For $x_{n+1}=x_n-x_n^3$, with $|x_1|>1$. What about the convergence? I see Let $\{ x_n \}_{n=1}^{\infty}$ such that $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$ that if $|x_1| \le 1$, then $x_n\to 0$. However, if $|x_1|>1$, what can we say about this sequence? It seems hard to find the tendency of it. For $x_1$ near $1$, it converges to $0$, but for $x_1$ large, it is not the case.
 A: So if
$$|x_{n}| > \sqrt{2}$$
The sequence diverges to infinity and negative infinity with the evens and odds.
If $$|x_{n}| = \sqrt{2}$$
The sequence is periodic with period 2.
If $$|x_{n}-x^3_{n}| < 1$$
Then the next step is in the attracting basin you already know about.
Finding the roots we get that we can expand the attracting basin out to
$$|x_{n}-x^3_{n}|<(\frac{2}{3 (9 - \sqrt{69}))})^{1/3} + \frac{(\frac{1}{2} (9 - \sqrt{69}))^{1/3}}{3^{2/3}}$$
There is a small unsolved area. I will try to repeat the process, expanding the attracting basin out to the $\sqrt{2}$.

So letting the above function be called g, we have a set of nested invariant attracting basins consisting of the intervals
$$A_{n}=[-a_{n},a_{n}]$$
With $$a_{1} = 1$$ and $$a_{n+1}=g(a_{n})$$
So that if $$x_{n} \in A_{k}$$ then $$x_{n+1} \in A_{k-1}$$
$$g(\sqrt{2}) = \frac{(5\sqrt{6}-9\sqrt{2})^{1/3}}{2^{1/3}3^{2/3}} - \frac{2^{1/3}3^{-1/3}}{(5\sqrt{6}-9\sqrt{2})^{1/3}}$$
So since g(1) > 1 and $g(\sqrt{2}) < \sqrt{2}$
g has an equilibrium less than $\sqrt{2}$. We can extend the attracting basin of [-1,1] out to [-e,e] where e is the equilibrium of g, however this still leaves a small unknown region between e and $\sqrt{2}$.
To find out what happens in the region where $|x_{n}|$ is between e and $\sqrt{2}$, consider that $|x_{n+1}| < |x_{n}|$, therefore the sequence $|x_{n}|$ converges since it is bounded and monotonically decreasing. It can either converge to zero or a positive value between e and $\sqrt{2}$.
If $|x_{n}|$ converges to zero then $x_{n}$ converges to zero. Otherwise x_{n} will converge to a periodic solution of period 2 consisting of the points p and -p where $$p \in [e,\sqrt{2}).$$
p must be a root of the polynomial $$x= (x-x^{3}) - (x - x^{3})^{3}.$$
The complex roots are $x = -(-1)^{1/6}, x = (-1)^{1/6}, x = -(-1)^{5/6}, x = (-1)^{5/6}$
The real roots are $x=0, x=\sqrt{2}, x=-\sqrt{2}.$
So there is no such p periodic. Therefore if $|x_{n}|<\sqrt{2}$ then $x_{n}$ converges to zero.
A: The essential nature of $x_{n+1}=x_n-x_n^3$ can be seen in the iterations of the function $f(x)=|x-x^3|$. That is, if $|x_n|\lt1$ then $x_{n+1}=\sigma(x_n)f(x_n)$, where $\sigma(x)$ is the sign of $x$, while if $|x_n|\gt1$ then $x_{n+1}=-\sigma(x_n)f(x_n)$.
Now the graph of $y=|x-x^3|$ lies below the line $y=x$ for $x\in(0,\sqrt2)$ and above it for $x\in(\sqrt2,\infty)$. Sketching a cobweb plot for the iterations of $f$ shows that $x_n\to0$ if $|x_1|\lt\sqrt2$ (that is, $0\lt x\lt\sqrt2$ implies $0\le f(x)\lt x$, so the sequence $|x_n|$ is a decreasing sequence bounded below, hence has a limit, which can only be $0$), while $|x_n|\to\infty$ if $|x_1|\gt\sqrt2$.
A: Let us first state some useful facts.

