Well here is my problem :
Problem :
Define :
$$f\left(x\right)=x^{(x!)^{(x)^{x!}}},f(1-x),g(x)=x,g(1-x)$$
These curves define a square (?) see the plot below :
I want to find the cartesian coordinate of the square's center .
As attempt and by symmetry I know it's something like $(0.5,y)$ . Unfortunetaly I cannot find the remainders .
If there is no closed form how to find it with a reasonable accuracy ?
Additional question :
I have a doubt on the geometric form so : Is it a square ?
partial answer as remarks :
This definitively not a square at least in plotting it and due to the equation of $f(x)$ .
To find the center of the pseudo-square we can use derivative at :
$$f'(a)=1$$
And next we have an exact square . Remains to show the closed form mention in eyeballfrog comment .
The nice thing is that we already know that the center of the square lies on $x = 1/2$. That means that the $y$ coordinate of the center of the square is halfway between $1/2$ and $f(1/2)$. Using $(1/2)! = \sqrt{\pi}/2$, we get $$ f(1/2) = (1/2)^{(\sqrt{\pi}/2)^{(1/2)^{\sqrt{\pi}/2}}} = 2^{-(\sqrt{\pi}/2)^{2^{-\sqrt{\pi}/2}}}. $$ Not the cleanest of expressions, but it works. We then use this to find $y$: $$ y = \frac{1}{2}\left[\frac{1}{2} + f\left(\frac{1}{2}\right)\right] = \frac{1}{4}\left[1 + 2^{1-(\sqrt{\pi}/2)^{2^{-\sqrt{\pi}/2}}}\right]. $$ If you wanted to approximate this value, note that $ \sqrt{\pi}/2 \approx 0.886$ is somewhat close to $1$, so you could write $\sqrt{\pi}/2 = 1 - \delta$ and then expand in power series in $\delta$. To first order you'd get $y \approx 1/2 + (\delta/8)\ln 2\approx 0.50986$, only 0.26% off the true value $0.51120$.