Tetration:Complex

From Tetration.net

1 Complex Tetration

Complex Tetration

The focus here is to derive ${}^{n}a_{(s_0, \ldots)}; a,n \in C \,$ . In general $a \rightarrow 0 \rightarrow k = 1 \,$ where $k>1 \,$ , therefore if $a \rightarrow z \rightarrow (k-1) = f(z) \,$ then $a \rightarrow n \rightarrow k \,$ can be found by taking $f^{n}(z) \,$ and setting $z= 1 \,$ giving $a \rightarrow n \rightarrow k = f^{n}(1) \,$ .

Hyperbolic Tetration (First three terms)

Consider how to define hyperbolic tetration for ${}^na \;$ where the function to be continuously iterated is $f(z) \equiv a^z \;$ and $a^{a_0}=a_0 \;$ .

Note: Sorry about the formatting, this site isn't properly parsing certain TeX commands. ${}^n a = {a_0} + \left( 1 - {a_0} \right) \, {\left( \log (a)\,{a_0} \right) }^n + \frac{{\log (a)}^2\,{\left( 1 - {a_0} \right) }^2\,{a_0}\, {\left( \log (a)\,{a_0} \right) }^{-1 + n}\, \left( -1 + {\left( \log (a)\,{a_0} \right) }^ n \right) }{2\,\left( -1 + \log (a)\,{a_0} \right) }$ $+ \frac{1}{6} {\left( 1 - {a_0} \right) }^3\, \left( \frac{{\log (a)}^3\,{a_0}\, {\left( \log (a)\,{a_0} \right) }^{-1 + n}\, \left( -1 + {\left( \log (a)\,{a_0} \right) }^n \right) \, \left( 1 + {\left( \log (a)\,{a_0} \right) }^n \right) }{\left( -1 + \log (a)\,{a_0} \right) \, \left( 1 + \log (a)\,{a_0} \right) } + \frac{3\,{\log (a)}^4\,{{a_0}}^2\, {\left( \log (a)\,{a_0} \right) }^{-2 + n}\, \left( -1 + {\left( \log (a)\,{a_0} \right) }^n \right) \, \left( -\left( \log (a)\,{a_0} \right) + {\left( \log (a)\,{a_0} \right) }^n \right) }{{\left( -1 + \log (a)\,{a_0} \right) }^2\, \left( 1 + \log (a)\,{a_0} \right) } \right)$

Parabolic Tetration (First three terms)

$e^{1/e}\approx 1.444667861 \,$

Using the equation for parabolic continuous iteration, $f^{\,n}(z)= f_0 +(z-f_0) + \frac{1}{2} n f_2 (z-f_0)^2 + \frac{1}{12} (3 (n^2-n) f_2^2 + 2 n f_3) (z-f_0)^3 + \ldots$ where $f(z)=\left ( e^{1/e} \right )^z$ and then setting $z={}^0 \left ( e^{1/e} \right ) = 1$ gives, $f^{\,n}(1)={}^n \left ( e^{1/e} \right )$

${}^n \left ( e^{1/e} \right ) = 1 + \frac{{\left( 1 - e \right) }^2\,e\,n\,{\log (e^{1/e})}^2}{2} + \frac{{\left( 1 - e \right) }^3\, \left( e\,n\,{\log (e^{1/e})}^3 + 3\,e^2\,\left( -n - \frac{\left( -1 + n \right) \,n}{2} + n^2 \right) \,{\log (e^{1/e})}^4 \right) }{6}$

Rationally Neutral Tetration

Period 2

Consider the bifurcation point between period one and period two at $e^{-e} \approx 0.065988 \;$ . The exponential map $\left (e^{-e} \right )^z$ has a derivative of $- 1$ at its fixed point. Therefore its second iterate ${\left (e^{-e} \right )}^{{\left (e^{-e} \right )}^z}$ has a derivative of $1$ at its fixed point and can be solved as a case of parabolic continuous iteration.

Period n

In general if an exponential map has a first derivative that is a n^th root of unity at its fixed point, then the n^th iterate of the exponential function can be solved using parabolic continuous iteration.

Superattracting Tetration

${}^z1 = \begin{cases}1 & \operatorname{Re}(z) \ge 0 \\ undefined & \operatorname{Re}(z) < 0 \end{cases}$