04/13/2009, 05:27 AM
So I tried using regular and natural tetration to fill out the rest of the graph. I almost ignored using bases where \( 1 < a < \eta \) because these bases would have 2 fixed points: (1) the upper fixed point would overlap the "blue" region which I already calculated with interpolation (but a good area for comparison), and (2) the lower fixed point would overlap the "green" region which I already calculated with natural iteration (also a good area for comparison).
The only part I used regular tetration for was the "red" region, which corresponds to bases where \( e^{-e} < a < 1 \). Using pure regular iteration from the primary fixed point of \( a^z \) the base-a tetrational is real-valued only for integers, and complex-valued everywhere else. So I fudged a little bit to make the graph, I solve the equation \( \text{Re}({}^{x}a) = x \) and these points are what you see in the "red" region of the graph. I'm pretty sure these are wrong, and that the graph would have to be complex here (if tetrationals for bases less than 1 form complex outputs for real inputs), so it's more of a heuristic than a continuation of the blue curve.
Andrew Robbins
The only part I used regular tetration for was the "red" region, which corresponds to bases where \( e^{-e} < a < 1 \). Using pure regular iteration from the primary fixed point of \( a^z \) the base-a tetrational is real-valued only for integers, and complex-valued everywhere else. So I fudged a little bit to make the graph, I solve the equation \( \text{Re}({}^{x}a) = x \) and these points are what you see in the "red" region of the graph. I'm pretty sure these are wrong, and that the graph would have to be complex here (if tetrationals for bases less than 1 form complex outputs for real inputs), so it's more of a heuristic than a continuation of the blue curve.
Andrew Robbins