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 $z\in D$: | + | 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$$. |
==== Proof. ==== | ==== Proof. ==== | ||
We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$. | We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$. | ||
| − | Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1: | + | Then we know that |
| + | $$\sup_{t\to\infty}\alpha_b(\sigma_b({\rm i}t))<\infty$$ | ||
| + | and the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1: | ||
$$\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)$$ | ||
| − | and is real for $z > -2$ (and can be continued to $\R$) and can | + | and is real for $z > -2$ (and can be continued to $\R$) and can so be developed into a real Fourier-Series ($A_k$, $\phi_k$ in $\R$): |
$$\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)$$ | ||
$$\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(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(x+{\rm i}y)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left({\rm i}(2\pi kx-\varphi_{k})-2\pi k y\right)+\exp\left({\rm i}(-2\pi kx+\varphi_{k})+2\pi k y\right)\right) | ||
| + | =\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\cos(2\pi kx-\varphi_{k})\left(\exp\left(-2\pi k y\right)+\exp(2\pi ky)\right)+{\rm i}\sin(2\pi kx-\varphi_{k})\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\right)$$ | ||
| + | |||
| + | $$\Im(\delta(x+{\rm i}y)+x+{\rm i}y)=y+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\sin(2\pi kx-\varphi_{k})$$ | ||
$$ | $$ | ||
| − | \delta( | + | \alpha_b(\sigma_b({\rm i}t))=\delta({\rm i}t)+{\rm i}t={\rm i}t+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right) |
$$ | $$ | ||
But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$. | But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$. | ||
Latest revision as of 08:04, 4 January 2017
This is work in progress, dont rely on it!
Proposition
For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$: $$\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$$.
Proof.
We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$. Then we know that $$\sup_{t\to\infty}\alpha_b(\sigma_b({\rm i}t))<\infty$$ and the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1: $$\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(x+{\rm i}y)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left({\rm i}(2\pi kx-\varphi_{k})-2\pi k y\right)+\exp\left({\rm i}(-2\pi kx+\varphi_{k})+2\pi k y\right)\right) =\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\cos(2\pi kx-\varphi_{k})\left(\exp\left(-2\pi k y\right)+\exp(2\pi ky)\right)+{\rm i}\sin(2\pi kx-\varphi_{k})\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\right)$$
$$\Im(\delta(x+{\rm i}y)+x+{\rm i}y)=y+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\sin(2\pi kx-\varphi_{k})$$
$$ \alpha_b(\sigma_b({\rm i}t))=\delta({\rm i}t)+{\rm i}t={\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 $\sigma_b(z)=\tau_b(z+c)$.