What I am looking for is, for example, $\sin(xy)=f(x)g(y)$, where $f$ and $y$ are some functions for $x$ and $y$.
It is "trivial" to see the answer is no, but not really able to come up with a reasonable reason.
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Martin Sleziak
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You would have to define what you mean by "separated functions". – Martin Argerami Sep 11 '17 at 23:42
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There are identities when $y=2$. Any generalization you might imagine would have to be consistent with those. – Ethan Bolker Sep 11 '17 at 23:44
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$$\sin (x y) = f(x)g(y) = \sin(xy + 2\pi) = \sin(x ( y + 2\pi /x)) = f(x) g(y + 2\pi /x).$$ If this should hold for all $x \neq 0$, then $g(y) = g(y + 2 \pi /x)$, so $g$ has arbitrary small period, implying that it is constant. Then $\sin(xy) = C f(x)$ which can not be true.
Slugger
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If $\sin(xy)=f(x)g(y)$ for all $x,y$, then for any $n \in \mathbb N$ and any real numbers $x_1,\dots,x_n,y_1,\dots,y_n$, the $n \times n$ matrix $$ \begin{bmatrix} \sin(x_1y_1) & \sin(x_1y_2) &\dots &\sin(x_1y_n) \\ \sin(x_2y_1) & \sin(x_2y_2) &\dots &\sin(x_2y_n) \\ \vdots & \vdots & \ddots & \vdots \\ \sin(x_ny_1) & \sin(x_ny_2) &\dots &\sin(x_ny_n) \end{bmatrix} $$ would have rank (at most) $1$. But in fact, that is (almost) never true. For example, the $2 \times 2$ matrix with $x_1=\pi/2,x_2=\pi/3,y_1 = 1,y_2=2$ is $$ \begin{bmatrix} 1 & 0 \\ \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} $$ with rank $2$.
GEdgar
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I guess there is a simpler reason: if we had such $f$, $g$ then $f(x) g(\frac{\pi/2}{x})$ would not be zero for every $x\ne 0$, while $f(x) g(\frac{2\pi}{x})$ would be zero, clearly not possible.
orangeskid
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Not really get you. So it is not possible for both functions be periodic? – Unavailable Sep 11 '17 at 23:49
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@Unavailable: I guess the hint does not work anymore. Have to come up with a new reason. – orangeskid Sep 11 '17 at 23:54