t e2xt . 17. Fðx; tÞFðy; tÞ ¼ Fðx þ y; tÞ þ t ðx þ y 1Þ Fðx þ y; tÞ t 18. Fðx; tÞFðy; tÞ ¼ 12 t þ GðtÞ Fðxþy ; 2 tÞ. 2
@ Fðx @t
þ y; tÞ.
3. Main results For convenience of the reader, we begin this section by giving the following useful results whose proofs will be omitted, for the sake of space requirement. Lemma 3.1. Let S ¼
Pn
f ðrÞ. If there exist a function gðrÞ such that f ðrÞ ¼ DgðrÞ, then
r¼1
S ¼ gðn þ 1Þ gð1Þ: Lemma 3.2. Let < a0 ; a1 ; a2 ; . . . > $ AðtÞ. Then, we have
< 0; a0 ; 2 a1 ; 3 a2 ; . . . > $ t AðtÞ;
and
k
< ak ; akþ1 ; akþ2 ; . . . > $
d
dt
k
AðtÞ;
k P 0:
Lemma 3.3. Let < a0 ðxÞ; a1 ðxÞ; a2 ðxÞ; . . . > $ Bðx; tÞ. Then, we have
< 0; a0 ðxÞ; 2 a1 ðxÞ; 3 a2 ðxÞ; . . . > $ t Bðx; tÞ; < ak ðxÞ; akþ1 ðxÞ; akþ2 ðxÞ; . . . > $
and
k @ Bðx; tÞ; @t
k P 0:
We are now ready to give the main results of this section. Theorem 3.1. The Bernoulli numbers Bn ; n P 0 and polynomials Bn ðxÞ; n P 0 satisfy
Bn ¼ Bn ð0Þ;
B2kþ1 ¼ 0;
ð1Þn Bn ¼ Bn ð1Þ;
n P 0: k P 1:
ð6Þ ð7Þ
n P 0:
ð8Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
ð1Þn Bn ¼
n X n k¼0
k
Bk ;
n P 0:
n1 X n Bk ¼ dn1 ; k k¼0
n1 X n k 2 Bk ¼ 2 ð1 2n Þ Bn ; k k¼0
DBn ðxÞ ¼ n xn1 ;
B0n ðxÞ ¼ n Bn1 ðxÞ;
Bn ðxÞ ¼
m X an ¼ a¼1
ð9Þ
n P 0:
ð10Þ
n P 0:
ð11Þ
n P 1:
ð12Þ
n P 1:
n X n Bk xnk ; k k¼0
521
ð13Þ
n P 0:
ð14Þ
1 ðBnþ1 ðm þ 1Þ Bnþ1 ð1ÞÞ: nþ1 n X n Bk ðxÞ ynk ; k k¼0
ð15Þ
Bn ðx þ yÞ ¼
n1 X n Bk ðxÞ ¼ n xn1 ; k k¼0
Bn ð1 xÞ ¼ ð1Þn Bn ðxÞ;
n1 X n k 2 Bk ðxÞ ¼ 2 Bn ð2xÞ 2n Bn ðxÞ ; k k¼0
1 Bn ð2xÞ ¼ 2n1 Bn ðx þ Þ þ Bn ðxÞ ; 2
n1 X n Bk ð2xÞ 2k Bk ðxÞ ¼ n 2n1 xn1 2 Bn ð2xÞ 2n Bn ðxÞ ; k k¼0
n X n Bk ðxÞBnk ðyÞ ¼ ð1 nÞBn ðx þ yÞ þ n ðx þ y 1ÞBn1 ðx þ yÞ: k k¼0
n P 0:
ð16Þ
n P 0:
ð17Þ
n P 0:
ð18Þ
n P 0:
ð19Þ
n P 0:
ð20Þ
n P 0:
ð21Þ
ð22Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
n1 X 2n k¼1
2k
22k1 B2k B2n2k ¼ ðn þ 22n1 Þ B2n ;
n P 2:
ð23Þ
Proof. To prove(6): From Property 1, we have 1 1 X tk X tk Bk ¼ Bk ð0Þ : k! k¼0 k! k¼0
Comparing the coefficients on both sides, we get
Bn ¼ Bn ð0Þ;
n P 0: j
To prove (7): From Property 2, we have 1 1 1 X tk X tk X tk ð1Þk Bk ¼ dk1 þ Bk : k! k! k! k¼0 k¼0 k¼0
Comparing the coefficients on both sides, we obtain
ð1Þn Bn ¼ dn1 þ Bn ; Therefore, we get B1 ¼
12
n P 0: and B2kþ1 ¼ 0;
k P 1, showing that the Bernoulli numbers with odd subscript P3 are zero. h
To prove (8): From Property 3, we have 1 1 X tk X tk ð1Þk Bk ¼ Bk ð1Þ : k! k¼0 k! k¼0
Comparing the coefficients on both sides, gives
ð1Þn Bn ¼ Bn ð1Þ;
n P 0:
From (6)–(8), we see that Bn ð1Þ ¼ Bn ð0Þ; n – 1.
j
To prove (9): From Property 4, we have 1 X tk ð1Þk Bk ¼ k! k¼0
1 k X t k¼0
!
k!
! tk : Bk k! k¼0
1 X
Using (2), yields
ð1Þn Bn ¼
n n X X n n ð1Þnk Bk ¼ Bk ; k k k¼0 k¼0
To prove (10): From Property 5, we obtain 1 k X t k¼0
!
k!
! 1 1 1 X X tk tk X tk Bk Bk ¼ dk1 : k! k! k! k¼0 k¼0 k¼0
By using (2), we get n X n Bk Bn ¼ dn1 : k k¼0
Consequently, we have
n P 0: j
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
523
n1 X n Bk ¼ dn1 : j k k¼0
To prove (11): Rewriting Property 6 in the form
ðet þ 1Þ ðGðtÞ Gð2tÞÞ ¼ t; yields ! 1 k 1 1 X X t tk X tk þ1 ð1 2k ÞBk ¼ dk1 : k! k! k! k¼0 k¼0 k¼0 Consequently, we have 1 k X t k¼0
k!
!
! 1 1 1 X X tk tk X tk k þ ð1 2 Þ Bk ð1 2k Þ Bk ¼ dk1 : k! k! k! k¼0 k¼0 k¼0
By using (2), we get n X n ð1 2k Þ Bk þ ð1 2n Þ Bn ¼ dn1 ; k k¼0
hence, n1 X n ð1 2k Þ Bk þ 2 ð1 2n Þ Bn ¼ dn1 : k k¼0
Therefore, we have n1 X n k 2 Bk ¼ 2 ð1 2n Þ Bn ; k k¼0
n P 0;
having used (10). The same result may also be obtained by using property (7).
j
To prove (12): From Property 8, we have 1 1 1 1 X X tk X tk X tkþ1 tk Bk ðx þ 1Þ Bk ðxÞ ¼ ðk þ 1Þ xk ¼ k xk1 : k! k¼0 k! k¼0 ðk þ 1Þ! k¼0 k! k¼0
Comparing the coefficients on both sides, gives
DBn ðxÞ ¼ Bn ðx þ 1Þ Bn ðxÞ ¼ n xn1 ;
n P 1: j
To prove (13): From Property 9, we have 1 @ X tk Bk ðxÞ @x k¼0 k!
! ¼
1 X ðk þ 1Þ Bk ðxÞ k¼0
tkþ1 : ðk þ 1Þ!
Consequently, we get 1 1 X tk X tk B0k ðxÞ ¼ k Bk1 ðxÞ : k! k! k¼0 k¼0
(Dash means differentiation with respect to x). Comparing the coefficients on both sides, yields
B0n ðxÞ ¼ n Bn1 ðxÞ;
n P 1:
ð24Þ
A more general result is given by
k d Bn ðxÞ ¼ ðnÞk Bnk ðxÞ; dx
k P 0: j
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
To prove (14): From Property 10, we have 1 X tk Bk ðxÞ ¼ k! k¼0
1 X tk xk k! k¼0
!
! 1 X tk : Bk k! k¼0
Using (2), we obtain
Bn ðxÞ ¼
n X n Bk xnk ; k k¼0
n P 0:
ð25Þ
Rewriting Property 10, in the form
GðtÞ ¼ ext Fðx; tÞ: Then, we can easily see that
Bn ¼
n X n k¼0
k
ð1Þnk Bk ðxÞ xnk ;
n P 0;
showing that the Bernoulli numbers can be expressed in terms of the Bernoulli polynomials. j To prove (15): From Property 10, we have
Fðx; tÞ ¼ GðtÞ ext : Integrating both sides with respect to x from x ¼ a to x ¼ a þ 1 (a is any real number), gives
Z
aþ1
Fðx; tÞ dx ¼ eat :
ð26Þ
a
Therefore,
Z
! 1 1 X X tk tk Bk ðxÞ ak : dx ¼ k! k! k¼0 k¼0
aþ1
a
Hence, 1 Z X
aþ1
Bk ðxÞ dx
a
k¼0
1 tk X tk ¼ ak : k! k¼0 k!
Comparing the coefficients on both sides, yields
Z
aþ1
Bn ðxÞ dx ¼ an ;
n P 0:
ð27Þ
a
Setting a ¼ 0 in (27), gives
Z
1
Bn ðxÞ dx ¼ dn0 ;
n P 0:
ð28Þ
0
Using (24), we obtain
Bnþ1 ð1Þ Bnþ1 ð0Þ ¼ ðn þ 1Þ dn0 ;
n P 0:
Consequently, we have
B1 ð1Þ ¼
1 2
and Bn ð1Þ ¼ Bn ð0Þ;
n > 1:
From (27), (13) and Definition 1.2, we have
1 DBnþ1 ðaÞ ¼ an : nþ1 Summing over a from 1 to m, on both sides gives the power sum
fn ðmÞ ¼
m X an ¼ a¼1
! m X 1 1 DBnþ1 ðaÞ ¼ ðBnþ1 ðm þ 1Þ Bnþ1 ð1ÞÞ: n þ 1 a¼1 nþ1
having used Lemma 3.1. j
ð29Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
To prove (16): From Property 11, we have 1 X tk Bk ðx þ yÞ ¼ k! k¼0
1 X tk yk k! k¼0
!
