Generalizations of Euler Numbers and Polynomials

IJMMS 2003:61, 3893–3901 PII. S016117120321108X http://ijmms.hindawi.com © Hindawi Publishing Corp.

GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS

QIU-MING LUO, FENG QI, and LOKENATH DEBNATH

Received 5 November 2002

The concepts of Euler numbers and Euler polynomials are generalized and some basic properties are investigated.

2000 Mathematics Subject Classiﬁcation: 11B68, 33E20.

1. Introduction. It is well known that the Euler numbers and polynomials can be deﬁned by the following deﬁnitions.

Deﬁnition 1.1 (see [1]). The Euler numbers Ek are deﬁned by the following expansion:

t ∞ 2e Ek secht = = tk, |t|≤π. (1.1) e2t +1 k! k=0

In [6, page 5], the Euler numbers are deﬁned by

∞ t/2 n 2n 2e t (−1) En t = sech = , |t|≤π. (1.2) et +1 2 (2n)! 2 n=0

Deﬁnition 1.2 (see [1, 6]). The Euler polynomials Ek(x) for x ∈ R are de- ﬁned by

xt ∞ 2e Ek(x) = tk, |t| <π. (1.3) et +1 k! k=0

Let N denote the set of all positive integers. It can also be shown that the polynomials Ei(t), i ∈ N, are uniquely determined by the following two properties: Ei(t) = iEi−1(t), E0(t) = 1, i (1.4) Ei(t +1)+Ei(t) = 2t .

Euler polynomials are related to the Bernoulli numbers. For information about Bernoulli numbers and polynomials, we refer to [1, 2, 3, 5, 6]. In this note, we give some generalizations of the concepts of Euler numbers and Euler polynomials and research their basic properties. In fact, motivations 3894 QIU-MING LUO ET AL. and ideas of this note and other articles, see, for example, [2, 3, 4], originate essentially from [5].

2. Generalizations of Euler numbers and polynomials. In this section, we give two deﬁnitions, the generalized Euler number and the generalized Euler polynomial, which generalize the concepts of Euler number and Euler polynomial.

Deﬁnition 2.1. For positive numbers a, b,andc, the generalized Euler numbers Ek(a,b,c) are deﬁned by

t ∞ 2c Ek(a,b,c) = tk. (2.1) b2t +a2t k! k=0

Deﬁnition 2.2. For any given positive numbers a, b, and c and x ∈ R,the generalized Euler polynomials Ek(x;a,b,c) are deﬁned by

xt ∞ 2c Ek(x;a,b,c) = tk. (2.2) bt +at k! k=0

Taking a = 1andb = c = e, then Definitions 1.1 and 1.2 can be deduced from Definitions 2.1 and 2.2, respectively. Thus, Definitions 2.1 and 2.2 generalize the concepts of Euler numbers and polynomials.

3. Some properties of the generalized Euler numbers. In this section, we study some basic properties of the generalized Euler numbers deﬁned in Deﬁnition 2.1.

Theorem 3.1. For positive numbers a, b, and c and real number x ∈ R, 1/2 x E0(a,b,c)= 1,Ek(1,e,e)= Ek,Ek 1,e ,e = Ek(x), (3.1) k k lnc −2lna Ek(a,b,c)= 2 (lnb −lna) Ek , (3.2) ( b − a) 2 ln ln k k j k−j Ek(a,b,c)= (lnb −lna) (lnc −lna−lnb) Ej . (3.3) j j=0

Proof. The formulas in (3.1) follow from Deﬁnitions 1.1, 1.2,and2.1 easily. By Deﬁnitions 1.2 and 2.1 and direct computation, we have ct ( c − a)/ ( b − a)· t( b − a) 2 = 2exp ln 2ln 2 ln ln 2 ln ln b2t +a2t exp 2t(lnb −lna) +1 ∞ k (3.4) k k lnc −2lna t = 2 (lnb −lna) Ek . 2(lnb −lna) k! k=0

Then, formula (3.2) follows. GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3895 E (x) = k −j k (x − / )k−j E Substituting k j=0 2 j 1 2 j into the formula (3.2)yields formula (3.3). The proof of the classical result for Ek(x) follows from the more general proof that will be given for (4.1).

Theorem 3.2. For k ∈ N,

k−1 1 k k−j k−j Ek(a,b,c)=− (2lnb −lnc) +(2lna−lnc) Ej (a,b,c), (3.5) 2 j j=0 Ek(a,b,c)= Ek(b,a,c), (3.6) α α α k Ek a ,b ,c = α Ek(a,b,c). (3.7)

Proof. By Deﬁnition 2.1, direct calculation yields

t t ∞ 1 b2 a2 tk 1 = + Ek(a,b,c) c c k 2 k= ! 0 ∞ k k ∞ 1 tk b2 a2 tk = ln + ln Ek(a,b,c) 2 k! c c k! (3.8) k=0 k=0   ∞ k 2 k−j 2 k−j k 1  k b a  t = ln + ln Ej (a,b,c) . 2 j c c k! k=0 j=0

Equating coeﬃcients of tk in (3.8)givesus

k k b2 k−j a2 k−j ln + ln Ej (a,b,c)= 0. (3.9) j c c j=0

Formula (3.5) follows. The other formulas follow from Deﬁnition 2.1 and formula (3.2).

