Basin of attraction of a fixed point

Question

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?

Answer 1

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).

Answer 2

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.