525
! 1 X tk : Bk ðxÞ k! k¼0
By using (2), we get
Bn ðx þ yÞ ¼
n X n Bk ðxÞ ynk ; k k¼0
n P 0:
ð30Þ
Setting x ¼ 0 in (30) gives (14) and setting y ¼ 1, then (30) yields
Bn ðx þ 1Þ ¼
n X n Bk ðxÞ; k k¼0
n P 0:
ð31Þ
If we put x ¼ y in (30), then we have
Bn ð2xÞ ¼
n X n Bk ðxÞ xnk ; k k¼0
n P 0: j
ð32Þ
To prove (17): From Property 12, we have 1 k X t k¼0
k!
! 1
! 1 1 X X tk tkþ1 ¼ Bk ðxÞ ðk þ 1Þ xk : k! ðk þ 1Þ! k¼0 k¼0
Consequently, we get 1 k X t k¼0
!
k!
! 1 1 1 X X tk tk X tk Bk ðxÞ Bk ðxÞ ¼ k xk1 : k! k! k! k¼0 k¼0 k¼0
Using (2), gives n X n Bk ðxÞ Bn ðxÞ ¼ n xn1 ; k k¼0
n P 0:
ð33Þ
Therefore, n1 X n Bk ðxÞ ¼ n xn1 : k k¼0
ð34Þ
In [26], Eq. (34) is the definition of the Bernoulli polynomials. From (31) and (34) we obtain
DBn ðxÞ ¼ Bn ðx þ 1Þ Bn ðxÞ ¼ n xn1 ;
n P 1: j
To prove (18): From Property 13, we have 1 1 X tk X tk Bk ð1 xÞ ¼ ð1Þk Bk ðxÞ : k! k¼0 k! k¼0
Comparing the coefficients on both sides, we obtain
Bn ð1 xÞ ¼ ð1Þn Bn ðxÞ;
n P 0:
ð35Þ
Setting x ¼ 0 in (35), yields
Bn ð1Þ ¼ ð1Þn Bn ð0Þ ¼ ð1Þn Bn ;
To prove (19): From Property 14, we have 1 k X t k¼0
k!
!
n P 0: j
! ! ! 1 1 1 1 X X X X tk tk tk tk þ : 2k Bk ðxÞ 2k Bk ðxÞ ¼ 2 xk Bk ðxÞ k! k! k! k! k¼0 k¼0 k¼0 k¼0
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
Using (2), gives
n n X X n k n 2 Bk ðxÞ þ 2n Bn ðxÞ ¼ 2 Bk ðxÞ xnk ; k k k¼0 k¼0
n P 0:
Hence,
n n X X n k n 2 Bk ðxÞ ¼ 2 Bk ðxÞ xnk 2n Bn ðxÞ; k k k¼0 k¼0
n P 0:
ð36Þ
Using (32) in (36), we obtain n1 X n k 2 Bk ðxÞ ¼ 2 Bn ð2xÞ 2n Bn ðxÞ ; k k¼0
n P 0:
Putting x ¼ 0 in the previous result, yields
Bn ¼
n1 X n k 1 2 Bk ; 2 ð1 2n Þ k¼0 k
n P 1: j
ð37Þ
To prove (20): From Property 15, we have
k 1 1 X X 1 t 1 tk Bk ðx þ Þ þ Bk ðxÞ ¼2 B ð2xÞ : k k 2 k! k! k¼0 k¼0 2
Comparing the coefficients on both sides, gives
1 1 þ Bn ðxÞ ¼ n1 Bn ð2xÞ; Bn x þ 2 2
n P 0:
Consequently, we get
1 þ Bn ðxÞ ; Bn ð2xÞ ¼ 2n1 Bn x þ 2
n P 0:
In particular, if x ¼ 0, then we get
Bn
1 ð1 2n1 Þ ¼ Bn ; 2 2n1
n P 0: j
ð38Þ
To prove (21): From Property 16, we have 1 k X t k¼0
!
k!
! 1 1 1 1 tk tk X X X X tkþ1 tk þ Bk ð2xÞ 2k Bk ðxÞ Bk ð2xÞ 2k Bk ðxÞ ¼ ðk þ 1Þ 2k xk ¼ k 2k1 xk1 : k! k! k¼0 ðk þ 1Þ! k¼0 k! k¼0 k¼0
Using (2), yields n X n Bk ð2xÞ 2k Bk ðxÞ þ ðBn ð2xÞ 2n Bn ðxÞÞ ¼ n 2n1 xn1 ; k k¼0
n P 0:
Consequently, we have n1 X n Bk ð2xÞ 2k Bk ðxÞ ¼ n 2n1 xn1 2 ðBn ð2xÞ 2n Bn ðxÞÞ; k k¼0
To prove (22): From Property 17, we have 1 X
tk Bk ðxÞ k! k¼0
!
1 X tk Bk ðyÞ k! k¼0
!
n P 0: j
1 1 1 X X tk tkþ1 @ X tk ¼ Bk ðx þ yÞ þ ðx þ y 1Þ ðk þ 1Þ Bk ðx þ yÞ t Bk ðx þ yÞ @t k! ðk þ 1Þ! k! k¼0 k¼0 k¼0
¼
1 1 1 X X tk tk X tk Bk ðx þ yÞ þ ðx þ y 1Þ k Bk1 ðx þ yÞ k Bk ðx þ yÞ : k! k! k! k¼0 k¼0 k¼0
!
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
527
By using (2), we get n X n Bk ðxÞBnk ðyÞ ¼ ð1 nÞ Bn ðx þ yÞ þ n ðx þ y 1ÞBn1 ðx þ yÞ: k k¼0
ð39Þ
Setting x ¼ y ¼ 0 in (39), we obtain the quadratic recurrence relation n X n Bk Bnk ¼ ð1 nÞ Bn n Bn1 ; k k¼0
n P 1;
which is also known in its equivalent form
n1 X 2n B2k B2n2k ¼ ð2n þ 1Þ B2n ; 2k k¼1
n P 2: j
ð40Þ
More than 250 years ago, Euler discovered the identity (40). It gives the sum of products of the Bernoulli numbers. To prove (23): From Property 18, we have 1 X tk Bk ðxÞ k! k¼0
!
1 X tk Bk ðyÞ k! k¼0
!
! ! 1 1 1 x þ y t kþ1 x þ y tk X X 1X tk ðk þ 1Þ 2k Bk þ Bk 2k B k 2 k¼0 2 2 ðk þ 1Þ! k! k! k¼0 k¼0 ! ! 1 1 1 X tk X k 1 X k1 x þ y tk x þ y tk : ¼ k 2 Bk1 þ Bk 2 Bk 2 k¼0 2 2 k! k! k! k¼0 k¼0
¼
By using (2), we obtain n n x þ y X X n n k x þ y Bk ðxÞBnk ðyÞ ¼ n 2n2 Bn1 2 Bk þ Bnk : 2 2 k k k¼0 k¼0
Setting x ¼ y in (41), yields n n X X n n k Bk ðxÞBnk ðxÞ ¼ n 2n2 Bn1 þ 2 Bk Bnk : k k k¼0 k¼0
But, Bn ðxÞ ¼ ð1Þn Bn ð1 þ xÞ. Therefore, n n X X n n k ð1Þnk Bk ðxÞBnk ð1 þ xÞ ¼ n 2n2 Bn1 þ 2 Bk Bnk : k k k¼0 k¼0
Putting x ¼ y ¼ 0 in (41), gives n X n ð1 2k ÞBk Bnk ¼ n 2n2 Bn1 : k k¼0
Replacing n by 2n, we get
2n X 2n ð1 2k ÞBk B2nk ¼ 0; k k¼0
n P 2:
Consequently, we have
n1 n1 X X 2n k 2n 4 B2k B2n2k ¼ ð1 4n Þ B2n þ B2k B2n2k ; 2k 2k k¼1 k¼1 By using (40), we end up with the quadratic identity
n1 X 2n 2k1 2 B2k B2n2k ¼ ðn þ 22n1 ÞB2n ; 2k k¼1
n P 2: j
n P 2:
ð41Þ
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4. Applications and further results 4.1. A connection between the Bernoulli polynomials, Bn ðxÞ and the Euler polynomials, En ðxÞ We begin this subsection by giving the following definition for the Euler polynomials. These polynomials are closely related to the Bernoulli polynomials [28]. Definition 4.1.1. ([12]). The sequence < E0 ðxÞ; E1 ðxÞ; E2 ðxÞ; . . . > of the Euler polynomials En ðxÞ; n P 0 is defined by the exponential generating function 1 X 2 ext tk ¼ Ek ðxÞ : t e þ 1 k¼0 k!
Hðx; tÞ ¼
ð42Þ
The basic properties of Euler numbers and polynomials can be found in [10,12,28,39]. The first 6 Euler polynomials are
E0 ðxÞ ¼ 1
E1 ðxÞ ¼ x
E4 ðxÞ ¼ x4 2x3 þ x
1 2
E2 ðxÞ ¼ x2 x
3 1 E3 ðxÞ ¼ x3 x2 þ 2 4
5 5 1 E5 ðxÞ ¼ x5 x4 þ x2 : 2 2 2
The exponential generating functions GðtÞ; Fðx; tÞ and Hðx; tÞ satisfy the relationship
Fðx; tÞ ¼
1 t þ GðtÞ Hðx; tÞ; 2
ð43Þ
as can be easily checked. We are now in a position to give an easy proof for the following theorem, which is the main result in [6], (see also [40]). Theorem 4.1.1. The Bernoulli polynomials Bn ðxÞ; n P 0 are related to the Euler polynomials En ðxÞ; n P 0 by
Bn ðxÞ ¼
n X n k¼0 k–1
k
Bk Enk ðxÞ ¼
n X n k¼0
k
ð1 dk1 Þ Bk Enk ðxÞ:
Proof. From (43), we have
! ! ! 1 1 1 X X tk 1X t kþ1 tk tk ¼ Ek ðxÞ ðk þ 1Þ Ek ðxÞ þ Bk Ek ðxÞ 2 k¼0 k! ðk þ 1Þ! k! k! k¼0 k¼0 k¼0 ! ! 1 1 1 X X 1X tk tk tk : ¼ k Ek1 ðxÞ þ Bk Ek ðxÞ 2 k¼0 k! k! k! k¼0 k¼0
1 X tk Bk ðxÞ ¼ k! k¼0
1 X 1 tk tþ Bk 2 k! k¼0
!
