Does $$\lim_{n\rightarrow \infty}\left (r+\frac{1}{n^2}\right)\uparrow \uparrow n=e$$ hold ?

$r$ is the number $e^{e^{-1}}$ , the largest real number for which the infinite power tower $r\uparrow r\uparrow r\uparrow \cdots$ converges. So we have a power tower of $n$ numbers having the value $e^{e^{-1}}+\frac{1}{n^2}$. Numerical examples I calculated with PARI/GP indicate that the error is about $\frac{3.614}{n}$.

  • $\begingroup$ Consider the related phenomena for Pi in the Mandelbrot set for $c=\frac{1}{4}+\epsilon$ which was explained by Gerald Edgar pi314.net/eng/mandelbrot.php One can put the iterations of $x \mapsto x^x+x+c$ into correspondance with $\tan^{-1}(y)$ More later ...limited time today. $\endgroup$ – Sheldon L Dec 9 '17 at 19:20
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    $\begingroup$ $r=\exp(1/e); b=r+\frac{1}{n^2}$ Then a more accurate estimate of n is $\text{slog}_b(e) \approx n \cdot \frac{\pi}{2}\sqrt{\frac{2r}{e}}$ $\endgroup$ – Sheldon L Dec 9 '17 at 21:13

From Gerald Edgar's 1991/1992 explanation of Pi in the Mandelbrot set, we learn that iterating a function of the type $x \mapsto x^2+x+\epsilon\;$ will take approximately $\frac{\pi}{\sqrt{\epsilon}}$ iterations to escape, where we start from the critical point at x=-0.5. From Gerald Edgar's Pi and the Mandelbrot set, "So our equation now reads $y'(n) = y^2 + \epsilon$. This has the solution $a\tan(an+c)$ where $a = \sqrt{\epsilon}$"

It turns out we can we put the equation iterating the Op's tetration expression into a similar form, for $$b>\exp\left(\frac{1}{e}\right);\;\;\; x \mapsto b^x$$

First we observe that if $\epsilon=\ln(\ln(b))+1$, and $y=x\cdot\ln(b)+(-1+\epsilon)$, then iterating $$y \mapsto \exp(y)-1+\epsilon\;\;\;\text{is exactly congruent to}\;\;\; x \mapsto b^x$$

Also notice that $\epsilon$ approaches zero as b approaches $\exp(1/e)$, and $\exp(y)-1=y+\frac{y^2}{2}+\frac{y^3}{6}...$ so this is also close to the desired form except it has $\frac{y^2}{2}$ instead of $y^2$

So instead, we need $$z=\frac{y}{2}= \frac{x\cdot\ln(b)+(-1+\epsilon)}{2}$$ and then we iterate $$z \mapsto \frac{\exp(2z)-1+\epsilon}{2}\;\;\;\text{is exactly congruent to}\;\;\; x \mapsto b^x$$

And $\frac{\exp(2z)-1}{2}=z+z^2+\frac{2z^3}{3}+\frac{z^4}{3}...$ has exactly the desired form to approximate $\frac{\pi}{\sqrt{\epsilon}}$ iterations by the tangent approximation.

So now with a little algebra, we need an equation for $\epsilon$ in terms of the Op's equation. $$\epsilon = \ln\left(\ln\left( \exp(1/e)+\frac{1}{n^2}\right)\right)+1$$ $$\epsilon = \frac{e}{n^2\exp(1/e)} + \mathcal{O}(\frac{1}{n^4})$$

We are iterating $z\mapsto f(z)+\epsilon/2\;\;$ so we expect the approximation for the total escape time to be $$\pi \cdot n \sqrt{\frac{2\exp(1/e)}{e}}$$

But the Op is interested in how long it takes to iterate until we get to the value of e, which is just about halfway... The halfway point is where the tangent approximation is centered at the inflection point which is 3 or 4 iterations before the halfway point since sexp(0)=1 and since the inflection point is 1.5 iterations to the left of e. So here is my final approximation. $$b=\exp\left(\frac{1}{e}\right)+\frac{1}{n^2};\;\;\;\text{slog}_b(e)\approx \frac{\pi\cdot n}{2}\sqrt{\frac{2\exp(1/e)}{e}}\approx 1.619465272666037 \cdot n$$

Here $n=\text{slog}_b(x)$ refers to the inverse function of $x = b \uparrow \uparrow n$ tetration notation extended to the reals. Lets use this approximation for $\frac{1}{n^2}=10^{-10}$, and then the equation gives 161946.5 iterations, and the iteration crosses $e$ between 161941..161942 iterations, so this is a very good estimate.

  • $\begingroup$ An amazing answer (+1) , so we can say that the above limit is actually correct, right ? $\endgroup$ – Peter Dec 11 '17 at 9:56
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    $\begingroup$ @Peter Yes the limit is correct; yes within about -4 using the equation I posted as n goes to infinity. $\endgroup$ – Sheldon L Dec 11 '17 at 13:07
  • $\begingroup$ @Peter Edgar's $a\tan(an+c)$ equation where $a=\sqrt{\epsilon}=\mathcal{O}\frac{1}{n}$ could be used to prove the limit also holds for n iterations of $b \uparrow \uparrow n$ approaching arbitrarily close to e even though it uses n iterations instead of the more accurate 1.619465272666037n iterations. $\endgroup$ – Sheldon L Dec 11 '17 at 20:56

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