Generalization of Bernoulli numbers and polynomials to the multiple case
Olivier Bouillot, Marne-la-Vall´eeUniversity, France
C.A.L.I.N. team seminary. Tuesday, 3th March 2015 . Introduction
Definition: The numbers Zes1,··· ,sr defined by
s1,··· ,sr X 1 Ze = s s , n1 1 ··· nr r 0 where s1, ··· , sr ∈ C such that <(s1 + ··· + sk ) > k, k ∈ [[1 ; r ]], are called multiple zeta values. Fact: There exist at least three different way to renormalize multiple zeta values at negative integers. 7 1 1 Ze0,−2(0) = , Ze0,−2(0) = , Ze0,−2 (0) = . MP 720 GZ 120 FKMT 18 Question: Is there a group acting on the set of all possible multiple zeta values renormalisations? Main goal: Define multiple Bernoulli numbers in relation with this. Outline 1 Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on Quasi-Symmetric Functions 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers Outline 1 Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on Quasi-Symmetric Functions Two Equivalent Definitions of Bernoulli Polynomials / Numbers Bernoulli numbers: Bernoulli polynomials: By a generating function: By a generating function: t X tn text X tn = b . = B (x) . et − 1 n n! et − 1 n n! n≥0 n≥0 By a recursive formula: By a recursive formula: b = 1 , B0(x) = 1 , 0 0 n ∀n ∈ N , Bn+1(x) = (n + 1)Bn(x) , X n + 1 Z 1 ∀n ∈ N , bk = 0 . ∗ k ∀n ∈ , Bn(x) dx = 0 . k=0 N 0 First examples: First examples: 1 1 1 1 b = 1, − , , 0, − , 0, , ··· B (x) = 1 , n 2 6 30 42 0 1 B (x) = x − , 1 2 2 1 B2(x) = x − x + , . 6 . Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b2n+1 = 0 if n > 0. P2 Bn(0) = Bn(1) if n > 1. m X m + 1 P3 b = 0, m > 0. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 n P4 X n n−k B (x + y) = B (x)y for all n. n k k k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. n P6 (−1) Bn(1 − x) = Bn(x), for all n. N−1 X n Bn+1(N) − Bn+1(0) P7 k = . n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Has to be extended, but too particular. Important property, but turns out to have a generalization with a corrective term... Has to be extended, but how??? Has to be extended, but how??? Does not depend of the Bernoulli numbers... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 n P4 X n n−k B (x + y) = B (x)y for all n. n k k k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. n P6 (−1) Bn(1 − x) = Bn(x), for all n. N−1 X p Bn+1(N) − Bn+1(0) P7 k = . n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Important property, but turns out to have a generalization with a corrective term... Has to be extended, but how??? Has to be extended, but how??? Does not depend of the Bernoulli numbers... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 n P4 X n n−k B (x + y) = B (x)y for all n. n k k k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. n P6 (−1) Bn(1 − x) = Bn(x), for all n. N−1 X p Bn+1(N) − Bn+1(0) P7 k = . n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Has to be extended, but how??? Has to be extended, but how??? Does not depend of the Bernoulli numbers... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 Important property, but turns n out to have a generalization P4 X n n−k B (x + y) = B (x)y for all n. n k k with a corrective term... k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. n P6 (−1) Bn(1 − x) = Bn(x), for all n. N−1 X p Bn+1(N) − Bn+1(0) P7 k = . n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Has to be extended, but how??? Does not depend of the Bernoulli numbers... