08/17/2007, 11:20 PM
If by my solution you mean \( {\ }^{y}(e^{1/e}) \) through parabolic iteration of \( e^x-1 \) and by Daniel's you mean \( {\ }^{y}(b^{1/b}) \) through hyperbolic iteration of \( b^x-1 \), then yes, I'd be happy to provide some high-quality graphics and numerical tables. I still need to re-read Jay's solution, since it seems dependent on the super-logarithmic constant which I don't understand yet. Also you brought up an interesting point that may prevent comparison of all methods, since some methods work over non-overlapping intervals.
Tetration can be divided into several intervals in terms of what values of x in \( {}^{y}x \) are valid for that definition: the parabolic point \( e^{1/e} \), the hyperbolic interval \( (e^{-e}, e^{1/e}) \), and I suppose would you could call the elliptic? interval \( (e^{1/e}, \infty) \), which is what my definition of the super-logarithm is valid for. These are all closely tied to the interval of convergence of the infinite hyper-power \( {}^{\infty}x \) which was found almost 230 years ago.
My how time has passed...
Andrew Robbins
Tetration can be divided into several intervals in terms of what values of x in \( {}^{y}x \) are valid for that definition: the parabolic point \( e^{1/e} \), the hyperbolic interval \( (e^{-e}, e^{1/e}) \), and I suppose would you could call the elliptic? interval \( (e^{1/e}, \infty) \), which is what my definition of the super-logarithm is valid for. These are all closely tied to the interval of convergence of the infinite hyper-power \( {}^{\infty}x \) which was found almost 230 years ago.
My how time has passed...Andrew Robbins