(08/29/2007, 05:04 AM)andydude Wrote: Well, I have given this a lot of thought, and I believe that it is easier to prove real-analytic tetration than complex-analytic tetration. The reason for this is that the real-valued tetration has a smaller domain than complex-valued tetration. The first realization is that tetration over real numbers can produce complex numbers. After this realization we can eliminate a great deal of the domain over which real-analycity must fail (because it is not continuous, and if its not continuous it can't be analytic). To show what I mean by this I have included a color-coded plot of the log(abs(b^^x)) where gray is a finite real number, blue is a complex number output, and red is indeterminate.

• Raw Domain
  
• Pretty Domain
  

I have also included a pretty-version of the domain where the circles indicate indeterminate outputs. The dark-gray quarter-plane is that largest domain over which real-analytic tetration can be defined. The medium-gray are regions which have real outputs, but the dotted line indicates a discontinuity, so this would make real-analycity fail if this were in the domain of real-tetration. The light-gray region is probably not real-valued, but it is real-valued with a first-approximation tetraiton (linear critical).

A mathematical definition of the dark-gray domain is:

\(
D = \left\{ (b, x) \text{ where } b > 0 \text{ and } \begin{cases}
x > -1 & \text{ if } b = 1, \\
x > -2 & \text{ otherwise}.
\end{cases}
\right\}
\)

So to summarize, I believe that what needs to be proven is that Tetration whose domain is the dark-gray region given above is real-analytic in both b and x. Once it is proven that real-analytic tetration exists over this domain, then we can worry about its uniqueness.

Andrew Robbins

(07/20/2010, 10:02 PM)tommy1729 Wrote: the links dont work andrew. ( message : forbidden )

so i cant understand what you meant.