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I was reading "Functional Equations and How to Solve Them" by Small and the following comment pops up without much justification on p. 13:

If $a(x)$ is an involution, then $f(a(x))=f(x)$ has as solutions $f(x) = T\,[x,a(x)]$, where $T$ is an arbitrary symmetric function of $u$ and $v$.

I was wondering why this was true (it works for examples I've tried, but I am not sure $(1)$ how to prove this and $(2)$ if there's anything obvious staring at me in the face here).

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    $\begingroup$ Write down what f(a(x)) is in terms of T. $\endgroup$ – Qiaochu Yuan Jul 30 '10 at 2:42
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Any function f(x) that is a solution to your functional equation f(a(x)) = f(x) must satisfy the property that it is unchanged when you plug in a(x) instead of x. In addition, f(x) clearly must be a function f(x) = T[x, a(x)] depending on the x and a(x); the question is to see why T must be symmetric.

Now, T[x, a(x)] = f(x) = f(a(x)) = T[a(x), a(a(x))] = T[a(x), x] since a(x) is an involution. In particular, this means that T must be symmetric in its two variables.

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    $\begingroup$ The same applies if you have a map $a$ satisfying $a^n = Id$ and there is a symmetric function in $x, a(x), ... , a^{n-1}(x)$. (Note that powers here refer to iterated composition.) $\endgroup$ – Akhil Mathew Jul 30 '10 at 2:52
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    $\begingroup$ LaTeX works here on MU? $\endgroup$ – A B Jul 30 '10 at 2:57
  • $\begingroup$ MU? What's that? The only possible interpretation which comes into my mind it "Math Underflow" (as opposed to "Math Overflow"). However that's not what this site is named (although it would have been a great choice IMHO). $\endgroup$ – celtschk Aug 23 '12 at 15:28

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