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I've been wondering about how it might be possible, given two compound exponential numbers generated by recurrence schemes equipped with addition multiplication and exponentiation (to some integer base), to systematically (not just muddling through on an ad-hoc basis) decide which is the greater.

This thought was prompted by my having looked at a couple of extremely interesting theorems (or demonstrations) that incur compound exponential numbers.

One is the overthrowing of the conjecture by Leo Moser that, given an arbitrary distance, a set of $n$ points on a sphere can be arranged such that there are at most $cn$ instances (out of the total of $n(n-1)/2$ distances of distance between them) of the given distance, where $c$ is some absolute constant. A group of researchers (I'll get the precise reference in a little while) exhibited a most ingenious construction whereby given any distance $\alpha<\pi$, and any number $k$, it is possible to construct an arrangement of points on the unit sphere such that every point has $k$ points at distance $\alpha$ from it. The trouble is ... the number of points required is given by $\operatorname{n}(k)$, where $\operatorname{n}(0)=1$, & $$\operatorname{n}(k+1)=\operatorname{n}(k)×2^{\operatorname{n}(k)} !$$ (That's an exclamation, by the way ... it isn't factorialed on top of that!!) So if you want to arrange points such that each has three others at a specified distance from it, you need $2^{11}=2048$ points; and if four occurences, then $2^{2059}$ points! (This is the number of Planck volumes comprised in the observable universe raised to some (small) power!).

Another (and this is better known, so I'll refrain from expounding this in fine detail) is Gijswijt's sequence - a sequence with a supremum that grows exceedingly slowly: the number 4 occurs first at index 220 ... and 5 has never been found - but it isn't expected to be seen explicitly either, as, though the sequence is unbounded, its been shown that the first occurence of a given number is at index about $$2^{2^{3^{\dots^{n-1}}}} .$$

So I've shown two very simple (simple in their underlying concept, at least) examples of recipes that disgourge compound exponential numbers. But it could be far from obvious, given two, which is the greater. Even in these two examples, the number produced is not a pure compound exponential. And there may simply not be enough sheer precision at hand to compute the uttermost indices and compare those. I have seen a principle that the author called the principle of insignificance expounded, that captures how very large changes in the lower indices correspond to absolutely miniscule changes in the uttermost index, such that if there is the least uncertainty in the uttermost index, it scarcely matters whether you make the lower ones $2$s or $10$s or $e$s ... or what! And also it might well be in another situation that our compound exponential number is generated by a much more complicated recipe.

So it would seem to me that the only way to compare two such numbers in general, would be to do it referencing purely the recipe itself, putting-by any notion that we might in any realistic sense apprehend the actual magnitude of them. I once gave the theory of ordinals & cardinals a going-over, and I have fragments of recollection that there might be something in that from which a system could be devised for making such a test as I have queried after here.

I can imagine that if the exponentiation is to a fixed base, common to both recipes, that there might be a relatively simple tree-based method ... but that if the base is not fixed, then it would be a very much more difficult matter. But that's just my imagination of it. And it's partly why I chose those two particular examples.

And that paper I mentioned is at

https://users.renyi.hu/~p_erdos/Erdos.html

1989-02 P. Erdős, D. Hickerson, J. Pach: A problem of Leo Moser about repeated distances on the sphere, Amer. Math. Monthly 96 (1989) no. 7, 569--575 (MR90h:52008; Zentralblatt 737.05006

And it's Leo Moser ... Moser with unumlauted o! I'll get the name right eventually!

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  • $\begingroup$ Much interesting thoughts! Did you come across Robert Munafo's "big numbers" and "hypercalc"-pages?(For a start mrob.com/pub/perl/hypercalc.html) In discussing his "hypercalc" he exposed some thoughts similar to yours in regard of comparision and comparatibility(? correct word?) of very large numbers. $\endgroup$ – Gottfried Helms Dec 29 '18 at 9:31
  • $\begingroup$ I've never seen that! It's clearly the same kind of thought. I do not know whether that program would handle numbers of such nature as I'm talking about, as they would not arise as a simple power-tower, but rather as an upside-down power-tree, with branches diverging as proceeding upwards, if you see what I mean. ¶ That business of points on a sphere & number of neighbours at an arbitrarily specified distance: have you seen that paper I mention? You might recall I said I think non-integer tetration would require a fundamentally new kind of thought ... that paper might be an instance! $\endgroup$ – AmbretteOrrisey Dec 29 '18 at 13:58
  • $\begingroup$ Ambrette- your thoughts are surely more general than that of R. Manufo: his implementation is just a procedure around one specific aspect. - That "points-on-a-sphere" sounds like a curious thing, perhaps next week I'll try to see what the paper tells me. - That request for a fundamentally rethinking rings positively some bells (as I've already commented earlier). I'd like much to see a (serious) new approach to that problem of fractal iteration - however I might no more be much creative there, my mind seems meanwhile to be a bit "worn-out", so-to-say... $\endgroup$ – Gottfried Helms Dec 29 '18 at 15:40
  • $\begingroup$ Maybe more general ... but not worked-out - which is why I'm posting on here! ¶ Is it fairly clear what I mean by using the algorithms by which the numbers are created directly to compare them? Say you go through them side-by-side & step-by-step: at each point you might say "doing this causes n to exceed m" - not because you are comparing the sizes as such, but rather because you know as a property of the algorithms themselves that that step causes n to exceed m; and further on, m might begin to exceed n. Something like ordinal arithmetic, as I hinted. $\endgroup$ – AmbretteOrrisey Dec 29 '18 at 18:19

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