1 X
Using (2), we obtain
Bn ðxÞ ¼
n n n X X X n n n n 1 Bk Enk ðxÞ ¼ B1 En1 ðxÞ þ Bk Enk ðxÞ ¼ Bk Enk ðxÞ n En1 ðxÞ þ 2 k 1 k k k¼0 k¼0 k¼0
n X n ¼ ð1 dk1 Þ Bk Enk ðxÞ: j k k¼0
k–1 ð44Þ
4.2. A Connection between the Bernoulli Polynomials, Bn ðxÞ and the Genocchi polynomials, Gn ðxÞ In recent years there has been a flurry of activity for doing research with Genocchi numbers and polynomials. See for instance, [3,21,23–25,30,32,35,37,38,40]. These numbers and polynomials are given by the following definition. Definition 4.2.1 [25]. The sequences < G0 ; G1 ; G2 ; . . . > of the Genocchi numbers, Gn ; n P 0 and < G0 ðxÞ; G1 ðxÞ; G2 ðxÞ; . . . > of the Genocchi polynomials, Gn ðxÞ; n P 0 are defined by the exponential generating functions
WðtÞ ¼
1 X 2t tn ¼ Gn : et þ 1 n¼0 n!
ð45Þ
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and 1 X 2 t ext tn ¼ Gn ðxÞ ; et þ 1 n¼0 n!
Q ðx; tÞ ¼
ð46Þ
respectively. The first 11 Genocchi numbers Gn ¼ Gn ð0Þ are
G0 ¼ 0
G1 ¼ 1
G2 ¼ 1
G3 ¼ 0
G4 ¼ 1
G5 ¼ 0
G6 ¼ 3
G7 ¼ 0
G8 ¼ 17
G9 ¼ 0
G10 ¼ 155:
The first few Genocchi polynomials are
G0 ðxÞ ¼ 0
G1 ðxÞ ¼ 1
G4 ðxÞ ¼ 4x3 6x2 þ 1
G2 ðxÞ ¼ 2x 1
G3 ðxÞ ¼ 3x2 3x
G5 ðxÞ ¼ 5x4 10x3 þ 5x
G6 ðxÞ ¼ 6x5 15x4 þ 15x2 3:
Consider
< E0 ðxÞ; E1 ðxÞ; E2 ðxÞ; . . . > $ Hðx; tÞ:
ð47Þ
Using Lemma 3.3, yields
< 0; E0 ðxÞ; 2 E1 ðxÞ; 3 E2 ðxÞ; . . . > $ t Hðx; tÞ ¼ Qðx; tÞ:
ð48Þ
(See (42) and (46)). On the other hand, we have from (46),
< G0 ðxÞ; G1 ðxÞ; G2 ðxÞ; . . . > $ Qðx; tÞ:
ð49Þ
From (48) and (49), we get
Gn ðxÞ ¼ n En1 ðxÞ;
n P 1:
ð50Þ
It is easy to see that the exponential generating functions GðtÞ; Fðx; tÞ and Q ðx; tÞ are related by
t Fðx; tÞ ¼
1 t þ GðtÞ Q ðx; tÞ: 2
ð51Þ
We are now ready to give a connection between the Bernoulli polynomials Bn ðxÞ; n P 0 and the Genocchi polynomials Gn ðxÞ; n P 0. This connection is given by the following result. Theorem 4.2.1. The Bernoulli polynomials Bn ðxÞ; n P 0 can be expressed by the Genocchi polynomials Gn ðxÞ; n P 0 as
Bn ðxÞ ¼
n n X nþ1 nþ1 1 1 X Bk Gnþ1k ðxÞ ¼ ð1 dk1 ÞBk Gnþ1k ðxÞ: nþ1 n þ 1 k¼0 k k k¼0
ð52Þ
k–1 First Proof. Using (51), we have 1 X ðk þ 1Þ Bk ðxÞ k¼0
tkþ1 ¼ ðk þ 1Þ!
1 X 1 tk tþ Bk 2 k! k¼0
!
1 X tk Gk ðxÞ k! k¼0
!
! ! 1 1 1 X X 1X t kþ1 tk tk ¼ : ðk þ 1Þ Gk ðxÞ þ Bk Gk ðxÞ 2 k¼0 ðk þ 1Þ! k! k! k¼0 k¼0
Consequently,
! ! 1 1 1 1 X X X tk 1 X tk tk tk k Bk1 ðxÞ ¼ k Gk1 ðxÞ þ Bk Gk ðxÞ : k! 2 k¼0 k! k! k! k¼0 k¼0 k¼0 Using (2), we obtain
n Bn1 ðxÞ ¼
n n n X X X n n n n 1 Bk Gnk ðxÞ ¼ B1 Gn1 ðxÞ þ Bk Gnk ðxÞ ¼ Bk Gnk ðxÞ n Gn1 ðxÞ þ 2 k 1 k k k¼0 k¼0 k¼0
n X n ð1 dk1 Þ Bk Gnk ðxÞ: ¼ k k¼0
Therefore, we have
k–1
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Bn ðxÞ ¼
n nþ1 1 X ð1 dk1 Þ Bk Gnþ1k ðxÞ; n þ 1 k¼0 k
since G0 ðxÞ ¼ 0: j Alternative proof. We have [6,40],
n X n
Bn ðxÞ ¼
k
k¼0
Bk Enk ðxÞ:
k–1 By using (50), we get
n n X nþ1 1 k Bk Gnþ1k ðxÞ: Bn ðxÞ ¼ Bk Gnþ1k ðxÞ ¼ nþ1 nþ1k k k¼0 k¼0 k–1 k–1 n X
ð53Þ
having used the identity
nþ1 k
¼
nþ1 nþ1k
n : j k
4.3. Further properties and identities It is worth mentioned that new more interesting identities can be obtained using the technique of this paper by adding new properties. Here we add the following 10 additional properties.
ð1 þ et Þ WðtÞ ¼ 2 t:
ð54Þ
WðtÞ WðtÞ ¼ 2 t:
ð55Þ
GðtÞ Gð2tÞ ¼
Qðx; tÞ ¼ ext WðtÞ:
Z 0
1
1 WðtÞ: 2
t Q ðx; tÞ dx ¼ 2 tanhð Þ: 2
ð56Þ
ð57Þ
ð58Þ
@ Q ðx; tÞ ¼ t Q ðx; tÞ: @x
ð59Þ
Qðx þ 1; tÞ þ Qðx; tÞ ¼ 2 t ext :
ð60Þ
GðtÞ WðtÞ ¼ t Gð2tÞ:
ð61Þ
ð2 t 2 WðtÞÞFðx; tÞ ¼ t Qðx; tÞ:
ð62Þ
@ Qðx; tÞQðy; tÞ ¼ 2 t Q ðx þ y; tÞ Q ðx þ y; tÞ t ðx þ y 1ÞQ ðx þ y; tÞ : @t
ð63Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
531
The property (54), together with (2), gives
Gn ¼ dn1
n1 n 1X Gk ; 2 k¼1 k
n P 0:
ð64Þ
From (55), one can see that
G1 ¼ 1;
G2nþ1 ¼ 0;
n P 1:
ð65Þ
Therefore using (50), we get
E2n ð0Þ ¼ 0;
n P 1:
ð66Þ
(See Theorem 3.2 in [20]). Similarly using (56), we obtain
Gn ¼ 2 ð1 2n Þ Bn ;
n P 0:
ð67Þ
From (11) and (67), we have the following curious identity between the Bernoulli and Genocchi numbers
Gn ¼
n1 X n k 2 Bk ; k k¼0
n P 1:
ð68Þ
Note that this identity can also be obtained from (37). The property (57), together with (2), yields
Gn ðxÞ ¼
n X n Gk xnk ; k k¼0
n P 0:
ð69Þ
Using (58), together with the following series expansion for tanhðtÞ
tanhðxÞ ¼
1 X 4n ð4n 1Þ B2n
2n!
