-2

Well, Bernoulli umbra is an umbra whose moments are the Bernoulli numbers.

But what is it philosophically?

For instance, we can consider imaginary unit $i$ an umbra with moments $\{1,0,−1,0,1,\ldots\}$, hyperbolic unity $j$ as an umbra with moments $\{1,0,1,0,1,0,\dots\}$.

Can we in the same way somehow think about Bernoulli umbra as some kind of a hypercomplex "number", vector or series or whatever other object that can be geometrically represented and imagined, that has algebraic properties besides having those moments?

Is there any geometric or set-theoretic object that represents Bernoulli umbra?

Daniele Tampieri
  • 5,630
Anixx
  • 9,302
  • 2
    Where is a definition of this concept? I looked at the Wikipedia page, but it doesn't really say much. – Ryan Budney Jan 28 '23 at 04:17
  • 2
    At first glance it looks a like "umbral calculus" is a type of formula mysticism. – Ryan Budney Jan 28 '23 at 04:22
  • 5
    I suspect the downvotes are by people who don't know what the umbral calculus is. The question seems perfectly fine to me, though it's true that it would have been good to include some references. For those unfamiliar with the umbral calculus, I recommend Gessel's Applications of the classical umbral calculus as a starting point. – Timothy Chow Jan 28 '23 at 13:54
  • 3
    I see now that the OP has already posted several related questions here on MO; e.g., here, here, here and here. – Timothy Chow Jan 28 '23 at 14:32
  • 2
    Also here and here. Plus some of these have been cross-posted to math.SE as well. – Timothy Chow Jan 28 '23 at 14:40
  • 1
    I think wikipedia says the (bernoulli) umbra can be expressed as a linear map with property $L(z^n)=B_n$. https://en.wikipedia.org/wiki/Umbral_calculus?wprov=sfla1 – Esa Pulkkinen Jan 28 '23 at 16:19
  • 1
    @EsaPulkkinen which is mentioned in the question. Bernoulli umbra's moments are Bernoulli numbers, obviously. – Anixx Jan 28 '23 at 17:07
  • 1
    @Anixx I think it's not obvious that the moments of the umbra, bernoulli numbers, can be expressed as a linear map. I think it was established by a theorem attributed to Gian-Carlo Rota. Anyway, the linear map, applied to polynomials on $z$, has algebraic properties of interest here. – Esa Pulkkinen Jan 28 '23 at 17:24
  • 1
    @EsaPulkkinen well, as the question states, complex numbers and hyperbolic numbers can be seen as umbrae as well, but they have much more properties than those which can be derived from the moments. Thus I was asking about a similar view on Bernoulli umbra, that is, the properties and representations beyond the moments. – Anixx Jan 28 '23 at 17:30
  • 1
    I'm thinking analyzing how the polynomials $P(z)$ behave produces info on how $L(P(z))$ behave, that is, you can lift many theorems from binomial coefficients etc. to the bernoulli numbers based on this. Extending those considerations to power series on $z$ should be simple, and you might be able to solve equations involving bernoulli numbers by lifting corresponding operations on power seriies. With the definitions on imaginary unit as moments I suppose this extends to complex and hyperbolic identities. I do wonder if repeated bernoulli $B_{B_i}$ is way to do repeated lifting of power series – Esa Pulkkinen Jan 28 '23 at 17:51
  • 1
    @EsaPulkkinen well, look at this question of mine: https://mathoverflow.net/questions/431080/is-there-an-accurate-representation-of-bernoulli-umbra?noredirect=1&lq=1 It seems, we can represent Bernoulli umbra with power series, but this works only for natural powers of the umbra. Once we go non-integer or negative, this does not work. – Anixx Jan 28 '23 at 18:16
  • 1
    Ues definitely negative and non-integer powers are more difficult. It's not obvious to me the Bernoulli numbers can be extended to those indices. I do remember Knuth's TAOCP had brief discussion on Bernoulli numbers, but probably not sufficient for solving this. – Esa Pulkkinen Jan 28 '23 at 18:47
  • 1
    @EsaPulkkinen actually, I did it with divergent integrals. If we define umbral multiplication this way: $$\int_0^\infty f(x)dx\cdot\int_0^\infty g(x)dx=\int_0^\infty D^2 \Delta^{-1} \left(\Delta D^{-2}f(x)\cdot\Delta D^{-2}g(x)\right) dx +D \Delta^{-1} \left(\Delta D^{-2}f(x)\cdot\Delta D^{-2}g(x)\right)|_{x=0}$$ then the resulting algebra is isomorphic to umbral calculus, with regularization of the integral working as index-lowering operator. https://mathoverflow.net/a/403727/10059 – Anixx Jan 28 '23 at 18:53
  • 1
    @EsaPulkkinen The problem with it, this multiplication is quite complicated. asically, this is what you get as a result: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Jan 28 '23 at 18:53
  • 1
    My experience on this is limited by my experience in computer implementation of linear algebra see https://esapulkkinen.github.io/cifl-math-library/ , where I never considered the umbral algebra linear map. But it does have support for power series and many operations. Divergent integrals are not however well supported (infinite sequences are better supported). – Esa Pulkkinen Jan 28 '23 at 19:17
  • Umbra? -- the full answer is in "Skumbrie w tomacie", by Konstanty Ildefons Gałczyński. – Wlod AA Jan 31 '23 at 23:19

1 Answers1

6

Since you used the terms hypercomplex and geometric, I interpret your question as asking whether people have found any connection between Bernoulli umbrae and Clifford algebra. I'm not too familiar with this area of research myself, but the answer seems to be yes. See for example Bernoulli and Euler Polynomials in Clifford Analysis by G.F. Hassan and L. Aloui (Advances in Applied Clifford Algebras 25 (2015), 351–376) or (Discrete) Almansi Type Decompositions: An umbral calculus framework based on $\mathfrak{osp}(1|2)$ symmetries by Nelson Faustino and Guangbin Ren (arXiv:1102.5434) and the references therein.

LSpice
  • 11,423
Timothy Chow
  • 78,129