Yuan, H.

Zhao, J.

English

MPG.PuRe

arXiv:1303.3607v1  [math.NT]  14 Mar 2013Restricted Sum Formula of Multiple Zeta ValuesHaiping YuanYork College of Pennsylvania, York, PA 17403Jianqiang ZhaoDepartment of Mathematics, Eckerd College, St. Petersburg, FL 337111 IntroductionFor fixed positive integer d and d-tuple of positive integers (s1, . . . , sd) with s1 > 1, themultiple zeta value ζ(s1, . . . , sd) is defined byζ(s1, . . . , sd) =∑k1>···>kd>0k−s11 · · · k−sdd , (1)where d is called the depth and s1 + · · · + sd the weight. The double zeta values werestudied by Euler [1] who derived many identities such as follows:2n−1∑k=2(−1)kζ(k, 2n− k) = 12ζ(2n),2n−1∑k=2ζ(k, 2n− k) = ζ(2n),from which we can easily get (see [2, Theorem 1])n−1∑k=1ζ(2k, 2n− 2k) = 34ζ(2n). (2)Using the stuﬄe relation ζ(2k)ζ(2n− 2k) = ζ(2k, 2n− 2k) + ζ(2n− 2k, 2k) + ζ(2n) wesee immediatelyn−1∑k=1ζ(2k)ζ(2n− 2k) = 2n+ 12ζ(2n). (3)Recently, Hoffman [3] extended (2) to arbitrary depths. Moreover, similar formulashave been obtained for some special type Hurwitz-zeta values [4] and alternating Eulersums [5]. In this paper we consider the following restricted sum of multiple zeta valuesQ(4n, d) =∑j1+···+jd=nj1,...,jd>0ζ(4j1, . . . , 4jd).Our main theorem is1Theorem 1.1. For any positive integers n ≥ d ≥ 3,Q(4n, d) =⌊ d−12⌋∑k=02k+1∑j=02k+2(−1)⌊k2 ⌋+j+d(2k + 1)!(2k + 1j)( j−24d)ζ(4n− 2k)pi2k+⌊ d−24⌋∑k=04k+2∑j=022k+5(−1)k+j+d(4k + 2)!(4k + 2j)( j−24d)(Q(4n− 4k, 2)− 78ζ(4n− 4k))pi4k.Remark 1.2. For d = 2, it’s easy to prove by stuﬄe relation thatQ(4n, 2) =12n−1∑k=1ζ(4k)ζ(4n− 4k)− n− 12ζ(4n)for n ≥ 2. However, it is an intriguing problem to find a compact formula similar to (3).Acknowledgement. Both authors would like to thank the Morningside Center ofMathematics, Chineses Academy of Science for hospitality when the paper was pre-pared. HY is partially supported by the Summer Research Grant from York Collegeof Pennsylvania and JZ is partially supported by NSF grant DMS1162116. They alsowant to thank the anonymous referee who pointed out a few inaccuracies and possibleimprovement in the original draft.2 The generating function of Q(4n, d)Recall that the symmetric function of the infinitely many variables x1, x2, · · · form asubring Sym of Q[x1, x2, · · · ] which is invariant under all the permutations of the vari-ables. Let ej =∑k1<···<kjxk1 . . . xkj be the j-th elementary function. Following Hoffman[3] let’s consider its generating functionE(t) =∞∏j=1(1 + txj) =∞∑j=0ejtjand define ε : Sym→ R to be the evaluation map such that ε(xj) = 1j4. LetF (s, t) =∞∏j=1(1 + tsxj + ts2x2j + · · · ).Then it is not hard to see that the generating function of Q(4n, d) is given byε(F (s, t))=∞∑n=0Q(4n, d)tdsn.First we need the following lemma.2Lemma 2.1. We haveε(F (s, t)) =sin pi 4√s(1− t) · sinh pi 4√s(1− t)√1− t sin pi 4√s · sinh pi 4√s .Proof. We have∞∏j=1(1 + tsxj + ts2x2j + · · · ) =∞∏j=1(1 + tsxj1− sxj)=∏∞j=1(1− s(1− t)xj)∏∞j=1(1− sxj)=E(−s(1 − t))E(−s) .Further,ε(E(−t)) =∞∏i=1(1− ti4)=∞∏i=1(1−√ti2)(1 +√ti2)=sin pi 4√t · sinh pi 4√tpi2√t.The lemma follows immediately.Let f(x) = sin x · sinh x/(2x2). The following lemma provides its series expansion.Lemma 2.2. We havef(x) =∞∑k=0(−1)k4k(4k + 2)!x4k.Proof. Using the well-known formula sin x = (eix − e−ix)/(2i) we obtainf(x) =12· eix − e−ix2ix· ex − e−x2x=e(i+1)x + e−(i+1)x − (e(i−1)x + e−(i−1)x)8ix2=14ix2(∞∑n=0(2i)nx2n(2n)!−∞∑n=0(−2i)nx2n(2n)!)=∞∑k=0(−1)k4k(4k + 2)!x4k,as desired.3 Proof of Theorem 1.1Let g(t) = f( 4√t). Theng(s(1− t))g(s)= ε(F (s/pi4, t)) =1g(s)∞∑k=0(−1)k4k(4k + 2)!sk(1− t)k.3Writeg(s(1− t))g(s)=∞∑d=0Gd(s)td.By the above expression, we haveGd(s) =(−s)dg(s)d!Ddg(s),where Dd denotes the d-th derivative with respect to s. SetGd(s) = Xd(s)4√s cot 4√s+ Yd(s)4√s coth 4√s+ Zd(s) cot4√s coth 4√s+Wd(s) (4)which yields easily(−1)sDdg(s)d!= Xd(s)s−d− 14 cos s14 sinh s14 + Yd(s)s−d− 14 sin s14 cosh s14+ Zd(s)s−d− 12 cos s14 cosh s14 +Wd(s)s−d− 12 sin s14 sinh s14 .To determine the coefficients Xd(s), Yd(s), Zd(s) and Wd(s) we differentiate the bothsides of the above equation to get the following system of recursive differential equations(d+ 1)Xd+1(s) = −sX ′d(s) +(d+14)Xd(s)− 14Zd(s)− 14Wd(s),(d+ 1)Yd+1(s) = −sY ′d(s) +(d+14)Yd(s) +14Zd(s)− 14Wd(s),(d+ 1)Zd+1(s) = −√s4Xd(s)−√s4Yd(s)− sZ ′d(s) +(d+12)Zd(s),(d+ 1)Wd+1(s) =√s4Xd(s)−√s4Yd(s) +(d+12)Wd(s)− sW ′d(s),with the initial conditions X0(s) = Y0(s) = Z0(s) = 0 and W0(s) = 1. Let xd(u) =Xd(u2), yd(u) = Yd(u2), zd(u) = Zd(u2) and wd(u) = Wd(u2). The above system ischanged into the following system:(d+ 1)xd+1(u) = −u2x′d(u) +(d+14)xd(u)−14zd(u)−14wd(u),(d+ 1)yd+1(u) = −u2y′d(u) +(d+14)yd(u) +14zd(u)−14wd(u),(d+ 1)zd+1(u) = −u4xd(u)−u4yd(u)−u2z′d(u) +(d+12)zd(u),(d+ 1)wd+1(u) =u4xd(u)− u4yd(u) +(d+12)wd(u)− u2w′d(u).(5)4Define α(u, v) =∑d≥0xd(u)vd =∑d≥0x˜d(v)ud,β(u, v) =∑d≥0yd(u)vd =∑d≥0y˜d(v)ud,γ(u, v) =∑d≥0zd(u)vd =∑d≥0z˜d(v)ud,δ(u, v) =∑d≥0wd(u)vd =∑d≥0w˜d(v)ud.(6)Multiplying the system (5) by vd and then taking the sum∑d≥0 we get:∂α∂v=v∂α∂v+14α− u2∂α∂u− 14γ − 14δ,∂β∂v=v∂β∂v+14β − u2∂β∂u+14γ − 14δ,∂γ∂v=v∂γ∂v+12γ − u2∂γ∂u− u4α− u4β,∂δ∂v=v∂δ∂v+12δ − u2∂δ∂u+u4α− u4β.Comparing the coefficients of un we getx˜′n(v) =vx˜′n(v) +14x˜n(v)−n2x˜n(v)−14z˜n(v)−14w˜n(v),y˜′n(v) =vy˜′n(v) +14y˜n(v)−n2y˜n(v) +14z˜n(v)−14w˜n(v),z˜′n(v) =vz˜′n(v) +12z˜n(v)−n2z˜n(v)−14x˜n−1(v)−14y˜n−1(v),w˜′n(v) =vw˜′n(v) +12w˜n(v)− n2w˜n(v) +14x˜n−1(v)− 14y˜n−1(v),(7)By definition (6), we see that the system has the following initial values: x˜n(0) =0, y˜n(0) = 0, z˜n(0) = 0 for all n ≥ 0 and w˜n(0) = 0 for all n ≥ 1. But for w˜0(v) we havefrom (5)w0(0) = 1, wd(0) =2d− 12dwd−1(0) ∀d ≥ 1.It follows that wd(0) =(2dd)/22d which yields easilyw˜0(v) =∑d≥0wd(0)vd = (1− v)− 12 .Similarly we see that z˜0(v) = 0. Solving (7) recursively starting from the first twoequations in (7) we find the following functions are the unique solution satisfying the5initial conditions:x˜n(v) =2n+1∑j=02n(−1)⌊n+22 ⌋+jj!