An example of generalization of the Bernoulli numbers and polynomials to any dimension


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1 An example of generalization of the Bernoulli numbers and polynomials to any dimension Olivier Bouillot, VillebonGeorges Charpak institute, France C.A.L.I.N. team seminary. Tuesday, 29 th March 26.
2 Introduction Definition: The numbers Ze s,,s r defined by Ze s,,s r = <n r < <n n s nr s r, where s,, s r C such that R(s + + s k ) > k, k [[ ; r ]], are called multiple zeta values. Fact: There exist at least three different way to renormalize multiple zeta values at negative integers. Ze, 2 MP () = 7 72, Ze, 2 GZ () = 2, Ze, 2 FKMT () = 8. 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.
3 Outline Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on QuasiSymmetric 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
4 Outline Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on QuasiSymmetric Functions
5 Two Equivalent Definitions of Bernoulli Polynomials / Numbers Bernoulli numbers: By a generating function: Bernoulli polynomials: By a generating function: t e t = t n b n n!. te xt e t = B n(x) tn n!. n n By a recursive formula: By a recursive formula: b =, n N, n k= ( ) n + b k k =. First examples: B (x) =, n N, B n+(x) = (n + )B n(x), n N, B n(x) dx =. First examples: b n =, 2, 6,, 3,, 42, B(x) =, B (x) = x 2, B 2(x) = x 2 x + 6,.
6 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. P2 B n() = B n() if n >. P3 m ( ) m + b k k =, m >. P4 k= B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= ) B k (x)y n k for all n. P5 B n(x + ) B n(x) = nx n, for all n. P6 ( ) n B n( x) = B n(x), for all n. P7 P8 P9 N k n = k= x Bn+(N) Bn+() n + Bn+(x) Bn+(a) B n(t) dt =. a n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
7 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= ) B k (x)y n k for all n. P5 B n(x + ) B n(x) = nx n, for all n. P6 ( ) n B n( x) = B n(x), for all n. P7 N k p = k= Bn+(N) Bn+() n +. P8 P9 x a B n(t) dt = Bn+(x) Bn+(a) n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
8 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. P5 B n(x + ) B n(x) = nx n, for all n. P6 ( ) n B n( x) = B n(x), for all n. P7 N k p = k= Bn+(N) Bn+() n +. P8 P9 x a B n(t) dt = Bn+(x) Bn+(a) n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
9 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. P5 B n(x + ) B n(x) = nx n, for all n. P6 ( ) n B n( x) = B n(x), for all n. Important property, but turns out to have a generalization with a corrective term... P7 N k p = k= Bn+(N) Bn+() n +. P8 P9 x a B n(t) dt = Bn+(x) Bn+(a) n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
10 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. Important property, but turns out to have a generalization with a corrective term... P5 B n(x + ) B n(x) = nx n, for all n. Has to be extended, but how??? P6 ( ) n B n( x) = B n(x), for all n. P7 N k p = k= Bn+(N) Bn+() n +. P8 P9 x a B n(t) dt = Bn+(x) Bn+(a) n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
11 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. Important property, but turns out to have a generalization with a corrective term... P5 B n(x + ) B n(x) = nx n, for all n. Has to be extended, but how??? P6 ( ) n B n( x) = B n(x), for all n. Has to be extended, but how??? P7 N k p = k= Bn+(N) Bn+() n +. P8 P9 x a B n(t) dt = Bn+(x) Bn+(a) n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
12 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. Important property, but turns out to have a generalization with a corrective term... P5 B n(x + ) B n(x) = nx n, for all n. Has to be extended, but how??? P6 ( ) n B n( x) = B n(x), for all n. Has to be extended, but how??? P7 N k p = k= Bn+(N) Bn+() n +. Does not depend of the Bernoulli numbers... P8 P9 x a B n(t) dt = Bn+(x) Bn+(a) n + m ( B n(mx) = m n B n x + k ) for all m > and n. m k=.