n¼1
x2n1 ;
we see that
Z
1
2 Gnþ1 : n P 1; nþ1
Gn ðxÞ dx ¼
0
ð70Þ
Therefore, we have
Z
1
En ðxÞ dx ¼
0
2 Enþ1 ð0Þ ; nþ1
n P 1:
In particular, by using (66), then we have
Z
1
E2n1 ðxÞ dx ¼ 0;
n P 1:
0
From (59), we obtain
G0n ðxÞ ¼ n Gn1 ðxÞ;
n P 1:
Using (60), gives
Gn ðx þ 1Þ þ Gn ðxÞ ¼ 2 n xn1 ;
n P 1:
ð71Þ
Replacing x by x and n by n þ 1 in (71), we see that
Gnþ1 ðxÞ þ Gnþ1 ð1 xÞ ¼ ð1Þn ðGnþ1 ðxÞ þ Gnþ1 ð1 þ xÞÞ ¼ ð1Þn 2 ðn þ 1Þ xn : Thus the sums Gn ðxÞ þ Gn ð1 þ xÞ and consequently, En ðxÞ þ En ð1 þ xÞ are even functions if n is odd. Putting x ¼ 0 in (71), yields
Gn ð1Þ þ Gn ¼ 2 dn1 :
ð72Þ
From (72), we see that
Gn ð1Þ ¼ Gn ð0Þ ¼ Gn ;
n P 2:
ð73Þ
Making use of (61), together with (2), we arrive at
Bn ¼
n1 X n 1 Bk Enk ð0Þ: ð2n 1Þ k¼0 k
ð74Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
Using (67) in (74) together with the fact that B1 ¼ 12, we get n1 X n
k
k¼0 k–1
Bk Enk ð0Þ ¼
1 dn1 ; 2
ð75Þ
on simplification. The following identity is a direct consequence of (75) n2 X n
k
k¼0
Bk Gnk ¼ dn2 :
ð76Þ
k–1 It is worth mentioned that the identity (76) may be also obtained by using (52), (28) and (70). Eq. (76) is also equivalent to
n
n X
k
k¼2
Gk Bnk ¼ dn2 : ð77Þ
k–n 1 Replacing n by 2n þ 2 and k by k þ 1 in (77), gives
2nþ1 X
2n þ 2
Gkþ1 B2nþ1k ¼ dn0 :
kþ1
k¼1 k–2n But
2n þ 2
kþ1
¼
2n þ 2 kþ1
2n þ 1
k
:
Hence,
2nþ1 X
2n þ 1 k
k¼1
ð2n þ 2Þ Ek ð0Þ B2nþ1k ¼ dn0 ;
k–2n having used (50). Consequently, we have 2nþ1 X
2n þ 1
k
k¼1
Ek ð0Þ B2nþ1k ¼
1 dn0 ; 2ðn þ 1Þ
ð78Þ
having used (66). The identity (78) is Theorem 3.3 in the very recent paper [20]. Using (72) and (7), we get
n1 X 2n k¼0
2k
B2k G2n2k ¼
n X 2n k¼1
2k
G2k B2n2k ¼ dn1 :
ð79Þ
By using (78), together with (66) we see that
n X 2n þ 1 k¼0
2k þ 1
E2kþ1 ð0Þ B2n2k ¼
1 dn0 : 2ðn þ 1Þ
ð80Þ
Using (50) in (80), gives
n X 2n k¼1
2k
E2k ð0Þ B2n2k ¼ dn1;0 :
ð81Þ
The property (62), together with (2), gives n1 X n 1 Ek ð0Þ Bnk ðxÞ ¼ Gn ðxÞ; 2 k k¼1
n P 1:
ð82Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
533
The property (63), together with (2), yields n1 X n Gk ðxÞGnk ðyÞ ¼ 2 ððn 1Þ Gn ðx þ yÞ n ðx þ y 1Þ Gn1 ðx þ yÞÞ; k k¼1
n P 2:
ð83Þ
It follows from (83) that if y ¼ 1 x, then we have n1 X n Gk ðxÞGnk ð1 xÞ ¼ 2 ðn 1Þ Gn ; n P 2; k k¼1
for any rael x;
ð84Þ
having used (73). Setting x ¼ y ¼ 0 in (83), we get the following quadratic identity n1 X n Gk Gnk ¼ 2 n Gn1 þ 2 ðn 1Þ Gn ; k k¼1
n P 2:
At this stage, let us consider the sequence < E0 ; E1 ; E2 ; . . . > of the Euler numbers En ; n P 0, defined by
EðtÞ ¼
1 X 2 1 tn ¼ ¼ En : et þ et coshðtÞ n¼0 n!
ð85Þ
Therefore,
< E0 ; E1 ; E2 ; . . . >$ EðtÞ ¼
1 X
En
n¼0
tn : n!
The odd-indexed Euler numbers are all zero. The even-indexed ones have alternating signs, and are given by the recursive recurrence relation
n1 X 2n E2k ; E2n ¼ 2k k¼0
E0 ¼ 1;
n P 1:
ð86Þ
The first few values for the Euler numbers are
E0 ¼ 1
E2 ¼ 1
E4 ¼ 5
E6 ¼ 61
E8 ¼ 1385
E10 ¼ 50521
The relationship between the two generating functions WðtÞ and EðtÞ is given by
et Wð2 tÞ ¼ 2 t EðtÞ:
ð87Þ
Using (87), together with (2), yields
En1 ¼
! n n k1 1 X 2 Gk þ 2n1 Gn ; n k¼1 k
n P 1:
ð88Þ
Consequently, we have
En ¼
n X n k¼0
k
1 2k Ek ð0Þ ¼ 2n En ð Þ; 2
ð89Þ
having used (50), together with the fact that the Euler polynomials En ðxÞ satisfy [8]
En ðxÞ ¼
n X n Ek ð0Þ xnk : k k¼0
Let us finish this subsection by giving some identities concerning the Euler numbers. Using (89) in (84) with x ¼ 12, gives the identity n1 X n k¼1
Since
k
k ðn kÞ Ek1 Enk1 ¼ 2n1 n ðn 1ÞEn1 ð0Þ;
n P 1:
n n2 k ðn kÞ ¼ n ðn 1Þ, we get, k k1
n1 X n2 k¼1
k1
Ek1 Enk1 ¼ 2n1 En1 ð0Þ;
n P 3;
ð90Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
on simplification. Replacing n by n þ 2 on both sides and k by k þ 1 in (90), gives n X n k¼0
k
Ek Enk ¼ 2nþ1 Enþ1 ð0Þ;
n P 1:
Consequently, showing that the Euler numbers, En satisfy the quadratic recurrence relation
n X 2n k¼0
2k
E2k E2n2k ¼ 22nþ1 E2nþ1 ð0Þ;
n P 1:
ð91Þ
5. Conclusion The current paper is addressed a generating function approach to obtain old and new identities. The approach is employed in a constructive way to obtain identities involving the Bernoulli, Euler and Genocchi polynomials and numbers. We established some properties for generating functions and two applications are given. Among the applications, we should stress that a simple proof for the main result in [6], (see also [40]) is given. Finally, some useful relationships between the Bernoulli, Euler and Genocchi polynomials and numbers are obtained. All properties and results of this paper are tested using the Computer Algebra System (CAS), Maple [33]. Acknowledgments The authors wish to thank anonymous referees, Prof. Dr. Melvin Scott, Editor-in-Chief and Prof. Dr. T.E.Simos, Senior Editor for useful comments that enhanced the quality of this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41]
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Derivation of identities involving some special polynomials and numbers via generating functions with applications q Moawwad El-Mikkawy ⇑, Faiz Atlan Mathematics Department, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
a r t i c l e
i n f o
Keywords: Bernoulli numbers Bernoulli polynomials Euler numbers Euler polynomials Genocchi numbers Genocchi polynomials Generating functions Maple
a b s t r a c t The current article focus on the ordinary Bernoulli, Euler and Genocchi numbers and polynomials. It introduces a new approach to obtain identities involving these special polynomials and numbers via generating functions. As an application of the new approach, an easy proof for the main result in [6] is given. Relationships between the Genocchi and the Bernoulli polynomials and numbers are obtained. Some interesting identities are discovered. Ó 2013 The Authors. Published by Elsevier Inc. All rights reserved.
1. Introduction and basic definitions The use of polynomials in many areas of science and engineering is quite remarkable: numerical analysis, operator theory, special functions, complex analysis, statistics, sorting and data compression, etc. Throughout this paper we shall focus on three special polynomials. These useful polynomials are the Bernoulli, Euler and Genocchi polynomials. The Bernoulli polynomials and the Bernoulli numbers, for example, are of fundamental importance in several parts of analysis and in the calculus of finite difference. These polynomials and numbers have applications in many fields. For example, in numerical analysis, statistics and combinatorics. The interested reader may refer to [1,2,4,6–9,11,15–17,19,29,31,41,44] and the references therein. The basic properties of Bernoulli numbers and polynomials are well known and are outlined, for example in [5,10,12,14,22,34,39,42]. The generating functions have an important role in many branches of mathematics, statistics and computer science, see for example [26,27,36,43]. The main object of the current paper is to show that the generating functions approach can be employed efficiently to obtain old and new identities involving the Bernoulli, Euler and Genocchi polynomials and numbers. The current paper is organized as follows: In the next section, we list some important properties, without proofs, for some exponential generating functions. The main results of this paper are given in Sections 3 and 4. Finally, a conclusion is given in Section 5. Throughout the paper, dnm is the kronecker symbol which is equal to 1 or 0 according as n ¼ m or not. Also empty summation is assumed equal to zero. Definition 1.1 [36]. The falling factorial of x; ðxÞn is defined by
q This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial-No Derivative Works License, which permits non-commercial use, distribution, and reproduction in any medium, provided the original author and source are credited. ⇑ Corresponding author. E-mail addresses: [email protected], [email protected] (M. El-Mikkawy), [email protected] (F. Atlan).
0096-3003/$ - see front matter Ó 2013 The Authors. Published by Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.06.014
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
ðxÞn ¼
1 if xðx 1Þðx 2Þ . . . ðx n þ 1Þ if
519
n ¼ 0; n P 1:
Definition 1.2 [13]. Let f is a real valued function. The forward difference operator, D of f is defined by
Df ðxÞ ¼ f ðx þ 1Þ f ðxÞ: Generating functions are one of the most surprising, useful, and clever tools in mathematics, computer science and statistics. By using generating functions, we can transform problems about sequences which they generate into problems about real valued functions. Definition 1.3 [36]. The ordinary generating function (OGF) of a sequence < a0 ; a1 ; a2 ; . . . > is defined by
f ðxÞ ¼
1 X an xn : n¼0
Definition 1.4 [26]. The exponential generating function (EGF) of a sequence < a0 ; a1 ; a2 ; . . . > is defined by
f ðxÞ ¼
1 X xn an : n! n¼0
Throughout this paper, we shall be concerned with the exponential generating functions. For these generating functions, we have 1 1 1 X xn X xn X xn an þ bn ¼ ðan þ bn Þ n! n¼0 n! n¼0 n! n¼0 1 X xn an n! n¼0
!