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 Important property, but turns n out to have a generalization P4 X n n−k B (x + y) = B (x)y for all n. n k k with a corrective term... k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. Has to be extended, but how??? n P6 (−1) Bn(1 − x) = Bn(x), for all n. N−1 X p Bn+1(N) − Bn+1(0) P7 k = . n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Does not depend of the Bernoulli numbers... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 Important property, but turns n out to have a generalization P4 X n n−k B (x + y) = B (x)y for all n. n k k with a corrective term... k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. Has to be extended, but how??? n P6 (−1) Bn(1 − x) = Bn(x), for all n. Has to be extended, but how??? N−1 X p Bn+1(N) − Bn+1(0) P7 k = . n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 Important property, but turns n out to have a generalization P4 X n n−k B (x + y) = B (x)y for all n. n k k with a corrective term... k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. Has to be extended, but how??? n P6 (−1) Bn(1 − x) = Bn(x), for all n. Has to be extended, but how??? N−1 X p Bn+1(N) − Bn+1(0) P7 k = . Does not depend of the Bernoulli numbers... n + 1 k=0 Z x Bn+1(x) − Bn+1(a) P8 Bn(t) dt = . a n + 1 m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 ??? Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 Important property, but turns n out to have a generalization P4 X n n−k B (x + y) = B (x)y for all n. n k k with a corrective term... k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. Has to be extended, but how??? n P6 (−1) Bn(1 − x) = Bn(x), for all n. Has to be extended, but how??? N−1 X p Bn+1(N) − Bn+1(0) P7 k = . Does not depend of the Bernoulli numbers... n + 1 k=0 Z x Bn+1(x) − Bn+1(a) Has a generalization using the derivative P8 Bn(t) dt = . of a multiple Bernoulli polynomial instead a n + 1 of the Bernoulli polynmials. m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. n n m k=0 Elementary properties satisfied by the Bernoulli polynomials and numbers P1 b = 0 if n > 0. 2n+1 Have to be extended, P2 Bn(0) = Bn(1) if n > 1. but is not restritive enough. m X m + 1 P3 b = 0, m > 0. Has to be extended, but too particular. k k k=0 B0 (z) = nB (z) if n > 0. n n−1 Important property, but turns n out to have a generalization P4 X n n−k B (x + y) = B (x)y for all n. n k k with a corrective term... k=0 n−1 P5 Bn(x + 1) − Bn(x) = nx , for all n. Has to be extended, but how??? n P6 (−1) Bn(1 − x) = Bn(x), for all n. Has to be extended, but how??? N−1 X p Bn+1(N) − Bn+1(0) P7 k = . Does not depend of the Bernoulli numbers... n + 1 k=0 Z x Bn+1(x) − Bn+1(a) Has a generalization using the derivative P8 Bn(t) dt = . of a multiple Bernoulli polynomial instead a n + 1 of the Bernoulli polynmials. m−1 n−1 X k P9 B (mx) = m B x + for all m > 0 and n ≥ 0. ??? n n m k=0 Outline 1 Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on Quasi-Symmetric Functions On the Hurwitz Zeta Function