(04/23/2015, 04:52 PM)marraco Wrote: 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
\( \left(imag(^{i.x}a)\right)^2 \,+\, \left({\frac{real(^{i.x}a)-c}{b}}\right)^2\,=\, r^2\\
\\
a=e^{e^{-1}} \\
b= e^{e^{-1}}\\
c= 2.655992036 \\
r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}\\
\\
W_{(x)}.e^{W_{(x)}}=x \,\Rightarrow \, W_{(1)}=0,56714329
\)
I've been trying to follow this thread, and finally I have something to contribute. For the bases \( 1 < a \le e^{1/e} \) it might be easier to use my expansion of this function. See http://arxiv.org/pdf/1503.07555v1.pdf
I came up with a holomorphic expression for \( ^^z a \) for these bases. It's a fast converging expression as well. It is not as messy as a Taylor series expansion of this same function. It's also a single holomorphic expression for all \( \Re(z) > 0 \), greatly reducing computational time.
This is for the periodic/pseudoperiodic extension of tetration (regular koenigs iteration) for bases \( 1 < a \le e^{1/e} \)
The expression isn't so easy to write out:
\( ^^z a= \frac{1}{\G(1-z)} (\sum_{n=0}^\infty (^^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty (^^{n+1}a) \frac{(-w)^n}{n!}) w^{-z}\,dw) \)
It is a periodic solution, except at eta.