Sinc and Jinc sums

In the previous post, we looked at an elegant equation involving integrals of the sinc function and computed the corresponding integrals for the jinc function.

\int_{-\infty}^\infty \mbox{sinc}(x) \, dx = \int_{-\infty}^\infty \mbox{sinc}^2(x) \, dx = \pi

It turns out the analogous equation holds for sums as well:

\sum_{n=-\infty}^\infty \mbox{sinc}(n) = \sum_{n=-\infty}^\infty \mbox{sinc}^2(n) = \pi

As before, we’d like to compute these two sums and see whether we can compute the corresponding sums for the jinc function.

The Poisson summation formula says that a function and its Fourier transform produce the same sums over the integers:

\sum_{n=-\infty}^\infty f(n) = \sum_{n=-\infty}^\infty \hat{f}(n)

Recall from the previous post that the Fourier transform of sinc is the function π box(π x) where the box function is 1 on [-1/2, 1/2] and zero elsewhere. The only integer n with πn inside [-1/2, 1/2] is 0, so the sum of sinc(n) over the integers equals π. A very similar argument shows that the sum of jinc(n) over the integers equals its Fourier transform at 0, which equals 2.

Let tri(x) be the triangle function, defined to be 1 – |x| for -1 < x < 1 and 0 otherwise. Then the Fourier transform of tri(x) is sinc2(π ω) and so π tri(π x) and sinc2 are Fourier transform pairs. The Poisson summation formula says the sum of sinc2 over the integers is the sum of π tri(π x) over the integers, which is π.

I don’t know the Fourier transform of jinc2 and doubt it’s easy to compute. Maybe the sum could be computed more easily without Fourier transforms, e.g. using contour integration.

3 thoughts on “Sinc and Jinc sums

  1. Haven’t worked this one out, but the jinc^2 sum, at least numerically, looks to be the same as the integral result, namely 8/(3*pi).

    BTW, similarly to the integrals, the sum of sinc(n-a)*sinc(n-b) is over all integers n is pi * sinc(p-q).

  2. The Fourier transform of jinc^2 is the self-convolution of the Fourier transform of jinc. The FT of jinc, per John’s previous post, is 2 sqrt(1-2pi|w|) except that the thing inside the sqrt is clipped at zero. Its support is therefore [-1/2pi,+1/2pi] and the support of its self-convolution is [-1/pi,+1/pi]. In particular, the only integer at which it’s non-zero is 0.

    So, sum jinc^2(n) = sum F[jinc^2](n) = F[jinc^2](0) which we already established is equal to the corresponding integral, and we’re done.

    In passing we’ve noticed a sufficient condition for the sum over integers to equal the integral over all x: the Fourier transform needs to be zero outside the open interval (-1,+1). In other words, you can integrate band-limited functions by summing them at regular intervals.

  3. The Fourier transform on jinc^2 can be written using elliptic integrals.

    (2/3)*(abs(phi)+2)*((abs(phi)^2+4)*E((-2+abs(phi))/(abs(phi)+2))-4*K((-2+abs(phi))/(abs(phi)+2))*abs(phi))
    for -2 <= phi <= 2.

    Found using Maple.

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