using sinh(x) ?

[bo198214]

Also, I think the superfunction of 2sinh is entire; it has no singularities.

Yes, generally the regular superfunction is entire at a repelling fixed point (because the inverse Schröder function is).

The math is much easier than the complex fixed point of base e, followed by a Riemann mapping...

Well but for the derived superfunction of b^x we dont know even whether it is analytic.
While we know that for the Kneser-solution.

My secret wish was that for the \( b^x-b^{-x} \) superfunction, the base change function would work without a wobble. I generated the superfunction for \( f(x)=2^x-2^{-x} \) with a base change back to the superfunction for 2sinh, \( f(x)=e^x-e^{-x} \). As expected, there was a small (magnitude of +/-0.02%) 1-cyclic wobble...

Hm, so even there! Actually I found out that you can do the basechange also for bases smaller \( \eta \): The limit
\( \mu_{a,b}(x)=\lim_{n\to\infty} \log_a^{[n]}(\exp_b^{[n]}(x)) \) exists for all \( 1<a<b \) and \( x\in (-\infty,\infty) \) and satisfies \( \mu_{a,b}\circ\exp_b = \exp_a\circ\mu_{a,b} \).
However it is constant for \( x\le a^+ \) (the upper fixed point of of \( a^x \)). BUT it is strictly increasing for \( x>a^+ \)! This means we can use the change of base to change bases of regular tetration at the *upper* fixed point.

But even then:
If \( \sigma_a \) and \( \sigma_b \) are the regular superfunctions to base \( a \) and \( b \) respectively at the *upper* fixed point, then
\( \mu_{a,b}\circ \sigma_b \) is a superfunction to base \( a \) but
\( \mu_{a,b}\circ \sigma_b \neq \sigma_a \).