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Just for curiosity I was trying to find a sequence of non-trivial continuous functions at $\mathbb{R}$ except finite many points such that $f_{n+1}(f_{n+1}(x))=f_{n}(x)$, $f_1(x)=x$ and by non trivial I mean that $f_n (x)\ne x , \ f_n(x) \ne \frac{1}{x}, \ f(x)\ne -x$.

Let $f_2(x)= \frac{1-x}{1+x}$ I tried to find $f_3(x)$ but bit was too hard for me to find so I am not sure if a closed form of $f_n$ exist and I don't think there is a unique solution if the closed form exits.

Note that $f_{n+1}\ne -f_n$ ,$f_{n+1}\ne -\frac{1}{f_n} , \ f_{n+1}\ne f_n$ these are the trivial cases.

Is there a closed form for $f_n$? As I said before I think there might be infinite many non trivial solutions but if that the case I want any solution.

pie
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    There is no unique solution. $f_{n+1}(x)=-f_n(x)$ is another simple example, while $f_{n+1}(x)=f_n(x)$ for $x$ rational but $=\frac1{f_n(x)}$ for $x$ irrational is another complicated example. These are called involutions – Henry Apr 16 '24 at 09:31
  • @Henry I forgot the trivial case $-x$ and I will update the question to make these functions continuous at $\mathbb{R}$ except finite many points – pie Apr 16 '24 at 09:34
  • You are going to need a more precise definition of simple and complicated: $f(x)=\sqrt{2}-x$ is a small step away from simple. You might want to look at https://math.stackexchange.com/questions/46635/examples-of-involutions-on-mathbbr – Henry Apr 16 '24 at 09:44
  • If you allow complex numbers then $$ f_3 (x) = - (1 + \sqrt 2 {\rm i)}\frac{{x - \frac{1}{3} + \frac{{\sqrt 2 }}{3}{\rm i}}}{{x + 1 - \sqrt 2 {\rm i}}} $$ satisfies $f_3(f_3(x))=\frac{1-x}{1+x}$. See also https://math.stackexchange.com/questions/3305915 – Gary Apr 17 '24 at 04:24
  • @Gary I begin to believe that maybe such function doesn't exist if it is restricted to $\mathbb{R}$ – pie Apr 17 '24 at 05:29
  • Given that $f_2(0)=1$, monotonically decreasing to $f_2(1)=0$, it should be straightforward to prove that there's no continuous $f_3()$ satisfying the conditions. – Steven Stadnicki Apr 17 '24 at 06:07
  • @StevenStadnicki I cann't see why is this straightforward, can you elaborate more ? – pie Apr 17 '24 at 06:09
  • It's a little trickier than I initiallly gave it credit for (there are a lot of little details) but the core of it is that since $f_2'(x)\lt 0$ for all $x$ in its domain, by expanding it out using the chain rule we have $f_3(f_3'(x))f_3'(x)\lt 0$ for all $x$ which implies that we can never have $f_3'(x)=0$ and never have $f_3(y)=0$ for all $y$ in the range of $f_3'()$. From there it comes down to just showing that these conditions are impossible to meet. (e.g., if we define $\varphi=f_3(1)$ then $f_3(\varphi)=0$ which implies that $\varphi$ isn't in the range of $f_3'()$.) – Steven Stadnicki Apr 17 '24 at 17:10

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