4

This question has crossed my mind, and I tried finding some solutions to that functional equation, then to find a pattern.

It's surprisingly hard to find real functional equation calculators online, so I only found 3 solutions.

$f(x)=\frac{1}{1-x}=f^{\circ4}$
$f(x)=\frac{1}{2-2x}=f^{\circ5}$
$f(x)=\frac{1}{3-3x}=f^{\circ7}$

I didn't really found more, nor find a pattern, nor anything online. So if anyone has an answer, feel free to share it!

GEdgar
  • 117,296
Pierre Carlier
  • 353
  • 2
    Well, if we restrict the discussion to the case where $f$ is injective too. Then we're looking for the solutions to $f^{\circ (n-1)}=id$. There are a lot of these, for instance things like $f(x)=\begin{cases}x+1&\text{if }x\in \Bbb Z\land 0\le x<n-2\ 0&\text{if }x=n-2\ x&\text{if }x\notin\Bbb Z\lor x<0\lor x>n-2\end{cases}$. – Sassatelli Giulio Nov 18 '23 at 12:57
  • 1
    What is the domain of these functions? E.g. if $f(x)=\frac{1}{1-x}$, then $1\not \in \operatorname{dom}(f)$. Is the intended domain $\mathbb{R}\setminus{1}$? – Joshua Tilley Nov 18 '23 at 15:51
  • A pattern? ... what is the corresponding thing for $\frac{1}{4-4x}$ ? In all of your cases, you have $f^{\circ(n-1)}(x) = x$. Are there solutions of yours that fail this? – GEdgar Nov 18 '23 at 17:52
  • 3
    Does $f^{\circ n} (x) $ mean the composition of $f$ with itself, $n$ times? – Max Lonysa Muller Nov 18 '23 at 17:55
  • Such a function $f:X\to X$ can be thought of as a group action on $X$. Specifying that such an action has finite order $n$ tells us that the size of the orbits of the action divide $n$. If we, for instance, pick $n=3$, and then specify a partition of $X$ into $3$ element sets, the existence of a function whose orbits are exactly that partition requires you to arbitrarily "choose infinitely many socks". And you cannot even write down one such function without imposing some additional structure on $X$. – ZKe Nov 18 '23 at 19:03
  • If you have a function whose iteration can be expressed through the mechanics of the Schroeder/Koenigs-solution, then for some iteration-height $h$ we get in the heart of the procedure an expression with $f°^h(x) = ... g(u^h \cdot c(x) ... ) ... $ where $u$ is the logarithm of the fixpoint of the function $f(x)$ . If the exponentiation of $u$ is periodic with $h$ , with period $n$ - say - then we have $f°^{h+n}(x) = f°^h(x) $ periodicity of $f(x)$. – Gottfried Helms Dec 08 '23 at 16:34
  • Thanks for you answer Gottfried. After reading some (more) papers about iterated functions that I could understand much, I realized I needed some basic knowledge in some fields of math that I don't currently possess. I know there's some things such as analytic continuation, cauchy integrals or more, but could you give me some kind of list of the fields/mathematical knowledge to have in order to understand the very basis of all of these papers? (Since the tetration forum "closed" it's hard to contact you guys, and mathstackexchange isn't the best. Is there a way to keep in touch with emails?) – Pierre Carlier Dec 19 '23 at 10:32
  • Wolfram|Alpha can find you more such solutions via the input Reduce[1/(a - a x) == Nest[1/(a - a#) &, x, n], {a, x}] ($n$ is the number of iterations (replace it by a natural number $> 2$) and $f\left( x \right) = \frac{1}{a - a \cdot x}$). Plotting these values via the curve $a\left( n \right)$ in a graph would look like this (graph from $n = 1$ to $n = 25$). – The Art Of Repetition Dec 23 '23 at 22:17
  • It's not exactly what you're looking for, but you can find there another class of functions satisfying your condition, see https://math.stackexchange.com/questions/4298238/fffx-x-with-fx-ne-x-and-ffx-ne-x/4298342#4298342 – Pavel R. Dec 23 '23 at 23:02
  • If you take $f^{-1}$ (inverse function) on both sides, you get $x = f^{\circ\left( n -1 \right)}\left( x \right)$. And there is already a question about this: Non-trivial examples of $n$-th iterative root of $h(x) = x$, i.e. solution of functional equation $f^{[n]}(x) = x$ – The Art Of Repetition Dec 29 '23 at 16:32
  • Helms, I currently struggle to rigorously define something relevant to iterated function. Could you try to help me? – Pierre Carlier Apr 27 '24 at 20:38

0 Answers0