Let the notation be $a^{\wedge\wedge}b = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_{b\,times}$ for tetration.

My mentor conjectured the following:

Let $n$ be a positive integer, then let $A(n)$ be any function that satisfies

$$\lim\limits_{n\to\infty} \left(e^{\frac1e} + \frac1n\right)^{\wedge\wedge}\left[(10 n)^{1/2} + n^{A(n)} + C+o(1)\right] - n = 0$$

where $C $ is a constant. Then $\lim\limits_{n\to\infty} A(n) = \frac1e $

Conjectured by tommy1729 Here : http://math.eretrandre.org/tetrationforum/showthread.php?tid=262&page=4

So could this be true ?

I have no idea how to do limits like this. I assume these type of limits are not in the books.

  • $\begingroup$ What's that $A(n)$? $\endgroup$ – user99914 May 31 '15 at 23:54
  • $\begingroup$ A(n) is just the function that makes the equation valid. $\endgroup$ – mick May 31 '15 at 23:57
  • $\begingroup$ Also, I believe that $C+o(1)=o(1)=c$ is a constant, so it is a little unnecessary. $\endgroup$ – Pauly B May 31 '15 at 23:58
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    $\begingroup$ $a^{\wedge\wedge}b$ is not defined for non-integers. I think the $o(1)$ is intended to account for this. It might be better wrapped in a ceiling function if this is the case, as he says that $0 \leq o(1) \leq 1$. $\endgroup$ – user208649 Jun 1 '15 at 0:04
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    $\begingroup$ I don't think that the limit given can be equal to zero. $e^{\frac{1}{e}} > 1$, and tetration definitely has greater than linear growth. $\endgroup$ – user208649 Jun 1 '15 at 0:14

Here is the correct formula, where $\eta=\exp(1/e)$ and $\alpha(x)$ is the upper repelling fixed point Abel function for iterating $x \mapsto \exp(x)-1$, which is generated using Ecalle's fps solution. For details, see posts by Will Jagy on $\alpha(x)$: How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$?

$$\lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[\pi\sqrt{\frac{2\eta\cdot n}{e}} -2\right] \approx 388.7874$$ I obtained the above limit value numerically, and then used that result to generate a corrected equation for the Op's question. $$\lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[\pi\sqrt{\frac{2\eta\cdot n}{e}} + \alpha(\frac{n}{e}-1) + C\right] -n = 0$$ $$C \approx -2 - \alpha(\frac{388.7874}{e}-1)$$

I was about to post a closely related question about Pi in the Mandelbrot set; it takes about $\pi \sqrt{n}$ iterations to escape near the parabolic cusp at c=0.25+1/n. Then I found this paper about the occurrence of Pi in the Mandelbrot set; although I haven't finished reading their paper, but presumably the same linear differential equation mechanisms can be used to justify the result, that it takes $\pi \sqrt{2n}$ iterations to "escape" for iterating $x \mapsto \exp(x)-1+\frac1n$, where we start at $x=-1+\frac1n$. Both iterations involve perturbations of $\frac1n$ near a parabolic fixed point.

http://www.doc.ic.ac.uk/~jb/teaching/jmc/pi-in-mandelbrot.pdf https://people.math.osu.edu/edgar.2/piand.html

It is simpler and mathematically equivalent to work with iterating $f(x)=\exp(x)-1+\frac1n$. Using the paper's methods, then one would want to prove it takes $\pi\sqrt{2n}$ iterations, for the function to begin growing, where growth would be defined as $f^{\circ \pi\sqrt{2n}}>2$; after that growth is superexponential.

n for iterating $f(x)=\exp(x)-1+\frac1n$ is equivalent to $n=\ln(\ln(\eta+1/m))+1\approx \frac{e}{\eta\cdot m} + \frac{\mathcal{O}}{m^2}$ for iterating $g(x) =(\eta+1/m)^x$ Then there is a simple linear conversion $f^{\circ k} = \frac{g^{\circ k}}{e}-1$ That is why I used $\alpha(\frac{n}{e}-1)$ in my corrected solution equation for the Op's question.

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    $\begingroup$ Also, notice that $\pi\sqrt{\frac{2\eta\cdot n}{e}} \approx \sqrt{10.49n}$, so this is close to the Op's original $\sqrt{10n}$ term. $\endgroup$ – Sheldon L Jun 15 '15 at 12:14

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