Gaussian Poisson summation identity ($S(a)=a^{-1/2}S(1/a),a>0$)

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 $b>0$ and a nonnegative integer $N$, define the symmetric partial sum and its positive remainder bySuccessive omitted terms have ratio $e^{-\pi b(2n+1)}\le e^{-\pi b(2N+3)}$ for $n\ge N+1$. Bounding them by a geometric series givesConsequently the physical approximation to $S(a)$ has error at most $B_N(a)$, while the dual approximation $a^{-1/2}{S}_N(1/a)$ has error at most $a^{-1/2}{B}_N(1/a)$. The prefactor must be included when setting an absolute error tolerance.

For a truncation error below $10^{-12}$, at $a=0.04$ the physical side first passes this bound with $N=15$: $31$ lattice points give $B_{15}(0.04)<2.172\times 10^{-14}$, whereas $N=14$ gives $B_{14}(0.04)>10^{-12}$. The dual side needs only $k=0$. At $a=25$ the physical side needs only $n=0$, whereas the dual side first passes the bound with $N=14$: $29$ lattice points give $25^{-1/2}{B}_{14}(0.04)<2.146\times 10^{-13}$. These counts measure symmetric lattice terms; using the even function symmetry reduces the distinct exponential function evaluations to $N+1$. Truncation bounds do not include rounding error in a numerical implementation.