Definition: The Hurwitz Zeta Function is defined, for Property: ∂ζ (s, z) = −sζ(s + 1, z). ∂z H1 X −s n ζ(s, x + y) = ζ(s + n, x)y . n n≥0 H2 ζ(s, z − 1) − ζ(s, z) = z −s . m−1 X k H3 ζ(s, mz) = m−s ζ s, z + if m ∈ ∗. m N k=0 Link between the Hurwitz Zeta Function and the Bernoulli polynomials Property: s 7−→ ζ(s, z) can be analytically extend to a meromorphic function on C, with a simple pole located at 1. B (z) Remark: ζ(−n, z) = − n+1 for all n ∈ . n + 1 N b ζ(−n, 0) = − n+1 for all n ∈ . n + 1 N Related properties: Hurwitz zeta function Bernoulli polynomials Derivative property H1 P4 Difference equation H2 P5 Multiplication theorem H3 P9 Consequence: The extension from Bernoulli to multiple Bernoulli polynomials will be done using a generalization of the Hurwitz zeta function: the Hurwitz multiple zeta functions. On Hurwitz Multiple Zeta Functions Definition of Hurwitz Multiple Zeta Functions s1,··· ,sr X 1 He (z) = s s , if z ∈ C − N<0 and (n1 + z) 1 ··· (nr + z) r 0 Lemma 1: (B., J. Ecalle, 2012) ∗ r For all sequences (s1, ··· , sr ) ∈ (N ) , s1 ≥ 2, we have: Hes1,··· ,sr (z − 1) − Hes1,··· ,sr (z) = Hes1,··· ,sr−1 (z) · z −sr . Lemma 2: ∗ The Hurwitz Multiple Zeta Functions multiply by the stuffle product (of N ). Reminder: If (Ω, +) is a semi-group, the stuffle is defined over Ω? by: ε u = u ε = u . ua vb = (u vb) a + (ua v) b + (u v)(a + b) . Outline 1 Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on Quasi-Symmetric Functions Reminders on Quasi-symmetric functions Definition: Let x = {x1, x2, x3, ···} be an infinite commutative alphabet. s1 sr A series is said to be quasi-symmetric when its coefficient of x1 ··· xr is equals s1 sr to this of x ··· x for all i1 < ··· < ir . i1 ir 2 2 2 2 Example : M2,1(x1, x2, x3,...) = x1 x2 + x1 x3 + ··· + x1 xn + ··· + x2 x3 + ··· 2 x1x2 is not in M2,1 but in M1,2. Fact 1: • Quasi-symmetric functions span a vector space: QSym. • A basis of QSym is given by the monomials MI , for composition I = (i1, ··· , ir ): X i1 ir Mi1,··· ,ir (X ) = xn1 ··· xnr 0 Fact 2: • QSym is an algebra whose product is the stuffle product. • QSym is also a Hopf algebra whose coproduct ∆ is given by: r X ∆ Mi1,··· ,ir (x) = Mi1,··· ,ik (x) ⊗ Mik+1,··· ,ir (x) . k=0 Outline 1 Reminders 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers Main Goal Heuristic: Bs1,··· ,sr (z) = Multiple (Divided) Bernoulli Polynomials = He−s1,··· ,−sr (z) . bs1,··· ,sr = Multiple (Divided) Bernoulli Numbers = He−s1,··· ,−sr (0) . We want to define Bs1,··· ,sr (z) such that: their properties are similar to Hurwitz Multiple Zeta Functions’ properties. their properties generalize these of Bernoulli polynomials. Main Goal: Find some polynomials Bs1,··· ,sr such that: n Bn+1(z) B (z) = , where n ≥ 0 , n + 1 Bn1,··· ,nr (z + 1) − Bn1,··· ,nr (z) = Bn1,··· ,nr−1 (z)z nr , for n , ··· , n ≥ 0 , 1 r the Bn1,··· ,nr multiply by the stuffle product. An algebraic construction Notation 1: Let X = {X1, ··· , Xn, ···} be a (commutative) alphabet of indeterminates. We denotes: n1 nr Y ,··· ,Y X n ,··· ,n Y Y B 1 r (z) = B 1 r (z) 1 ··· r , n1! nr ! n1,··· ,nr ≥0 ∗ for all r ∈ N , Y1, ··· , Yr ∈ X. Remark: BY1,··· ,Yr (z + 1) − BY1,··· ,Yr (z) = BY1,··· ,Yr−1 (z)ezYr . Notation 2: Let A = {a1, ··· , an, ···} be