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Template:Short description Template:For In mathematics, the multiple zeta functions are generalizations of the Riemann zeta function, defined by

ζ(s1,,sk)=n1>n2>>nk>0 1n1s1nksk=n1>n2>>nk>0 i=1k1nisi,

and converge when Re(s1) + ... + Re(si) > i for all i. Like the Riemann zeta function, the multiple zeta functions can be analytically continued to be meromorphic functions (see, for example, Zhao (1999)). When s1, ..., sk are all positive integers (with s1 > 1) these sums are often called multiple zeta values (MZVs) or Euler sums. These values can also be regarded as special values of the multiple polylogarithms.[1][2]

The k in the above definition is named the "depth" of a MZV, and the n = s1 + ... + sk is known as the "weight".[3]

The standard shorthand for writing multiple zeta functions is to place repeating strings of the argument within braces and use a superscript to indicate the number of repetitions. For example,

ζ(2,1,2,1,3)=ζ({2,1}2,3).

Definition

Multiple zeta functions arise as special cases of the multiple polylogarithms

Lis1,,sd(μ1,,μd)=k1>>kd>0μ1k1μdkdk1s1kdsd

which are generalizations of the polylogarithm functions. When all of the μi are nth roots of unity and the si are all nonnegative integers, the values of the multiple polylogarithm are called colored multiple zeta values of level n. In particular, when n=2, they are called Euler sums or alternating multiple zeta values, and when n=1 they are simply called multiple zeta values. Multiple zeta values are often written

ζ(s1,,sd)=k1>>kd>01k1s1kdsd

and Euler sums are written

ζ(s1,,sd;ε1,,εd)=k1>>kd>0ε1k1εkdk1s1kdsd

where εi=±1. Sometimes, authors will write a bar over an si corresponding to an εi equal to 1, so for example

ζ(a,b)=ζ(a,b;1,1).

Integral structure and identities

It was noticed by Kontsevich that it is possible to express colored multiple zeta values (and thus their special cases) as certain multivariable integrals. This result is often stated with the use of a convention for iterated integrals, wherein

0xf1(t)dtfd(t)dt=0xf1(t1)(0t1f2(t2)(0t2(0tdfd(td)dtd))dt2)dt1

Using this convention, the result can be stated as follows:[2]

Lis1,,sd(μ1,,μd)=01(dtt)s11dta1t(dtt)sd1dtadt where aj=i=1jμi1 for j=1,2,,d.

This result is extremely useful due to a well-known result regarding products of iterated integrals, namely that

(0xf1(t)dtfn(t)dt)(0xfn+1(t)dtfm(t)dt)=σ𝔖𝔥n,m0xfσ(1)(t)fσ(m)(t) where 𝔖𝔥n,m={σSmσ(1)<<σ(n),σ(n+1)<<σ(m)} and Sm is the symmetric group on m symbols.

To utilize this in the context of multiple zeta values, define X={a,b}, X* to be the free monoid generated by X and 𝔄 to be the free -vector space generated by X*. 𝔄 can be equipped with the shuffle product, turning it into an algebra. Then, the multiple zeta function can be viewed as an evaluation map, where we identify a=dtt, b=dt1t, and define

ζ(𝐰)=01𝐰 for any 𝐰X*,

which, by the aforementioned integral identity, makes

ζ(as11basd1b)=ζ(s1,,sd).

Then, the integral identity on products gives[2]

ζ(w)ζ(v)=ζ(w ⧢ v).

Two parameters case

In the particular case of only two parameters we have (with s > 1 and n, m integers):[4]

ζ(s,t)=n>m1 1nsmt=n=21nsm=1n11mt=n=11(n+1)sm=1n1mt
ζ(s,t)=n=1Hn,t(n+1)s where Hn,t are the generalized harmonic numbers.

Multiple zeta functions are known to satisfy what is known as MZV duality, the simplest case of which is the famous identity of Euler:

n=1Hn(n+1)2=ζ(2,1)=ζ(3)=n=11n3,

where Hn are the harmonic numbers.

