Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1.
Say \( \beta \) is an Abel function of e^x-1, then
\( \alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0)) \)
is an Abel function of \( e^x \). This should also work for beta being the Abel function of \( 2\sinh(x) \).
This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2].
[1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502.
[2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733.
Say \( \beta \) is an Abel function of e^x-1, then
\( \alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0)) \)
is an Abel function of \( e^x \). This should also work for beta being the Abel function of \( 2\sinh(x) \).
This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2].
[1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502.
[2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733.