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 elementary superfunctions bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/23/2009, 01:25 PM (This post was last modified: 05/11/2009, 09:56 PM by bo198214.) Triggered by the interesting finding of Andrew I open this thread for the further investigation of elementary superfunctions, i.e. functions $F$ that are expressible with elementary functions and operations such that $F(x+1)=f(F(x))$ for a given elementary function $f$. Our first example is: $f(x)=2x^2-1$ with a superfunction $F(x)=\cos(2^x)$. Now the $\cosh$ has the same property $\cosh(2x)=2\cosh(x)^2 -1$ as the $\cos$. Hence $F(x)=\cosh(2^x)$ is another superfunction of $2x^2-1$. Indeed $f^{[t]}(x)=F(t+F^{-1}(x))$ exists and is differentiable at $x=1$. But it does not exist at the other fixed point $-\frac{1}{2}$, because $\operatorname{arccosh}\left(-\frac{1}{2}\right)$ is not defined. Edit: both are regular super-functions at fixed point 1. $\lim_{x\to-\infty} F(x)=1$. So if we are at polynomials $f$, we can also give an elementary superfunction for $f(x)=x^a$, i.e. $F(x)=c^{a^x}$. Because $F(x+1)=c^{a^xa}=F(x)^a$. Edit: these are the regular super-exponentials at 1. $\lim_{x\to-\infty} F(x)=1$. Generally for Chebyshev polynomials, these are the polynomials $T_n$ such that $\cos(nx)=T_n(\cos(x))$ - for example above we used $T_2(x)=2x^2-1$ -, we know already two elementary superfunctions of $f(x)=T_n(x)$, these are $F(x)=cos(n^x)$ and $F(x)=\cosh(n^x)$. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/23/2009, 02:23 PM (This post was last modified: 05/11/2009, 10:00 PM by bo198214.) Peeping a bit into the Szekeres-Seminar somewhere on the forum mentioned by Andrew, and rearranging the elementary Schröder functions found by Schröder himself, I can add some more elementary superfunctions: 1. $f(x)=2(x+x^2)$, $F(x)=\frac{e^{2^x}-1}{2}$. Let us check: $F(x+1)=\frac{\left(e^{2^x}\right)^2 -1 }{2}$ and on the other hand $2(F(x)+F(x)^2)=e^{2^x}-1 + \frac{(e^{2^x}-1)^2}{2}=\frac{\left(e^{2^x}\right)^2 -1 }{2}$ So its indeed a superfunction. Edit: It is regular at fixed point $0$: $\lim_{x\to-\infty} \frac{e^{2^{x}}-1}{2} = 0$ 2. $f(x)=4(x+x^2)$, $F(x)=\sinh\left(2^x\right)^2$ Let us check: $F(x+1)=\sinh\left(22^x\right)^2=4\sinh(2^x)^2\cosh(2^x)^2=4\sinh(2^x)^2(1 + \sinh(2^x)^2)=4(F(x)+F(x)^2)$ It is again regular at 0. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/23/2009, 03:46 PM (This post was last modified: 05/11/2009, 10:04 PM by bo198214.) Next function $f(x)=cx^a$, the natural numbered iterates are: $f^{[1]}(x)=cx^a$ $f^{[2]}(x)=c(cx^a)^a=c^{a+1} x^{a^2}$ $f^{[3]}(x)=c\left(c^{a+1}x^{a^2}\right)^a = c^{a^2+a+1} x^{a^3}$ $f^{[n]}(x)=c^{\sum_{k=0}^{n-1}a^k} x^{a^n}=c^{\frac{a^n-1}{a-1}} x^{a^n}$ So a super-function would be $F(t)=c^{\frac{a^t-1}{a-1}}\exp(a^t)=\exp(a^t+\ln( c)\frac{a^t-1}{a-1})$ (*) $F(t)=\exp\left(\left(1+\frac{\ln( c)}{a-1}\right)a^t -\frac{\ln( c)}{a-1}\right)$ For the regular iteration we need to find a fixed point $\lambda$ $c\lambda^a = \lambda$ $c\lambda^{a-1}=1$ or $\lambda=0$ $\lambda=c^{-\frac{1}{a-1}}=c^{\frac{1}{1-a}}$, $a\neq 1$. Then (*) looks like: $F(x) = (e /\lambda)^{a^x} \lambda$ If we translate $F$ along the x-axis we can get $F(x)=\exp(a^x)\lambda$ Check: $cF(x)^a=\exp(a^x)^a c\lambda^a=\exp(a^{x+1})\lambda=F(x+1)$ Edit: It is regular at $\lambda$. Summary: $F(x)=\exp(a^x)c^{\frac{1}{1-a}}$ is the at $\lambda=c^{\frac{1}{1-a}}$ regular superfunction of $f(x)=cx^a$, for $a\neq 1$. tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 04/27/2009, 11:16 PM bo198214 Wrote:Summary: $F(x)=\exp(a^x)c^{\frac{1}{1-a}}$ is an elementary superfunction of $f(x)=cx^a$, for $a\neq 1$. this result is far from new. --- 'in general' super-functions are of hypergeometric or inverse hypergeometric " kind " , where " kind " mainly denotes nested structures. and with ' in general ' i mean usually if the (original) function is elementary. i advocated the concept of inverse hypergeometric functions before , as e.g. on sci.math but without much results. usually , if we arrive at an integral expression for our super-function its hypergeometric or inverse hypergeometric. and that can often be reduced to elementary by using 'integral calculus'. i tried to related all of this to half iterations of exp(x) but nothing worked. mainly because exp(x) lacks a real fixpoint and a real zero at the same time .... regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/28/2009, 08:33 AM This thread was not intended to make a lot of wise comments, but rather to establish a collection of examples of elementary super-functions. So if you have one which is not already mentioned, please post. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/28/2009, 09:09 AM Ansus Wrote:What you're speaking about is simply a fractional iteration. What you mean is probably regular iteration. Everything is a fractional iteration that satisfies: $f^{[1]} = f$ and $f^{[s+t]}= f^{[s]}\circ f^{[t]}$ While regular iteration at a fixed point p is a certain fractional iteration that satisfies that $f^{[t]}$ is differentiable at the fixed point $p$. Regular iteration is unique by this demand. Quote:You know that fractional iteration can be expressed in terms of Newton's and Lagrange's series. There are also methods to solve iterational equations that can be extended to fractional iterations. Ansus, again, the title of this thread is *elementary* *superfunctions*. That implies 1. I am seeking *elementary* functions, while all methods of regular iteration dont return elementary functions per se. 2. I am seeking *superfunctions*, they dont have necessarily to be regular (=the result of regular iteration). So if you have an example of an elementary superfunction that is not by regular iteration at some fixed point, I would highly appreciate it! bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/28/2009, 04:25 PM 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. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/28/2009, 06:12 PM Ansus Wrote:Mathematica simply finds these formulas. I do not know how it does. You mean by solving the recurrence F(z+1)=f(F(z))? Please take the time, simplify the results and verify them. I mean its a quite strange looking term $\left(-\frac{1}{\sqrt{a}}\right)^x$, isnt it? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/28/2009, 08:17 PM (This post was last modified: 04/28/2009, 08:24 PM by bo198214.) Thanks, for simplifying. Ansus Wrote:What's wrong with it? Well the function is supposed to be real and this would give complex values, if they do not cancel out, or something. But I see the functions are all decreasing, so one would not expect a real solution. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/28/2009, 08:25 PM bo198214 Wrote:Well the function is supposed to be real and this would give complex values, if they do not cancel out, or something. Yeah, also, Mathematica is very pedantic about exponents. For example, it does not assume $(a^b)^c = a^{(bc)}$ because this is only true for complex numbers where $-\pi < \text{Im}(b\ln(a)) < \pi$ or something. « Next Oldest | Next Newest »

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