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I started playing with the functional equation $x^n f(1/x) = f(x)$, with $n \in \mathbb N$. I found two solutions: $g(x)=a x^{n/2}$ and $h(x)=b(1+x+...+x^n)=b S_n(x)$.

I also found that a linear combination of $g,h$ is also a solution.

Any other solutions? More general ideas?

Giuseppe
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    Since $x\neq 0$ then $f(1/x)=1/x^{n-1}$ and so $f(x)=x^{n-1}$ by substitution. Which is the only possible solution. Or am I missing something? I'm not sure where you got your functions from. – freakish Mar 26 '24 at 08:42
  • Perhaps you meant $x^n f(1/x) = f(x)$ ? – Martin R Mar 26 '24 at 08:50
  • Yes, I am an idiot. I meant that. I will try to fix it. – Giuseppe Mar 26 '24 at 08:51
  • I apologize for the confusion. Writing in the subway is not a good idea. – Giuseppe Mar 26 '24 at 08:53
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    Every $n$-th degree polynomial with “symmetric coefficients” $a_k = a_{n-k}$ is a solution. – Martin R Mar 26 '24 at 08:53
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    Also every function $f:(0, 1] \to \Bbb R$ with $f(1) = 1$ can be extended to a solution $f:(0, \infty) \to \Bbb R$ via the functional equation, so there are many solutions. – Martin R Mar 26 '24 at 08:56
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    ... The condition $f(1) = 1$ is not necessary. – You need to restrict the problem to certain classes of functions (polynomials, differentiable/smooth functions on $(0, \infty)$, ...) in order to get nontrivial results. – Martin R Mar 26 '24 at 09:07

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Clearly, $x\neq 0$ for the functional equation to even make sense. Let $P(x)=\frac{f(x)}{x^{n/2}}$. Now, $$x^{n/2}f(1/x)=\frac{f(x)}{x^{n/2}}$$ $$P(1/x)=P(x)$$ Thus $P(1/x)=P(x)$ is necessary. Plugging in $f(x)=x^{n/2}$ to the functional equation, this condition is also sufficient.

We now have our answer: all functions $P$ for which $P(x)=P(1/x)$ will yield a valid function $f$ under $f(x)=x^{n/2}P(x)$, as this maintains a necessary and sufficient condition. Notice that since $f$ is not specified to be continuous, $P$ could even be a point wise function satisfying the condition. There isn't much restriction.

In the results you provided, the first has $P(x)=a$ and the second has $P(x)=b\sum_{i=-n/2}^{n/2}x^i$, both of which satisfy the given condition.

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