On multiple zeta values of even arguments

arXiv:1205.7051v2 [math.NT] 8 Jun 2012

DESY 12-091

arXiv:1205.7051 [math.NT]

On Multiple Zeta Values of Even Arguments Michael E. Hoffman1 Dept. of Mathematics, U. S. Naval Academy Annapolis, MD 21402 USA and Deutsches Elektronen-Synchrotron DESY Platanenalle 6, D-15738 Zeuthen, Germany [email protected] June 8, 2012 Keywords: multiple zeta values, symmetric functions, Bernoulli numbers MR Classifications: Primary 11M32; Secondary 05E05, 11B68

Abstract For k ≤ n, let E(2n, k) be the sum of all multiple zeta values with even arguments whose weight is 2n and whose depth is k. Of course E(2n, 1) is the value ζ(2n) of the Riemann zeta function at 2n, and it is well known that E(2n, 2) = 34 ζ(2n). Recently Z. Shen and T. Cai gave formulas for E(2n, 3) and E(2n, 4) in terms ζ(2n) and ζ(2)ζ(2n − 2). We give two formulas for E(2n, k), both valid for arbitrary k ≤ n, one of which generalizes the Shen-Cai results; by comparing the two we obtain a Bernoulli-number identity. We also give an explicit generating function for the numbers E(2n, k).

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Supported by a grant from the German Academic Exchange Service (DAAD) during the preparation of this paper. The author also thanks DESY for providing facilities and financial support for travel.

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Introduction and Statement of Results

For positive integers i1 , . . . , ik with i1 > 1, we define the multiple zeta value ζ(i1 , . . . , ik ) by X 1 ζ(i1, . . . , ik ) = (1) ik . i1 n · · · n 1 k n1 >···>nk ≥1 The multiple zeta value (1) is said to have weight i1 + · · · + ik and depth k. Many remarkable identities have been proved about these numbers, but in this note we will concentrate on the case where the ij are even integers. Let E(2n, k) be the sum of all the multiple zeta values of even-integer arguments having weight 2n and depth k, i.e., X E(2n, k) = ζ(i1 , . . . , ik ). i1 , . . . , ik even i1 +···+ik =2n

Of course

(−1)n−1 B2n (2π)2n , (2) 2(2n)! where B2n is the 2nth Bernoulli number, by the classical formula of Euler. Euler also studied double zeta values (i.e., multiple zeta values of depth 2) and in his paper [2] gave two identities which read E(2n, 1) = ζ(2n) =

2n−1 X

1 (−1)i ζ(i, 2n − i) = ζ(2n) 2 i=2 2n−1 X i=2

ζ(i, 2n − i) = ζ(2n)

in modern notation. From these it follows that 3 E(2n, 2) = ζ(2n), 4 though Gangl, Kaneko and Zagier [3] seem to be the first to have pointed it out in print. Recently Shen and Cai [10] proved the formulas 1 5 E(2n, 3) = ζ(2n) − ζ(2)ζ(2n − 2), n ≥ 3 8 4 35 5 E(2n, 4) = ζ(2n) − ζ(2)ζ(2n − 2), n ≥ 4. 64 16 1

(3) (4)

Identity (3) was also proved by Machide [9] using a different method. This begs the question whether there is a general formula of this type for E(2n, k). The pattern 3 , 4

3 5 5 · = , 4 6 8

3 5 7 35 · · = 4 6 8 64

of the leading coefficients makes one curious. In fact, the general result is as follows. Theorem 1. For k ≤ n,   1 2k − 1 ζ(2n) E(2n, k) = 2(k−1) k 2 ⌋ ⌊ k−1 2

−

X j=1

  1 2k − 2j − 1 ζ(2j)ζ(2n − 2j). k 22k−3 (2j + 1)B2j

The next two cases after (4) are 63 ζ(2n) − 128 231 ζ(2n) − E(2n, 6) = 512 E(2n, 5) =

21 ζ(2)ζ(2n − 2) + 64 21 ζ(2)ζ(2n − 2) + 64

3 ζ(4)ζ(2n − 4) 64 21 ζ(4)ζ(2n − 4). 256

We prove Theorem 1 in §3 below, using the generating function X F (t, s) = 1 + E(2n, k)tn sk . n≥k≥1

In §2 we establish the following explicit formula. Theorem 2.

