Talk:Constant-recursive sequence

Be more explicit about eventually-periodic case

Two possible improvements to this article:

• First, I think we should update the article to be clearer about the case of sequences like $1,0,0,0,\ldots$. This is allowed as a constant-recursive sequence according to the current text (see Eventually periodic sequences) but sections like Characterization in terms of exponential polynomials do not apply for such sequences.

• Second, the article should clearly place itself relative to linear difference equation. These are basically the same concept. I think the latter article is excluding the eventually-periodic case, though. And maybe we should here too... best idea would be to dig up a reference textbook and see how they define it.

Thoughts? Happy to make some of these changes when I get the chance. Caleb Stanford (talk) 19:00, 7 November 2021 (UTC)

Eventually periodic sequences can only be excluded artificially, since "$s(n+\ell )=s(n)$ for all $n\geq N$" is equivalent to "$s(n+N+\ell )=s(n+N)$ for all $n\geq 0$", which satisfies the definition of being constant-recursive. I agree it's worth discussing this in the article, as well as the fact that an "eventually constant-recursive" sequence is constant-recursive, for the same reason. The sequence $1,0,0,0,\dots$ is described by an exponential polynomial, namely $0^{n}$ since zero to the power of zero is $1$ when the exponent only takes on integer values. Eric Rowland (talk) 23:31, 7 November 2021 (UTC)

Thanks User:Eric Rowland, but how would you plan to represent $0,1,0,0,\ldots$ as an exponential polynomial -- since $0^{n-1}$ doesn't work? To exclude eventually-periodic sequences non-artificially, we just have to require that in the satisfied linear recurrence $s(n)=c_{1}s(n-1)+c_{2}s(n-2)+\dots +c_{d}s(n-d),$ $c_{d}\neq 0$. Caleb Stanford (talk) 23:41, 7 November 2021 (UTC)

Good point. I checked The Concrete Tetrahedron (by Kauers and Paule), and they do require $c_{d}\neq 0$. This is artificial from the perspective of generating series because then the characterization as rational series needs an extra requirement on the degree of the numerator. But on the other hand it does fix the characterization as exponential polynomials. Eric Rowland (talk) 00:47, 8 November 2021 (UTC)

Thanks for looking into another reference! My preference would then be to have $c_{d}\neq 0$. Three Two One other justifications for that criterion: (i) it allows the sequence to be uniquely extended in the negative direction as well, (ii) it's implied by the "alternate definition" under "Definition" in the article, namely "the set of sequences $\{s(n+r)_{n\geq 0}:r\geq 0\}$ is contained in a vector space whose dimension is finite." For $1,0,0,0,...$ we get the set of all finite-support sequences, which is infinite-dimensional. Caleb Stanford (talk) 01:06, 8 November 2021 (UTC)

I'm fine with requiring $c_{d}\neq 0$, if you want to make the change. I agree with your point (i). For your point (ii), if $s(n)_{n\geq 0}$ is the sequence $1,0,0,0,...$, then $s(n+1)_{n\geq 0}$ and all higher shifts are the zero sequence, right? In that case $\{s(n+r)_{n\geq 0}:r\geq 0\}$ is contained in the $1$-dimensional space of sequences of the form $a,0,0,0,...$. Eric Rowland (talk) 01:17, 8 November 2021 (UTC)

Oh you are right about (ii), my mistake. OK, thanks for the discussion. Caleb Stanford (talk) 01:33, 8 November 2021 (UTC)

Given that some definitions (see (ii) and the generating series definition) point to eventually-periodic sequences naturally being considered constant-recursive, I think I'm now somewhat inclined to allow $c_{d}=0$. Also, it leads to a more general definition, in terms of many of the results (e.g. closure properties). For now, I edited the article with some clarifications to the definition. Caleb Stanford (talk) 01:57, 8 November 2021 (UTC)