I'm studying some basic behaviour of the logistic map $x_{n+1}=r x_n(1-x_n)$, with $x_0\in (0,1]$ and $r\in (0,4]$, for a project. I can't seem to figure out why the map does not converge (i.e. has no fixed points) for $r>3$ or how I could prove that it doesn't.
Edit: striked a part of the text that, apparently, wasn't really what I meant to ask but to make sure it can still be read, to potentially make sense of some answers which might have been given to the original question.
It still has a fixed point in $(0,1]$ for $r > 3$, namely $(r-1)/r$. But if $f(x) = r x (1-x)$, $f'((r-1)/r) = 2 - r$. If $r > 3$ this is less than $-1$, which implies that the fixed point is unstable. The other fixed point $0$ is unstable if $r > 1$.
Since both fixed points are unstable, the sequence $x_n$ can only converge if it hits a fixed point exactly. This can happen, e.g. if $x_0 = 1/r$ you get $x_1 = 1 - 1/r$, and then $x_n = 1 - 1/r$ for all $n \ge 1$.