1 X
xn bn n! n¼0
! ¼
1 X xn cn ; n! n¼0
and
ð1Þ
where cn ¼
n X n k¼0
k
ak bnk ;
n P 0:
ð2Þ
If gðxÞ is the generating function of the sequence < a0 ; a1 ; a2 ; . . . >, then we may write the correspondence between the sequence and its generating function by using a double-sided arrow as follows
< a0 ; a1 ; a2 ; . . . > $ gðxÞ ¼ a0 þ a1 x þ a2 x2 þ
ð3Þ
Definition 1.5 [18]. The sequences < B0 ; B1 ; B2 ; . . . > of the Bernoulli numbers, Bn ; n P 0 and < B0 ðxÞ; B1 ðxÞ; B2 ðxÞ; . . . > of the Bernoulli polynomials, Bn ðxÞ; n P 0 are defined by the exponential generating functions
GðtÞ ¼
et
1 X t tn ¼ Bn 1 n¼0 n!
ð4Þ
1 X tex t tn ¼ B ðxÞ ; n et 1 n¼0 n!
ð5Þ
and
Fðx; tÞ ¼
respectively. The first 6 Bernoulli polynomials are
B0 ðxÞ ¼ 1 B1 ðxÞ ¼ x
1 2
B4 ðxÞ ¼ x4 2 x3 þ x2
B2 ðxÞ ¼ x2 x þ
1 30
B5 ðxÞ ¼ x5
1 6
B3 ðxÞ ¼ x3
5 4 5 3 1 x þ x x: 2 3 6
The first 11 Bernoulli numbers are
B0 ¼ 1 B1 ¼ B6 ¼
1 42
1 2
B2 ¼
1 6
B7 ¼ 0 B8 ¼
1 30
B3 ¼ 0 B4 ¼
1 30
B9 ¼ 0 B10 ¼
5 : 66
B5 ¼ 0
3 2 1 x þ x 2 2
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
2. Properties of the generating functions G (t) and F (x,t) In this section we are going to list some properties, without proofs, for the exponential generating functions GðtÞ and Fðx; tÞ given in the previous section.
Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property Property
1. GðtÞ ¼ Fð0; tÞ. 2. GðtÞ ¼ t þ GðtÞ. 3. GðtÞ ¼ Fð1; tÞ. 4. GðtÞ ¼ et GðtÞ. 5. ðet 1Þ GðtÞ ¼ t. t 6. GðtÞ Gð2tÞ ¼ et þ1 . t 7. ðe þ 1Þ Gð2tÞ ¼ 2 GðtÞ. 8. Fðx þ 1; tÞ Fðx; tÞ ¼ t ext . @ 9. @x Fðx; tÞ ¼ t Fðx; tÞ. 10. Fðx; tÞ ¼ ext GðtÞ. 11. Fðx þ y; tÞ ¼ eyt Fðx; tÞ. 12. ðet 1Þ Fðx; tÞ ¼ t ext . 13. Fð1 x; tÞ ¼ Fðx; tÞ. 14. ðet þ 1Þ Fðx; 2tÞ ¼ 2 ext Fðx; tÞ. 15. Fðx þ 12 ; tÞ þ Fðx; tÞ ¼ 2 Fð2 x; 2t Þ. 16. ðet þ 1ÞðFð2x; tÞ Fðx; 2 tÞÞ ¼ t e2xt . 17. Fðx; tÞFðy; tÞ ¼ Fðx þ y; tÞ þ t ðx þ y 1Þ Fðx þ y; tÞ t 18. Fðx; tÞFðy; tÞ ¼ 12 t þ GðtÞ Fðxþy ; 2 tÞ. 2
@ Fðx @t
þ y; tÞ.
3. Main results For convenience of the reader, we begin this section by giving the following useful results whose proofs will be omitted, for the sake of space requirement. Lemma 3.1. Let S ¼
Pn
f ðrÞ. If there exist a function gðrÞ such that f ðrÞ ¼ DgðrÞ, then
r¼1
S ¼ gðn þ 1Þ gð1Þ: Lemma 3.2. Let < a0 ; a1 ; a2 ; . . . > $ AðtÞ. Then, we have
< 0; a0 ; 2 a1 ; 3 a2 ; . . . > $ t AðtÞ;
and
k
< ak ; akþ1 ; akþ2 ; . . . > $
d
dt
k
AðtÞ;
k P 0:
Lemma 3.3. Let < a0 ðxÞ; a1 ðxÞ; a2 ðxÞ; . . . > $ Bðx; tÞ. Then, we have
< 0; a0 ðxÞ; 2 a1 ðxÞ; 3 a2 ðxÞ; . . . > $ t Bðx; tÞ; < ak ðxÞ; akþ1 ðxÞ; akþ2 ðxÞ; . . . > $
and
k @ Bðx; tÞ; @t
k P 0:
We are now ready to give the main results of this section. Theorem 3.1. The Bernoulli numbers Bn ; n P 0 and polynomials Bn ðxÞ; n P 0 satisfy
Bn ¼ Bn ð0Þ;
B2kþ1 ¼ 0;
ð1Þn Bn ¼ Bn ð1Þ;
n P 0: k P 1:
ð6Þ ð7Þ
n P 0:
ð8Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
ð1Þn Bn ¼
n X n k¼0
k
Bk ;
n P 0:
n1 X n Bk ¼ dn1 ; k k¼0
n1 X n k 2 Bk ¼ 2 ð1 2n Þ Bn ; k k¼0
DBn ðxÞ ¼ n xn1 ;
B0n ðxÞ ¼ n Bn1 ðxÞ;
Bn ðxÞ ¼
m X an ¼ a¼1
ð9Þ
n P 0:
ð10Þ
n P 0:
ð11Þ
n P 1:
ð12Þ
n P 1:
n X n Bk xnk ; k k¼0
521
ð13Þ
n P 0:
ð14Þ
1 ðBnþ1 ðm þ 1Þ Bnþ1 ð1ÞÞ: nþ1 n X n Bk ðxÞ ynk ; k k¼0
ð15Þ
Bn ðx þ yÞ ¼
n1 X n Bk ðxÞ ¼ n xn1 ; k k¼0
Bn ð1 xÞ ¼ ð1Þn Bn ðxÞ;
n1 X n k 2 Bk ðxÞ ¼ 2 Bn ð2xÞ 2n Bn ðxÞ ; k k¼0
1 Bn ð2xÞ ¼ 2n1 Bn ðx þ Þ þ Bn ðxÞ ; 2
n1 X n Bk ð2xÞ 2k Bk ðxÞ ¼ n 2n1 xn1 2 Bn ð2xÞ 2n Bn ðxÞ ; k k¼0
n X n Bk ðxÞBnk ðyÞ ¼ ð1 nÞBn ðx þ yÞ þ n ðx þ y 1ÞBn1 ðx þ yÞ: k k¼0
n P 0:
ð16Þ
n P 0:
ð17Þ
n P 0:
ð18Þ
n P 0:
ð19Þ
n P 0:
ð20Þ
n P 0:
ð21Þ
ð22Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
n1 X 2n k¼1
2k
22k1 B2k B2n2k ¼ ðn þ 22n1 Þ B2n ;
n P 2:
ð23Þ
Proof. To prove(6): From Property 1, we have 1 1 X tk X tk Bk ¼ Bk ð0Þ : k! k¼0 k! k¼0
Comparing the coefficients on both sides, we get
Bn ¼ Bn ð0Þ;
n P 0: j
To prove (7): From Property 2, we have 1 1 1 X tk X tk X tk ð1Þk Bk ¼ dk1 þ Bk : k! k! k! k¼0 k¼0 k¼0
Comparing the coefficients on both sides, we obtain
ð1Þn Bn ¼ dn1 þ Bn ; Therefore, we get B1 ¼
12
n P 0: and B2kþ1 ¼ 0;
k P 1, showing that the Bernoulli numbers with odd subscript P3 are zero. h
To prove (8): From Property 3, we have 1 1 X tk X tk ð1Þk Bk ¼ Bk ð1Þ : k! k¼0 k! k¼0
Comparing the coefficients on both sides, gives
ð1Þn Bn ¼ Bn ð1Þ;
n P 0:
From (6)–(8), we see that Bn ð1Þ ¼ Bn ð0Þ; n – 1.
j
To prove (9): From Property 4, we have 1 X tk ð1Þk Bk ¼ k! k¼0
1 k X t k¼0
!
k!
! tk : Bk k! k¼0
1 X
Using (2), yields
ð1Þn Bn ¼
n n X X n n ð1Þnk Bk ¼ Bk ; k k k¼0 k¼0
To prove (10): From Property 5, we obtain 1 k X t k¼0
!
k!
! 1 1 1 X X tk tk X tk Bk Bk ¼ dk1 : k! k! k! k¼0 k¼0 k¼0
By using (2), we get n X n Bk Bn ¼ dn1 : k k¼0
Consequently, we have
n P 0: j
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
523
n1 X n Bk ¼ dn1 : j k k¼0
To prove (11): Rewriting Property 6 in the form
ðet þ 1Þ ðGðtÞ Gð2tÞÞ ¼ t; yields ! 1 k 1 1 X X t tk X tk þ1 ð1 2k ÞBk ¼ dk1 : k! k! k! k¼0 k¼0 k¼0 Consequently, we have 1 k X t k¼0
k!
!
! 1 1 1 X X tk tk X tk k þ ð1 2 Þ Bk ð1 2k Þ Bk ¼ dk1 : k! k! k! k¼0 k¼0 k¼0
By using (2), we get n X n ð1 2k Þ Bk þ ð1 2n Þ Bn ¼ dn1 ; k k¼0
hence, n1 X n ð1 2k Þ Bk þ 2 ð1 2n Þ Bn ¼ dn1 : k k¼0
Therefore, we have n1 X n k 2 Bk ¼ 2 ð1 2n Þ Bn ; k k¼0
n P 0;
having used (10). The same result may also be obtained by using property (7).
j
To prove (12): From Property 8, we have 1 1 1 1 X X tk X tk X tkþ1 tk Bk ðx þ 1Þ Bk ðxÞ ¼ ðk þ 1Þ xk ¼ k xk1 : k! k¼0 k! k¼0 ðk þ 1Þ! k¼0 k! k¼0
Comparing the coefficients on both sides, gives
DBn ðxÞ ¼ Bn ðx þ 1Þ Bn ðxÞ ¼ n xn1 ;
n P 1: j
To prove (13): From Property 9, we have 1 @ X tk Bk ðxÞ @x k¼0 k!
