Derivative of exp^[1/2] at the fixed point?

[andydude]

And the formal half iterate begins with
\( \exp^{0.5}(z+L) = L + \sqrt{L}z + \frac{L\cdot z^2}{2(L+\sqrt{L})} + ... \)

Ok, so I replaced y with 1/2 and log(L) with L in the regular iteration power series to get this:
\( \exp^{1/2}(z + L)
= L
+ \sqrt{L} z
+ \frac{\sqrt{L}}{2(1 + \sqrt{L})} z^2
+ \frac{\sqrt{L} - 3L + 4L^{3/2} - 3L^2 + L^{5/2}}{6(1 + L)(1 - L)^2} z^3
+ \cdots
\)
as expected it's the same power series.

I wanted to highlight one of my findings in this paper (page 12) that is related but separate from this, which is a power series for \( f^{1/x}(x) \) for any analytic function \( f \) with a parabolic fixed point at 0.

\( f^{1/x}(x)
= \frac{x}{1 - f_2}
+ \left(f_2 - \frac{f_3}{f_2}\right)\frac{\log(1 - f_2)}{(1 - f_2)^2} x^2
+ \cdots
\)

Substituting in \( f(z) = \exp_{\eta}(z) = \exp(z/e) \) we get

\( \exp_{\eta}^{1/z}(z)
= e
+ 2(z-e)
- \frac{2 \log(2)}{3e} (z - e)^2
+ \frac{(1 + \log(4))^2}{18e^2} (z - e)^3
+ \cdots
\)

which I realize is a different base, but still interesting.

Using a similar technique, we might be able to find a comparable power series for \( \exp^{1/z}(z) \), at which point \( z=2 \) would at the very least intersect with this function.