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Show that $\lim_{n\to \infty}\sin{n\pi x} =0$ if $x\in \mathbb{Z},$ but the limit fails to exist if $x\notin \mathbb{Z}.$

1st part

If $x\in \mathbb{Z}$ then $\sin{n\pi x}=0$ for all $n,$ giving the first part.

Edit:

2nd part

If $x\notin \mathbb{Z},$ I want to show that the limit doesn't exist. How to do that?

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  • $\begingroup$ Did you mean $\,\lim_{n\to\infty}\sin (n!\pi x)$? $\endgroup$ – Yiorgos S. Smyrlis Mar 13 '17 at 18:09
  • $\begingroup$ not factorial, only n i.e $\sin{(n\pi x)}$ $\endgroup$ – user1942348 Mar 13 '17 at 18:10
  • $\begingroup$ If $x$ is not an integer, ain't the limit even not exist? $\endgroup$ – Megadeth Mar 13 '17 at 18:11
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    $\begingroup$ question says that f(x) is not defined then $\endgroup$ – user1942348 Mar 13 '17 at 18:13
  • $\begingroup$ The title has a logical problem. You define $f(x)$ for every $x$ and then claim it's undefined for some $x.$ It would be better to leave out the notation $f(x)$ altogether. $\endgroup$ – zhw. Mar 20 '17 at 23:02
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If $x = \frac{p}{q} \in \mathbb{Q} \setminus \mathbb{Z}$

Let $k$ be an integer.

  • If $n = 2qk$, then $\sin(n \pi x) = \sin(2\pi pk) = 0$.
  • If $n = 2qk+1$, then $\sin(n \pi x) = \sin(2\pi pk+\pi x) = \sin(\pi x) \neq 0$

So the sequence $n \to \sin(n \pi x)$ contains two different numbers infinitely often, so it has no limit.

If $x \in \mathbb{R} \setminus \mathbb{Q}$

For convenience we define the irrational number $y = x/2$.

The sequence $a_n = ny \operatorname{mod} 1$ is dense on $[0, 1)$. (The equidistribution theorem gives the stronger result that it's uniformly distributed on that interval, but we don't need that for our proof)

This implies that for every $n_0 \in \mathbb{N}$ and $0 \le a < b \le 1$, you can find an $n>n_0$ such that $a_n \in (a, b)$. (because otherwise, there would only be a finite number of elements in the interval $(a, b)$, and then it wouldn't be dense)

The idea of this proof is that for every $n_0 \in \mathbb{N}$, we can always find an $n > n_0$ such that $\sin(n \pi x) > c_1$, and we can also find an $n > n_0$ such that $\sin(n \pi x) < c_2$, where $c_1 > c_2$.

We take $c_1 = \sin(2 \pi \cdot 0.1)$ and $c_2 = \sin(2 \pi \cdot 0.6)$.

Given an $n_0$, we can find an $n > n_0$ such that $a_n \in (0.1, 0.2)$. We have:

$$n \pi x \operatorname{mod} 2\pi = 2\pi n y \operatorname{mod} 2\pi = 2\pi(n y \operatorname{mod} 1) = 2\pi a_n \\ \sin(n \pi x) = \sin(2\pi a_n) > c_1 $$ Similarly, for every $n_0$ we can find an $n > n_0$ such that $a_n \in (0.6, 0.7)$ and $$\sin(n \pi x) < c_2$$

So, the sequence $n \to \sin(n \pi x)$ has no limit.

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Take $x=1/2$, then $$ \sin(n\pi x)=\left\{ \begin{array}{rlc} 0 & \text{if $n$ is even},\\ 1 & \text{if $n=4k+1$ for some $k\in\mathbb N$,} \\ -1 & \text{if $n=4k+3$ for some $k\in\mathbb N$.} \end{array} \right. $$ Thus the sequence $\sin(n\pi x)$ DOES NOT converge, even if $x$ is rational!

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    $\begingroup$ It is a particular case you have taken. If you have taken x=p/q as a general case, then it might be interesting. Also case for x as irrational also required. Would you give a general proof for the problem. $\endgroup$ – user1942348 Mar 14 '17 at 17:15
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For $x \in \mathbb{R}$, for $n \in \mathbb{N}$, we have $$\sin ((n+1) \pi x) - \sin (n \pi x) = \sin (n \pi x) \big( \cos(\pi x) - 1) + \cos(n \pi x)\sin(\pi x).$$

Denote $A = \cos(\pi x)-1$ and $B = \sin(\pi x)$. We have $A \neq 0$ and $B \neq 0$, and $$\sin \big( (n+1) \pi x \big) - \sin (n \pi x) = A \sin(n \pi x) + B \cos(n \pi x).$$

