quantifying the basin of attraction for $g(x) = 2.5x(1-x)$ I'm trying to estsblish algebraically that the the logistic map
$$
g(x) = 2.5x(1-x)
$$
has a basin of attraction --- for the fixed point at $x = \frac{3}{5}$ --- of $0 < x < 1$. I have verified with result using plots.
The usual way to do this is to find the values of $x$ so that 
$$
\frac{|f(x) - p|}{|x-p|} < 1
$$
is true for a fixed point at $x = p$. We set up the inequality:
$$
\frac{|2.5x(1-x) - 0.6|}{|x-0.6|} < 1
$$
It can be show that this reduces to
$$
\frac{2.5|x-0.6||x-0.4|}{|x-0.6|} < 1
$$
so that
$$
2.5|x-0.4| < 1
$$
This inequailty is solved by $x > 0$ and $x < 0.8$. The first solution is fine, but the second does not make sense. How come we don't get the correct condition of $x < 1$ out of the algebra? For example, the textbook shows that this method works for the logistic map of
$$
g(x) = 2x(1-x)
$$
for the sink located at $x = 0.5$.
 A: The criterion you are using $(\left|f(x) - p \right | < \left|x-p\right|)$ means that $x$ is mapped closer to $p$. This is neither a sufficient nor a necessary condition for a point to be in the basin of attraction. 
Since $x$ may move away from the fixed point temporarily, the condition is not a sufficient one. This is what you observed for $0.8<x<1$. For example, for $x=0.9$ we have $\left|f(x)-p \right| = 0.375 > 0.3 = \left|x-p\right|$. However, if $(0,0.8)$ is indeed part of the basin of attraction (see below), then everything that gets mapped into it, is a part of it as well.
To see that this is not even a sufficient condition, look at the map:
$$h(x) = \begin{cases}
  2.5·x·(1-x) & \text{if } x≤1 \\
        1+\tfrac{1}{2} (x-1) & \text{otherwise}
\end{cases} $$
This behaves like your map for $x≤1$, but if $x>1$, the map just halves its distance to $1$. Consequently, the map also moves points closer to $p$ and your condition is fulfilled. Yet, these points are obviously not in $p$’s basin of attraction.
For a continuous $f$, you can turn your condition into a sufficient one if you additionally ensure that $\left|x-p\right|$ does not have a fixed point other than $p$, i.e., there is no $x≠p$ such that $\left|f(x)-p\right| = \left|x-p\right|$. This should be easy in your case.
So, to show that $(0,1)$ is the basin of attraction $B$, you could proceed as follows:


*

*Use the sufficient condition to show that $(0,0.8) ⊂ B$.

*Show that every point in $[0.8,1)$ gets mapped to a point in $(0,0.8)$ and hence $[0.8,1) ⊂ B$.

*Show that every non-positive point gets mapped to a non-positive point and hence $(−∞,0] ∩ B = ∅$.

*Show that every $x≥1$ gets mapped to a non-positive point and hence $[1,∞) ∩ B = ∅$.

