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IJMMS 2003:61, 3893–3901PII. S016117120321108X
http://ijmms.hindawi.com© Hindawi Publishing Corp.
GENERALIZATIONS OF EULER NUMBERSAND POLYNOMIALS
QIU-MING LUO, FENG QI, and LOKENATH DEBNATH
Received 5 November 2002
The concepts of Euler numbers and Euler polynomials are
generalized and somebasic 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 followingexpansion:
secht = 2et
e2t+1 =∞∑k=0
Ekk!tk, |t| ≤π. (1.1)
In [6, page 5], the Euler numbers are defined by
2et/2
et+1 = secht2=
∞∑n=0
(−1)nEn(2n)!
(t2
)2n, |t| ≤π. (1.2)
Definition 1.2 (see [1, 6]). The Euler polynomials Ek(x) for x
∈R are de-fined by
2ext
et+1 =∞∑k=0
Ek(x)k!
tk, |t|
-
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, and c, the
generalized Eulernumbers Ek(a,b,c) are defined by
2ct
b2t+a2t =∞∑k=0
Ek(a,b,c)k!
tk. (2.1)
Definition 2.2. For any given positive numbers a, b, and c and x
∈R, thegeneralized Euler polynomials Ek(x;a,b,c) are defined by
2cxt
bt+at =∞∑k=0
Ek(x;a,b,c)k!
tk. (2.2)
Taking a= 1 and b = c = e, then Definitions 1.1 and 1.2 can be
deduced fromDefinitions 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,
E0(a,b,c)= 1, Ek(1,e,e)= Ek, Ek(1,e1/2,ex
)= Ek(x), (3.1)Ek(a,b,c)= 2k(lnb− lna)kEk
(lnc−2lna
2(lnb− lna)), (3.2)
Ek(a,b,c)=k∑j=0
(kj
)(lnb− lna)j(lnc− lna− lnb)k−jEj. (3.3)
Proof. The formulas in (3.1) follow from Definitions 1.1, 1.2,
and 2.1 easily.
By Definitions 1.2 and 2.1 and direct computation, we have
2ct
b2t+a2t =2exp
((lnc−2lna)/2(lnb− lna)·2t(lnb− lna))
exp(2t(lnb− lna))+1
=∞∑k=0
2k(lnb− lna)kEk(
lnc−2lna2(lnb− lna)
)tk
k!.
(3.4)
Then, formula (3.2) follows.
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GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3895
Substituting Ek(x)=∑kj=0 2−j
(kj
)(x−1/2)k−jEj into the formula (3.2) yields
formula (3.3). The proof of the classical result for Ek(x)
follows from the moregeneral proof that will be given for
(4.1).
Theorem 3.2. For k∈N,
Ek(a,b,c)=−12
k−1∑j=0
(kj
)[(2lnb− lnc)k−j+(2lna− lnc)k−j]Ej(a,b,c), (3.5)Ek(a,b,c)=
Ek(b,a,c), (3.6)
Ek(aα,bα,cα
)=αkEk(a,b,c). (3.7)Proof. By Definition 2.1, direct calculation
yields
1= 12
[(b2
c
)t+(a2
c
)t] ∞∑k=0
tk
k!Ek(a,b,c)
= 12
∞∑k=0
tk
k!
[(lnb2
c
)k+(
lna2
c
)k] ∞∑k=0
tk
k!Ek(a,b,c)
= 12
∞∑k=0
k∑j=0
(kj
)[(lnb2
c
)k−j+(
lna2
c
)k−j]Ej(a,b,c)
tkk!.
(3.8)
Equating coefficients of tk in (3.8) gives us
k∑j=0
(kj
)[(lnb2
c
)k−j+(
lna2
c
)k−j]Ej(a,b,c)= 0. (3.9)
Formula (3.5) follows.
The other formulas follow from Definition 2.1 and formula
(3.2).
Remark 3.3. For positive numbers a, b, and c, we have
E0(a,b,c)= 1,E1(a,b,c)= lnc− lna− lnb,E2(a,b,c)=
(lnc−2lna)(lnc−2lnb),E3(a,b,c)=
[(lnc− lna− lnb)2−3(lnb− lna)2](lnc− lna− lnb).
(3.10)
Since it is well known and easily established that the Ek are
integers, Ej = 0if j is odd, and Ej(0) = 0 if j is positive and
even, it follows from (3.3) and(3.2) that Ek(a,b,c) is an integer
polynomial in lna, lnb, and lnc which ishomogeneous 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.