Now denote $C = \sqrt{A^2+B^2}$ ; classically, there exists $\phi \in \mathbb{R}$ such that $$\forall n \in \mathbb{N},\ \sin \big( (n+1) \pi x \big) - \sin (n \pi x) = C \sin (n \pi x + \phi).$$

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Now we assume that $\lim \limits_{n \to +\infty} \sin(n \pi x)$ exists. Thus $C \sin (n \pi x + \phi) \underset{n \to +\infty}{\longrightarrow} 0$. Then, you can find here a short proof that $$\forall y \in \mathbb{R},\ |\sin (y)| \ge \frac{2}{\pi}d(y,\pi \mathbb{Z})$$ where $d(t,A) = \inf \{ |t-a|,\ a\in A\}$ stands for the distance to the set $A$.

As $C > 0$, we have that $d(n\pi x + \phi, \pi \mathbb{Z}) \underset{n \to +\infty}{\longrightarrow} 0$. Using the continuity of the distance yields $d\big( ((n+1)\pi x + \phi)-(n\pi x + \phi), \pi \mathbb{Z} \big) \underset{n \to +\infty}{\longrightarrow} 0$, i.e. $d(\pi x, \mathbb{Z}) \underset{n \to +\infty}{\longrightarrow} 0$, so $d(\pi x, \pi \mathbb{Z}) = 0$. As $\pi \mathbb{Z}$ is closed in $\mathbb{R}$, we conclude $\pi x \in \pi \mathbb{Z}$, and thus $x \in \mathbb{Z}$.

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Hence, if $\lim \limits_{n \to +\infty} \sin(n \pi x)$ exists, then $x \in \mathbb{Z}$ (and thus, if $x$ is not an integer, then $\big( \sin (n \pi x) \big)_{n \ge 0}$ does not converge).

Note that if $x$ is irrational, you can even prove that $\big( \sin (n \pi x) \big)_{n \ge 0}$ is dense in $[0,1]$.

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  • $\begingroup$ @user1942348 is there a problem with my proof ? Too long, not detailed enough ? (just asking, since you were requiring one from Y. Smyrlis, I thought so). Don't hesitate to ask if there is a problem to point out, or if you want additional details (btw there is a much shorter proof in the case $x \in \mathbb{Q} \backslash \mathbb{Z}$ if you need it) $\endgroup$ – charmd Mar 14 '17 at 18:35
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Hint: Suppose the limit exists, fix its value as $f(x)$, then use the formal definition of limit and find two values of $n$ that satisfy the definition of limit buthave a different enough value.

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Fix $0<x<1.$ Define $A$ to be the closed arc on the unit circle centerd at $(0,1),$ having arc length $\pi x.$ Good to draw a picture here, because it makes things transparent. A little geometry shows that if $e^{it}\in A,$ then $\sin t \ge \sin [(\pi/2)(1-x)] >0.$

Claim: $e^{in\pi x}\in A$ for infinitely many $n.$

Proof of claim: The points $e^{in\pi x}, n=1,2,\dots$ travel around the circle infinitely many times, in steps of arc length $\pi x.$ The closed arc $A$ has arc length $\pi x.$ Thus there is no way $e^{in \pi x}$ can "jump over" $A$ in this process, so it will land in $A$ at least once in every orbit of the circle. That proves the claim.

The claim implies $\sin (n\pi x) \ge \sin [(\pi/2)(1-x)]$ for infinitely many $n.$

But exactly the same kind of reasoning applies to the arc $B,$ the closed arc on the unit circle centerd at $(0,-1),$ having arc length $\pi x.$ The conclusion of the claim holds for $B,$ hence $\sin (n\pi x) \le - \sin [(\pi/2)(1-x)]$ for infinitely many $n.$

Now in general, if $a>0$ and $y_n$ is a sequence such that $y_n\ge a$ for infinitely many $n,$ and $y_n \le -a$ for infinitely many $n,$ then $y_n$ cannot converge. Since that's the situation with $\sin (n\pi x)$ here, we see $\sin (n\pi x)$ diverges.

Now $0<x<1$ above, but $\pi$-periodicity, or negative $\pi$-periodicity, shows the same result holds if $x\in (m,m+1)$ for any $m\in \mathbb Z.$ Thus $\sin (n\pi x)$ diverges for all $x\notin \mathbb Z.$

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If x is a rational the sequence rotation $n\pi x$ radians of the unit circle will form an equidistributed (see equidistribution theorems) periodic orbit. The length of the orbit will be exactly two times the denominator. Hence will form a periodic orbit of order 2 if x is an integer, luckily these two positions both give sin=0. If x is irrational, then notice $xn\pi$ is dense and topologically transitive to use some terminology from dynamical systems, hence will never stabilize.

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