(2n+ 1− j)!(1− v)j−24 ;y˜n(v) =2n+1∑j=02n(−1)⌊n+32 ⌋+jj!(2n+ 1− j)!(1− v)j−24 ;z˜n(v) = (1− (−1)n)2n∑j=02n−1(−1)n−12 +jj!(2n− j)! (1− v)j−24 ;w˜n(v) = (1 + (−1)n)2n∑j=02n−1(−1)n2+jj!(2n− j)! (1− v)j−24 .Using (6) we can solve xn(v), yn(v), zn(v) and wn(v) and getxd(u) =⌊ d−12⌋∑n=02n+1∑j=02n(−1)⌊n+22 ⌋+j+d(2n+ 1)!(2n+ 1j)( j−24d)un;yd(u) =⌊ d−12⌋∑n=02n+1∑j=02n(−1)⌊n+32 ⌋+j+d(2n+ 1)!(2n+ 1j)( j−24d)un;zd(u) =2⌊ d−24⌋+1∑n=02n∑j=0(1− (−1)n)2n−1(−1)n−12 +j+d(2n)!(2nj)( j−24d)un;wd(u) =2⌊ d4⌋∑n=02n∑j=0(1 + (−1)n)2n−1(−1)n2 +j+d(2n)!(2nj)( j−24d)un.ThusXd(s) =⌊ d−12⌋∑n=02n+1∑j=02n(−1)⌊n+22 ⌋+j+d(2n+ 1)!(2n+ 1j)( j−24d)sn2 ;Yd(s) =⌊ d−12⌋∑n=02n+1∑j=02n(−1)⌊n+32 ⌋+j+d(2n+ 1)!(2n+ 1j)( j−24d)sn2 ;Zd(s) =⌊ d−24⌋∑n=04n+2∑j=022n+1(−1)n+j+d(4n+ 2)!(4n+ 2j)( j−24d)sn+1/2;Wd(s) =⌊ d4⌋∑n=04n∑j=022n(−1)n+j+d(4n)!(4nj)( j−24d)sn.6By the well-known formulasz cot z = −2∞∑n=0ζ(2n)pi2nz2n, z coth z = −2∞∑n=0(−1)n ζ(2n)pi2nz2n,we obtain4√s cot 4√s = −2∞∑n=0ζ(2n)pi2nsn2 , 4√s coth 4√s = −2∞∑n=0(−1)n ζ(2n)pi2nsn2 ,and√s cot 4√s·coth 4√s = 4∞∑k=0∑m+l=k(−1)m ζ(2m)ζ(2l)pi2ksk2 = 4∞∑k=0∑m+l=2k(−1)m ζ(2m)ζ(2l)pi4ksk.Here by exchanging m and l we notice that the inner sum vanishes if k is odd. Hencethe coefficient of sn in Gd(pi4s) isQ(4n, d) =2⌊ d−12⌋∑k=02k+1∑j=02k(−1)⌊k2 ⌋+j+d(2k + 1)!(2k + 1j)( j−24d)ζ(4n− 2k)pi2k+2⌊ d−12⌋∑k=02k+1∑j=0(−1)k 2k(−1)⌊k+12 ⌋+j+d(2k + 1)!(2k + 1j)( j−24d)ζ(4n− 2k)pi2k+4⌊ d−24⌋∑k=04k+2∑j=022k+1(−1)k+j+d(4k + 2)!(4k + 2j)( j−24d) ∑m,l≥0,m+l=2n−2k(−1)mζ(2m)ζ(2l)pi4ksince Wd(s) has degree less than n. Observe that the first two lines are the same andfor any positive integer w∑m,l≥0,m+l=2w(−1)mζ(2m)ζ(2l) =2w−1∑l=1ζ(4l)ζ(4w − 4l)−2w−1∑l=1ζ(2l)ζ(4w − 2l)− ζ(4w)=4Q(4w, 2) + (2w − 3)ζ(4w)− 4w + 12ζ(4w)=4Q(4w, 2)− 72ζ(4w)by stuﬄe relation ζ(4m)ζ(4l) = ζ(4m, 4l) + ζ(4l, 4m) + ζ(4m + 4l) and equation (3).Therefore we finally getQ(4n, d) = 4⌊ d−12⌋∑k=02k+1∑j=02k(−1)⌊k2 ⌋+j+dζ(4n− 2k)pi2k(2k + 1)!(2k + 1j)( j−24d)ζ(4n− 2k)pi2k+ 4⌊ d−24⌋∑k=04k+2∑j=022k+1(−1)k+j+d(4k + 2)!(4k + 2j)( j−24d)(4Q(4n− 4k, 2)− 72ζ(4n− 4k))pi4k.7This concludes the proof of Theorem 1.1 and this paper.References[1] L. Euler, Meditationes circa singulare serierum genus, Novi Comm. Acad. Sci.Petropol. 20 (1775), 140–186; reprinted in Opera Omnia, ser. I, vol. 15, B. G.Teubner, Berlin, 1927, pp. 217–267.[2] H. Gangl, M. Kaneko, and D. Zagier, Double zeta values and modular forms, in:Automorphic Forms and Zeta Functions, S. Bo¨cherer et. al. (eds.), World Scientific,Singapore, 2006, pp. 71-106.[3] M. E. Hoffman, On multiple zeta values of even arguments, arxiv: 1205.7051.[4] J. Zhao, Sum formula of multiple Hurwitz zeta values, to appear in Forum Mathe-maticum, arxiv: 1207.2368.[5] J. Zhao, Restricted sum Formula of alternating Euler sums, arxiv: 1207.5366.8

Restricted sum formula of multiple zeta values

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