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The title says it all. I was plotting random functions on my phone and noticed this graph. I don't think this function is periodic (WA also agrees). Is there a way to prove if a function like this is periodic or not?

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  • $\begingroup$ If by periodic you mean that there don't exist any two $x_0,x_1, x_0\neq x_1$ where $\sin\lfloor x_0\rfloor=\sin\lfloor x_1\rfloor$ then I think it might be easy to prove, although I don't immediately see a trick to it... $\endgroup$ – abiessu Feb 7 '15 at 14:44
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    $\begingroup$ I mean that the graph doesn't repeat like $\sin$ does after every $2\pi$. Though, what you're saying can imply the same thing but I'm not sure. $\endgroup$ – AvZ Feb 7 '15 at 14:47
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Suppose $f(x+n)=f(x)$ for all $x$.
The discontinuities are at integers for $f(x)$, so must be for $f(x+n)$, so $n$ is an integer.
$\sin n=f(n)=f(0)=0$ so $n$ is a multiple of $\pi$, $n=m\pi$.
$\pi$ is irrational, so $n=0$.
So $f(x)$ is not periodic.

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    $\begingroup$ I didn't think of that. Nice one (+1). Suppose if $f(x)=\sin\lfloor\pi x\rfloor$, can this be periodic? $\endgroup$ – AvZ Feb 7 '15 at 14:58
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    $\begingroup$ $\sin\lfloor\pi x\rfloor$ has to be zero again, but the sine of an integer is never zero again. $\endgroup$ – Empy2 Feb 7 '15 at 15:31

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