Special values of double zeta functions, with s > 0 and even, t > 1 and odd, but s+t = 2N+1 (taking if necessary ζ(0) = 0):[4]

ζ(s,t)=ζ(s)ζ(t)+12[(s+ts)1]ζ(s+t)r=1N1[(2rs1)+(2rt1)]ζ(2r+1)ζ(s+t12r)
s t approximate value explicit formulae OEIS
2 2 0.811742425283353643637002772406 34ζ(4) Template:OEIS link
3 2 0.228810397603353759768746148942 3ζ(2)ζ(3)112ζ(5) Template:OEIS link
4 2 0.088483382454368714294327839086 (ζ(3))243ζ(6) Template:OEIS link
5 2 0.038575124342753255505925464373 5ζ(2)ζ(5)+2ζ(3)ζ(4)11ζ(7) Template:OEIS link
6 2 0.017819740416835988362659530248 Template:OEIS link
2 3 0.711566197550572432096973806086 92ζ(5)2ζ(2)ζ(3) Template:OEIS link
3 3 0.213798868224592547099583574508 12((ζ(3))2ζ(6)) Template:OEIS link
4 3 0.085159822534833651406806018872 17ζ(7)10ζ(2)ζ(5) Template:OEIS link
5 3 0.037707672984847544011304782294 5ζ(3)ζ(5)14724ζ(8)52ζ(6,2) Template:OEIS link
2 4 0.674523914033968140491560608257 2512ζ(6)(ζ(3))2 Template:OEIS link
3 4 0.207505014615732095907807605495 10ζ(2)ζ(5)+ζ(3)ζ(4)18ζ(7) Template:OEIS link
4 4 0.083673113016495361614890436542 12((ζ(4))2ζ(8)) Template:OEIS link

Note that if s+t=2p+2 we have p/3 irreducibles, i.e. these MZVs cannot be written as function of ζ(a) only.[5]

Three parameters case

In the particular case of only three parameters we have (with a > 1 and n, j, i integers):

ζ(a,b,c)=n>j>i1 1najbic=n=11(n+2)aj=1n1(j+1)bi=1j1(i)c=n=11(n+2)aj=1nHj,c(j+1)b

Euler reflection formula

The above MZVs satisfy the Euler reflection formula:

ζ(a,b)+ζ(b,a)=ζ(a)ζ(b)ζ(a+b) for a,b>1

Using the shuffle relations, it is easy to prove that:[5]

ζ(a,b,c)+ζ(a,c,b)+ζ(b,a,c)+ζ(b,c,a)+ζ(c,a,b)+ζ(c,b,a)=ζ(a)ζ(b)ζ(c)+2ζ(a+b+c)ζ(a)ζ(b+c)ζ(b)ζ(a+c)ζ(c)ζ(a+b) for a,b,c>1

This function can be seen as a generalization of the reflection formulas.

Symmetric sums in terms of the zeta function

Let S(i1,i2,,ik)=n1n2nk11n1i1n2i2nkik, and for a partition Π={P1,P2,,Pl} of the set {1,2,,k}, let c(Π)=(|P1|1)!(|P2|1)!(|Pl|1)!. Also, given such a Π and a k-tuple i={i1,...,ik} of exponents, define s=1lζ(jPsij).

The relations between the ζ and S are: S(i1,i2)=ζ(i1,i2)+ζ(i1+i2) and S(i1,i2,i3)=ζ(i1,i2,i3)+ζ(i1+i2,i3)+ζ(i1,i2+i3)+ζ(i1+i2+i3).

Theorem 1 (Hoffman)

For any real i1,,ik>1,, σΣkS(iσ(1),,iσ(k))=partitions Π of {1,,k}c(Π)ζ(i,Π).

Proof. Assume the ij are all distinct. (There is no loss of generality, since we can take limits.) The left-hand side can be written as σn1n2nk11ni1σ(1)ni2σ(2)nikσ(k). Now thinking on the symmetric

group Σk as acting on k-tuple n=(1,,k) of positive integers. A given k-tuple n=(n1,,nk) has an isotropy group

Σk(n) and an associated partition Λ of (1,2,,k): Λ is the set of equivalence classes of the relation given by ij iff ni=nj, and Σk(n)={σΣk:σ(i)i}. Now the term 1ni1σ(1)ni2σ(2)nikσ(k) occurs on the left-hand side of σΣkS(iσ(1),,iσ(k))=partitions Π of {1,,k}c(Π)ζ(i,Π) exactly |Σk(n)| times. It occurs on the right-hand side in those terms corresponding to partitions Π that are refinements of Λ: letting denote refinement, 1ni1σ(1)ni2σ(2)nikσ(k) occurs ΠΛ(Π) times. Thus, the conclusion will follow if |Σk(n)|=ΠΛc(Π) for any k-tuple n={n1,,nk} and associated partition Λ. To see this, note that c(Π) counts the permutations having cycle type specified by Π: since any elements of Σk(n) has a unique cycle type specified by a partition that refines Λ, the result follows.[6]

