第 49 卷 第 1 期
2013 年 2 月
兰 州 大 学 学 报（自然科学版）
Journal of Lanzhou University (Natural Sciences)
Vol. 49 No. 1
Feb. 2013
Articlcal ID: 0455-2059(2013)01-0100-04
Sixfold restricted sum formula for multiple zeta values at
even integers
LEI Peng, GUO Li
School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, China
Abstract: Identities for the multiple zeta functions
P
ζ(2n1 , 2n2 , · · · , 2nk ), are known for k = 2,
n1 +n2 +···+nk =n
n1 , n2 , ··· , nk >1
3, 4, 5. The multiple zeta values was given for k = 6 by way of recursion
P
231
ζ(2n) −
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 , 2n6 ) = 512
21
ζ(2)ζ(2n
64
− 2) +
21
ζ(4)ζ(2n
256
− 4).
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
Key words: multiple zeta value; shuffle product; recursion
CLC number: O156.2
Document code: A
AMS Subject Classifications(2000): 11M99
多重 Zeta 偶数值的六重求和公式
雷 鹏, 郭 锂
兰州大学 数学与统计学院, 兰州 730000
摘
P
要: 对于多重 Zeta 函数
ζ(2n1 , 2n2 , · · · , 2nk ), 当 k = 2, 3, 4, 5 时等式是已知的. 利用递归的
n1 +n2 +···+nk =n
n1 , n2 , ··· , nk >1
方法给出了 k = 6 时的多重 Zeta 值
P
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 , 2n6 ) = 231
ζ(2n) −
512
21
ζ(2)ζ(2n
64
− 2) +
21
ζ(4)ζ(2n
256
− 4).
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
关键词: 多重 Zeta 值; 洗牌乘积; 递归
文献标识码: A
中图分类号: O156.2
The multiple zeta values (MZVs) are special values
A principle goal in the theoretical study of MZVs
of the multi-variable analytic function
is to determine all algebraic relations among them.
ζ(n1 , n2 , · · · , nk ) =
X
Conjecturally all such relations come from the so-
m1 >m2 >···>mk
1
mn1 1 mn2 2 · · · mnk k
>0
,
(1)
at integers n1 > 2, ni > 1, 1 6 i 6 k. Their study
in the two variable cases went back to Goldbach and
Euler[1] . The general concept was introduced in the
1990’s, with motivation from both mathematics[2]
and physics[3] . Since then the subject has been studied intensively with interactions from a broad range
called extended double shuffle relations. But there
is no definite way to exhaust all of them and new
identities of MZVs are being found steadily[9−10] .
One of Euler’s major discoveries on double zeta
values is his sum formula
n−1
X
ζ(i, n − i) = ζ(n).
i=2
metic geometry, combinatorics, number theory, knot
Reference [11] proved
n−1
X
3
ζ(2m, 2n − 2m) = ζ(2n),
4
m=1
theory, Hopf algebra, quantum field theory and mir-
for n > 2.
of areas in mathematics and physics, including arith-
ror symmetry
[3−8]
.
(2)
Reference [12] proved the following results: for
Received date: 2012-10-08
Biography: LEI Peng(1977−), male, born in Zhongwei, Ningxia Hui Autonomous Region, lecturer, e-mail: [email protected],
majoring in semigroup algebra and algebraic number theory.
No. 1
LEI Peng, et al: Sixfold restricted sum formula for multiple zeta values at even integers
n > 3,
X
By symmetry, we have
Proof
1
5
ζ(2a, 2b, 2c) = ζ(2n) − ζ(2)ζ(2n − 2),
8
4
a+b+c=n
X
A=
X
(3)
X
ζ(2a, 2b, 2c, 2d) =
X
35
5
ζ(2n) − ζ(2)ζ(2n − 2).
(4)
64
16
In [13], the following result for the fivefold restrict-
X
5
a=1
values at even integers by the method of recursion,
a=1
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 , 2n6 ) =
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
A=
5
′
b , c , d , e be positive integers. In order to prove
the theorem, we need the following lemmas.
Lemma 1
For n > 2, we have
n−1
X
2n + 1
ζ(2m)ζ(2n − 2m) =
ζ(2n).
2
m=1
X
X
(b′ − 1)ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ) =
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
5
X
(c′ − 1)ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ) =
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
For n > 6, let
X
ζ(2a + 2f, 2b, 2c, 2d, 2e)+
5
X
(d′ − 1)ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ).
