For a full <Euclidean lattice> $\Lambda\subseteq\mathbb R^n$, its <dual lattice> is
$$
\Lambda^\vee=\{y:\langle x,y\rangle\in\mathbb Z\text{ for every }x\in\Lambda\}.
$$
For a Schwartz function, the <Poisson summation formula for a Euclidean lattice> states
$$
\sum_{\lambda\in\Lambda}f(\lambda)
=m(\Lambda)^{-1}\sum_{\mu\in\Lambda^\vee}\widehat f(\mu),
\qquad
\widehat f(y)=\int_{\mathbb R^n}f(x)e^{-2\pi i\langle x,y\rangle}\,dx.
$$
To prove it, periodize $f$ over $\Lambda$. The resulting function on $\mathbb R^n/\Lambda$ has Fourier coefficient $m(\Lambda)^{-1}\widehat f(\mu)$ at $\mu\in\Lambda^\vee$. Evaluating its absolutely convergent Fourier series at zero gives the identity. The same proof applies under the usual weaker hypotheses ensuring convergence of both sides.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 137 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-137/4/a/solution