13 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. Important property, but turns out to have a generalization with a corrective term... P5 B n(x + ) B n(x) = nx n, for all n. Has to be extended, but how??? P6 ( ) n B n( x) = B n(x), for all n. Has to be extended, but how??? P7 P8 P9 N k p = k= x a B n(t) dt = Bn+(N) Bn+() n + m B n(mx) = m n Bn+(x) Bn+(a) n + k=. Does not depend of the Bernoulli numbers.... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ( B n x + k ) for all m > and n. m
14 Elementary properties satisfied by the Bernoulli polynomials and numbers P b 2n+ = if n >. Have to be extended, P2 B n() = B n() if n >. but is not restritive enough. P3 P4 m k= ( m + k ) b k =, m >. B n(z) = nb n (z) if n >. n ( n B n(x + y) = k k= Has to be extended, but too particular. ) B k (x)y n k for all n. Important property, but turns out to have a generalization with a corrective term... P5 B n(x + ) B n(x) = nx n, for all n. Has to be extended, but how??? P6 ( ) n B n( x) = B n(x), for all n. Has to be extended, but how??? P7 P8 P9 N k p = k= x a B n(t) dt = Bn+(N) Bn+() n + m B n(mx) = m n Bn+(x) Bn+(a) n + k=. Does not depend of the Bernoulli numbers.... Has a generalization using the derivative of a multiple Bernoulli polynomial instead of the Bernoulli polynmials. ( B n x + k ) for all m > and n.??? m
15 Outline Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on QuasiSymmetric Functions
16 On the Hurwitz Zeta Function Definition: The Hurwitz Zeta Function is defined, for Re s >, and z C N <, by: ζ(s, z) = n> (n + z) s. Property: ζ z H (s, z) = sζ(s +, z). ( s n ζ(s, x + y) = n ) ζ(s + n, x)y n. H2 ζ(s, z ) ζ(s, z) = z s. m ( H3 ζ(s, mz) = m s ζ s, z + k ) if m N. m k=
17 Link between the Hurwitz Zeta Function and the Bernoulli polynomials Property: s ζ(s, z) can be analytically extend to a meromorphic function on C, with a simple pole located at. Remark: ζ( n, z) = Bn+(z) n + Related properties: for all n N. ζ( n, ) = bn+ for all n N. n + Hurwitz zeta function Bernoulli polynomials Derivative property H 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.
18 On Hurwitz Multiple Zeta Functions Definition of Hurwitz Multiple Zeta Functions He s,,s r (z) =, if z C N (n + z) s < and <n r < <n (nr + z) sr (s,, s r ) (N ) r, such that s 2. Lemma : (B., J. Ecalle, 22) For all sequences (s,, s r ) (N ) r, s 2, we have: He s,,s r (z ) He s,,s r (z) = He s,,s r (z) z sr. Lemma 2: The Hurwitz Multiple Zeta Functions multiply by the stuffle product (of N ). Reminder: If (Ω, +) is a semigroup, the stuffle is defined over Ω by: { ε u = u ε = u. ua vb = (u vb) a + (ua v) b + (u v)(a + b).
19 Outline Reminders Reminders on Bernoulli Polynomials and Numbers Reminders on Hurwitz Zeta Function and Hurwitz multiple zeta functions Reminders on QuasiSymmetric Functions
20 Reminders on Quasisymmetric functions Definition: Let x = {x, x 2, x 3, } be an infinite commutative alphabet. A series is said to be quasisymmetric when its coefficient of x s x r sr to this of x s i x sr i r for all i < < i r. is equals Example : M 2,(x, x 2, x 3,...) = x 2 x 2 + x 2 x x 2 x n + + x 2 2 x 3 + x x 2 2 is not in M 2, but in M,2. Fact : Quasisymmetric functions span a vector space: QSym. A basis of QSym is given by the monomials M I, for composition I = (i,, i r ): M i,,i r (X ) = x i n xn ir r <n < <n r Fact 2: QSym is an algebra whose product is the stuffle product. QSym is also a Hopf algebra whose coproduct is given by: ( M i,,i r (x) ) = r M i,,i k (x) M ik+,,i r (x). k=
21 Outline 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
22 Main Goal Heuristic: B s,,s r (z) = Multiple (Divided) Bernoulli Polynomials = He s,, s r (z). b s,,s r = Multiple (Divided) Bernoulli Numbers = He s,, s r (). We want to define B s,,s r (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 B s,,s r such that: B n (z) = Bn+(z) n +, where n, B n,,n r (z + ) B n,,n r (z) = B n,,n r (z)z nr, for n,, n r, the B n,,n r multiply by the stuffle product.