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
a+b+c+d+e+f =n
a, b, c, d, e, f >1
Adding in the above five equations and noting that
ζ(2a, 2b + 2f, 2c, 2d, 2e)+
a′ +b′ +c′ +d′ +e′ = n, we get
X
5A = (n − 5)
ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ).
a+b+c+d+e+f =n
a, b, c, d, e, f >1
ζ(2a, 2b, 2c + 2f, 2d, 2e)+
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
a+b+c+d+e+f =n
a, b, c, d, e, f >1
ζ(2a, 2b, 2c, 2d + 2f, 2e)+
Multiplying
a+b+c+d+e+f =n
a, b, c, d, e, f >1
1
5
(6)
on the two sides of the above equation,
we complete the proof.
ζ(2a, 2b, 2c, 2d, 2e + 2f ).
a+b+c+d+e+f =n
a, b, c, d, e, f >1
X
(a′ − 1)ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ) =
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
5
[15]
A = (n − 5)
d=1
d′ > 1. The above equation equals
X
5
(e′ − 1)ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ).
Throughout this paper, let a, b, c, d, e, f and a′ ,
Then we have
c=1
b=1
By symmetry, we also have
Preliminaries
X
d=1
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
21
21
231
ζ(2n)− ζ(2)ζ(2n − 2) +
ζ(4)ζ(2n − 4).
512
64
256
X
c=1
b=1
Let a′ = a, b′ = b, c′ = c, d′ = d, e′ = n − a′ − b′ − c′−
For n > 6, we have
X
e=1
ζ(2a, 2b, 2c, 2d, 2(n − a − b − c − d)).
which differs from the method in [14].
X
d=1
ζ(2a, 2b, 2c, 2d, 2(n − a − b − c − d)) =
n−4
X n−a−3
X n−a−b−2
X n−a−b−c−1
X
5
(n−a−b−c−d−1)·
the sixfold restricted sum formula for multiple zeta
A=
ζ(2a,
c=1
b=1
2b, 2c, 2d, 2(n−a−b−c−d)) =
n−5
X n−a−b−3
X
X n−a−4
X n−a−b−c−2
(n−a−b−c−d−1)·
5
21
3
63
ζ(2n) − ζ(2)ζ(2n − 2) + ζ(4)ζ(2n − 4).
128
64
64
In this paper, we obtain the following result for
Lemma 2
n−5
n−a−b−c−d−1
X n−a−4
X n−a−b−3
X n−a−b−c−2
X
X
a=1
a+b+c+d+e=n
a, b, c, d, e>1
′
ζ(2a, 2b, 2c, 2d, 2e + 2f ) =
a+b+c+d+e+f =n
a, b, c, d, e, f >1
tegers was announced.
X
ζ(2a, 2b, 2c, 2d, 2e) =
′
ζ(2a, 2b, 2c, 2e + 2f, 2d)+
a+b+c+d+e+f =n
a, b, c, d, e, f >1
ed sum formula for multiple zeta values at even in-
′
ζ(2a, 2b, 2e + 2f, 2c, 2d)+
a+b+c+d+e+f =n
a, b, c, d, e, f >1
a+b+c+d=n
a, b, c, d>1
1
ζ(2a, 2e + 2f, 2b, 2c, 2d)+
a+b+c+d+e+f =n
a, b, c, d, e, f >1
X
Theorem 1
X
ζ(2e + 2f, 2a, 2b, 2c, 2d)+
a+b+c+d+e+f =n
a, b, c, d, e, f >1
a, b, c>1
and, for n > 4,
101
2
ζ(2a′ , 2b′ , 2c′ , 2d′ , 2e′ ).
a′ +b′ +c′ +d′ +e′ =n
a′ , b′ , c′ , d′ , e′ >1
(5)
The proof of theorem 1
Proof Using the quasi-shuffle relation and Lemma 2, we have
X
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 )ζ(2n6 ) =
102
Journal of Lanzhou University (Natural Sciences)
X
6
X
we get
ζ(2n1 + 2n6 , 2n2 , 2n3 , 2n4 , 2n5 )+
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
X
ζ(2n1 , 2n2 + 2n6 , 2n3 , 2n4 , 2n5 )+
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
X
ζ(2n1 , 2n2 , 2n3 + 2n6 , 2n4 , 2n5 )+
X
ζ(2n1 , 2n2 , 2n3 , 2n4 + 2n6 , 2n5 )+
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
X
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 + 2n6 ) =
X
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 , 2n6 )+
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
(n − 5)
ζ(2a, 2b, 2c, 2d, 2e).
a+b+c+d+e=n
a, b, c, d, e>1
Applying (5) to the second part of the above equation, we obtain
(n − 5)(
21
63
ζ(2n) − ζ(2)ζ(2n − 2)+
128
64
3
ζ(4)ζ(2n − 4)).
