Timeline for Is the following sum irrational?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 16, 2014 at 6:05 | comment | added | Gerry Myerson | If you have access to Maple, then after you do with(IntegerRelations), you can do PSLQ([Zeta(3),Pi^4]) and PSLQ([Zeta(3),Pi^6]) to try to express your zeta quotients as rationals. Probably a good idea to do something like Digits:=2000 first, and then use more digits check to see whether the answers are spurious. | |
| Jul 16, 2014 at 4:33 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
added 654 characters in body
|
| Jun 29, 2014 at 20:58 | answer | added | Diego Marques | timeline score: 4 | |
| Jun 17, 2014 at 8:00 | comment | added | user25199 | According to mathematica, the continued fraction does not terminate before the first million digits. | |
| Jun 17, 2014 at 4:04 | comment | added | Ryan Reich | Of course; I just meant that on its own the result was incomparable with Apery's. | |
| Jun 17, 2014 at 3:38 | comment | added | Stanley Yao Xiao | Thanks, but it is an improvement to Apery given that we know $\zeta(3)$ is not rational. | |
| Jun 17, 2014 at 3:37 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
added 68 characters in body
|
| Jun 17, 2014 at 1:57 | comment | added | Ryan Reich | Not exactly "stronger" (it could be 1, for instance). | |
| Jun 17, 2014 at 1:38 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
[Edit removed during grace period]
|
| Jun 17, 2014 at 1:32 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |