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Take the Gaussian function $f_a(x)=e^{-\pi a{x}^2}$ with $a>0$. Every derivative is a polynomial times the same Gaussian function, so $f_a$ is a Schwartz function. The Fourier transform of a Gaussian gives, in the cycles convention,Thus the Poisson summation formula gives the exact identityThe Gaussian Fourier transform is given in scaled form in Exercise 41 of Terence Tao's 245C, Notes 2: The Fourier transform.^[ref] The identity is also the imaginary-axis transformation of the Jacobi theta function. At $a=1$ the Gaussian function is its own Fourier transform; at reciprocal values of $a$ its spatial and spectral widths exchange.

For a short derivation of the Gaussian Fourier transform, write $H(\xi )={\hat{f}}_a(\xi )$. Differentiation under the integral and integration by parts, using $f_a^{\prime }(x)=-2\pi ax{f}_a(x)$, giveThe Gaussian integral gives $H(0)=a^{-1/2}$, so this ordinary differential equation yields the claimed formula. The decay of the Gaussian function justifies these steps and removes the boundary term.^[ref]

For $a=0.04$, the physical Gaussian function has width parameter $1/\sqrt{a}=5$, while its Fourier transform has width parameter $\sqrt{a}=0.2$. Many physical samples matter, but the dual zero sample already givesFor $a=25$, use the physical side instead:These are bounds on the full positive tails, obtained below, rather than estimates based only on the first omitted term. Ordinary floating-point arithmetic rounds these sums to $5$ and $1$; the exact real numbers are slightly larger. A Gaussian sum truncation bound lets us choose the cheaper side and certify its truncation error before evaluating it.

For a Schwartz function on the real numbers, its periodization $P_f$ is a smooth periodic function of period of a function one. Rapid decay gives uniform convergence on $[0,1]$ of the defining series and each differentiated series. Its $k$th Fourier coefficient isThe change of variable $u=x+n$ uses $e^{2\pi ikn}=1$. Rapidly decreasing Fourier coefficients give the absolutely and uniformly convergent Fourier series $P_f(x)={\sum }_k\hat{f}(k)e^{2\pi ikx}$. Evaluation at $x=0$ proves the Poisson summation formula.^[ref]

The same identity explains aliasing in the Nyquist–Shannon sampling theorem: applying the Poisson summation formula to the Schwartz function $x\mapsto f(x)e^{-2\pi i\xi x}$ givesThe sampled spectrum is periodic in frequency, and shifted copies of the continuous Fourier transform add together. When those copies overlap, the samples cannot separate their contributions. The Nyquist–Shannon sampling theorem adds a bandlimited function hypothesis to make reconstruction possible.