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 Universal uniqueness criterion II bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/16/2008, 05:27 PM (This post was last modified: 11/18/2008, 11:32 AM by bo198214.) Lets summarize what we have so far: Proposition. Let $S$ be a vertical strip somewhat wider than $1$, i.e. $S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z) for some $x_1\in\mathbb{R}$ and $\epsilon>0$. Let $D=\mathbb{C}\setminus (-\infty,x_0]$ for some $x_0 and let $G\subseteq G'$ be two domains (open and connected) for values, and let $F$ be holomorphic on $G'$. Then there is at most one function $f$ that satisifies (1) $f$ is holomorphic on $D$ and $f(S)\subseteq G\subseteq f(D)=G'$ (2) $f$ is real and strictly increasing on $\mathbb{R}\cap S$ (3) $f(z+1)=F(f(z))$ for all $z\in D$ and $f(x_1)=y_1$ (4) There exists an inverse holomorphic function $f^{-1}$ on $G$, i.e. a holomorphic function such that $f(f^{-1}(z))=z=f^{-1}(f(z))$ for all $z\in G$. Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (4)) and satisfies $h(z)=g(\delta(z))$. By (3) and (4) $\delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1$ and $\delta(0)=0$. So $\delta$ can be continued from $S$ to an entire function and is real and strictly increasing on the real axis. $\delta(\mathbb{C})=\mathbb{C}$ by our previous considerations. By Big Picard every real value of $\delta$ is taken on infinitely often if $\delta$ is not a polynomial, but every real value is only taken on once on the real axis, thatswhy still $\delta(\mathbb{C}\setminus\mathbb{R})=\mathbb{C}$. But this is in contradiction to $g^{-1}:G\to D=\mathbb{C}\setminus [x_0,\infty)$. So $\delta$ must be a polynomial that takes on every real value at most once. This is only possible for $\delta(x)=x+c$ with $c=0$ because $\delta(0)=0$.$\boxdot$. In the case of tetration one surely would chose $x_0=-2$ and $x_1=0$ or $x_1=-1$. However I am not sure about the domain $G$ which must contain $f(S)$ and hence give some bijection $f:G\leftrightarrow S'$, with some $S'\supseteq S$. Of course in the simplest case one just chooses $G=f(S)$ if one has some function $f$ in mind already. However then we can have a different function $f_2$ with $f(S)\neq f_2(S)$ but our intention was to have a criterion that singles out other solutions. So we need an area $G$ on which every slog should be defined at least and satisfy $\text{sexp}(\text{slog})=\text{id}=\text{slog}(\text{sexp})$ as well as $\text{sexp}(S)\subseteq G$. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/16/2008, 06:13 PM (This post was last modified: 11/16/2008, 06:23 PM by bo198214.) I just see that we can essentially simplify our conditions: Proposition. Let $S$ be a vertical strip somewhat wider than $1$, i.e. $S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z) for some $x_1\in\mathbb{R}$ and $\epsilon>0$. Let $D$, $G$, $G'$ be three domains (open and connected) such that $S\subseteq D$, $G\subseteq G'$ and let $F$ be holomorphic on $G'$, let $y_1\in G$. Then there exist at most one function $f$ that satisifies (1) $f$ is holomorphic on $D$ and $f(S)\subseteq G\subseteq f(D)=G'$ (2) $f(z+1)=F(f(z))$ for all $z\in D$ and $f(x_1)=y_1$ (3) There is a $D'\subseteq D$ such that $f: D' \leftrightarrow G$ is biholomorphic. Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (3)) and satisfies $h(z)=g(\delta(z))$. By (3) and (4) $\delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1$ and $\delta(0)=0$. So $\delta$ can be continued from $S$ to an entire function. But the same is also true for $\delta_2=h^{-1}(g(z))$ by the same reasoning. But as $\delta\circ\delta_2=g^{-1}\circ h\circ h^{-1}\circ g =\text{id}$ and $\delta_2\circ \delta = \text{id}$, we see that $\delta:\mathbb{C}\leftrightarrow\mathbb{C}$ is a bijection! Again with Picard's big theorem we conclude that $\delta(x)=x+c$ with $c=0$.$\boxdot$ Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/17/2008, 01:27 AM (This post was last modified: 11/20/2008, 09:30 AM by bo198214.) bo198214 Wrote:...Let $D=\mathbb{C}\setminus [x_0,\infty)$ for some $x_0...Henryk: For tetration, I would define $D=\mathbb{C}\setminus (-\infty, x_0]$ for some $x_0... then I like your proof. I include below the small part of http://math.eretrandre.org/tetrationforu...77#pid2477 which is picture of slog(S), assuming, $x_1=-1$ and that $\epsilon$ is small and not seen. Vertical lines correspond to $\Re(\mathrm{sexp}(z))=0$ and $\Re(\mathrm{sexp}(z))=1$ Horisontal lines correspond to $\Im(\mathrm{sexp}(z))=0$ and $\Im(\mathrm{sexp}(z))=1$ The curvilinear mesh is produced by images of lines $\Re(z)= 0.2, 0.4, 0.6, 0.8, 1 ; \Im(z)>0$ and $\Im(z)= 0.2, 0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.4, 2.6, 2.8, 4; 0\le /Re(z) \le 1$. The pink cutline corresponds to $\Re(z)= -2$. The images of lines with integer $\Im(z)$ are a little bit extended. Also, images of lines $\Re(z)=2$ and $\Re(z)=3$are shown. There is biholomorphizm $\mathrm{sexp}(S): \leftrightarrow S$.     