For k=3, the theorem says σΣ3S(iσ(1),iσ(2),iσ(3))=ζ(i1)ζ(i2)ζ(i3)+ζ(i1+i2)ζ(i3)+ζ(i1)ζ(i2+i3)+ζ(i1+i3)ζ(i2)+2ζ(i1+i2+i3) for i1,i2,i3>1. This is the main result of.[7]

Having ζ(i1,i2,,ik)=n1>n2>nk11n1i1n2i2nkik. To state the analog of Theorem 1 for the ζs, we require one bit of notation. For a partition

Π={P1,,Pl} of {1,2,k}, let c~(Π)=(1)klc(Π).

Theorem 2 (Hoffman)

For any real i1,,ik>1, σΣkζ(iσ(1),,iσ(k))=partitions Π of {1,,k}c~(Π)ζ(i,Π).

Proof. We follow the same line of argument as in the preceding proof. The left-hand side is now σn1>n2>>nk11ni1σ(1)ni2σ(2)nikσ(k), and a term 1n1i1n2i2nkik occurs on the left-hand since once if all the ni are distinct, and not at all otherwise. Thus, it suffices to show ΠΛc~(Π)={1, if |Λ|=k0, otherwise . (1)

To prove this, note first that the sign of c~(Π) is positive if the permutations of cycle type Π are even, and negative if they are odd: thus, the left-hand side of (1) is the signed sum of the number of even and odd permutations in the isotropy group Σk(n). But such an isotropy group has equal numbers of even and odd permutations unless it is trivial, i.e. unless the associated partition Λ is {{1},{2},,{k}}.[6]

The sum and duality conjectures

Source:[6]

We first state the sum conjecture, which is due to C. Moen.[8]

Sum conjecture (Hoffman). For positive integers k and n, i1++ik=n,i1>1ζ(i1,,ik)=ζ(n), where the sum is extended over k-tuples i1,,ik of positive integers with i1>1.

Three remarks concerning this conjecture are in order. First, it implies i1++ik=n,i1>1S(i1,,ik)=(n1k1)ζ(n). Second, in the case k=2 it says that ζ(n1,1)+ζ(n2,2)++ζ(2,n2)=ζ(n), or using the relation between the ζs and Ss and Theorem 1, 2S(n1,1)=(n+1)ζ(n)k=2n2ζ(k)ζ(nk).

This was proved by Euler[9] and has been rediscovered several times, in particular by Williams.[10] Finally, C. Moen[8] has proved the same conjecture for k=3 by lengthy but elementary arguments. For the duality conjecture, we first define an involution τ on the set of finite sequences of positive integers whose first element is greater than 1. Let T be the set of strictly increasing finite sequences of positive integers, and let Σ:T be the function that sends a sequence in to its sequence of partial sums. If Tn is the set of sequences in T whose last element is at most n, we have two commuting involutions Rn and Cn on Tn defined by Rn(a1,a2,,al)=(n+1al,n+1al1,,n+1a1) and Cn(a1,,al) = complement of {a1,,al} in {1,2,,n} arranged in increasing order. The our definition of τ is τ(I)=Σ1RnCnΣ(I)=Σ1CnRnΣ(I) for I=(i1,i2,,ik) with i1++ik=n.

For example, τ(3,4,1)=Σ1C8R8(3,7,8)=Σ1(3,4,5,7,8)=(3,1,1,2,1). We shall say the sequences (i1,,ik) and τ(i1,,ik) are dual to each other, and refer to a sequence fixed by τ as self-dual.[6]

Duality conjecture (Hoffman). If (h1,,hnk) is dual to (i1,,ik), then ζ(h1,,hnk)=ζ(i1,,ik).

This sum conjecture is also known as Sum Theorem, and it may be expressed as follows: the Riemann zeta value of an integer n ≥ 2 is equal to the sum of all the valid (i.e. with s1 > 1) MZVs of the partitions of length k and weight n, with 1 ≤ k ≤ n − 1. In formula:[3]

s1>1s1++sk=nζ(s1,,sk)=ζ(n).

