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.
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. (Though you have to watch out (!) and not be misled by numerical or convergence error (is that a type of numerical error?) esp. considering slowdown of convergence as \( e^{1/e} \) is approached, and also the rounding errors of numerical differentiation if you're differentiating it numerically.)
3. Wouldn't it be \( \frac{\log(a)^2}{2} \)? Or do you mean the series for f expanded about the fixed point?
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. (Though you have to watch out (!) and not be misled by numerical or convergence error (is that a type of numerical error?) esp. considering slowdown of convergence as \( e^{1/e} \) is approached, and also the rounding errors of numerical differentiation if you're differentiating it numerically.)
3. Wouldn't it be \( \frac{\log(a)^2}{2} \)? Or do you mean the series for f expanded about the fixed point?
