4 ms·
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 contribu
by 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 agoDo 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.