a non-commutative alphabet. We denotes: X X X ,··· ,X k1 kr B(z) = 1 + B (z)ak1 ··· akr ∈ C[z][[X]]hhAii . r>0 k1,··· ,kr >0 X zXk Remark: B(z + 1) = B(z) · 1 + e ak k>0 The abstract construction in the case of quasi-symmetric functions Let see an analogue of B(z) where the multiple Bernoulli polynomials are replaced with the monomial functions MI (x) of QSym: n1 nr Y1,··· ,Yr X Y1 Yr M (x) := Mn1+1,··· ,nr +1(x) ··· , for all Y1, ··· , Yr ∈ X . n1! nr ! n1,··· ,nr ≥0 X X X ,··· ,X k1 kr M := 1 + M (x) ak1 ··· akr r>0 k1,··· ,kr >0 r X X Y X xnXk Xk ,··· ,Xk = 1 + 1 + xne ak M 1 r (x) r>0 0 Computation of the coproduct of M: (which does not act on the X ’s) r X Y ,...,Y Y ,...,Y ∆MY1,··· ,Yr (x) = M 1 k (x) ⊗ M k+1 r (x) . k=0 ∆M = M ⊗ M . Transcription of the multiplication by the stuffle Property: (J. Y. Thibon, F. Chapoton, J. Ecalle, F. Menous, D. Sauzin, ...) n1,··· ,nr A family of objects (B )n1,n2,n3,···≥0 multiply by the stuffle product if, and only if, there exists a character χz of QSym such that n1,··· ,nr χz Mn1+1,··· ,nr +1(x) = B (z) (1) Consequences: 1. χz can be extended to QSym[[X]], applying it terms by terms. Y1,··· ,Yr Y1,··· ,Yr χz M (x) = B (z) , for all Y1, ··· , Yr ∈ X . n1,··· ,nr 2. If B multiply the stuffle, B = χz (M) is “group-like” in C[z][[X]]hhAii. Reformulation of the main goal Reformulation of the main goal Find some polynomials Bn1,··· ,nr such that: zX e k 1 hB(z)|a i = − , k X e k − 1 Xk X zXk B(z + 1) = B(z) · E(z) , where E(z) = 1 + e ak , k>0 B is a “group-like” element of C[z][[X]]hhAii . Outline 1 Reminders 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers A singular solution X zXk Remainder: E(z) = 1 + e ak . k>0 From a false solution to a singular solution... ←− z(X +···+X ) Y X X e k1 kr S(z) = E(z − n) = 1 + r ak1 ··· akr is a n>0 r>0 Y X +···+X k1,··· ,kr >0 (e k1 ki − 1) i=1 zX e k 1 hB(z)|a i = − , k X e k − 1 Xk false solution to system B(z + 1) = B(z) · E(z) , B is a “group-like” element of C[z][[X]]hhAii . Explanations: 1. B(z) = ··· = B(z − n) · E(z − n) ··· E(z − 1) ←− Y = ··· = lim B(z) · E(z − n) . n−→−∞ n>0 2. S(z) ∈ C[z]((X))hhAii, S(z) 6∈ C[z][[X]]hhAii. Heuristic: Find a correction of S, to send it into C[z][[X]]hhAii. Another solution Fact: If ∆(f )(z) = f (z − 1) − f (z), ker ∆ ∩ zC[z] = {0}. Consequence: There exist a unique family of polynomials such that: ( n1,··· ,nr n1,··· ,nr n1,··· ,nr−1 nr B0 (z + 1) − B0 (z) = B0 (z)z . n1,··· ,nr B0 (0) = 0 . This produces a series B0 ∈ C[z][[X]]hhAii defined by: X ,··· ,X X X k1 kr B0(z) = 1 + B0 (z) ak1 ··· akr . r>0 k1,··· ,kr >0 Lemma: (B., 2013) 1 The noncommutative series B0 is a “group-like” element of C[z][[X]]hhAii. 2 The coefficients of B0(z) satisfy a recurence relation: zY1 Y1 e − 1 B0 (z) = , Y1 ∈ X . eY1 − 1 Y1+Y2,Y3,··· ,Yr Y2,Y3,··· ,Yr Y1,··· ,Yr B0 (z) − B0 (z) B0 (z) = , Y1, ··· , Yr ∈ X . eY1 − 1 −1 3 The series B0 can be expressed in terms of S: B0(z) = S(0) ·S(z). Characterization of the set of solutions Reminder: A family of multiple Bernoulli polynomials produces a series B such that: X zXk B(z + 1) = B(z) · E(z) , where E(z) = 1 + e ak , k>0 B is a “group-like” element of C[z][[X]]hhAii , ezXk 1 hB(z)|ak i = X − . e k − 1 Xk Proposition: (B. 2013) Any