If $x$ is a real number greater than $e^{e^{-1}}$ , then $x\uparrow \uparrow n$ (A power tower of $n$ $x's$) tends to $\infty$, if $n$ tends to $\infty$. Therefore, there must be a number $n$, such that $x\uparrow\uparrow n>10^{100}$

Can I determine the smallest number $n$ satisfying this inequality without applying the iteration $x_1=x$ , $x_{n+1}=x^{x_n}$ ?

For example, for $\color\red {x=e^{e^{-1}}+10^{-10}}$, we have $\color\green {n=323\ 892}$

  • 2
    A generalization of Kneser's method allows to handle real bases greater $\eta = e^{1/e}$ and from this there is an $\text{SLOG}()$-function (I'd like it to be called (iteration)-"height"-function $\text{HGH}()$ btw.) There is a program for Pari/GP where this generalization is implemented. (see Sheldon Levenstein, Tetration-forum). If I handle this correctly, then for $x=2$ it seems to give $n \approx 4.6579446897813824$. Reading your question a second time, you want a solution without iteration of the exponentiation/iterated log itself. But it cannot work with such large numbers without iter.) – Gottfried Helms Dec 8 '17 at 7:19
  • There are two different questions here: (1) how to approximate the slog_b(10^100) for b near some arbitrary value, say b=3. (2) how to approximate the slog_b(10^100) near the singularity at b=exp(1/e). The two questions have very different answers. For (1), surprisingly there is an analytic function. But it has a singularity at b=exp(1/e). For (2), I would start with my answer to math.stackexchange.com/questions/2558545/… ... – Sheldon L Jul 27 at 23:19

Consider the function $f(t)=x^t/t$. The issue with computing such for $x$ extremely close to $e^{e^{-1}}$ is a region where $f(t)\approx1$. And so the challenge is to try and "skip over" this region, as I attempt to do:

We have $f'(t)=0$ when $t=\frac1{\ln(x)}$, and hence for $x>e^{e^{-1}}$, we have

$$f(t)\ge f\left(\frac1{\ln(x)}\right)=e\ln(x)>0$$

Hence, the inequality

$$x^t\ge te\ln(x)$$

From which we may deduce results such as

$$x_{n+k}\ge x_k(e\ln(x))^n$$

Allowing faster computation of $x_n$ when $f(x_n)$ gets extremely close to $1$.

Further, if we let $e\ln(x)=1+y$, then

$$(e\ln(x))^n=(1+y)^k=\exp(n\ln(1+y))=\exp(n(y\pm y^2/2))$$


For example, with $x=e^{e^{-1}}+10^{-10}$, and given $x_{55}\approx2.626~381~821~026~68$, we may deduce that

$$x_{n+55}\ge2.626~381~821~026~68\times\exp(1.881~596~387~466~52n\times10^{-10})$$

Hence, if we want $x_{n+55}\ge2.8123$, we get

$$n\approx363~497~128$$

From there you can brute force a few more iterations until you exceed $10^{100}$.

which admittedly is a terrible approximation, but I guess what'd you expect?

  • Note: numbers at the end may be inaccurate due to floating point... – Simply Beautiful Art Jul 20 at 2:23

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; with the Op using x as the base: find n such that $x \uparrow \uparrow n = 10^{100}$, where $10^{100}$ is a good approximation for escapes to infinity. With a little algebra, we need an equation for $\epsilon$ in terms of x from the Op's equation. Here; I suggest using the shorthand $\eta=\exp(1/e)$, and then if $x>\eta$, but only a little bigger than $\eta$.

$$\epsilon = \ln\left(\ln\left(x + \eta-\eta \right)\right)+1$$ $$\epsilon = \frac{e\cdot \left(x-\eta\right)}{\eta} + \mathcal{O} \left(x-\eta\right)^2$$

We are iterating $z\mapsto f(z)+\epsilon/2\;\;$ so we expect the approximation for the total escape time to be

$$n \approx \frac{\pi}{\sqrt{x-\eta}} \cdot \sqrt{\frac{2\eta}{e}}$$

$$n \approx \frac{\pi}{\sqrt{x-\eta}}\sqrt{\frac{2\eta}{e}};\;\;\;x\uparrow \uparrow n >\approx 10^{100}$$

For $x=\eta+10^{-10}\;\;\;n\approx 323893.054$ so I would use this approximation. I give both the equation for n in terms of x, and for x in terms of n. $$n \approx -0.94 + \frac{\pi}{\sqrt{x-\eta}}\sqrt{\frac{2\eta}{e}};\;\;\;x\approx \eta+\frac{2\eta\pi^2}{e}\left(\frac{1}{n+0.94}\right)^2;\;\;\;x\uparrow \uparrow n \approx 10^{100}$$

This approximation seems to work remarkably well for values $x=\eta+\epsilon$ where $\epsilon<2.5\cdot 10^{-3}$. This approximation for the number of iterations required is far more accurate than one would naively expect. For example, it is accurate with an error of 0.11 or less iterations when $60 \leq n \leq 10^8$; For example, when used with the correct value of x for n=10^7, (found by iteration), the equation for n predicts 10000000.00141 iterations. The error 10^6,10^7, and 10^8 are almost the same.

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