User talk:Dmitrii Kouznetsov/Analytic Tetration

Henryk Trappmann 's theorems

Theorem T1.

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.

Theorem T2

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

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

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

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