Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Multiple Zeta Values of Even Arguments
(and Some Analogous Results)
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Michael E. Hoffman
Guessing the
General
Theorem
Symmetric
Functions
USNA Mathematics Colloquium
12 November 2014
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Outline
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
1 Introduction
2 Why E (2n, k) is a Rational Times π 2n
Outline
Introduction
3 Guessing the General Theorem
Why E (2n, k)
is a Rational
Times π 2n
4 Symmetric Functions
Guessing the
General
Theorem
5 Proving the Formula
Symmetric
Functions
Proving the
Formula
6 Analogues
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Multiple Zeta Values
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
The multiple zeta values (MZVs) are defined by
ζ(i1 , . . . , ik ) =
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
i1
n1 >···>nk ≥1 n1
1
· · · nkik
,
where i1 , . . . , ik are positive integers with i1 > 1. We call k the
depth of the MZV, and i1 + · · · + ik its weight. There are many
remarkable relations among MZVs, starting with ζ(2, 1) = ζ(3)
(which is often rediscovered and posed as a problem, but was
known to Euler). One interesting fact is that all known
relations are homogeneous by weight, but we are very far from
proving this must be true.
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Ancient History
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Euler studied MZVs of depth 1 (which are just the values of
the “Riemann” zeta function) and obtained the famous formula
(−1)n−1 B2n (2π)2n
.
2(2n)!
Somewhat less famously, he also studied MZVs of depth 2, and
of the many formulas he gave two are of interest here. First, he
found that the sum of all the depth-2 MZVs of weight m is just
ζ(m), that is
m−1
ζ(i , m − i ) = ζ(m).
ζ(2n) =
i =2
Second, he gave an alternating sum formula for even weights:
2n−1
Analogues
i =2
1
(−1)i ζ(i , 2n − i ) = ζ(2n).
2
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Recent History
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
When MZVs of arbitrary depth started to be studied around
1988, one of the first results to attract interest was the
generalization of Euler’s sum theorem
ζ(i1 , . . . , ik ) = ζ(m).
i1 ,...,ik =m, i1 >1
This was conjectured by C. Moen (who proved it for k = 3)
and proved in general by A. Granville and D. Zagier. More
recently there has been interest in the sums of MZVs of fixed
weight with even arguments, i.e.,
ζ(2i1 , . . . , 2ik ).
E (2n, k) =
i1 +···+ik =n
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Formulas for Even-Argument Sums
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
If you add Euler’s sum theorem in weight 2n to his alternating
sum theorem in the same weight, you get
ME Hoffman
Outline
E (2n, 2) =
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
3
ζ(2n).
4
(Though implicit in Euler, this result seems to have been first
stated in print by H. Gangl, M. Kaneko and D. Zagier in 2006.)
Now it is true that E (2n, k) is always a rational multiple of
ζ(2n) (as I will shortly explain), but for k > 2 this multiple
depends on the weight 2n. Nevertheless, it is possible to get
formulas with weight-free coefficients if extra terms are allowed.
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
The Shen-Cai Formulas
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
In 2012 Z. Shen and T. Cai published the following formulas
for even-argument sums of depth 3 and 4:
1
5
E (2n, 3) = ζ(2n) − ζ(2)ζ(2n − 2)
8
4
5
35
E (2n, 4) = ζ(2n) − ζ(2)ζ(2n − 2).
64
16
What caught my eye was the coefficient of ζ(2n). Along with
E (2n, 1) = ζ(2n) and E (2n, 2) = 34 ζ(2n) we have a sequence
1,
3
,
4
5
,
8
35
, ...
64
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
The Shen-Cai Formulas
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
In 2012 Z. Shen and T. Cai published the following formulas
for even-argument sums of depth 3 and 4:
1
5
E (2n, 3) = ζ(2n) − ζ(2)ζ(2n − 2)
8
4
5
35
E (2n, 4) = ζ(2n) − ζ(2)ζ(2n − 2).
64
16
What caught my eye was the coefficient of ζ(2n). Along with
E (2n, 1) = ζ(2n) and E (2n, 2) = 34 ζ(2n) we have a sequence
Proving the
Formula
Analogues
3
,
4
1,
Symmetric
Functions
1,
3
1· ,
4
1·
ME Hoffman
5
,
8
3 5
· ,
4 6
35
, ...
