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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 $x = 0.5$.

Victoria
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  • The inequality you seek is only valid for $0<x<4/5$. However, any $x\in[4/5,1)$ maps into $(0,2/5]$ and is subsequently in the basin of attraction. – Mark McClure Aug 12 '16 at 23:25
  • Thanks - is there any significance to the upper bound of 4/5 in the inequality? Does this method not work in general for finding the entire basin of attraction? – Victoria Aug 12 '16 at 23:30
  • I don't think there's any particular significance to the $4/5$ in this particular example; that's just the way the algebra works out. In general, it's unusual for a basin of attraction to coincide exactly with the set of all points that move closer to the fixed point under one application of $f$. The basin of attraction can be quite complicated. For $x^2-x-1$, the basin of attraction is the complement of a Cantor set, as explained in this question. – Mark McClure Aug 12 '16 at 23:37
  • Thank you. Can one prove that the set $(0, 1)$ will contain all points that are attracted to this sink? – Victoria Aug 12 '16 at 23:42
  • Ah - good question. Yes, I think that is much easier as the image of any negative number is again a negative number. – Mark McClure Aug 12 '16 at 23:45
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    Many thanks for your help. I will go away and do some more algebra! – Victoria Aug 12 '16 at 23:47

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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:

  1. Use the sufficient condition to show that $(0,0.8) ⊂ B$.
  2. Show that every point in $[0.8,1)$ gets mapped to a point in $(0,0.8)$ and hence $[0.8,1) ⊂ B$.
  3. Show that every non-positive point gets mapped to a non-positive point and hence $(−∞,0] ∩ B = ∅$.
  4. Show that every $x≥1$ gets mapped to a non-positive point and hence $[1,∞) ∩ B = ∅$.
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