some questions about sexp

[mike3]

(question 1)
does the following hold :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k ?

Though tempting for uniqueness this seems not to be the case, the half iterate of exp(x)-1 (which is conjugate to \( e^{x/e} \)) has the following powerseries coefficients at 0:

Code:

0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120,...

That Taylor series does not converge (I've seen this one before, it's from the parabolic regular iteration, right?), though. So \( 0 \) is actually a singularity of some kind of \( \mathrm{dxp}^{1/2}_{e^{1/e}}(x) \) (\( \mathrm{dxp}_b(x) = \exp_b(x) - 1 \)) and so there is no Taylor expansion there.
moderators note: corrected latex expression

What about (here for base \( e^{1/e} \))

\( \frac{d^n}{dx^n} \mathrm{sexp}_{e^{1/e}}(x) \) does not change sign for all \( x > 0 \)

?