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

Question

Let $n=\underbrace{11\dots1}_{1996\text{ figures}}$. Does there exist such a function $f(x)$ that for all real $x \ne 0, x \ne 1$ holds $$\underbrace{f \bigg ( f \Big( \dots \big( f}_{n\text{ iteratoins}}(x) \big) \Big) \bigg)=\left (1-\frac {1}{\sqrt[n]{x}} \right)^n?$$

Answer 1

The answer is yes. Define $ g : \mathbb R \setminus \{ 0 , 1 \} \to \mathbb R \setminus \{ 0 , 1 \} $ with $ g ( x ) = 1 - \frac 1 x $. Then you will have $ g \big( g ( x ) \big) = \frac 1 { 1 - x } $ and $ g \Big( g \big( g ( x ) \big) \Big) = x $ for all $ x \in \mathbb R \setminus \{ 0 , 1 \} $. Thus inductively we'll have $$ g ^ m ( x ) = \begin {cases} x & m \equiv 0 \pmod 3 \\ 1 - \frac 1 x & m \equiv 1 \pmod 3 \\ \frac 1 { 1 - x } & m \equiv 2 \pmod 3 \end {cases} $$ for every nonnegative integer $ m $, where $ g ^ m $ denotes $ m $'th iteration of $ g $.

Now, define $ f : \mathbb R \setminus \{ 0 , 1 \} \to \mathbb R \setminus \{ 0 , 1 \} $ with $ f ( x ) = g ( \sqrt [ n ] x ) ^ n = \left( 1 - \frac 1 { \sqrt [ n ] x } \right) ^ n $. It's easy to inductively show that $ f ^ m ( x ) = g ^ m ( \sqrt [ n ] x ) ^ n $ for every nonnegative integer $ m $. Thus we have $$ f ^ m ( x ) = \begin {cases} x & m \equiv 0 \pmod 3 \\ \left( 1 - \frac 1 { \sqrt [ n ] x } \right) ^ n & m \equiv 1 \pmod 3 \\ \frac 1 { ( 1 - \sqrt [ n ] x ) ^ n } & m \equiv 2 \pmod 3 \end {cases} $$ and in particular, since $ n \equiv 1996 \equiv 1 \pmod 3 $, $ f ^ n ( x ) = \left( 1 - \frac 1 { \sqrt [ n ] x } \right) ^ n $.