64
1·
3 5 7
· · , ...
4 6 8
Multiple Zeta Values of Even Arguments (& Analogues)
Hooked
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Once I saw the formula for the leading coefficient, I was
hooked. I wouldn’t be satisfied until I had a general formula for
E (2n, k). In fact I worked the whole thing out in the ten days
from 21 May 2012 (when I started working in earnest) to 31
May 2012 (when I submitted a preprint to the arXiv). It’s the
fastest I’ve ever written a paper, and it was the excellent
working environment at DESY Zeuthen that enabled me to
concentrate so fully on a single problem. I thank DESY
Zeuthen (and particularly Johannes Blümlein) for providing the
environment, and the DAAD (German Academic Exchange
Service) for providing the funding that made my visit possible.
Vielen Dank!
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Symmetric Sums
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Now I’ll explain why E (2n, k) ∈ Qπ 2n .
Theorem (Hoffman, 1992)
Let i1 , . . . , ik be integers larger than 1. If I = (i1 , . . . , ik ) and
the symmetric group Sk acts on I in the obvious way, then
ζ(σ · I ) =
c(Π)ζ(I , Π),
σ∈Sk
partitions Π of {1, . . . , k}
where for the partition Π = {P1 , . . . , Pt } we define
c(Π) = (−1)k−t (|P1 | − 1)! · · · (|Pt | − 1)!
Proving the
Formula
and ζ(I , Π) =
Analogues
t
s=1
ME Hoffman
⎛
ζ⎝
⎞
ij ⎠ .
j∈Ps
Multiple Zeta Values of Even Arguments (& Analogues)
Symmetric Sums cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
This theorem shows that any symmetric sum of MZVs (i.e.,
invariant under permutation of the exponent string) is a
rational linear combination of products of ordinary MZVs.
Thus ζ(6, 2) + ζ(2, 6) = ζ(6)ζ(2) − ζ(8), but it is unknown if
ζ(6, 2) is such a linear combination. In my 1992 paper I used
the theorem to prove that
π 2n
.
E (2n, n) = ζ(2, 2, . . . , 2) =
(2n + 1)!
n
Clearly any E (2n, k) is a symmetric sum, and although the
details look messy it must be linear combination of products of
even zetas homogeneous by weight, and so a rational multiple
of π 2n (hence of ζ(2n)).
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
But Which Rational Multiple?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
The problem with using my theorem to actually compute
things is that it’s a sum over set partitions, which increase
according to Bell numbers
1, 2, 5, 15, 52, 203, 877, 4140 . . .
Further, the E (2n, k) get to be more complicated symmetric
sums as k gets smaller than n:
E (2n, n−1) = ζ(4, 2, . . . , 2)+ζ(2, 4, 2, . . . , 2)+· · ·+ζ(2, . . . , 2, 4)
E (2n, n − 2) = ζ(4, 4, 2, . . . , 2) + · · · + ζ(2, 2, . . . , 2, 4, 4)+
ζ(6, 2, . . . , 2) + · · · + ζ(2, . . . , 2, 6)
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
But Which Rational Multiple? cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
For example, it takes a fairly heroic calculation to get E (14, 5)
directly from my theorem: one has to add two sums of 52
products each, which eventually reduces to
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
1
3
1
1
ζ(2)4 ζ(6) + ζ(2)3 ζ(4)2 − ζ(2)2 ζ(4)ζ(6) − ζ(2)3 ζ(8)
24
12
4
4
19
7
5
1
− ζ(2)ζ(4)3 + ζ(2)ζ(4)ζ(8) + ζ(4)2 ζ(6) + ζ(2)ζ(6)2
4
4
24
6
5
7
2
+ ζ(2) ζ(10) − ζ(6)ζ(8) − 2ζ(4)ζ(10) − ζ(2)ζ(12) + 3ζ(14)
4
2
π 14
.