! ¼
1 X ðk þ 1Þ Bk ðxÞ k¼0
tkþ1 : ðk þ 1Þ!
Consequently, we get 1 1 X tk X tk B0k ðxÞ ¼ k Bk1 ðxÞ : k! k! k¼0 k¼0
(Dash means differentiation with respect to x). Comparing the coefficients on both sides, yields
B0n ðxÞ ¼ n Bn1 ðxÞ;
n P 1:
ð24Þ
A more general result is given by
k d Bn ðxÞ ¼ ðnÞk Bnk ðxÞ; dx
k P 0: j
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
To prove (14): From Property 10, we have 1 X tk Bk ðxÞ ¼ k! k¼0
1 X tk xk k! k¼0
!
! 1 X tk : Bk k! k¼0
Using (2), we obtain
Bn ðxÞ ¼
n X n Bk xnk ; k k¼0
n P 0:
ð25Þ
Rewriting Property 10, in the form
GðtÞ ¼ ext Fðx; tÞ: Then, we can easily see that
Bn ¼
n X n k¼0
k
ð1Þnk Bk ðxÞ xnk ;
n P 0;
showing that the Bernoulli numbers can be expressed in terms of the Bernoulli polynomials. j To prove (15): From Property 10, we have
Fðx; tÞ ¼ GðtÞ ext : Integrating both sides with respect to x from x ¼ a to x ¼ a þ 1 (a is any real number), gives
Z
aþ1
Fðx; tÞ dx ¼ eat :
ð26Þ
a
Therefore,
Z
! 1 1 X X tk tk Bk ðxÞ ak : dx ¼ k! k! k¼0 k¼0
aþ1
a
Hence, 1 Z X
aþ1
Bk ðxÞ dx
a
k¼0
1 tk X tk ¼ ak : k! k¼0 k!
Comparing the coefficients on both sides, yields
Z
aþ1
Bn ðxÞ dx ¼ an ;
n P 0:
ð27Þ
a
Setting a ¼ 0 in (27), gives
Z
1
Bn ðxÞ dx ¼ dn0 ;
n P 0:
ð28Þ
0
Using (24), we obtain
Bnþ1 ð1Þ Bnþ1 ð0Þ ¼ ðn þ 1Þ dn0 ;
n P 0:
Consequently, we have
B1 ð1Þ ¼
1 2
and Bn ð1Þ ¼ Bn ð0Þ;
n > 1:
From (27), (13) and Definition 1.2, we have
1 DBnþ1 ðaÞ ¼ an : nþ1 Summing over a from 1 to m, on both sides gives the power sum
fn ðmÞ ¼
m X an ¼ a¼1
! m X 1 1 DBnþ1 ðaÞ ¼ ðBnþ1 ðm þ 1Þ Bnþ1 ð1ÞÞ: n þ 1 a¼1 nþ1
having used Lemma 3.1. j
ð29Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
To prove (16): From Property 11, we have 1 X tk Bk ðx þ yÞ ¼ k! k¼0
1 X tk yk k! k¼0
!
525
! 1 X tk : Bk ðxÞ k! k¼0
By using (2), we get
Bn ðx þ yÞ ¼
n X n Bk ðxÞ ynk ; k k¼0
n P 0:
ð30Þ
Setting x ¼ 0 in (30) gives (14) and setting y ¼ 1, then (30) yields
Bn ðx þ 1Þ ¼
n X n Bk ðxÞ; k k¼0
n P 0:
ð31Þ
If we put x ¼ y in (30), then we have
Bn ð2xÞ ¼
n X n Bk ðxÞ xnk ; k k¼0
n P 0: j
ð32Þ
To prove (17): From Property 12, we have 1 k X t k¼0
k!
! 1
! 1 1 X X tk tkþ1 ¼ Bk ðxÞ ðk þ 1Þ xk : k! ðk þ 1Þ! k¼0 k¼0
Consequently, we get 1 k X t k¼0
!
k!
! 1 1 1 X X tk tk X tk Bk ðxÞ Bk ðxÞ ¼ k xk1 : k! k! k! k¼0 k¼0 k¼0
Using (2), gives n X n Bk ðxÞ Bn ðxÞ ¼ n xn1 ; k k¼0
n P 0:
ð33Þ
Therefore, n1 X n Bk ðxÞ ¼ n xn1 : k k¼0
ð34Þ
In [26], Eq. (34) is the definition of the Bernoulli polynomials. From (31) and (34) we obtain
DBn ðxÞ ¼ Bn ðx þ 1Þ Bn ðxÞ ¼ n xn1 ;
n P 1: j
To prove (18): From Property 13, we have 1 1 X tk X tk Bk ð1 xÞ ¼ ð1Þk Bk ðxÞ : k! k¼0 k! k¼0
Comparing the coefficients on both sides, we obtain
Bn ð1 xÞ ¼ ð1Þn Bn ðxÞ;
n P 0:
ð35Þ
Setting x ¼ 0 in (35), yields
Bn ð1Þ ¼ ð1Þn Bn ð0Þ ¼ ð1Þn Bn ;
To prove (19): From Property 14, we have 1 k X t k¼0
k!
!
n P 0: j
! ! ! 1 1 1 1 X X X X tk tk tk tk þ : 2k Bk ðxÞ 2k Bk ðxÞ ¼ 2 xk Bk ðxÞ k! k! k! k! k¼0 k¼0 k¼0 k¼0
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
Using (2), gives
n n X X n k n 2 Bk ðxÞ þ 2n Bn ðxÞ ¼ 2 Bk ðxÞ xnk ; k k k¼0 k¼0
n P 0:
Hence,
n n X X n k n 2 Bk ðxÞ ¼ 2 Bk ðxÞ xnk 2n Bn ðxÞ; k k k¼0 k¼0
n P 0:
ð36Þ
Using (32) in (36), we obtain n1 X n k 2 Bk ðxÞ ¼ 2 Bn ð2xÞ 2n Bn ðxÞ ; k k¼0
n P 0:
Putting x ¼ 0 in the previous result, yields
Bn ¼
n1 X n k 1 2 Bk ; 2 ð1 2n Þ k¼0 k
n P 1: j
ð37Þ
To prove (20): From Property 15, we have
k 1 1 X X 1 t 1 tk Bk ðx þ Þ þ Bk ðxÞ ¼2 B ð2xÞ : k k 2 k! k! k¼0 k¼0 2
Comparing the coefficients on both sides, gives
1 1 þ Bn ðxÞ ¼ n1 Bn ð2xÞ; Bn x þ 2 2
n P 0:
Consequently, we get
1 þ Bn ðxÞ ; Bn ð2xÞ ¼ 2n1 Bn x þ 2
n P 0:
In particular, if x ¼ 0, then we get
Bn
1 ð1 2n1 Þ ¼ Bn ; 2 2n1
n P 0: j
ð38Þ
To prove (21): From Property 16, we have 1 k X t k¼0
!
k!
! 1 1 1 1 tk tk X X X X tkþ1 tk þ Bk ð2xÞ 2k Bk ðxÞ Bk ð2xÞ 2k Bk ðxÞ ¼ ðk þ 1Þ 2k xk ¼ k 2k1 xk1 : k! k! k¼0 ðk þ 1Þ! k¼0 k! k¼0 k¼0
Using (2), yields n X n Bk ð2xÞ 2k Bk ðxÞ þ ðBn ð2xÞ 2n Bn ðxÞÞ ¼ n 2n1 xn1 ; k k¼0
n P 0:
Consequently, we have n1 X n Bk ð2xÞ 2k Bk ðxÞ ¼ n 2n1 xn1 2 ðBn ð2xÞ 2n Bn ðxÞÞ; k k¼0
To prove (22): From Property 17, we have 1 X
tk Bk ðxÞ k! k¼0
!
1 X tk Bk ðyÞ k! k¼0
!
n P 0: j
1 1 1 X X tk tkþ1 @ X tk ¼ Bk ðx þ yÞ þ ðx þ y 1Þ ðk þ 1Þ Bk ðx þ yÞ t Bk ðx þ yÞ @t k! ðk þ 1Þ! k! k¼0 k¼0 k¼0
¼
1 1 1 X X tk tk X tk Bk ðx þ yÞ þ ðx þ y 1Þ k Bk1 ðx þ yÞ k Bk ðx þ yÞ : k! k! k! k¼0 k¼0 k¼0
!
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
527
By using (2), we get n X n Bk ðxÞBnk ðyÞ ¼ ð1 nÞ Bn ðx þ yÞ þ n ðx þ y 1ÞBn1 ðx þ yÞ: k k¼0
ð39Þ
Setting x ¼ y ¼ 0 in (39), we obtain the quadratic recurrence relation n X n Bk Bnk ¼ ð1 nÞ Bn n Bn1 ; k k¼0
n P 1;
which is also known in its equivalent form
n1 X 2n B2k B2n2k ¼ ð2n þ 1Þ B2n ; 2k k¼1
n P 2: j
ð40Þ
More than 250 years ago, Euler discovered the identity (40). It gives the sum of products of the Bernoulli numbers. To prove (23): From Property 18, we have 1 X tk Bk ðxÞ k! k¼0
!
1 X tk Bk ðyÞ k! k¼0
!
