# Basin of attraction of a fixed point

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?

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.

• I have two fixed points, so do you mean on each side of each fixed point? Sep 18 '15 at 15:48
• One fixed point is attracting, the other is repelling. A repelling fixed point has no basin of attraction. Sep 18 '15 at 17:03
• I found the basin of attraction (through graphing trial-and-error) to be $(-1,\infty)$. Is it possible to prove this rigorously? Sep 18 '15 at 17:37
• 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 $(-1,+\infty)$. Sep 19 '15 at 0:12
• 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. Sep 19 '15 at 0:13

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).

• You forgot about the singularity. Sep 18 '15 at 17:10