I wonder, can anyone describe an expression or formula of a transform that converts

$$\sum_{k=0}^\infty \frac{a_k x^k}{k!}$$


$$\sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$$

where $B_k(x)$ are Bernoulli polynomials.

For instance,

$$x^2\, \to\, x^2-x+1/6$$

$$e^x \,\to \, \frac{e^x}{e-1}$$


  • $\begingroup$ There is no such a mapping except a trivial matching of coefficients, I think. $\endgroup$ – Henry.L Mar 31 '17 at 21:47

An operator performing the mapping is $$O= D/(e^D-1)=e^{B.(0)D},$$

where $D=d/dx $ and $(B.(0))^n=B_n(x)|_{x=0}$, since the Bernoulli polynomials are an Appell sequence.

Edit (6/20/2017):

This operator is essentially the Todd operator. See the discussions on pg. 30 and Appendix B of "Permutohedra, associahedra, and beyond" by Postnikov of the Todd operator as a transform of the homogeneous volume polynomials for classes of polytopes into a generalized Ehrhart polynomial coding the number of lattice points in the polytopes.

(Edit 8/2018)

For some idea of the importance of this Todd operator in modern mathematics and physics, see New Models for Veneziano Amplitudes: Combinatorial, Symplectic and Supersymmetric Aspects by Kholodenko.

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  • $\begingroup$ This is a general result with the Bernoulli numbers replaced by any moment/fundamental- base sequence $a_n = A_n (0)$ of any Appell sequence $A_n(x)=(a.+x)^n $. $\endgroup$ – Tom Copeland Apr 1 '17 at 0:57
  • $\begingroup$ This is also Umbral Calculus. $\endgroup$ – T. Amdeberhan Apr 1 '17 at 1:40
  • $\begingroup$ The inverse transformation is simply $(e^D-1)/D=e^{R.(0)D}$ with $R_n(0)= 1/(n+1)$ and associated Appell sequence $R_n(x)=(R.(0)+x)^n$; therefore, $R_n(B.(x))=x^n=B_n(R.(x))$, and the two Appell sequences form an umbral compositional inverse pair. $\endgroup$ – Tom Copeland Apr 2 '17 at 18:02
  • $\begingroup$ In this notation, what would be the same but applied to the integral of a function? Also, I wonder what would be non-operator expression for this transform. $\endgroup$ – Anixx Apr 8 '17 at 10:05
  • $\begingroup$ Is there a name for such operator anywhere? $\endgroup$ – Anixx Apr 8 '17 at 10:34

Another way, somewhat related with the above answers, is the $p$-adic Volkenborn integral. You can find this, for example, in Schikhof's or in Alain Robert's books on $p$-adic calculus, or Henri Cohen vol. 2 of his books on number theory. This approach is useful because of the relation of Bernoulli numbers and L-functions: one can easily define good and elementary $p$-adic zeta functions using the Volkenborn integral (actually, this was Kubota and Leopoldt's original approach).

Let $\mathbb{Z}_p$ be the ring of $p$-adic integers and let $\mathbb{C}_p$ be the topological completion of an algebraic closure of the field of fractions $\mathbb{Q}_p$ of $\mathbb{Z}_p$ (a nice and large field for doing $p$-adic analysis). Let $f:\mathbb{Z}_p\to\mathbb{C}_p$ be an analytic function, that is, $f$ is of the form

$$f(x)=\sum_{n\ge0}a_n\frac{x^n}{n!},\qquad a_n\in\mathbb{C}_p,\quad \frac{a_n}{n!}\to0.$$

(We suppose $f$ analytic for simplicity and because of what you are asking). Then the Volkenborn integral of $f$ is defined by the following $p$-adic limit:


Then, one has the following relation with Bernoulli numbers and polynomials:

$$\int_{\mathbb{Z}_p}t^ndt=B_n$$ and $$\int_{\mathbb{Z}_p}(x+t)^ndt=B_n(x).$$

This Volkenborn integral is a continuous linear operator on a Banach space of functions (see the books mentioned above). Hence, with $f$ as above, one obtains:

$$\int_{\mathbb{Z}_p}f(t)dt=\sum_{n\ge0}a_n\frac{B_n}{n!}$$ and $$\int_{\mathbb{Z}_p}f(x+t)dt=\sum_{n\ge0}a_n\frac{B_n(x)}{n!}.$$

Hope this helps.

Note: This integral is a special case of "$p$-adic distributions", which are one of the main tools that are now used to define $p$-adic zeta functions attached to arthmetic objects. See, for example, Washington or Lang books on cyclotomic fields for a nice introduction.

PS: For a nice "general zeta functions" interpretation of your question, see Lemma 2.4 in this article by Friedman and Pereira https://arxiv.org/abs/1105.2603 It was published in the IJNT, but the arxiv version is the same as the published version.

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The transfert operator describe an expression or formula of a transform that $\sum_{k=0}^\infty \frac{a_k x^k}{k!}$ into $ \sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$ is the p-adic operator such that :The eigenvalues of the p-adic transfer operator are the Bernoulli polynomials,and are associated with the eigenvalues $p^{-n}$ , Try to check this paper by LINAS VEPŠTAS, page 8. Theorem with proof show that

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