! ! 1 1 1 x þ y t kþ1 x þ y tk X X 1X tk ðk þ 1Þ 2k Bk þ Bk 2k B k 2 k¼0 2 2 ðk þ 1Þ! k! k! k¼0 k¼0 ! ! 1 1 1 X tk X k 1 X k1 x þ y tk x þ y tk : ¼ k 2 Bk1 þ Bk 2 Bk 2 k¼0 2 2 k! k! k! k¼0 k¼0
¼
By using (2), we obtain n n x þ y X X n n k x þ y Bk ðxÞBnk ðyÞ ¼ n 2n2 Bn1 2 Bk þ Bnk : 2 2 k k k¼0 k¼0
Setting x ¼ y in (41), yields n n X X n n k Bk ðxÞBnk ðxÞ ¼ n 2n2 Bn1 þ 2 Bk Bnk : k k k¼0 k¼0
But, Bn ðxÞ ¼ ð1Þn Bn ð1 þ xÞ. Therefore, n n X X n n k ð1Þnk Bk ðxÞBnk ð1 þ xÞ ¼ n 2n2 Bn1 þ 2 Bk Bnk : k k k¼0 k¼0
Putting x ¼ y ¼ 0 in (41), gives n X n ð1 2k ÞBk Bnk ¼ n 2n2 Bn1 : k k¼0
Replacing n by 2n, we get
2n X 2n ð1 2k ÞBk B2nk ¼ 0; k k¼0
n P 2:
Consequently, we have
n1 n1 X X 2n k 2n 4 B2k B2n2k ¼ ð1 4n Þ B2n þ B2k B2n2k ; 2k 2k k¼1 k¼1 By using (40), we end up with the quadratic identity
n1 X 2n 2k1 2 B2k B2n2k ¼ ðn þ 22n1 ÞB2n ; 2k k¼1
n P 2: j
n P 2:
ð41Þ
528
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
4. Applications and further results 4.1. A connection between the Bernoulli polynomials, Bn ðxÞ and the Euler polynomials, En ðxÞ We begin this subsection by giving the following definition for the Euler polynomials. These polynomials are closely related to the Bernoulli polynomials [28]. Definition 4.1.1. ([12]). The sequence < E0 ðxÞ; E1 ðxÞ; E2 ðxÞ; . . . > of the Euler polynomials En ðxÞ; n P 0 is defined by the exponential generating function 1 X 2 ext tk ¼ Ek ðxÞ : t e þ 1 k¼0 k!
Hðx; tÞ ¼
ð42Þ
The basic properties of Euler numbers and polynomials can be found in [10,12,28,39]. The first 6 Euler polynomials are
E0 ðxÞ ¼ 1
E1 ðxÞ ¼ x
E4 ðxÞ ¼ x4 2x3 þ x
1 2
E2 ðxÞ ¼ x2 x
3 1 E3 ðxÞ ¼ x3 x2 þ 2 4
5 5 1 E5 ðxÞ ¼ x5 x4 þ x2 : 2 2 2
The exponential generating functions GðtÞ; Fðx; tÞ and Hðx; tÞ satisfy the relationship
Fðx; tÞ ¼
1 t þ GðtÞ Hðx; tÞ; 2
ð43Þ
as can be easily checked. We are now in a position to give an easy proof for the following theorem, which is the main result in [6], (see also [40]). Theorem 4.1.1. The Bernoulli polynomials Bn ðxÞ; n P 0 are related to the Euler polynomials En ðxÞ; n P 0 by
Bn ðxÞ ¼
n X n k¼0 k–1
k
Bk Enk ðxÞ ¼
n X n k¼0
k
ð1 dk1 Þ Bk Enk ðxÞ:
Proof. From (43), we have
! ! ! 1 1 1 X X tk 1X t kþ1 tk tk ¼ Ek ðxÞ ðk þ 1Þ Ek ðxÞ þ Bk Ek ðxÞ 2 k¼0 k! ðk þ 1Þ! k! k! k¼0 k¼0 k¼0 ! ! 1 1 1 X X 1X tk tk tk : ¼ k Ek1 ðxÞ þ Bk Ek ðxÞ 2 k¼0 k! k! k! k¼0 k¼0
1 X tk Bk ðxÞ ¼ k! k¼0
1 X 1 tk tþ Bk 2 k! k¼0
!
1 X
Using (2), we obtain
Bn ðxÞ ¼
n n n X X X n n n n 1 Bk Enk ðxÞ ¼ B1 En1 ðxÞ þ Bk Enk ðxÞ ¼ Bk Enk ðxÞ n En1 ðxÞ þ 2 k 1 k k k¼0 k¼0 k¼0
n X n ¼ ð1 dk1 Þ Bk Enk ðxÞ: j k k¼0
k–1 ð44Þ
4.2. A Connection between the Bernoulli Polynomials, Bn ðxÞ and the Genocchi polynomials, Gn ðxÞ In recent years there has been a flurry of activity for doing research with Genocchi numbers and polynomials. See for instance, [3,21,23–25,30,32,35,37,38,40]. These numbers and polynomials are given by the following definition. Definition 4.2.1 [25]. The sequences < G0 ; G1 ; G2 ; . . . > of the Genocchi numbers, Gn ; n P 0 and < G0 ðxÞ; G1 ðxÞ; G2 ðxÞ; . . . > of the Genocchi polynomials, Gn ðxÞ; n P 0 are defined by the exponential generating functions
WðtÞ ¼
1 X 2t tn ¼ Gn : et þ 1 n¼0 n!
ð45Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
and 1 X 2 t ext tn ¼ Gn ðxÞ ; et þ 1 n¼0 n!
Q ðx; tÞ ¼
ð46Þ
respectively. The first 11 Genocchi numbers Gn ¼ Gn ð0Þ are
G0 ¼ 0
G1 ¼ 1
G2 ¼ 1
G3 ¼ 0
G4 ¼ 1
G5 ¼ 0
G6 ¼ 3
G7 ¼ 0
G8 ¼ 17
G9 ¼ 0
G10 ¼ 155:
The first few Genocchi polynomials are
G0 ðxÞ ¼ 0
G1 ðxÞ ¼ 1
G4 ðxÞ ¼ 4x3 6x2 þ 1
G2 ðxÞ ¼ 2x 1
G3 ðxÞ ¼ 3x2 3x
G5 ðxÞ ¼ 5x4 10x3 þ 5x
G6 ðxÞ ¼ 6x5 15x4 þ 15x2 3:
Consider
< E0 ðxÞ; E1 ðxÞ; E2 ðxÞ; . . . > $ Hðx; tÞ:
ð47Þ
Using Lemma 3.3, yields
< 0; E0 ðxÞ; 2 E1 ðxÞ; 3 E2 ðxÞ; . . . > $ t Hðx; tÞ ¼ Qðx; tÞ:
ð48Þ
(See (42) and (46)). On the other hand, we have from (46),
< G0 ðxÞ; G1 ðxÞ; G2 ðxÞ; . . . > $ Qðx; tÞ:
ð49Þ
From (48) and (49), we get
Gn ðxÞ ¼ n En1 ðxÞ;
n P 1:
ð50Þ
It is easy to see that the exponential generating functions GðtÞ; Fðx; tÞ and Q ðx; tÞ are related by
t Fðx; tÞ ¼
1 t þ GðtÞ Q ðx; tÞ: 2
ð51Þ
We are now ready to give a connection between the Bernoulli polynomials Bn ðxÞ; n P 0 and the Genocchi polynomials Gn ðxÞ; n P 0. This connection is given by the following result. Theorem 4.2.1. The Bernoulli polynomials Bn ðxÞ; n P 0 can be expressed by the Genocchi polynomials Gn ðxÞ; n P 0 as
Bn ðxÞ ¼
n n X nþ1 nþ1 1 1 X Bk Gnþ1k ðxÞ ¼ ð1 dk1 ÞBk Gnþ1k ðxÞ: nþ1 n þ 1 k¼0 k k k¼0
ð52Þ
k–1 First Proof. Using (51), we have 1 X ðk þ 1Þ Bk ðxÞ k¼0
tkþ1 ¼ ðk þ 1Þ!
1 X 1 tk tþ Bk 2 k! k¼0
!
1 X tk Gk ðxÞ k! k¼0
!
! ! 1 1 1 X X 1X t kþ1 tk tk ¼ : ðk þ 1Þ Gk ðxÞ þ Bk Gk ðxÞ 2 k¼0 ðk þ 1Þ! k! k! k¼0 k¼0
Consequently,
! ! 1 1 1 1 X X X tk 1 X tk tk tk k Bk1 ðxÞ ¼ k Gk1 ðxÞ þ Bk Gk ðxÞ : k! 2 k¼0 k! k! k! k¼0 k¼0 k¼0 Using (2), we obtain
n Bn1 ðxÞ ¼
n n n X X X n n n n 1 Bk Gnk ðxÞ ¼ B1 Gn1 ðxÞ þ Bk Gnk ðxÞ ¼ Bk Gnk ðxÞ n Gn1 ðxÞ þ 2 k 1 k k k¼0 k¼0 k¼0
n X n ð1 dk1 Þ Bk Gnk ðxÞ: ¼ k k¼0
Therefore, we have
k–1
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
Bn ðxÞ ¼
n nþ1 1 X ð1 dk1 Þ Bk Gnþ1k ðxÞ; n þ 1 k¼0 k
since G0 ðxÞ ¼ 0: j Alternative proof. We have [6,40],
n X n
Bn ðxÞ ¼
k
k¼0
Bk Enk ðxÞ:
k–1 By using (50), we get
n n X nþ1 1 k Bk Gnþ1k ðxÞ: Bn ðxÞ ¼ Bk Gnþ1k ðxÞ ¼ nþ1 nþ1k k k¼0 k¼0 k–1 k–1 n X
ð53Þ
having used the identity
nþ1 k
¼
nþ1 nþ1k
n : j k
4.3. Further properties and identities It is worth mentioned that new more interesting identities can be obtained using the technique of this paper by adding new properties. Here we add the following 10 additional properties.
