Define Poisson kernel as $$ P_r ( \theta) := \frac{1}{2\pi} \frac{1-r^2}{1- 2r \cos \theta + r^2} $$ Then I want to prove the Poisson summation formula which is $$ P_r (2\pi x) = \sum_{n=-\infty}^\infty P_y (x+n)\;\;\;\;\text{(here $r = e^{-2 \pi y} $}) $$
Poisson summation formula (in general)
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real-analysis
functional-analysis
special-functions
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0Where are you stuck in the computations? – 2012-07-31
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0What are $x$ and $y$? – 2012-07-31
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0@Norbert I think $x,y \in \Bbb R$. – 2012-07-31
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0@DavideGiraudo I just wrote down by the definition above($P_r$), but I couldn't handle the summation, it's too complicated. – 2012-07-31
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2In what sense is the right-hand side expected to converge? – 2012-07-31
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0The rhs depends on $y$, while lhs doesn't. This is strange – 2012-07-31
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0@Norbert $r = e^{-2 \pi y}$ as written above. – 2012-07-31
1 Answers
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We will use the result that $$\sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2}= \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}} = P_y(2\pi\,x)\,, $$
Recalling Poisson formula in the upper half plane for $y>0$, $$ P_{y}(x) = \frac{y}{x^2+y^2}\,. $$
We construct the sum
$$ \sum_{n=-\infty}^{\infty} P_y(x+n) = \sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2} =\frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos( 2 \pi x ) + e^{-4 \pi y}} = P_y(2\pi\,x) $$
Substituting $r= {\rm e}^{-2\pi y}$ in the above result gives the desired result.
$$ P_r(2\pi x) = \frac{1}{2} \frac{1 - r^2}{1 - 2 r \cos( 2 \pi x ) + r^2} $$
The whole idea was to exploit the Poisson integral formula in the upper half plane.