RESEARCH Open Access
Some results for the q-Bernoulli, q-Euler numbers
and polynomials
Daeyeoul Kim
1
and Min-Soo Kim
2*
* Correspondence:

2
Department of Mathematics,
KAIST, 373-1 Guseong-dong,
Yuseong-gu, Daejeon 305-701,
South Korea
Full list of author information is
available at the end of the article
Abstract
The q-analogues of many well known formulas are derived by using sev eral results of
q-Bernoulli, q-Euler numbers and polynomials. The q-analogues of ζ-type functions
are given by using generating functions of q-Bernoulli, q-Euler numbers and
polynomials. Finally, their values at non-positive integers are also been computed.
2010 Mathematics Subject Classification: 11B68; 11S40; 11S80.
Keywords: Bosonic p-adic integrals, Fermionic p-adic integrals, q-Bernoulli polyno-
mials, q-Euler polynomials, generating functions, q-analogues of ζ-type functions,
q-analogues of the Dirichlet’s L-functions
1. Introduction
Carlitz [1,2] introduced q-analogues of the Bernoulli numbers and polynomials. F rom
that time on these and other related subjects have been studied by various authors
(see, e.g., [3-10]). Many recent studies on q-anal ogue of the Ber noulli, Euler nu mbers,
and polynomials can be found in Choi et al. [11], Kamano [3], Kim [5,6,12], Luo [7],
Satoh [9], Simsek [13,14] and Tsumura [10].


n
k

B
k
x
n−k
, n ≥ 0,
(1:1)
where B
k
are the Bernoulli numbers given by the coefficients in the power series
t
e
t
− 1
=
∞

k=0
B
k
t
k
k!
.
(1:2)
From the above definition, we see B
k

a
k
(1:3)
(see [15,16]). Notice that, from the definition B
k
Î ℚ, and these integrals are inde-
pendent of the prime p which used to compute them. The examples of (1.3) are:
lim
N→∞
1
p
N
p
N
−1

a=0
a = lim
N→∞
1
p
N
p
N
(p
N
− 1)
2
= −
1

1
6
= B
2
.
(1:4)
Euler numbers E
k
, k ≥ 0 are integers given by (cf. [17-19])
E
0
=1, E
k
= −
k−1

i=0
2|k−i

k
i

E
i
for k =1,2,
(1:5)
The Euler polynomial E
k
(x) is defined by (see [[20], p. 25]):
E

k

1
2

,
(1:7)
where E
0
=1,E
2
=-1,E
4
=5,E
6
= -61, Therefore, E
k
≠ E
k
(0), in fact ([[19], p. 374
(2.1)])
E
k
(0) =
2
k +1
(1 − 2
k+1
)B
k+1

Let UD(ℤ
p
, ℂ
p
) denote the space of all uniformly (or strictly) differentiable ℂ
p
-valued
functions on ℤ
p
. The p-adic q-integral of a function f Î UD(ℤ
p
)onℤ
p
is defined by
I
q
(f ) = lim
N→∞
1
[p
N
]
q
p
N
−1

a=0
f (a)q
a

1
p
N
p
N
−1

a=0
f (a)=

p
f (z)dμ
1
(z).
(2:3)
From (2.1), we have the fermionic p-adic integral on ℤ
p
as follows:
I
−1
(f ) = lim
q→−1
I
q
(f ) = lim
N→∞
p
N
−1


(q)=

p
[z]
k
q
dμ
1
(z) = lim
N→∞
1
p
N
p
N
−1

a=0
[a]
k
q
,
(2:5)
E
k
(q)=2
k

p


B
k
(q) → B
k
=

p
z
k
dμ
1
(z), E
k
(q) → E
k
=

p
(2z +1)
k
dμ
−1
(z).
For |q -1|
p
<1 and z Î ℤ
p
, we have
q
iz

N
ℤ
p
(see [16]).
By (2.3) and (2.7), we obtain
I
1
(q
iz
)=
1
q
i
− 1
lim
N→∞
(q
i
)
p
N
− 1
p
N
=
1
q
i
− 1
lim


m=1

p
N
m

(q
i
− 1)
m
=
1
q
i
− 1
lim
N→∞
∞

m=1
1
m

p
N
− 1
m − 1

(q

m−1
(q
i
− 1)
m
m
=
i log q
q
i
− 1
(2:8)
since the series log
log(1 + x)=

∞
m=1
(−1)
m−1
x
m
/m
converges at |x|
p
<1. Similarl y,
by (2.4), we obtain (see [[4], p. 4, (2.10)])
I
−1
(q
iz