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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,
Ek(x;a,b,c)=k∑j=0
(kj
)(lnc)k−j
2j
(x− 1
2
)k−jEj(a,b,c), (4.1)
Ek(x;a,b,c)=k∑j=0
(kj
)(lnc)k−j
(lnba
)j(x− 1
2
)k−jEj(
lnc−2lna2(lnb− lna)
), (4.2)
Ek(x;a,b,c)=k∑j=0
j∑�=0
(kj
)(j�
)(lnc)k−j
2j
[lnba
]�[ln
cab
]j−�[x− 1
2
]k−jE�,
(4.3)
Ek(a,b,c)= 2kEk(
12
;a,b,c), (4.4)
Ek(x)= Ek(x;1,e,e). (4.5)
Proof. By Definitions 2.1 and 2.2, we have
2c2xt
b2t+a2t =∞∑k=0
2kEk(x;a,b,c)tk
k!,
2c2xt
b2t+a2t =2ct
b2t+a2t ·c(2x−1)t
= ∞∑k=0
tk
k!Ek(a,b,c)
∞∑k=0
tk
k!(2x−1)k(lnc)k
=∞∑k=0
k∑j=0
(kj
)(lnc)k−j(2x−1)k−jEj(a,b,c)
tkk!.
(4.6)
Equating the coefficients of tk/k! in (4.6) yields
2kEk(x;a,b,c)=k∑j=0
(kj
)(lnc)k−j(2x−1)k−jEj(a,b,c). (4.7)
Formula (4.1) follows.
The other formulas follow directly from substituting formulas
(3.2) and (3.3)
into (4.1) and taking x = 1/2 in (4.1), respectively.
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GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3897
Theorem 4.2. For positive integer 1≤ p ≤ k,
∂p
∂xpEk(x;a,b,c)= k!(k−p)! (lnc)
pEk−p(x;a,b,c), (4.8)∫ xβEk(t;a,b,c)dt = 1(k+1) lnc
[Ek+1(x;a,b,c)−Ek+1(β;a,b,c)
]. (4.9)
Proof. Differentiating equation (2.2) with respect to x
yields
∂∂xEk(x;a,b,c)= k(lnc)Ek−1(x;a,b,c). (4.10)
Using formula (4.10) and by mathematical induction, formula
(4.8) follows.
Rearranging formula (4.10) produces
Ek(x;a,b,c)= 1(k+1) lnc∂∂xEk+1(x;a,b,c). (4.11)
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,
Ek(x+1;a,b,c)=k∑j=0
(kj
)(lnc)k−jEj(x;a,b,c), (4.12)
Ek(x+1;a,b,c)= 2xk(lnc)k
+k∑j=0
(kj
)[(lnc)k−j−(lnb)k−j−(lna)k−j]Ej(x;a,b,c),
(4.13)
Ek(x+1;a,b,c)= Ek(x;ac,bc,c). (4.14)
Proof. From Definition 2.2 and straightforward calculation, we
have
2cxt
bt+at ·ct =
∞∑k=0
tk
k!Ek(x;a,b,c)
∞∑k=0
tk
k!(lnc)k
=∞∑k=0
k∑j=0
(kj
)(lnc)k−jEj(x;a,b,c)
tkk!,
2cxt
bt+at ·ct = 2c
(x+1)t
bt+at =∞∑k=0
tk
k!Ek(x+1;a,b,c).
(4.15)
Therefore, from equating the coefficients of tk/k! in (4.15),
formula (4.12) fol-lows.
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3898 QIU-MING LUO ET AL.
Similarly, we obtain
2c(x+1)t
bt+at =∞∑k=0
tk
k!Ek(x+1;a,b,c)= 2cxt+ 2c
xt
bt+at(ct−bt−at)
= 2∞∑k=0
tk
k!xk(lnc)k
+ ∞∑k=0
tk
k!Ek(x;a,b,c)
∞∑k=0
((lnc)k−(lnb)k−(lna)k) tk
k!
=∞∑k=0
[2xk(lnc)k
+k∑j=0
(kj
)[(lnc)k−j−(lnb)k−j−(lna)k−j]Ej(x;a,b,c)
]tk
k!.