Remark 3.3. For positive numbers a, b,andc, we have

E0(a,b,c)= 1,

E1(a,b,c)= lnc −lna−lnb, (3.10) E2(a,b,c)= (lnc −2lna)(lnc −2lnb), 2 2 E3(a,b,c)= (lnc −lna−lnb) −3(lnb −lna) (lnc −lna−lnb).

Since it is well known and easily established that the Ek are integers, Ej = 0 if j is odd, and Ej (0) = 0ifj is positive and even, it follows from (3.3)and (3.2)thatEk(a,b,c) is an integer polynomial in lna,lnb, and lnc which is homogeneous of degree k and which is divisible by lnc −lna−lnb if k is odd, and divisible by (lnc −2lna)(lnc −2lnb) if k is even and positive. 3896 QIU-MING LUO ET AL.

4. Some properties of the generalized Euler polynomials. In this section, we investigate properties of the generalized Euler polynomials deﬁned by Deﬁnition 2.2.

Theorem 4.1. For any given positive numbers a,b, and c and x ∈ R,

k k−j k (lnc)k−j 1 Ek(x;a,b,c) = x − Ej (a,b,c), (4.1) j 2j 2 j=0 k j k−j k k−j b 1 lnc −2lna Ek(x;a,b,c) = (lnc) ln x − Ej , (4.2) j a 2 2(lnb −lna) j=0 k j j− k−j k j (lnc)k−j b c 1 Ek(x;a,b,c) = ln ln x − E, j 2j a ab 2 j=0 =0 (4.3) k 1 Ek(a,b,c)= 2 Ek ;a,b,c , (4.4) 2

Ek(x) = Ek(x;1,e,e). (4.5)

Proof. By Deﬁnitions 2.1 and 2.2, we have

∞ 2xt k 2c k t = 2 Ek(x;a,b,c) , b2t +a2t k! k=0 2c2xt 2ct = ·c(2x−1)t b2t +a2t b2t +a2t    ∞ tk ∞ tk (4.6)   k k = Ek(a,b,c) (2x −1) (lnc) k! k! k=0 k=0   ∞ k k tk  k−j k−j  = (lnc) (2x −1) Ej (a,b,c) . j k! k=0 j=0

Equating the coeﬃcients of tk/k!in(4.6)yields

k k k k−j k−j 2 Ek(x;a,b,c) = (lnc) (2x −1) Ej (a,b,c). (4.7) j j=0

Formula (4.1) follows. The other formulas follow directly from substituting formulas (3.2) and (3.3) into (4.1) and taking x = 1/2in(4.1), respectively. GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3897

Theorem 4.2. For positive integer 1 ≤ p ≤ k,

∂p k E (x a,b,c) = ! ( c)pE (x a,b,c), p k ; ln k−p ; (4.8) ∂x (k−p)! x 1 Ek(t;a,b,c)dt = Ek+1(x;a,b,c)−Ek+1(β;a,b,c) . (4.9) β (k+1)lnc

Proof. Diﬀerentiating equation (2.2) with respect to x yields

∂ Ek(x;a,b,c) = k(lnc)Ek− (x;a,b,c). (4.10) ∂x 1

Using formula (4.10) and by mathematical induction, formula (4.8) follows. Rearranging formula (4.10) produces

1 ∂ Ek(x;a,b,c) = Ek+ (x;a,b,c). (4.11) (k+1)lnc ∂x 1

Formula (4.9) follows from integration on both sides of formula (4.11).

Theorem 4.3. For positive numbers a, b, and c and x ∈ R, k k k−j Ek(x +1;a,b,c) = (lnc) Ej (x;a,b,c), (4.12) j j=0

k k Ek(x +1;a,b,c) = 2x (lnc) k k k−j k−j k−j + (lnc) −(lnb) −(lna) Ej (x;a,b,c), j j=0 (4.13) a b Ek(x +1;a,b,c) = Ek x; , ,c . (4.14) c c

Proof. From Deﬁnition 2.2 and straightforward calculation, we have    xt ∞ k ∞ k 2c t  t  t k ·c = Ek(x;a,b,c) (lnc) bt +at k! k! k=0 k=0   ∞ k k  k k−j  t = (lnc) Ej (x;a,b,c) , (4.15) j k! k=0 j=0 ∞ xt (x+1)t k 2c t 2c t ·c = = Ek(x +1;a,b,c). bt +at bt +at k! k=0

Therefore, from equating the coeﬃcients of tk/k!in(4.15), formula (4.12)follows. 3898 QIU-MING LUO ET AL.

