Timeline for Expected number of lines meeting four given lines or "what is 1.72..."
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2020 at 15:35 | history | edited | Moritz Firsching | CC BY-SA 4.0 |
using new formula
|
| Nov 4, 2020 at 9:33 | answer | added | Leo | timeline score: 4 | |
| Mar 23, 2017 at 7:31 | vote | accept | Moritz Firsching | ||
| Feb 20, 2017 at 11:55 | history | edited | Moritz Firsching | CC BY-SA 3.0 |
added the new formulation of the integral to the question
|
| Jan 28, 2017 at 0:48 | answer | added | Adam P. Goucher | timeline score: 11 | |
| Jan 27, 2017 at 8:25 | comment | added | Moritz Firsching | @user35593: I am not quite sure what you mean by "brute force". | |
| Jan 27, 2017 at 8:02 | comment | added | Adam P. Goucher | 1.726230867078 with 401^6 points. I'm now running 409^6 points to estimate the error between this and the actual integral. | |
| Jan 27, 2017 at 4:27 | comment | added | user35593 | Is it known where the determinant is zero? did someone try mathematica? I dont see why it shouldnt be possible to solve this integral by brute force. | |
| Jan 26, 2017 at 22:03 | comment | added | Moritz Firsching | @AdamP.Goucher: great, thanks! that should give a few more digits.. | |
| Jan 26, 2017 at 21:58 | comment | added | Adam P. Goucher | I'm using Gauss-Legendre quadrature, so the grid is non-uniform. It's unlikely (but not impossible) to be any worse than Monte Carlo, and can sometimes be significantly better (e.g. for smooth functions). Also, this works out faster since I don't need to generate random numbers or perform trig functions on the fly. By the way, my latest result is 1.726230356553 with 299^6 points (in 95 minutes). I'm doing an overnight run with 401^6 points, starting now... | |
| Jan 26, 2017 at 21:51 | comment | added | Moritz Firsching | @AdamP.Goucher: I didn't use a grid, but choose the points uniformly random, so I guess it doesn't make much of a difference?! | |
| Jan 26, 2017 at 20:07 | comment | added | Adam P. Goucher | 1.726228333055 using 199^6 points. Now trying 299^6 points... | |
| Jan 26, 2017 at 19:47 | comment | added | Adam P. Goucher | Note that the integrand is periodic in each direction with period pi, so you can cut your amount of work by a factor of 64. There's another factor of 6 from the symmetry of permuting (s1, t1), (s2, t2) and (s3, t3). A preliminary run using Gauss-Legendre quadrature with ~ 99^6/6 points gave an approximation of 1.726209603407 within 12 seconds; I'll try a larger net (maybe 199^6) reasonably soon. | |
| Jan 26, 2017 at 17:22 | history | edited | user44143 | CC BY-SA 3.0 |
fixed typos
|
| Jan 26, 2017 at 17:05 | history | asked | Moritz Firsching | CC BY-SA 3.0 |