Taylor series of upx function

[sheldonison]

There is EXACT Tailor series, if we insist that upx(z^*) = upx(z)^*
and upx(z) is holomorphic outside the part of the negative part of axis,
id est, holomorphic everywhere except \( z\le 2 \).
Then the solution is unique.
You can calculate so many derivatives as you like in any regular point, and, in particular, at z=0. The algorithm of evaluation is described in
http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf

Ok so far. My higher math skills aren't as good as I'd like them to be, so I'm not able to follow Dimitrii's paper. I'm looking at the pdf paper, and in section 2.1, Dimitrii writes:

(1.4) F(z + 1) = exp(F(z))
....
(2.1) ln(F(z + 1)) = F(z)
....
In this paper I assume that logarithm ln is single-valued function, that is analytic at the complex plane with cut along the negative part of the real axis. Then, all the solutions of equation (2.1) are solutions of (1.4), but it is not obvious, that a solution F of (1.4) satisfies also (2.1). One could add term 2\( \pi \)i to the right hand side of equation (2.1), and the solution of the resulting equation will be also solution of (1.4). Searching for the simple tetration, I begin with analytic solutions, which satisfy also equation (2.1).

Shortly after thereafter, Dimitrii generates the value L for the critical point, but by then, I was already lost. I think the key is that in the complex plane, exp and ln, equations (1.4) and (2.1) are no longer inverses of each other, or perhaps more likely, that f(z) is an approximation function, which may not hold when adding increments of 2pi*i. Do you need the taylor series for f(z) to generate L? Is the key adding increments of 2*pi*i to the taylor series expansion for approximations of F(z)? Or does that require the contour integrals, in section (4)?
- Sheldon