Lanczos sigma factor

Remarks

Let $S_{m-1} f(x) = \sum_{\abs{k} \le m-1} \hat f(k) e^{ikx}$ be the partial sum of the Fourier series. Since $\frac{m}{2\pi} \int_{x - \pi/m}^{x + \pi/m} e^{ikt} \, dt = \frac{\sin(\pi k/m)}{\pi k/m} e^{ikx}$ for $k \ne 0$, $$\Sigma_m f(x) = \frac{m}{2\pi} \int_{x - \pi/m}^{x + \pi/m} S_{m-1} f(t) \, dt.$$ So the sigma approximation is the moving average of the partial sum over a window of length $2\pi/m$. In the notation $\operatorname{sinc} u = \frac{\sin \pi u}{\pi u}$, the factors are $\sigma_{m,k} = \operatorname{sinc}(k/m)$.

Example: for $m = 2$ and $f(x) = \cos x$, the factor is $\sigma_{2, \pm 1} = \frac{\sin(\pi/2)}{\pi/2} = \frac{2}{\pi}$, so $\Sigma_2 f(x) = \frac{2}{\pi} \cos x$. This agrees with the window average $\frac{1}{\pi} \int_{x - \pi/2}^{x + \pi/2} \cos t \, dt = \frac{2}{\pi} \cos x$.

Non-example: the Fejér weights $1 - \abs{k}/m$ also fade out high frequencies, but they are not the sigma factors; for $m = 2$ and $k = 1$ they give $1/2$, not $2/\pi$.