Timeline for Two versions of Sylvester identity
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2020 at 19:18 | vote | accept | Honza | ||
| Aug 29, 2020 at 19:10 | comment | added | Honza | Thanks again. I believe that you are correct, which implies that MathWorld is making a false statement - what an incredible carelessness for such a popular and well respected source. This also means that the second formulation implies the special case (one row, one column) of the first one (as you have shown), but their equivalence is then out of the question. | |
| Aug 29, 2020 at 17:27 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 29, 2020 at 17:20 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 29, 2020 at 17:15 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 29, 2020 at 17:08 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 29, 2020 at 15:23 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 29, 2020 at 15:10 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 29, 2020 at 13:32 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 28, 2020 at 21:04 | comment | added | Honza | Thanks for a neat (special-case) proof. How would one proceed with a fully general proof? Now, I would be quite happy to see how the special case of FIRST version (with r, s, p, q each being a single row/column) implies the more general (r,s,p,q being sets) case. Or is it better to ask that as a separate question? | |
| Aug 28, 2020 at 18:56 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 28, 2020 at 18:00 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 28, 2020 at 17:49 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 28, 2020 at 17:34 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 28, 2020 at 17:29 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |