4 ms·

I'm familiar with the integral. But why is that sum a particularly good approximation to the integral? It seems better than one has a right to expect, althoug

I'm familiar with the integral. But why is that sum a particularly good approximation to the integral? It seems better than one has a right to expect, although it's been a while since I did any numerical analysis.

messe 8y ago
Riemann sums. EDIT: The short answer is that it's a good approximation because e^(-n^2/c) drops off very vast as n approaches zero, meaning most of the contributions will come from small (on the order of sqrt(c)). Other than that, it's just a Riemann sum presented in an aesthetically pleasing manner.
- TimonKnigge 8y ago
  Do you mean 'approaches infinity'? Because at n=0 the expression is maximal :-) But the point is not to drop small terms at the extremes - it is still an infinite sum. The point is to approximate sections of the function by rectangles, and those rectangles are a bad approximation around 0 too.
madcaptenor 8y ago
Replying to my own comment. See (4.4) in the article and the discussion after it. The error in this approximation is like exp(-pi^2 c). (4.10) in the article translates this to that we should expect about 4.2c digits of accuracy, since log_10 exp(-pi^2) is about 4.2. This falls out of Poisson's summation formula, according to the OP. Generally I'd expect a numerical integration scheme to have an error like h^k for some constant k - for example for Riemann sums the error goes like h^2 and for Simpson's rule like h^5. h is the distance between the sampling points and so is analogous to 1/c. So this doesn't just follow from Riemann summation.
- tanderson92 8y ago
  Trapezoidal rule quadrature is better than you would expect (2nd order) for periodic (even better for analytic and indeed entire, in this case) integrands. See this paper: http://eprints.maths.ox.ac.uk/1734/ http://eprints.maths.ox.ac.uk/1734/

4 ms·

I'm familiar with the integral. But why is that sum a particularly good approximation to the integral? It seems better than one has a right to expect, althoug

by madcaptenor 8y ago

messe 8y ago
Riemann sums. EDIT: The short answer is that it's a good approximation because e^(-n^2/c) drops off very vast as n approaches zero, meaning most of the contributions will come from small (on the order of sqrt(c)). Other than that, it's just a Riemann sum presented in an aesthetically pleasing manner.
- TimonKnigge 8y ago
  Do you mean 'approaches infinity'? Because at n=0 the expression is maximal :-) But the point is not to drop small terms at the extremes - it is still an infinite sum. The point is to approximate sections of the function by rectangles, and those rectangles are a bad approximation around 0 too.
madcaptenor 8y ago
Replying to my own comment. See (4.4) in the article and the discussion after it. The error in this approximation is like exp(-pi^2 c). (4.10) in the article translates this to that we should expect about 4.2c digits of accuracy, since log_10 exp(-pi^2) is about 4.2. This falls out of Poisson's summation formula, according to the OP. Generally I'd expect a numerical integration scheme to have an error like h^k for some constant k - for example for Riemann sums the error goes like h^2 and for Simpson's rule like h^5. h is the distance between the sampling points and so is analogous to 1/c. So this doesn't just follow from Riemann summation.
- tanderson92 8y ago
  Trapezoidal rule quadrature is better than you would expect (2nd order) for periodic (even better for analytic and indeed entire, in this case) integrands. See this paper: http://eprints.maths.ox.ac.uk/1734/ http://eprints.maths.ox.ac.uk/1734/