Timeline for Using higher-order Bring radicals to solve arbitrary polynomials
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2023 at 2:17 | answer | added | CarP24 | timeline score: 0 | |
| Mar 1, 2023 at 13:22 | comment | added | Тyma Gaidash | There is only one question, so how does one know which question the answers consider? | |
| Apr 16, 2019 at 19:59 | answer | added | weux082690 | timeline score: 2 | |
| Apr 17, 2011 at 13:21 | vote | accept | Daniel Miller | ||
| Apr 13, 2011 at 15:51 | answer | added | David E Speyer | timeline score: 13 | |
| Apr 13, 2011 at 14:12 | answer | added | Dylan Thurston | timeline score: 8 | |
| Apr 13, 2011 at 12:41 | history | edited | Daniel Miller | CC BY-SA 3.0 |
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| Apr 12, 2011 at 18:18 | comment | added | J.C. Ottem | Ah, I thought the statement was 'any n-th degree polynomial can be solved in terms of some fixed function $B_n(t)$..' which seems false. | |
| Apr 12, 2011 at 17:59 | comment | added | Alan Wilder | @JC define the Bring radical to be the polynomial itself. | |
| Apr 12, 2011 at 17:29 | comment | added | J.C. Ottem | How is it obvious that every degree n polynomial can be solved in terms of one $B_n(t)$? It is well-known that the general 7th degree polynomial requires functions depending on two parameters - this is related to Hilbert's thirteenth problem.. | |
| Apr 12, 2011 at 15:15 | history | edited | Daniel Miller | CC BY-SA 3.0 |
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| Apr 12, 2011 at 15:14 | comment | added | Daniel Miller | You're right, I meant to write "finite collection." | |
| Apr 12, 2011 at 15:12 | comment | added | wood | wouldnt a positive answer to the first question imply a positive answer to the second as well. | |
| Apr 12, 2011 at 15:07 | history | asked | Daniel Miller | CC BY-SA 3.0 |