For example, with length k = 2 and weight n = 7:

ζ(6,1)+ζ(5,2)+ζ(4,3)+ζ(3,4)+ζ(2,5)=ζ(7).

Euler sum with all possible alternations of sign

The Euler sum with alternations of sign appears in studies of the non-alternating Euler sum.[5]

Notation

n=1Hn(b)(1)(n+1)(n+1)a=ζ(a¯,b) with Hn(b)=+1+12b+13b+ are the generalized harmonic numbers.
n=1H¯n(b)(n+1)a=ζ(a,b¯) with H¯n(b)=1+12b13b+
n=1H¯n(b)(1)(n+1)(n+1)a=ζ(a¯,b¯)
n=1(1)n(n+2)an=1H¯n(c)(1)(n+1)(n+1)b=ζ(a¯,b¯,c¯) with H¯n(c)=1+12c13c+
n=1(1)n(n+2)an=1Hn(c)(n+1)b=ζ(a¯,b,c) with Hn(c)=+1+12c+13c+
n=11(n+2)an=1Hn(c)(1)(n+1)(n+1)b=ζ(a,b¯,c)
n=11(n+2)an=1H¯n(c)(n+1)b=ζ(a,b,c¯)

As a variant of the Dirichlet eta function we define

ϕ(s)=12(s1)2(s1)ζ(s) with s>1
ϕ(1)=ln2

Reflection formula

The reflection formula ζ(a,b)+ζ(b,a)=ζ(a)ζ(b)ζ(a+b) can be generalized as follows:

ζ(a,b¯)+ζ(b¯,a)=ζ(a)ϕ(b)ϕ(a+b)
ζ(a¯,b)+ζ(b,a¯)=ζ(b)ϕ(a)ϕ(a+b)
ζ(a¯,b¯)+ζ(b¯,a¯)=ϕ(a)ϕ(b)ζ(a+b)

if a=b we have ζ(a¯,a¯)=12[ϕ2(a)ζ(2a)]

Other relations

Using the series definition it is easy to prove:

ζ(a,b)+ζ(a,b¯)+ζ(a¯,b)+ζ(a¯,b¯)=ζ(a,b)2(a+b2) with a>1
ζ(a,b,c)+ζ(a,b,c¯)+ζ(a,b¯,c)+ζ(a¯,b,c)+ζ(a,b¯,c¯)+ζ(a¯,b,c¯)+ζ(a¯,b¯,c)+ζ(a¯,b¯,c¯)=ζ(a,b,c)2(a+b+c3) with a>1

A further useful relation is:[5]

ζ(a,b)+ζ(a¯,b¯)=s>0(a+bs1)![Za(a+bs,s)(as)!(b1)!+Zb(a+bs,s)(bs)!(a1)!]

where Za(s,t)=ζ(s,t)+ζ(s¯,t)[ζ(s,t)+ζ(s+t)]2(s1) and Zb(s,t)=ζ(s,t)2(s1)

Note that s must be used for all value >1 for which the argument of the factorials is 0

Other results

For all positive integers a,b,,k:

n=2ζ(n,k)=ζ(k+1) or more generally:
n=2ζ(n,a,b,,k)=ζ(a+1,b,,k)
n=2ζ(n,k¯)=ϕ(k+1)
n=2ζ(n,a¯,b)=ζ(a+1,b)
n=2ζ(n,a,b¯)=ζ(a+1,b¯)
n=2ζ(n,a¯,b¯)=ζ(a+1,b¯)
limkζ(n,k)=ζ(n)1
1ζ(2)+ζ(3)ζ(4)+=|12|
ζ(a,a)=12[(ζ(a))2ζ(2a)]
ζ(a,a,a)=16(ζ(a))3+13ζ(3a)12ζ(a)ζ(2a)

Mordell–Tornheim zeta values

The Mordell–Tornheim zeta function, introduced by Template:Harvtxt who was motivated by the papers Template:Harvtxt and Template:Harvtxt, is defined by

ζMT,r(s1,,sr;sr+1)=m1,,mr>01m1s1mrsr(m1++mr)sr+1

It is a special case of the Shintani zeta function.

References

Notes

Template:Reflist