Taylor series of upx function

[sheldonison]

Dear Zagreus. If you are still interested...
I have expressed the holomorphic function which satisfies the equation you wrote through the Cauchi integral. It allows the straigihtforward differentiation. I claim that such a function is unique, if we insist that
\( \mathrm{upx}(z^*)= \mathrm{upx}(z)^* \)
\( \mathrm{upx'}(x)>0 \forall x>-2 \)
upx is holonorphic in the complex plane except real values smaller or equal 2.
I have evaluated several coefficients in the Tailor expansion:
\( \mathrm{upx}(z)=\sum_{n=0}^\infty c_n z^n
~\forall z\in \mathbb{C}: |z|<2 \)
Here is the table of the coefficients \( c_0, c_1,c_1,.. \):
.....

Thank you for your table! About as far as I could go was to plug the taylor series coefficients into excel and show that the function and its first derivative match for several values for upx(x) and upx(x+1).

What does this equation mean?
\( \mathrm{upx}(z^*)= \mathrm{upx}(z)^* \)
I was expecting more along the lines of
upx(x+1) = exp ( upx(x) )

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? 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?

Thanks again for your post!
- Sheldon