Is there something deeper here (like this is the truncation of an exact formula or something else going on)?


When the tower is $0$ deep, $1$ deep, $2$ deep, …, $8$ deep, you get: $${1., 0.693147, 1.06736, 1.41713, 1.96284, 3.14158, 8.09658, 375.216, 1.89078*10^{128}}$$ So it is probably just coincidence; I don't see any particular pattern there, I'm afraid.


Just an amusing coincidence.

Wolfy says it is 3.141577387...


I. Facts:



II. Taylor series:

$ln(x)=\frac{x-1}{x}+\frac{1}{2}(\frac{x-1}{x})^2+\frac{1}{3}(\frac{x-1}{x})^3+... ; x\geq\frac{1}{2} \tag{F1}$

$A^y=e^{ylnA}=1+\frac{ylnA}{1!}+\frac{(ylnA)^2}{2!}+...;y\in R \tag{F2}$

Via trial and error I rough approximate (F1) and (F2) into:

$ln(x)\approx\frac{x-1}{x}+\frac{1}{2}(\frac{x-1}{x})^2+Q(\frac{x-1}{x})^3 ; x\geq\frac{1}{2} \tag{F3}$

Where Q: \begin{cases} \frac{2}{3} & \quad \ \text{ $x\leq\pi$}\\ \frac{4}{5} & \quad \text{ $\pi<x\leq\frac{3}{2}\pi$}\\ 1 & \quad \text{ $\frac{3}{2}\pi<x$} \end{cases} $A^y\approx1+ylnA+\frac{2}{3}(ylnA)^2;y\in R \tag{F4}$

III. Approximations via (F3):





$ln6\approx1.76$ this reminds $\sqrt{3}$ and the fact $\pi>3$

IV. Let's define:






V. So:


We have to show that : $d\cong2$ (see: topic)

We gonna now use (F4),(F3) and the approximation(s) from point III.:


Via (F3):$ln(1.632)\approx0.52\approx0.5$ since in real $ln(1.632)\approx0.489806\approx0.5$ pretty close.

Hense: $d\cong1+\frac{c}{2}+\frac{2}{3}(\frac{c}{2})^2$

Because we know what we need for $d$- we can use this quadratic formula* to predict $c$:

$c\approx1.37$ which 'is' my $ln4$.

Cause we have define $c=(ln4)^b$ , we need here to show that $b\cong1$:


VI. Summary:

Via the Taylor series and rough approximations we could transform this power-tower and check some equations with a deviation into 7%

*$0=c^2+3c-6$; only the positive $c$ like in the power-tower.


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