The equation $f^{\circ k}(x) = \mathrm{Id}$ for $x\in E$ is called the Babbage equation and the general solution is given in the following way [M. Kuczma, Functional equations in a single variable]:
Let $1 = n_0 < \cdots < n_r = k$ be the complete set of divisors of $k$ and let $$ E = \cup_{i=0}^{r}\cup_{j=1}^{n_i}U^i_{j} $$ be an arbitrary decomposition of $E$ into disjoint sets such that for every $i$ the sets $U^i_{1},\ldots,U^i_{i_n}$ have the same cardinality. For $1\leq i\leq r$ and $1\leq j\leq n_i-1$, let $f_{ij}$ be an arbitrary one-to-one map of $U^i_j$ onto $U^i_{j+1}$. Then the general solution is given by:

  • $f(x) = x$ for $x\in U^0_1$;
  • $f(x) = f_{ij}(x)$ for $x\in U^i_j$, $j=1,\ldots,n_i-1$, $i\ge 1$;
  • $f(x) = f_{i1}^{-1}(\cdots(f_{i,n_i-1}^{-1}(x))\cdots)$ for $x\in U_{n_i}^i$, $i\geq 1$.

This solution is also valid when $f$ is a function of more that one variable.

In dimension 1, the solution $f$ is continuous if and only if $f$ is monotone, and in this case either $f(x)=x$ or $f$ is a decreasing function such that $f^{\circ 2} = \mathrm{Id}$.

Question: In dimension $n\ge 2$, assuming $f$ is continuous, what are the general solutions of the Babbage equation?

We can of course give some special solutions, such as when $n=2$:

  • $f(x,y)=(y,x)$ for even $k$.
  • $f(x,y)=(f_1(x),f_2(y))$ such that $(f_1)^{\circ k}(x)=x$ and $(f_2)^{\circ k}(y)=y$.
  • $\begingroup$ Of course, the question is what you expect $f$ to be. If you don't assume, say, continuity (regarding the domain as just a set), then certainly the answer below describes everything: the whole set splits into orbits, which should be of bounded length. $\endgroup$ Apr 3, 2014 at 19:14
  • $\begingroup$ Can you tell us what's the "general solution" for a function of one variable? Just curious. $\endgroup$ Apr 3, 2014 at 20:59
  • $\begingroup$ I modified my question with a general solution. $\endgroup$
    – Lucy
    Apr 4, 2014 at 12:25
  • $\begingroup$ @AlexDegtyarev You are right, I assume the solutions to be continuous or differentiable. $\endgroup$
    – Lucy
    Apr 4, 2014 at 12:26

1 Answer 1


There are certainly many other solutions. Here is a recipe for constructing them. If $S$ is a set (say infinite to simplify the presentation). We split $S$ into a disjoint union of $k$-element subsets $S_\alpha$. On each such subset we consider a cyclic permutation. We then fit them together to a mapping on the whole space. This clearly satisfies your condition.

  • $\begingroup$ Yes. I considered permutations before. Maybe I should modify my question to the subclass of functions up to permutations rotations and reflections. $\endgroup$
    – Lucy
    Apr 3, 2014 at 18:12
  • $\begingroup$ Actually I think that all solutions arise essentially in this manner (I didn't include this fact in my response since I hadn't sat down to write it out). In this case, this kind of modification might not be very useful. $\endgroup$
    – barcelos
    Apr 3, 2014 at 18:20
  • $\begingroup$ I suspect that the article "On certain real solutions of Babbage's functional equation" by Ritt (Ann.Math. 79 (1972) 113-122) is being referred to. By the way, using his simple observation that a function which is conjugate to a solution is also a solution, it is easy to get a plethora of CONTINUOUS solutions on the plane. Just take a function which is conjugate to a Möbius function. Many of the latter are periodic in the above sense. $\endgroup$
    – barcelos
    Apr 4, 2014 at 8:59
  • $\begingroup$ Thank you, @barcelos. I read the paper. I am trying to characterize the "complete" continuous solutions and hope to find interesting results as in dimension 1, where continuity forces all solutions to be either trivial($f(x)=x$) or a involutory, i.e. $f^{\circ 2}(x)=x$. Moreover, a involutory on a real interval is nothing but a decreasing monotone function with some fixed points. $\endgroup$
    – Lucy
    Apr 4, 2014 at 15:07

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