# Why is $(\ln 6)^{(\ln 5)^{(\ln 4)^{(\ln 3)^{(\ln 2)}}}} \approx\pi$ (matches upto 4 digits).

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:

$$ln6\approx1.792$$

$$\sqrt{\pi}\approx1.772$$

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

III. Approximations via (F3):

$$ln2\approx0.708\approx\frac{1}{\sqrt{2}}$$

$$ln3\approx1.09$$

$$ln4\approx1.37$$

$$ln5\approx1.632$$

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

IV. Let's define:

$$a=ln2$$

$$b=(ln3)^a$$

$$c=(ln4)^b$$

$$d=(ln5)^c$$

$$f=(ln6)^d$$

V. So:

$$f\approx(\sqrt{\pi})^d=\pi^\frac{d}{2}$$

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

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

$$d=(ln5)^c\approx(1.632)^c\approx1+cln(1.632)+\frac{2}{3}(cln(1.632))^2$$

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 $$b\cong1$$:

$$b=(ln3)^a=(ln3)^\frac{1}{\sqrt{2}}\approx1+\frac{1}{\sqrt{2}}ln(1.09)+\frac{2}{3}(\frac{1}{\sqrt{2}}ln(1.09))^2\approx1+\frac{0.086}{1.41}+\frac{0.086^2}{3}\approx1.0635$$

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.