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 elementary superfunctions tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/28/2009, 09:37 PM bo198214 Wrote:Yes exactly those formula I was looking for! However can you shorten them a bit by gathering terms or introducing constants for repeatedly occuring terms? If possible it would be very preferable to indicate the fixed point, if the super-function is obtained by regular iteration. Would anyway be good if you could explain how you obtained the formulas or what the idea behind is. just like ansus you dont seem to realize mathematica just uses a handful solutions and all others are special cases. notice for instance that EVERY f(x) in this thread are 1) f(x) = polynomial ( see also "logistic map" and " inverse hypergeo " ) 2) f(x) = moebius = (a x + b)/(c x + d) 3) f(x) = a + b x^c and 2) has a closed form solution so all that one needs to do to find the superfunction of e.g. 1 / a x + b is to fill in some variables in the general formula. at this point , its just that simple. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/29/2009, 04:58 PM Ansus Wrote:The superfunction for 1/x is probably indeed complex. Why do you suppose it to be real? Well I was rather after real functions. Input strictly increasing, output real. What does mathematica say about $f(x)=\frac{x-1}{x+1}$? This is strictly increasing and has no real fixed point. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/29/2009, 05:46 PM Ansus Wrote:But I wonder why Mathematica did not find something like $F(x)=C^{(-1)^x}$ which is a case of $F(x)=C^{b^x}$ Interestingly Maple finds exactly that solution. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/29/2009, 06:59 PM 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 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/11/2009, 02:02 PM (This post was last modified: 05/12/2009, 09:23 AM by Kouznetsov.) Introduction. I begin with this introduction in order to indicate, how do I understand the super-functions and our role about them. I remember that the Moderator has an ability to remove this introduction, together with all the lyrics around (and I appreciate his good will to keep all the posts so dry as his gunpowder); however, I hope the definitions below do not contradict those he suggested for our joint paper (which is "yet to be finished" during several last months); so, some definitions have some chance to survive. Terminology '''Superfunction''' $S$ comes from iteration of some given function $h$, called "base-function" or "transfer function". There is some analogy with fiber optics, which explains why this $h$ should be called "transfer function". Those who hate any physics (and, especially, the phenomenological fiber optics), may imagine that the function $h$ transfers the value of function $S$ at some point $z$ to the value at the point $z+1$, as the basic equation suggests: $S(z+1)=f(S(z))$ This equation is very basic; so, the only given function $h$ may also be called the "base-function". Iterations Roughly, for some function $h$ and for some constant $t$, the super-function could be defined with expression ${{S(z)} \atop \,} {= \atop \,} {{\underbrace{h\Big(h\big(... h(t)...\big)\Big)}} \atop {z \mathrm{~evaluations~of~function~}h\!\!\!\!\!\!}}$ then $S$ can be interpreted as superfunction of function $h$. Such definition is valid only for positive integer $z$. In particular, $S(1)=h(t)$. The most research and applications around the superfunctions are related with various extensions of super-function: analysis of the existence, uniqueness and ways of the evaluation. For some functions $h$, such as addition of a constant or multiplication by a constant, the superfunction can be expressed in terms of elementary function. Namely such examples were motivation of this message. History and Lowstory Analysis of superfunctions cames from the application to the evaluation of fractional iterations of functions. Super-functions and their inverse functions allow evaluation of not only minus-first power of a function (inverse function), but also any real and even complex iteration of the function. Historically, the first function of such kind considered was $\sqrt{\exp}~$; then, function $\sqrt{!~}~$ was used as logo of the Physics department of the Moscow State University, see http://zhurnal.lib.ru/img/g/garik/dubinu...ndex.shtml http://ofvp.phys.msu.ru/pdf/Kandidov_70.pdf http://nauka.relis.ru/11/0412/11412002.htm (bitte, all 3 in Russian). That time, researchers did not have computational facilities for evaluation of such functions, but the $\sqrt{\exp}$ was more lucky than the $~\sqrt{!