Difference between revisions of "Uniqueness of Tetration"

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== Proposition ==
 
== Proposition ==
For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which satisfies for all $z\in D$:
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For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$:
$$\overline{\sigma_b(z)}=\sigma_b(\overline{z})$$, $$\sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b(s+{\rm i}t)\right| <\infty$$ for each $s>-2$.
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$$\sigma_b(0)=1, \sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b({\rm i}t)\right| <\infty$$.
  
 
==== Proof. ====
 
==== Proof. ====
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Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1:
 
Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1:
 
$$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$
 
$$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$
and is real for $z > -2$ (and can be continued to $\R$) and can to developed into a real Fourier-Series ($A_k$,  $\phi_k$ in $\R$):
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and is real for $z > -2$ (and can be continued to $\R$) and can so be developed into a real Fourier-Series ($A_k$,  $\phi_k$ in $\R$):
  
 
$$\delta(t)=\sum_{k=0}^{\infty}A_{k}\cos\left(2\pi kt-\varphi_{k}\right)$$
 
$$\delta(t)=\sum_{k=0}^{\infty}A_{k}\cos\left(2\pi kt-\varphi_{k}\right)$$
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$$
 
$$
\delta(s+{\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}ks-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}ks+2\pi kt+{\rm i}\varphi_{k}\right)\right)
+
\delta({\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right)
 
$$
 
$$
  
 
But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.
 
But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.

Revision as of 16:42, 3 January 2017

Proposition

For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$: $$\sigma_b(0)=1, \sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b({\rm i}t)\right| <\infty$$.

Proof.

We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$. Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1: $$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$ and is real for $z > -2$ (and can be continued to $\R$) and can so be developed into a real Fourier-Series ($A_k$, $\phi_k$ in $\R$):

$$\delta(t)=\sum_{k=0}^{\infty}A_{k}\cos\left(2\pi kt-\varphi_{k}\right)$$

$$\delta(z)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}kz-{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}kz+{\rm i}\varphi_{k}\right)\right)$$

$$ \delta({\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right) $$

But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.