1

I am wondering if it exits a way to find "easely" the solutions of an equation about a function $f$ such that $$f^{(n)}(x)=x$$

where $f^{(n)}$ is the n-th composition of $f$ itself.

Obviously the identity is a trivial solution, I'm asking for all solution depending on $n$.

For example I know that for $n=4$, if $$f(x)=\frac{1+x}{1-x}$$ then $f^{(4)}(x)=x$

Any hints would be helpfull, thank you in advance.

ParaH2
  • 1,672
  • I do not think that an "easy" solution exists for that. – Peter Mar 21 '16 at 22:49
  • Well, for composite $n$ you can construct factor solutions. Maybe if those are all the solutions for $n$ you can restate the problem as finding solutions for $f^{(p)}(x)=x$, $p$ prime. – abnry Mar 21 '16 at 22:52
  • Sorry, I actually think partition numbers come into play. – abnry Mar 21 '16 at 22:53
  • I dislike partition numbers. So it looks to be more difficult than I thought. Maybe instead of find them it exists a way to know a upper bond of how much solutions there are for any n ? – ParaH2 Mar 21 '16 at 22:59
  • What are the domain and range of $f$? ((For any given $n > 1$, there are infinitely many functions $f: \Bbb{N} \to \Bbb{N}$ such that $f^n(x) = x$ for all $x$). – Rob Arthan Mar 21 '16 at 23:11
  • The solutions are infinite. For example, for $n=4$ you can use $f(x)=\frac{(k^2+1)/2+x}{k-x}$ for an arbitrary $k$. A similar parametric functon can be find easily other small $n$. For example, for $n=5$ I got $$f(x)=\frac{\frac{1}{2} \left(3 b^2+4 b+\sqrt{5} (b+1)^2+3\right)+x}{b-x}$$ for arbitrary $b$. – Giovanni Resta Mar 22 '16 at 13:46
  • Hum interesting and if n=100! ? :D – ParaH2 Mar 22 '16 at 17:36

1 Answers1

3

Here's a general answer that might help you.

Suppose we have a rational function $$\rho(x)=\frac{ax+b}{cx+d}$$ and we define the "coefficient matrix" of $\rho$ to be $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$ then the coefficient matrix of $\rho \circ \rho$ or $\rho^2$ is $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} ^2$$ This theorem is rather trivial, and I will leave the proof to you.

If you know about rotation matrices, then you should know that $$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix} ^n=\begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \\ \end{pmatrix}$$ Meaning that if we let $$\theta=\frac{2\pi}{n}$$ Then we have $$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix} ^n=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ And so, since $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ is the coefficient matrix of the identity function $x$, if we take $\rho$ to be the rational function defined by the coefficient matrix $$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}$$ Then $$\rho^n(x)=x$$ and so $\rho$ can produce a solution to any functional equation in the form you asked about (though I'm sure it cannot be the only solution to any of them).

Does this help?

Franklin Pezzuti Dyer
  • 39,754
  • 9
  • 73
  • 166