Call a function $h: \Bbb R^2 \to \Bbb R$ simple if there are functions $f,g: \Bbb R \to \Bbb R $ such that $h(x,y) = f(x)g(y)$. In fancy words, $h$ factors under the tensor product.

Is the function $h(x,y) = \sin(xy)$ equal to a finite sum of simple functions?

The Taylor expansion of $h$ gives an infinite sum of simple functions equal to $h$. Testing for simplicity can be done by diving by the partial derivatives and checking if the result is independent of one variable.


If $h(x,y):=\sin(xy)$ satisfies $h(x,y)=f(x)g(y)$ then the matrix $$\begin{bmatrix}h(x_1,y_1) & h(x_1,y_2) \\ h(x_2,y_1) & h(x_2,y_2)\end{bmatrix}$$ would be of rank at most 1. In general, if $h(x,y)$ were a sum of $k$ simple functions, then the rank of the similar $(k+1)\times(k+1)$ matrix would be at most $k$ and hence its determinant would be $0$.

In the case of $k=1,$ with $\;x_1=x,\;x_2=2x,\;y_1=1,\;y_2=2,\;$ the determinant is equal to $\sin(x)\sin(4x)-\sin(2x)^2$ which is nonzero in general. Therefore $h(x,y)$ is not a simple function.


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