Talk:Constant-recursive sequence

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Be more explicit about eventually-periodic case

Latest comment: 3 years ago3 comments2 people in discussion

Two possible improvements to this article:

• First, I think we should update the article to be clearer about the case of sequences like ${\displaystyle 1,0,0,0,\ldots }$ . 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)Reply

Eventually periodic sequences can only be excluded artificially, since "${\displaystyle s(n+\ell )=s(n)}$  for all ${\displaystyle n\geq N}$ " is equivalent to "${\displaystyle s(n+N+\ell )=s(n+N)}$  for all ${\displaystyle n\geq 0}$ ", 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 ${\displaystyle 1,0,0,0,\dots }$  is described by an exponential polynomial, namely ${\displaystyle 0^{n}}$  since zero to the power of zero is ${\displaystyle 1}$  when the exponent only takes on integer values. Eric Rowland (talk) 23:31, 7 November 2021 (UTC)Reply

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