5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Welcome to the website. In future, please typeset your equations using Mathjax for better presentation. $\endgroup$ – Shubham Johri Dec 25 '18 at 20:35
  • $\begingroup$ I think you mean $x_n\to 0$ or $\sin x_n\to 1$. $\endgroup$ – J.G. Dec 25 '18 at 20:41
  • $\begingroup$ 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. $\endgroup$ – tch Dec 25 '18 at 20:43
  • 1
    $\begingroup$ @TylerChen "If the limit exists, it should satisfy $x_{n+1} = x_n$". What the heck does this mean $\endgroup$ – mathworker21 Dec 25 '18 at 20:45
  • 4
    $\begingroup$ @Cesareo There are infinitely many solutions. Any $\pi/2+2k\pi$ works. $\endgroup$ – ImNotTheGuy Dec 25 '18 at 21:12
7
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.