# How can I prove that $x_{n+1} = x_n \sin(x_n)$ converges for any $x_0$?

I have already prove that it is true if exist $$\sin(x_m) = 1$$ or $$\sin(x_m) = -1$$. So I need make proof only for absolute-decreasing progression. But I am afraid that $$\sin(x_n)$$ can has limit $$1$$ and can go to $$1$$ very fast. Is it a good way to count limit $$\sin(x_n)$$? And what solution is correct?

• Welcome to the website. In future, please typeset your equations using Mathjax for better presentation. – Shubham Johri Dec 25 '18 at 20:35
• I think you mean $x_n\to 0$ or $\sin x_n\to 1$. – J.G. Dec 25 '18 at 20:41
• If the limit exists it should satisfy $x_{n+1} = x_n$. Think about what this tells you about possible limit points, and then look at what happens if $x_n$ is above these points or below these points. – tch Dec 25 '18 at 20:43
• @TylerChen "If the limit exists, it should satisfy $x_{n+1} = x_n$". What the heck does this mean – mathworker21 Dec 25 '18 at 20:45
• @Cesareo There are infinitely many solutions. Any $\pi/2+2k\pi$ works. – ImNotTheGuy Dec 25 '18 at 21:12

Let $$y_n = |x_n|$$

Then $$y_{n+1} = y_n |\sin y_n| \leq y_n \forall n$$ (*)

So $$y_n$$ is decreasing and bounded below by $$0$$. Therefore it converges to some limit $$l$$. By passing to the limit in (*) we get that either $$l=0$$ or $$sin l= 1$$ giving the family of solutions discussed in the comments, namely $$l=2k\pi+\frac{\pi}{2}$$

Now note that if $$y_n$$ converges to $$0$$ then $$x_n$$ also converges to $$0$$. If $$y_n$$ converges to a nonzero value $$l$$ then a little more work will also prove that $$x_n$$ also converges to either $$l$$ or $$-l$$ because for n large enough $$y_n$$ will be between $$l$$ and $$l+\frac{\pi}{2}$$ hence $$x_n$$ will have constant sign

Either way, $$x_n$$ will also be convergent

Note that $$0\le|x_{n+1}|=|x_n\sin x_n|\le|x_n|$$, so the sequence of absolute values, $$|x_0|,|x_1|,|x_2|,\ldots$$ definitely converges to a nonnegative limit $$L(x_0)$$ satisfying the equation $$L(x_0)=L(x_0)|\sin L(x_0)|$$. If $$L(x_0)=0$$, we're done: $$|x_n|\to0$$ implies $$x_n\to0$$. If $$L(x_0)\gt0$$, we need only worry about the possibility that $$x_n$$ approaches both $$L(x_0)$$ and $$-L(x_0)$$. But this can't happen: If $$x_n\approx\sigma L(x_0)$$ where $$\sigma=\sin L(x_0)\in\{1,-1\}$$, then, using the fact that $$\sin(\sigma u)=\sigma\sin u$$ for all $$u$$ and $$\sigma^2=1$$, we have

$$x_{n+1}=x_n\sin x_n\approx\sigma L(x_0)\sin(\sigma L(x_0))=\sigma L(x_0)(\sigma\sin L(x_0))=\sigma L(x_0)(\sigma^2)=\sigma L(x_0)\approx x_n$$

Remarks: As ImNotTheGuy observes in a comment beneath the OP, $$\pi/2+2k\pi$$ (with $$k\in\mathbb{Z}$$) is a fixed point for the mapping $$f(x)=x\sin x$$, which means there are sequences that tend to each such limit, as well as sequences that tend to $$0$$. There are, in fact, uncountably many sequences tending to each of the possible limits: If $$k\ge0$$, for example, and $$x_n=\pi/2+2k\pi+\epsilon$$ with $$\epsilon\gt0$$ but very small, then $$x_{n+1}=x_n\cos\epsilon =\pi/2+2k\pi+\epsilon'$$ with

$$\epsilon'=\epsilon\cos\epsilon-(1-\cos\epsilon)(\pi/2+2k\pi)\approx\epsilon-(\epsilon^2/2)(\pi/2+2k\pi)\lt\epsilon$$

(but still positive), so each positive limit has a "basin of attraction" that includes an open inteval of values for $$x_0$$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $$|x_{n+1}|=|x_n\sin x_n|\le|x_n|$$, successive terms in any sequence always get closer to the origin).

In general If $$x_0$$ is an initial approximation of a ﬁxed point , the ﬁxed point iteration generates a sequence of approximants by $$x_{n+1}=\varphi (x_n),$$ where $$\varphi$$ is a continuous function. Suppose that at the ﬁxed point that we indicate with $$\alpha$$ we have $$\varphi '(\alpha)=\varphi ''(\alpha)=\cdots\varphi ^{p-1}(\alpha)=0, \quad \varphi ^p(\alpha)\neq 0 \qquad (*);$$ then we have the following theorem:

Let $$\alpha$$ be a fixed point of $$\varphi$$ and $$I_{\epsilon}:=\{x\in\Bbb R: |x-\alpha|<\epsilon\}$$. Asuume that $$\varphi \in \Bbb C^p[I_{\epsilon}]$$ satisfes $$(*)$$. If $$M(\epsilon):=|\varphi '(t)|<1$$ then the fixed point iteration converges to $$\alpha$$ for every $$x_0\in I_{\epsilon}$$ and the order the convergence is $$p$$.

Note that to use this theorem you have to know tha fxed point $$\alpha$$ (you can plot the function for example and obtain an approximation of point). You can find this argument in this book.