familly of polynomials which are solution of the previous system comes from a noncommutative series B ∈ C[z][[X]]hhAii such that there exists b ∈ C[[X]]hhAii satisfying: 1 1 1. hb|Ak i = X − 2. b is “group-like” e k − 1 Xk −1 3. B(z) = b · B0 = b · S(0) ·S(z). Theorem: (B., 2013) The subgroup of “group-like” series of C[z][[X]]hhAii, with vanishing coefficients in length 1, acts on the set of all possible multiple Bernoulli polynomials, i.e. on the set of all possible algebraic renormalization. Outline 1 Reminders 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers Some notations New Goal: From B(z) = b · B0, determine a suitable series b such that the reflexion formula n (−1) Bn(1 − z) = Bn(z) , n ∈ N has a nice generalization. For a generic series s ∈ C[z][[X ]]hhAii, X X X ,··· ,X k1 kr s(z) = s (z) ak1 ··· akr , r∈N k1,··· ,kr >0 we consider: X X X ,··· ,X kr k1 s(z) = s (z) ak1 ··· akr r∈N k1,··· ,kr >0 X X −X ,··· ,−X s(z) = s k1 kr (z) a ··· a e k1 kr r∈N k1,··· ,kr >0 The reflection equation for B0(z) Proposition: (B. 2014) −1 X X r X Let sg = 1 + (−1) ak1 ··· akr = 1 + an . Then, r>0 k1,··· ,kr >0 n>0 −1 −1 Se(0) = S(0) · sg and Se(1 − z) = S(z) . Corollary 1: (B. 2014) −1 For all z ∈ C, we have: sg · Be 0(1 − z) = B0(z) . Example: −X ,−Y ,−Z X ,Y ,Z X +Y ,Z X ,Y +Z B0 (1 − z) = −B0 (z) − B0 (z) − B0 (z) X +Y +Z Y ,Z Y +Z −B0 (z) + B0 (z) + B0 (z) . The generalization of the reflection formula Corollary 2: (B. 2014) −1 Be (1 − z) · B(z) = eb · sg · b . (2) Remark: Se(0) · sg −1 · S(0) = 1. Heuristic: A reasonable candidate for a multi-Bernoulli polynomial comes from the coefficients of a series B(z) = b · B0(z) where b satisfies: 1 1 1. hb|ak i = X − 2. b is “group-like” e k − 1 Xk 3. eb · sg −1 · b = 1 . Outline 1 Reminders 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers Resolution of an equation u · sg −1 · u = 1 . Goal: Characterise the solutions of e u is “group-like” . Proposition: (B., 2014) p X X (−1)r 2r Let us denote sg −1 = 1 + a ··· a .. 22r r k1 kr r>0 k1,··· ,kr >0 −1 Any “group-like” solution u of eu · sg · u = 1 comes from a “primitive” series v satisfying v + ev = 0 , and is given by: √ u = exp(v) · sg . 1 1 If moreover hu|ak i = X − , then necessarily, we have: e k − 1 Xk 1 1 1 hv|ak i = X − + := f (Xk ) . e k − 1 Xk 2 The choice of a series v New goal: Find a nice series v satisfying: 1 1 1 1. v is “primitive”. 2. v + ev = 0. 3. hv|ak i = X − + = f (Xk ). e k − 1 Xk 2 Remark: hv|ak i is an odd formal series in Xk ∈ X. Generalization: ev = −v , so v = v . 1 =⇒ hv|a a i = − f (X + X ), but does not determine hv|a a a i . k1 k2 2 k1 k2 k1 k2 k3 A restrictive condition: A natural condition is to have: there exists αr ∈ C such that hv|ak1 ··· akr i = αr f (Xk1 + ··· + Xkr ) . Now, there is a unique “primitive” series v satisfying this condition and the new goal: (−1)r−1 hv|a ··· a i = f (X + ··· + X ) . k1 kr r k1 kr Definition Definition : (B., 2014) The series B(z) and b defined by ( √ B(z) = exp(v) · Sg · (S(0))−1 ·S(z) √ b = exp(v) · Sg are noncommutative series of C[z][[X]]hhAii whose coefficients are respectively the exponential generating functions of multiple Bernoulli polynomials and multiple Bernoulli numbers. Example: The exponential generating function of bi-Bernoulli polynomials and numbers are respectively: n1 n2 X n ,n X Y 1 1 1 3 B 1 2 (z) = − f (X + Y ) + f (X )f (Y ) − f (X ) + n1! n2! 