Why does this sequence of operations give $x^{\frac{1}{x-1}}$? I found (purely from experimentation) that if you start with a random number and successively:


*

*Exponentiate, 

*Raise to the power of $x$,

*Take the log with the same base as step one,

*Take the $x$-th root


(Repeat)
The base for steps one and three is arbitrary.
It will converge to $x^{\frac{1}{x-1}}$. Why?
 A: Let $r$ be the random number, and let $r_n$ be the number obtained after repeating the 4 step procedure $n$ times. Let $b$ be the base used in steps 1 and 3. Starting with $r_n$, we do the following: 
1 Exponentiate: $b^{r_n}$
2 Raise to the power $x$: $(b^{r_n})^x = b^{xr_n}$
3 Take the log (same base): $\log_b(b^{xr_n}) = xr_n$
4 Take the $x$-th root: $\sqrt[x]{xr_n}$
Thus, we have the recurrence relation: $r_{n+1} = \sqrt[x]{xr_n}$ with the initial condition $r_0 = r$. 
If $\displaystyle\lim_{n \to \infty}r_n = L$ exists, then we must have: 
$L = \sqrt[x]{xL}$ 
$L^x = xL$ 
$L^{x-1} = x$ OR $L = 0$ 
$L = x^{\tfrac{1}{x-1}}$ OR $L = 0$. 
To show that the limit does infact exist and isn't $0$, we need some conditions on $x$. Did you impose the restrictions $x > 1$ and $r > 0$? Assuming so, with some simple algebra and induction, we can show that: 
If $r_0 \ge x^{\tfrac{1}{x-1}}$, then $r_n$ is non-increasing and bounded below by $x^{\tfrac{1}{x-1}}$. 
If $0 < r_0 \le x^{\tfrac{1}{x-1}}$, then $r_n$ is non-decreasing and bounded above by $x^{\tfrac{1}{x-1}}$. 
From that, we can conclude that the limit exists, and isn't $0$. Hence, $\displaystyle\lim_{n \to \infty}r_n = x^{\tfrac{1}{x-1}}$.