=
29189160000
Fortunately I didn’t persist in this foolishness for too long.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
But Which Rational Multiple? cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
By May 23rd I could prove that
E (2k, k) =
ME Hoffman
Outline
π 2k
(2k + 1)!
has some only slightly more complicated cousins, e.g.,
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
kπ 2k+2
3(2k + 1)!(2k + 3)
k(7k + 13)π 2k+4
,
E (2k + 4, k) =
90(2k + 1)!(2k + 3)(2k + 5)
E (2k + 2, k) =
Symmetric
Functions
Proving the
Formula
Analogues
the latter of which makes the calculation of E (14, 5) much
faster. I’ll say where these formulas came from later, but first
I’ll explain what I did with them.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Guessing the General Theorem
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
My goal was to come up with extensions of the Shen-Cai
formulas. Following the pattern of the leading coefficients, I
decided the formula for k = 5 should look like
E (2n, 5) =
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
63
ζ(2n) + aζ(2)ζ(n − 2) + bζ(4)ζ(n − 4)
128
for constants a and b. Setting n = 5 and n = 6 gave me two
equations in two unknowns, which I solved to get a = − 21
64 and
3
. So my guess was now
b = 64
E (2n, 5) =
21
3
63
ζ(2n) − ζ(2)ζ(n − 2) + ζ(4)ζ(n − 4),
128
64
64
which I tested at n = 7: this is why I wanted to know E (14, 5).
The formula passed the test, so I worked out a conjectural
formula for E (2n, 6) in a similar way.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Accumulating Data
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
I now assumed that the general formula looked like
E (2n, k) =
ME Hoffman
k−1 1
22(k−1)
2
2k − 1
ck,j ζ(2j)ζ(2n − 2j),
ζ(2n) +
k
j=1
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
and by solving systems with k close to n I got the following
coefficients ck,j (starting with k = 3)
k
k
k
k
k
k
= 3 : c3,1 = − 14
5
= 4 : c4,1 = − 16
= 5 : c5,1 = − 21
64
= 6 : c6,1 = − 21
64
= 7 : c7,1 = − 165
512
= 8 : c8,1 = − 1287
4096
ME Hoffman
3
c5,2 = 64
21
c6,2 = 256
27
c7,2 = 256
495
c8,2 = 4096
3
c7,3 = − 1024
27
c8,3 = − 4096
Multiple Zeta Values of Even Arguments (& Analogues)
Guessing Coefficients
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
All these coefficients are consistent with the formulas
22(k−2)
ck,2 =
22(k−2)
ck,3 =
22(k−2)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
−1
ck,1 =
3
−3
2k − 3
k
2k − 5
k
2k − 7
k
so I decided that
ck,j =
aj
2(k−2)
2
2k − 2j − 1
k
for some sequence a1 , a2 , . . . .
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
What is the Next Term of this Sequence?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now I pressed ahead, just solving for aj by using the equation
E (2k, k) = π 2k /(2k + 1)! repeatedly. So my sequence was
a1 = −1, a2 = 3, a3 = −3,
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
What is the Next Term of this Sequence?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now I pressed ahead, just solving for aj by using the equation
E (2k, k) = π 2k /(2k + 1)! repeatedly. So my sequence was
a1 = −1, a2 = 3, a3 = −3, a4 =
5
,
3
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
What is the Next Term of this Sequence?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now I pressed ahead, just solving for aj by using the equation
E (2k, k) = π 2k /(2k + 1)! repeatedly. So my sequence was
a1 = −1, a2 = 3, a3 = −3, a4 =
5
3
, a5 = − ,
3
5
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
What is the Next Term of this Sequence?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now I pressed ahead, just solving for aj by using the equation
E (2k, k) = π 2k /(2k + 1)! repeatedly. So my sequence was
a1 = −1, a2 = 3, a3 = −3, a4 =
5
3
105
, a5 = − , a6 =
3
5
691
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
What is the Next Term of this Sequence?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now I pressed ahead, just solving for aj by using the equation
E (2k, k) = π 2k /(2k + 1)! repeatedly. So my sequence was
a1 = −1, a2 = 3, a3 = −3, a4 =
5
3
105
, a5 = − , a6 =
3
5
691
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
It did not escape my notice that 691 is the numerator of the
Bernoulli number B12 . In fact all the data was consistent with
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
What is the Next Term of this Sequence?
