Given a sequence $a_{n}=\sin nx$,$n=1,2,...$,where $x\in(0,\pi)$,what is the limit point of the sequence?
It's non-empty by Weierstrass theorem,but is there more information we know about it? Is it finite or infinite?Or even dense in $[-1,1]$?
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1Depends on $x$. – Thomas Andrews Dec 19 '13 at 05:05
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@ThomasAndrewsIs there possibility that it's dense in $[-1,1]$?I mean, for some fixed $x$? – Chris Wayne Dec 19 '13 at 05:20
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It's mostly certainly possible for it to be dense - if $x/\pi$ is not rational, then $\sin nx$ is dense in $[-1,1]$. – Thomas Andrews Dec 19 '13 at 06:37
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@ThomasAndrewsMmm...how can i find a rigorous proof for this? – Chris Wayne Dec 19 '13 at 06:49
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You can find a proof that ${\sin(n)}$ for $n \in \mathbb{N}$ is dense in $[-1,1]$ here: http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 The methods at that answer suffice to prove Thomas Andrews' claim above. – Benjamin Dickman Dec 19 '13 at 08:42
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@BenjaminDickmanIt's really helpful,much appreciate bro!By the way it's a big surprise that you had been here in Nanjing! – Chris Wayne Dec 19 '13 at 10:17
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Note that the set of limit points of a sequence is always closed. Thus, saying that it is "dense in..." is the same as saying "equal to...". – Post No Bulls Dec 21 '13 at 20:03
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Hints:
$$\sin n\frac\pi2=\begin{cases}\;\;\,0&,\;\;n\;\;\text{is even}\\{}\\-1&,\;\;n\;\;\text{is}\;\;3\pmod 4\\{}\\\;\;\,1&,\;\;n\;\;\text{is}\;\;1\pmod 4\end{cases}$$
and you already have three cluster points. Check what happens with some other values of $\;x\;$ ...
DonAntonio
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Depends - he the original question asked for cluster points of a set, not cluster points of a sequence. Also, it is unclear from the question whether $x$ is allowed to vary, or is fixed. – Thomas Andrews Dec 19 '13 at 05:06
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Well, I'm not sure whether $;x;$ varies or not, either, but for the above particular value of $;x;$ there is no one cluster point, but several. – DonAntonio Dec 19 '13 at 05:08
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No, in that case, $A={-1,0,+1}$ which has zero cluster points. The sequence $0,1,0,-1,....$ has three cluster points. That was my point about the distinction between limit points of a set and a sequence. – Thomas Andrews Dec 19 '13 at 05:10
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@ThomasAndrewsOh it's my fault,I mean the limit points of a sequence – Chris Wayne Dec 19 '13 at 05:11
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@CWeid Yeah, I guessed as much, was just pointing out the problem as stated wouldn't give any cluster (limit) points. – Thomas Andrews Dec 19 '13 at 06:38
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