23 An algebraic construction Notation : Let X = {X,, X n, } be a (commutative) alphabet of indeterminates. We denotes: B Y,,Y r (z) = for all r N, Y,, Y r X. n,,n r B n,,n r (z) Y n n Y r nr! n r!, Remark: B Y,,Y r (z + ) B Y,,Y r (z) = B Y,,Y r (z)e zyr. Notation 2: Let A = {a,, a n, } be a noncommutative alphabet. We denotes: B(z) = + B X k,,x kr (z)a k a kr C[z][[X]] A. r> k,,k r > ( Remark: B(z + ) = B(z) + ) e zx k a k k>
24 The abstract construction in the case of quasisymmetric functions Let see an analogue of B(z) where the multiple Bernoulli polynomials are replaced with the monomial functions M I (x) of QSym: M Y,,Y r (x) := n,,n r M n +,,n r +(x) Y n n Y r nr! n r!, for all Y,, Y r X. M := + M X k,,x kr (x) a k a kr r> k,,k r > = + r ( + x ne xnx k a k )M X k,,x kr (x) r> <p < <p r i= k> ( = + x ne xnx k a k ) C[[x]][[X]] A. n> k> Computation of the coproduct of M: (which does not act on the X s) M Y,,Y r (x) = M = M M. r M Y,...,Y k (x) M Y k+,...,y r (x). k=
25 Transcription of the multiplication by the stuffle Property: (J. Y. Thibon, F. Chapoton, J. Ecalle, F. Menous, D. Sauzin,...) A family of objects (B n,,n r ) n,n 2,n 3, multiply by the stuffle product if, and only if, there exists a character χ z of QSym such that χ z ( Mn +,,n r +(x) ) = B n,,n r (z) () Consequences:. χ z can be extended to QSym[[X]], applying it terms by terms. ) χ z (M Y,,Y r (x) = B Y,,Y r (z), for all Y,, Y r X. 2. If B n,,n r multiply the stuffle, B = χ z(m) is grouplike in C[z][[X]] A.
26 Reformulation of the main goal Reformulation of the main goal Find some polynomials B n,,n r such that: B(z) a k = ezx k, e X k X k B(z + ) = B(z) E(z), where E(z) = + k> e zx k a k, B is a grouplike element of C[z][[X]] A.
27 Outline 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
28 A singular solution Remainder: E(z) = + k> e zx k a k. From a false solution to a singular solution... S(z) = n> E(z n) = + r> k,,k r > B(z) a k = false solution to system i= e z(x k + +X kr ) a r k a kr is a (e X k + +X ki ) ezx k e X k X k, B(z + ) = B(z) E(z), B is a grouplike element of C[z][[X]] A. Explanations:. B(z) = = B(z n) E(z n) E(z ) ( ) = = lim B(z n) E(z n). n + 2. S(z) C[z]((X)) A, S(z) C[z][[X]] A. Heuristic: Find a correction of S, to send it into C[z][[X]] A. n>
29 Another solution Fact: If (f )(z) = f (z ) f (z), ker zc[z] = {}. Consequence: There exist a unique family of polynomials such that: { B n,,n r (z + ) B n,,n r (z) = B n,,n r (z)z nr. B n,,n r () =. This produces a series B C[z][[X]] A defined by: B (z) = + B X k,,x kr (z) a k a kr. r> Lemma: (B., 23) k,,k r > The noncommutative series B is a grouplike element of C[z][[X]] A. 2 The coefficients of B (z) satisfy a recurence relation: B Y (z) = ezy, Y e Y X. B Y,,Y r (z) = BY +Y 2,Y 3,,Y r (z) B Y 2,Y 3,,Y r (z), Y e Y,, Y r X. 3 The series B can be expressed in terms of S: B (z) = ( S() ) S(z).
