08/18/2007, 09:35 AM
I just wanted to clarify that my super-logarithm solution only works for \( x > e^{1/e} \) even though my pdf says \( x > 1 \). I think that the reason for this is that tetration for bases between 1 and \( e^{1/e} \) has an upper limit (due to the convergence of \( {}^{\infty}x \)), and thus the super-logarithm has a limited domain over which it is defined. This upper limit of tetration also means that there will be a singularity on the boundary of this domain, which might explain why I have had such poor results as \( x \rightarrow e^{1/e} \) with experiments with my method.
When I first discovered the super-logarithm solution, the only singularities in the matrix equation (rather than the series expansion) seemed to be less than 1, this is what led me to the requirement \( x > 1 \), which I later learned to be not strict enough. Sorry. Hope this helps.
For an abstraction of my super-logarithm method, see http://math.eretrandre.org/tetrationforu...181#pid181
Andrew Robbins
When I first discovered the super-logarithm solution, the only singularities in the matrix equation (rather than the series expansion) seemed to be less than 1, this is what led me to the requirement \( x > 1 \), which I later learned to be not strict enough. Sorry. Hope this helps.
For an abstraction of my super-logarithm method, see http://math.eretrandre.org/tetrationforu...181#pid181
Andrew Robbins