Dmitrii Kouznetsov's Tetration Extension

[Kouznetsov]

See Analytic solution of F(z+1)=exp(F(z)) in complex z-plane for more information.

About the uniqueness: It is well known that if we have a solution \( \alpha \) of the Abel equation \( \alpha(z+1)=f(\alpha(z)) \) then for any 1-periodic function \( \phi \) also \( \beta(z)=\alpha(z+\phi(z)) \) is a solution to the Abel equation. (Because \( \beta(z+1)=\alpha(z+1+\phi(z+1))=\alpha(z+\phi(z)+1)=f(\alpha(z+\phi(z))=f( \beta( z)) \)).

So let \( F \) be one solution of
(*) \( F(z+1)=\exp(F(z)) \) with
(**) \( \lim_{y\to\infty} F(x+iy) = L \) and \( \lim_{y\to -\infty} F(x+iy)=L^\ast \)
then \( G(z)=G(z+\sin(z)) \) is another solution of (*). Let us now consider (**). We know that \( \sin(z)=-i\frac{e^{iz}-e^{-iz}}{2} \) and \( \sin(x+iy)=i\frac{-e^{ix}e^{-y}+e^{-ix}e^y}{2} \)

\( G(x+iy)=F\left(x+iy+i\frac{-e^{ix}e^{-y}+e^{-ix}e^y}{2}\right) \). As \( e^{-y}\to 0 \)
at least for x=0 also
\( \lim_{y\to\infty} G(iy)=\lim_{y\to\infty} F(i(y+e^y))=L \)

By the way, your deduction gives the hint, how to prove the Theorem 1.

Perhaps. I am not convinced yet that it is true, but if it was, this would be great. We even know vice versa if we have two solutions \( F \) and \( G \) of the Abel equation then \( G^{-1}(F(x+1))-(x+1)=G^{-1}(\exp(F(x)))-(x+1)=G^{-1}(\exp(G(G^{-1}(F(x)))-(x+1)=G^{-1}(F(x))+1-(x+1)=G^{-1}(F(x))-x \), meaning that \( \phi=G^{-1}\circ F-\text{id} \) is a 1-periodic function, so we know already that each other solution \( G \) of the Abel equation must be of the form \( F(z+\phi(z)) \) for some 1-periodic \( \phi \). To prove theorem 1 everything depends on the behaviour of those 1-periodic functions for \( \Im(z)\to\infty \). For uniqueness roughly the real value must be go to infinity for the imaginary argument going to infinity.

(However, we have to scale the argument of sin function.)

Yes, my negligence. The \( z \) in \( \sin(z) \) has to be replaced by \( \frac{z}{2\pi} \) in the previous post.

How about the collaboration?

That would be great.

Can you please compute values on the real axis for bases \( b<e^{1/e} \) for example for \( b=\sqrt{2} \)? I would like to compare your solution with the regular tetration developed at the lower real fixed point (which would be 2 in the case of \( b=\sqrt{2} \)) of \( b^x \).

Bo, I got your message about base b=e^(1/e) and b=sqrt(2). In these cases, the real part of quasiperiod is zero, and I cannot run my algorithm as is. I need to adopt it. It will take time. I do not think that b=sqrt(2) is of specific interest (just integer L(b)=4); we need to consider the general case.
You may advance faster than I do. You may begin with the plot of the asymptotic period T(b) and analysis of its limiting behavior in vicinity of b=1 and b=e^(1/e). Please, provide the good approximation for (at least) the leading terms.

P.S. you may also correct misprints in your post:
invert the scaling factor for the argument of sin, and
delete the expression with unmatched parenthesis.