√ √ sin(π 1 − s t) √ . F (t, s) = √ 1 − s sin(π t)

Our proof uses symmetric functions. We define a homomorphism Z : Sym → R, where Sym is the algebra of symmetric functions, and a family Nn,k ∈ Sym such that Z sends Nn,k to E(2n, k). We then obtain a formula for the generating functions X Nn,k tn sk ∈ Sym[[t, s]] F(t, s) = 1 + n≥k≥1

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and apply Z to get Theorem 2. From the form of F(t, s) we show that it satisfies a partial differential equation (Proposition 1 below), which is equivalent to a recurrence for the Nn,k . From the latter we obtain a formula for the Nn,k in terms of complete and elementary symmetric functions, to which Z can be applied to give the following alternative formula for E(2n, k). Theorem 3. For k ≤ n,   n−k  (−1)n−k−1π 2n X n − i 2n + 1 E(2n, k) = 2(22i−1 − 1)B2i . 2i k (2n + 1)! i=0 Note that the sum given by Theorem 3 has n−k+1 terms, while that given by Theorem 1 has ⌊ k−1 ⌋ + 1 terms. Yet another explicit formula for E(2n, k) 2 can be obtained by setting d = 1 in Theorem 7.1 of Komori, Matsumoto and Tsumura [7]. That formula expresses E(2n, k) as a sum over partitions of k, and it is not immediately clear how it relates to our two formulas. Comparison of Theorems 1 and 3 establishes the following Bernoullinumber identity. Theorem 4. For k ≤ n, ⌋ ⌊ k−1 2

X i=0

2k − 2i − 1 k

  2n + 1 B2n−2i = 2i + 1 k 2k−2n

(−1) 2

  n−k  X n − i 2n + 1 (22i−1 − 1)B2i . 2i k i=0

It is interesting to contrast this result with the Gessel-Viennot identity (see [1, Theorem 4.2]) valid on the complementary range: ⌊ k−1 ⌋ 2

X i=0

2k − 2i − 1 k



   2n + 1 2k − 2n 2n + 1 , B2n−2i = k 2 2i + 1

k > n.

Note that the right-hand side of equation (5) is zero unless k ≥ 2n.

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(5)

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Symmetric Functions

We think of Sym as the subring of Q[[x1 , x2 , . . . ]] consisting of those formal power series of bounded degree that are invariant under permutations of the xi . A useful reference is the first chapter of Macdonald [8]. We denote the elementary, complete, and power-sum symmetric functions of degree i by ei , hi , and pi respectively. They have associated generating functions E(t) = H(t) = P (t) =

∞ X

j=0 ∞ X

j=0 ∞ X

j

ej t = hj tj =

∞ Y

(1 + txi )

i=1 ∞ Y

1 = E(−t)−1 1 − txi

i=1 ∞ X

pj tj−1 =

j=1

i=1

H ′ (t) xi = . 1 − txi H(t)

As explained in [5] and in greater detail in [6], there is a homomorphism ζ : Sym0 → R, where Sym0 is the subalgebra of Sym generated by p2 , p3 , p4 , . . . , such that ζ(pi ) is the value ζ(i) of the Riemann zeta function at i, for i ≥ 2 (in [5, 6] this homomorphism is extended to all of Sym, but we do not need the extension here). Let D : Sym → Sym be the degree-doubling map that sends xi to x2i . Then D(Sym) ⊂ Sym0 , so the composition Z = ζD is defined on all of Sym. (Alternatively, we can simply think of Z as sending xi to 1/i2 : see [8, Ch. I, §2, ex. 21].) Note that Z(pi ) = ζ(2i) ∈ R. Further, Z sends the monomial symmetric function mi1 ,...,ik to the symmetrized sum of multiple zeta values X 1 ζ(2iσ(1) , 2iσ(2) , . . . , 2iσ(k) ), | Iso(i1 , . . . , ik )| σ∈S k

where Sk is the symmetric group on k letters and Iso(i1 , . . . , ik ) is the subgroup of Sk that fixes (i1 , . . . , ik ) under the obvious action. Now let Nn,k be the sum of all the monomial symmetric functions corresponding to partitions of n having length k. Of course Nn,k = 0 unless k ≤ n, and Nk,k = ek . Then Z sends Nn,k to E(2n, k). Also, if we define (as in the introduction) X F(t, s) = 1 + Nn,k tn sk , n≥k≥1

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then Z sends F(t, s) to the generating function F (t, s). We have the following simple description of F(t, s). Lemma 1. F(t, s) = E((s − 1)t)H(t). Proof. Evidently F(t, s) has the formal factorization ∞ ∞ Y Y 1 + (s − 1)txi 2 2 (1 + stxi + st xi + · · · ) = = E((s − 1)t)H(t). 1 − txi i=1 i=1

Proof of Theorem 2. Using the well-known formula for ζ(2, 2, . . . , 2) [4, Cor. 2.3], π 2n Z(en ) = ζ(2, 2, . . . , 2) = . (6) | {z } (2n + 1)! n

Hence

√ sinh(π t) √ , Z(E(t)) = π t

and the image of H(t) = E(−t)−1 is √ √ π −t π t √ √ . Z(H(t)) = = sinh(π −t) sin(π t) Thus from Lemma 1 F (t, s) = Z(F(t, s)) is p p √ sinh(π (s − 1)t) π t sin(π (1 − s)t) √ . p √ =√ Z(E((s − 1)t)H(t)) = 1 − s sin(π t) π (s − 1)t sin(π t) Taking limits as s → 1 in Theorem 2, we obtain √ π t √ F (t, 1) = sin π t and so, taking the coefficient of tn , the following result.