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now I pressed ahead, just solving for aj by using the equation
E (2k, k) = π 2k /(2k + 1)! repeatedly. So my sequence was
a1 = −1, a2 = 3, a3 = −3, a4 =
5
3
105
, a5 = − , a6 =
3
5
691
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
It did not escape my notice that 691 is the numerator of the
Bernoulli number B12 . In fact all the data was consistent with
aj =
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
and so
ck,j =
−1
(4j + 2)B2j
−1
2k − 2j − 1
.
+ 1)B2j
k
22k−3 (2j
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
The Formula at Last
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
On May 26th I had my formula:
E (2n, k) =
ζ(2n) 2k − 1
22k−2
k
k−1
2
Outline
−
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
j=1
Since analytic continuation gives ζ(0) = − 12 , this can be
written
Symmetric
Functions
Proving the
Formula
Analogues
ζ(2j)ζ(2n − 2j) 2k − 2j − 1
.
22k−3 (2j + 1)B2j
k
k−1
2
E (2n, k) = −
ζ(2j)ζ(2n − 2j) 2k − 2j − 1
.
22k−3 (2j + 1)B2j
k
j=0
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Symmetric Functions
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Proving this formula (which took me four more days) was a
quite different exercise from discovering it. It involved a detour
through the algebra Sym of symmetric functions. Here are
some definitions.
Let x1 , x2 , . . . all have degree 1, and let P ⊂ Q[[x1 , x2 , . . . ]] be
the set of formal power series in the xi of bounded degree.
Then P is a graded Q-algebra. An element f ∈ P is symmetric
if the coefficient of any term
xin11 · · · xinkk
Guessing the
General
Theorem
Symmetric
Functions
with i1 , . . . , ik distinct, agrees with that of
x1n1 · · · xknk .
Proving the
Formula
Analogues
The set of such f forms an algebra Sym ⊂ P.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Bases for Sym
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
For any integer partition λ = (λ1 , λ2 , . . . , λk ), the monomial
symmetric function mλ is the “smallest” symmetric function
containing the monomomial x1λ1 x2λ2 · · · xkλk . For example,
m21 = x12 x2 + x12 x3 + · · · + x22 x3 + · · · + x1 x22 + x1 x32 + · · ·
The set {mλ |λ is a partition} is an integral basis for Sym.
The elementary symmetric functions are
ek = m(1k ) ,
where (1k ) is the partition of k into 1’s.
The complete symmetric functions are
mλ .
hk =
λ is a partition of k
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Bases for Sym cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
The power-sum symmetric functions are
pk = m(k) .
If for a partition λ = (λ1 , λ2 , . . . , λk ) we let
Outline
eλ = eλ1 eλ2 · · · eλk
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
and similarly hλ and pλ , then it is well known that
{eλ |λ is a partition} and {hλ |λ is a partition}
are integral bases for Sym, while
Proving the
Formula
{pλ |λ is a partition}
Analogues
is a rational basis.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Generating Functions
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
The generating function of the elementary symmetric
functions is
ei t i =
(1 + txi ),
E (t) =
i ≥0
ME Hoffman
Outline
while that for the complete symmetric functions is
Introduction
H(t) =
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
i ≥1
hi t i =
i ≥0
i ≥1
1
= E (−t)−1 .
1 − txi
The logarithmic derivative of the latter is the generating
function P(t) for the power sums:
P(t) =
d
d
log H(t) = −
log(1 − txi ) =
pi t i −1 .
dt
dt
i ≥1
Analogues
ME Hoffman
i ≥1
Multiple Zeta Values of Even Arguments (& Analogues)
Even MZVs as Homomorphic Images
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
There is a homomorphism Z : Sym → R induced by sending xi
to 1/i 2 . Then
ME Hoffman
Z(pk ) = ζ(2k),
Outline
Z(ek ) = E (2k, k) =
π 2k
(2k + 1)!
