User talk:Dmitrii Kouznetsov/Analytic Tetration

Henryk Trappmann 's theorems

This is approach to the Second part of the Theorem 0, which is still absent in the main text.

Copypast from http://math.eretrandre.org/tetrationforum/showthread.php?tid=165&pid=2458#pid2458

Theorem T1. (about Gamma function)

Let $~F~$ be holomorphic on the right half plane let $~F(z+1)=zF(z)~$ for all $~z~$ such that $~\Re (z)>0~$.
Let $~F(1)=1~$.
Let $F$ be bounded on the strip $~1\leq \Re (z)<2~$.
Then $~F~$ is the gamma function.

Proof, see in Reinhold Remmert, "Funktionnentheorie", Springer, 1995. As reference is given: H. Wielandt 1939. (Mein Gott, so old reference!)

Consider function $~v=F-\Gamma ~$ on the right half plane, it also satisfies equation $~v(z+1)~=~z~v(z)~$ Hence, $~v~$ has a meromorphic continuation to $~\mathbb {C} ~$; and the poles are allowed only at non–positive integer values of the argument.

While $~v(1)=0~$, we have $~\lim _{z\rightarrow 0}z~v(z)=0~$, hence, $~v~$ has a holomorphic continuation to 0 and also to each $~-n~$, $~n\in \mathbb {N}$ by $~v(z+1)=z~v(z)~$.

In the range $~1\leq \Re (z)<2~$, $~v(z)~$ is pounded. It is because function $~\Gamma ~$ is bounded there.

Then $~v(z)~$ is also restricted on $~\mathbb {S} ~$, because $~v(z)!$ and $~v(1-z)!$ have the same values on $~\mathbb {S} ~$. Now $~q(z+1)=-q(z)~$, hence $~q~$ is bounded on whole $~\mathbb {C} ~$, and by the Liouville Theorem, $~q(z)=q(1)=0$. Hence, $~v=0~$ and $~F=\Gamma ~$.

(end of proof)

Theorem T2 (about exponential)

Let $~E~$ be solution of $~E(z+1)=bE(x)~$, $~E(1)=b~$, bounded in the strip $~0\leq \Re (z)<1~$.

Then $~E~$ is exponential on base $~b~$, id est, $~E=\exp _{b}~$.

Proof. We know that every other solution must be of the form $~g(z)=f(z+p(z))~$ where $~p~$ is a 1-periodic holomorphic function. This can roughly be seen by showing periodicity of $~h(z)=f^{-1}(g(z))-z~$.

$~f(z+p(z))=b^{z+p(z)}=b^{p(z)}b^{z}=b^{p}(z)f(z)~=q(z)f(z)~$,

where $~q(z)=b^{p}(z)~$ is also a 1-periodic function,

While each of $~f~$ and $~g~$ is bounded on $~\mathbb {S} ~$, $~q~$ must be bounded too.

Theorem T3 (about Fibbonachi)

Let $~\phi ={\frac {1+{\sqrt {5}}}{2}}~$.
Let $~F(z+1)=F(z)+F(z-1)~$ Let $~F(0)=1~$ Let $~F(1)=1~$

Then $~F(z)={\frac {\phi ^{z}-(1-\phi )^{z}}{\sqrt {5}}}$

Discussion. Id est, $~F~$ is Fibbonachi function.

Theorem T4 (about tetration)

First intent to formulate

Let $~b>\exp(1/\mathrm {e} )~$.
Let each of $~F_{1}~$ and $~F_{2}~$ satisfies conditions

$~\exp _{b}(F(z))=F(z+1)~$ for $\Re (z)>-2$

$~F(z)~$ is holomorphic function, bounded in the strip $~|\Re (z)|\leq 1~$ .

Then $~F_{1}=F_{2}~$

Second intent to formulate

(0) Let $~b>\exp(1/\mathrm {e} )~$.

(1) Let each of $~F_{1}~$ and $~F_{2}~$ is holomorphic function on $~\mathbb {C} ^{\prime }=\mathbb {C} \backslash (-\infty ,-2]$, satisfying conditions

(2) $F(0)=1$

(3) $~\exp _{b}(F(z))=F(z+1)~$ for $\Re (z)>-2$

(4) $~F~$ is bounded on $~\mathbb {S} =~\{x+\mathrm {i} y|-2<x\leq 1,y\in \mathbb {R} \}$

Then $~F_{1}=F_{2}~$

Proof of Theorem T4

Lemma 1

(0) Let $~b>\exp(1/\mathrm {e} )~$.

(1) Let $~f~$ be holomorphic function on $~\mathbb {C} ^{\prime }=\mathbb {C} \backslash (-\infty ,-2]~$, such that

(2) $f(0)=1$

(3) $~\exp _{b}(f(z))=f(z+1)~$ for $\Re (z)>-2~$

(4) $~f~$ is bounded on $~\mathbb {S} =~\{x+\mathrm {i} y|-2<x\leq 1,y\in \mathbb {R} \}~$

Let $~\mathbb {D} =~\{x+\mathrm {i} y|-2<x,y\in \mathbb {R} \}~$

Then $~f(\mathbb {D} )=\mathbb {C} ~$

Proof of theorem T4

Henryk, I cannot copypast your proof here: I do not see, where do you use condition

$~b>\exp(1/\mathrm {e} )$

?

From Lemma 1, the ...

Discussion

Such $~F~$ is unique tetration on the base $~b~$.