(1 − q)
k
k

i=0

k
i

(−1)
i
i
q
i
− 1
,
(2:10)
E
k
(q)=
2
k+1
(1 − q)
k
k

i=0

k
i

dμ
1
(z)andE
k
(x, q)=

p
[x + z]
k
q
dμ
−1
(z),
(2:12)
where q Î ℂ
p
with 0 <|q -1|
p
<1 and x Î ℤ
p
, respectively. We will rewrite the above
equations in a slightly different way. By (2.5), (2.6), and (2.12), after some elementary
calculations, we get
Kim and Kim Advances in Difference Equations 2011, 2011:68
/>Page 4 of 16
B
k
(x, q)=
k


E
i
(q)
2
i

x −
1
2

k−i
q
q
i(x−
1
2
)
=
⎛
⎝
q
x−
1
2
2
E(q)+

x −
1
2

)
k
and
[x + z]
k
q
=

1
2

k
q

[2x − 1]
q
1
2
+ q
x−
1
2

1
2

−1
q

z +

2

−i
q

z +
1
2

i
q
(2:15)
(cf. [4,5] ). T he above formulas can be found in [7 ] which are the q-analogues of the
corresponding cla ssical formulasin[[17],(1.2)]and[23],etc.Obviously,put
x =
1
2
in
(2.14). Then
E
k
(q)=2
k
E
k

1
2
, q


q
(k ≥ 0),
E
k
(x + y, q)=
k

i=0

k
i

q
iy
E
i
(x, q)[y]
k−i
q
(k ≥ 0).
Proof. Applying the relationship
[x + y −
1
2
]
q
=[y]
q
+ q
y

q
y
⎛
⎝
q
x−
1
2
2
E(q)+

x −
1
2

q
⎞
⎠
+[y]
q
⎞
⎠
k
=
k

i=0

k
i


q
iy
E
i
(x, q)[y]
k−i
q
.
Similarly, the first identity follows.□
Kim and Kim Advances in Difference Equations 2011, 2011:68
/>Page 5 of 16
Remark 2.3. From (2.12), we obtain the not completely trivial identities
lim
q→1
B
k
(x + y, q)=
k

i=0

k
i

B
i
(x)y
k−i
=(B(x)+y)

noulli and Euler polynomials, see [17,15] and see also the references c ited in each of
these earlier works.
Lemma 2.4. Let n be any positive integer. Then
k

i=0

k
i

q
i
[n]
i
q
B
i
(x, q
n
)=[n]
k
q
B
k

x +
1
n
, q
n

.
Proof. Use Lemma 2.2, the proof can be obtained by the simi lar way to [[7], Lemm a
2.3]. □
We note here that similar expressions to those of Lemma 2.4 ar e given by Luo [[7],
Lemma 2.3]. Obviously, Lemma 2.4 are the q-analogues of
k

i=0

k
i

n
i
B
i
(x)=n
k
B
k

x +
1
n

,
k

i=0


(z)
= lim
N→∞
1
np
N
np
N
−1

a=0
[nx + a]
k
q
=
1
n
lim
N→∞
1
p
N
n−1

i=0
p
N
−1

a=0

[n]
k
q
n
n−1

i=0
B
k

x + i
n
, q
n

.
(2:18)
Kim and Kim Advances in Difference Equations 2011, 2011:68
/>Page 6 of 16
If we put x = 0 in (2.18) and use (2.13), we find easily that
B
k
(q)=
[n]
k
q
n
n−1

i=0

ij
B
j
(q
n
)
=
1
n
k

j=0
[n]
j
q

k
j

B
j
(q
n
)
n−1

i=0
q
ij
[i]

From (2.4), we see that
E
k
(nx, q)=

p
[nx + z]
k
q
dμ
−1
(z)
= lim
N→∞
n−1

i=0
p
N
−1

a=0
[nx + na + i]
k
q
(−1)
na+i
=[n]
k
q

(−1)
i
E
k

x + i
n
, q
n

if n odd.
(2:21)
From (2.18) and (2.21), we can obtain Proposition 2.5 below.
Proposition 2.5 (Multiplication formulas). Let n be any positive integer. Then
B
k
(x, q)=
[n]
k
q
n
n−1

i=0
B
k

x + i
n
, q

F
q
(t )=
∞

k=0
B
k
(q)
t
k
k!
= e
B(q)t
and G
q
(t )=
∞

k=0
E
k
(q)
t
k
k!
= e
E(q)t
,
(3:1)

e
[m]
q
t
,
G
q
(t )=2
∞

m=0
(−1)
m
e
2[m+
1
2
]
q
t
.
Proof. Combining (2.10) and (3.1), F
q
(t) may be written as
F
q
(t )=
∞

