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\end{equation}  For $f(x)=x$, $x_i=(-1)^i\pi$  ? We then have \begin{equation}  \sum_{i=-\infty}^{\infty} (-1)^i\pi\prod_{j=-\infty}^\infty\frac{x-(-1)^j\pi}{(-1)^i\pi-(-1)^j\pi} = cos(x)  \end{equation}