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Let $n=11...1$ (1996 figures). Does there exist such a function $f(x)$ that for all real $x \not =0, x \not =1$ holds $$f \left ( f\left (...\left (f(x) \right) \right) \right)=\left (1-\frac {1}{\sqrt[n]{x}} \right)^n?$$ (in the left side the $n$-th iteratoin of $f$ is written )

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    $\begingroup$ That means $f (1 - 1/\sqrt [n]x)^n) = (1-/\sqrt [n+1])^{n+1} $. Can you solve that. $\endgroup$ – fleablood Feb 5 '16 at 8:16
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    $\begingroup$ Actually it doesn't have to mean that. But it's still a way to find the funtion. $\endgroup$ – fleablood Feb 5 '16 at 8:25
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    $\begingroup$ @fleablood Are you sure? It seems to me by reading the question that the relationship is true for $n=1996$, not necessarly $\forall n \in \mathbb{N}$. $\endgroup$ – Martigan Feb 5 '16 at 8:54
  • $\begingroup$ No, I'm not sure. Finding such an iterative function would work but I'm not sure one can. I made a small error in calculating. $\endgroup$ – fleablood Feb 5 '16 at 16:45

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