3
$\begingroup$

I want to find the basin of attraction of a fixed point.

For example, I have $f(x)=\frac 1{x+1}$, whose fixed points are $\frac{-1\pm \sqrt{5}}{2}$. Now, I must create a neighborhood around the $x$ point that would consist of all points around it that would attract to it. How can I figure out the radius of a neighborhood?

$\endgroup$
3
$\begingroup$

For the endpoints $a,b$ of the immediate basin of attraction of an attracting fixed point, the possibilities are:

  1. $\pm \infty$.

  2. A singularity (i.e. where $f$ is undefined).

  3. A repelling fixed point.

  4. $(a,b)$ is a $2$-cycle, i.e. $f(a) = b$ and $f(b) = a$.

    Take the closest of these possibiities on each side of your fixed point.

$\endgroup$
5
  • $\begingroup$ I have two fixed points, so do you mean on each side of each fixed point? $\endgroup$
    – user271990
    Sep 18 '15 at 15:48
  • 1
    $\begingroup$ One fixed point is attracting, the other is repelling. A repelling fixed point has no basin of attraction. $\endgroup$ Sep 18 '15 at 17:03
  • $\begingroup$ I found the basin of attraction (through graphing trial-and-error) to be $(-1,\infty)$. Is it possible to prove this rigorously? $\endgroup$
    – user271990
    Sep 18 '15 at 17:37
  • 1
    $\begingroup$ Yes. From what I wrote above, the candidates for endpoints (for the attracting fixed point $(-1+\sqrt{5})/2$ are $-\infty$, the repelling fixed point $(-1-\sqrt{5})/2$, the singularity $-1$, and $+\infty$ (there are no real $2$-cycles). The closest candidate to the left of the attracting fixed point is $-1$, the closest to the right is $+\infty$. So the immediate basin of attraction is $(-1,+\infty)$. $\endgroup$ Sep 19 '15 at 0:12
  • 2
    $\begingroup$ Note that this is the immediate basin of attraction, not the full basin of attraction: there are also points not in $(-1,+\infty)$ which are attracted to the attracting fixed point. $\endgroup$ Sep 19 '15 at 0:13
0
$\begingroup$

You have four possible regions here:

  1. the region to the right of the right fixed point $(x > \frac{-1+\sqrt{5}}{2})$,
  2. the region between the fixed points $(\frac{-1-\sqrt{5}}{2} < x < \frac{-1+\sqrt{5}}{2})$
  3. the region to the left of the singularity $(x < -1)$.
  4. the region between the left fixed point and the singularity $(-1 < x < \frac{-1-\sqrt{5}}{2})$

If you find the derivative in these three regions, you should be able to see what the basin is for each point (if it exists).

$\endgroup$
1
  • 2
    $\begingroup$ You forgot about the singularity. $\endgroup$ Sep 18 '15 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.