Tetration:Complex

From Tetration.net

Contents

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 nth root of unity at its fixed point, then the nth 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}