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The following is, I assume, a known fact---however, the only proof I know uses the axiom of choice (then a theorem about well-ordered sets as well as transfinite recursion). I was wondering if anyone knew a more elementary solution.

Let $a_0$, $a_1$, ..., $a_n$ be a sequence of real numbers, $a_n \neq 0$. Then the functional equation $$ a_0x + a_1f(x) + a_2 f^{(2)}(x) + ... + a_n f^{(n)}(x) = 0 $$ always has a solution.

Here $f^{(n)}$ denotes function $f(f(...(x)...))$, that is, the $n$-fold composition of function $f$ with itself. And "function" here means an absolutely arbitrary mapping from $\mathbb R$ to $\mathbb R$---it does not have to be continuous or bijective etc.

Of course, if polynomial $P(t) = \sum_{k=0}^n a_kt^k$ has a real root $r$, then function $f(x) = rx$ is such a solution. However, when all roots are complex, the matter becomes quite non-trivial (but solving this problem for an arbitrary quadratic polynomial $Q(t)$ would be enough, as any functional root of $Q$ would also be a functional root of any polynomial divisible by $Q$).

JimT
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  • There is another solution (also using axiom of choice as it is needed for Hamel bases) which involves algebraic number fields and linear spaces over them. It is shorter than the one I had in mind, but by no means less elementary. – JimT Feb 19 '22 at 18:03
  • Related: https://math.stackexchange.com/questions/312385/find-a-real-function-f-mathbbr-to-mathbbr-such-that-ffx-x which contains an in-depth discussion of the specific instance for the quadratic polynomial $1+t^2;$ this instance of the problem reduces to finding $f\colon \mathbb R \to \mathbb R$ such that $f(f(x))=-x.$ – Mitchell Spector Feb 21 '22 at 04:56
  • Yes, I saw that one. However, such special cases -- when the polynomial's roots are unity roots -- are kind of easy. The general case is significantly harder, and as far as I know, does not allow for the explicit construction. That's why I created this, in hope that someone perhaps knows an elementary solution for the general case. – JimT Feb 23 '22 at 20:30

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