Convergence of iterative formulae I have noticed that the the iterative formula: $$x_{n+1}=a^{\frac1{x_n}} $$ has some interesting behaviour starting from $x_0=1$ I began with $a=10$ for which the formula converges to a solution to $x=10^\frac{1}{x} $ However, when it comes to $a=20$ the formula does not converge, it approaches two different values that it bounces between - so not a solution to the equation to $x=20^\frac{1}{x} $. Moving the value of $a$ 10 and 20 shows that past ~15 seems to be when convergence to one value stops happening. But this transition is a smooth one, the convergence becomes slower approaching 15 and the two values become further apart passing 15. Could someone explain this behaviour, specifically what value near 15 is the epicentre of this behaviour and why the convergence stops in the first place. An unjustified thought was: $e^e=15.15426224..$.

**You guess was right, it is $\mathbf{e^e}$.**

Now to the point. Say you iterate some function, like $x_{n+1}=f(x_n)$, and you know there is a point $x_0$ such that $f(x_0)=x_0$; will the iterative process converge to that point?

Roughly speaking, it will if $|f'(x_0)|<1$ and won't otherwise. (You prove this by assuming a small deviation from $x_0$ and checking whether it gets smaller yet.)

Now to the point. In our case $f(x)=a^{1/x}$, so $f'(x)=-\dfrac{\ln a}{x^2}\cdot a^{1/x}$. Now, at the fixed point we have $x_0=a^{1/x_0}$, so $|f'(x_0)|=\dfrac{\ln a}{x_0^2}\cdot x_0=\dfrac{\ln a}{x_0}$. For this to equal $1$, we need $x_0=\ln a$. By plugging that back into the equation, we get:

$$x_0=a^{1/x_0}=a^{1/\ln a} = e$$

Consequently, the critical value of parameter is

$$a_{crit}=x_0^{x_0}=e^e$$

Q.e.d.