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 elementary superfunctions bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/11/2009, 09:31 PM (This post was last modified: 05/11/2009, 09:32 PM by bo198214.) (05/11/2009, 09:12 PM)BenStandeven Wrote: Actually, $\cos(2^x) = \cosh(i 2^x) = \cosh(2^{x + \frac{\pi i}{2 \ln 2}})$, so they are translations of each other, albeit along the imaginary axis instead of the real axis. Well observed! So this is not even the worst non-uniqueness. This is generally the case, that there are these two regular super-functions at a fixed point. And one is the other translated by some imaginary value (up to real translations along $x$) which is half of the period of both functions. tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 05/13/2009, 05:48 PM (04/29/2009, 06:59 PM)bo198214 Wrote: Ansus Wrote:I've verified both and both indeed correct solutions. Mathematica finds $F(x)=C^{b^x}$ for the general case of $f(x)=x^b$. What Maple gives for $f(x)=\frac{x+a}{x-a}$? Nothing hasnt this been solved before ? regarding my critisism ( in a previous post in this thread ) that only ( elementary or other ) superfunctions of polynomials , a x ^ b or moebius functions are known ; a challange : what is the superfunction of (a x ^ 2 + b x + c) / ( A x^2 + B x + C ) ( yes, it can in some cases be reduced , common factors divided away , different cases occur ( different amount and position of fixpoints ) , i know that ... ) i sometimes call this superfunction " generalized sine / cosine " ( on sci.math e.g. ) regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 07/16/2009, 04:42 AM (This post was last modified: 07/17/2009, 09:49 AM by Kouznetsov.) (05/11/2009, 07:39 PM)bo198214 Wrote: (05/11/2009, 05:17 PM)Ansus Wrote: It should be noted that superfunction is not unique in most cases. For example, for $f(x)=2 x^2-1$, superfunction is $F(x)=\cos(2^x C)$ Ya, this is the simple kind of non-uniqueness, its just a translation along the x-axis. However there are also more severe types of non-uniques, as I already introduced in my first post, we have two solutions (which are not translations of each other): $F(x)=\cos(2^x)$ and $F(x)=\cosh(2^x)$. 1. Sorry, Henryk, they are translations of each other. $\cos(2^x)=\cosh\Big(2^{x+\pi\cdot i\cdot \ln(2)/2}\Big)$. We already had similar discussion with respect to tetration on base $\sqrt{2}$, http://www.ils.uec.ac.jp/~dima/PAPERS/2009sqrt2.pdf , figure 3. the growing up SuperExponential (red) and the tetration (blue), at the appropriate translations formula (5.7) and formula (5. become very similar and bounded along the real axis functions (green). (I do not know why the number of forumla that follows (5.7) becomes some strange "smile". I mean just the number of formula, nothing more.) 2. There are many ways to extend the table of superfunctions. I suggest the group of transforms of the pairs (TransferFunciton, SuperFunctions). Theorem. Let $f(F(z)=F(z+1)$, Let $q(p(z))=z$, Let $h(z)=p(h(q(z)))$, Let $E(z)=p(F(z))$. Then $h(E(z))=E(z+1)$. Proof: $h(E(z))=p(f(q(p(F(z)))))=p(f(F(z)))=p(F(z+1))=E(z+1)$ (end of proof). With transform $p$, from the pair (f,F) we get the pair (h,E). 2.1. Also, in the right hand side of the expression $f(z)=F(1+F^{-1}(z))$ we can swap $F$ and $F^{-1}$; this gives the new transfer function $h(z)=F^{-1}(1+F(z))$ with known superfunction $F^{-1}$. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/27/2010, 10:27 PM (This post was last modified: 03/27/2010, 10:44 PM by bo198214.) (04/29/2009, 04:58 PM)bo198214 Wrote: What does mathematica say about $f(x)=\frac{x-1}{x+1}$? This is strictly increasing and has no real fixed point. (04/29/2009, 05:07 PM)Ansus Wrote: It gives $ F(x)=\frac{\left(-\frac{1}{2}-\frac{i}{2}\right)^x(i C-1) + \left(-\frac{1}{2}+\frac{i}{2}\right)^x(i C+1)}{\left(-\frac{1}{2}-\frac{i}{2}\right)^x(i+ C)+\left(-\frac{1}{2}+\frac{i}{2}\right)^x(i-C) }$ Well we can do better: $f^{\circ u}(z)=\frac{\cos(\frac{\pi}{4}u)z-\sin(\frac{\pi}{4}u) }{\sin(\frac{\pi}{4}u)z+\cos(\frac{\pi}{4}u)}$ more detailed