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Corollary 1. For all n ≥ 1, n X

E(2n, k) =

k=1

2(22n−1 − 1)(−1)n−1 B2n π 2n . (2n)!

Another consequence of Lemma 1 is the following partial differential equation. Proposition 1. t

∂F ∂F (t, s) + (1 − s) (t, s) = tP (t)F(t, s). ∂t ∂s

Proof. From Lemma 1 we have ∂F (t, s) = (s − 1)E ′ ((s − 1)t)H(t) + E((s − 1)t)H ′(t) ∂t ∂F (t, s) = tE ′ ((s − 1)t)H(t) ∂s from which the conclusion follows. Now examine the coefficient of tn sk in Proposition 1 to get the following. Proposition 2. For n ≥ k + 1, p1 Nn−1,k + p2 Nn−2,k + · · · + pn−k Nk,k = (n − k)Nn,k + (k + 1)Nn,k+1. It is also possible to prove this result directly via a counting argument like that used to prove the lemma of [6, p. 16]. The preceding result allows us to write Nn,k explicitly in terms of complete and elementary symmetric functions as follows. Lemma 2. For r ≥ 0, Nk+r,k

  r X i k+i hr−i ek+i . = (−1) i i=0

Proof. We use induction on r, the result being evident for r = 0. Proposition 2 gives r+1 X

pi Nk+r+1−i,k = (r + 1)Nk+r+1,k + (k + 1)Nk+r+1,k+1,

i=1

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which after application of the induction hypothesis becomes r+1 r+1−j X X i=1

j=0

  k+j hr+1−i−j Nk+j,k+j = (−1) pi j  r  X k+1+j hr−j Nk+1+j,k+1+j . (r + 1)Nk+r+1,k + (k + 1) j j=0 j

The latter equation can be rewritten   r+1−j X k+j pi hr+1−i−j = Nk+j,k+j (−1) j i=1 j=0

r X

j

  k+j hr+1−j Nk+j,k+j . (r + 1)Nk+r+1,k − (k + 1) (−1) j−1 j=1 r+1 X

j

Now the inner sum on the left-hand side is (r +1−j)hr+1−j by the recurrence relating the complete and power-sum symmetric functions, so we have (r + 1)Nk+r+1,k − (r + 1)Nk,k hr+1 =      r+1 X k+j k+j j , + (k + 1) (−1) hr+1−j Nk+j,k+j (r + 1 − j) j−1 j j=1 and the conclusion follows after the observation that (k + 1) Proof of Theorem 3. Rewrite Lemma 2 in the form Nn,k =

 n−k  X n−i i=0

k

(−1)n−k−i hi en−i

and apply the homomorphism Z, using equation (6) and Z(hi ) =

2(22i−1 − 1)(−1)i−1 B2i π 2i . (2i)!

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k+j j−1




=j

k+j j




.

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Proof of Theorems 1 and 4

From the introduction we recall the statement of Theorem 1:   1 2k − 1 ζ(2n) E(2n, k) = 2(k−1) k 2 ⌊ k−1 ⌋   2 X 2k − 2j − 1 1 − ζ(2j)ζ(2n − 2j). 2k−3 (2j + 1)B k 2 2j j=1 We note that Euler’s formula (2) can be used to write the result in the alternative form ⌊ k−1 ⌋ 2

X (−1)j π 2j ζ(2n − 2j) 2k − 2j − 1 E(2n, k) = k 22k−2j−2 (2j + 1)! j=0

(7)

which avoids mention of Bernoulli numbers. We now expand out the generating function F (t, s). We have F (t, s) = √

where

√√ 1 √ sin(π t 1 − s) 1 − s sin π t √ ∞ ∞ X π t X (−1)j π 2j tj (1 − s)j √ = = sk Gk (t), (2j + 1)! sin π t j=0 k=0 √   t X (−1)j π 2j tj j π k √ . Gk (t) = (−1) sin π t j≥k (2j + 1)! k

