Timeline for Do convolution and multiplication satisfy any nontrivial algebraic identities?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2018 at 7:44 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
added MathJax (the question has been bumped anyway)
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| S Jun 28, 2015 at 21:11 | history | suggested | Boris Burkov | CC BY-SA 3.0 |
added some missing TeX dollar signs
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| Jun 28, 2015 at 20:35 | review | Suggested edits | |||
| S Jun 28, 2015 at 21:11 | |||||
| Nov 15, 2009 at 6:55 | comment | added | Darsh Ranjan | I think I have a proof now (for real vector spaces), which I've put in a separate community wiki post. Since getting it down to multilinear identities was the key, it's fair to mark this as "accepted." | |
| Nov 15, 2009 at 6:53 | vote | accept | Darsh Ranjan | ||
| Nov 7, 2009 at 12:11 | comment | added | Darsh Ranjan | Oh, duh. Yes, you're right: every homogeneous form can be depolarized that way into an equivalent multilinear one. | |
| Nov 7, 2009 at 0:55 | comment | added | Terry Tao | A putative identity like $f \cdot f = f * f$ would depolarise to $f \cdot g + g \cdot f = f * g + g * f$ (apply the initial identity to $f+g$ and $f-g$, subtract, and divide by 4). | |
| Nov 6, 2009 at 8:49 | comment | added | Darsh Ranjan | I don't see how to carry out the depolarisation, though. How would that apply to something like, say, f.f or f*f? | |
| Nov 1, 2009 at 21:18 | history | answered | Terry Tao | CC BY-SA 2.5 |