I was dealing with a problem on tetration and am supposed to explain why this problem was challenging to me- obviously, difficulties stemmed from the amazing growth of $^{n}3$. The question now is: Is there a way to estimate this tetration in form of "elementary" functions, as the factorial is approximated by the Stirling formula, $$\mathcal{O}\left(\sqrt{2{\pi}n}\left(\frac{n}{e}\right)^n\right).$$ Is there an elementary representation of the form $\mathcal{O}(f(n))$ for this tetration as a function of $n$?
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I feared so, thanks anway- is there any more, say "tanglible" fucniton which does the trick as well? – IMOPUTFIE Mar 13 '20 at 06:19
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1@PeterForeman is it? – Simply Beautiful Art Mar 14 '20 at 20:01
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@PeterForeman The wording there is a bit unclear. One does not consider tetration as a composition of finitely many exponential functions because the amount of compositions is itself a function of $n$. As "amount of compositions" is not one of the listed operations, tetration is not considered elementary. – Simply Beautiful Art Mar 14 '20 at 22:30
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@PeterForeman You can see here for a similar discussion on why the factorial is not a polynomial (for a similar reason: factorial's degree does not exist, which is analogous to my answer). – Simply Beautiful Art Mar 14 '20 at 22:37
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@SimplyBeautifulArt Ah yes, I was assuming the function being considered was $3^{3^{\dots^{3^x}}}$ for some reason. – Peter Foreman Mar 14 '20 at 22:39
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By most definitions of elementary functions, the answer is no.
Every elementary $\mathbb N\to\mathbb R$ function in $n$ is bounded by ${}^kn$ for some natural $k$, since they are defined by a finite combination of operations and functions which are at most exponential.
As we have ${}^kn\ll{}^n3$, it follows that tetration grows faster than all elementary functions.
Simply Beautiful Art
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