k=0

1
log q
+
k

i=1

k
i

(−1)
i
i
q
i
− 1

.
Here, the term with i = 0 is understood to be 1/log q (the limiting value of the sum-
mand in the limit i ® 0). Specifically, by making use of the following well-known bino-
mial identity
k

k − 1
i − 1

= i

k
i

i
− 1

=
∞

k=0
1
(1 − q)
k
t
k
k!
+logq
∞

k=1
k
(1 − q)
k
t
k
k!
∞

m=0
q
m
k−1


m
)
k−1
= e
t
1−q
+
t log q
1 − q
∞

m=0
q
m
∞

k=0

1 − q
m
1 − q

k
t
k
k!
.
Next, by (2.11) and (3.1), we obtain the result
G
q


k=0
2
k
∞

m=0
(−1)
m
⎛
⎝
1 − q
m+
1
2
1 − q
⎞
⎠
k
t
k
k!
=2
∞

m=0
(−1)
m
∞


From (2.13)and (2.14), we define the q-Bernoulli and q-Euler polynomials by
F
q
(t , x)=
∞

k=0
B
k
(x, q)
t
k
k!
=
∞

k=0
(q
x
B(q)+[x]
q
)
k
t
k
k!
,
(3:2)
G
q

t
k
k!
.
(3:3)
Hence, we have
Lemma 3.3.
F
q
(t , x)=e
[x]
q
t
F
q
(q
x
t)=e
t
1−q
+
t log q
1 − q
∞

m=0
q
m+x
e
[m+x]

[x]
q
t
= e
[x]
q
t
F
q
(q
x
t).
The second identity leads at once to Lemma 3.1. Hence, the lemma follows. □
Lemma 3.4.
G
q
(t , x)=e
[x−
1
2
]
q
t
G
q
⎛
⎝
q
x−
1

q
(k ≥ 0),
E
k
(x +1,q)+E
k
(x, q)=2[x]
k
q
(k ≥ 0).
Proof. By applying (3.2) and Lemma 3.3, we obtain (3.4)
F
q
(t , x)=
∞

k=0
B
k
(x, q)
t
k
k!
=1+
∞

k=0

1
(1 − q)

log q
1 − q
∞

m=0
q
m+x
[m + x]
k−1
q
(k ≥ 1).
(3:5)
Hence,
B
k
(x +1,q) − B
k
(x, q)=k
q
x
log q
q − 1
[x]
k−1
q
(k ≥ 1).
Similarly we prove the second part by (3.3) and Lemma 3.4. This proof is complete.
□
From Lemma 2.2 and Corollary 3.5, we obtain for any integer k ≥ 0,
[x]

1
2

k

i=0

k
i

q
i
E
i
(x, q)+E
k
(x, q)

which are the q-analogues of the following familiar expansions (see, e.g., [[7], p. 9]):
x
k
=
1
k +1
k

i=0

k +1
i

n
, q
n

− B
k+1

x +
1 − n
n
, q
n

=
nq
n(x−1)+1
log q
q − 1
k +1
[n]
k+1
q
(1 + q[nx − n]
q
)
k
,
E
k


k

x +
1
n

− B
k

x +
1 − n
n

= k

x +
1 − n
n

k−1
(k ≥ 1, n ≥ 1),
(3:6)
E
k

x +
1
n

+ E

Kim and Kim Advances in Difference Equations 2011, 2011:68
/>Page 10 of 16
In Corollary 3.5, let x = 0. We arrive at the following proposition.
Proposition 3.7.
B
0
(q)=1, (qB(q)+1)
k
− B
k
(q)=

log
p
q
q−1
if k =1
0 if k > 1,
E
0
(q)=1,

q
−
1
2
E(q)
2
+


q
−
1
2
2
E(q)+

−
1
2

q
⎞
⎠
k
and E
k
(1, q)=
⎛
⎝
q
1
2
2
E(q)+

1
2

q

Now, by evaluating the kth derivative of both sides of Lemma 3.1 at t = 0, we obtain
the following
B
k
(q)=

d
dt

k
F
q
(t )





t=0
=

1
1 − q

k
−
k log q
q − 1
∞


(−1)
m

m +
1
2

k
q
(4:2)
for k ≥ 0.
Definition 4.1 (q-analogues of the Riemann’s ζ-functions). For s Î ≤, define
ζ
q
(s)=
1
s − 1
1

