= Gaussian Poisson summation identity
{c}
{title2=$S(a)=a^{-1/2}S(1/a),\quad a>0$}
Take the <Gaussian function> $f_a(x)=e^{-\pi ax^2}$ 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,
$$
\widehat f_a(\xi)=a^{-1/2}e^{-\pi\xi^2/a}.
$$
Thus the <Poisson summation formula> gives the exact identity
$$
S(a):=\sum_{n\in\mathbb Z}e^{-\pi an^2}
=\frac1{\sqrt a}\sum_{k\in\mathbb Z}e^{-\pi k^2/a}
=\frac1{\sqrt a}S(1/a),\qquad a>0.
$$
The <Gaussian Fourier transform> is given in scaled form in Exercise 41 of <Terence Tao>'s \i[245C, Notes 2: The Fourier transform].<245C, Notes 2: The Fourier transform by Terence Tao>[§3, Exercise 41: transform of a scaled Gaussian; put r=1/sqrt(a) and adjust its prefactor]{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 $H(\xi)=\widehat f_a(\xi)$. Differentiation under the <integral> and <integration by parts>, using $f_a'(x)=-2\pi axf_a(x)$, give
$$
H'(\xi)=-2\pi i\int_{\mathbb R}xf_a(x)e^{-2\pi i\xi x}\,dx
=-\frac{2\pi\xi}{a}H(\xi).
$$
The <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.<Number Theory I, MIT 18.785, Fall 2021/Lecture 17: The functional equation>[Lecture 17, §17.1, Lemma 17.9 and proof, printed p. 3: differential-equation proof for the unit Gaussian; scaled here to a>0]{ref}
\Image[/poisson-gaussian-physical-dual-weights.png]
{title=Gaussian weights at integer physical and dual points for reciprocal parameters 0.04 and 25}
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 gives
$$
S(0.04)=5\left(1+2e^{-25\pi}+2e^{-100\pi}+\cdots\right),\qquad
0<S(0.04)-5<7.774\times10^{-34}.
$$
For $a=25$, use the physical side instead:
$$
S(25)=1+2e^{-25\pi}+2e^{-100\pi}+\cdots,\qquad
0<S(25)-1<1.555\times10^{-34}.
$$
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.
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