Timeline for Why is it the case that $\sum\limits_{x \in \mathbb{Z}} e^{-(x + \frac{1}{4})^2} = \sqrt{\pi}$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1 at 2:40 | vote | accept | dotdashdashdash | ||
| Jun 11 at 2:04 | comment | added | dotdashdashdash | (Hmm... perhaps the surprising thing with Sum 12 isn't that the approximation to $\sqrt{\pi}$ is merely quite good, but that it's 42 billion digits good, and to get that good you need some special properties of Jacobi theta fns?) | |
| Jun 11 at 1:59 | comment | added | dotdashdashdash | Hmm... Sum 12 in that (very cool) article is the following: $$\left(\frac{1}{10^5} \sum_{n=-\infty}^{\infty} e^{-\left(n^2 / 10^{10}\right)}\right)^2 \doteq \pi$$ This seems different -- here they're essentially just taking a Riemann series with a very small rectangle width (compared to the width of the Gaussian), so it's no surprise it has low error. In the case I presented, the Riemann series has a rectangle width that is comparable to the width of the Gaussian, so it seems surprising that it would have such low error. | |
| Jun 11 at 1:54 | vote | accept | dotdashdashdash | ||
| Jun 11 at 1:58 | |||||
| Jun 11 at 1:52 | vote | accept | dotdashdashdash | ||
| Jun 11 at 1:54 | |||||
| Jun 10 at 21:24 | history | answered | Dave Benson | CC BY-SA 4.0 |