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
and
Z(hk ) =
k
E (2k, i ).
i =1
For k ≤ n, we define symmetric functions Nn,k so that
Z(Nn,k ) = E (2n, k): that is, Nn,k is the sum of all monomial
symmetric functions corresponding to partitions of n having
length k.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Recurrences
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Now one of the first things I proved about the Nn,k was the
recurrence
p1 Nn−1,k +p2 Nn−2,k +· · ·+pn−k Nk,k = (n−k)Nn,k +(k+1)Nn,k+1 ,
which somewhat resembles the standard recurrence
Introduction
Why E (2n, k)
is a Rational
Times π 2n
p1 hn−1 + p2 hn−2 + · · · + pn−1 h1 + pn = nhn
Guessing the
General
Theorem
of power-sum and complete symmetric functions. In particular,
p1 Nk,k = Nk+1,k + (k + 1)Nk+1,k+1 ,
Symmetric
Functions
Proving the
Formula
Analogues
or
Nk+1,k = p1 Nk,k − (k + 1)Nk+1,k+1 .
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Recurrences cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
Apply Z to the latter equation to get
E (2k + 2, k) = ζ(2)E (2k, k) − (k + 1)E (2k + 2, k + 1)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
leading to the formula
(k + 1)π 2k+2
π 2m+2
−
6(2k + 1)!
(2k + 3)!
2k+2
kπ
=
3(2k + 1)!(2k + 3)
E (2k + 2, k) =
we saw earlier; the formula for E (2k + 4, k) arises in a similar
way. In fact, the desire for such formulas is what led me to
prove the recurrence.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Generating Function for the Nn,k
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
It’s actually easy to write the generating function
Nn,k t n s k ,
F(t, s) = 1 +
n≥k≥1
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
if we remember that the power of t keeps track of degree,
while s must keep track of the number of distinct xi ’s involved
in a symmetric function:
F(t, s) =
i =1
Symmetric
Functions
Proving the
Formula
Analogues
∞
(1 + stxi + st 2 xi2 + st 3 xi3 + · · · )
=
∞
1 + (s − 1)txi
i =1
1 − txi
ME Hoffman
= H(t)E ((s − 1)t).
Multiple Zeta Values of Even Arguments (& Analogues)
Generating Function for the E (2n, k)
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
Now let’s apply Z to this result: if
E (2n, k)t n s k
F (t, s) = 1 +
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
n≥k≥1
is the generating function for the E (2n, k), we have
F (t, s) = Z(F(t, s)) = Z(H(t))Z(E ((s − 1)t)
From our formula for Z(ei ) we have
Z(E (t)) =
∞
i =0
√
sinh(π t)
π 2i t i
√
.
=
(2i + 1)!
π t
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Generating Function for the E (2n, k) cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
But H(t) = E (−t)−1 , so
√
√
π t
π −t
√
√
=
Z(H(t)) =
sinh π −t
sin π t
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
and thus
F (t, s) =Z(H(t))Z(E (−(1 − s)t))
√
√
√
π t sin(π 1 − s t)
√
√
√
=
sin π t π 1 − s t
√ √
sin(π t 1 − s)
√ .
=√
1 − s sin π t
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Using the Generating Function
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Armed with the generating function, one can use any computer
algebra system that does series (like PARI or Maple) to spit out
as many of the E (2n, k) as desired. This makes it easy to
compute not only
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
E (14, 5) =
π 14
29189160000
but also, say,
E (28, 9) =
19697π 28
.
142327470280408148736000000
(In practice one finds that the hardest thing is finding the right
term in the output and reading out all the digits correctly.)
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Expanding out the Generating Function
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
There’s still the problem of proving the conjecture. We want a
formula for E (2n, k) for fixed k, which means we want to look
at the coefficient of s k in F (t, s). Writing
s k Gk (t),
F (t, s) = 1 +
k≥1
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
then it is easy to see from the explicit formula that
√
(−1)j π 2j t j j
k π t
√
Gk (t) =(−1)
sin π t j≥k (2j + 1)! k
√
√
k
k
sin π t
k π t (−t) d
√
√
=(−1)
sin π t k! dt k
π t
.
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
A Little Calculus Problem
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
So we have the small
problem of finding the nth
√
√ calculus
derivative of sin(π t)/π t. Evidently
√
√
√
(−t)n d n sin π t
2
2 sin π t
√
= Pn (π t) cos π t + Qn (π t) √
n! dt n
π t
π t
Outline
Introduction
for some polynomials Pn , Qn .
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
A Little Calculus Problem
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
So we have the small
problem of finding the nth
√
√ calculus
derivative of sin(π t)/π t. Evidently
√
√
√
(−t)n d n sin π t
2
2 sin π t
√
= Pn (π t) cos π t + Qn (π t) √
n! dt n
π t
π t
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
for some polynomials Pn , Qn . With a little help from Maple I
found that
n−1
2
Pn (x) = −
j=0
Symmetric
Functions
Proving the
Formula
Analogues
2n − 2j − 1
(−4x)j
2n−1
2
(2j + 1)!
n
n2 Qn (x) =
(−4x)j 2n − 2j
.
22n (2j)!
n
j=0
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Expanding out the Generating Function
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Making use of our calculus exercise, we can now write
k−1
2
Gk (t) = −
j=0
√
2k − 2j − 1 √
(−4π 2 t)j
π
t
cot
π
t
22k−1 (2j + 1)!
k
k
2
(−4π 2 t)j 2k − 2j
+
22k (2j)!
k
j=0
k−1 2
√
√ 2k − 2j − 1
(−4π 2 t)j
1
= (1 − π t cot π t)
2
22k−2 (2j + 1)!
k
j=0
Proving the
Formula
+ terms of degree < k.
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Restating the Conjecture
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Now going back to the statement of our conjecture, we can
write it a bit more simply if we use Euler’s formula for ζ(2j);
this cancels the Bernoulli number in the denominator, leaving
us with the cleaner version
k−1
2
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
E (2n, k) =
(−1)j π 2j ζ(2n − 2j) 2k − 2j − 1
.
22k−2j−2 (2j + 1)
k
j=0
Thus, we will be done if we can show Gk (t) is equal to
n≥k
tk
k−1
2
(−1)j π 2j ζ(2n − 2j) 2k − 2j − 1
.
22k−2j−2 (2j + 1)
k
j=0
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Restating the Conjecture cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
Write the latter sum in the form
k−1
2
j=0
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
2k − 2j − 1 (−4π 2 t)j
ζ(2n − 2j)t n−j ,
22k−2 (2j + 1)!
k
n≥k
and note that
ζ(2n − 2j)t n−j =
ζ(2n − 2j)t n−j −
n≥j+1
n≥k
=
m≥2
k−1
ζ(2n − 2j)t n−j
n=j+1
ζ(m)t m −
k−1
ζ(2n − 2j)t n−j
n=j+1
k−1
√
√
1
= (1 − π t cot π t) −
ζ(2n − 2j)t n−j .
2
n=j+1
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Conclusion
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Hence the conjecture reads
k−1
2 √
√ 2k − 2j − 1
1
(−4π 2 t)j
Gk (t) = (1−π t cot π t)
2k−2
2
2
(2j + 1)!
k
j=0
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
+ terms of degree < k,
where the “terms of degree < k” are exactly those necessary to
cancel the terms of degree < k in the preceeding sum. But this
agrees with the result we got by expanding out the generating
function, so we’re done.
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Another Formula for E (2n, k)
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
We can get another sum for E (2n, k) as follows. Using the
recurrence mentioned earlier, we can obtain an explicit formula
for Nn,k = |π|=n, length(π)=k mπ in terms of elementary and
complete symmetric functions:
Outline
Nn,k =
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
n−k
i =0
n−i
(−1)n−k−i hi en−i .
k
Apply Z to get
E (2n, k) =
n−k
(−1)n−k−1 π 2n n − i
(2n + 1)!
k
i =0
2n + 1
2(22i −1 −1)B2i .
2i
Notice that this has n − k + 1 terms rather than k−1
2 +1
terms as the other formula does.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Imitation is the Sincerest Form of Flattery
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
It wasn’t long after I posted my paper on the arXiv that the
industrious Jianqing Zhao produced an analogue. (His preprint
was posted on July 10.) Define the multiple t-value
t(i1 , . . . , ik ) to be the sum of all the terms in ζ(i1 , . . . , ik ) with
odd denominators. Then t(n) = (1 − 2−n )ζ(n) for n ≥ 2. We
can define
t(2i1 , . . . , 2ik ).