Similarly, we obtain

∞ (x+1)t k xt 2c t xt 2c t t t = Ek(x +1;a,b,c) = 2c + c −b −a bt +at k! bt +at k=0 ∞ tk = 2 xk(lnc)k k k= !  0   ∞ k ∞ k  t  k k k t  + Ek(x;a,b,c) (lnc) −(lnb) −(lna) k k k= ! k= ! 0 0 ∞ = 2xk(lnc)k k=0 k k k k−j k−j k−j t + (lnc) −(lnb) −(lna) Ej (x;a,b,c) . j k! j=0 (4.16)

By equating coeﬃcients of tk/k!, we obtain formula (4.13). Since

∞ tk 2c(x+1)t 2cxt Ek(x +1;a,b,c) = = k! bt +at b/c t + a/c t k=0 (4.17) ∞ tk a b = Ek x; , ,c , k! c c k=0 by equating coeﬃcients, we obtain formula (4.14). The proof is complete.

Corollary 4.4. The following formulas are valid for positive numbers a, b, and c and real number x:

k Ek(x +1)+Ek(x) = 2x , (4.18) k k Ek(x +1) = Ej (x), (4.19) j j=0 k k Ek(x +1;1,b,b)+Ek(x;1,b,b)= 2x (lnb) , (4.20) k k k−j Ek(x +1;1,b,b)= Ej (x;1,b,b)(lnb) , (4.21) j j=0 k−1 k k−j k k Ej (x;1,b,b)(lnb) +2Ek(x;1,b,b)= 2x (lnb) , (4.22) j j=0 x+1 k 1 k+1 k−j Ek(t;a,b,c)dt = (lnc) Ej (x;a,b,c). (4.23) x (k+1)lnc j j=0 GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3899

Theorem 4.5. For positive numbers a,b,c > 0, x ∈ R, and nonnegative integer k, k c c Ek(1−x;a,b,c) = (−1) Ek x; , ,c , (4.24) a b a b Ek(1−x;a,b,c) = Ek −x; , ,c . (4.25) c c

Proof. From Deﬁnition 2.2 and easy computation, we have

∞ tk 2c(1−x)t 2ct ·c−xt 2c−xt Ek(1−x;a,b,c) = = = k! bt +at bt +at c/b −t + c/a −t k=0 (4.26) ∞ k t k c c = (−1) Ek x; , ,c . k! a b k=0

Equating coeﬃcients of tk above leads to formula (4.24). By the same procedure, we can establish formula (4.25).

Theorem 4.6. For positive numbers a,b,c > 0, nonnegative natural number k, and x,y ∈ R, k k k−j k−j Ek(x +y;a,b,c) = (lnc) y Ej (x;a,b,c), j j=0 (4.27) k k k−j k−j Ek(x +y;a,b,c) = (lnc) x Ej (y;a,b,c). j j=0

Proof. These two formulas can be deduced from the following calculation and considering symmetry of x and y:

∞ tk 2c(x+y)t 2cxt ·cyt Ek(x +y;a,b,c) = = k bt +at bt +at k= ! 0    ∞ k ∞ k  t  t k k = Ek(x;a,b,c) (lnc) y (4.28) k! k! k=0 k=0   ∞ k k  k k−j k−j  t = (lnc) y Ej (x;a,b,c) . j k! k=0 j=0

The proof is complete.

Theorem 4.7. For natural numbers k and m and positive number b,

m k 1 m (−1) = (−1) Ek(m+1;1,b,b)−Ek(1;1,b,b) . (4.29) 2(lnb)k =1 3900 QIU-MING LUO ET AL.

Proof. Rearranging formula (4.20)givesus

k 1 x = Ek(x +1;1,b,b)+Ek(x;1,b,b) . (4.30) 2(lnb)k

Replacing x by ∈ N and summing up from 1 to m yields

m m k 1 (−1) = (−1) Ek(+1;1,b,b)+Ek(;1,b,b) 2(lnb)k =1 =1 (4.31) 1 m = (−1) Ek(m+1;1,b,b)−Ek(1;1,b,b) . 2(lnb)k

The proof is complete.

Remark 4.8. Finally, we give several concrete formulas as follows:

E (x a,b,c) = , 0 ; 1 1 1 E1(x;a,b,c) = x − lnc + (lnc −lna−lnb), 2 2 1 2 1 (4.32) E (x;a,b,c) = x − (lnc)2 + x − (lnc −lnb −lna)lnc 2 2 2 1 + (lnc −2lna)(lnc −2lnb). 4

Acknowledgments. The authors would like to express many thanks to the anonymous referees for their valuable comments. The ﬁrst two authors were supported in part by NNSF of China, Grant 10001016, SF for the Promi- nent Youth of Henan Province, Grant 0112000200, SF of Henan Innovation Talents at Universities, NSF of Henan Province, Grant 004051800, Doctor Fund of Jiaozuo Institute of Technology, China. The third author was partially supported by a grant of the Faculty Research Council of the University of Texas-Pan American.

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Qiu-Ming Luo: Department of Broadcast-Television Teaching, Jiaozuo University, Jiaozuo City, Henan 454002, China E-mail address: [email protected]

Feng Qi: Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan 454000, China E-mail address: [email protected] URL: http://rgmia.vu.edu.au/qi.html

Lokenath Debnath: Department of Mathematics, University of Texas-Pan American, Edinburg, TX 78539, USA E-mail address: [email protected] Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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