1
1−q

s−1
+
log q
q − 1
∞

m=1
q

,E(s) is a analytic function on ≤.
Also, we have
lim
q→1
ζ
q
(s)=
∞

m=1
1
m
s
= ζ (s) and lim
q→1
ζ
q,E
(s)=2
∞

m=0
(−1)
m
(m +1)
s
= ζ
E
(s).
(4:3)
(In [[26], p. 1070], our ζ

define
ζ
q
(s, x)=
1
s − 1
1
(
1
1−q
)
s−1
+
log q
q − 1
∞

m=0
q
m+x
[m + x]
s
q
,
ζ
q,E
(s, x)=2
∞

m=0

(x, q).
Proof. From Lemma 3.3 and Definition 4.3, we have

d
dt

k
F
q
(t , x)





t=0
= −kζ
q
(1 − k, x)
for k ≥ 1. We obtain the desired result by (3 .2). Similarly the second form follows by
Lemma 3.4 and (3.3). □
Proposition 4.5. Let d be any positive integer. Then
F
q
(t , x)=
1
d
d−1

i=0

Kim and Kim Advances in Difference Equations 2011, 2011:68
/>Page 12 of 16
Proof. Substituting m = nd + i wit h n = 0, 1, and i = 0, , d - 1 into Lemma 3.3, we
have
F
q
(t , x)=e
t
1−q
+
t log q
1 − q
∞

m=0
q
m+x
e
[m+x]
q
t
= e
[d]
q
t
1−q
d
+
1
d

d
+
[d]
q
t log q
d
1 − q
d
∞

n=0
(q
d
)
n+
x+i
d
e
[n+
x+i
d
]
q
d
[d]
q
t

,
where we use

,
ζ
q,E
(−k, x)=[d]
k
q
d−1

i=0
(−1)
i
ζ
q
d
,E

−k,
x + i
d

if d odd.
Let c be a primitive Dirichlet character o f conductor f Î N.Wedefinethegenerat-
ing function F
q,c
(x, t)andG
q,c
(x, t) of the generali zed q-Bernoulli and q-Euler polyno-
mials as follows:
F
q,χ

(t , x)=
∞

k=0
E
k,χ
(x, q)
t
k
k!
=
f

a=1
(−1)
a
χ(a)G
q
f

[f ]
q
t,
a + x
f

if f odd,
(4:5)
where B
k,c

q
t
if f odd,
(4:7)
Kim and Kim Advances in Difference Equations 2011, 2011:68
/>Page 13 of 16
respectively. As q ® 1 in (4.6) and (4.7), we have F
q,c
(t, x) ® F
c
(t, x)andG
q,c
(t, x)
® G
c
(t, x), where F
c
(t, x) and G
c
(t, x) are the usual generating function of generalized
Bernoulli and Euler numbers, respectively, which are defined as follows [13]:
F
χ
(t , x)=
f

a=1
χ(a)te
(a+x)t
e

k=0
G
k,χ
(x)
t
k
k!
if f odd.
(4:9)
From (3.2), (3.3), (4.4) and (4.5), we can easily see that
B
k,χ
(x, q)=
[f ]
k
q
f
f

a=1
χ(a)B
k

a + x
f
, q
f

,
(4:10)

log q
q − 1
∞

m=0
χ(m)q
m+x
[m + x]
s
q
,

q
(s, x, χ)=2
∞

m=0
(−1)
m
χ(m)
[m + x]
s
q
.
Similarly, we can compute the values of L
q
(s, x, c) at non-positive integers.
Theorem 4.8. For k ≥ 1, we have
L
q

q
t
1−q
f
+
[f ]
q
t log q
f
1 − q
f
∞

n=0
(q
f
)
n+
x+a
f
e
[n+
x+a
f
]
q
f
[f ]
q
t

obtain
B
k,χ
(x, q)=

d
dt

k

∞

k=0
B
k,χ
(x, q)
t
k
k!






t=0
=
k log q
1 − q
∞

Similarly the second identity follows. This completes the proof. □
Acknowledgements
This study was supported by the National Research Foundation of Korea (NRF) grant fu nded by the Korea
government (MEST) (2011-0001184).
Author details
1
National Institute for Mathematical Sciences, Doryong-dong, Yuseong-gu Daejeon 305-340, South Korea
2
Department
of Mathematics, KAIST, 373-1 Guseong-dong, Yuseong-gu, Daejeon 305-701, South Korea
Authors’ contributions
The authors have equal contributions to each part of this paper. All the authors read and approved the final
manuscript.
Competing interests
The authors declare that they have no competing interests.
Received: 2 September 2011 Accepted: 23 December 2011 Published: 23 December 2011
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doi:10.1186/1687-1847-2011-68
Cite this article as: Kim and Kim: Some results for the q-Bernoulli, q-Euler numbers and polynomials. Advances in
Difference Equations 2011 2011:68.
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