Talk:Principle of Recursive Definition

Perhaps this page should be move to Finite Recursion so as to distinguish it from Principle of Transfinite Recursion Andrew Salmon 21:45, 24 July 2012 (UTC)


 * Don't think so. Could add a "also known as" section, if you like. But this result is generally known as the "Principle of Recursive Definition" (at least in the books I got) so I don't see the need to rename. --prime mover 22:06, 24 July 2012 (UTC)


 * Perhaps add a redirect from that page to this? I don't know how. --Andrew Salmon 22:13, 24 July 2012 (UTC)

Presentation issue
It should be noted somewhere in the presentation that we are actually just defining the function $n \mapsto f^n(a)$. Thoughts on how to do this? &mdash; Lord_Farin (talk) 13:36, 19 April 2015 (UTC)


 * $g^n (a)$ don't you mean? Except it's not, it's $g^{n - p} (a)$ which is sufficiently un-intuitive (I had to get a piece of paper and work it out carefully) to require that it ought to be a page in its own right.


 * It's interesting that none of the sources I've seen make this point explicit.


 * Suggest we establish that result, then in each of the pages (outside the included section) make the statement that: "From (that result) we note that $g (p+n) = g^n (p)$". Then again on the parent page -- so as to avoid having this same statement made 4 times on the parent page. --prime mover (talk) 14:20, 19 April 2015 (UTC)


 * It's actually $f(p+n) = g^{n-p}(a)$, and indeed, it is not required that $T = \N$.


 * Slowly, ideas for restructuring these pages are coming together. I am however having trouble with the $p$-based things. Since the only imaginable cases are $p = 0$ and $p = 1$, I would suggest that we prove it for $0$, and add $1$ as a corollary (easy, through Image of Successor Mapping forms Peano Structure). What do you say? In the mean time, I will try to get started in my sandbox. &mdash; Lord_Farin (talk) 15:11, 19 April 2015 (UTC)


 * I've seen a few cases in number theory where you start from somewhere that isn't $0$ or $1$, can't remember exactly where now. So the "generalised" version where you start at an arbitrary point $p$ is useful, otherwise you're going to have to jump through some extra logical hoops before you can get starter. The $p$-based approach is the treatment taken by Warner, and I would recommend we keep it. --prime mover (talk) 15:31, 19 April 2015 (UTC)


 * Fair enough. I will make the $p$-version a corollary, then. The important argument doesn't depend on the starting element, and I don't want the proofs littered about over multiple pages just because authors have made different choices as to the level of generality. &mdash; Lord_Farin (talk) 15:35, 19 April 2015 (UTC)


 * Being brutally awkward as usual, I would suggest that the corollaries are for $p = 0$ and $p = 1$ -- they are applications of the truly more general version where $n$ starts at $p$. --prime mover (talk) 18:18, 19 April 2015 (UTC)

I have thought about that, but paedagogic considerations really make me favour the $0$-case on Principle of Recursive Definition &mdash; I think it aids the reader's understanding to present it in this way. &mdash; Lord_Farin (talk) 18:25, 19 April 2015 (UTC)


 * We'll have to agree to disagree on this point, then. No worries. --prime mover (talk) 18:33, 19 April 2015 (UTC)


 * Reading some further, I have decided to make it a "General Result" rather than a corollary. Also saves me the effort of writing a connecting proof. &mdash; Lord_Farin (talk) 11:35, 5 May 2015 (UTC)