Timeline for A Putnam problem with a twist
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2022 at 23:37 | history | edited | LSpice | CC BY-SA 4.0 |
Link to @darijgrinberg's answer
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| Dec 7, 2018 at 18:11 | comment | added | Gjergji Zaimi | @darijgrinberg Yes, for the postscript, I was thinking the nonempty subsets with $\cup$. We can take the function $f(A,B)=q^{-|A\cup B|}$ and then multiply the row(or column) of A by $q^{|A|}$. | |
| Dec 7, 2018 at 16:36 | comment | added | Sam Hopkins | @darijgrinberg: that one might be different, I see. | |
| Dec 7, 2018 at 16:28 | comment | added | darij grinberg | @SamHopkins: Do you really get zeros for the postscript problem? | |
| Dec 7, 2018 at 16:21 | comment | added | Sam Hopkins | @darijgrinberg: another way to resolve that might be to add a row of all $1$s to the bottom of the matrix (thinking of the last row/column as corresponding to the empty set) by choosing the corresponding $f_i$ to be the constant function $1$. When I expand the determinant about the last row, I will just get the determinant of the $(2^n-1)\times(2^n-1)$ submatrix in the upper left because every other $(2^n-1)\times(2^n-1)$ determinant will have a column of all zeros. | |
| Dec 7, 2018 at 16:15 | comment | added | darij grinberg | @SamHopkins: Because $\cap$ is not always a nonempty subset. This is what confused me, too. But I think $\cup$ can be used perfectly well. | |
| Dec 7, 2018 at 16:04 | comment | added | Sam Hopkins | @darijgrinberg: isn't he using $\cap$ as $\wedge$ in the obvious way? | |
| Dec 7, 2018 at 15:17 | comment | added | darij grinberg | Oh, are you applying Lindström's result with $\cup$ as $\wedge$? | |
| Dec 7, 2018 at 11:03 | history | answered | Gjergji Zaimi | CC BY-SA 4.0 |