Taylor series of upx function

[Kouznetsov]

...
To me, it would seem as if the first couple of derivatives at x=-1,0 were the same for both solutions, than it is likely that they're the same solution, though the two could still diverge in the higher derivatives.

I had submitted the evaluation of the first derivative with 14 (I hope) decimal digits.
\( \mathrm{tet}'(-1)=\mathrm{tet}'(0)\approx 1.09176735125832 \)
In the similar way, the second derivative can be evaluated, and so on.
But I like the approach by Bo: it is better to prove, than to compare the numerical evaluations. We already have proof that no other tetration is allowed to be holomorphic in the complex plane with cutted out set \( \{x~\in~ \mathbb{R} ~:~x\le -2 \} \).

By the way, how about to add some small amount of finction
\( d(z)= \left\{
\begin{array}{ccc}
\exp\!\Big(-0.1/x^2 - 0.1/(z+1)^2\big) &,& (x+1)x \ne 0\\
0 &,& (x+1)x = 0
\end{array}\right. \)?THis function has all the derivatives along the real axis, and all of them are zero at points 0 and -1.
Consider the modification of tetration \( \mathrm{tet}(z) \) to
\( \mathrm{tem}(z)=\mathrm{tet}(z)+d(z) \) with corresponding extension from the range [-1,0] usind the recurrent equation. At the real axis, the modified tetration tem has the same "expansion" as the tetration in these points; however, such a modification destroys the holomorphism.