Then Theorem 1 is equivalent to the statement that ⌊ k−1 ⌋ 2

X (−1)j π 2j ζ(2n − 2j) 2k − 2j − 1 Gk (t) = t k 22k−2j−2(2j + 1)! j=0 n≥k X

n

8

(8)

for all k. We can write the latter sum as ⌊ k−1 ⌋ 2

X j=0

  2k − 2j − 1 X (−4π 2 t)j ζ(2n − 2j)tn−j − k 22k−2 (2j + 1)! n≥j+1 ⌊ k−1 ⌋ 2

X j=0

 k−1  2k − 2j − 1 X (−4π 2 t)j ζ(2n − 2j)tn−j = k 22k−2 (2j + 1)! n=j+1 k−1

⌊ 2 ⌋   √ √ X 1 2k − 2j − 1 (−4π 2 t)j − (1 − π t cot π t) 2k−2 (2j + 1)! k 2 2 j=0 ⌊ k−1 ⌋ 2

X j=0

 k−1  2k − 2j − 1 X (−4π 2 t)j ζ(2n − 2j)tn−j , (9) k 22k−2 (2j + 1)! n=j+1

where we have used the generating function ∞

X √ √ 1 (1 − π t cot π t) = ζ(2i)ti . 2 i=1 Note that the last sum in (9) is a polynomial that cancels exactly those terms in ⌋ ⌊ k−1   2 √ √ X 2k − 2j − 1 (−4π 2 t)j 1 (10) (1 − π t cot π t) k 2 22k−2 (2j + 1)! j=0 of degree less than k. Thus, to prove Theorem 1 it suffices to show that Gk (t) = terms of degree ≥ k in expression (10). From equation (8) it is evident that √ √   π t (−t)k dk sin π t √ · √ Gk (t) = . · k k! dt sin π t π t We use this to obtain an explicit formula for Gk (t). Lemma 3. For k ≥ 0,

√ √ Gk (t) = Pk (π 2 t)π t cot π t + Qk (π 2 t), 9

(11)

where Pk , Qk are polynomials defined by ⌊ k−1 ⌋ 2

X

Pk (x) = −

j=0

  2k − 2j − 1 (−4x)j k 22k−1 (2j + 1)!

⌊ k2 ⌋

X (−4x)j 2k − 2j  Qk (x) = . k 22k (2j)! j=0 Proof. In view of equation (11), the conclusion is equivalent to √ f (k) (t) = (−1)k k!t−k Pk (π 2 t) cos π t + (−1)k k!t−k Qk (π 2 t)f (t), √ √ where f (t) = sin π t/π t. Differentiating, one sees that the polynomials Pk and Qk are determined by the recurrence 1 (k + 1)Pk+1(x) = kPk (x) − xPk′ (x) − Qk (x) 2 x 2k + 1 Qk (x) − xQ′k (x) + Pk (x) (k + 1)Qk+1 (x) = 2 2 together with the initial conditions P0 (x) = 0, Q0 (x) = 1. The recurrence and initial conditions are satisfied by the explicit formulas above. Proof of Theorem 1. Using Lemma 3, we have ⌊ k−1 ⌋ 2

Gk (t) = −

X j=0

  √ 2k − 2j − 1 √ (−4π 2 t)j π t cot π t 2k−1 k 2 (2j + 1)! k

+

⌊2⌋   X (−4π 2 t)j 2k − 2j j=0

22k (2j)!

k

=

k−1

⌊ 2 ⌋   √ √ X 2k − 2j − 1 1 (−4π 2 t)j (1 − π t cot π t) k 2 22k−2 (2j + 1)! j=0

+ terms of degree < k, and this completes the proof.

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Proof of Theorem 4. Using Theorem 1 in the form of equation (7), eliminate ζ(2n − 2j) using Euler’s formula (2) and then compare with Theorem 3 to get ⌊ k−1 ⌋ 2

X j=0

  2k − 2j − 1 (−1)n−1 π 2n B2n−2j = k 22k−2n−1 (2n − 2j)!(2j + 1)!   n−k  (−1)n−k−1 π 2n X n − i 2n + 1 2(22i−1 − 1)B2i . 2i k (2n + 1)! i=0

Now multiply both sides by (−1)n−1 22k−2n−1 π −2n (2n + 1)! and rewrite the factorials on the left-hand side as a binomial coefficient.

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[8] I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Oxford Univ. Press, New York, 1995. [9] T. Machide, Extended double shuffle relations and the generating function of triple zeta values of any fixed weight, preprint arXiv:1204.4085. [10] Z. Shen and T. Cai, Some formulas for multiple zeta values, J. Number Theory 132 (2012), 314-323.

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