~}~~$; at least the existence of holomorphic function $\varphi$ such that $\varphi(\varphi(z))=\exp(z)$ has been reported in 1950 by Helmuth Kneser (H.Kneser. “Reelle analytische L¨osungen der Gleichung $\varphi(\varphi(x)) = e^x$ und verwandter Funktionalgleichungen”. Journal fur die reine und angewandte Mathematik, 187 (1950), 56-67.) Extensions The recurrence above can be written as equations $S(z\!+\!1)=h(S(z)) ~ \forall z\in \mathbb{N} : z>0$ $S(1)=h(t)$. Instead of the last equation, one could write $S(0)=t$ and extend the range of definition of superfunction $S$ to the non-negative integers. Then, one may postulate $S(-1)=h^{-1}(t)$ and extend the range of validity to the integer values larger than $-2$. The following extension, for example, $S(-2)=h^{-2}(t)$ is not trivial, because the inverse function may happen to be not defined for some values of $t$. In particular, [[tetration]] can be interpreted as super-function of exponential for some real base $b$; in this case, $h=\exp_{b}$ then, at $t=1$, $S(-1)=\log_b(1)=0$. but $S(-2)=\log_b(0)~ \mathrm{is~ not~ defined}$. For extension to non-integer values of the argument, superfunction should be defined in different way. Definitions. For connected domains $C \subseteq \mathbb{C}$ and $D \subseteq \mathbb{C}, and two numbers [tex]a\in C$ and $d\in D$, the $(a \!\mapsto\! d)$ super-function of a transfer function $~h~$ is function $S$, holomorphic on $C$, such that $S(z\!+\!1)=h(S(z)) ~ \forall z\in C : S(z)\in D$ and $S(a)=b$. If $h=\exp_b [tex] for some [tex]b\in \mathbb{C}$, then the $(a \!\mapsto\! d)$ super-function of a transfer function $~h~$ is called $(a \!\mapsto\! d)$ super-exponential on the base $b$. If $a=0$ and $d=1$, then such a super-exponential is called tetration and justify the appearance of this post at this Forum. As it was already mentioned in this forum, in general, the super-function is not unique. For a given transfer function $h$, from given $(a\mapsto d)$ super-funciton $F$, another $(a\mapsto d)$ super-function $G$ could be constructed as $G(z)=F(z+\mu(z))$ where $\mu$ is any 1-periodic function, holomorphic at least in some vicinity of the real axis, such that $\mu(a)=0$. The modified super-function may have narrowed range of holomorphism. The challenging task is to specify some domain $C$ such that $(C, a \mapsto d)$ super-function is unique. In particular, the $(C, 0\mapsto 1)$ super-function of $\exp_b$, for $b>1$, is called [[tetration]] and is believed to be unique at least for $C= \{ z \in \mathbb{C} ~:~\Re(z)>-2 \}$; for the case $b>\exp(1/e)$ Examples Oh, en fin, I touch the goal of this post. Sorry for the long introduction above. Below, I consider various simple base-functions $h$ . Elementary increment Let $h(z)=z+1=++z$. Then, the identity function $I$ such that $I(z)=z \forall z \in \mathbb{C}$ is $(\mathbb{C}, 0\mapsto 0)$ superfunction of $h$. Addition Chose a $b$ and define function $\mathrm{add}_b$ such that $\mathrm{add}_b(z)=b\!+\!z ~ \forall z \in \mathbb{C}$ Define function $\mathrm{mul_b}$ such that $\mathrm{mul_b}(z)=b\!\cdot\! z ~ \forall z \in \mathbb{C}$. Then, function $~\mathrm{mul_b}~$ is $(\mathbb{C}, 0 \mapsto b )$ superfunction of $h$. Multiplication Exponential $\exp_b$ is $(\mathbb{C}, 0 \mapsto 1 )$ super-function of function $\mathrm{mul}_b$, defined in the previous example. Quadratic polynomial Let $h(z)=2 z^2-1$. Those, who like some Quantum Mechanics, may treat this function as a scaled second Hermitian polynomial, justifying the letter, used to denote the transfer function. Then, $f(z)=\cos( \pi \cdot 2^z)$ is a $(\mathbb{C},~ 0\! \rightarrow\! 1)$ superfunction of $H$. Indeed, $f(z+1)=\cos(2 \pi \cdot 2^z)=2\cos(\pi \cdot 2^z)^2 -1 =H(f(z))$ and $f(0)=\cos(2\pi)=1$ In this case, the superfunction $f$ is periodic; its period $T=\frac{2\pi \mathrm {i}}{\ln(2)} \mathrm{i}\approx 9.0647202836543876194 \!~i$. Such super-function approaches unity in the negative direction of the real axis, $\lim_{x\rightarrow -\infty} f(x)=1$ The example above and the two examples below are suggested at Mueller. Problems in Mathematics. http://www.math.tu-berlin.de/~mueller/projects.html Rational function. In general, the transfer function $h$ has no need to be entire function. Here is the example with meromorphic function $h$. Let $C= \{z \in \mathbb{C}: 2^z\ne n+1/2 \forall n\in \mathbb{Z})\}$ $D= \mathbb{C}\backslash \{1,-1\}$ $h(z)=\frac{2z}{1-z^2} ~ \forall z\in D~$ $S(z)=\tan(\pi 2^z) \forall z\in C$ Tthen, $S$ is $(C, 0\! \mapsto\! 