Henrik, 1. Do you plan to polish this proof or I may include this into the paper? 2. Can we claim, that some of singularities of a modified tetration are at $\Re(z)>-2$ ? (In this case, we can include the case $1 at once). bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/17/2008, 02:45 PM Kouznetsov Wrote:bo198214 Wrote:...Let $D=\mathbb{C}\setminus [x_0,\infty)$ for some $x_0...Henryk: For tetration, I would define $D=\mathbb{C}\setminus (-\infty, x_0]$ for some $x_0... then I like your proof. Yes that was a mistype, it should read $\mathbb{C}\setminus(-\infty,x_0]$. I change that in the original post. Quote:I include below the small part of http://math.eretrandre.org/tetrationforu...77#pid2477 which is picture of slog(S), thanks, for the illustration. Quote:1. Do you plan to polish this proof or I may include this into the paper? Its not yet finished, we still have no universal domain $G$ for the slog, which we need to have uniqueness independent of the specific domain $\text{sexp}(S)$. Quote:2. Can we claim, that some of singularities of a modified tetration are at $\Re(z)>-2$ ? Dmitrii, modified tetration is not all, if you want to consider $g(z)=f(J(z))$ then you need to have a $J$ first, a $J=f^{-1}\circ g$ that is holomorphic on $S$ which in turn imposes conditions on the cut, set in the domain of definition of $f^{-1}$. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/18/2008, 01:25 AM bo198214 Wrote:Kouznetsov Wrote:2. Can we claim, that some of singularities of a modified tetration are at $\Re(z)>-2$ ?Dmitrii, modified tetration is not all, if you want to consider $g(z)=f(J(z))$ then you need to have a $J$ first, a $J=f^{-1}\circ g$ that is holomorphic on $S$ which in turn imposes conditions on the cut, set in the domain of definition of $f^{-1}$.Sorry, I try again: If $h$ is entire and 1-periodic and $J(z)=z+h(z)$, then can we claim that $\exists z \in \mathbb{C} ~:~ \Re(z)>-2 ~,~ J(z)\!=\!-2~.~$ ? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/18/2008, 11:31 AM Kouznetsov Wrote:I try again: If $h$ is entire and 1-periodic and $J(z)=z+h(z)$, then can we claim that $\exists z \in \mathbb{C} ~:~ \Re(z)>-2 ~,~ J(z)\!=\!-2~.~$ ? Hm, sorry, I dont know. Wherefore do you need such a statement? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/19/2008, 02:59 AM bo198214 Wrote:Kouznetsov Wrote:I try again: If $h$ is entire and 1-periodic and $J(z)=z+h(z)$, then can we claim that $\exists z \in \mathbb{C} ~:~ \Re(z)>-2 ~,~ J(z)\!=\!-2~.~$ ?Hm, sorry, I dont know..ok, I do know. Conjectiure: Let $h$ be entire 1-periodic non constant function, $h(0)=0$, $|h'(z)| <1$ for all $z$ in some vicinity of the real axis $J(z)=z+h(z)$ for all complex $z$ Then there exist $z \in \mathbb{C}$ such that $\Re(z)>0$ and $J(z)=-2$ bo198214 Wrote:Wherefore do you need such a statement?Yes, I do. Then we have beautiful and general proof of uniqueness of tetration as it is defined at http://en.citizendium.org/wiki/Tetration P.S. For other participants, I repeat here the essense from one of our previous discussions. Many times I tried to build-up an example to negate the conjecture above. Therefore I claim this conjecture. I agree with you, that it is not sufficient reason for such a claim. (Although I never met a simple condition for an existing function, such that I could not provide an example.) It would be interesting to construct the sufficient reason. (then the conjecture becomes Theorem) Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/19/2008, 03:14 AM bo198214 Wrote:Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (3)) and satisfies $h(z)=g(\delta(z))$...Why $h$? There was no $h$ above. Should not be $f$? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/19/2008, 01:16 PM Kouznetsov Wrote:bo198214 Wrote:Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (3)) and satisfies $h(z)=g(\delta(z))$...Why $h$? There was no $h$ above. Should not be $f$? $h$ and $g$ are two functions that satisfy the above conditions, as I wrote. Perhaps write better: "Let $f=g$ and $f=h$ be two solutions of the above conditions." « Next Oldest | Next Newest »

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