(4.16)
By equating coefficients of tk/k!, we obtain formula
(4.13).Since
∞∑k=0
tk
k!Ek(x+1;a,b,c)= 2c
(x+1)t
bt+at =2cxt(
b/c)t+(a/c)t
=∞∑k=0
tk
k!Ek(x;ac,bc,c),
(4.17)
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:
Ek(x+1)+Ek(x)= 2xk, (4.18)
Ek(x+1)=k∑j=0
(kj
)Ej(x), (4.19)
Ek(x+1;1,b,b)+Ek(x;1,b,b)= 2xk(lnb)k, (4.20)
Ek(x+1;1,b,b)=k∑j=0
(kj
)Ej(x;1,b,b)(lnb)k−j, (4.21)
k−1∑j=0
(kj
)Ej(x;1,b,b)(lnb)k−j+2Ek(x;1,b,b)= 2xk(lnb)k, (4.22)
∫ x+1x
Ek(t;a,b,c)dt = 1(k+1) lnck∑j=0
(k+1j
)(lnc)k−jEj(x;a,b,c). (4.23)
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GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3899
Theorem 4.5. For positive numbers a,b,c > 0, x ∈ R, and
nonnegative in-teger k,
Ek(1−x;a,b,c)= (−1)kEk(x;ca,cb,c), (4.24)
Ek(1−x;a,b,c)= Ek(−x; a
c,bc,c). (4.25)
Proof. From Definition 2.2 and easy computation, we have
∞∑k=0
tk
k!Ek(1−x;a,b,c)= 2c
(1−x)t
bt+at =2ct ·c−xtbt+at =
2c−xt(c/b
)−t+(c/a)−t=
∞∑k=0
tk
k!(−1)kEk
(x;ca,cb,c).
(4.26)
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 numberk, and x,y ∈R,
Ek(x+y ;a,b,c)=k∑j=0
(kj
)(lnc)k−jyk−jEj(x;a,b,c),
Ek(x+y ;a,b,c)=k∑j=0
(kj
)(lnc)k−jxk−jEj(y ;a,b,c).
(4.27)
Proof. These two formulas can be deduced from the following
calculation
and considering symmetry of x and y :
∞∑k=0
tk
k!Ek(x+y ;a,b,c)= 2c
(x+y)t
bt+at =2cxt ·cytbt+at
= ∞∑k=0
tk
k!Ek(x;a,b,c)
∞∑k=0
tk
k!(lnc)kyk
=∞∑k=0
k∑j=0
(kj
)(lnc)k−jyk−jEj(x;a,b,c)
tkk!.
(4.28)
The proof is complete.
Theorem 4.7. For natural numbers k and m and positive number
b,
m∑�=1(−1)��k = 1
2(lnb)k[(−1)mEk(m+1;1,b,b)−Ek(1;1,b,b)
]. (4.29)
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3900 QIU-MING LUO ET AL.
Proof. Rearranging formula (4.20) gives us
xk = 12(lnb)k
[Ek(x+1;1,b,b)+Ek(x;1,b,b)
]. (4.30)
Replacing x by � ∈N and summing up � from 1 to m
yieldsm∑�=1(−1)��k = 1
2(lnb)k
m∑�=1(−1)�[Ek(�+1;1,b,b)+Ek(�;1,b,b)]
= 12(lnb)k
[(−1)mEk(m+1;1,b,b)−Ek(1;1,b,b)
].
(4.31)
The proof is complete.
Remark 4.8. Finally, we give several concrete formulas as
follows:
E0(x;a,b,c)= 1,E1(x;a,b,c)=
(x− 1
2
)lnc+ 1
2(lnc− lna− lnb),
E2(x;a,b,c)=(x− 1
2
)2(lnc)2+
(x− 1
2
)(lnc− lnb− lna) lnc
+ 14(lnc−2lna)(lnc−2lnb).
(4.32)
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 sup-
ported by a grant of the Faculty Research Council of the
University of Texas-Pan
American.
References
[1] M. Abramowitz and I. A. Stegun (eds.), Handbook of
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[2] B.-N. Guo and F. Qi, Generalization of Bernoulli
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http://rgmia.vu.edu.au/v4n4.htmlhttp://rgmia.vu.edu.au/v4n4.htmlhttp://rgmia.vu.edu.au/v4n4.html
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GENERALIZATIONS OF EULER NUMBERS AND POLYNOMIALS 3901
[6] Zh.-X. Wang and D.-R. Guo, Introduction to Special Function,
The Series of Ad-vanced Physics of Peking University, Peking
University Press, Beijing, 2000(Chinese).
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 ofTechnology, Jiaozuo City, Henan 454000,
China
E-mail address: [email protected]:
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]
mailto:[email protected]:[email protected]://rgmia.vu.edu.au/qi.htmlmailto:[email protected]
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