# Identity from repeated function composition

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

• 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. – Eevee Trainer Mar 7 at 0:06

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$. If $$\omega$$ is an n-th root of unity then $$f(x)=\omega x$$ satisfies $$f^{n} (x)=x$$. • 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. – Eevee Trainer Mar 7 at 0:08 • Moreover, I think the OP has in mind functions$\mathbb{R} \to \mathbb{R}$. – Jean Marie Mar 7 at 0:10 • @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. – John Omielan Mar 7 at 0:11
• True, that's a fair point. – Eevee Trainer Mar 7 at 0:12
• 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. – GossipM Mar 7 at 0:20