# Does there exist such a function $f(x)$ that $f(f(…(f(x))))=\left (1-\frac {1}{\sqrt[n]{x}} \right)^n?$

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 )

• That means $f (1 - 1/\sqrt [n]x)^n) = (1-/\sqrt [n+1])^{n+1}$. Can you solve that. – fleablood Feb 5 '16 at 8:16
• Actually it doesn't have to mean that. But it's still a way to find the funtion. – fleablood Feb 5 '16 at 8:25
• @fleablood Are you sure? It seems to me by reading the question that the relationship is true for $n=1996$, not necessarly $\forall n \in \mathbb{N}$. – Martigan Feb 5 '16 at 8:54
• 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. – fleablood Feb 5 '16 at 16:45