0)$ superfunction of $h$. For the proof, the trigonometric formula $\tan(2 \alpha)=\frac{2 \tan(\alpha)}{1-\tan(\alpha)^2}~~ \forall \alpha \in \mathbb{C} \backslash \{\alpha\in \mathbb{C} : \cos(\alpha)=0 || \sin(\alpha)=\pm \cos(\alpha) \}$ can be used at $\alpha=\pi 2^z$, that gives $ h(S(z))=\frac{2 \tan(\pi 2^z)}{1-\tan(\pi 2^z)} = \tan(2 \pi 2^z)=S(z+1)$ Algebraic transfer function. However, the transfer function has no need to be even meromorphic. Let $C= \{z \in \mathbb{C}: |Arg(cos(\pi 2^z))| < \pi \}$ $D= \{z \in \mathbb{C}: |Arg(1-z^2)| < \pi \}$ $h(z)=2z \sqrt{1-z^2} \forall z \in D$ $S(z)=\sin(\pi 2^z) \forall z \in C$ Then, $S is is [tex](C,~ 0\!\rightarrow \!0)$ superfunction of $H$ for $C= \{z\in \mathbb C : \Re( \cos(\pi 2^z))>0 \}$. The proof is similar to the previous two cases. Exponential transfer function. Let $b>1$, $H(z)= \exp_b(z) \forall z \in \mathbb{C}$, $C= \{ z \in \mathbb{C} : \Re(z)>-2 \}$. Then, tetrational $\mathrm{tet}_b$ is a $(C,~ 0\! \rightarrow\! 1)$ super-function of $\exp_b$. more extensions. In general, we may take any special function $S$, such that $S(z+1)$ can be expressed through $S(z)$ with holomorphic elementary functions, then we may declare this expression as transfer function $h$, and then, function $S$ appears as super-function. I invite participants to construct more super-functions that can be easy represented through some already known special functions. P.S. Oh, mein Gott! I just realized the correct tread for this post. It repeats a lot of staff already posted here... Sorry... I see, there are already replies, so, I ssto to edit; the only correct obvious misprints... bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 05/11/2009, 07:39 PM (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 $F(x)=\cosh(2^x)$. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 05/11/2009, 08:19 PM (05/11/2009, 08:09 PM)Ansus Wrote: And also as we had seen, for 1/x there are also two different solutions. It is an open question thus how much independent superfunctions has a given function. Its not an open question, there are infinitely many (even for real-analytic solutions which $1/x$ does not have). If you have one solution $F$ just take any 1-periodic function $\theta$ and then $F(x+\theta(x))$ is another solution. Even elementary if $\theta$ is elementary (say linear combination of some $\sin(2\pi k x)$). Thatswhy I always try to find for elementary solutions whether they are regular at some fixed point because this reduces the number of real analytic solutions to two at one fixed point (analogously to $\exp_{\sqrt{2}}$) up to x-translation. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 05/11/2009, 09:09 PM (05/11/2009, 08:37 PM)Ansus Wrote: count all solutions that differ only by periodic term as one solution. Then we have only one strictly increasing solution. If F and G are two super-functions of f then $F(x)=G(x+\theta(x))$ for $\theta(x)=G^{-1}(F(x))-x$. BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 05/11/2009, 09:12 PM (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 $F(x)=\cosh(2^x)$. 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. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 05/11/2009, 09:27 PM (This post was last modified: 05/11/2009, 09:28 PM by bo198214.) (05/11/2009, 08:19 PM)bo198214 Wrote: Thatswhy I always try to find for elementary solutions whether they are regular at some fixed point because this reduces the number of real analytic solutions to two at one fixed point (analogously to $\exp_{\sqrt{2}}$) up to x-translation. I want to illustrate this phenomenon with a picture of the two regular super-functions $F_{1,+}(x)=\cosh(2^x)$ and $F_{1,-}(x)=\cos(2^x)$ of $f(x)=2*x^2-1$ at the fixed point 1.     The upper curve is $\cosh(2^x)$ and the lower curve is $\cos(2^x)$. We see that they have both same asymptote to the left, which is the fixed point 1. Compare this with the both super-exponentials at 4 (these are $F_{4,5}$ and $F_{4,3}$) in the picture in this post. This is a general behaviour of the two real regular super-functions at one fixed point: Either to the left or to the right (depending whether the derivative at the fixed point is bigger or smaller than 1) they both approach the fixed point, one from above the other from below. « Next Oldest | Next Newest »

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