5,...\}$?

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Dec 10, 2023 at 0:26	comment	added	Anixx		@SidharthGhoshal well, yes. Moreover, this anti-umbra can be represented as set of integrable functions on interval $(0,1)$, with evaluation operator being $\int_0^1 f(x)dx$ and $\overline{B}=x$, see my recent answer. Unfortunately, for normal, Bernoulli umbra there is no such easy representation.
Dec 10, 2023 at 0:13	comment	added	Sidharth Ghoshal		Ah there probably is some connection with what you’re finding and this idea. $\ln(1+x)$ is the function inverse of $e^x -1$. $\frac{1}{e^x-1}$ has a Laurent series that generates the Bernoulli numbers. $\frac{1}{\ln(1+x)}$ also has a rational Taylor series which encodes some kind of arithmetic information in it. It seems reasonable to wonder if there’s a connection with that and some nice operator of the form $D^{\pm} \Delta^{\pm}$
Dec 9, 2023 at 23:24	comment	added	Anixx		@SidharthGhoshal $D$ is derivative and $\Delta^{-1}$ is antidifference.
Dec 9, 2023 at 23:16	comment	added	Sidharth Ghoshal		If $\Delta^{-1}$ is a forward difference then what is $D$? I’m trying to interpret the symbol $D\Delta^{-1}$? Is that just a derivative?
Dec 9, 2023 at 20:18	answer	added	Anixx		timeline score: 0
May 27, 2023 at 13:49	comment	added	Anixx		@TomCopeland ah, inderstood, thanks
May 27, 2023 at 13:49	history	edited	Anixx	CC BY-SA 4.0	deleted 10 characters in body
May 27, 2023 at 13:45	comment	added	Tom Copeland		'It seems' = 'apparently' are usually understood as flagging a conjecture or hypothesis, not a proven statement. I simply note that it is fact, so you can drop the qualification 'it seems'.
May 27, 2023 at 13:21	comment	added	Anixx		@TomCopeland what quaifier do you mean to drop? I did not get it.
Feb 3, 2023 at 18:29	history	edited	Tom Copeland		Added two more tags
Feb 3, 2023 at 17:09	comment	added	Tom Copeland		From the sliding average formula in my answer, in my notation, $\psi(\bar{b}.+x) = \psi(\bar{B}.(x)) = \int_x^{x+1} \psi(t) dt = \ln(x)$ (from Wolfram Alpha).
Feb 3, 2023 at 16:58	comment	added	Tom Copeland		In your notation, $\frac{x}{1+\bar{B}x} = x - \bar{B}^1 x^2 + \bar{B}^2 x^3 - \bar{B}^3 x^4 + \cdots$ umbrally evaluated gives $x - \bar{B}_1 x_2 + \bar{B}_2 x^3 - \bar{B}_3 x^4 + \cdots = x - x_2/2 + x^3/3 - x^4 /4+ \cdots = \ln(1+x)$, so you can drop the qualifier 'it seems'.
Feb 2, 2023 at 17:05	history	edited	YCor		edited tags
Feb 2, 2023 at 16:44	answer	added	Tom Copeland		timeline score: 4
Jan 29, 2023 at 12:33	comment	added	Anixx		Why is the downvote?
Jan 28, 2023 at 15:56	comment	added	Anixx		Also posted here on Jan, 24: math.stackexchange.com/questions/4624584/…
Jan 28, 2023 at 15:55	history	asked	Anixx	CC BY-SA 4.0

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