2 2 2 8 n1,n2≥0 ezY − 1 1 ezY − 1 +f (X ) − eY − 1 2 eY − 1 ez(X +Y ) − 1 ezY − 1 + − . (eX − 1)(eX +Y − 1) (eX − 1)(eY − 1) Examples of explicit expression for multiple Bernoulli numbers: Consequently, we obtain explicit expressions like, for n1, n2, n3 > 0: 1 bn +1 bn +1 bn +n +1 bn1,n2 = 1 2 − 1 2 . 2 n1 + 1 n2 + 1 n1 + n2 + 1 1 bn +1 bn +1 bn +1 bn1,n2,n3 = + 1 2 3 6 n1 + 1 n2 + 1 n3 + 1 1 b b b b − n1+n2+1 n3+1 + n1+1 n2+n3+1 4 n1 + n2 + 1 n3 + 1 n1 + 1 n2 + n3 + 1 1 b + n1+n2+n3+1 . 3 n1 + n2 + n3 + 1 Remark: If n1 = 0, n2 = 0 or n3 = 0, the expressions are not so simple... Table of Multiple Bernoulli Numbers in length 2 bp,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 3 1 1 1 q = 0 − 0 0 − 0 8 12 120 252 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 Outline 1 Reminders 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers one out of two antidiagonals is “constant”. “symmetrie” relatively to p = q . cross product around the zeros are equals : 28800 · 127008 = 604802 . Table of Multiple Bernoulli Numbers in length 2 bp,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 3 1 1 1 q = 0 − 0 0 − 0 8 12 120 252 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 one out of four Multiple Bernoulli Numbers is null. “symmetrie” relatively to p = q . cross product around the zeros are equals : 28800 · 127008 = 604802 . Table of Multiple Bernoulli Numbers in length 2 bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 3 1 1 1 q = 0 − 0 0 − 0 8 12 120 252 1 1 1 1 1 1 1 q = 1 − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 one out of four Multiple Bernoulli Numbers is null. one out of two antidiagonals is “constant”. cross product around the zeros are equals : 28800 · 127008 = 604802 . Table of Multiple Bernoulli Numbers in length 2 bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 3 1 1 1 q = 0 − 0 0 − 0 8 12 120 252 1 1 1 1 1 1 1 q = 1 − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 one out of four Multiple Bernoulli Numbers is null. one out of two antidiagonals is “constant”. “symmetrie” relatively to p = q . Table of Multiple Bernoulli Numbers in length 2 bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 3 1 1 1 q = 0 − 0 0 − 0 8 12 120 252 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 one out of four Multiple Bernoulli Numbers is null. one out of two antidiagonals is “constant”. “symmetrie” relatively to p = q . cross product around the zeros are equals : 28800 · 127008 = 604802 . Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bp,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 q = 2 q = 3 q = 4 q = 5 q = 6 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 q = 2 0 0 0 0 q = 3 q = 4 0 0 0 0 q = 5 q = 6 0 0 0 0 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 q = 2 0 0 0 0 240 1 q = 3 − 2880 1 q = 4 0 − 0 0 0 504 1 q = 5 6048 1 q = 6 0 0 0 0 480 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 q = 3 − − − 240 2880 504 480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 691 q = 5 − − 504 6048 480 264 65520 1 1 691 q = 6 0 0 − 0 0 480 264 65520 