7
$\begingroup$

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

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

into

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

where $B_k(x)$ are Bernoulli polynomials.

For instance,

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

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

etc.

$\endgroup$
  • $\begingroup$ There is no such a mapping except a trivial matching of coefficients, I think. $\endgroup$ – Henry.L Mar 31 '17 at 21:47
7
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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:

$$\int_{\mathbb{Z}_p}f(t)dt=\lim_{m\to\infty}p^{-m}\sum_{k=0}^{p^m-1}f(k).$$

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.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.