T (2n, k) =
i1 +···+ik =n
Zhao showed that
Symmetric
Functions
Proving the
Formula
Analogues
k−1
2
T (2n, k) =
(−1)j π 2j t(2n − 2j) 2k − 2j − 2
.
22k−2 (2j)!k
k −1
j=0
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Flattery cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
Zhao deduces his formula from the generating function
cos π2 (1 − s)t
√ ,
T (2n, k)t n s k =
1+
cos π2 t
n≥k≥1
(1)
whose proof is entirely analogous to that for
sin(π (1 − s)t)
n k
√ .
E (2n, d )t s = √
1+
1 − s sin(π t)
n≥k≥1
1
The homomorphism T : Sym → R sending xi to (2i −1)
2 sends
pk to t(2k), from which it is easy to see that
π √ π√
t tan
t .
T(P(t)) =
4
2
From this one finds T(H(t)) and T(E (t)), and equation (1)
follows by applying T to F(s, t) = H(t)E ((s − 1)t).
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Another Extension
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Not content with the t-value extension, Zhao wrote another
paper jointly with H. Yuan examining the sum of multiple zeta
values whose arguments are multiples of 4. Define
ζ(4i1 , 4i2 , . . . , 4ik ).
Q(4n, k) =
i1 +···+ik =n
The generating function
G (t, s) = 1 +
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
Q(4n, k)t n s k
n≥k≥1
is actually not hard to obtain: it can be written
sin π 4 (1 − s)t sinh π 4 (1 − s)t
√ √ √
.
G (t, s) =
1 − s sin π 4 t sinh π 4 t
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Another Extension cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
The formula for G (t, s) is again obtained by applying an
appropriate homomorphism to the identity
F(t, s) = H(t)E ((s − 1)t).
Here we need Q : Sym → R defined by sending sending xi to
1
. Then it is well known that
i4
√
√
sinh(π 4 t) sin(π 4 t)
√
,
Q(E (−t)) =
π2 t
from which the formula for G (t, s) follows. The hard part of
the Yuan-Zhao result is their formula for Q(4n, k), which is not
easy to fit on a slide. Their version of the “calculus exercise”
required to prove it requires four functions rather than two.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Another Extension cont’d
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
ME Hoffman
Nevertheless, the first few cases are rather nice. First,
n−1
Q(4n, 2) =
k=1
Outline
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
n−1
1
ζ(4n).
ζ(4k)ζ(4n − 4k) −
2
2
Using this, one has two formulas reminiscent of those proved
by Shen and Cai:
1
5
5
ζ(4n) + ζ(4n − 2)ζ(2) + Q(4n, 2)
16
8
4
13
85
115
ζ(4n) + ζ(4n − 2)ζ(2) + Q(4n, 2).
Q(4n, 4) = −
64
64
64
Q(4n, 3) = −
Analogues
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)
Further Extensions
Multiple Zeta
Values of
Even
Arguments
(&
Analogues)
Further extensions are certainly possible. For example, if we let
ζ(6i1 , . . . , 6ik ),
S(6n, k) =
i1 +···+ik =n
ME Hoffman
Outline
then from known results it is easy to show that
Introduction
Why E (2n, k)
is a Rational
Times π 2n
Guessing the
General
Theorem
Symmetric
Functions
Proving the
Formula
Analogues
1+
S(6n, k)t n s k =
n≥k≥1
sin(π
6
√ t(1 − s))(cosh(π 3 6 t(1 − s)) − cos(π 6 t(1 − s)))
√ √
.
√
√
√
1 − s sin(π 6 t)(cosh(π 3 6 t) − cos(π 6 t))
But the complexity of the explicit formula Zhao and Yuan got
for Q(4n, k) suggests that an explicit formula for S(6n, k)
would be a real mess.
ME Hoffman
Multiple Zeta Values of Even Arguments (& Analogues)