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Sketch of an alternative proof, more DSP: Use the fact that the approximation in the paper is equal to the inner product of e^{-x^2} with a Dirac comb [1]. The
by man-and-laptop 8y ago
Sketch of an alternative proof, more DSP:
Use the fact that the approximation in the paper is equal to the inner product of e^{-x^2} with a Dirac comb [1]. The amazing thing about the Gaussian function and the Dirac comb is that they're both preserved by the Fourier transform. So apply the Fourier transform to both of them, observe that the Fourier transform is unitary (essentially a rotation), and therefore doesn't affect the inner product; then expand the inner product of the Fourier transforms.
Essentially, it's the same proof, but not in the language of Theta functions. In fact, I'd argue it's a better proof, because it generalises.
Generalisation: This technique applies to all Riemann sums, as long as you can compute the Fourier transform of the function. The thinner the tails of the function's Fourier transform, the faster its Riemann sums will converge.
[1] - https://en.wikipedia.org/wiki/Dirac_comb https://en.wikipedia.org/wiki/Dirac_comb
- abenedic 8y agoMyself, I would say this is 80% correct. The bigger deal is the spectral convergence of the trapezoidal sum for periodic entire functions.
- man-and-laptop 8y agoI gave my generalisation in terms of frequency, and not in terms of "number of rectangles" in the Riemann sum, which I conveniently assumed to be infinite. A better generalisation: if a function is thin-tailed, and its Fourier transform is thin-tailed, then it's Riemann sums converge to its integral at an exponential rate. In particular, this applies to the Gaussian function. I'll check out what you said.