Consider the logistic map $f(x) = kx(1-x)$ defined on $\mathbb{R}$. We already know $\frac{k-1}{k}$ is a fixed point of $f$, but my issue is showing it's an attractor when $k \in [1,3]$. There is an equivalent definition of attracting fixed points involving the derivative of $f$, but the definition I'm using is that the fixed point $a$ is an attractor when there is an open ball centered at $a$ such that all iterated maps based on points in that open ball converge to $a$.
I presume $B_{0.1} (\frac{k-1}{k})$ is a valid choice of an open ball, based on some initial iterations with differing values of $k$. Though finding the largest valid radius might lead to a cleaner proof. But how would I go about proving all iterated maps in the ball converge to the fixed point? Like most converging sequence problems, we let $\varepsilon > 0$ and try to find some $N \in \mathbb{N}$ such that if $n > N$, $|f^n(x) - \frac{k-1}{k}| < \varepsilon$. I believe the biggest problem is that there is no simple way to express the $n^{th}$ iteration of $x$. I also don't know how I could find a good $N$.