= Poisson summation formula
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{title2=$\sum_{n\in\mathbb Z}f(n)=\sum_{k\in\mathbb Z}\widehat f(k)$}
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= Poisson resummation
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{synonym}
= Poisson summation
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{synonym}
The <Poisson summation formula> equates sample sums of a <function> and its <Fourier transform>. Exchanging broad and narrow sums can save work. Use the convention
$$
\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi i\xi x}\,dx.
$$
For every <Schwartz function> $f\in\mathcal S(\mathbb R)$,
$$
\boxed{\sum_{n\in\mathbb Z}f(n)=\sum_{k\in\mathbb Z}\widehat f(k).}
$$
Both <series> have <absolute convergence>. A sufficient hypothesis is that $f$ is <smooth> and $\sup_x(1+|x|)^M|f^{(r)}(x)|<\infty$ for all nonnegative <integers> $M,r$; its <Fourier transform> is then also a <Schwartz function>.<245C, Notes 2: The Fourier transform by Terence Tao>[§3, Definition 32 and Exercise 54: Schwartz hypotheses and Poisson summation on the integer lattice]{ref} This assumption controls both decay and regularity; mere membership of an <L1 space> does not control values at individual sampling points, as <periodization of an integrable function> illustrates.
Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in <Fourier analysis>, $\widetilde f(\omega)=\int f(x)e^{-i\omega x}\,dx$, so $\widehat f(\xi)=\widetilde f(2\pi\xi)$ and the right-hand side becomes $\sum_k\widetilde f(2\pi k)$. Keeping the <Fourier transform> convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the <Poisson summation formula for a Euclidean lattice>.
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