Gaussian Poisson Summation Duality

The Poisson summation formula establishes a profound connection between a function and its Fourier transform by summing their values over discrete lattices. For Gaussian functions, this manifests as a remarkable identity linking sums over the integer lattice and its reciprocal.

Mathematical Framework

The fundamental identity is expressed as:
\sum_(n=-∞)^∞ e^(-π a n²) = 1/√a \sum_(k=-∞)^∞ e^(-(π k²)/a)

Here, a > 0 represents the Gaussian rate. The left side is a sum over the direct integer lattice (n ∈ ℤ), corresponding to narrow Gaussians when a is large. The right side is a sum over the reciprocal lattice (k ∈ ℤ), corresponding to the Fourier transforms of the direct terms.

Because the Fourier transform of a Gaussian is another Gaussian with an inverse rate, one side of this identity will always converge faster than the other (unless a = 1, where both converge equally). If a is very small, the direct sum involves many significant terms, but its reciprocal sum involves terms that decay extremely rapidly, making it computationally much more efficient.

Experiment

Explore the duality by manipulating the Gaussian Poisson Summation Duality.

Adjust the Gaussian rate control. Observe how a small rate (e.g., 0.25) makes the direct sum converge slowly (leaving a large omitted tail) but makes the reciprocal sum converge very rapidly.
Set the Gaussian rate to a large value (e.g., 4). Notice that the situation reverses: the direct sum now converges almost immediately, while the reciprocal sum requires more terms.
Modify the Truncation limit. See how increasing N reduces the omitted tails on both sides, while the mathematically infinite sum and the identity residual (the difference between the infinite limits computed on both sides) remain robustly accurate.

Worked Example: Rate 0.5 with N=1

Let us calculate the partial sums for a = 0.5 with truncation N = 1. The formula requires summing up to N=1. Let's see the terms.

Direct Partial Sum:

For n = 0: e^(-π (0.5) (0)²) = 1
For n = 1: e^(-π (0.5) (1)²) = e^(-0.5π) ≈ 0.20787957635
For n = -1: e^(-π (0.5) (-1)²) = e^(-0.5π) ≈ 0.20787957635

Summing these: 1 + 2(0.20787957635) = 1.4157591527.

Reciprocal Partial Sum:

For k = 0: 1/(√(0.5)) e^(-(π (0)²)/0.5) = √2 ≈ 1.41421356237
For k = 1: 1/(√(0.5)) e^(-(π (1)²)/0.5) = √2 e^(-2π) ≈ 0.00264096285
For k = -1: 1/(√(0.5)) e^(-(π (-1)²)/0.5) = √2 e^(-2π) ≈ 0.00264096285

Summing these: 1.41421356237 + 2(0.00264096285) = 1.41949548807.

The infinite sum converges to approximately 1.41949548808. The reciprocal partial sum is much closer to the infinite sum (omitted tail ~ 3.44e-11) compared to the direct partial sum (omitted tail ~ 0.0037).

Limitations

The module restricts the Gaussian rate to values between 0.25 and 4. While the Poisson summation formula holds for all a > 0, extreme values of a can cause floating-point underflow or overflow for the rapidly decaying terms, necessitating specialized high-precision arithmetic beyond standard 64-bit representations. Additionally, we only compute up to N=80 terms for the "infinite" limits due to IEEE 754 precision limits and computational constraints, which is typically well beyond where convergence occurs for this subset.

References

Gressman, P. T. "Poisson Summation Formula". University of Pennsylvania. Available at: https://www2.math.upenn.edu/~gressman/analysis/14-poisson.html

This reference provides a mathematical foundation for the Poisson summation formula, demonstrating its applicability to Gaussian distributions and exploring its properties in continuous Fourier transform spaces.

Gaussian Poisson Summation Duality