Taylor series of upx function

[Kouznetsov]

What does this equation mean?
\( \mathrm{upx}(z^*)= \mathrm{upx}(z)^* \)

The asterisc * means the complex conjugation. You can evaluate function at the complex–conjugated argument, and the result is the conjugated value of the function. In particular, for real values of the argument, the function is real.

I was expecting more along the lines of
upx(x+1) = exp ( upx(x) )

??
These are not just lines, this is all the range of complex values of x, except part of the real axis \( x\le -2 \).

The next line is also intriguing....
\( \mathrm{upx'}(x)>0 \forall x>-2 \)
Does this mean that other holomorphic solutions exist, but they all must have a negative derivative at some point x>-2?

No, but condition that the derivative is positive simplifies my proof of the uniqueness. Recently, Henryk Trappmann suggested the proof through the inverse function; it does not use the requirement that, at the real argument, the derivative is positive. I see no errors in his proof.
So, we may remove this claim about the first derivative.

I started working on the algebra to generate the general equations for all 5th and 6th order upx approximations that have smooth 1st, 2nd, and 3rd derivatives, but not necessarily smooth 4th and 5th derivatives. You would claim that all members of that set, if extended with higher order coefficients so that all derivatives would be smooth, would have a negative 1st derivative somewhere?

Your question is not clear.
Instead of "somewhere" you could say "at some real value of the argument, larger, than \( -2 \)".
Then the answer is "not".
You may consider the function \(
f(x)=\mathrm{upx}(x+s\cdot\sin(2 \pi x)) \)
at small real values of parameter \( s \), id est, \( -1<s<1 \).
Smoothness of all the derivatives at the real axis is not sufficient for the uniqueness. We have to claim the holomorphism at least for the domain with real part of the argument larger than \( -2 \).
I have pictures for the singular solutions too.