30 Characterization of the set of solutions Reminder: A family of multiple Bernoulli polynomials produces a series B such that: B(z + ) = B(z) E(z), where E(z) = + e zx k a k, k> B is a grouplike element of C[z][[X]] A, B(z) a k = Proposition: (B. 23) ezx k e X k X k. Any familly of polynomials which are solution of the previous system comes from a noncommutative series B C[z][[X]] A such that there exists b C[[X]] A satisfying:. b A k = 2. b is grouplike e X k X k 3. B(z) = b B = b (S() ) S(z). Theorem: (B., 23) The subgroup of grouplike series of C[z][[X]] A, with vanishing coefficients in length, acts on the set of all possible multiple Bernoulli polynomials, i.e. on the set of all possible algebraic renormalization.
31 Outline 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
32 Some notations New Goal: From B(z) = b B, determine a suitable series b such that the reflexion formula ( ) n B n( z) = B n(z), n N has a nice generalization. For a generic series s C[z][[X ]] A, s(z) = r N s X k,,x kr (z) a k a kr, k,,k r > we consider: s(z) = r N s(z) = r N s X kr,,x k (z) ak a kr k,,k r > k,,k r > s X k,, X kr (z) a k a kr
33 The reflection equation for B (z) Proposition: (B. 24) Let sg = + r> k,,k r > ( ) r a k a kr = ( + n> a n ). Then, S() = ( S() ) sg and S( z) = ( S(z) ). Corollary : (B. 24) For all z C, we have: sg B ( z) = ( B (z) ). Example: X, Y, Z B X,Y,Z X +Y,Z X,Y +Z ( z) = B (z) B (z) B (z) X +Y +Z B (z) + B Y,Z (z) + B Y +Z (z).
34 The generalization of the reflection formula Corollary 2: (B. 24) B( z) B(z) = b sg b. (2) Remark: S() sg S() =. Heuristic: A reasonable candidate for a multibernoulli polynomial comes from the coefficients of a series B(z) = b B (z) where b satisfies:. b a k = 2. b is grouplike e X k X k 3. b sg b =.
35 Outline 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
36 Resolution of an equation Goal: Characterise the solutions of { ũ sg u =. u is grouplike. Proposition: (B., 24) ( ) r 2 2r ( ) 2r a r k a kr.. Let us denote sg = + r> k,,k r > Any grouplike solution u of ũ sg u = comes from a primitive series v satisfying v + ṽ =, and is given by: u = exp(v) sg. If moreover u a k = e X k X k, then necessarily, we have: v a k = e X k X k + 2 := f (X k).
37 The choice of a series v New goal: Find a nice series v satisfying:. v is primitive. 2. v + ṽ =. 3. v a k = Remark: v a k is an odd formal series in X k X. Generalization: ṽ = v, so v = v. e X k X k + 2 = f (X k). = v a k a k2 = 2 f (X k + X k2 ), but does not determine v a k a k2 a k3. A restrictive condition: A natural condition is to have: there exists α r C such that v a k a kr = α r f (X k + + X kr ). Now, there is a unique primitive series v satisfying this condition and the new goal: v a k a kr = ( )r f (X k + + X kr ). r
38 Definition Definition : (B., 24) The series B(z) and b defined by { B(z) = exp(v) Sg (S()) S(z) b = exp(v) Sg are noncommutative series of C[z][[X]] A whose coefficients are respectively the exponential generating functions of multiple Bernoulli polynomials and multiple Bernoulli numbers. Example: The exponential generating function of bibernoulli polynomials and numbers are respectively: n,n 2 B n,n 2 (z) X n Y n 2 n! n 2! = 2 f (X + Y ) + 2 f (X )f (Y ) 2 f (X ) f (X ) ezy e Y 2 e zy e Y e z(x +Y ) + (e X )(e X +Y ) e zy (e X )(e Y ).
