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 prop- erties: 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 polyno- mial. 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)fol- lows. 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) .