information in this article To be complete I will also give Gottfried's solution for $f(z)=\frac{1}{z+1}$ which has two real fixed points at $\pm\frac{1}{2}\sqrt{5}-\frac{1}{2}$, and singularity at -1: $f^{\circ u}(z)=\frac{\operatorname{fib}_u-\operatorname{fib}_{u-1}z}{\operatorname{fib}_{u+1}-\operatorname{fib}_{u}z}$ $\operatorname{fib}_u=\frac{\phi^u-(1-\phi)^u}{\sqrt{5}}$ $\phi=\frac{1+\sqrt{5}}{2}$ bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/18/2010, 01:17 PM (This post was last modified: 04/18/2010, 01:19 PM by bo198214.) Did we mention already the tangent? It has this nice addition theorem: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ which brings us the superfunction: $\sigma(z)=\tan(z\cdot\arctan( c))$ $\sigma(z+1)=\frac{\sigma(z)+c}{1-c\sigma(z)}=f(\sigma(z))$ for the function $f(z)=\frac{z+c}{1-cz}$ $f$ is another particular case of a linear fraction (where the regular iteration at both fixed points coincide). The two (non-real) fixed points are: $\frac{z+c}{1-cz}=z$, $z+c=z-cz^2$ $z=\pm i$ for $c\neq 0$ tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 04/18/2010, 11:10 PM (04/18/2010, 01:17 PM)bo198214 Wrote: Did we mention already the tangent? It has this nice addition theorem: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ which brings us the superfunction: $\sigma(z)=\tan(z\cdot\arctan( c))$ $\sigma(z+1)=\frac{\sigma(z)+c}{1-c\sigma(z)}=f(\sigma(z))$ for the function $f(z)=\frac{z+c}{1-cz}$ $f$ is another particular case of a linear fraction (where the regular iteration at both fixed points coincide). The two (non-real) fixed points are: $\frac{z+c}{1-cz}=z$, $z+c=z-cz^2$ $z=\pm i$ for $c\neq 0$ i cant help to say that this is the velocity addition formula in relativity with lightspeed = 1. tan(x) = u , tan(y) = v tan(x+y) = (u + v)/(1 - uv). an intresting connection. well at least a pretty one. tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 08:22 AM Another one: $f(x)=x^2+(1+\sqrt{5})x+1$ $F(x)=e^{2^x} - \frac{1+\sqrt{5}}{2}$ $f$ has two fixed points with the derivations: $f'(\frac{1-\sqrt{5}}{2}) = 2$ and $f'(\frac{-1-\sqrt{5}}{2}) = 0$. The above superfunction $F$ is the regular iteration at the upper fixed point $\frac{1-\sqrt{5}}{2}$ Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 04/25/2010, 09:11 AM (04/25/2010, 08:22 AM)bo198214 Wrote: Another one: $f(x)=x^2+(1+\sqrt{5})x+1$ $F(x)=e^{2^x} - \frac{1+\sqrt{5}}{2}$ $f$ has two fixed points with the derivations: $f'(\frac{1-\sqrt{5}}{2}) = 2$ and $f'(\frac{-1-\sqrt{5}}{2}) = 0$. The above superfunction $F$ is the regular iteration at the upper fixed point $\frac{1-\sqrt{5}}{2}$ Henryk, it seems to me that such a case can be obtained from the example 5 of the Table 1 of our article D.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12 (English version), p.8-14 (Russian version; http://www.ils.uec.ac.jp/~dima/PAPERS/2009superfae.pdf http://www.ils.uec.ac.jp/~dima/PAPERS/2010superfar.pdf with transformation at the bottom of that table at $P(z)=z - \frac{1+\sqrt{5}}{2}$ I suspect, with that short table we have covered the most of simple elementary superfunctions... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 09:23 AM (This post was last modified: 04/25/2010, 09:43 AM by bo198214.) (04/25/2010, 09:11 AM)Kouznetsov Wrote: Henryk, it seems to me that such a case can be obtained from the example 5 of the Table 1 of our article Yes that is right, $f$ is polynomially conjugated to $x^2$, i.e. there exists a polynomial $P$ such that $f(P(x))=P(x^2)$, and as we know a superfunction of $x^2$ is $e^{2^x}$, we know that a superfunction of $f$ is $P(e^{2^x})$. But we dont have a decision criteria when a given polynomial $f$ is conjugated to some $x^n$. Particularly all polynomials without real fixed point, e.g. $x^2+1$, seem not be (real) conjugated to some $x^n$. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 10:48 AM Challenge: Is there an elementary superfunction of a polynomial that has no real fixed point? « Next Oldest | Next Newest »

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