11/13/2009, 08:51 AM
(11/13/2009, 02:22 AM)mike3 Wrote: 1. The limit formula will simply fail to converge when the fixpoint is no longer attracting, no? (unless you reverse it, but then you get a different function so this cannot be interpreted as continuation of the formula) If so, that just marks the region of convergence of that limit formula, it does not imply that it is a natural boundary, just as the failure of the convergence of a Taylor series beyond its radius of convergence does not imply that such radius is a natural boundary.
yes, yes, I know my argumentation is very weak. Its more a feeling. Lets see whether sometime I can make this more precise, or whether it just turns out to be misleading.
Quote:2. If \( e^{1/e} \) is a singularity, as you seem to suspect, the derivative should explode at that point -- consider \( \frac{d}{dx} \sqrt{x} \) as \( x \rightarrow 0^{+} \). Though I just realized, it need not be the first derivative, but a higher one may do it. It can't be a smooth point.
I dont think so, if all derivatives from one approaching sector exists the function can still be non-analytic at that point. In this case one says f has an asymptotic powerseries development. This is particularly true for the non-integer iteration of \( e^x-1 \). You can derive a powerseries at 0 and you have an (or rather two) analytic function(s) whose derivatives are compatible with the powerseries development at 0 when approaching 0 from one direction. But the function itself is not analytic, because the two analytic functions defined in opposite sectors are not continuations of each other.
Quote:3. Wouldn't it be \( \frac{\log(a)^2}{2} \)? Or do you mean the series for f expanded about the fixed point?
Mike, I explained in my previous posts with what functions I work. The function \( f \) in the powerseries formula (which is \( g \) in my original post) is \( \ln(a)(e^x-1) \) where \( a \) is the lower fixed point.
This is a linear conjugation of \( b^x \) which moves the lower fixed point to 0.