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 q = 3 − − − 240 2880 504 28800 480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 691 q = 5 − − 504 6048 480 264 65520 1 1 691 q = 6 0 0 − 0 0 480 264 65520 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 691 q = 5 − − 504 6048 480 264 65520 1 1 691 q = 6 0 0 − 0 0 480 264 65520 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 65520 1 1 691 q = 6 0 0 − 0 0 480 264 65520 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 q = 0 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 1 1 691 q = 6 0 0 − 0 0 480 264 65520 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bi-Bernoulli numbers. bep,q p = 0 p = 1 p = 2 p = 3 p = 4 p = 5 p = 6 3 1 1 1 q = 0 − 0 0 − 0 8 12 120 252 1 1 1 1 1 1 1 q = 1 − − − 24 288 240 2880 504 6048 480 1 1 1 q = 2 0 0 − 0 0 240 504 480 1 1 1 1 1 1 1 q = 3 − − − − 240 2880 504 28800 480 60480 264 1 1 1 q = 4 0 − 0 0 − 0 504 480 264 1 1 1 1 1 1 691 q = 5 − − − 504 6048 480 60480 264 127008 65520 1 1 691 q = 6 0 0 − 0 0 480 264 65520 Outline 1 Reminders 2 Algebraic reformulation of the problem 3 The Structure of a Multiple Bernoulli Polynomial 4 The General Reflexion Formula of Multiple Bernoulli Polynomial 5 An Example of Multiple Bernoulli Polynomial 6 An algorithm to compute the double Bernoulli Numbers 7 Properties satisfied by Bernoulli polynomials and numbers Properties satisfied by multiple Bernoulli polynomials 1 Proposition: (B., 2013) The multiple Bernoulli polynomials Bn1,··· ,nr multiply the stuffle. Theorem: (B., 2014) P’1 All multiple Bernoulli numbers satisfy : b2p1,··· ,2pr = 0 . n1,··· ,nr n1,··· ,nr P’2 If nr > 0, B (0) = B (1) . P’5 Bn1,··· ,nr (z + 1) − Bn1,··· ,nr (z) = Bn1,··· ,nr−1 (z) · z nr . P’6 There exists a reflexion formula for multiple Bernoulli polynoms: Be (1 − z) · B(z) = 1 . s1,··· ,sr P’7 The truncated multiple sums of powers SN , defined by s1,··· ,sr X s1 sr SN = n1 ··· nr 0≤nr <··· are given by the coefficients of B0(N). Properties satisfied by multiple Bernoulli polynomials 2 Proposition: (B. 2014) n1,··· ,nr nr ,··· ,n1 For all positive integers n1, ··· , nr , b = b . Proposition: (B., 2014) For a series s(z) ∈ C[z][[X ]]hhAii, let us define ∆(s)(z) by: X X ∆(s)(z) = Xk1 ··· Xkr hs(z)|ak1 ··· akr i ak1 ··· akr . r∈N k1,··· ,kr >0 ∆ is a derivation, and : P’4 The derivation of multiple Bernoulli polynomials are given by: −1 −1 −1 ∂z B(z) = ∆ b ·S(0) · b ·S(0) · B(z) + ∆(B(z)) . Properties satisfied by multiple Bernoulli polynomials 3 Proposition: (B., 2015) P’3 The recurrence relation of bi-Bernoulli numbers is (partially) given by: p q ! ! ! X X p q k,l p,q 2 be − be = k l k=0 l=0 1 1 1 1 − − 2 p + 1 2 q + 1 1 1 1 1 + − beq + bep − 2 p + 1 2 q + 1 1 1 1 1 − − − − 2 p + 1 2 p + q + 1 3 −bep + 4 if p, q > 0 . Conclusion 1. We have respectively defined the Multiple (divided) Bernoulli Polynomials and Multiple (divided) Bernoulli Numbers by: ( √ B(z) = exp(v) · Sg · (S(0))−1 ·S(z) √ b = exp(v) · Sg They both multiply the stuffle. 2. The Multiple Bernoulli Polynomials satisfy a nice generalization of: the nullity of b2n+1 if n > 0. the symmetry Bn(1) = Bn(0) if n > 1. n−1 the difference equation ∆(Bn)(x) = nx . n the reflection formula (−1) Bn(1 − x) = Bn(x). THANK YOU FOR YOUR ATTENTION !