39 Examples of explicit expression for multiple Bernoulli numbers: Consequently, we obtain explicit expressions like, for n, n 2, n 3 > : b n,n 2 = 2 ( bn + ) b n2 + n + n 2 + bn +n 2 +. n + n 2 + b n + b n2 + b n3 + b n,n 2,n 3 = + 6 n + n 2 + n 3 + ( ) bn +n 2 + b n3 + 4 n + n 2 + n bn + b n2 +n 3 + n + n 2 + n 3 + b n +n 2 +n n + n 2 + n 3 +. Remark: If n =, n 2 = or n 3 =, the expressions are not so simple...
40 Table of Multiple Bernoulli Numbers in length 2 b p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 3 q = 8 2 q = q = 2 24 q = q = 4 54 q =
41 Outline 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
42 Table of Multiple Bernoulli Numbers in length 2 b p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 3 q = q = q = q = q = q = one out of four Multiple Bernoulli Numbers is null
43 Table of Multiple Bernoulli Numbers in length 2 be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = q = q = q = one out of four Multiple Bernoulli Numbers is null. one out of two antidiagonals is constant
44 Table of Multiple Bernoulli Numbers in length 2 be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = q = q = q = one out of four Multiple Bernoulli Numbers is null. one out of two antidiagonals is constant. symmetrie relatively to p = q
45 Table of Multiple Bernoulli Numbers in length 2 be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 3 q = q = q = q = q = q = 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 : =
46 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. b p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 q = 3 q = 4 q = 5 q = 6
47 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 q = 3 q = 4 q = 5 q = 6 48
48 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 24 q = q = 4 54 q = q =
49 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 24 q = q = 4 54 q = q =
50 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 24 q = q = 4 54 q = q =
51 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 24 q = q = 4 54 q = q =
52 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 24 q = q = 4 54 q = q =
53 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 q = q = q = 2 24 q = q = 4 54 q = q =
54 Construction Table of Multiple Bernoulli Numbers in length 2 From these previous remarks, let us present an algorithm to compute bibernoulli numbers. be p,q p = p = p = 2 p = 3 p = 4 p = 5 p = 6 3 q = 8 2 q = q = 2 24 q = q = 4 54 q = q =
55 Outline 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
56 Properties satisfied by multiple Bernoulli polynomials Proposition: (B., 23) The multiple Bernoulli polynomials B n,,n r multiply the stuffle. Theorem: (B., 24) P All multiple Bernoulli numbers satisfy : b 2p,,2p r =. P 2 If n r >, B n,,n r () = B n,,n r (). P 5 B n,,n r (z + ) B n,,n r (z) = B n,,n r (z) z nr. P 6 There exists a reflexion formula for multiple Bernoulli polynoms: B( z) B(z) =. P 7 The truncated multiple sums of powers S s,,s r N, defined by S s,,s r s s N = n n r r n r < <n <N are given by the coefficients of B (N).
57 Properties satisfied by multiple Bernoulli polynomials 2 Proposition: (B. 24) For all positive integers n,, n r, b n,,n r = b nr,,n. Proposition: (B., 24) For a series s(z) C[z][[X ]] A, let us define (s)(z) by: (s)(z) = X k X kr s(z) a k a kr a k a kr. r N is a derivation, and : k,,k r > P 4 The derivation of multiple Bernoulli polynomials are given by: ( zb(z) = b S() ) ( b S() ) B(z) + (B(z)).
58 Properties satisfied by multiple Bernoulli polynomials 3 Proposition: (B., 25) P 3 ( The recurrence relation of bibernoulli numbers is (partially) given by: p ( )( ) q p q 2 )be k,l be p,q = k l k= l= ( 2 ) ( p + 2 ) q + ( + 2 ) ( be q + be p p + 2 ) q + ( 2 ) ( ) p + 2 p + q + if p, q >. be p + 3 4
59 Conclusion. We have respectively defined the Multiple (divided) Bernoulli Polynomials and Multiple (divided) Bernoulli Numbers by: { B(z) = exp(v) Sg (S()) S(z) b = exp(v) Sg They both multiply the stuffle. 2. The Multiple Bernoulli Polynomials satisfy a nice generalization of: the nullity of b 2n+ if n >. the symmetry B n() = B n() if n >. the difference equation (B n)(x) = nx n. the reflection formula ( ) n B n( x) = B n(x). THANK YOU FOR YOUR ATTENTION!
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