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Do functions exist such that $f^n(x)=x$ for values $n > 2$? For $n=2$ we have $-x$ and $1/x$, and for $n=3$ we can show $1/(1-x)$ is a solution.

I assume that $n$ must be prime and preferably solutions are differentiable over some interval.

Thanks

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    $\begingroup$ Well, for all $n \in \Bbb Z^+$, it's obvious $f(x)=x$ is a solution, albeit a trivial solution. For $n$ even, it is immediately clear that $-x$ and $1/x$ still work, too. Interestingly the solution for $n=3$ doesn't work for $n=5$, so there's something to be discussed there. $\endgroup$ – Eevee Trainer Mar 7 at 0:06
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Function defined by

$$f(x)=\dfrac{x \cos(a) - \sin(a)}{x \sin(a) + \cos(a)} \ \ \text{with} \ \ a=\frac{2 \pi}{n} \tag{1}$$

called an "homographic function" (or "fractional linear function") is such that

$$f^n(x)=x\tag{2}$$

Proof :

1) Coefficients used in (1) are exactly the entries of the rotation matrix $R_a$ with angle $a$.

2) The correspondence between functions

$$f(x)=\dfrac{x p + q}{x r + s} \leftrightarrow \binom{p \ \ q}{r \ \ s} \tag{3}$$

(see https://users.math.msu.edu/users/sen/math_840_2005/lectures/lec_11s.pdf where it is presented for complex values of the variable) is a "nice" homomorphism putting compositions of such "homographies" in correspondence with products of their associated matrices.

Thus (2) is equivalent to $R_a^n=Id$.

For example $y=f(x):\frac{1}{x}=\frac{0x+1}{1x+0}$ corresponds to the case $a=\pi$ in (1).

Remark : Correspondence (3) can be turned into an isomorphism if entries $a,b,c,d$ are considered "up to a factor" (we define in this way the "projective linear group" $PGL(2,\mathbb{R}). Another approach would be to constraint entries $a,b,c,d$ to be such that $\det(\binom{p \ \ q}{r \ \ s})=1$.

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If $\omega$ is an n-th root of unity then $f(x)=\omega x$ satisfies $f^{n} (x)=x$.

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  • $\begingroup$ Question: how exactly does the solution $f(x) = 1/(1-x)$ (to $f^3(x) = x$) fit into this? I can see your answer is correct but based on this solution I don't think this encompasses every solution, unless I'm missing something obvious. $\endgroup$ – Eevee Trainer Mar 7 at 0:08
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    $\begingroup$ Moreover, I think the OP has in mind functions $\mathbb{R} \to \mathbb{R}$. $\endgroup$ – Jean Marie Mar 7 at 0:10
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    $\begingroup$ @EeveeTrainer Apart from the trivial solution, which you state in your question comment, the answer provided here shows it's true for all integer $n$. Note the OP asks "Do functions exist ...", and specifically, the OP doesn't ask for every possible solution. $\endgroup$ – John Omielan Mar 7 at 0:11
  • $\begingroup$ True, that's a fair point. $\endgroup$ – Eevee Trainer Mar 7 at 0:12
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    $\begingroup$ Yeah could have been more clear. Looking for specific solutions and real functions. There’s a connection to cycles in Newton’s method of finding roots, which is why I wanted differentiability. $\endgroup$ – GossipM Mar 7 at 0:20

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