*

*If some $x_m = -1, 0, 1$, then all $x_n = 0$ for $n > m$ and thus $x_n \to 0$.


*$x_{n+1} = x_n(1 - x_n^2) = x_n(1 - \lvert x_n \rvert^2)$.


*If $\lvert x_n \rvert \in (0,1)$, then $\lvert x_{n+1} \rvert = \lvert x_n \rvert (1 - \lvert x_n \rvert^2) \in (0, \lvert x_n \rvert)$:
Obvious, for $\lvert x_n \rvert \in (0,1)$ we have $\lvert x_{n+1} \rvert = \lvert x_n \rvert (1 - \lvert x_n \rvert^2)$ and $0 <  \lvert x_n \rvert (1 - \lvert x_n \rvert^2)  < \lvert x_n \rvert$.


*Hence, if some $\lvert x_m \rvert \in (0,1)$, then the sequence $(\lvert x_n \rvert)_{n \ge m}$ is strictly decreasing with values in $(0,1)$.


*If some $\lvert x_m \rvert \in [0,1]$, then $x_n \to 0$:
If $\lvert x_m \rvert = 0,1$, then 1. applies. If $\lvert x_m \rvert \in (0,1)$, then by 4. the sequence $(\lvert x_n \rvert)$ converges to some $\xi \in [0,1)$. But then $\xi = \xi(1- \xi^2)$ by 3. which has the unique solution $\xi = 0$.


*If all $\lvert x_n \rvert > 1$, then $x_{n+1} = -x_n(\lvert x_n \rvert^2 -1)$ for all $n$. Thus the sequence $(x_n)$ has alternating signs and thus does not converge.


*Note that if $\lvert x_n \rvert > 1$, then $\lvert x_{n+1} \rvert= \lvert x_n \rvert(\lvert x_n \rvert^2 -1)$.
Let us prove the follwing result which gives a complete answer of the question:


*All $\lvert x_n \rvert > 1$ if and only if $\lvert x_1 \rvert \ge \sqrt 2$.

If $\lvert x_1 \rvert = \sqrt 2$, then we see by induction that $\lvert x_n \rvert = \sqrt 2$ for all $n$.
If $\lvert x_1 \rvert > \sqrt 2$, then $(\lvert x_n \rvert)$ is strictly increasing:
By induction using 7. we get $\lvert x_{n+1} \rvert= \lvert x_n \rvert(\lvert x_n \rvert^2 -1) > \lvert x_n \rvert (\lvert x_1 \rvert^2 -1) > \lvert x_n \rvert$. One can moreover show that $\lvert x_n \rvert \to +\infty$, but that is irrelevant.
Let us finally look at $1 < \lvert x_1 \rvert < \sqrt 2$:
Assume that all $\lvert x_n \rvert > 1$. Then by 7. (using induction) we get $1 < \lvert x_{n+1} \rvert < \lvert x_n \rvert < \sqrt 2$ for all $n$. Therefore $(\lvert x_n \rvert)$ is strictly decreasing. Thus $(\lvert x_n \rvert)$ converges to some $\xi \in [1, \sqrt 2)$. But then $\xi = \xi(\xi^2 -1)$ which has solutions $\xi = 0, \pm \sqrt 2$. None of them lies in $[1, \sqrt 2)$, thus the above assumption was wrong and we get some $\lvert x_n \rvert \in [0,1]$.
A: We only care about when $|x|>1$. The condition for it to get further away from the origin occurs when,
$$|x-x^3|>|x|$$
We can divide by $|x|$ since it's nonzero, and the inequality is unchanged because it's positive.
$$|1-x^2|>1$$
Since $|x|>1$ we know $x^2>1$ and $x^2-1$ is positive, and so we can replace the absolute value with it.
$$x^2-1 > 1$$
$$x^2 > 2$$
This means when $x$ is outside $[-\sqrt{2}, \sqrt{2}]$ it is repelled further away from the origin.
Similarly, working through with the reverse inequality gives that it is attracted to the origin when $x$ is inside $(-\sqrt{2}, \sqrt{2})$ (knowing that when $|x|\le 1$ it goes to the origin as well as given in the original question.)
