Timeline for Variations on a theme of O'Bryant, Cooper and Eichhorn concerning power series over $\mathbb Z/2\mathbb Z$
Current License: CC BY-SA 2.5
18 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 8, 2010 at 11:41 | vote | accept | paul Monsky | ||
| Jul 8, 2010 at 11:40 | vote | accept | paul Monsky | ||
| Jul 8, 2010 at 11:41 | |||||
| S Jul 8, 2010 at 11:40 | vote | accept | paul Monsky | ||
| Jul 8, 2010 at 11:40 | |||||
| Jul 8, 2010 at 11:40 | vote | accept | paul Monsky | ||
| S Jul 8, 2010 at 11:40 | |||||
| Jul 7, 2010 at 22:16 | answer | added | paul Monsky | timeline score: 1 | |
| Jul 7, 2010 at 21:20 | answer | added | paul Monsky | timeline score: 1 | |
| Jul 5, 2010 at 12:43 | comment | added | Wadim Zudilin | Paul, as one of the interested viewers I would be glad to learn about your simple argument. (Your comment is like Fermat's "I have discovered a truly marvelous proof that...") Thanks! | |
| Jul 5, 2010 at 6:22 | answer | added | Wadim Zudilin | timeline score: 4 | |
| Jul 5, 2010 at 2:00 | comment | added | Wadim Zudilin | The differential operator $D=\operatorname{id}+x(d/dx)$ "kills" the unwanted odd powers, so that the original problem is equivalent to $D(fg)\equiv D(f)g^2\equiv D(f)g(x^2)\pmod{2}$ where the congruence is applied to all coefficients in the power series expansions. I wonder whether this is helpful, but $f$ and $g$ are related to the thetanulls, $2f=x^{-1/4}\vartheta_2(x)$ and $2g=1+\vartheta_3(x)$ where $x=\exp(\pi i\tau)$, and there exist DEs for the latter ones. | |
| Jul 5, 2010 at 0:27 | comment | added | Will Jagy | Hi, Folks. I can certainly believe the Question, for the quotient I get, by hand, exponents 0, 3, 6, 9, 10, 11, 13, 15, 17, 19, 23, 28, 33, 36, 37, 41, 47, 49, 55, 57, 59, 65, 66, 71,... Maybe something will come to me, but meanwhile it appears Paul thought of a solution. Have you seen Greg Kuperberg's take on the original project? mathoverflow.net/questions/26839/… | |
| Jul 4, 2010 at 23:00 | history | edited | Will Jagy | CC BY-SA 2.5 |
field
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| Jul 4, 2010 at 13:07 | comment | added | paul Monsky | I now think I can answer my own question (in the affirmative) with a rather simple argument. But I'll leave things as they are for the time being for the interest of viewers. | |
| Jul 4, 2010 at 12:39 | comment | added | paul Monsky | Wadim, You miscalculated--the quotient is 1+x^3-2*x^4+2*x^5-x^6+.... Of course over Z/2 the quotient is 1+x^3+x^6... Paul | |
| Jul 4, 2010 at 11:51 | history | edited | Wadim Zudilin | CC BY-SA 2.5 |
typo corrected; links added
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| Jul 4, 2010 at 11:40 | history | edited | Wadim Zudilin | CC BY-SA 2.5 |
TeXified
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| Jul 4, 2010 at 11:39 | history | edited | Robin Chapman | CC BY-SA 2.5 |
TeXed up
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| Jul 4, 2010 at 11:06 | history | asked | paul Monsky | CC BY-SA 2.5 |