64
On the other hand, by (5), we get
X
(7)
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 )ζ(2n6 ) =
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
n−5
X
(
n6 =1
X
21
63
ζ(2n − 2n6 ) −
ζ(2)ζ(2n − 2 − 2n6 )+
128
64
3
ζ(4)ζ(2n − 4 − 2n6 ))ζ(2n6 ) =
64
n−1
63 X
(
ζ(2n − 2n6 )ζ(2n6 ) − ζ(2)ζ(2n − 2)−
128 n =1
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 , 2n6 ) =
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
X
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 )ζ(2n6 )−
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
(n − 5)
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
X
7
5
ζ(4), ζ(2)ζ(4) = ζ(6),
2
4
5
7
2
ζ(2)ζ(6) = ζ(8), ζ (4) = ζ(8).
3
6
Using the above formulas and (7), we get
ζ 2 (2) =
6
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
6
ζ(6) = π 6 /945, ζ(8) = π 6 /9450,
ζ(2n1 , 2n2 , 2n3 , 2n4 , 2n5 , 2n6 )+
n1 +n2 +n3 +n4 +n5 +n6 =n
n1 , n2 , n3 , n4 , n5 , n6 >1
Vol. 49
X
ζ(2a, 2b, 2c, 2d, 2e) =
a+b+c+d+e=n
a, b, c, d, e>1
63 2n + 1
(
ζ(2n) − ζ(2)ζ(2n − 2) − ζ(4)ζ(2n − 4)−
128
2
ζ(6)ζ(2n − 6) − ζ(8)ζ(2n − 8))−
2n − 1
21
ζ(2)(
ζ(2n − 2) − ζ(2)ζ(2n − 4)−
64
2
3
2n − 3
ζ(4)ζ(2n − 6) − ζ(6)ζ(2n − 8)) +
ζ(4)(
ζ(2n − 4)−
64
2
63
ζ(2n)−
ζ(2)ζ(2n − 6) − ζ(4)ζ(2n − 8)) − (n − 5)(
128
3
21
ζ(2)ζ(2n − 2) +
ζ(4)ζ(2n − 4)) =
64
64
693
126
63
21 5
ζ(2n) −
ζ(2)ζ(2n − 2) + (−
+
· +
256
64
128
64 2
3
3 2n − 3
·
−
(n − 5))ζ(4)ζ(2n − 4)+
64
2
64
18 7
63
+
· )ζ(6)ζ(2n − 6)+
(−
128
64 4
63
21 5
3 7
(−
+
· −
· )ζ(8)ζ(2n − 8) =
128
64 3
64 6
126
63
693
ζ(2n) −
ζ(2)ζ(2n − 2) +
ζ(4)ζ(2n − 4).
256
64
128
Multiplying
1
6
to two sides of the above equation,
we obtain the theorem.
6
ζ(4)ζ(2n − 4) − ζ(6)ζ(2n − 6) − ζ(8)ζ(2n − 8))−
n−2
X
21
ζ(2n − 2 − 2n6 )ζ(2n6 ) − ζ(2)ζ(2n − 4)−
ζ(2)(
64
n =1
6
ζ(4)ζ(2n − 6) − ζ(6)ζ(2n − 8))+
n−3
X
3
ζ(4)(
ζ(2n − 4 − 2n6 )ζ(2n6 ) − ζ(2)ζ(2n − 6)−
64
n =1
6
ζ(4)ζ(2n − 8)).
Reference
[1] Juskevic A P, Winter E. Leonhard Euler and
Christian Goldbach[M]. Berlin: Akademie-Verlag,
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J Math, 1992, 152(2): 275−290.
Applying Lemma 1 to the first, sixth and tenth
[3] Broadhurt D J, Kreimer D. Association of mul-
terms of the above sum, we have
63 2n + 1
(
ζ(2n) − ζ(2)ζ(2n − 2) − ζ(4)ζ(2n − 4)−
128
2
ζ(6)ζ(2n − 6) − ζ(8)ζ(2n − 8))−
21
2n − 1
ζ(2)(
ζ(2n − 2) − ζ(2)ζ(2n − 4)−
64
2
ζ(4)ζ(2n − 6) − ζ(6)ζ(2n − 8))+
3
2n − 3
ζ(4)(
ζ(2n − 4)−
64
2
ζ(2)ζ(2n − 6) − ζ(4)ζ(2n − 8)).
tiple zeta values with positive knots via Feynman di-
Noting that
ζ(2) = π 2 /6, ζ(4) = π 4 /90,
agrams up to 9 loops[J]. Phys Lett B, 1997, 393(3/4):
403−412.
[4] Bowman D, Bradley D. Resolution of some open
problems concerning multiple zeta evaluations of arbitrary depth[J]. Composition Math, 2003, 139(1):
85−100.
[5] Borwein J M, Broadhurst D J, Bradley D M,
et al. Special values of multiple polylogarithms[J].
Trans Amer Math Soc, 2001, 353(3): 907−941.
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