How it looks (i.θ)ₐ

[marraco]

and sexp(10) is a humongous number. The largest number pari-gp can represent is about \( \text{sexp}_e\left(4.08\right)\approx 7.24676\cdot 10^{131475196{ \)

That was my dumb mistake. I didn't needed sexp(10), but sexp(I*10).

¿Do you think that the tiny spirals at the end of the lines ͥ·ˣa (as x→∞) are accurate, or just a numerical artifact?

¿Did you attempted to find the base for which the lines turn into an "ellipse" (or something close)?. I guess a=e^(e^-1); that would "explain" the change of state.


Now, if the bases\( \\[15pt]

{1<a<e^{e^{-1}} } \)really turn into ellipses, then it should be easy to find an algebraic expression for tetration to real exponents, or at least an important insight for \( \\[20pt]

{^{\frac{1}{x}}a} \)

The base \( \\[15pt]

{a=e^{e^{-1}} } \) seems to match an ellipse with center near c=2.65599203615835 (Don't take that precision as accurate. I got it from Excel), relation of axis b=a, and radius \( \\[20pt]

{r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}} \), or

\( \left(imag(^{i.x}a)\right)^2 \,+\, \left({\frac{real(^{i.x}a)-c}{b}}\right)^2\,=\, r^2\\
\\
a=e^{e^{-1}} \\
b= e^{e^{-1}}\\
c= 2.655992036 \\
r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}\\
\\
W_{(x)}.e^{W_{(x)}}=x \,\Rightarrow \, e={\frac{1}{W_{(1)}}}^{\frac{1}{W_{(1)}}} \,\Rightarrow \, W_{(1)}=0,56714329
\)