For what starting points $x_1$ will the iterations eventually return to $x_1$?

Question

Consider the function $$f(x)=1+x^2$$ and some starting point $x_1>0$.

If we repeatedly apply Newton's Method

$$x_{n+1}=x_n-\frac{f \left ( x_n \right )}{f' \left ( x_n \right ) }$$

we observe this oscillatory behaviour where the $x_n$'s bounce back and forth between the stationary point at $x=0$.

I am wondering if there is a way to find all possible starting points $x_1$ such that there exists some $x_n=x_1$. Plainly speaking, is there a starting point that repeated iterations will eventually return to?

Intuitively, due to the oscillatory behaviour of the $x_n$'s, it seems that this might be possible. But I am not sure how to find it exactly.

Answer 1

If I understood your question correctly, you can take $x_1:=1/\sqrt{3}$, then $x_2:=-1/\sqrt{3}$ and $x_3=x_1$. In order to find the value for $x_1$, just solve the equation $$g(g(x))=x\quad \mbox{where}\quad g(x):=x-\frac{f(x)}{f'(x)}=x-\frac{x^2+1}{2x},$$ that is $$x^4-6x^2+1-4x^2(x^2-1)=0\Leftrightarrow (3x^2-1)(1+x^2)=0.$$

Answer 2

Let your function be $f(x)=x-\frac {x^2+1}{2x}=\frac {x^2-1}{2x}$. It is easy to show there are no points where $f(x)=x$ because you would need $\frac {x^2+1}{2x}=0$. If $x_2=x_0$ you need $f(f(x))=x$. If you do the algebra, this is $x=\frac {(x^2-1)^2-4x^2}{4x(x^2-1)}$ which has solutions $\pm \frac 1{\sqrt 3}$ and oscillates between them. You can then look for solutions to $x=f(f(f(x)))$. As you look for each longer iteration the degree of the polynomial gets higher, so the solution gets harder. You know that all even polynomials will have a factor $x^2-\frac 13$, so you can divide that out, but it still isn't easy.