ð1 þ et Þ WðtÞ ¼ 2 t:
ð54Þ
WðtÞ WðtÞ ¼ 2 t:
ð55Þ
GðtÞ Gð2tÞ ¼
Qðx; tÞ ¼ ext WðtÞ:
Z 0
1
1 WðtÞ: 2
t Q ðx; tÞ dx ¼ 2 tanhð Þ: 2
ð56Þ
ð57Þ
ð58Þ
@ Q ðx; tÞ ¼ t Q ðx; tÞ: @x
ð59Þ
Qðx þ 1; tÞ þ Qðx; tÞ ¼ 2 t ext :
ð60Þ
GðtÞ WðtÞ ¼ t Gð2tÞ:
ð61Þ
ð2 t 2 WðtÞÞFðx; tÞ ¼ t Qðx; tÞ:
ð62Þ
@ Qðx; tÞQðy; tÞ ¼ 2 t Q ðx þ y; tÞ Q ðx þ y; tÞ t ðx þ y 1ÞQ ðx þ y; tÞ : @t
ð63Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
531
The property (54), together with (2), gives
Gn ¼ dn1
n1 n 1X Gk ; 2 k¼1 k
n P 0:
ð64Þ
From (55), one can see that
G1 ¼ 1;
G2nþ1 ¼ 0;
n P 1:
ð65Þ
Therefore using (50), we get
E2n ð0Þ ¼ 0;
n P 1:
ð66Þ
(See Theorem 3.2 in [20]). Similarly using (56), we obtain
Gn ¼ 2 ð1 2n Þ Bn ;
n P 0:
ð67Þ
From (11) and (67), we have the following curious identity between the Bernoulli and Genocchi numbers
Gn ¼
n1 X n k 2 Bk ; k k¼0
n P 1:
ð68Þ
Note that this identity can also be obtained from (37). The property (57), together with (2), yields
Gn ðxÞ ¼
n X n Gk xnk ; k k¼0
n P 0:
ð69Þ
Using (58), together with the following series expansion for tanhðtÞ
tanhðxÞ ¼
1 X 4n ð4n 1Þ B2n
2n!
n¼1
x2n1 ;
we see that
Z
1
2 Gnþ1 : n P 1; nþ1
Gn ðxÞ dx ¼
0
ð70Þ
Therefore, we have
Z
1
En ðxÞ dx ¼
0
2 Enþ1 ð0Þ ; nþ1
n P 1:
In particular, by using (66), then we have
Z
1
E2n1 ðxÞ dx ¼ 0;
n P 1:
0
From (59), we obtain
G0n ðxÞ ¼ n Gn1 ðxÞ;
n P 1:
Using (60), gives
Gn ðx þ 1Þ þ Gn ðxÞ ¼ 2 n xn1 ;
n P 1:
ð71Þ
Replacing x by x and n by n þ 1 in (71), we see that
Gnþ1 ðxÞ þ Gnþ1 ð1 xÞ ¼ ð1Þn ðGnþ1 ðxÞ þ Gnþ1 ð1 þ xÞÞ ¼ ð1Þn 2 ðn þ 1Þ xn : Thus the sums Gn ðxÞ þ Gn ð1 þ xÞ and consequently, En ðxÞ þ En ð1 þ xÞ are even functions if n is odd. Putting x ¼ 0 in (71), yields
Gn ð1Þ þ Gn ¼ 2 dn1 :
ð72Þ
From (72), we see that
Gn ð1Þ ¼ Gn ð0Þ ¼ Gn ;
n P 2:
ð73Þ
Making use of (61), together with (2), we arrive at
Bn ¼
n1 X n 1 Bk Enk ð0Þ: ð2n 1Þ k¼0 k
ð74Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
Using (67) in (74) together with the fact that B1 ¼ 12, we get n1 X n
k
k¼0 k–1
Bk Enk ð0Þ ¼
1 dn1 ; 2
ð75Þ
on simplification. The following identity is a direct consequence of (75) n2 X n
k
k¼0
Bk Gnk ¼ dn2 :
ð76Þ
k–1 It is worth mentioned that the identity (76) may be also obtained by using (52), (28) and (70). Eq. (76) is also equivalent to
n
n X
k
k¼2
Gk Bnk ¼ dn2 : ð77Þ
k–n 1 Replacing n by 2n þ 2 and k by k þ 1 in (77), gives
2nþ1 X
2n þ 2
Gkþ1 B2nþ1k ¼ dn0 :
kþ1
k¼1 k–2n But
2n þ 2
kþ1
¼
2n þ 2 kþ1
2n þ 1
k
:
Hence,
2nþ1 X
2n þ 1 k
k¼1
ð2n þ 2Þ Ek ð0Þ B2nþ1k ¼ dn0 ;
k–2n having used (50). Consequently, we have 2nþ1 X
2n þ 1
k
k¼1
Ek ð0Þ B2nþ1k ¼
1 dn0 ; 2ðn þ 1Þ
ð78Þ
having used (66). The identity (78) is Theorem 3.3 in the very recent paper [20]. Using (72) and (7), we get
n1 X 2n k¼0
2k
B2k G2n2k ¼
n X 2n k¼1
2k
G2k B2n2k ¼ dn1 :
ð79Þ
By using (78), together with (66) we see that
n X 2n þ 1 k¼0
2k þ 1
E2kþ1 ð0Þ B2n2k ¼
1 dn0 : 2ðn þ 1Þ
ð80Þ
Using (50) in (80), gives
n X 2n k¼1
2k
E2k ð0Þ B2n2k ¼ dn1;0 :
ð81Þ
The property (62), together with (2), gives n1 X n 1 Ek ð0Þ Bnk ðxÞ ¼ Gn ðxÞ; 2 k k¼1
n P 1:
ð82Þ
M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
533
The property (63), together with (2), yields n1 X n Gk ðxÞGnk ðyÞ ¼ 2 ððn 1Þ Gn ðx þ yÞ n ðx þ y 1Þ Gn1 ðx þ yÞÞ; k k¼1
n P 2:
ð83Þ
It follows from (83) that if y ¼ 1 x, then we have n1 X n Gk ðxÞGnk ð1 xÞ ¼ 2 ðn 1Þ Gn ; n P 2; k k¼1
for any rael x;
ð84Þ
having used (73). Setting x ¼ y ¼ 0 in (83), we get the following quadratic identity n1 X n Gk Gnk ¼ 2 n Gn1 þ 2 ðn 1Þ Gn ; k k¼1
n P 2:
At this stage, let us consider the sequence < E0 ; E1 ; E2 ; . . . > of the Euler numbers En ; n P 0, defined by
EðtÞ ¼
1 X 2 1 tn ¼ ¼ En : et þ et coshðtÞ n¼0 n!
ð85Þ
Therefore,
< E0 ; E1 ; E2 ; . . . >$ EðtÞ ¼
1 X
En
n¼0
tn : n!
The odd-indexed Euler numbers are all zero. The even-indexed ones have alternating signs, and are given by the recursive recurrence relation
n1 X 2n E2k ; E2n ¼ 2k k¼0
E0 ¼ 1;
n P 1:
ð86Þ
The first few values for the Euler numbers are
E0 ¼ 1
E2 ¼ 1
E4 ¼ 5
E6 ¼ 61
E8 ¼ 1385
E10 ¼ 50521
The relationship between the two generating functions WðtÞ and EðtÞ is given by
et Wð2 tÞ ¼ 2 t EðtÞ:
ð87Þ
Using (87), together with (2), yields
En1 ¼
! n n k1 1 X 2 Gk þ 2n1 Gn ; n k¼1 k
n P 1:
ð88Þ
Consequently, we have
En ¼
n X n k¼0
k
1 2k Ek ð0Þ ¼ 2n En ð Þ; 2
ð89Þ
having used (50), together with the fact that the Euler polynomials En ðxÞ satisfy [8]
En ðxÞ ¼
n X n Ek ð0Þ xnk : k k¼0
Let us finish this subsection by giving some identities concerning the Euler numbers. Using (89) in (84) with x ¼ 12, gives the identity n1 X n k¼1
Since
k
k ðn kÞ Ek1 Enk1 ¼ 2n1 n ðn 1ÞEn1 ð0Þ;
n P 1:
n n2 k ðn kÞ ¼ n ðn 1Þ, we get, k k1
n1 X n2 k¼1
k1
Ek1 Enk1 ¼ 2n1 En1 ð0Þ;
n P 3;
ð90Þ
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M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 220 (2013) 518–535
on simplification. Replacing n by n þ 2 on both sides and k by k þ 1 in (90), gives n X n k¼0
k
Ek Enk ¼ 2nþ1 Enþ1 ð0Þ;
n P 1:
Consequently, showing that the Euler numbers, En satisfy the quadratic recurrence relation
n X 2n k¼0
2k
E2k E2n2k ¼ 22nþ1 E2nþ1 ð0Þ;
n P 1:
ð91Þ
5. Conclusion The current paper is addressed a generating function approach to obtain old and new identities. The approach is employed in a constructive way to obtain identities involving the Bernoulli, Euler and Genocchi polynomials and numbers. We established some properties for generating functions and two applications are given. Among the applications, we should stress that a simple proof for the main result in [6], (see also [40]) is given. Finally, some useful relationships between the Bernoulli, Euler and Genocchi polynomials and numbers are obtained. All properties and results of this paper are tested using the Computer Algebra System (CAS), Maple [33]. Acknowledgments The authors wish to thank anonymous referees, Prof. Dr. Melvin Scott, Editor-in-Chief and Prof. Dr. T.E.Simos, Senior Editor for useful comments that enhanced the quality of this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41]
M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions, Dover Publications, NewYork, 1972. T. Agoh, K. Dilcher, Convolution identities and lacunary recurrences for Bernoulli numbers, J. Number Theory 124 (2007) 105–122. A. Bayad, T. Kim, Identities for the Bernoulli, the Euler and the Genocchi numbers and polynomials, Adv. Stud. Contemp. Math. 20 (2) (2010) 247–253. J. Bernoulli, Ars Conjectandi, Thurnisiorum, Basel, 1713. Bernoulli polynomials, entry in:
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[42] B. Sury, Bernoulli numbers and the Riemann Zeta function, Resonance, July 2003, available at: www.ias.ac.in/resonance/July2003/July2003p5462.htm+. [43] H.S. Wilf, Generatingfunctionology, A K Peters 1994, second/ third ed., Second edition available from