Sums of powers of consecutive q-integers

0.08 MB·

by  Taekyun Kim

SUMS OF POWERS OF CONSECUTIVE q-INTEGERS
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0 Taekyun Kim
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n Institute of Science Education,
a Kongju National University, Kongju 314-701, S. Korea
J
e-mail: [email protected]
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T Abstract. Let n,k be thepositive integers(k >1), and letSn,q(k) be the sums of the
N n-thpowersofpositiveq-integersuptok−1: Sn,q(k)= lk=−01ql[l]nq , where[l]q = qql−−11.
.
h Following an idea due to J. Bernoulli, we explore a formPula for Sn,q(k) as follows:
t
a
[m Sn+1,q(k)= n+1 1 n n+i 1 βi,qqki[k]nq+1−i+ (1−q(nn++1)k1)βn+1,q,
Xi=0(cid:16) (cid:17)
1
v where βi,q are the q-Bernoulli numbers.
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1. Introduction
0
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0 Let n be a natural numbers. There are formulas such as
/
h n2 +n
t 1+2+···+n = ,
a 2
m
2n3 +3n2 +n
: 12 +22 +···+n2 = ,
v
6
i
X n4 +2n3 +n2
13 +23 +···+n3 = ,··· .
r
a 4
In Korea, these are subjects of the high school mathematics. J. Bernoulli (1713) first
discovered the method which one can produce those formulae for the sum n lk for
l=1
2000 Mathematics Subject Classification 11S80, 11B68 P
Key words and phrases: q-Bernoulli number, p-adic q-integrals, zeta function, Dirichlet series
Typeset by AMS-TEX
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2 SUMS OF POWERS OF CONSECUTIVE q-INTEGERS
any natural numbers. Let q be an indeterminate which can be considered in complex
number field, and for any integer k define the q-integer as [k] = qk−1, cf. [1,2]. Note
q q−1
that limq→1[k]q = k. In this paper we evaluatesums of powers of consecutive q-integers.
For any positive integers n,k(> 1), let S (k) = k−1ql[l]n. Following an idea due to
n,q l=0 q
J. Bernoulli, we explore a formula for S (k) as follows:
n,q
P
1 n n+1 (1−q(n+1)k)β
S (k) = β qki[k]n+1−i + n+1,q,
n+1,q n+1 i i,q q n+1
i=0(cid:18) (cid:19)
X
where β are the q-Bernoulli numbers.
i,q
2. Sum of the n-th powers of positive q-integers up to k −1
Let j be the positive integers. Then we easily see that
(1) [j +1]2 −[j]2 = ([j] +qj)2 −[j]2 = qj(2[j] +qj).
q q q q q
By (1), we have
k−1 k−1
[2k]
(2) [k]2 = ([j +1]2 −[j]2) = 2 qj[j] + q.
q q q q [2]
q
j=0 j=0
X X
Hence,
k−1
[2k] 1
(3) qj[j] = [k]2 − q .
q q [2] 2
q
j=0 (cid:18) (cid:19)
X
By the same method of Eq.(1), we easily see that
(4) [j +1]3 −[j]3 = 3[j]2qj+1 +3[j] qj +q3j.
q q q q
Thus we have
k−1
1 1 [2k] 1[3k]
(5) [j]2qj+1 = [k]3 − [k]2 − q − q.
q 3 q 2 q [2] 3 [3]
q q
j=0 (cid:18) (cid:19)
X
TAEKYUN KIM 3
Example. For q = 9 in (5), we note that
10
2 2
9 9 9 9 9 9
( )2 +( )3 1+ +···+( )k 1+ +···+( )k−2 +···
10 10 10 10 10 10
(cid:18) (cid:19) (cid:18) (cid:19)
3 2
1 1 1 1 1 1 1
= − − − .
3 1− 9 2 1− 9 1−( 9 )2 3 1−( 9 )3
!
(cid:18) 10(cid:19) (cid:18) 10(cid:19) 10 (cid:18) 10 (cid:19)
Let n,k be the positive integers (k > 1) and let S (k) = k−1ql[l]n. Then we note
n,q l=0 q
that
P
n−1 k−1
n
(6) i Si,qn−i(k) = [l+1]nq −[l]nq = [k]nq.
i=0 (cid:18) (cid:19) l=0
X X(cid:0) (cid:1)
By replacing n by n+1, we see that
n n−1
n+1 n+1
(7) [k]nq+1 = i Si,qn+1−i(k) = i Si,qn+1−i(k)+(n+1)Sn,q(k).
i=0(cid:18) (cid:19) i=0 (cid:18) (cid:19)
X X
Therefore we obtain the following:
n−1
n+1
(8) (n+1)Sn,q(k) = [k]nq+1 − i Si,qn+1−i(k).
i=0 (cid:18) (cid:19)
X
3. q-analogs of Bernoulli polynomials
In this section, we assume q ∈ C with |q| < 1. The q-Bernoulli polynomials β (x)
n,q
are defined by means of the generating function F (t) as follows:
q
∞ ∞
q −1 β (x)
(9) Fq(t) = e1−tq −t qn+xe[n+x]qt = n,q tn, |q| < 1, |t| < 1 .
logq n!
n=0 n=0
X X
In the case x = 0, β (= β (0)) will be called the q-Bernoulli numbers, cf. [1, 2]. By
n,q n,q
(9), we easily see that
n n n
n 1 n k
(10) β (x) = qjxβ [x]n−j = qkx(−1)k.
n,q j j,q q 1−q k [k]
q
j=0(cid:18) (cid:19) (cid:18) (cid:19) k=0(cid:18) (cid:19)
X X
4 SUMS OF POWERS OF CONSECUTIVE q-INTEGERS
In (9), (10), the q-Bernoulli numbers can be rewritten as
q −1
(11) β = , (qβ +1)k −β = δ ,
0,q k,q k,1
logq
where δ is Kronecker symbol and we use the usual convention about replacing βi by
k,1
β , cf. [1,2]. By (9), (10), and (11), we easily see that
i,q
∞ ∞ ∞ k−1 tn−1
− ql+ke[l+k]qt + qle[l]qt = n ql[l]n−1 .
n!
!
l=0 l=0 n=1 l=0
X X X X
Thus we have
k−1
(12) β (k)−β = n ql[l]n−1.
n,q n,q q
l=0
X
By (12), we obtain the following:
k−1 1 n n+1 (1−q(n+1)k)
(13) ql[l]n = qklβ [k]n+1−l + β .
q n+1 l l,q q n+1 n+1,q
l=0 l=0 (cid:18) (cid:19)
X X
In Eq.(10), we know that
n k
n k
(14) β (x) = [x]n−kβ [x]l(q−1)l.
n,q k q k,q l q
k=0(cid:18) (cid:19) l=0 (cid:18) (cid:19)
X X
By (14), we easily see that
k 1
(15) β (x)d[x] = (β (k)−β ) = S (k).
n,q q n+1,q n+1,q n,q
n+1
Z0
References
1. T. Apstol, Introduction to analytic number theory, Undergraduate Texts in Math., Springer-
Verlag, New York, 1986.
2. T. Kim, q-Volkenborn Integration, Russian J. Math. Phys. 9 (2002), 288-299.